Vector Mechanics for Engineers - Manual Solutions Dynamics 10th edition

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SOLUTION MANUAL

CHAPTER 11

PROBLEM 11.CQ1 A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h. The average speed of the bus for the entire 200-mile trip is: (a) more than 60 mi/h (b) equal to 60 mi/h (c) less than 60 mi/h

SOLUTION The time required for the bus to travel from A to B is 2 h and from B to C is 100/70 = 1.43 h, so the total time is 3.43 h and the average speed is 200/3.43 = 58 mph. Answer: (c) 

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PROBLEM 11CQ2 Two cars A and B race each other down a straight road. The position of each car as a function of time is shown. Which of the following statements are true (more than one answer can be correct)? (a) At time t2 both cars have traveled the same distance (b) At time t1 both cars have the same speed (c) Both cars have the same speed at some time t < t1 (d) Both cars have the same acceleration at some time t < t1 (e) Both cars have the same acceleration at some time t1 < t < t2

SOLUTION The speed is the slope of the curve, so answer c) is true. The acceleration is the second derivative of the position. Since A’s position increases linearly the second derivative will always be zero. The second derivative of curve B is zero at the pont of inflection which occurs between t1 and t2. Answers: (c) and (e) 

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PROBLEM 11.1 The motion of a particle is defined by the relation x = t 4 − 10t 2 + 8t + 12 , where x and t are expressed in inches and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when t = 1 s.

SOLUTION x = t 4 − 10t 2 + 8t + 12

At t = 1 s,

v=

dx = 4t 3 − 20t + 8 dt

a=

dv = 12t 2 − 20 dt

x = 1 − 10 + 8 + 12 = 11

x = 11.00 in. 

v = 4 − 20 + 8 = −8

v = −8.00 in./s 

a = 12 − 20 = −8 

a = −8.00 in./s 2 

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PROBLEM 11.2 The motion of a particle is defined by the relation x = 2t 3 − 9t 2 + 12t + 10, where x and t are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v = 0.

SOLUTION x = 2t 3 − 9t 2 + 12t + 10

Differentiating,

v=

a=

dx = 6t 2 − 18t + 12 = 6(t 2 − 3t + 2) dt = 6(t − 2)(t − 1) dv = 12t − 18 dt

So v = 0 at t = 1 s and t = 2 s. At t = 1 s,

x1 = 2 − 9 + 12 + 10 = 15 a1 = 12 − 18 = −6

t = 1.000 s  x1 = 15.00 ft  a1 = −6.00 ft/s 2 

At t = 2 s, x2 = 2(2)3 − 9(2) 2 + 12(2) + 10 = 14

t = 2.00 s  x2 = 14.00 ft 

a2 = (12)(2) − 18 = 6

a2 = 6.00 ft/s 2 

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PROBLEM 11.3 The vertical motion of mass A is defined by the relation x = 10 sin 2t + 15cos 2t + 100, where x and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity and acceleration of A when t = 1 s, (b) the maximum velocity and acceleration of A.

SOLUTION x = 10sin 2t + 15cos 2t + 100 v=

dx = 20 cos 2t − 30sin 2t dt

a=

dv = −40sin 2t − 60 cos 2t dt

For trigonometric functions set calculator to radians: (a) At t = 1 s.

x1 = 10sin 2 + 15cos 2 + 100 = 102.9 v1 = 20cos 2 − 30sin 2 = −35.6 a1 = −40sin 2 − 60 cos 2 = −11.40

x1 = 102.9 mm  v1 = −35.6 mm/s  a1 = −11.40 mm/s 2 

(b) Maximum velocity occurs when a = 0. −40sin 2t − 60cos 2t = 0 tan 2t = −

60 = −1.5 40

2t = tan −1 (−1.5) = −0.9828 and −0.9828 + π

Reject the negative value. 2t = 2.1588 t = 1.0794 s t = 1.0794 s for vmax

so

vmax = 20cos(2.1588) − 30sin(2.1588) vmax = −36.1 mm/s 

= −36.056

Note that we could have also used vmax = 202 + 302 = 36.056

by combining the sine and cosine terms. For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms. amax = 402 + 602 = 72.1 mm/s 2

amax = 72.1 mm/s 2 

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PROBLEM 11.4 A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is defined by the relation x = 60e−4.8t sin16t where x and t are expressed in mm and seconds, respectively. Determine the position, the velocity and the acceleration of the railroad car when (a) t = 0, (b) t = 0.3 s.

SOLUTION x = 60e−4.8t sin16t dx = 60(−4.8)e −4.8t sin16t + 60(16)e−4.8t cos16t dt v = −288e−4.8t sin16t + 960e −4.8t cos16t v=

a=

dv = 1382.4e−4.8t sin16t − 4608e−4.8t cos16t dt − 4608e−4.8t cos16t − 15360e−4.8t sin16t

a = −13977.6e −4.8t sin16t − 9216e−4.8 cos16t

(a) At t = 0,

x0 = 0

x0 = 0 mm 

v0 = 960 mm/s

v0 = 960 mm/s

a0 = −9216 mm/s 2

(b) At t = 0.3 s,

a0 = 9220 mm/s 2

 

e−4.8t = e −1.44 = 0.23692 sin16t = sin 4.8 = −0.99616 cos16t = cos 4.8 = 0.08750 x0.3 = (60)(0.23692)(−0.99616) = −14.16

x0.3 = 14.16 mm



v0.3 = 87.9 mm/s



v0.3 = −(288)(0.23692)(−0.99616) + (960)(0.23692)(0.08750) = 87.9 a0.3 = −(13977.6)(0.23692)( −0.99616) − (9216)(0.23692)(0.08750) = 3108

a0.3 = 3110 mm/s 2 

or 3.11 m/s 2



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PROBLEM 11.5 The motion of a particle is defined by the relation x = 6t 4 − 2t 3 − 12t 2 + 3t + 3, where x and t are expressed in meters and seconds, respectively. Determine the time, the position, and the velocity when a = 0.

SOLUTION We have

x = 6t 4 − 2t 3 − 12t 2 + 3t + 3

Then

v=

dx = 24t 3 − 6t 2 − 24t + 3 dt

and

a=

dv = 72t 2 − 12t − 24 dt

When a = 0:

72t 2 − 12t − 24 = 12(6t 2 − t − 2) = 0 (3t − 2)(2t + 1) = 0

or t=

or At t =

2 s: 3

2 1 s and t = − s (Reject) 3 2 4

3

2

2 2 2 2 x2/3 = 6   − 2   − 12   + 3   + 3 3 3 3 3 3

t = 0.667 s  or

x2/3 = 0.259 m 

2

2 2 2 v2/3 = 24   − 6   − 24   + 3 3 3 3

or v2/3 = −8.56 m/s 

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PROBLEM 11.6 The motion of a particle is defined by the relation x = t 3 − 9t 2 + 24t − 8, where x and t are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION We have

x = t 3 − 9t 2 + 24t − 8

Then

v=

dx = 3t 2 − 18t + 24 dt

and

a=

dv = 6 t − 18 dt

(a)

When v = 0:

3 t 2 − 18t + 24 = 3(t 2 − 6t + 8) = 0 (t − 2)(t − 4) = 0 t = 2.00 s and t = 4.00 s 

(b)

When a = 0:

6t − 18 = 0 or t = 3 s x3 = (3)3 − 9(3)2 + 24(3) − 8

At t = 3 s: First observe that 0 ≤ t < 2 s:

v>0

2 s < t ≤ 3 s:

v0

4 s < t ≤ 7 s:

v0

t AB = 20.7 s 

vB = 51.8 mi/h 

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PROBLEM 11.44 An elevator is moving upward at a constant speed of 4 m/s. A man standing 10 m above the top of the elevator throws a ball upward with a speed of 3 m/s. Determine (a) when the ball will hit the elevator, (b) where the ball will hit the elevator with respect to the location of the man.

SOLUTION Place the origin of the position coordinate at the level of the standing man, the positive direction being up. The ball undergoes uniformly accelerated motion. yB = ( yB )0 + (vB )0 t −

1 2 gt 2

with ( yB )0 = 0, (vB )0 = 3 m/s, and g = 9.81 m/s 2 . yB = 3t − 4.905t 2

The elevator undergoes uniform motion. y E = ( y E ) 0 + vE t

with ( yE )0 = −10 m and vE = 4 m/s. (a)

Set yB = yE

Time of impact.

3t − 4.905t 2 = −10 + 4t 4.905t 2 + t − 10 = 0 t = 1.3295 and −1.5334

(b)

t = 1.330 s 

Location of impact. yB = (3)(1.3295) − (4.905)(1.3295)2 = −4.68 m yE = −10 + (4)(1.3295) = −4.68 m

(checks)

4.68 m below the man 

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PROBLEM 11.45 Two rockets are launched at a fireworks display. Rocket A is launched with an initial velocity v0 = 100 m/s and rocket B is launched t1 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 300 m as A is falling and B is rising. Assuming a constant acceleration g = 9.81 m/s2 , determine (a) the time t1, (b) the velocity of B relative to A at the time of the explosion.

SOLUTION Place origin at ground level. The motion of rockets A and B is Rocket A:

v A = (v A )0 − gt = 100 − 9.81t y A = ( y A ) 0 + (v A ) 0 t −

Rocket B:

1 2 gt = 100t − 4.905t 2 2

(1) (2)

vB = (vB )0 − g (t − t1 ) = 100 − 9.81(t − t1 )

(3)

1 g (t − t1 ) 2 2 = 100(t − t1 ) − 4.905(t − t1 )2

(4)

yB = ( yB )0 + (vB )0 (t − t1 ) −

Time of explosion of rockets A and B. y A = yB = 300 ft From (2),

300 = 100t − 4.905t 2 4.905t 2 − 100t + 300 = 0 t = 16.732 s and 3.655 s

From (4),

300 = 100(t − t1 ) − 4.905(t − t12 ) t − t1 = 16.732 s and 3.655 s

Since rocket A is falling, Since rocket B is rising, (a)

Time t1:

(b)

Relative velocity at explosion.

t = 16.732 s t − t1 = 3.655 s t1 = t − (t − t1 )

From (1),

v A = 100 − (9.81)(16.732) = −64.15 m/s

From (3),

vB = 100 − (9.81)(16.732 − 13.08) = 64.15 m/s

Relative velocity:

vB/A = vB − v A

t1 = 13.08 s 

vB/A = 128.3 m/s 

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PROBLEM 11.46 Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 60 mi/h. At t = 0, A starts and accelerates at a constant rate a A , while at t = 5 s, B begins to slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other x = 294 ft and v A = vB , determine (a) the acceleration a A , (b) when the vehicles pass each other, (c) the distance d between the vehicles at t = 0.

SOLUTION

For t ≥ 0:

v A = 0 + a At xA = 0 + 0 +

1 a At 2 2

0 ≤ t < 5 s:

xB = 0 + (vB )0 t (vB )0 = 60 mi/h = 88 ft/s

At t = 5 s:

xB = (88 ft/s)(5 s) = 440 ft

For t ≥ 5 s:

vB = (vB )0 + aB (t − 5)

1 aB = − a A 6

xB = ( xB ) S + (vB )0 (t − 5) +

1 aB (t − 5) 2 2

Assume t > 5 s when the cars pass each other. At that time (t AB ), v A = vB :

a At AB = (88 ft/s) −

x A = 294 ft:

294 ft =

Then or

a A( 76 t AB − 65 ) 1 2

2 a At AB

=

aA (t AB − 5) 6

1 2 a At AB 2 88 294

2 44t AB − 343t AB + 245 = 0

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PROBLEM 11.46 (Continued)

Solving (a)

With t AB > 5 s,

t AB = 0.795 s and t AB = 7.00 s 294 ft =

1 a A (7.00 s) 2 2 a A = 12.00 ft/s 2 

or (b)

t AB = 7.00 s 

From above

Note: An acceptable solution cannot be found if it is assumed that t AB ≤ 5 s. (c)

We have

d = x + ( xB )t AB = 294 ft + 440 ft + (88 ft/s)(2.00 s) 1 1  +  − × 12.00 ft/s 2  (2.00 s)2 2 6  d = 906 ft 

or

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PROBLEM 11.47 The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d ) the relative velocity of the counterweight W with respect to the elevator.

SOLUTION Choose the positive direction downward. (a)

Velocity of cable C. yC + 2 yE = constant vC + 2vE = 0 vE = 4 m/s

But, or (b)

vC = −2vE = −8 m/s

vC = 8.00 m/s 

Velocity of counterweight W. yW + yE = constant vW + vE = 0 vW = −vE = −4 m/s

(c)

vW = 4.00 m/s 

Relative velocity of C with respect to E. vC/E = vC − vE = (−8 m/s) − ( +4 m/s) = −12 m/s vC/E = 12.00 m/s 

(d )

Relative velocity of W with respect to E. vW /E = vW − vE = (−4 m/s) − (4 m/s) = −8 m/s vW /E = 8.00 m/s 

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PROBLEM 11.48 The elevator shown starts from rest and moves upward with a constant acceleration. If the counterweight W moves through 30 ft in 5 s, determine (a) the acceleration of the elevator and the cable C, (b) the velocity of the elevator after 5 s.

SOLUTION We choose positive direction downward for motion of counterweight. yW =

At t = 5 s,

1 aW t 2 2

yW = 30 ft 30 ft =

1 aW (5 s) 2 2

aW = 2.4 ft/s 2

(a)

Accelerations of E and C. Since Thus: Also, Thus:

(b)

aW = 2.4 ft/s 2

yW + yE = constant vW + vE = 0, and aW + aE = 0 aE = − aW = −(2.4 ft/s 2 ),

a E = 2.40 ft/s 2 

yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0 aC = −2aE = −2(−2.4 ft/s 2 ) = +4.8 ft/s 2 ,

aC = 4.80 ft/s 2 

Velocity of elevator after 5 s. vE = (vE )0 + aE t = 0 + (−2.4 ft/s 2 )(5 s) = −12 ft/s

( v E )5 = 12.00 ft/s 

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PROBLEM 11.49 Slider block A moves to the left with a constant velocity of 6 m/s. Determine (a) the velocity of block B, (b) the velocity of portion D of the cable, (c) the relative velocity of portion C of the cable with respect to portion D.

SOLUTION From the diagram, we have x A + 3 yB = constant

Then

v A + 3vB = 0

(1)

and

a A + 3aB = 0

(2)

(a)

Substituting into Eq. (1)

6 m/s + 3vB = 0 v B = 2.00 m/s 

or (b)

From the diagram

yB + yD = constant

Then

vB + vD = 0



v D = 2.00 m/s 

(c)

From the diagram

x A + yC = constant

Then

v A + vC = 0

Now

vC = −6 m/s

vC/D = vC − vD = (−6 m/s) − (2 m/s) = −8 m/s vC/D = 8.00 m/s 

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PROBLEM 11.50 Block B starts from rest and moves downward with a constant acceleration. Knowing that after slider block A has moved 9 in. its velocity is 6 ft/s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 2 s.

SOLUTION From the diagram, we have x A + 3 yB = constant

Then

v A + 3vB = 0

(1)

and

a A + 3aB = 0

(2)

(a)

Eq. (2): a A + 3aB = 0 and a B is constant and positive  a A is constant and negative Also, Eq. (1) and (vB )0 = 0  (v A )0 = 0 v A2 = 0 + 2a A [ x A − ( x A )0 ]

Then When |Δ x A | = 0.4 m:

(6 ft/s) 2 = 2a A (9/12 ft) a A = 24.0 ft/s 2

or



Then, substituting into Eq. (2): −24 ft/s 2 + 3aB = 0 aB =

or

24 ft/s 2 3

a B = 8.00 ft/s 2 

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PROBLEM 11.50 (Continued)

(b)

We have

vB = 0 + a B t

At t = 2 s:

 24  vB =  ft/s 2  (2 s)  3 

v B = 16.00 ft/s 

or yB = ( yB )0 + 0 +

Also At t = 2 s: or

y B − ( y B )0 =

1 aB t 2 2

1  24  ft/s 2  (2 s) 2  2  3 

y B − (y B )0 = 16.00 ft 



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PROBLEM 11.51 Slider block B moves to the right with a constant velocity of 300 mm/s. Determine (a) the velocity of slider block A, (b) the velocity of portion C of the cable, (c) the velocity of portion D of the cable, (d ) the relative velocity of portion C of the cable with respect to slider block A.

SOLUTION

From the diagram

xB + ( xB − x A ) − 2 x A = constant

Then

2vB − 3v A = 0

(1)

and

2aB − 3a A = 0

(2)

Also, we have

vD + v A = 0

Then (a)

− xD − x A = constant

Substituting into Eq. (1)

2(300 mm/s) − 3v A = 0

or (b)

From the diagram Then Substituting

(3)

v A = 200 mm/s



vC = 600 mm/s



xB + ( xB − xC ) = constant 2vB − vC = 0 2(300 mm/s) − vC = 0

or

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PROBLEM 11.51 (Continued)

(c)

From the diagram Then Substituting

( xC − x A ) + ( xD − x A ) = constant vC − 2v A + vD = 0 600 mm/s − 2(200 mm/s) + vD = 0

v D = 200 mm/s

or (d)

We have



vC/A = vC − v A = 600 mm/s − 200 mm/s

vC/A = 400 mm/s

or



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PROBLEM 11.52 At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slider block A has moved 240 mm to the right its velocity is 60 mm/s, determine (a) the accelerations of A and B, (b) the acceleration of portion D of the cable, (c) the velocity and change in position of slider block B after 4 s.

SOLUTION

xB + ( xB − x A ) − 2 x A = constant

From the diagram Then

2vB − 3v A = 0

(1)

and

2aB − 3a A = 0

(2)

(a)

First observe that if block A moves to the right, v A → and Eq. (1)  v B → . Then, using Eq. (1) at t = 0 2(150 mm/s) − 3(v A )0 = 0 (v A )0 = 100 mm/s

or

Also, Eq. (2) and aB = constant  a A = constant v A2 = (v A )02 + 2a A [ x A − ( x A )0 ]

Then When x A − ( x A )0 = 240 mm:

(60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm)

or

aA = −

40 mm/s 2 3

a A = 13.33 mm/s 2

or



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PROBLEM 11.52 (Continued)

Then, substituting into Eq. (2)  40  2aB − 3  − mm/s 2  = 0  3  aB = −20 mm/s 2

or (b)

a B = 20.0 mm/s 2



From the diagram, − xD − x A = constant vD + v A = 0

Then Substituting

aD + a A = 0  40  aD +  − mm/s 2  = 0 3  

or (c)

We have

vB = ( vB ) 0 + a B t

At t = 4 s:

vB = 150 mm/s + ( −20.0 mm/s 2 )(4 s)

or Also At t = 4 s:

x B = ( xB ) 0 + ( vB ) 0 t +

a D = 13.33 mm/s 2



v B = 70.0 mm/s



1 aB t 2 2

xB − ( xB )0 = (150 mm/s)(4 s) +

1 (−20.0 mm/s 2 )(4 s) 2 2

x B − (x B )0 = 440 mm

or



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PROBLEM 11.53 Collar A starts from rest and moves upward with a constant acceleration. Knowing that after 8 s the relative velocity of collar B with respect to collar A is 24 in./s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 6 s.

SOLUTION From the diagram 2 y A + yB + ( yB − y A ) = constant

Then

v A + 2 vB = 0

(1)

and

a A + 2aB = 0

(2)

(a)

Eq. (1) and (v A )0 = 0  (vB )0 = 0 Also, Eq. (2) and a A is constant and negative  a B is constant and positive. Then Now From Eq. (2) So that

v A = 0 + a At

vB = 0 + a B t

vB/A = vB − v A = (aB − a A )t 1 aB = − a A 2 3 vB/A = − a At 2

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PROBLEM 11.53 (Continued)

At t = 8 s:

3 24 in./s = − a A (8 s) 2

a A = 2.00 in./s 2 

or and then

1 aB = − (−2 in./s 2 ) 2

a B = 1.000 in./s 2 

or (b)

At t = 6 s:

vB = (1 in./s 2 )(6 s)

v B = 6.00 in./s 

or 

Now At t = 6 s:

yB = ( yB )0 + 0 + y B − ( y B )0 =

1 aB t 2 2

1 (1 in./s 2 )(6 s) 2 2

y B − (y B )0 = 18.00 in. 

or

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PROBLEM 11.54 The motor M reels in the cable at a constant rate of 100 mm/s. Determine (a) the velocity of load L, (b) the velocity of pulley B with respect to load L.

SOLUTION

Let xB and xL be the positions, respectively, of pulley B and load L measured downward from a fixed elevation above both. Let xM be the position of a point on the cable about to enter the reel driven by the motor. Then, considering the lengths of the two cables, xM + 3xB = constant

vM + 3vB = 0

xL + ( xL − xB ) = constant

2v L + v B = 0

vM = 100 mm/s

with

vB = − vL =

vM = −33.333 m/s 3

vB = −16.667 mm/s 2

v L = 16.67 mm/s 

(a)

Velocity of load L.

(b)

Velocity of pulley B with respect to load L. vB/L = vB − vL = −33.333 − (−16.667) = −16.667 v B/L = 16.67 mm/s 

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PROBLEM 11.55 Block C starts from rest at t = 0 and moves downward with a constant acceleration of 4 in./s2. Knowing that block B has a constant velocity of 3 in./s upward, determine (a) the time when the velocity of block A is zero, (b) the time when the velocity of block A is equal to the velocity of block D, (c) the change in position of block A after 5 s.

SOLUTION From the diagram: Cord 1:

2 y A + 2 y B + yC = constant

Then

2v A + 2vB + vC = 0

and

2a A + 2aB + aC = 0

Cord 2:

(1)

( y D − y A ) + ( y D − y B ) = constant

Then

2vD − v A − v B = 0

and

2aD − a A − aB = 0

(2)

Use units of inches and seconds. Motion of block C:

vC = vC 0 + aC t where aC = −4 in./s 2

= 0 + 4t

Motion of block B:

vB = −3 in./s;

Motion of block A:

From (1) and (2),

aB = 0

1 1 v A = −vB − vC = 3 − (4t ) = 3 − 2t in./s 2 2 1 1 a A = −aB − aC = 0 − (4) = −2 in./s 2 2 2

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PROBLEM 11.55 (Continued)

(a)

Time when vB is zero. 3 − 2t = 0

Motion of block D:

From (3), vD =

(b)

t = 1.500 s 

1 1 1 1 v A + vB = (3 − 2t ) − (3) = −1t 2 2 2 2

Time when vA is equal to v0. 3 − 2t = −t

(c)

t = 3.00 s 

Change in position of block A (t = 5 s). 1 a At 2 2 1 = (3)(5) + (−2)(5)2 = −10 in. 2

Δy A = ( v A ) 0 t +

Change in position = 10.00 in. 

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PROBLEM 11.56 Block A starts from rest at t = 0 and moves downward with a constant acceleration of 6 in./s2. Knowing that block B moves up with a constant velocity of 3 in./s, determine (a) the time when the velocity of block C is zero, (b) the corresponding position of block C.

SOLUTION

The cable lengths are constant. L1 = 2 yC + 2 yD + constant L 2 = y A + yB + ( yB − yD ) + constant

Eliminate yD. L1 + 2 L 2 = 2 yC + 2 yD + 2 y A + 2 yB + 2( yB − yD ) + constant 2( yC + y A + 2 yB ) = constant

Differentiate to obtain relationships for velocities and accelerations, positive downward. vC + v A + 2vB = 0

(1)

aC + a A + 2aB = 0

(2)

Use units of inches and seconds. Motion of block A:

v A = a At + 6t Δy A =

Motion of block B:

1 1 a At 2 = (6)t 2 = 3t 2 2 2

v B = 3 in./s

vB = −3 in./s

ΔyB = vB t = −3t

Motion of block C:

From (1), vC = −v A − 2vB = −6t − 2(−3) = 6 − 6t ΔyC =

(a)

Time when vC is zero.

(b)

Corresponding position.



t

0

vC dt = 6t − 3t 2

6 − 6t = 0

t = 1.000 s 

ΔyC = (6)(1) − (3)(1)2 = 3 in.

ΔyC = 3.00 in. 

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PROBLEM 11.57 Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s 2 . Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

SOLUTION

From the diagram 3 y A + 4 yB + xC = constant

Then

3v A + 4vB + vC = 0

(1)

and

3a A + 4aB + aC = 0

(2)

(vB ) = 0,

Given:

a A = constent (aC ) = 75 mm/s 2

At t = 2 s,

v B = 480 mm/s vC = 280 mm/s

(a)

Eq. (2) and a A = constant and aC = constant  aB = constant vB = 0 + a B t

Then At t = 2 s:

480 mm/s = aB (2 s) aB = 240 mm/s 2



or

a B = 240 mm/s 2 

or

a A = 345 mm/s 2 

Substituting into Eq. (2) 3a A + 4(240 mm/s 2 ) + (75 mm/s 2 ) = 0 a A = −345 mm/s

 

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PROBLEM 11.57 (Continued)

(b)

vC = (vC )0 + aC t

We have At t = 2 s:

280 mm/s = (vC )0 + (75 mm/s)(2 s) vC = 130 mm/s



or

( vC )0 = 130.0 mm/s



Then, substituting into Eq. (1) at t = 0 3(v A )0 + 4(0) + (130 mm/s) = 0 v A = −43.3 mm/s

(c)

We have At t = 3 s:

xC = ( xC )0 + (vC )0 t +

or 1 aC t 2 2

xC − ( xC )0 = (130 mm/s)(3 s) + = 728 mm

(v A )0 = 43.3 mm/s 

1 (75 mm/s 2 )(3 s) 2 2

or

xC − (xC )0 = 728 mm



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PROBLEM 11.58 Block B moves downward with a constant velocity of 20 mm/s. At t = 0, block A is moving upward with a constant acceleration, and its velocity is 30 mm/s. Knowing that at t = 3 s slider block C has moved 57 mm to the right, determine (a) the velocity of slider block C at t = 0, (b) the accelerations of A and C, (c) the change in position of block A after 5 s.

SOLUTION

From the diagram 3 y A + 4 yB + xC = constant

Then

3v A + 4vB + vC = 0

(1)

and

3a A + 4aB + aC = 0

(2)

v B = 20 mm/s ;

Given:

( v A )0 = 30 mm/s

(a)

Substituting into Eq. (1) at t = 0 3(−30 mm/s) + 4(20 mm/s) + (vC )0 = 0 (vC )0 = 10 mm/s

(b)

We have At t = 3 s:

( vC )0 = 10.00 mm/s

or

xC = ( xC )0 + (vC )0 t +

1 aC t 2 2

57 mm = (10 mm/s)(3 s) +

1 aC (3 s)2 2

aC = 6 mm/s 2



Now



or

aC = 6.00 mm/s 2



v B = constant → aB = 0

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PROBLEM 11.58 (Continued)

Then, substituting into Eq. (2) 3a A + 4(0) + (6 mm/s 2 ) = 0 a A = −2 mm/s 2

(c)

We have At t = 5 s:

y A = ( y A )0 + (v A )0 t +

or

a A = 2.00 mm/s 2 

1 a At 2 2

y A − ( y A )0 = (−30 mm/s)(5 s) +

1 (−2 mm/s 2 )(5 s)2 2

= −175 mm y A − (y A )0 = 175.0 mm 

or

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PROBLEM 11.59 The system shown starts from rest, and each component moves with a constant acceleration. If the relative acceleration of block C with respect to collar B is 60 mm/s 2 upward and the relative acceleration of block D with respect to block A is 110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the change in position of block D after 5 s.

SOLUTION From the diagram 2 y A + 2 yB + yC = constant

Cable 1: Then

2v A + 2vB + vC = 0

(1)

and

2a A + 2aB + aC = 0

(2)

Cable 2:

( yD − y A ) + ( yD − yB ) = constant

Then and

− v A − v B + 2 vD = 0

(3)

− a A − aB + 2aD = 0

(4)

Given: At t = 0, v = 0; all accelerations constant; aC/B = 60 mm/s 2 , aD /A = 110 mm/s 2 (a)

We have

aC/B = aC − aB = −60 or aB = aC + 60

and

aD/A = aD − a A = 110 or a A = aD − 110

Substituting into Eqs. (2) and (4) Eq. (2):

2(aD − 110) + 2( aC + 60) + aC = 0 3aC + 2aD = 100

or Eq. (4):

−(aD − 110) − ( aC + 60) + 2aD = 0 − aC + aD = −50

or

(5)

(6)

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PROBLEM 11.59 (Continued)

Solving Eqs. (5) and (6) for aC and aD aC = 40 mm/s 2 aD = −10 mm/s 2

Now

vC = 0 + aC t

At t = 3 s:

vC = (40 mm/s 2 )(3 s) vC = 120.0 mm/s 

or (b)

We have At t = 5 s:

yD = ( yD )0 + (0)t + yD − ( yD )0 =

1 aD t 2 2

1 ( −10 mm/s 2 )(5 s) 2 2 y D − (y D )0 = 125.0 mm 

or

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PROBLEM 11.60* The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if aB > 10 mm/s 2 , (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

SOLUTION From the diagram 2 y A + 2 yB + yC = constant

Cable 1: Then

2v A + 2vB + vC = 0

(1)

and

2a A + 2aB + aC = 0

(2)

Cable 2: ( yD − y A ) + ( yD − yB ) = constant Then

− v A − vB − 2 vD = 0

(3)

and

− a A − aB + 2aD = 0

(4)

t =0

Given: At

v=0 ( y A )0 = ( yB )0 = ( yC )0

All accelerations constant. At t = 2 s yC /A = 280 mm vB /A = 80 mm/s

When

y A − ( y A )0 = 160 mm yB − ( yB )0 = 320 mm aB > 10 mm/s 2

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PROBLEM 11.60* (Continued)

(a)

We have

y A = ( y A )0 + (0)t +

1 a At 2 2

and

yC = ( yC )0 + (0)t +

1 aC t 2 2

Then

yC/A = yC − y A =

1 (aC − a A )t 2 2

At t = 2 s, yC/A = −280 mm: −280 mm =

or

1 (aC − a A )(2 s) 2 2

aC = a A − 140

(5)

Substituting into Eq. (2) 2a A + 2aB + (a A − 140) = 0

or

1 a A = (140 − 2aB ) 3

Now

vB = 0 + a B t

(6)

v A = 0 + a At vB/A = vB − v A = (aB − a A )t

Also When



yB = ( yB )0 + (0)t + v B /A = 80 mm/s :

1 aB t 2 2 80 = (aB − a A )t

Δy A = 160 mm :

160 =

1 a At 2 2

ΔyB = 320 mm :

320 =

1 aB t 2 2

(7)

1 (aB − a A )t 2 2

Then

160 =

Using Eq. (7)

320 = (80)t or t = 4 s

Then

160 =

1 a A (4)2 2

or

a A = 20.0 mm/s 2 

and

320 =

1 aB (4) 2 2

or

a B = 40.0 mm/s 2 

Note that Eq. (6) is not used; thus, the problem is over-determined.

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PROBLEM 11.60* (Continued)

(b)

Substituting into Eq. (5) aC = 20 − 140 = −120 mm/s 2

and into Eq. (4) −(20 mm/s 2 ) − (40 mm/s 2 ) + 2aD = 0

or

aD = 30 mm/s 2

Now

vC = 0 + aC t

When vC = −600 mm/s: or Also At t = 5 s:

−600 mm/s = ( −120 mm/s 2 )t t =5s yD = ( yD )0 + (0)t + yD − ( yD )0 =

1 aD t 2 2

1 (30 mm/s 2 )(5 s)2 2 y D − (y D )0 = 375 mm 

or

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PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate.

SOLUTION

v0 = −14 ft/s

Change in v = area under a−t curve.

t = 0 to t = 2 s:

v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s

v2 = −8 ft/s

t = 2 s to t = 5 s:

v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s

v5 = +16 ft/s

t = 5 s to t = 8 s:

v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s

v8 = +25 ft/s

t = 8 s to t = 10 s:

v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s

v10 = +15 ft/s

t = 10 s to t = 15 s:

v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s

v15 = −10 ft/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83

PROBLEM 11.61 (Continued)

Plot v−t curve. Then by similar triangles Δ’s find t for v = 0.

x0 = 0

Change in x = area under v−t curve

t = 0 to t = 2 s:

1 x2 − x0 = (−14 − 8)(2) = −22 ft 2

x2 = −22 ft

t = 2 s to t = 3 s:

1 x3 − x2 = (−8)(1) = −4 ft 2

x8 = −26 ft

t = 3 s to t = 5 s:

1 x5 − x3 = (+16)(2) = +16 ft 2

x5 = −10 ft

t = 5 s to t = 8 s:

1 x8 − x5 = (+16 + 25)(3) = +61.5 ft 2

x8 = +51.6 ft

t = 8 s to t = 10 s:

1 x10 − x8 = (+25 + 15)(2) = + 40 ft 2

x10 = +91.6 ft

t = 10 s to t = 13 s:

1 x13 − x10 = (+15)(3) = +22.5 ft 2

x13 = +114 ft

t = 13 s to t = 15 s:

1 x15 − x13 = (−10)(2) = −10 ft 2

x15 = +94 ft

(a)

Maximum velocity: When t = 8 s,

(b)

Maximum x: When t = 13 s,

vm = 25.0 ft/s  xm = 114.0 ft 

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PROBLEM 11.62 For the particle and motion of Problem 11.61, plot the v−t and x−t curves for 0 < t < 15 s and determine the velocity of the particle, its position, and the total distance traveled after 10 s. PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate.

SOLUTION

v0 = −14 ft/s

Change in v = area under a−t curve.

t = 0 to t = 2 s:

v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s

v2 = −8 ft/s

t = 2 s to t = 5 s:

v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s

v5 = +16 ft/s

t = 5 s to t = 8 s:

v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s

v8 = +25 ft/s

t = 8 s to t = 10 s:

v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s

v10 = +15 ft/s

t = 10 s to t = 15 s:

v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s

v15 = −10 ft/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 85

PROBLEM 11.62 (Continued)

Plot v−t curve. Then by similar triangles Δ’s find t for v = 0.

x0 = 0

Change in x = area under v−t curve

t = 0 to t = 2 s:

1 x2 − x0 = (−14 − 8)(2) = −22 ft 2

x2 = −22 ft

t = 2 s to t = 3 s:

1 x3 − x2 = (−8)(1) = −4 ft 2

x8 = −26 ft

t = 3 s to t = 5 s:

1 x5 − x3 = (+16)(2) = +16 ft 2

x5 = −10 ft

t = 5 s to t = 8 s:

1 x8 − x5 = (+16 + 25)(3) = +61.5 ft 2

x8 = +51.6 ft

t = 8 s to t = 10 s:

1 x10 − x8 = (+25 + 15)(2) = + 40 ft 2

x10 = +91.6 ft

t = 10 s to t = 13 s:

1 x13 − x10 = (+15)(3) = +22.5 ft 2

x13 = +114 ft

t = 13 s to t = 15 s:

1 x15 − x13 = (−10)(2) = −10 ft 2

x15 = +94 ft

when t = 10 s:

v10 = +15 ft/s  x10 = +91.5 ft/s 

Distance traveled: t = 0 to t = 105 t = 0 to t = 3 s:

Distance traveled = 26 ft

t = 3 s to t = 10 s

Distance traveled = 26 ft + 91.5 ft = 117.5 ft Total distance traveled = 26 + 117.5 = 143.5 ft 

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PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a −t and x−t curves for 0 < t < 50 s, and determine (b) the total distance traveled by the particle when t = 50 s, (c) the two times at which x = 0.

SOLUTION (a)

at = slope of v −t curve at time t

From t = 0 to t = 10 s:

v = constant  a = 0 −20 − 60 = −5 m/s 2 26 − 10

t = 10 s to t = 26 s:

a=

t = 26 s to t = 41 s:

v = constant  a = 0

t = 41 s to t = 46 s:

a=

t = 46 s:

−5 − (−20) = 3 m/s 2 46 − 41

v = constant  a = 0

x2 = x1 + (area under v −t curve from t1 to t2 )

At t = 10 s:

x10 = −540 + 10(60) + 60 m

Next, find time at which v = 0. Using similar triangles tv = 0 − 10 60

At

t = 22 s: t = 26 s: t = 41 s: t = 46 s: t = 50 s:

=

26 − 10 80

or

tv = 0 = 22 s

1 x22 = 60 + (12)(60) = 420 m 2 1 x26 = 420 − (4)(20) = 380 m 2 x41 = 380 − 15(20) = 80 m  20 + 5  x46 = 80 − 5   = 17.5 m  2  x50 = 17.5 − 4(5) = −2.5 m

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PROBLEM 11.63 (Continued)

(b)

From t = 0 to t = 22 s: Distance traveled = 420 − (−540) = 960 m t = 22 s to t = 50 s: Distance traveled = |− 2.5 − 420| = 422.5 m Total distance traveled = (960 + 422.5) ft = 1382.5 m Total distance traveled = 1383 m 

(c)

Using similar triangles Between 0 and 10 s:

(t x = 0 )1 − 0 10 = 540 600 (t x = 0 )1 = 9.00 s 

Between 46 s and 50 s:

(t x = 0 )2 − 46 4 = 17.5 20 (t x =0 )2 = 49.5 s 

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PROBLEM 11.64 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a −t and x −t curves for 0 < t < 50 s, and determine (b) the maximum value of the position coordinate of the particle, (c) the values of t for which the particle is at x = 100 m.

SOLUTION (a)

at = slope of v −t curve at time t v = constant  a = 0

From t = 0 to t = 10 s:

−20 − 60 = −5 m/s 2 26 − 10

t = 10 s to t = 26 s:

a=

t = 26 s to t = 41 s:

v = constant  a = 0

t = 41 s to t = 46 s:

a=

−5 − (−20) = 3 m/s 2 46 − 41

v = constant  a = 0

t = 46 s:

x2 = x1 + (area under v −t curve from t1 to t2 )

At t = 10 s:

x10 = −540 + 10(60) = 60 m

Next, find time at which v = 0. Using similar triangles tv = 0 − 10 60

At

t = 22 s: t = 26 s: t = 41 s: t = 46 s: t = 50 s:

=

26 − 10 80

or

tv = 0 = 22 s

1 x22 = 60 + (12)(60) = 420 m 2 1 x26 = 420 − (4)(20) = 380 m 2 x41 = 380 − 15(20) = 80 m  20 + 5  x46 = 80 − 5   = 17.5 m  2  x50 = 17.5 − 4(5) = −2.5 m

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PROBLEM 11.64 (Continued)

(b)

Reading from the x −t curve

(c)

Between 10 s and 22 s

xmax = 420 m 

100 m = 420 m − (area under v −t curve from t , to 22 s) m 100 = 420 −

1 (22 − t1 )(v1 ) 2

(22 − t1 )(v1 ) = 640

Using similar triangles v1 60 = 22 − t1 12

Then

v1 = 5(22 − t1 )

or

(22 − t1 )[5(22 − t1 )] = 640 t1 = 10.69 s and t1 = 33.3 s 10 s < t1 < 22 s 

We have

t1 = 10.69 s 

Between 26 s and 41 s: Using similar triangles 41 − t2 15 = 20 300 t2 = 40.0 s 

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PROBLEM 11.65 During a finishing operation the bed of an industrial planer moves alternately 30 in. to the right and 30 in. to the left. The velocity of the bed is limited to a maximum value of 6 in./s to the right and 12 in./s to the left; the acceleration is successively equal to 6 in./s2 to the right, zero 6 in./s2 to the left, zero, etc. Determine the time required for the bed to complete a full cycle, and draw the v−t and x−t curves.

SOLUTION We choose positive to the right, thus the range of permissible velocities is −12 in./s < v < 6 in./s since acceleration is −6 in./s 2 , 0, or + 6 in./s 2 . The slope the v−t curve must also be −6 in./s2, 0, or +6 in./s2.

Planer moves = 30 in. to right: +30 in. = 3 + 6t1 + 3

t1 = 4.00 s

Planer moves = 30 in. to left: −30 in. = −12 − 12t2 + 12

t2 = 0.50 s

Total time = 1 s + 4 s + 1 s + 2 s + 0.5 s + 2 s = 10.5 s

ttotal = 10.50 s 

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PROBLEM 11.66 A parachutist is in free fall at a rate of 200 km/h when he opens his parachute at an altitude of 600 m. Following a rapid and constant deceleration, he then descends at a constant rate of 50 km/h from 586 m to 30 m, where he maneuvers the parachute into the wind to further slow his descent. Knowing that the parachutist lands with a negligible downward velocity, determine (a) the time required for the parachutist to land after opening his parachute, (b) the initial deceleration.

SOLUTION Assume second deceleration is constant. Also, note that 200 km/h = 55.555 m/s, 50 km/h = 13.888 m/s

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PROBLEM 11.66 (Continued)

(a)

Now Δ x = area under v −t curve for given time interval Then  55.555 + 13.888  (586 − 600) m = −t1   m/s 2  

t1 = 0.4032 s

(30 − 586) m = −t2 (13.888 m/s) t2 = 40.0346 s 1 (0 − 30) m = − (t3 )(13.888 m/s) 2 t3 = 4.3203 s ttotal = (0.4032 + 40.0346 + 4.3203) s

(b)

We have

ainitial = =

ttotal = 44.8 s 

Δ vinitial t1 [−13.888 − (−55.555)] m/s 0.4032 s

= 103.3 m/s 2 ainitial = 103.3 m/s 2 

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PROBLEM 11.67 A commuter train traveling at 40 mi/h is 3 mi from a station. The train then decelerates so that its speed is 20 mi/h when it is 0.5 mi from the station. Knowing that the train arrives at the station 7.5 min after beginning to decelerate and assuming constant decelerations, determine (a) the time required for the train to travel the first 2.5 mi, (b) the speed of the train as it arrives at the station, (c) the final constant deceleration of the train.

SOLUTION Given: At t = 0, v = 40 mi/h, x = 0; when x = 2.5 mi, v = 20 mi/h; at t = 7.5 min, x = 3 mi; constant decelerations. The v −t curve is first drawn as shown. (a)

A1 = 2.5 mi

We have

1h  40 + 20  (t1 min)  mi/h × = 2.5 mi  60 min  2  t1 = 5.00 min 

(b)

A2 = 0.5 mi

We have  20 + v2 (7.5 − 5) min ×   2

1h   mi/h × 60 min = 0.5 mi  v2 = 4.00 mi/h 

(c)

We have

afinal = a12 =

(4 − 20) mi/h 5280 ft 1 min 1h × × × (7.5 − 5) min mi 60 s 3600 s afinal = −0.1564 ft/s 2 

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PROBLEM 11.68 A temperature sensor is attached to slider AB which moves back and forth through 60 in. The maximum velocities of the slider are 12 in./s to the right and 30 in./s to the left. When the slider is moving to the right, it accelerates and decelerates at a constant rate of 6 in./s2; when moving to the left, the slider accelerates and decelerates at a constant rate of 20 in./s2. Determine the time required for the slider to complete a full cycle, and construct the v – t and x −t curves of its motion.

SOLUTION The v −t curve is first drawn as shown. Then ta =

vright aright

=

12 in./s =2s 6 in./s 2

vleft 30 in./s = aleft 20 in./s = 1.5 s

td =

A1 = 60 in.

Now

[(t1 − 2) s](12 in./s) = 60 in.

or

t1 = 7 s

or

A2 = 60 in.

and or

{[(t2 − 7) − 1.5] s}(30 in./s) = 60 in.

or

t2 = 10.5 s tcycle = t2

Now

tcycle = 10.5 s 

We have xii = xi + (area under v −t curve from ti to tii ) At

t = 5 s:

1 (2) (12) = 12 in. 2 x5 = 12 + (5 − 2)(12)

t = 7 s:

= 48 in. x7 = 60 in.

t = 2 s:

t = 8.5 s: t = 9 s: t = 10.5 s:

x2 =

1 x8.5 = 60 − (1.5)(30) 2 = 37.5 in. x9 = 37.5 − (0.5)(30)

= 22.5 in. x10.5 = 0

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PROBLEM 11.69 In a water-tank test involving the launching of a small model boat, the model’s initial horizontal velocity is 6 m/s, and its horizontal acceleration varies linearly from −12 m/s 2 at t = 0 to −2 m/s 2 at t = t1 and then remains equal to −2 m/s 2 until t = 1.4 s. Knowing that v = 1.8 m/s when t = t1 , determine (a) the value of t1 , (b) the velocity and the position of the model at t = 1.4 s.

SOLUTION Given:

v0 = 6 m/s; for 0 < t < t1 ,

for

t1 < t < 1.4 s a = −2 m/s 2 ;

at

t = 0 a = −12 m/s 2 ;

at t = t1

a = −2 m/s 2 , v = 1.8 m/s 2

The a −t and v −t curves are first drawn as shown. The time axis is not drawn to scale. (a)

vt1 = v0 + A1

We have

 12 + 2  1.8 m/s = 6 m/s − (t1 s)  m/s2   2  t1 = 0.6 s 

(b)

v1.4 = vt1 + A2

We have

v1.4 = 1.8 m/s − (1.4 − 0.6) s × 2 m/s 2 v1.4 = 0.20 m/s 

Now x1.4 = A3 + A4 , where A3 is most easily determined using integration. Thus, a=

for 0 < t < t1:

−2 − (−12) 50 t − 12 = t − 12 0.6 3

dv 50 = a = t − 12 dt 3

Now

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PROBLEM 11.69 (Continued)

At t = 0, v = 6 m/s: or



v 6

dv =



t

50  t − 12  dt  0 3 

v=6+

25 2 t − 12t 3

We have

dx 25 = v = 6 − 12t + t 2 dt 3

Then

A3 =



xt1 0

dx =



0.6 0

(6 − 12t +

25 2 t )dt 3

0.6

25   = 6t − 6t 2 + t 3  = 2.04 m 9 0 

Also

 1.8 + 0.2  A4 = (1.4 − 0.6)   = 0.8 m 2  

Then

x1.4 = (2.04 + 0.8) m x1.4 = 2.84 m 

or

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PROBLEM 11.70 The acceleration record shown was obtained for a small airplane traveling along a straight course. Knowing that x = 0 and v = 60 m/s when t = 0, determine (a) the velocity and position of the plane at t = 20 s, (b) its average velocity during the interval 6 s < t < 14 s.

SOLUTION

Geometry of “bell-shaped” portion of v −t curve

The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of v −t diagram is: = Δx = 6 m

(a)

When t = 20 s:

v20 = 60 m/s  x20 = (60 m/s) (20 s) − (shaded area)

= 1200 m − 6 m (b)

From t = 6 s to t = 14 s:

x20 = 1194 m 

Δt = 8 s

Δ x = (60 m/s)(14 s − 6 s) − (shaded area) = (60 m/s)(8 s) − 6 m = 480 m − 6 m = 474 m vaverage =

Δ x 474 m = Δt 8s

vaverage = 59.25 m/s 

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PROBLEM 11.71 In a 400-m race, runner A reaches her maximum velocity v A in 4 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25 s. Runner B reaches her maximum velocity vB in 5 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25.2 s. Both runners then run the second half of the race with the same constant deceleration of 0.1 m/s 2. Determine (a) the race times for both runners, (b) the position of the winner relative to the loser when the winner reaches the finish line.

SOLUTION Sketch v −t curves for first 200 m. Runner A:

t1 = 4 s, t2 = 25 − 4 = 21 s A1 =

1 (4)(v A )max = 2(vA ) max 2

A2 = 21(v A )max A1 + A2 = Δ x = 200 m 23(v A ) max = 200

Runner B:

or

t1 = 5 s, A1 =

(v A ) max = 8.6957 m/s

t2 = 25.2 − 5 = 20.2 s

1 (5)(vB ) max = 2.5(vB ) max 2

A2 = 20.2(vB ) max A1 + A2 = Δ x = 200 m 22.7(vB ) max = 200

Sketch v −t curve for second 200 m. A3 = vmaxt3 − t3 =

Runner A:

or

(vB ) max = 8.8106 m/s

Δv = | a |t3 = 0.1t3 1 Δvt3 = 200 2

or

0.05t32 − vmaxt3 + 200 = 0

(

vmax ± (vmax ) 2 − (4)(0.05)(200) = 10 vmax ± (vmax )2 − 40 (2)(0.05)

(vmax ) A = 8.6957,

Reject the larger root. Then total time

(t3 ) A = 146.64 s

and

)

27.279 s

t A = 25 + 27.279 = 52.279 s

t A = 52.2 s 

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PROBLEM 11.71 (Continued)

Runner B: (vmax ) B = 8.8106, (t3 ) B = 149.45 s

and

26.765 s t B = 25.2 + 26.765 = 51.965 s

Reject the larger root. Then total time

t B = 52.0 s 

Velocity of A at t = 51.965 s: v1 = 8.6957 − (0.1)(51.965 − 25) = 5.999 m/s

Velocity of A at t = 51.279 s: v2 = 8.6957 − (0.1)(52.279 − 25) = 5.968 m/s

Over 51.965 s ≤ t ≤ 52.965 s, runner A covers a distance Δ x Δ x = vave (Δt ) =

1 (5.999 + 5.968)(52.279 − 51.965) 2

Δ x = 1.879 m 

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PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h? Draw the v −t curve.

SOLUTION Relative to truck, car must move a distance:

Allowable increase in speed:

Δ x = 16 + 40 + 50 + 40 = 146 ft Δvm = 50 − 35 = 15 mi/h = 22 ft/s

Acceleration Phase:

t1 = 22/5 = 4.4 s

A1 =

1 (22)(4.4) = 48.4 ft 2

Deceleration Phase:

t3 = 22/20 = 1.1 s

A3 =

1 (22)(1.1) = 12.1 ft 2

But: Δ x = A1 + A2 + A3 :

146 ft = 48.4 + (22)t2 + 12.1 ttotal = t1 + t2 + t3 = 4.4 s + 3.89 s + 1.1 s = 9.39 s

t2 = 3.89 s t B = 9.39 s 

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PROBLEM 11.73 Solve Problem 11.72, assuming that the driver of the car does not pay any attention to the speed limit while passing and concentrates on reaching position B and resuming a speed of 35 mi/h in the shortest possible time. What is the maximum speed reached? Draw the v −t curve. PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h? Draw the v −t curve.

SOLUTION Relative to truck, car must move a distance: Δ x = 16 + 40 + 50 + 40 = 146 ft

Δvm = 5t1 = 20t2 ; Δ x = A1 + A2 :

t2 =

146 ft =

1 (Δvm )(t1 + t2 ) 2

146 ft =

1 1   (5t1 )  t1 + t1  2 4  

t12 = 46.72

1 t1 4

t1 = 6.835 s

t2 =

1 t1 = 1.709 4

ttotal = t1 + t2 = 6.835 + 1.709

t B = 8.54 s 

Δvm = 5t1 = 5(6.835) = 34.18 ft/s = 23.3 mi/h Speed vtotal = 35 mi/h, vm = 35 mi/h + 23.3 mi/h

vm = 58.3 mi/h 

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PROBLEM 11.74 Car A is traveling on a highway at a constant speed (v A )0 = 60 mi/h, and is 380 ft from the entrance of an access ramp when car B enters the acceleration lane at that point at a speed (vB )0 = 15 mi/h. Car B accelerates uniformly and enters the main traffic lane after traveling 200 ft in 5 s. It then continues to accelerate at the same rate until it reaches a speed of 60 mi/h, which it then maintains. Determine the final distance between the two cars.

SOLUTION Given:

(v A )0 = 60 mi/h, (vB )0 = 1.5 mi/h; at t = 0, ( x A )0 = −380 ft, ( xB ) 0 = 0; at t = 5 s, xB = 200 ft; for 15 mi/h < vB ≤ 60 mi/h, aB = constant; for vB = 60 mi/h, aB = 0

First note

60 mi/h = 88 ft/s 15 mi/h = 22 ft/s

The v −t curves of the two cars are then drawn as shown. Using the coordinate system shown, we have at t = 5 s, xB = 200 ft:

 22 +(vB )5  (5 s)   ft/s = 200 ft 2   (vB )5 = 58 ft/s

or Then, using similar triangles, we have

(88 − 22) ft/s (58 − 22) ft/s = ( = aB ) t1 5s

or

t1 = 9.16 67 s

Finally, at t = t1    22 + 88  ft/s  xB/A = xB − x A =  (9.1667 s)    2    − [−380 ft + (9.1667 s) (88 ft/s)]

xB/A = 77.5 ft 

or

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PROBLEM 11.75 An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2 until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the elevator begins to move, a man standing 12 m above the initial position of the top of the elevator throws a ball upward with an initial velocity of 20 m/s. Determine when the ball will hit the elevator.

SOLUTION Given:

At t = 0 vE = 0; For 0 < vE ≤ 7.8 m/s, aE = 1.2 m/s 2 ↑; For vE = 7.8 m/s, aE = 0; At t = 2 s, vB = 20m/s ↑

The v −t curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve for the elevator is 1.2 m/s2 , while the slope of the curve for the ball is − g (−9.81 m/s 2 ).

The time t1 is the time when vE reaches 7.8 m/s. Thus,

vE = (0) + aE t

or

7.8 m/s = (1.2 m/s 2 )t1

or

t1 = 6.5 s

The time ttop is the time at which the ball reaches the top of its trajectory. Thus, or or

vB = (vB )0 − g (t − 2)

0 = 20 m/s − (9.81 m/s 2 ) (t top − 2) s ttop = 4.0387 s

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PROBLEM 11.75 (Continued)

Using the coordinate system shown, we have 0 < t < t1 :

1  yE = −12 m +  aE t 2  m 2 

At t = ttop :

yB =

and

y E = −12 m +

At and at t = t1 ,

1 (4.0387 − 2) s × (20 m/s) 2 = 20.387 m 1 (1.2 m/s 2 )(4.0387 s) 2 2 = −2.213 m

t = [2 + 2(4.0387 − 2)] s = 6.0774 s, yB = 0 yE = −12 m +

1 (6.5 s) (7.8 m/s) = 13.35 m 2

The ball hits the elevator ( yB = yE ) when ttop ≤ t ≤ t1. For t ≥ ttop :

1  yB = 20.387 m −  g (t − ttop )2  m 2  

Then, when

yB = yE 1 20.387 m − (9.81 m/s 2 ) (t − 4.0387)2 2 1 = −12 m + (1.2 m/s 2 ) (t s)2 2

or Solving

5.505t 2 − 39.6196t + 47.619 = 0 t = 1.525 s and t = 5.67 s

t = 5.67 s 

Since 1.525 s is less than 2 s,

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PROBLEM 11.76 Car A is traveling at 40 mi/h when it enters a 30 mi/h speed zone. The driver of car A decelerates at a rate of 16 ft/s2 until reaching a speed of 30 mi/h, which she then maintains. When car B, which was initially 60 ft behind car A and traveling at a constant speed of 45 mi/h, enters the speed zone, its driver decelerates at a rate of 20 ft/s2 until reaching a speed of 28 mi/h. Knowing that the driver of car B maintains a speed of 28 mi/h, determine (a) the closest that car B comes to car A, (b) the time at which car A is 70 ft in front of car B.

SOLUTION (v A )0 = 40 mi/h; For 30 mi/h < v A ≤ 40 mi/h, a A = −16 ft/s 2 ; For v A = 30 mi/h, a A = 0;

Given:

( x A /B )0 = 60 ft; (vB )0 = 45 mi/h; When xB = 0, aB = −20 ft/s 2 ; For vB = 28 mi/h, aB = 0

First note

40 mi/h = 58.667 ft/s 30 mi/h = 44 ft/s 45 mi/h = 66 ft/s 28 mi/h = 41.067 ft/s

At t = 0

The v −t curves of the two cars are as shown. At

t = 0:

Car A enters the speed zone.

t = (tB )1:

Car B enters the speed zone.

t = tA:

Car A reaches its final speed.

t = tmin :

v A = vB

t = (t B )2 :

Car B reaches its final speed.

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PROBLEM 11.76 (Continued)

(a)

(v A )final − (v A )0 tA

aA =

We have

(44 − 58.667) ft/s tA

−16 ft/s 2 =

or

t A = 0.91669 s

or Also

60 ft = (tB )1 (vB )0

or

60 ft = (tB )1 (66 ft/s) aB =

and

−20 ft/s 2 =

or

or

(t B )1 = 0.90909 s

(vB )final − (vB )0 (t B ) 2 − (t B )1 (41.067 − 66) ft/s [(t B )2 − 0.90909] s

Car B will continue to overtake car A while vB > v A . Therefore, ( x A/B ) min will occur when v A = vB , which occurs for (tB )1 < tmin < (tB ) 2

For this time interval v A = 44 ft/s vB = (vB )0 + aB [t − (tB )1 ]

Then

at t = tmin :

44 ft/s = 66 ft/s + (−20 ft/s2 )(tmin − 0.90909) s tmin = 2.00909 s

or

Finally ( x A/B )min = ( x A )tmin − ( xB )tmin   (v ) + (v A )final   = t A  A 0 + (tmin − t A )(v A )final   2       (v ) + (v A )final   − ( xB )0 + (t B )1 (vB )0 + [tmin − (t B )1 ]  B 0  2   

   58.667 + 44  = (0.91669 s)  ft/s + (2.00909 − 0.91669) s × (44 ft/s)   2       66 + 44   −  −60 ft + (0.90909 s)(66 ft/s) + (2.00909 − 0.90909) s ×   ft/s   2    = (47.057 + 48.066) ft − (−60 + 60.000 + 60.500) ft

= 34.623 ft

or ( x A/B ) min = 34.6 ft 

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PROBLEM 11.76 (Continued)

(b)

Since ( x A/B ) ≤ 60 ft for t ≤ tmin , it follows that x A/B = 70 ft for t > (tB ) 2 [Note (tB ) 2  tmin ]. Then, for t > (tB ) 2 x A/B = ( x A/B )min + [(t − tmin )(vA )final ]   + (vB )final   (v ) − [(t B )2 − (tmin )]  A final + [t − (t B )2 ](vB )final   2    

or

70 ft = 34.623 ft + [(t − 2.00909) s × (44 ft/s)]    44 + 41.06  − (2.15574 − 2.00909) s ×   ft/s + (t − 2.15574) s × (41.067) ft/s  2     t = 14.14 s 

or

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PROBLEM 11.77 An accelerometer record for the motion of a given part of a mechanism is approximated by an arc of a parabola for 0.2 s and a straight line for the next 0.2 s as shown in the figure. Knowing that v = 0 when t = 0 and x = 0.8 ft when t = 0.4 s, (a) construct the v −t curve for 0 ≤ t ≤ 0.4 s, (b) determine the position of the part at t = 0.3 s and t = 0.2 s.

SOLUTION Divide the area of the a −t curve into the four areas A1, A2 , A3 and A4. 2 (8)(0.2) = 1.0667 ft/s 3 A2 = (16)(0.2) = 3.2 ft/s A1 =

1 (16 + 8)(0.1) = 1.2 ft/s 2 1 A4 = (8)(0.1) = 0.4 ft/s 2 A3 =

Velocities: v0 = 0 v0.2 = v0 + A1 + A2

v0.2 = 4.27 ft/s 

v0.3 = v0.2 + A3

v0.3 = 5.47 ft/s 

v0.4 = v0.3 + A4

v0.4 = 5.87 ft/s 

Sketch the v −t curve and divide its area into A5 , A6 , and A7 as shown. 0.8 0.4  x dx = 0.8 − x =  t vdt

2 (0.4)(0.1) = 0.0267 ft, 3

x0.3 = 0.227 ft 

x0.2 = 0.8 − ( A5 + A6 ) − A7

At t = 0.2 s, With A5 + A6 = and

0.4

x = 0.8 −  t vdt

x0.3 = 0.8 − A5 − (5.47)(0.1)

At t = 0.3 s, With A5 =

or

2 (1.6)(0.2) = 0.2133 ft, 3

A7 = (4.27)(0.2) = 0.8533 ft

x0.2 = 0.8 − 0.2133 − 0.8533

x0.2 = −0.267 ft 

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PROBLEM 11.78 A car is traveling at a constant speed of 54 km/h when its driver sees a child run into the road. The driver applies her brakes until the child returns to the sidewalk and then accelerates to resume her original speed of 54 km/h; the acceleration record of the car is shown in the figure. Assuming x = 0 when t = 0, determine (a) the time t1 at which the velocity is again 54 km/h, (b) the position of the car at that time, (c) the average velocity of the car during the interval 1 s ≤ t ≤ t1.

SOLUTION Given:

At

t = 0, x = 0, v = 54 km/h;

For t = t1 ,

v = 54 km/h

First note (a)

54 km/h = 15 m/s vb = va + (area under a −t curve from ta to tb )

We have Then

at

t = 2 s:

v = 15 − (1)(6) = 9 m/s

t = 4.5 s:

1 v = 9 − (2.5)(6) = 1.5 m/s 2

t = t1 :

15 = 1.5 +

1 (t1 − 4.5)(2) 2 t1 = 18.00 s 

or (b)

Using the above values of the velocities, the v −t curve is drawn as shown.

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PROBLEM 11.78 (Continued)

Now

x at t = 18 s x18 = 0 + Σ (area under the v −t curve from t = 0 to t = 18 s)  15 + 9  = (1 s)(15 m/s) + (1 s)   m/s  2  1   + (2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s)  3   2   + (13.5 s)(1.5 m/s) + (13.5 s)(13.5 m/s)  3   = [15 + 12 + (3.75 + 6.25) + (20.25 + 121.50)] m

= 178.75 m (c)

First note

or

x18 = 178.8 m 

x1 = 15 m x18 = 178.75 m

Now

vave =

Δ x (178.75 − 15) m = = 9.6324 m/s Δt (18 − 1) s vave = 34.7 km/h 

or

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PROBLEM 11.79 An airport shuttle train travels between two terminals that are 1.6 mi apart. To maintain passenger comfort, the acceleration of the train is limited to ±4 ft/s2, and the jerk, or rate of change of acceleration, is limited to ±0.8 ft/s2 per second. If the shuttle has a maximum speed of 20 mi/h, determine (a) the shortest time for the shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.

SOLUTION xmax = 1.6 mi; | amax | = 4 ft/s 2

Given:

 da  = 0.8 ft/s2 /s; vmax = 20 mi/h    dt max

First note (a)

20 mi/h = 29.333 ft/s 1.6 mi = 8448 ft

To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the time for which v = vmax . The time Δ t required for the train to have an acceleration of 4 ft/s2 is found from a  da  = max  dt  Δt  max

or or

4 ft/s2 0.8 ft/s2 /s Δt = 5 s Δt =

Now, da    since dt = constant   

after 5 s, the speed of the train is

v5 =

1 (Δt )(amax ) 2

or

v5 =

1 (5 s)(4 ft/s 2 ) = 10 ft/s 2

Then, since v5 < vmax , the train will continue to accelerate at 4 ft/s2 until v = vmax . The a −t curve must then have the shape shown. Note that the magnitude of the slope of each inclined portion of the curve is 0.8 ft/s2/s.

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PROBLEM 11.79 (Continued)

Now

at t = (10 + Δt1 ) s, v = vmax : 1  2  (5 s)(4 ft/s2 )  + (Δt1 )(4 ft/s2 ) = 29.333 ft/s 2 

or

Δt1 = 2.3333 s

Then

at t = 5 s: t = 7.3333 s:

1 (5)(4) = 10 ft/s 2 v = 10 + (2.3333)(4) = 19.3332 ft/s

v =0+

1 t = 12.3333 s: v = 19.3332 + (5)(4) = 29.3332 ft/s 2

Using symmetry, the v −t curve is then drawn as shown.

Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have   10 + 19.3332   2 (2.3333 s)   ft/s  2     + (10 + Δt2 ) s × (29.3332 ft/s) = 8448 ft

or

Δt2 = 275.67 s

Then

tmin = 4(5 s) + 2(2.3333 s) + 275.67 s

= 300.34 s tmin = 5.01 min 

or (b)

We have

vave =

Δx 1.6 mi 3600 s = × Δt 300.34 s 1h vave = 19.18 mi/h 

or

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PROBLEM 11.80 During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ±4.8 ft/s2 per second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average values of the velocity of the belt during that time.

SOLUTION Given:

(a)

At

t = 0, x = 0, v = 0; xmax = 1.2 ft;

when

 da  x = xmax , v = 0;   = 4.8 ft/s 2  dt max

Observing that vmax must occur at t = 12 tmin , the a −t curve must have the shape shown. Note that the magnitude of the slope of each portion of the curve is 4.8 ft/s2/s.

We have

at t = Δt : t = 2Δt :

1 1 v = 0 + (Δt )(amax ) = amax Δt 2 2 vmax =

1 1 amax Δt + ( Δt )( amax ) = amax Δt 2 2

Using symmetry, the v −t is then drawn as shown.

Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have (2Δt )(vmax ) = xmax vmax = amax Δt  2amax Δt 2 = xmax

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PROBLEM 11.80 (Continued)

Now

amax = 4.8 ft/s2 /s so that Δt

2(4.8Δt ft/s3 )Δt 2 = 1.2 ft or Then

Δt = 0.5 s tmin = 4Δt tmin = 2.00 s 

or (b)

We have

vmax = amax Δt = (4.8 ft/s 2 /s × Δt)Δt = 4.8 ft/s 2 /s × (0.5 s) 2

vmax = 1.2 ft/s 

or Also

vave =

Δx 1.2 ft = Δttotal 2.00 s vave = 0.6 ft/s 

or

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PROBLEM 11.81 Two seconds are required to bring the piston rod of an air cylinder to rest; the acceleration record of the piston rod during the 2 s is as shown. Determine by approximate means (a) the initial velocity of the piston rod, (b) the distance traveled by the piston rod as it is brought to rest.

SOLUTION a −t curve; at

Given: 1.

t = 2 s, v = 0

The a −t curve is first approximated with a series of rectangles, each of width Δt = 0.25 s. The area (Δt)(aave) of each rectangle is approximately equal to the change in velocity Δv for the specified interval of time. Thus, Δv ≅ aave Δt

where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table. 2.

v(2) = v0 +

Now and approximating the area



2 0



2 0

a dt = 0

a dt under the a −t curve by Σaave Δt ≈ ΣΔv, the initial velocity is then

equal to v0 = −ΣΔv

Finally, using v2 = v1 + Δv12

where Δv12 is the change in velocity between times t1 and t2, the velocity at the end of each 0.25 interval can be computed; see column 3 of the table and the v −t curve. 3.

The v −t curve is then approximated with a series of rectangles, each of width 0.25 s. The area (Δt )(vave ) of each rectangle is approximately equal to the change in position Δ x for the specified interval of time. Thus Δ x ≈ vave Δt

where vave and Δ x are given in columns 4 and 5, respectively, of the table.

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PROBLEM 11.81 (Continued)

4.

With x0 = 0 and noting that x2 = x1 + Δ x12

where Δ x12 is the change in position between times t1 and t2, the position at the end of each 0.25 s interval can be computed; see column 6 of the table and the x−t curve.

(a)

We had found

(b)

At t = 2 s

v0 = 1.914 m/s  x = 0.840 m 

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PROBLEM 11.82 The acceleration record shown was obtained during the speed trials of a sports car. Knowing that the car starts from rest, determine by approximate means (a) the velocity of the car at t = 8 s, (b) the distance the car has traveled at t = 20 s.

SOLUTION Given: a −t curve; at 1.

t = 0, x = 0, v = 0

The a −t curve is first approximated with a series of rectangles, each of width Δt = 2 s. The area (Δt )(aave ) of each rectangle is approximately equal to the change in velocity Δv for the specified interval of time. Thus, Δv ≅ aave Δt

where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table. 2.

Noting that v0 = 0 and that v2 = v1 + Δv12

where Δv12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s interval can be computed; see column 3 of the table and the v −t curve. 3.

The v −t curve is next approximated with a series of rectangles, each of width Δt = 2 s. The area (Δt )(vave ) of each rectangle is approximately equal to the change in position Δx for the specified interval of time. Thus,

Δx ≅ vave Δt

where vave and Δx are given in columns 4 and 5, respectively, of the table. 4.

With x0 = 0 and noting that x2 = x1 + Δ x12

where Δ x12 is the change in position between times t1 and t2, the position at the end of each 2 s interval can be computed; see column 6 of the table and the x −t curve.

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PROBLEM 11.82 (Continued)

(a)

At t = 8 s, v = 32.58 m/s

(b)

At t = 20 s

v = 117.3 km/h 

or

x = 660 m 

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PROBLEM 11.83 A training airplane has a velocity of 126 ft/s when it lands on an aircraft carrier. As the arresting gear of the carrier brings the airplane to rest, the velocity and the acceleration of the airplane are recorded; the results are shown (solid curve) in the figure. Determine by approximate means (a) the time required for the airplane to come to rest, (b) the distance traveled in that time.

SOLUTION Given: a −v curve: v0 = 126 ft/s

The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on the figure). For uniformly accelerated motion v22 = v12 + 2a( x2 − x1 ) v2 = v1 + a(t2 − t1 )

v22 − v12 2a v −v Δt = 2 1 a

Δx =

or

For the five regions shown above, we have Region

v1 , ft/s

v2 , ft/s

a, ft/s 2

Δx, ft

Δt , s

1

126

120

−12.5

59.0

0.480

2

120

100

−33

66.7

0.606

3

100

80

−45.5

39.6

0.440

4

80

40

−54

44.4

0.741

5

40

0

−58

13.8

0.690

223.5

2.957

Σ

(a)

From the table, when v = 0

t = 2.96 s 

(b)

From the table and assuming x0 = 0, when v = 0

x = 224 ft 

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PROBLEM 11.84 Shown in the figure is a portion of the experimentally determined v − x curve for a shuttle cart. Determine by approximate means the acceleration of the cart (a) when x = 10 in., (b) when v = 80 in./s.

SOLUTION Given: v − x curve

First note that the slope of the above curve is

dv . dx

a=v

(a)

Now

dv dx

When

x = 10 in., v = 55 in./s

Then

 40 in./s  a = 55 in./s    13.5 in.  a = 163.0 in./s 2 

or (b)

When v = 80 in./s, we have  40 in./s  a = 80 in./s    28 in.  a = 114.3 in./s 2 

or

Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary that the same scale be used for the x and v axes (e.g., 1 in. = 50 in., 1 in. = 50 in./s). In the above solution, Δv and Δ x were measured directly, so different scales could be used. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 121

PROBLEM 11.85 Using the method of Section 11.8, derive the formula x = x0 + v0t + 12 at 2 for the position coordinate of a particle in uniformly accelerated rectilinear motion.

SOLUTION The a −t curve for uniformly accelerated motion is as shown.

Using Eq. (11.13), we have x = x0 + v0t + (area under a −t curve) (t − t )

 1  = x0 + v0t + (t × a)  t − t   2  1 = x0 + v0 t + at 2 Q.E.D. 2



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PROBLEM 11.86 Using the method of Section 11.8 determine the position of the particle of Problem 11.61 when t = 8 s. PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate.

SOLUTION

x0 = 0 v0 = −14 ft/s

when t = 8s: x = x0 + v0t + Σ A(t1 − t ) = 0 − (14 ft/s)(8 s) + [(3 ft/s2 )(2 s)](7 s) + [(8 ft/s2 )(3 s)](4.5 s) + [(3 ft/s)(3 s)](1.5 s) x8 = −112 ft + 42 ft + 108 ft + 13.5 ft

x8 = 51.5 ft 

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PROBLEM 11.87 The acceleration of an object subjected to the pressure wave of a large explosion is defined approximately by the curve shown. The object is initially at rest and is again at rest at time t1. Using the method of section 11.8, determine (a) the time t1, (b) the distance through which the object is moved by the pressure wave.

SOLUTION

(a)

Since v = 0 when t = 0 and when t = t1 the change in v between t = 0 and t = t1 is zero. Thus, area under a−t curve is zero A1 + A2 + A3 = 0

1 1 1 (30)(0.6) + ( −10)(0.2) + ( −10)(t1 − 0.8) = 0 2 2 2 9 − 1 − 5t1 + 4 = 0

(b)

t1 = 2.40 s 

Position when t = t1 = 2.4 s 2 x = x0 + v0t1 + A1 (t1 − 0.2) + A2 (t1 − 0.733) + A3   (t1 − 0.8) 3 1 2 = 0 + 0 + (9)(2.4 − 0.2) + ( −1)(2.4 − 0.733) +  ( −10)(2.4 − 0.8)  (2.4 − 0.8) 2 3 = 19.8 m − 1.667 m − 8.533 m

x = 9.60 m 

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PROBLEM 11.88 For the particle of Problem 11.63, draw the a −t curve and determine, using the method of Section 11.8, (a) the position of the particle when t = 52 s, (b) the maximum value of its position coordinate. PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a −t and x −t curves for 0 < t < 50 s, and determine (b) the total distance traveled by the particle when t = 50 s, (c) the two times at which x = 0.

SOLUTION a=

We have where

dv dt

dv dt

is the slope of the v −t curve. Then t=0

from

to t = 10 s:

v = constant  a = 0

t = 10 s to t = 26 s: a = t = 26 s to t = 41 s:

v = constant  a = 0

t = 41 s to t = 46 s: a = t > 46 s:

−20 − 60 = −5 m/s 2 26 − 10

−5 − (−20) = 3 m/s 2 46 − 41

v = constant  a = 0

The a −t curve is then drawn as shown. (a)

From the discussion following Eq. (11.13), we have

x = x0 + v0t1 + ΣA(t1 − t )

where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s x = −540 m + (60 m/s)(52 s) + {−[(16 s)(5 m/s 2 )](52 − 18) s + [(5 s)(3 m/s 2 )](52 − 43.5)s} = [−540 + (3120) + (−2720 + 127.5)] m x = −12.50 m 

or

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PROBLEM 11.88 (Continued)

(b)

Noting that xmax occurs when v = 0 ( dx = 0 ), it is seen from the v–t curve that xmax occurs for dt 10 s < t < 26 s. Although similar triangles could be used to determine the time at which x = xmax (see the solution to Problem 11.63), the following method will be used. For

10 s < t1 < 26 s, we have x = −540 + 60t1 1  − [(t1 − 10)(5)]  (t1 − 10)  m 2  5 = −540 + 60t1 − (t1 − 10) 2 2

When x = xmax : or Then

dx = 60 − 5(t1 − 10) = 0 dt

(t1 ) xmax = 22 s 5 xmax = −540 + 60(22) − (22 − 10)2 2 xmax = 420 m 

or

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PROBLEM 11.CQ3 Two model rockets are fired simultaneously from a ledge and follow the trajectories shown. Neglecting air resistance, which of the rockets will hit the ground first? (a) A (b) B (c) They hit at the same time. (d) The answer depends on h.

SOLUTION The motion in the vertical direction depends on the initial velocity in the y-direction. Since A has a larger initial velocity in this direction it will take longer to hit the ground. Answer: (b) 

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PROBLEM 11.CQ4 Ball A is thrown straight up. Which of the following statements about the ball are true at the highest point in its path? (a) The velocity and acceleration are both zero. (b) The velocity is zero, but the acceleration is not zero. (c) The velocity is not zero, but the acceleration is zero. (d) Neither the velocity nor the acceleration are zero.

SOLUTION At the highest point the velocity is zero. The acceleration is never zero. Answer: (b) 

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PROBLEM 11.CQ5 Ball A is thrown straight up with an initial speed v0 and reaches a maximum elevation h before falling back down. When A reaches its maximum elevation, a second ball is thrown straight upward with the same initial speed v0. At what height, y, will the balls cross paths? (a)

y=h

(b)

y > h/2

(c)

y = h/2

(d)

y < h/2

(e)

y=0

SOLUTION When the ball is thrown up in the air it will be constantly slowing down until it reaches its apex, at which point it will have a speed of zero. So, the time it will take to travel the last half of the distance to the apex will be longer than the time it takes for the first half. This same argument can be made for the ball falling from the maximum elevation. It will be speeding up, so the first half of the distance will take longer than the second half. Therefore, the balls should cross above the half-way point. Answer: (b) 

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PROBLEM 11.CQ6 Two cars are approaching an intersection at constant speeds as shown. What velocity will car B appear to have to an observer in car A? (a)

(b)

(c)

(d )

(e)

SOLUTION Since vB = vA+ vB/A we can draw the vector triangle and see

v B = v A + v B/A

Answer: (e) 

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PROBLEM 11.CQ7 Blocks A and B are released from rest in the positions shown. Neglecting friction between all surfaces, which figure below best indicates the direction α of the acceleration of block B? (a)

(b)

(c)

(d)

(e)

SOLUTION    Since aB = a A + aB/A we get

Answer: (d ) 

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PROBLEM 11.89 A ball is thrown so that the motion is defined by the equations x = 5t and y = 2 + 6t − 4.9t 2 , where x and y are expressed in meters and t is expressed in seconds. Determine (a) the velocity at t = 1 s, (b) the horizontal distance the ball travels before hitting the ground.

SOLUTION Units are meters and seconds. Horizontal motion:

vx =

dx =5 dt

Vertical motion:

vy =

dy = 6 − 9.8t dt

(a)

Velocity at t = 1 s.

vx = 5 v y = 6 − 9.8 = −3.8

v = vx2 + v y2 = 52 + 3.82 = 6.28 m/s tan θ =

(b)

vy vx

=

−3.8 5

Horizontal distance:

θ = −37.2°

v = 6.28 m/s

37.2° 

( y = 0) y = 2 + 6t − 4.9t 2 t = 1.4971 s x = (5)(1.4971) = 7.4856 m

x = 7.49 m 

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PROBLEM 11.90 The motion of a vibrating particle is defined by the position vector r = 10(1 − e−3t )i + (4e −2t sin15t ) j, where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0, (b) t = 0.5 s.

SOLUTION r = 10(1 − e−3t )i + (4e−2t sin15t ) j

Then

v=

and

a=

(a)

dr = 30e−3t i + [60e−2t cos15t − 8e−2t sin15t ]j dt

dv = −90e −3t i + [−120e −2t cos15t − 900e−2t sin15t − 120e−2t cos15t + 16e−2t sin15t ]j dt = −90e −3t i + [−240e −2t cos15t − 884e −2t sin15t ] j

When t = 0: v = 30i + 60 j mm/s

v = 67.1 mm/s

63.4° 

a = −90i − 240 j mm/s 2

a = 256 mm/s 2

69.4° 

v = 8.29 mm/s

36.2° 

a = 336 mm/s 2

86.6° 

When t = 0.5 s: v = 30e −1.5 i + [60e−1 cos 7.5 − 8e−1 sin 7.5] = 6.694i + 4.8906 j mm/s

a = 90e−1.5 i + [ −240e −1 cos 7.5 − 884e−1 sin 7.5 j] = −20.08i − 335.65 j mm/s 2

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PROBLEM 11.91 The motion of a vibrating particle is defined by the position vector r = (4sin π t )i − (cos 2π t ) j, where r is expressed in inches and t in seconds. (a) Determine the velocity and acceleration when t = 1 s. (b) Show that the path of the particle is parabolic.

SOLUTION r = (4sin π t )i − (cos 2π t ) j v = (4π cos π t )i + (2π sin 2π t ) j a = −(4π 2 sin π t )i + (4π 2 cos 2π t ) j

(a)

When t = 1 s: v = (4π cos π )i + (2π sin 2π ) j

v = −(4π in/s)i 

a = −(4π 2 sin π )i − (4π 2 cos π ) j

(b)

a = −(4π 2 in/s 2 ) j 

Path of particle: Since r = xi + y j ;

x = 4sin π t ,

y = − cos 2π t

Recall that cos 2θ = 1 − 2sin 2 θ and write y = − cos 2π t = −(1 − 2sin 2 π t )

But since x = 4sin π t or sin π t =

(1)

1 x, Eq.(1) yields 4

2  1   y = − 1 − 2  x    4   

y=

1 2 x − 1 (Parabola)  8

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PROBLEM 11.92 The motion of a particle is defined by the equations x = 10t − 5sin t and y = 10 − 5cos t , where x and y are expresed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval 0 ≤ t ≤ 2π , and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the corresponding times, positions, and directions of the velocities.

SOLUTION Sketch the path of the particle, i.e., plot of y versus x. Using x = 10t − 5sin t , and y = 10 − 5cos t obtain the values in the table below. Plot as shown. t(s)

x(ft)

y(ft)

0

0.00

5

10.71

10

31.41

15

52.12

10

62.83

5

π 2

π 3

π 2



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PROBLEM 11.92 (Continued)

(a)

Differentiate with respect to t to obtain velocity components. vx =

dx = 10 − 5cos t and v y = 5sin t dt

v 2 = vx2 + v y2 = (10 − 5cos t )2 + 25sin 2 t = 125 − 100cos t d (v ) 2 = 100sin t = 0 t = 0. ± π . ± 2π  ± N π dt

(b)

When t = 2 N π .

cos t = 1.

and

v2 is minimum.

When t = (2 N + 1)π .

cos t = −1.

and

v2 is maximum.

(v 2 )min = 125 − 100 = 25(ft/s) 2

vmin = 5 ft/s 

(v 2 )max = 125 + 100 = 225(ft/s) 2

vmax = 15 ft/s 

When v = vmin . When N = 0,1, 2,

x = 10(2π N ) − 5sin(2π N )

x = 20π N ft 

y = 10 − 5cos(2π N )

y = 5 ft 

vx = 10 − 5cos(2π N )

vx = 5 ft/s 

v y = 5sin(2π N )

tan θ =

vy vx

vy = 0 

θ =0 

= 0,

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PROBLEM 11.92 (Continued)

t = (2 N + 1)π s 

When v = vmax . x = 10[2π ( N − 1)] − 5sin[2π ( N + 1)]

x = 20π ( N + 1) ft 

y = 10 − 5cos[2π ( N + 1)]

y = 15 ft 

vx = 10 − 5cos[2π ( N + 1)]

vx = 15 ft/s 

v y = 5sin[2π ( N + 1)]

tan θ =

vy vx

vy = 0 

θ =0 

= 0,

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PROBLEM 11.93 The damped motion of a vibrating particle is defined by the position vector r = x1[1 − 1/(t + 1)]i + ( y1e−π t/2 cos 2π t ) j, where t is expressed in seconds. For x1 = 30 mm and y1 = 20 mm, determine the position, the velocity, and the acceleration of the particle when (a) t = 0, (b) t = 1.5 s.

SOLUTION We have

1   −π t/2 r = 30 1 − cos 2π t ) j  i + 20(e  t +1

Then

v=

= 30

1  π  i + 20  − e −π t/2 cos 2π t − 2π e−π t/2 sin 2π t  j 2 (t + 1)  2 

= 30

 1 1  i − 20π e −π t/ 2  cos 2π t + 2 sin 2π t   j 2 2 (t + 1)   

a=

and

dr dt

dv dt

 π  2 1  i − 20π  − e−π t/2  cos 2π t + 2 sin 2π t  + e−π t/2 (−π sin 2π t + 4 cos 2π t )  j 3 (t + 1) 2   2  60 =− i + 10π 2 e−π t/2 (4 sin 2π t − 7.5 cos 2π t ) j (t + 1)3 = −30

(a)

At t = 0:

 1 r = 30 1 −  i + 20(1) j  1 r = 20 mm 

or  1 1  v = 30   i − 20π (1)  + 0   j 1   2

v = 43.4 mm/s

or a=−

46.3° 

60 i + 10π 2 (1)(0 − 7.5) j (1) a = 743 mm/s 2

or

85.4° 

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PROBLEM 11.93 (Continued)

(b)

At t = 1.5 s:

1   −0.75π r = 30 1 − (cos 3π ) j  i + 20e 2.5   = (18 mm)i + ( −1.8956 mm) j

or

r = 18.10 mm

6.01° 

v = 5.65 mm/s

31.8° 

a = 70.3 mm/s 2

86.9° 

30 1  i − 20π e−0.75π  cos 3π + 0  j 2 (2.5) 2   = (4.80 mm/s)i + (2.9778 mm/s) j

v=

or a=−

60 i + 10π 2 e−0.75π (0 − 7.5 cos 3π ) j (2.5)3

= (−3.84 mm/s 2 )i + (70.1582 mm/s 2 ) j

or

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PROBLEM 11.94 The motion of a particle is defined by the position vector r = A(cos t + t sin t )i + A(sin t − t cos t ) j, where t is expressed in seconds. Determine the values of t for which the position vector and the acceleration are (a) perpendicular, (b) parallel.

SOLUTION We have

r = A(cos t + t sin t )i + A(sin t − t cos t ) j

Then

v=

and

a=

(a)

dr = A( − sin t + sin t + t cos t )i dt + A(cos t − cos t + t sin t ) j = A(t cos t )i + A(t sin t ) j dv = A(cos t − t sin t )i + A(sin t + t cos t ) j dt

When r and a are perpendicular, r ⋅ a = 0 A[(cos t + t sin t )i + (sin t − t cos t ) j] ⋅ A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0

(b)

or

(cos t + t sin t )(cos t − t sin t ) + (sin t − t cos t )(sin t + t cos t ) = 0

or

(cos 2 t − t 2 sin 2 t ) + (sin 2 t − t 2 cos 2 t ) = 0

or

1− t2 = 0

or

t =1 s 

or

t=0 

When r and a are parallel, r × a = 0 A[(cos t + t sin t )i + (sin t − t cos t ) j] × A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0

or Expanding

[(cos t + t sin t )(sin t + t cos t ) − (sin t − t cos t )(cos t − t sin t )]k = 0 (sin t cos t + t + t 2 sin t cos t ) − (sin t cos t − t + t 2 sin t cos t ) = 0 2t = 0

or

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PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

SOLUTION We have

r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k

Then

v=

and

a=

dr = R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k dt

dv dt = R(−ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t )i + R(ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t )k = R(−2ωn sin ωn t − ωn2 t cos ωn t )i + R(2ωn cos ωn t − ωn2 t sin ωn t )k

Now

v 2 = vx2 + v 2y + vz2 = [ R(cos ωn t − ωn t sin ωn t )]2 + (c)2 + [ R(sin ωn t + ωn t cos ωn t )]2

(

)

= R 2  cos 2 ωn t − 2ωn t sin ωn t cos ωn t + ωn2 t 2 sin 2 ωn t  + sin 2 ωn t + 2ωn t sin ωn t cos ωn t + ωn2 t 2 cos 2 ωn t  + c 2 

(

(

)

)

= R 2 1 + ωn2 t 2 + c 2

(

Also,

)

v = R 2 1 + ωn2 t 2 + c 2 

or a 2 = ax2 + a 2y + az2

(

) cos ω t − ω t sin ω t )  

2

=  R −2ωn sin ωn t − ωn2 t cos ωn t  + (0) 2  

( (

2

2 +  R 2ωn n n n  = R 2  4ωn2 sin 2 ωn t + 4ωn3t sin ωn t cos ωn t + ωn4t 2 cos 2 ωn t  + 4ωn2 cos 2 ωn t − 4ωn3t sin ωn t cos ωn t + ωn4 t 2 sin 2 ωn t  

(

(

= R 2 4ωn2 + ωn4 t 2

)

)

)

a = Rωn 4 + ωn2 t 2 

or

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PROBLEM 11.96 The three-dimensional motion of a particle is defined by the position vector r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k , where r and t are expressed in feet and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid (y/A)2 − (x/A)2 − (z/B)2 = 1. For A = 3 and B = 1, determine (a) the magnitudes of the velocity and acceleration when t = 0, (b) the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other.

SOLUTION We have

r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k

or

x = At cos t cos t =

Then

x At

y = A t 2 + 1 z = Bt sin t sin t =

2

 y t2 =   −1  A

z Bt 2

2

2

2

2

x z t2 =   +    A  B

or 2

Then

 y x z  A  −1 =  A  +  B       

or

 y  x  z  A  −  A  −  B  =1      

2

(a)

2

 x   z  cos 2 t + sin 2 t = 1    +   = 1  At   Bt 

Now

2

2



Q.E.D.

With A = 3 and B = 1, we have v=

dr t j + (sin t + t cos t )k = 3(cos t − t sin t )i + 3 2 dt t +1

( ) t

and

t 2 + 1 − t t 2 +1 dv a= j = 3( − sin t − sin t − t cos t )i + 3 dt (t 2 + 1) + (cos t + cos t − t sin t )k 1 j + (2 cos t − t sin t )k = −3(2 sin t + t cos t )i + 3 2 (t + 1)3/2

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PROBLEM 11.96 (Continued)

At t = 0:

v = 3(1 − 0)i + (0) j + (0)k v = vx2 + v y2 + vz2 v = 3 ft/s 

or and Then

a = −3(0)i + 3(1) j + (2 − 0)k a 2 = (0) 2 + (3) 2 + (2) 2 = 13 a = 3.61 ft/s 2 

or (b)

If r and v are perpendicular, r ⋅ v = 0 t   [(3t cos t )i + (3 t 2 + 1) j + (t sin t )k ] ⋅ [3(cos t − t sin t )i +  3  j + (sin t + t cos t )k ] = 0 2  t +1 

or Expanding

t   (3t cos t )[3(cos t − t sin t )] + (3 t 2 + 1)  3  + (t sin t )(sin t + t cos t ) = 0 2 t 1 +   (9t cos 2 t − 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0 10 + 8 cos 2 t − 8t sin t cos t = 0

or (with t ≠ 0)

7 + 2 cos 2t − 2t sin 2t = 0

or Using “trial and error” or numerical methods, the smallest root is

t = 3.82 s 

Note: The next root is t = 4.38 s. 

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PROBLEM 11.97 An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 300 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B.

SOLUTION First note

v0 = 180 km/h = 264 ft/s

Place origin of coordinates at Point A. Vertical motion. (Uniformly accelerated motion) y = 0 + (0)t −

At B: or

1 2 gt 2

1 −300 ft = − (32.2 ft/s 2 )t 2 2 t B = 4.31666 s

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

At B:

d = (264 ft/s)(4.31666 s) d = 1140 ft 

or

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PROBLEM 11.98 A helicopter is flying with a constant horizontal velocity of 180 km/h and is directly above Point A when a loose part begins to fall. The part lands 6.5 s later at Point B on an inclined surface. Determine (a) the distance d between Points A and B, (b) the initial height h.

SOLUTION Place origin of coordinates at Point A. Horizontal motion:

(vx )0 = 180 km/h = 50 m/s x = x0 + (vx )0 t = 0 + 50t m

At Point B where t B = 6.5 s, (a)

Distance AB. 325 cos10°

From geometry

d=

Vertical motion:

y = y0 + ( v y ) 0 t −

At Point B (b)

xB = (50)(6.5) = 325 m

d = 330 m  1 2 gt 2

1 − xB tan10° = h + 0 − (9.81)(6.5) 2 2 h = 149.9 m 

Initial height.

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PROBLEM 11.99 A baseball pitching machine “throws” baseballs with a horizontal velocity v0. Knowing that height h varies between 788 mm and 1068 mm, determine (a) the range of values of v0, (b) the values of α corresponding to h = 788 mm and h = 1068 mm.

SOLUTION

(a)

y0 = 1.5 m, (v y )0 = 0

Vertical motion: t =

2( y0 − y ) g

or

tB =

2( y0 − h) g

When h = 788 mm = 0.788 m,

tB =

(2)(1.5 − 0.788) = 0.3810 s 9.81

When h = 1068 mm = 1.068 m,

tB =

(2)(1.5 − 1.068) = 0.2968 s 9.81

y = y0 + (v y )0 t −

At Point B,

Horizontal motion:

1 2 gt 2

or y =h

x0 = 0, (vx )0 = v0 , x = v0t

or

v0 =

x x = B t tB

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PROBLEM 11.99 (Continued)

v0 =

12.2 = 32.02 m/s 0.3810

and

v0 =

12.2 = 41.11 m/s 0.2968

32.02 m/s ≤ v0 ≤ 41.11 m/s

or

Vertical motion:

v y = (v y )0 − gt = − gt

Horizontal motion:

vx = v0

With xB = 12.2 m,

(b)

we get

tan α = −

115.3 km/h ≤ v0 ≤ 148.0 km/h 

(v y ) B dy gt =− = B dx (vx ) B v0

For h = 0.788 m,

tan α =

(9.81)(0.3810) = 0.11673, 32.02

α = 6.66° 

For h = 1.068 m,

tan α =

(9.81)(0.2968) = 0.07082, 41.11

α = 4.05° 

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PROBLEM 11.100 While delivering newspapers, a girl throws a newspaper with a horizontal velocity v0. Determine the range of values of v0 if the newspaper is to land between Points B and C.

SOLUTION Vertical motion. (Uniformly accelerated motion) y = 0 + (0)t −

1 2 gt 2

Horizontal motion. (Uniform) x = 0 + (vx )0 t = v0 t

At B:

y:

t B = 0.455016 s

or Then

x:

y:

or

1 −2 ft = − (32.2 ft/s 2 )t 2 2 tC = 0.352454 s

or Then

7 ft = (v0 ) B (0.455016 s) (v0 ) B = 15.38 ft/s

or At C:

1 1 −3 ft = − (32.2 ft/s 2 )t 2 3 2

x:

1 12 ft = (v0 )C (0.352454 s) 3 (v0 )C = 35.0 ft/s 15.38 ft/s < v0 < 35.0 ft/s 

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PROBLEM 11.101 Water flows from a drain spout with an initial velocity of 2.5 ft/s at an angle of 15° with the horizontal. Determine the range of values of the distance d for which the water will enter the trough BC.

SOLUTION First note

(vx )0 = (2.5 ft/s) cos 15° = 2.4148 ft/s (v y )0 = −(2.5 ft/s) sin 15° = −0.64705 ft/s

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y ) 0 t −

1 2 gt 2

At the top of the trough 1 −8.8 ft = (−0.64705 ft/s) t − (32.2 ft/s 2 ) t 2 2

or

t BC = 0.719491 s

(the other root is negative)

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

In time t BC

xBC = (2.4148 ft/s)(0.719491 s) = 1.737 ft

Thus, the trough must be placed so that xB < 1.737 ft or xC ≥ 1.737 ft 0 < d < 1.737 ft 

Since the trough is 2 ft wide, it then follows that

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PROBLEM 11.102 Milk is poured into a glass of height 140 mm and inside diameter 66 mm. If the initial velocity of the milk is 1.2 m/s at an angle of 40° with the horizontal, determine the range of values of the height h for which the milk will enter the glass.

SOLUTION First note (vx )0 = (1.2 m/s) cos 40° = 0.91925 m/s (v y )0 = −(1.2 m/s) sin 40° = −0.77135 m/s

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

Vertical motion. (Uniformly accelerated motion) y = y0 + ( v y ) 0 t −

1 2 gt 2

Milk enters glass at B. x:

0.08 m = (0.91925 m/s) t or tB = 0.087028 s

y : 0.140 m = hB + (−0.77135 m/s)(0.087028 s) 1 − (9.81 m/s 2 )(0.087028 s) 2 2

or

hB = 0.244 m

Milk enters glass at C. x : 0.146 m = (0.91925 m/s) t or tC = 0.158825 s y : 0.140 m = hC + (−0.77135 m/s)(0.158825 s) 1 − (9.81 m/s 2 )(0.158825 s) 2 2

or

hC = 0.386 m 0.244 m < h < 0.386 m 



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PROBLEM 11.103 A volleyball player serves the ball with an initial velocity v0 of magnitude 13.40 m/s at an angle of 20° with the horizontal. Determine (a) if the ball will clear the top of the net, (b) how far from the net the ball will land.

SOLUTION First note

(vx )0 = (13.40 m/s) cos 20° = 12.5919 m/s (v y )0 = (13.40 m/s) sin 20° = 4.5831 m/s

(a)

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

At C

9 m = (12.5919 m/s) t or tC = 0.71475 s

Vertical motion. (Uniformly accelerated motion) y = y0 + ( v y ) 0 t −

At C:

1 2 gt 2

yC = 2.1 m + (4.5831 m/s)(0.71475 s) 1 − (9.81 m/s 2 )(0.71475 s)2 2 = 2.87 m yC > 2.43 m (height of net)  ball clears net 

(b)

At B, y = 0:

1 0 = 2.1 m + (4.5831 m/s)t − (9.81 m/s 2 )t 2 2

Solving

t B = 1.271175 s (the other root is negative)

Then

d = (vx )0 t B = (12.5919 m/s)(1.271175 s) = 16.01 m b = (16.01 − 9.00) m = 7.01 m from the net 

The ball lands

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PROBLEM 11.104 A golfer hits a golf ball with an initial velocity of 160 ft/s at an angle of 25° with the horizontal. Knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B where the ball first lands.

SOLUTION (vx )0 = (160 ft/s) cos 25°

First note

(v y )0 = (160 ft/s) sin 25° xB = d cos 5°

and at B Now

yB = − d sin 5°

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t d cos 5° = (160 cos 25°)t or t B =

At B

cos 5° d 160 cos 25°

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t −

1 2 gt 2

( g = 32.2 ft/s 2 ) 1 2 gt B 2

At B:

− d sin 5° = (160 sin 25°)t B −

Substituting for t B

 cos 5°  1  cos 5°  2 − d sin 5° = (160 sin 25°)  d − g   d 2  160 cos 25°   160 cos 25° 

2

or

2 (160 cos 25°) 2 (tan 5° + tan 25°) 32.2 cos 5° = 726.06 ft

d=

d = 242 yd 

or

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PROBLEM 11.105 A homeowner uses a snowblower to clear his driveway. Knowing that the snow is discharged at an average angle of 40° with the horizontal, determine the initial velocity v0 of the snow.

SOLUTION First note

(vx )0 = v0 cos 40° (v y )0 = v0 sin 40°

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

At B:

14 = (v0 cos 40°) t or t B =

14 v0 cos 40°

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t −

At B:

1 2 gt 2

1.5 = (v0 sin 40°) t B −

( g = 32.2 ft/s 2 )

1 2 gt B 2

Substituting for t B   1   14 14 1.5 = (v0 sin 40°)  − g   v0 cos 40°  2  v0 cos 40° 

or

v02

=

1 2

2

(32.2)(196)/ cos 2 40° −1.5 + 14 tan 40° v0 = 22.9 ft/s 

or

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PROBLEM 11.106 At halftime of a football game souvenir balls are thrown to the spectators with a velocity v0. Determine the range of values of v0 if the balls are to land between Points B and C.

SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below Point A. The coordinates of Point A are x0 = 0, y0 = 2m. The components of initial velocity are (vx )0 = v0 cos 40° m/s and (v y )0 = v0 sin 40°. Horizontal motion:

x = x0 + (vx )0 t = (v0 cos 40°)t

Vertical motion:

y = y0 + ( v y ) 0 t =

From (1), Then

1 2 gt 2 1 = 2 + (v0 sin 40°) = − (9.81)t 2 2

v0 t =

(2)

x cos 40°

(3)

y = 2 + x tan 40° − 4.905t 2 t2 =

Point B:

(1)

2 + x tan 40° − y 4.905

(4)

x = 8 + 10 cos 35° = 16.1915 m y = 1.5 + 10sin 35° = 7.2358 m 16.1915 v0 t = = 21.1365 m cos 40° 2 + 16.1915 tan 40° − 7.2358 t2 = 4.905 21.1365 v0 = 1.3048

t = 1.3048 s v0 = 16.199 m/s

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PROBLEM 11.106 (Continued)

Point C:

x = 8 + (10 + 7) cos 35° = 21.9256 m y = 1.5 + (10 + 7)sin 35° = 11.2508 m v0 t =

21.9256 = 28.622 m cos 40°

t2 =

2 + 21.9256 tan 40° − 11.2508 4.905

v0 =

28.622 1.3656

t = 1.3656 s v0 = 20.96 m/s 16.20 m/s < v0 < 21.0 m/s 

Range of values of v0.

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PROBLEM 11.107 A basketball player shoots when she is 16 ft from the backboard. Knowing that the ball has an initial velocity v0 at an angle of 30° with the horizontal, determine the value of v0 when d is equal to (a) 9 in., (b) 17 in.

SOLUTION First note

(vx )0 = v0 cos 30°

Horizontal motion. (Uniform)

(v y )0 = v0 sin 30°

x = 0 + (v x ) 0 t

(16 − d ) = (v0 cos 30°) t or t B =

At B:

16 − d v0 cos 30°

y = 0 + (v y )0 t −

Vertical motion. (Uniformly accelerated motion)

1 2 gt 2

1 2 gt B 2

At B:

3.2 = (v0 sin 30°) t B −

Substituting for tB

 16 − d  1  16 − d  3.2 = (v0 sin 30°)  − g   v0 cos 30°  2  v0 cos 30° 

or

v02 =

(a)

d = 9 in.:

v02

d = 17 in.:

v02

(b)

=

=

( g = 32.2 ft/s 2 )

2

2 g (16 − d ) 2 3 

1 3

(16 − d ) − 3.2  

2(32.2) (16 − 129 )

3 

1 3

(16 − 129 ) − 3.2

2(32.2) (16 − 17 12 )

3 

1 3

2

v0 = 29.8 ft/s 

2

(16 − 1217 ) − 3.2

v0 = 29.6 ft/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 156

PROBLEM 11.108 A tennis player serves the ball at a height h = 2.5 m with an initial velocity of v0 at an angle of 5° with the horizontal. Determine the range for which of v0 for which the ball will land in the service area which extends to 6.4 m beyond the net.

SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below the point where the racket impacts the ball. The coordinates of this impact point are x0 = 0, y0 = h = 2.5 m. The components of initial velocity are (vx )0 = v0 cos 5° and (v y )0 = v0 sin 5°. Horizontal motion:

x = x0 + (vx )0 t = (v0 cos 5°)t

Vertical motion:

y = y0 + ( v y ) 0 t =

(1)

1 2 gt 2

1 = 2.5 − (v0 sin 5°)t = − (9.81)t 2 2

From (1), Then

v0 t =

x cos 5°

(2) (3)

y = 2.5 − x tan 5° − 4.905t 2 t2 =

2.5 − x tan 5° − y 4.905

(4)

At the minimum speed the ball just clears the net. x = 12.2 m,

y = 0.914 m

12.2 = 12.2466 m cos 5° 2.5 − 12.2 tan 5° − 0.914 t2 = 4.905 12.2466 v0 = 0.32517

v0 t =

t = 0.32517 s v0 = 37.66 m/s

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PROBLEM 11.108 (Continued)

At the maximum speed the ball lands 6.4 m beyond the net. x = 12.2 + 6.4 = 18.6 m y=0 18.6 v0 t = = 18.6710 m cos 5° 2.5 − 18.6 tan 5° − 0 t2 = t = 0.42181 s 4.905 18.6710 v0 = v0 = 44.26 m/s 0.42181 37.7 m/s < v0 < 44.3 m/s 

Range for v0.

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PROBLEM 11.109 The nozzle at A discharges cooling water with an initial velocity v0 at an angle of 6° with the horizontal onto a grinding wheel 350 mm in diameter. Determine the range of values of the initial velocity for which the water will land on the grinding wheel between Points B and C.

SOLUTION First note (vx )0 = v0 cos 6° (v y )0 = −v0 sin 6°

Horizontal motion. (Uniform) x = x0 + (vx )0 t

Vertical motion. (Uniformly accelerated motion) y = y0 + (v y )0 t −

1 2 gt 2

( g = 9.81 m/s 2 )

x = (0.175 m) sin 10° y = (0.175 m) cos 10°

At Point B:

x : 0.175 sin 10° = −0.020 + (v0 cos 6°)t tB =

or

0.050388 v0 cos 6°

y : 0.175 cos 10° = 0.205 + (−v0 sin 6°)t B −

1 2 gtB 2

Substituting for t B  0.050388  1  0.050388  −0.032659 = (−v0 sin 6°)   − (9.81)    v0 cos 6°  2  v0 cos 6° 

2

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PROBLEM 11.109 (Continued)

v02 =

or

1 2

(9.81)(0.050388)2

cos 2 6°(0.032659 − 0.050388 tan 6°)

(v0 ) B = 0.678 m/s

or

x = (0.175 m) cos 30° y = (0.175 m) sin 30°

At Point C:

x : 0.175 cos 30° = −0.020 + (v0 cos 6°)t tC =

or

0.171554 v0 cos 6°

y : 0.175 sin 30° = 0.205 + (−v0 sin 6°)tC −

1 2 gtC 2

Substituting for tC  0.171554  1  0.171554  −0.117500 = (−v0 sin 6°)   − (9.81)    v0 cos 6°  2  v0 cos 6° 

or or

v02 =

1 2

2

(9.81)(0.171554) 2

cos 2 6°(0.117500 − 0.171554 tan 6°)

(v0 )C = 1.211 m/s 0.678 m/s < v0 < 1.211 m/s 



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PROBLEM 11.110 While holding one of its ends, a worker lobs a coil of rope over the lowest limb of a tree. If he throws the rope with an initial velocity v0 at an angle of 65° with the horizontal, determine the range of values of v0 for which the rope will go over only the lowest limb.

SOLUTION First note (vx )0 = v0 cos 65° (v y )0 = v0 sin 65°

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

At either B or C, x = 5 m s = (v0 cos 65°)t B,C

or

t B ,C =

5 (v0 cos 65°)

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t −

1 2 gt 2

( g = 9.81 m/s 2 )

At the tree limbs, t = tB ,C   1   5 5 yB ,C = (v0 sin 65°)  − g   v0 cos 65°  2  v0 cos 65° 

2

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PROBLEM 11.110 (Continued)

or

v02 = =

1 2

(9.81)(25)

2

cos 65°(5 tan 65° − yB , C ) 686.566 5 tan 65° − yB, C

At Point B:

v02 =

686.566 5 tan 65° − 5

or

(v0 ) B = 10.95 m/s

At Point C:

v02 =

686.566 5 tan 65° − 5.9

or

(v0 )C = 11.93 m/s 10.95 m/s < v0 < 11.93 m/s 



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PROBLEM 11.111 The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle α with the horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle α, (b) the angle θ that the velocity of the ball at Point B forms with the horizontal.

SOLUTION First note v0 = 72 km/h = 20 m/s (vx )0 = v0 cos α = (20 m/s) cos α

and

(v y )0 = v0 sin α = (20 m/s) sin α

(a)

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (20 cos α ) t

At Point B:

14 = (20 cos α )t or t B =

7 10 cos α

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t −

At Point B:

1 2 1 gt = (20 sin α )t − gt 2 2 2

0.08 = (20 sin α )t B −

( g = 9.81 m/s 2 )

1 2 gt B 2

Substituting for t B   1   7 7 0.08 = (20 sin α )  − g   10 cos α  2  10 cos α 

or Now

8 = 1400 tan α −

2

1 49 g 2 cos 2 α

1 = sec2 α = 1 + tan 2 α 2 cos α

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PROBLEM 11.111 (Continued)

Then or Solving

8 = 1400 tan α − 24.5 g (1 + tan 2 α ) 240.345 tan 2 α − 1400 tan α + 248.345 = 0

α = 10.3786° and α = 79.949°

Rejecting the second root because it is not physically reasonable, we have

α = 10.38°  (b)

We have

vx = (vx )0 = 20 cos α

and

v y = (v y )0 − gt = 20 sin α − gt

At Point B:

(v y ) B = 20 sin α − gt B = 20 sin α −

7g 10 cos α

Noting that at Point B, v y < 0, we have tan θ = = =

|(v y ) B | vx 7g 10 cos α

− 20 sin α

20 cos α 7 9.81 200 cos 10.3786°

− sin 10.3786°

cos 10.3786°

θ = 9.74° 

or

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PROBLEM 11.112 A model rocket is launched from Point A with an initial velocity v0 of 75 m/s. If the rocket’s descent parachute does not deploy and the rocket lands a distance d = 100 m from A, determine (a) the angle α that v0 forms with the vertical, (b) the maximum height above Point A reached by the rocket, and (c) the duration of the flight.

SOLUTION Set the origin at Point A.

x0 = 0,

y0 = 0

Horizontal motion:

x = v0 t sin α

Vertical motion:

y = v0t cos α − cos α = sin 2 α + cos 2 α =

sin α =

x v0t

1 2 gt 2

1  1  y + gt 2   v0 t  2 

(2)

2 1  2  1 2  x y gt + +     =1 2 (v0 t )2    

x 2 + y 2 + gyt 2 +

1 24 g t = v02t 2 4

1 24 g t − v02 − gy t 2 + ( x 2 + y 2 ) = 0 4

(

At Point B,

(1)

)

(3)

x 2 + y 2 = 100 m, x = 100 cos 30° m y = −100 sin 30° = −50 m 1 (9.81)2 t 4 − [752 − (9.81)(−50)] t 2 + 1002 = 0 4 24.0590 t 4 − 6115.5 t 2 + 10000 = 0 t 2 = 252.54 s 2 t = 15.8916 s

and 1.6458 s 2 and 1.2829 s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165

PROBLEM 11.112 (Continued)

Restrictions on α :

0 < α < 120° tan α =

x 2

y + gt 1 2

=

100 cos 30° = 0.0729 −50 + (4.905)(15.8916) 2

α = 4.1669° 100 cos 30° = −2.0655 −50 + (4.905)(1.2829)2 α = 115.8331°

and

Use α = 4.1669° corresponding to the steeper possible trajectory. (a)

Angle α .

(b)

Maximum height.

α = 4.17°  v y = 0 at

y = ymax

v y = v0 cos α − gt = 0 t=

v0 cos α g

ymax = v0 t cos α − =

(c)

Duration of the flight.

v 2 cos 2 α 1 gt = 0 2 2g

(75) 2 cos 2 4.1669° (2)(9.81)

ymax = 285 m 

(time to reach B) t = 15.89 s 



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PROBLEM 11.113 The initial velocity v0 of a hockey puck is 105 mi/h. Determine (a) the largest value (less than 45°) of the angle α for which the puck will enter the net, (b) the corresponding time required for the puck to reach the net.

SOLUTION First note

v0 = 105 mi/h = 154 ft/s (vx )0 = v0 cos α = (154 ft/s) cos α

and

(v y )0 = v0 sin α = (154 ft/s) sin α

(a)

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (154 cos α )t

At the front of the net, Then or

x = 16 ft 16 = (154 cos α )t tenter =

8 77 cos α

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (154 sin α ) t − gt 2 2

y = 0 + (v y )0 t −

( g = 32.2 ft/s 2 )

At the front of the net, yfront = (154 sin α ) tenter −

1 2 gtenter 2

  1   8 8 = (154 sin α )  − g  77 cos 2 77 cos α α     32 g = 16 tan α − 5929 cos 2 α

2

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PROBLEM 11.113 (Continued)

Now Then or

1 = sec2 α = 1 + tan 2 α cos 2 α yfront = 16 tan α − tan 2 α −

32 g (1 + tan 2 α ) 5929

 5929  5929 tan α + 1 + yfront  = 0 2g 32 g   5929 2g

(

 ±  − 5929 2g 

)

2

(

)

1/2

 − 4 1 + 5929 y 32 g front   2

Then

tan α =

or

 5929 2  5929 5929  ±  − tan α = yfront    − 1 + 4 × 32.2  4 × 32.2   32 × 32.2  

or

tan α = 46.0326 ± [(46.0326) 2 − (1 + 5.7541 yfront )]1/2

1/2

Now 0 < yfront < 4 ft so that the positive root will yield values of α > 45° for all values of yfront. When the negative root is selected, α increases as yfront is increased. Therefore, for α max , set yfront = yC = 4 ft

(b)

Then

tan α = 46.0326 − [(46.0326) 2 − (1 + 5.7541 + 4)]1/ 2

or

α max = 14.6604°

We had found

α max = 14.66° 

8 77 cos α 8 = 77 cos 14.6604°

tenter =

tenter = 0.1074 s 

or

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PROBLEM 11.114 A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the worker can wash from his position at A, (b) the corresponding angle α.

SOLUTION First note

(vx )0 = v0 cos α = (11.5 m/s) cos α (v y )0 = v0 sin α = (11.5 m/s) sin α

By observation, d max occurs when

ymax = 1.1 m.

Vertical motion. (Uniformly accelerated motion) v y = (v y )0 − gt = (11.5 sin α ) − gt y = ymax

When Then

at

1 2 gt 2 1 = (11.5 sin α ) t − gt 2 2

y = 0 + (v y ) 0 t −

B , (v y ) B = 0

(v y ) B = 0 = (11.5 sin α ) − gt 11.5 sin α g

( g = 9.81 m/s 2 )

or

tB =

and

yB = (11.5 sin α ) t B −

1 2 gt B 2

Substituting for t B and noting yB = 1.1 m  11.5 sin α 1.1 = (11.5 sin α )  g  1 = (11.5) 2 sin 2 α 2g

or

sin 2 α =

2.2 × 9.81 11.52

 1  11.5 sin α  − g  g  2  

2

α = 23.8265°

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169

PROBLEM 11.114 (Continued)

(a)

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (11.5 cos α ) t

At Point B: where Then

x = d max tB =

and t = t B

11.5 sin 23.8265° = 0.47356 s 9.81

d max = (11.5)(cos 23.8265°)(0.47356) d max = 4.98 m 

or (b)

α = 23.8° 

From above

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PROBLEM 11.115 An oscillating garden sprinkler which discharges water with an initial velocity v0 of 8 m/s is used to water a vegetable garden. Determine the distance d to the farthest Point B that will be watered and the corresponding angle α when (a) the vegetables are just beginning to grow, (b) the height h of the corn is 1.8 m.

SOLUTION First note

(vx )0 = v0 cos α = (8 m/s) cos α (v y )0 = v0 sin α = (8 m/s) sin α

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (8 cos α ) t

At Point B: or

x = d : d = (8 cos α ) t d tB = 8 cos α

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (8 sin α ) t − gt 2 ( g = 9.81 m/s 2 ) 2 1 0 = (8 sin α ) t B − gt B2 2 y = 0 + (v y )0 t −

At Point B: Simplifying and substituting for t B

0 = 8 sin α − d=

or (a)

1  d  g  2  8 cos α 

64 sin 2α g

(1)

When h = 0, the water can follow any physically possible trajectory. It then follows from Eq. (1) that d is maximum when 2α = 90°

α = 45° 

or Then

d=

64 sin (2 × 45°) 9.81 d max = 6.52 m 

or

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PROBLEM 11.115 (Continued)

(b)

Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as α increases in value from 0 to 45° and then d decreases as α is further increased. Thus, d max occurs for the value of α closest to 45° and for which the water just passes over the first row of corn plants. At this row, xcom = 1.5 m tcorn =

so that

1.5 8 cos α

Also, with ycorn = h, we have h = (8 sin α ) tcorn −

1 2 gtcorn 2

Substituting for tcorn and noting h = 1.8 m,  1.5  1  1.5  1.8 = (8 sin α )  − g   8 cos α  2  8 cos α 

or Now Then or

1.8 = 1.5 tan α −

2

2.25 g 128 cos 2 α

1 = sec2 α = 1 + tan 2 α cos 2 α 1.8 = 1.5 tan α −

2.25(9.81) (1 + tan 2 α ) 128

0.172441 tan 2 α − 1.5 tan α + 1.972441 = 0

α = 58.229° and α = 81.965°

Solving

From the above discussion, it follows that d = d max when

α = 58.2°  Finally, using Eq. (1) d=

64 sin (2 × 58.229°) 9.81 d max = 5.84 m 

or

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PROBLEM 11.116* A mountain climber plans to jump from A to B over a crevasse. Determine the smallest value of the climber’s initial velocity v0 and the corresponding value of angle α so that he lands at B.

SOLUTION First note

(vx )0 = v0 cos α (v y )0 = v0 sin α

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (v0 cos α ) t

At Point B: or

1.8 = (v0 cos α )t tB =

1.8 v0 cos α

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (v0 sin α ) t − gt 2 2

y = 0 + (v y )0 t −

At Point B:

−1.4 = (v0 sin α ) t B −

( g = 9.81 m/s 2 )

1 2 gt B 2

Substituting for t B  1.8 −1.4 = (v0 sin α )   v0 cos α

or

 1  1.8  − g   2  v0 cos α 

v02 =

1.62 g cos α (1.8 tan α + 1.4)

=

1.62 g 0.9 sin 2α + 1.4 cos 2 α

2

2

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PROBLEM 11.116* (Continued)

Now minimize v02 with respect to α. We have

dv02 −(1.8 cos 2α − 2.8 cos α sin α ) = 1.62 g =0 dα (0.9 sin 2α + 1.4 cos 2 α ) 2 1.8 cos 2α − 1.4 sin 2α = 0

or

tan 2α =

or

18 14

α = 26.0625° and α = 206.06°

or

Rejecting the second value because it is not physically possible, we have

α = 26.1°  Finally,

v02 =

1.62 × 9.81 cos 26.0625°(1.8 tan 26.0625° + 1.4) 2

(v0 )min = 2.94 m/s 

or

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PROBLEM 11.117 The velocities of skiers A and B are as shown. Determine the velocity of A with respect to B.

SOLUTION We have

vA = vB + vA/B

The graphical representation of this equation is then as shown. Then

v 2A/B = 302 + 452 − 2(30)(45) cos 15°

or

vA /B = 17.80450 ft/s

and or

30 17.80450 = sin α sin 15°

α = 25.8554° α + 25° = 50.8554°

vA /B = 17.8 ft/s

50.9° 

Alternative solution. vA /B = v A − vB

= 30 cos 10° i − 30 sin 10° j − (45 cos 25° i − 45° sin 25° j) = 11.2396i + 13.8084 j = 5.05 m/s = 17.8 ft/s

50.9°

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PROBLEM 11.118 The three blocks shown move with constant velocities. Find the velocity of each block, knowing that the relative velocity of A with respect to C is 300 mm/s upward and that the relative velocity of B with respect to A is 200 mm/s downward.

SOLUTION From the diagram Cable 1:

y A + yD = constant

Then

v A + vD = 0

Cable 2:

(1)

( yB − yD ) + ( yC − yD ) = constant vB + vC − 2vD = 0

Then

(2)

Combining Eqs. (1) and (2) to eliminate vD , 2v A + vB + vC = 0

(3)

Now

v A/C = v A − vC = −300 mm/s

(4)

and

vB/A = vB − v A = 200 mm/s

(5)

(3) + (4) − (5) 

Then

(2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200) v A = 125 mm/s 

or vB − (−125) = 200

and using Eq. (5)

v B = 75 mm/s 

or −125 − vC = −300

Eq. (4)

vC = 175 mm/s 

or

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PROBLEM 11.119 Three seconds after automobile B passes through the intersection shown, automobile A passes through the same intersection. Knowing that the speed of each automobile is constant, determine (a) the relative velocity of B with respect to A, (b) the change in position of B with respect to A during a 4-s interval, (c) the distance between the two automobiles 2 s after A has passed through the intersection.

SOLUTION

v A = 45 mi/h = 66 ft/s vB = 30 mi/h = 44 ft/s

Law of cosines vB2/A = 662 + 442 − 2(66)(44) cos110° vB/A = 90.99 ft/s

vB = v A + vB/A

Law of sines sin β sin110° = 66 90.99

β = 42.97°

α = 90° − β = 90° − 42.97° = 47.03° v B/A = 91.0 ft/s

(a)

Relative velocity:

(b)

Change in position for Δt = 4 s. ΔrB/A = vB/A Δt = (91.0 ft/s)(4 s)

(c)

rB/A = 364 ft

47.0° 

47.0° 

Distance between autos 2 seconds after auto A has passed intersection. Auto A travels for 2 s. v A = 66 ft/s

20°

rA = v At = (66 ft/s)(2 s) = 132 ft rA = 132 ft

20°

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PROBLEM 11.119 (Continued)

v B = 44 ft/s

Auto B

rB = v B t = (44 ft/s)(5 s) = 220 ft rB = rA + rB/A

Law of cosines rB2/A = (132) 2 + (220) 2 − 2(132)(220) cos110° rB/A = 292.7 ft

Distance between autos = 293 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178

PROBLEM 11.120 Shore-based radar indicates that a ferry leaves its slip with a velocity v = 18 km/h 70°, while instruments aboard the ferry indicate a speed of 18.4 km/h and a heading of 30° west of south relative to the river. Determine the velocity of the river.

SOLUTION

We have

v F = v R + v F/R

or

v F = v F/R + v R

The graphical representation of the second equation is then as shown. We have

vR2 = 182 + 18.42 − 2(18)(18.4) cos 10°

or

vR = 3.1974 km/h

and or

18 3.1974 = sin α sin 10°

α = 77.84°

Noting that v R = 3.20 km/h

17.8° 

Alternatively one could use vector algebra.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179

PROBLEM 11.121 Airplanes A and B are flying at the same altitude and are tracking the eye of hurricane C. The relative velocity of C with respect to A is vC/A = 350 km/h 75°, and the relative velocity of C with respect to B is vC/B = 400 km/h 40°. Determine (a) the relative velocity of B with respect to A, (b) the velocity of A if ground-based radar indicates that the hurricane is moving at a speed of 30 km/h due north, (c) the change in position of C with respect to B during a 15-min interval.

SOLUTION (a)

We have

v C = v A + v C/ A

and

v C = v B + v C/ B

Then

v A + vC/A = v B + v C/B

or

v B − v A = v C/A − vC/B

Now

v B − v A = v B/A

so that

v B/A = vC/A − vC/B

or

vC/A = vC/B + v B/A

The graphical representation of the last equation is then as shown. We have

vB2/A = 3502 + 4002 − 2(350)(400) cos 65°

or

vB/A = 405.175 km/h

and or

400 405.175 = sin α sin 65°

α = 63.474° 75° − α = 11.526°

v B/A = 405 km/h

11.53° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180

PROBLEM 11.121 (Continued)

(b)

We have

v C = v A + v C/ A

or

v A = (30 km/h) j − (350 km/h)(− cos 75°i − sin 75° j ) v A = (90.587 km/h)i + (368.07 km/h) j v A = 379 km/h

or (c)

76.17° 

Noting that the velocities of B and C are constant, we have rB = (rB )0 + v B t

Now

rC = (rC )0 + v C t

rC/B = rC − rB = [(rC )0 − (rB )0 ] + ( vC − v B )t

= [(rC )0 − (rB )0 ] + v C/B t

Then

ΔrC/B = (rC/B )t2 − (rC/B )t1 = v C/B (t2 − t1 ) = vC/B Δt

For Δt = 15 min:

1  ΔrC/B = (400 km/h)  h  = 100 km 4 

ΔrC/B = 100 km

40° 

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PROBLEM 11.122 Pin P moves at a constant speed of 150 mm/s in a counterclockwise sense along a circular slot which has been milled in the slider block A shown. Knowing that the block moves downward at a constant speed 100 mm/s determine the velocity of pin P when (a) θ = 30°, (b) θ = 120°.

SOLUTION v P = v A + v P/A v P = 100 ms (− j) + 150(cos θ i + sin θ j)mm/s

(a)

For θ = 30°

v P = −100 mm/s ( j) + 150( − cos(30°)i + sin(30°) j)mm/s v P = (−75i + 29.9038 j)mm/s v P = 80.7 mm/s

(b)

For θ = 120°

21.7° 

v P = −100 mm/s ( j) + 150(− cos(120°)i + sin(120°) j)mm/s v P = (−129.9038i + −175 j) mm/s v P = 218 mm/s

53.4° 

Alternative Solution (a)

For θ = 30°, vP /A = 7.5 in./s

30°

vP = v A + vP/A

Law of cosines vP2 = (150) 2 + (100) 2 − 2(100)(150) cos 30° vP = 80.7418 mm/s

Law of sines sin β sin 30° = 150 80.7418

β = 111.7° vP = 80.7 mm/s

21.7° 

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PROBLEM 11.122 (Continued)

(b)

For θ = 120°, vP/A = 150 mm/s

30°

Law of cosines vP2 = (150) 2 + (100) 2 − 2(100)(150) cos120° vP = 217.9449 mm/s

Law of sines sin β sin120° = β = 36.6° 150 217.9449 α = 90 − β = 90° − 36.6 = 53.4°

vP = 218 mm/s

53.4° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183

PROBLEM 11.123 Knowing that at the instant shown assembly A has a velocity of 9 in./s and an acceleration of 15 in./s2 both directed downward, determine (a) the velocity of block B, (b) the acceleration of block B.

SOLUTION Length of cable = constant L = x A + 2 xB/A = constant v A + 2vB/A = 0

(1)

a A + 2aB/A = 0

(2)

a A = 15 in./s 2

Data:

v A = 9 in./s

Eqs. (1) and (2)

a A = −2aB/A

v A = −2vB/A

15 = −2aB/A

9 = −2vB/A

aB/A = −7.5 in./s 2 a B/A = 7.5 in./s 2

(a)

vB/A = −4.5 in./s

40°

v B/A = −4.5 in./s

Velocity of B.

v B = v A + v B/A

Law of cosines:

vB2 = (9) 2 + (4.5) 2 − 2(9)(4.5) cos 50°

40°

vB = 7.013 in./s

Law of sines:

sin β sin 50° = 4.5 7.013

β = 29.44°

α = 90° − β = 90° − 29.44° = 60.56° v B = 7.01 in./s

60.6° 

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PROBLEM 11.123 (Continued)

(b)

Acceleration of B. a B may be found by using analysis similar to that used above for vB . An alternate method is a B = a A + a B/A a B = 15 in./s 2 ↓ +7.5 in./s 2

40°

= −15 j − (7.5 cos 40°)i + (7.5 sin 40°) j = −15 j − 5.745i + 4.821j a B = −5.745i − 10.179 j a B = 11.69 in./s 2

60.6° 

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PROBLEM 11.124 Knowing that at the instant shown block A has a velocity of 8 in./s and an acceleration of 6 in./s2 both directed down the incline, determine (a) the velocity of block B, (b) the acceleration of block B.

SOLUTION From the diagram

2 x A + xB/A = constant

Then

2v A + vB/A = 0 | vB/A | = 16 in./s

or

2a A + aB/A = 0

and

| aB/A | = 12 in./s 2

or

Note that v B/A and a B/A must be parallel to the top surface of block A. (a)

We have

v B = v A + v B/A

The graphical representation of this equation is then as shown. Note that because A is moving downward, B must be moving upward relative to A. We have

vB2 = 82 + 162 − 2(8)(16) cos 15°

or

vB = 8.5278 in./s

and or

8 8.5278 = sin α sin 15°

α = 14.05° v B = 8.53 in./s

(b)

54.1° 

The same technique that was used to determine v B can be used to determine a B . An alternative method is as follows. We have

a B = a A + a B/A

= (6i ) + 12(− cos 15°i + sin 15° j)* = −(5.5911 in./s 2 )i + (3.1058 in./s 2 ) j a B = 6.40 in./s 2

or

54.1° 

* Note the orientation of the coordinate axes on the sketch of the system. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 186

PROBLEM 11.125 A boat is moving to the right with a constant deceleration of 0.3 m/s2 when a boy standing on the deck D throws a ball with an initial velocity relative to the deck which is vertical. The ball rises to a maximum height of 8 m above the release point and the boy must step forward a distance d to catch it at the same height as the release point. Determine (a) the distance d, (b) the relative velocity of the ball with respect to the deck when the ball is caught.

SOLUTION Horizontal motion of the ball:

vx = (vx )0 ,

Vertical motion of the ball:

v y = (v y )0 − gt yB = (v y )0 t −

At maximum height,

xball = (vx )0 t

1 2 gt , (v y ) 2 − (v y )02 = −2 gy 2 vy = 0 and y = ymax

(v y )2 = 2 gymax = (2)(9.81)(8) = 156.96 m 2 /s 2 (v y )0 = 12.528 m/s

At time of catch,

tcatch = 2.554 s

or Motion of the deck:

1 (9.81)t 2 2 and v y = 12.528 m/s

y = 0 = 12.528 −

vx = (vx )0 + aDt ,

xdeck = (vx )0 t +

1 a Dt 2 2

Motion of the ball relative to the deck: (vB/D ) x = (vx )0 − [(vx )0 + aDt ] = −aDt 1 1   xB/D = (vx )0 t − (vx )0 t + aDt 2  = − aDt 2 2 2   (vB/D ) y = (v y )0 − gt , yB/D = yB

(a) (b)

At time of catch,

1 d = xD/B = − (− 0.3)(2.554) 2 2 (vB/D ) x = −(− 0.3)(2.554) = + 0.766 m/s (vB/D ) y = 12.528 m/s

d = 0.979 m  or 0.766 m/s v B/D = 12.55 m/s

86.5° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187

PROBLEM 11.126 The assembly of rod A and wedge B starts from rest and moves to the right with a constant acceleration of 2 mm/s2. Determine (a) the acceleration of wedge C, (b) the velocity of wedge C when t = 10 s.

SOLUTION (a)

We have

a C = a B + a C/ B

The graphical representation of this equation is then as shown. First note

α = 180° − (20° + 105°) = 55°

Then

aC 2 = sin 20° sin 55° aC = 0.83506 mm/s 2

(b)

aC = 0.835 mm/s 2

75° 

vC = 8.35 mm/s

75° 

For uniformly accelerated motion vC = 0 + aC t

At t = 10 s:

vC = (0.83506 mm/s 2 )(10 s) = 8.3506 mm/s

or

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PROBLEM 11.127 Determine the required velocity of the belt B if the relative velocity with which the sand hits belt B is to be (a) vertical, (b) as small as possible.

SOLUTION A grain of sand will undergo projectile motion. vsx = vsx = constant = −5 ft/s 0

vs y = 2 gh = (2)(32.2 ft/s 2 )(3 ft) = 13.90 ft/s ↓

y-direction.

v S/B = v S − v B

Relative velocity. (a)

(1)

If vS/B is vertical, −vS /B j = −5i − 13.9 j − ( −vB cos 15°i + vB sin 15° j) = −5i − 13.9 j + vB cos 15°i − vB sin 15° j

Equate components.

i : 0 = −5 + vB cos 15°

vB =

5 = 5.176 ft/s cos 15° v B = 5.18 ft/s

(b)

15° 

vS/C is as small as possible, so make vS/B ⊥ to vB into (1). −vS/B sin 15°i − vS/B cos 15° j = − 5i − 13.9 j + vB cos 15°i − vB sin 15° j

Equate components and transpose terms. (sin 15°) vS/B + (cos 15°) vB = 5 (cos 15°) vS/B − (sin 15°) vB = 13.90

Solving,

vS/B = 14.72 ft/s

vB = 1.232 ft/s v B = 1.232 ft/s

15° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189

PROBLEM 11.128 Conveyor belt A, which forms a 20° angle with the horizontal, moves at a constant speed of 4 ft/s and is used to load an airplane. Knowing that a worker tosses duffel bag B with an initial velocity of 2.5 ft/s at an angle of 30° with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.

SOLUTION First determine the velocity of the bag as it lands on the belt. Now [(vB ) x ]0 = (vB )0 cos 30° = (2.5 ft/s) cos 30° [(vB ) y ]0 = (vB )0 sin 30° = (2.5 ft/s) sin 30°

Horizontal motion. (Uniform) x = 0 + [(vB ) x ]0 t

(vB ) x = [(vB ) x ]0

= (2.5 cos 30°) t

= 2.5 cos 30°

Vertical motion. (Uniformly accelerated motion) y = y0 + [(vB ) y ]0 t −

1 2 gt 2

= 1.5 + (2.5 sin 30°) t −

(vB ) y = [(vB ) y ]0 − gt

1 2 gt 2

= 2.5 sin 30° − gt

The equation of the line collinear with the top surface of the belt is y = x tan 20°

Thus, when the bag reaches the belt 1.5 + (2.5 sin 30°) t −

1 2 gt = [(2.5 cos 30°) t ] tan 20° 2

or

1 (32.2) t 2 + 2.5(cos 30° tan 20° − sin 30°) t − 1.5 = 0 2

or

16.1t 2 − 0.46198t − 1.5 = 0

Solving

t = 0.31992 s and t = −0.29122 s (Reject)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190

PROBLEM 11.128 (Continued)

The velocity v B of the bag as it lands on the belt is then v B = (2.5 cos 30°)i + [2.5 sin 30° − 32.2(0.319 92)] j

= (2.1651 ft/s)i − (9.0514 ft/s) j

Finally or

v B = v A + v B/A v B/A = (2.1651i − 9.0514 j) − 4(cos 20°i + sin 20° j)

= −(1.59367 ft/s)i − (10.4195 ft/s) j v B/A = 10.54 ft/s

or

81.3° 

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PROBLEM 11.129 During a rainstorm the paths of the raindrops appear to form an angle of 30° with the vertical and to be directed to the left when observed from a side window of a train moving at a speed of 15 km/h. A short time later, after the speed of the train has increased to 24 km/h, the angle between the vertical and the paths of the drops appears to be 45°. If the train were stopped, at what angle and with what velocity would the drops be observed to fall?

SOLUTION vrain = vtrain + vrain/train

Case :

vT = 15 km/h

;

vR / T

30°

Case :

vT = 24 km/h

;

vR / T

45°

Case :

(vR ) y tan 30 = 15 − (vR ) x

(1)

Case :

(vR ) y tan 45° = 24 − (vR ) x

(2)

Substract (1) from (2)

(vR ) y (tan 45° − tan 30°) = 9 (vR ) y = 21.294 km/h

Eq. (2):

21.294 tan 45° = 25 − (vR ) x (vR ) x = 2.706 km/h

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PROBLEM 11.129 (Continued)

3.706 21.294 β = 7.24° 21.294 = 21.47 km/h = 5.96 m/s vR = cos 7.24°

tan β =

vR = 5.96 m/s

82.8° 

Alternate solution Alternate, vector equation

v R = vT + v R / T

For first case,

v R = 15i + vR / T −1 (− sin 30°i − cos 30° j)

For second case,

v R = 24i + vR /T − 2 (− sin 45°i − cos 45° j)

Set equal 15i + vR /T −1 (− sin 30°i − cos 30° j) = 24i + vR /T − 2 (− sin 45°i − cos 45° j)

Separate into components: i:

15 − vR /T −1 sin 30° = 24 − vR /T − 2 sin 45° −vR /T −1 sin 30° + vR /T − 2 sin 45° = 9

(3)

−vR /T −1 cos 30° = −vR / T − 2 cos 45°

j:

vR / T −1 cos 30° + vR /T − 2 cos 45° = 0

(4)

Solving Eqs. (3) and (4) simultaneously, vR /T −1 = 24.5885 km/h

vR /T − 2 = 30.1146 km/h

Substitute v R /T − 2 back into equation for v R . v R = 24i + 30.1146(− sin 45°i − cos 45° j) v R = 2.71i − 21.29 j

 −21.29   = −82.7585°  2.71 

θ = tan −1 

v R = 21.4654 km/hr = 5.96 m/s v R = 5.96 m/s

82.8° 

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PROBLEM 11.130 As observed from a ship moving due east at 9 km/h, the wind appears to blow from the south. After the ship has changed course and speed, and as it is moving north at 6 km/h, the wind appears to blow from the southwest. Assuming that the wind velocity is constant during the period of observation, determine the magnitude and direction of the true wind velocity.

SOLUTION v wind = v ship + v wind/ship v w = v s + v w/s

Case  v s = 9 km/h →; v w /s

Case  v s = 6 km/h ↑ ; v w /s 15 = 1.6667 9 α = 59.0°

tan α =

vw = 92 + 152 = 17.49 km/h

v w = 17.49 km/h

59.0° 

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PROBLEM 11.131 When a small boat travels north at 5 km/h, a flag mounted on its stern forms an angle θ = 50° with the centerline of the boat as shown. A short time later, when the boat travels east at 20 km/h, angle θ is again 50°. Determine the speed and the direction of the wind.

SOLUTION We have

vW = v B + vW/B

Using this equation, the two cases are then graphically represented as shown.

With vW now defined, the above diagram is redrawn for the two cases for clarity.

Noting that

θ = 180° − (50° + 90° + α ) = 40° − α We have

vW 5 = sin 50° sin (40° − α )

φ = 180° − (50° + α ) = 130° − α

vW 20 = sin 50° sin (130° − α )

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PROBLEM 11.131 (Continued)

Therefore or or

5 20 = sin (40° − α ) sin (130° − α ) sin 130° cos α − cos 130° sin α = 4(sin 40° cos α − cos 40° sin α ) tan α =

sin 130° − 4 sin 40° cos 130° − 4 cos 40°

or

α = 25.964°

Then

vW =

5 sin 50° = 15.79 km/h sin (40° − 25.964°)

vW = 15.79 km/h

26.0° 

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PROBLEM 11.132 As part of a department store display, a model train D runs on a slight incline between the store’s up and down escalators. When the train and shoppers pass Point A, the train appears to a shopper on the up escalator B to move downward at an angle of 22° with the horizontal, and to a shopper on the down escalator C to move upward at an angle of 23° with the horizontal and to travel to the left. Knowing that the speed of the escalators is 3 ft/s, determine the speed and the direction of the train.

SOLUTION v D = v B + v D/B

We have

v D = vC + v D/C

The graphical representations of these equations are then as shown. Then

vD 3 = sin 8° sin (22° + α )

Equating the expressions for

vD 3 = sin 7° sin (23° − α )

vD 3

sin 8° sin 7° = sin (22° + α ) sin (23° − α )

or

or or Then

sin 8° (sin 23° cos α − cos 23° sin α ) = sin 7° (sin 22° cos α + cos 22° sin α ) tan α =

sin 8° sin 23° − sin 7° sin 22° sin 8° cos 23° + sin 7° cos 22°

α = 2.0728° vD =

3 sin 8° = 1.024 ft/s sin (22° + 2.0728°)

v D = 1.024 ft/s

2.07° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 197

PROBLEM 11.132 (Continued)

Alternate solution using components. v B = (3 ft/s)

30° = (2.5981 ft/s)i + (1.5 ft/s) j

vC = (3 ft/s)

30° = (2.5981 ft/s)i − (1.5 ft/s) j

v D/B = u1

22° = −(u1 cos 22°)i − (u1 sin 22°) j

v D/C = u2

23° = −(u2 cos 23°)i + (u2 sin 23°) j

v D = vD

α = −(vD cos α )i + (vD sin α ) j

v D = v B + v D/B = vC + v D/C 2.5981i + 1.5 j − (u1 cos 22°)i − (u1 sin 22°) j = 2.5981i − 1.5 j − (u2 cos 23°)i + (u2 sin 23°) j

Separate into components, transpose, and change signs. u1 cos 22° − u2 cos 23° = 0 u1 sin 22° + u1 sin 23° = 3

Solving for u1 and u2 ,

u1 = 3.9054 ft/s

u2 = 3.9337 ft/s

v D = 2.5981i + 1.5 j − (3.9054 cos 22°)i − (3.9054 sin 22°) j = −(1.0229 ft/s)i + (0.0370 ft/s) j

or

v D = 2.5981i − 1.5 j − (3.9337 cos 23°)i + (3.9337 sin 23°) j = −(1.0229 ft/s)i + (0.0370 ft/s) j

v D = 1.024 ft/s

2.07° 

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PROBLEM 11.CQ8 The Ferris wheel is rotating with a constant angular velocity ω. What is the direction of the acceleration of Point A? (a) (b) (c) (d ) (e) The acceleration is zero.

SOLUTION The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration pointed upwards. Answer: (b) 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 199

PROBLEM 11.CQ9 A racecar travels around the track shown at a constant speed. At which point will the racecar have the largest acceleration? (a) A (b) B (c) C (d ) The acceleration will be zero at all the points.

SOLUTION The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration. The normal acceleration will be maximum where the radius of curvature is a minimum, that is at Point A. Answer: (a) 

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PROBLEM 11.CQ10 A child walks across merry-go-round A with a constant speed u relative to A. The merry-go-round undergoes fixed axis rotation about its center with a constant angular velocity ω counterclockwise.When the child is at the center of A, as shown, what is the direction of his acceleration when viewed from above. (a) (b) (c) (d ) (e) The acceleration is zero.

SOLUTION Polar coordinates are most natural for this problem, that is,  a = ( r − rθ 2 )er + (rθ + 2rθ)eθ

(1)

r = 0, θ = 0, r = 0, θ = ω, r = -u. When we substitute From the information given, we know  these values into (1), we will only have a term in the −θ direction. Answer: (d ) 

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PROBLEM 11.133 Determine the smallest radius that should be used for a highway if the normal component of the acceleration of a car traveling at 72 km/h is not to exceed 0.8 m/s 2 .

SOLUTION an =

v2

an = 0.8 m/s 2

ρ

v = 72 km/h = 20 m/s 0.8 m/s 2 =

(20 m/s) 2

ρ

ρ = 500 m  

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PROBLEM 11.134 Determine the maximum speed that the cars of the roller-coaster can reach along the circular portion AB of the track if ρ is 25 m and the normal component of their acceleration cannot exceed 3 g.

SOLUTION We have

an =

v2

ρ

Then

(vmax ) 2AB = (3 × 9.81 m/s 2 )(25 m)

or

(vmax ) AB = 27.124 m/s (vmax ) AB = 97.6 km/h 

or

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PROBLEM 11.135 A bull-roarer is a piece of wood that produces a roaring sound when attached to the end of a string and whirled around in a circle. Determine the magnitude of the normal acceleration of a bull-roarer when it is spun in a circle of radius 0.9 m at a speed of 20 m/s.

SOLUTION an =

v2

ρ

=

(20 m/s) 2 = 444.4 m/s 2 0.9 m an = 444 m/s 2 



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PROBLEM 11.136 To test its performance, an automobile is driven around a circular test track of diameter d. Determine (a) the value of d if when the speed of the automobile is 45 mi/h, the normal component of the acceleration is 11 ft/s 2 , (b) the speed of the automobile if d = 600 ft and the normal component of the acceleration is measured to be 0.6 g.

SOLUTION (a)

First note Now

v = 45 mi/h = 66 ft/s an =

ρ=

v2

ρ (66 ft/s) 2 = 396 ft 11 ft/s 2

d = 2ρ

(b)

d = 792 ft 

v2

We have

an =

Then

1  v 2 = (0.6 × 32.2 ft/s 2 )  × 600 ft  2 

ρ

v = 76.131 ft/s v = 51.9 mi/h 

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PROBLEM 11.137 An outdoor track is 420 ft in diameter. A runner increases her speed at a constant rate from 14 to 24 ft/s over a distance of 95 ft. Determine the magnitude of the total acceleration of the runner 2 s after she begins to increase her speed.

SOLUTION We have uniformly accelerated motion v22 = v12 + 2at Δs12

Substituting or

(24 ft/s)2 = (14 ft/s)2 + 2at (95 ft) at = 2 ft/s 2

Also

v = v1 + at t

At t = 2 s:

v = 14 ft/s + (2 ft/s 2 )(2 s) = 18 ft/s v2

Now

an =

At t = 2 s:

an =

Finally

a 2 = at2 + an2

At t = 2 s:

a 2 = 22 + 1.542862

ρ (18 ft/s) 2 = 1.54286 ft/s 2 210 ft

a = 2.53 ft/s 2 

or

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PROBLEM 11.138 A robot arm moves so that P travels in a circle about Point B, which is not moving. Knowing that P starts from rest, and its speed increases at a constant rate of 10 mm/s2, determine (a) the magnitude of the acceleration when t = 4 s, (b) the time for the magnitude of the acceleration to be 80 mm/s2.

SOLUTION Tangential acceleration: Speed:

at = 10 mm/s 2 v = at t v2

an =

where

ρ = 0.8 m = 800 mm

(a)

When t = 4 s

ρ

v = (10)(4) = 40 mm/s an =

Acceleration:

ρ

=

at2 t 2

Normal acceleration:

(40)2 = 2 mm/s 2 800

a = at2 + an2 = (10) 2 + (2) 2 a = 10.20 mm/s 2 

(b)

Time when a = 80 mm/s 2 a 2 = an2 + at2 2

 (10) 2 t 2  2 (80) =   + 10 800   2

t 4 = 403200 s 4 t = 25.2 s 

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PROBLEM 11.139 A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the maximum total acceleration of the train must not exceed 1.5 m/s 2 , determine (a) the shortest distance in which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at .

SOLUTION When v = 72 km/h = 20 m/s and ρ = 400 m, an =

v2

ρ

=

(20)2 = 1.000 m/s 2 400

a = an2 + at2

But

at = a 2 − an2 = (1.5) 2 − (1.000) 2 = ± 1.11803 m/s 2

Since the train is accelerating, reject the negative value. (a)

Distance to reach the speed. v0 = 0

Let

x0 = 0 v12 = v02 + 2at ( x1 − x0 ) = 2at x1 x1 =

(b)

v12 (20) 2 = 2at (2)(1.11803)

x1 = 178.9 m 

Corresponding tangential acceleration. at = 1.118 m/s 2 



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PROBLEM 11.140 A motorist starts from rest at Point A on a circular entrance ramp when t = 0, increases the speed of her automobile at a constant rate and enters the highway at Point B. Knowing that her speed continues to increase at the same rate until it reaches 100 km/h at Point C, determine (a) the speed at Point B, (b) the magnitude of the total acceleration when t = 20 s.

SOLUTION v0 = 0

Speeds:

v1 = 100 km/h = 27.78 m/s s =

Distance:

π 2

(150) + 100 = 335.6 m v12 = v02 + 2at s

Tangential component of acceleration: at =

At Point B,

v12 − v02 (27.78) 2 − 0 = = 1.1495 m/s 2 2s (2)(335.6)

vB2 = v02 + 2at sB

where

sB =

π 2

(150) = 235.6 m

vB2 = 0 + (2)(1.1495)(235.6) = 541.69 m 2 /s 2 vB = 23.27 m/s

(a)

At t = 20 s,

vB = 83.8 km/h 

v = v0 + at t = 0 + (1.1495)(20) = 22.99 m/s

ρ = 150 m

Since v < vB , the car is still on the curve.

(b)

Normal component of acceleration:

an =

Magnitude of total acceleration:

|a| =

v2

ρ

=

(22.99)2 = 3.524 m/s 2 150

at2 + an2 =

(1.1495) 2 + (3.524) 2

| a | = 3.71 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 209

PROBLEM 11.141 Racecar A is traveling on a straight portion of the track while racecar B is traveling on a circular portion of the track. At the instant shown, the speed of A is increasing at the rate of 10 m/s2, and the speed of B is decreasing at the rate of 6 m/s2. For the position shown, determine (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

SOLUTION v A = 240 km/h = 66.67 m/s

Speeds:

vB = 200 km/h = 55.56 m/s

vA = 66.67 m/s

Velocities:

vB = 55.56 m/s

(a)

50°

vB/A = vB − vA

Relative velocity:

vB/A = (55.56 cos 50°) ← + 55.56sin 50° ↓ + 66.67 = 30.96 → + 42.56 = 52.63 m/s

53.96° vB /A = 189.5 km/h

Tangential accelerations:

(aA )t = 10 m/s 2 (aB )t = 6 m/s 2 an =

Normal accelerations:

ρ

( ρ = ∞)

Car B:

( ρ = 300 m)

Acceleration of B relative to A:

50°

v2

Car A:

(aB ) n =

(b)

54.0° 

(aA ) n = 0

(55.56)2 = 10.288 300

(aB ) n = 10.288 m/s 2

40°

aB/A = aB − aA a B/A = (a B )t + (a B ) n − (a A )t − (a A ) n =6

50° + 10.288

40° + 10 → + 0

= (6cos 50° + 10.288cos 40° + 10) + (6sin 50° − 10.288sin 40°) = 21.738 → + 2.017

a B/A = 21.8 m/s 2

5.3° 

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PROBLEM 11.142 At a given instant in an airplane race, airplane A is flying horizontally in a straight line, and its speed is being increased at the rate of 8 m/s 2 . Airplane B is flying at the same altitude as airplane A and, as it rounds a pylon, is following a circular path of 300-m radius. Knowing that at the given instant the speed of B is being decreased at the rate of 3 m/s 2 , determine, for the positions shown, (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

SOLUTION First note

v A = 450 km/h vB = 540 km/h = 150 m/s

(a)

v B = v A + v B/A

We have

The graphical representation of this equation is then as shown. We have

vB2 /A = 4502 + 5402 − 2(450)(540) cos 60° vB/A = 501.10 km/h

and

540 501.10 = sin α sin 60°

α = 68.9° v B/A = 501 km/h

(b)

First note Now

a A = 8 m/s 2 ( aB ) n =

v 2B

ρB

=

Then

60°

(150 m/s) 2 300 m

(a B ) n = 75 m/s 2



(a B )t = 3 m/s 2

68.9° 

30° 

a B = (a B )t + (a B )n = 3(− cos 60° i + sin 60° j) + 75(−cos 30° i − sin 30° j) = −(66.452 m/s 2 )i − (34.902 m/s 2 ) j

Finally

a B = a A + a B/A

a B/A = ( −66.452i − 34.902 j) − (8i ) = −(74.452 m/s 2 )i − (34.902 m/s 2 ) j

a B/A = 82.2 m/s 2

25.1° 

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PROBLEM 11.143 From a photograph of a homeowner using a snowblower, it is determined that the radius of curvature of the trajectory of the snow was 30 ft as the snow left the discharge chute at A. Determine (a) the discharge velocity v A of the snow, (b) the radius of curvature of the trajectory at its maximum height.

SOLUTION (a) The acceleration vector is 32.2 ft/s . At Point A, tangential and normal components of a are as shown in the sketch. an = a cos 40° = 32.2 cos 40° = 24.67 ft/s 2 v A2 = ρ A (a A ) n = (30)(24.67) = 740.0 ft 2 /s 2 v A = 27.2 ft/s

40° 

vx = 27.20 cos 40° = 20.84 ft/s

(b)

At maximum height,

v = vx = 20.84 ft/s an = g = 32.2 ft/s 2 ,

ρ =

v2 (20.84) 2 = 32.2 an

ρ = 13.48 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212

PROBLEM 11.144 A basketball is bounced on the ground at Point A and rebounds with a velocity v A of magnitude 2.5 m/s as shown. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory.

SOLUTION (a)

(a A ) n =

We have

ρA =

or

v 2A

ρA (2.5 m/s) 2 (9.81 m/s 2 ) sin 15°

ρ A = 2.46 m 

or (b)

( aB ) n =

We have

vB2

ρB

where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A sin 15°



ρB =

Then

[(2.5 m/s) sin 15°]2 = 0.0427 m 9.81 m/s 2

ρ B = 42.7 mm 

or

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 213

PROBLEM 11.145 A golfer hits a golf ball from Point A with an initial velocity of 50 m/s at an angle of 25° with the horizontal. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory.

SOLUTION (a)

We have or

(a A ) n =

ρA =

v 2A

ρA (50 m/s)2 (9.81 m/s 2 ) cos 25°

ρ A = 281 m 

or (b)

We have

( aB ) n =

vB2

ρB

where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A cos 25°

Then

ρB =

[(50 m/s) cos 25°]2 9.81 m/s 2

ρ B = 209 m 

or

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 214

PROBLEM 11.146 Three children are throwing snowballs at each other. Child A throws a snowball with a horizontal velocity v0. If the snowball just passes over the head of child B and hits child C, determine the radius of curvature of the trajectory described by the snowball (a) at Point B, (b) at Point C.

SOLUTION The motion is projectile motion. Place the origin at Point A. Horizontal motion:

v x = v0

x = v0 t

Vertical motion:

y0 = 0,

(v y ) = 0

v y = − gt

1 y = − gt 2 2

2h , g

t=

where h is the vertical distance fallen.

| v y| = 2 gh

Speed:

v 2 = vx2 + v 2y = v02 + 2 gh

Direction of velocity. cos θ =

v0 v

Direction of normal acceleration. an = g cos θ =

Radius of curvature: At Point B,

ρ=

gv0 v 2 = v ρ

v3 gv0

hB = 1 m; xB = 7 m tB =

(2)(1 m) = 0.45152 s 9.81 m/s 2

xB = v0t B

v0 =

xB 7m = = 15.504 m/s t B 0.45152 s

vB2 = (15.504) 2 + (2)(9.81)(1) = 259.97 m 2 /s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215

PROBLEM 11.146 (Continued)

(a)

Radius of curvature at Point B.

ρB = At Point C

(259.97 m 2 /s 2 )3/ 2 (9.81 m/s 2 )(15.504 m/s)

ρ B = 27.6 m 

hC = 1 m + 2 m = 3 m

vC2 = (15.504) 2 + (2)(9.81)(3) = 299.23 m 2 /s 2

(b)

Radius of curvature at Point C.

ρC =

(299.23 m 2 /s 2 )3/2 (9.81 m/s 2 )(15.504 m/s)

ρC = 34.0 m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 216

PROBLEM 11.147 Coal is discharged from the tailgate A of a dump truck with an initial velocity v A = 2 m/s 50° . Determine the radius of curvature of the trajectory described by the coal (a) at Point A, (b) at the point of the trajectory 1 m below Point A.

SOLUTION a A = g = 9.81 m/s 2

(a) At Point A.

Sketch tangential and normal components of acceleration at A. (a A ) n = g cos 50°

ρA =

vA2 (2) 2 = (a A ) n 9.81cos 50°

ρ A = 0.634 m 

(b) At Point B, 1 meter below Point A. Horizontal motion: (vB ) x = (v A ) x = 2 cos 50° = 1.286 m/s Vertical motion:

(vB ) 2y = (v A ) 2y + 2a y ( yB − y A ) = (2 cos 40°)2 + (2)(−9.81)(−1) = 21.97 m 2 /s 2 (vB ) y = 4.687 m/s

tan θ =

(vB ) y (vB ) x

=

4.687 , 1.286

or

θ = 74.6°

aB = g cos 74.6°

(vB ) 2x + (vB ) 2y vB 2 ρB = = (aB )n g cos 74.6° =

(1.286) 2 + 21.97 9.81cos 74.6°

ρ B = 9.07 m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217

PROBLEM 11.148 From measurements of a photograph, it has been found that as the stream of water shown left the nozzle at A, it had a radius of curvature of 25 m. Determine (a) the initial velocity vA of the stream, (b) the radius of curvature of the stream as it reaches its maximum height at B.

SOLUTION (a)

We have

(a A ) n =

v 2A

ρA

or

4  v A2 =  (9.81 m/s 2 )  (25 m) 5  

or

v A = 14.0071 m/s vA = 14.01 m/s

(b)

We have Where

Then

( aB ) n =

vB2

ρB

vB = ( v A ) x =

ρB

36.9° 

4 vA 5

( 4 × 14.0071 m/s ) = 5

2

9.81 m/s 2

ρ B = 12.80 m 

or

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218

PROBLEM 11.149 A child throws a ball from Point A with an initial velocity v A of 20 m/s at an angle of 25° with the horizontal. Determine the velocity of the ball at the points of the trajectory described by the ball where the radius of curvature is equal to three-quarters of its value at A.

SOLUTION Assume that Points B and C are the points of interest, where yB = yC and vB = vC . Now

(a A ) n =

v 2A

ρA

or

ρA =

v 2A g cos 25°

Then

ρB =

v A2 3 3 ρA = 4 4 g cos 25° vB2

We have

( aB ) n =

where

(aB ) n = g cos θ

so that

v A2 vB2 3 = 4 g cos 25° g cos θ

or

vB2 =

ρB

3 cos θ 2 vA 4 cos 25°

(1)

Noting that the horizontal motion is uniform, we have ( v A ) x = ( vB ) x

where Then or

(v A ) x = v A cos 25°

(vB ) x = vB cos θ

v A cos 25° = vB cos θ cos θ =

vA cos 25° vB

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 219

PROBLEM 11.149 (Continued) Substituting for cos θ in Eq. (1), we have

or

vB2 =

 v A2 3  vA  cos 25°  4  vB  cos 25°

vB3 =

3 3 vA 4

3 vB = 3 v A = 18.17 m/s 4 4 cos 25° 3 θ = ± 4.04°

cos θ = 3

and

v B = 18.17 m/s

4.04° 

v B = 18.17 m/s

4.04° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 220

PROBLEM 11.150 A projectile is fired from Point A with an initial velocity v 0 . (a) Show that the radius of curvature of the trajectory of the projectile reaches its minimum value at the highest Point B of the trajectory. (b) Denoting by θ the angle formed by the trajectory and the horizontal at a given Point C, show that the radius of curvature of the trajectory at C is ρ = ρ min /cos3 θ .

SOLUTION For the arbitrary Point C, we have (aC ) n =

ρC =

or

vC2

ρC vC2 g cos θ

Noting that the horizontal motion is uniform, we have (v A ) x = (vC ) x (v A ) x = v0 cos α

where

v0 cos α = vC cos θ

Then

cos α v0 cos θ

or

vC =

so that

1 ρC = g cos θ

(a)

2

 cos α  v 2 cos 2 α v0  = 0  g cos3 θ  cos θ 

In the expression for ρC , v0 , α , and g are constants, so that ρC is minimum where cos θ is maximum. By observation, this occurs at Point B where θ = 0.

ρ min = ρ B = (b)

(vC ) x = vC cos θ

ρC =

1 cos3 θ

ρC =

ρ min cos3 θ

v02 cos 2 α g

 v02 cos 2 α  g 

Q.E.D.



Q.E.D.



  

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PROBLEM 11.151* Determine the radius of curvature of the path described by the particle of Problem 11.95 when t = 0. PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

SOLUTION We have

v=

dr = R(cos ωn t − ωn t sin ωn t )i + cj + R (sin ωn t + ωn t cos ωn t )k dt

and

a=

dv = R − ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t i dt

(

)

(

)

+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k

or Now

a = ωn R [−(2 sin ωn t + ωn t cos ωn t )i + (2 cos ωn t − ωn t sin ωn t ) k ]

v 2 = R 2 (cos ωn t − ωn t sin ωn t )2 + c 2 + R 2 (sin ωn t + ωn t cos ωn t )2

(

)

= R 2 1 + ωn2 t 2 + c 2

Then and

Now At t = 0:

(

)

1/2

v =  R 2 1 + ωn2 t 2 + c 2   

R 2ωn2 t dv = 1/ 2 dt  2 R 1 + ωn2 t 2 + c 2   

(

2

a =

at2

+

an2

)

2 2  dv   v  =   +    dt   ρ 

2

dv =0 dt a = ωn R(2 k ) or a = 2ωn R v2 = R2 + c2

Then, with we have or

dv = 0, dt a= 2ωn R =

v2

ρ R 2 + c2

ρ=

ρ

R 2 + c2  2ωn R

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PROBLEM 11.152* Determine the radius of curvature of the path described by the particle of Problem 11.96 when t = 0, A = 3, and B = 1.

SOLUTION With

A = 3,

B =1

we have

r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k

Now

v=

and

  2 dv   t t + 1 − a= = 3(− sin t − sin t − t cos t )i + 3   dt 2 t + 1 

)

(

 3t  dr = 3(cos t − t sin t )i +  2 j + (sin t + t cos t )k  t + 1  dt   t t2 + 1

  j  

+ (cos t + cos t − t sin t )k = − 3(2sin t + t cos t )i + 3

1 j 2 t + ( 1)1/2

+ (2 cos t − t sin t )k v 2 = 9 (cos t − t sin t )2 + 9

Then

t2 + (sin t + t cos t )2 t2 + 1

Expanding and simplifying yields v 2 = t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t

Then

v = [t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t ]1/ 2

and

dv 4t 3 + 38t + 8(−2 cos t sin t + 4t 3sin 2 t + 2t 4 sin t cos t ) − 8[(3t 2 + 1)sin 2t + 2(t 3 + t ) cos 2t ] = dt 2[t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t ) sin 2t ]1/ 2

Now

2 2  dv   v  a 2 = at2 + an2 =   +    dt   ρ 

2

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PROBLEM 11.152* (Continued)

At t = 0:

a = 3j + 2k

or

a = 13 ft/s 2 dv =0 dt v 2 = 9 (ft/s) 2

Then, with

dv = 0, dt

we have

a=

or

ρ=

v2

ρ 9 ft 2 /s 2

ρ = 2.50 ft 

13 ft/s 2

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PROBLEM 11.153 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R /r ) 2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s 2 , determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. Earth: (umean )orbit = 107 Mm/h.

SOLUTION g = 274 m/s 2 ,

For the sun, R=

and

Given that an =

1 1 D =   (1.39 × 109 ) = 0.695 × 109 m 2 2

gR 2 v2 and that for a circular orbit = a n r r2 r =

Eliminating an and solving for r,

v = 107 × 106 m/h = 29.72 × 103 m/s

For the planet Earth, Then

gR 2 v2

r =

(274)(0.695 × 109 ) 2 = 149.8 × 109 m (29.72 × 103 ) 2

r = 149.8 Gm 

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PROBLEM 11.154 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R /r ) 2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s 2 , determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. Saturn: (umean )orbit = 34.7 Mm/h.

SOLUTION g = 274 m/s 2

For the sun, R=

and

Given that an =

1 1 D =   (1.39 × 109 ) = 0.695 × 109 m 2 2

gR 2 v2 .and that for a circular orbit: = a n r r2 gR 2 v2

Eliminating an and solving for r,

r =

For the planet Saturn,

v = 34.7 × 106 m/h = 9.639 × 103 m/s

Then,

r =

(274)(0.695 × 109 ) 2 = 1.425 × 1012 m (9.639 × 103 )2

r = 1425 Gm 

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PROBLEM 11.155 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Venus: g = 29.20 ft/s 2 , R = 3761 mi.

SOLUTION From Problems 11.153 and 11.154,

an =

gR 2 r2

For a circular orbit,

an =

v2 r

v= R

Eliminating an and solving for v, For Venus,

g r

g = 29.20 ft/s 2 R = 3761 mi = 19.858 × 106 ft. r = 3761 + 100 = 3861 mi = 20.386 × 106 ft

Then,

v = 19.858 × 106

29.20 = 23.766 × 103 ft/s 20.386 × 106 v = 16200 mi/h 

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PROBLEM 11.156 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Mars: g = 12.17 ft/s 2 , R = 2102 mi.

SOLUTION From Problems 11.153 and 11.154,

an =

gR 2 r2

For a circular orbit,

an =

v2 r

v= R

Eliminating an and solving for v, For Mars,

g r

g = 12.17 ft/s 2 R = 2102 mi = 11.0986 × 106 ft r = 2102 + 100 = 2202 mi = 11.6266 × 106 ft

Then,

v = 11.0986 × 106

12.17 = 11.35 × 103 ft/s 11.6266 × 106 v = 7740 mi/h 

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PROBLEM 11.157 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Jupiter: g = 75.35 ft/s 2 , R = 44, 432 mi.

SOLUTION From Problems 11.153 and 11.154,

an =

gR 2 r2

For a circular orbit,

an =

v2 r

v= R

Eliminating an and solving for v, For Jupiter,

g r

g = 75.35 ft/s 2 R = 44432 mi = 234.60 × 106 ft r = 44432 + 100 = 44532 mi = 235.13 × 106 ft

Then,

v = (234.60 × 106 )

75.35 = 132.8 × 103 ft/s 6 235.13 × 10 v = 90600 mi/h 

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PROBLEM 11.158 A satellite is traveling in a circular orbit around Mars at an altitude of 300 km. After the altitude of the satellite is adjusted, it is found that the time of one orbit has increased by 10 percent. Knowing that the radius of Mars is 3382 km, determine the new altitude of the satellite. (See information given in Problems 11.153–11.155.)

SOLUTION an = g

We have Then

g

R2 r2

and an =

v2 r

R2 v2 = r r2 v=R

g r

where

r =R+h

The circumference s of a circular orbit is equal to s = 2π r

Assuming that the speed of the satellite in each orbit is constant, we have s = vtorbit

Substituting for s and v 2π r = R torbit =

=

Now

g torbit r

2π r 3/2 R g 2π ( R + h)3/2 R g

(torbit ) 2 = 1.1(torbit )1 2π ( R + h2 )3/2 2π ( R + h1 )3/2 = 1.1 R R g g h2 = (1.1)2/3 ( R + h1 ) − R

= (1.1)2/3 (3382 + 300) km − (3382 km) h2 = 542 km 

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PROBLEM 11.159 Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth. (See information given in Problems 11.153–11.155.)

SOLUTION an = g

We have Then or

g

R2 r2

and an =

v2 r

R2 v2 = r r2 v=R

g r

where

r =R+h

The circumference s of the circular orbit is equal to s = 2π r

Assuming that the speed of the telescope is constant, we have s = vtorbit

Substituting for s and v 2π r = R

or

torbit = =

g torbit r

2π r 3/2 R g 2π [(6370 + 590) km]3/2 1h × −3 2 1/2 6370 km [9.81 × 10 km/s ] 3600 s torbit = 1.606 h 

or

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PROBLEM 11.160 Satellites A and B are traveling in the same plane in circular orbits around the earth at altitudes of 120 and 200 mi, respectively. If at t = 0 the satellites are aligned as shown and knowing that the radius of the earth is R = 3960 mi, determine when the satellites will next be radially aligned. (See information given in Problems 11.153–11.155.)

SOLUTION an = g

We have Then

g

R2 r2

R2 v2 = r r2

and an = or

v2 r

v=R

g r

r =R+h

where The circumference s of a circular orbit is equal to

s = 2π r

Assuming that the speeds of the satellites are constant, we have s = vT

Substituting for s and v 2π r = R

or Now

T=

g T r

2π r 3/ 2 2π ( R + h)3/2 = R g R g

hB > hA  (T ) B > (T ) A

Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B completes N orbits, satellite A must complete ( N + 1) orbits. Thus, TC = N (T ) B = ( N + 1)(T ) A

or

 2π ( R + hB )3/2   2π ( R + hA )3/2  ( 1) N = N +    g g  R   R 

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PROBLEM 11.160 (Continued)

N=

or

=

Then

( R + hA )3/2 ( R + hB )

3/ 2

− ( R + hA )

1

(

3960 + 200 3960 +120

)

3/2

TC = N (T ) B = N

−1

3/2

=

1

(

R + hB R + hA

)

3/2

−1

= 33.835 orbits

2π ( R + hB )3/2 R g

[(3960 + 200) mi] × 1 h 2π 3960 mi 32.2 ft/s 2 × 1 mi 1/2 3600s 5280 ft 3/2

= 33.835

(

)

TC = 51.2 h 

or Alternative solution

From above, we have (T ) B > (T ) A . Thus, when the satellites are next radially aligned, the angles θ A and θ B swept out by radial lines drawn to the satellites must differ by 2π . That is,

θ A = θ B + 2π For a circular orbit

s = rθ

From above

s = vt and v = R

Then

θ=

At time TC : or

R g ( R + hA )

3/2

TC = TC = =

g r

R g R g s vt 1  g = = R t  t = 3/2 t =  r r r r  r ( R + h)3/2 R g ( R + hB )3/ 2

TC + 2π 2π

 1 1 R g −  ( R + hA )3/ 2 ( R + hB )3/ 2  2π

(

1 mi (3960 mi) 32.2 ft/s 2 × 5280 ft

× ×

1 [(3960 + 120) mi ]3/ 2

1 −

)

1/2

1

[(3960 + 200) mi ]3/ 2

1h 3600 s TC = 51.2 h 

or

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PROBLEM 11.161 The oscillation of rod OA about O is defined by the relation θ = (3/π )(sin π t ), where θ and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r = 6(1 − e−2t ) where r and t are expressed in inches and seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the collar relative to the rod.

SOLUTION Calculate the derivatives with respect to time. 3

r = 6 − 6e −2t in.

θ=

r = 12e −2t in/s

θ = 3cosπ t rad/s

r = −24e−2t in/s 2

θ = −3π sin π t rad/s2

π

sin π t rad

At t = 1 s,

(a)

3

r = 6 − 6e −2 = 5.1880 in.

θ=

r = 12e −2 = 1.6240 in/s

θ = 3cos π = −3 rad/s

 r = −24e−2 = −3.2480 in/s 2

θ = −3π sin π = 0

π

sin π = 0

Velocity of the collar. v = rer + rθeθ = 1.6240 e r + (5.1880)(−3)eθ v = (1.624 in/s)er + (15.56 in/s)eθ 

(b)

Acceleration of the collar. a = (r − rθ 2 )e r + (rθ + 2rθ)eθ = [ −3.2480 − (5.1880)( −3) 2 ]er + (5.1880)(0) + (2)(1.6240)(−3)]eθ (−49.9 in/s 2 )er + (−9.74 in/s 2 )eθ 

(c)

Acceleration of the collar relative to the rod. a B /OA = (−3.25 in/s 2 )er 

a B /OA =  re r

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PROBLEM 11.162 The rotation of rod OA about O is defined by the relation θ = t 3 − 4t , where θ and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r = 2.5t 3 − 5t 2 , where r and t are expressed in inches and seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the acceleration of the collar, (c) the radius of curvature of the path of the collar.

SOLUTION Calculate the derivatives with respect to time. r = 2.5t 3 − 5t 2

θ = t 3 − 4t

r = 7.5t 2 − 10t

θ = 3t 2 − 4

 r = 15t − 10

θ = 6t

At t = 1 s,

(a)

r = 2.5 − 5 = −2.5 in.

θ = 1 − 4 = −3 rad

r = 7.5 − 10 = −2.5 in./s

θ = 3 − 4 = −1 rad/s

r = 15 − 10 = 5 in./s 2

θ = 6 rad/s 2

Velocity of the collar. v = rer + rθeθ = −2.5er + (−2.5)( −1)eθ v = ( −2.50 in./s)er + (2.50 in./s)eθ  v = (2.50) 2 + (2.50) 2 = 3.5355 in./s

Unit vector tangent to the path. et =

(b)

v = −0.70711e r + 0.70711eθ v

Acceleration of the collar. a = (r − rθ 2 )e r + (rθ + 2rθ)eθ = [5 − (−2.5)(−1)2 ]er + [(−2.5)(6) + (2)(−2.5)(−1)]eθ

a = (7.50 in/s 2 )er + (−10.00 in/s 2 )eθ 

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PROBLEM 11.162 (Continued) Magnitude:

a = (7.50) 2 + (10.00) 2 = 12.50 in./s 2 at = aet

Tangential component:

at = (7.50)(−0.70711) + (−10.00)(0.70711) = −12.374 in./s 2

Normal component: (c)

an = a 2 − at2 = 1.7674 in./s 2

Radius of curvature of path. an =

ρ=

v2

ρ v 2 (3.5355 in./s) 2 = an 1.7674 in./s 2

ρ = 7.07 in. 

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PROBLEM 11.163 The path of particle P is the ellipse defined by the relations r = 2/(2 − cos π t ) and θ = π t , where r is expressed in meters, t is in seconds, and θ is in radians. Determine the velocity and the acceleration of the particle when (a) t = 0, (b) t = 0.5 s.

SOLUTION We have

r=

2 2 − cos π t

θ = πt

Then

r =

−2π sin π t (2 − cos π t )2

θ = π

and

r = −2π

π cos π t (2 − cos π t ) − sin π t (2π sin π t ) (2 − cos π t )3

= −2π 2

(a)

At t = 0:

Now

2cos π t − 1 − sin 2 π t (2 − cos π t )3

r=2m

θ =0

r = 0

θ = π rad/s

r = −2π 2 m/s 2

θ = 0

v = rer + rθeθ = (2)(π )eθ v = (2π m/s)eθ 

or and

θ = 0

a = (r − rθ 2 )e r + (rθ + 2rθ)eθ = [ −2π 2 − (2)(π )2 ]er

a = −(4π 2 m/s 2 )er 

or (b)

At t = 0.5 s:

θ=

r =1 m r =

−2π π = − m/s 2 2 (2)

r = −2π 2

−1 − 1 π 2 = m/s 2 3 2 (2)

π rad 2

θ = π rad/s θ = 0

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PROBLEM 11.163 (Continued)

Now

 π v = rer + rθeθ =  −  e r + (1)(π )eθ  2 π  v = −  m/s  er + (π m/s)eθ  2  

or and

a = (r − rθ 2 )e r + (rθ + 2rθ)eθ

π 2    π  = − (1)(π ) 2  e r +  2  −  (π )  eθ   2   2 

π2  a = −  m/s 2  er − (π 2 m/s 2 )eθ   2 

or

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PROBLEM 11.164 The two-dimensional motion of a particle is defined by the relations r = 2a cos θ and θ = bt 2 /2, where a and b are constants. Determine (a) the magnitudes of the velocity and acceleration at any instant, (b) the radius of curvature of the path. What conclusion can you draw regarding the path of the particle?

SOLUTION (a)

We have

r = 2a cos θ

1 θ = bt 2 2

Then

r = −2aθ sin θ

θ = bt

and

 r = −2a (θ sin θ + θ 2 cos θ )

θ = b

Substituting for θ and θ r = −2abt sin θ  r = −2ab(sin θ + bt 2 cos θ )

Now Then

vθ = rθ = 2abt cos θ

vr = r = −2abt sin θ

v = vr2 + vθ2 = 2abt[( − sin θ ) 2 + (cos θ ) 2 ]1/2 v = 2abt 

or Also

ar = r − rθ 2 = −2ab(sin θ + bt 2 cos θ ) − 2ab 2 t 2 cos θ = −2ab(sin θ + 2bt 2 cos θ )

and

aθ = rθ + 2rθ = 2ab cos θ − 4ab 2 t 2 sin θ = −2ab(cos θ − 2bt 2 sin θ )

Then

a = ar2 + aB2 = 2ab[(sin θ + 2bt 2 cos θ ) 2 + (cos θ − 2bt 2 sin θ ) 2 ]1/2 a = 2ab 1 + 4b 2 t 4 

or (b)

2

at2

+

an2

2 2  dv   v  =   +    dt   ρ 

Now

a =

Then

dv d = (2abt ) = 2ab dt dt

2

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PROBLEM 11.164 (Continued)

so that or

(

2ab 1 + 4b 2 t 4

)

2

= (2ab) 2 + an2

4a 2b 2 (1 + 4b 2t 4 ) = 4a 2 b 2 + an2

or

an = 4ab 2 t 2

Finally

an =

v2 (2abt ) 2 ρ= ρ 4ab 2 t 2

ρ =a 

or Since the radius of curvature is a constant, the path is a circle of radius a.



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PROBLEM 11.165 As rod OA rotates, pin P moves along the parabola BCD. Knowing that the equation of this parabola is r = 2b /(1 + cos θ ) and that θ = kt , determine the velocity and acceleration of P when (a) θ = 0, (b) θ = 90°.

SOLUTION 2b θ = kt 1 + cos kt 2bk sin kt θ = k θ = 0 r = (1 + cos kt ) 2 2bk [(1 + cos kt )2 k cos kt + (sin kt )2(1 + cos kt )( k sin kt )] r = (1 + cos kt ) 4 r=

(a)

When θ = kt = 0: 2bk 1 [(2)2 k (1) + 0] = bk 2 4 2 (2)

r =b

r = 0

 r=

θ =0

θ = k

θ = 0

vθ = rθ = bk

vr = r = 0

v = bk eθ 

1 1  ar =  r − rθ 2 = bk 2 − bk 2 = − bk 2  2 2   aθ = rθ + 2rθ = b(0) + 2(0) = 0 

(b)

1 a = − bk 2 er  2

When θ = kt = 90°: r = 2b

r = 2bk

θ = 90°

θ = k

vr = r = 2bk

2bk [0 + 2k ] = 4bk 2 19 θ = 0 r =

vθ = rθ = 2bk

v = 2bk er + 2bk eθ 

ar = r − rθ 2 = 4bk 2 − 2bk 2 = 2bk 2 a = rθ + 2rθ = 2b(0) + 2(2bk )k = 4bk 2 θ

a = 2bk 2 e r + 4bk 2 eθ 

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PROBLEM 11.166 The pin at B is free to slide along the circular slot DE and along the rotating rod OC. Assuming that the rod OC rotates at a constant rate θ, (a) show that the acceleration of pin B is of constant magnitude, (b) determine the direction of the acceleration of pin B.

SOLUTION From the sketch: r = 2b cos θ r = −2b sin θ θ

Since θ = constant, θ = 0 r = −2b cos θ θ 2 ar = r − rθ 2 = −2b cos θ θ 2 − (2b cos θ )θ 2 a = −4b cos θ θ 2 r

aθ = rθ + 2rθ = (2b cos θ )(0) + 2(−2b sin θ )θ 2 a = −4b sin θ θ 2 θ

a = ar2 + aθ2 = 4bθ 2 (− cos θ ) 2 + (− sin θ )2 a = 4bθ 2

Since both b and θ are constant, we find that a = constant 

γ = tan −1

 −4b sin θ θ 2 aθ = tan −1  2 ar  −4b cos θ θ

  

γ = tan −1 (tan θ ) γ =θ Thus, a is directed toward A 

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PROBLEM 11.167 To study the performance of a racecar, a high-speed camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine (a) the speed of the car in terms of b, θ , and θ, (b) the magnitude of the acceleration in terms of b, θ , θ, and θ.

SOLUTION (a)

We have

r=

b cos θ

Then

r =

bθ sin θ cos 2 θ

We have

v 2 = vr2 + vθ2 = (r)2 + (rθ) 2 2

or

 bθ sin θ   bθ 2 =   +   2  cos θ   cos θ   b2θ 2 b2θ 2  sin 2θ = + 1  = 2  2 4 cos θ  cos θ  cos θ bθ v=± cos 2θ

For the position of the car shown, θ is decreasing; thus, the negative root is chosen.

v=−

bθ  cos 2θ

v=−

bθ  cos 2θ

Alternative solution. From the diagram or

r = −v sin θ bθ sin θ = −v sin θ cos 2θ

or   

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PROBLEM 11.167 (Continued) (b)

For rectilinear motion

a=

dv dt

Using the answer from Part a v=−

Then

a=

bθ cos 2θ

d  bθ  − dt  cos 2θ

= −b

  

θ cos 2θ − θ(−2θ cos θ sin θ ) cos 4θ a=−

or

b (θ + 2θ 2 tan θ )  cos 2θ

Alternative solution From above Then

Now where

and

b bθ sin θ r = cos θ cos 2 θ (θ sin θ + θ 2 cos θ )(cos 2θ ) − (θ sin θ )(−2θ cos θ sin θ ) r = b cos 4θ  θ sin θ θ 2 (1 + sin 2 θ )  = b +  2 cos3θ  cos θ  r=

a 2 = ar2 + aθ2  θ sin θ θ 2 (1 + sin 2 θ )  bθ 2 + ar = r − rθ 2 = b  − 2 cos 2θ  cos θ  cos θ b   2θ 2 sin 2 θ  = + θ sin θ   cos θ  cos 2θ  b sin θ  ar = (θ + 2θ 2 tan θ ) 2 cos θ aθ = rθ + 2rθ = =

Then

bθ bθ 2 sin θ +2 cos θ cos 2θ

b cos θ  (θ + 2θ tan θ ) cos 2θ

a=±

b (θ + 2θ 2 tan θ )[(sin θ ) 2 + (cos θ )2 ]1/ 2 2 cos θ

For the position of the car shown, θ is negative; for a to be positive, the negative root is chosen. b a=− (θ + 2θ 2 tan θ )  cos 2 θ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244

PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle β . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ.

SOLUTION From the diagram r d = sin (180° − β ) sin ( β − θ ) or

d sin β = r (sin β cos θ − cos β sin θ ) tan β tan β cos θ − sin θ

or

r=d

Then

r = d tan β

−(− tan β sin θ − cos θ )  θ (tan β cos θ − sin θ )2

tan β sin θ + cos θ = dθ tan β (tan β cos θ − sin θ ) 2

From the diagram vr = v cos ( β − θ )

where

vr = r

Then dθ tan β

tan β sin θ + cos θ = v(cos β cos θ + sin β sin θ ) (tan β cos θ − sin θ )2 = v cos β (tan β sin θ + cos θ ) v=

or

dθ tan β sec β  (tan β cosθ − sin θ )2

Alternative solution. We have

v 2 = vr2 + vθ2 = (r) 2 + ( rθ) 2

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PROBLEM 11.168 (Continued) Using the expressions for r and r from above  tan β sin θ + cos θ  v =  dθ tan β  (tan β cos θ − sin θ ) 2  

2

1/ 2

or

 (tan β sin θ + cos θ )2  dθ tan β v=± + 1  2 (tan β cos θ − sin θ )  (tan β cos θ − sin θ )  1/ 2

  tan 2 β + 1 dθ tan β =±  2 (tan β cos θ − sin θ )  (tan β cos θ − sin θ ) 

Note that as θ increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen. v=

dθ tan β sec β  (tan β cos θ − sin θ )2

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PROBLEM 11.169 At the bottom of a loop in the vertical plane, an airplane has a horizontal velocity of 315 mi/h and is speeding up at a rate of 10 ft/s2. The radius of curvature of the loop is 1 mi. The plane is being tracked by radar at O. What are r , θ and θ for this the recorded values of r,  instant?

SOLUTION Geometry. The polar coordinates are r = (2400) 2 + (1800)2 = 3000 ft

Velocity Analysis.

 1800   = 36.87°  2400 

θ = tan −1 

v = 315 mi/h = 462 ft/s

vr = 462 cos θ = 369.6 ft/s vθ = −462sin θ = −277.2 ft/s vr = r vθ = rθ

θ =

r = 370 ft/s 

vθ 277.2 =− r 3000

θ = −0.0924 rad/s  Acceleration analysis.

at = 10 ft/s 2 an =

v2

ρ

=

(462) 2 = 40.425 ft/s 2 5280

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PROBLEM 11.169 (Continued)

ar = at cos θ + an sin θ = 10 cos 36.87° + 40.425 sin 36.87° = 32.255 ft/s 2 aθ = − at sin θ + an cos θ = −10 sin 36.87° + 40.425 cos 36.87° = 26.34 ft/s 2 ar =  r − rθ 2  r = ar + rθ 2  r = 32.255 + (3000)( −0.0924) 2

 r = 57.9 ft/s 2 

aθ = rθ + 2rθ a 2rθ θ = θ − r r 26.34 (2)(369.6)(−0.0924) = − 3000 3000

θ = 0.0315 rad/s 2 

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PROBLEM 11.170 Pin C is attached to rod BC and slides freely in the slot of rod OA which rotates at the constant rate ω. At the instant when β = 60°, r and θ. Express your answers in terms determine (a) r and θ, (b)  of d and ω.

SOLUTION

Looking at d and β as polar coordinates with d = 0, vβ = d β = d ω,

vd = d = 0

aβ = d β + 2d β = 0,

r = d 3 for angles shown.

Geometry analysis: (a)

Velocity analysis:

ad = d − d β 2 = −d ω 2

Sketch the directions of v, er and eθ. vr = r = v ⋅ er = d ω cos120° 1 r = − d ω  2 vθ = rθ = v ⋅ eθ = d ω cos 30°

θ = (b)

Acceleration analysis:

3 dω cos 30° dω 2 = r d 3

θ =

1 ω 2

Sketch the directions of a, er and eθ. ar = a ⋅ e r = a cos150° = −

3 dω 2 2

3  r − rθ2 = − d ω2 2  r =−

3 3 1  dω 2 + rθ 2 = − dω 2 + d 3  ω  2 2 2 

2

r = −

3 dω 2  4

1 aθ = a ⋅ eθ = d ω 2 cos120° = − d ω 2 2 a = rθ + 2rθ θ

θ =

1 (aθ − 2rθ) = r

1 3d

 1  1  1   2  − d ω − (2)  − d ω  ω   2 2   2   

θ = 0 

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PROBLEM 11.171 For the racecar of Problem 11.167, it was found that it took 0.5 s for the car to travel from the position θ = 60° to the position θ = 35°. Knowing that b = 25 m, determine the average speed of the car during the 0.5-s interval. PROBLEM 11.167 To study the performance of a racecar, a high-speed camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine (a) the speed of the car in terms of b, θ , and θ, (b) the magnitude of the acceleration in terms of b, θ , θ, and θ.

SOLUTION From the diagram: Δr12 = 25 tan 60° − 25 tan 35° = 25.796 m

Now

vave = =

Δr12 Δt12 25.796 m 0.5 s

= 51.592 m/s vave = 185.7 km/h 

or

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PROBLEM 11.172 For the helicopter of Problem 11.168, it was found that when the helicopter was at B, the distance and the angle of elevation of the helicopter were r = 3000 ft and θ = 20°, respectively. Four seconds later, the radar station sighted the helicopter at r = 3320 ft and θ = 23.1°. Determine the average speed and the angle of climb β of the helicopter during the 4-s interval. PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle β . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ.

SOLUTION We have

r0 = 3000 ft r4 = 3320 ft

θ0 = 20° θ 4 = 23.1°

From the diagram: Δr 2 = 30002 + 33202 − 2(3000)(3320) cos (23.1° − 20°)

or

Δr = 362.70 ft

Now

vave =

Δr Δt 362.70 ft = 4s = 90.675 ft/s vave = 61.8 mi/h 

or Also, or

Δr cos β = r4 cos θ 4 − r0 cos θ0 cos β =

3320 cos 23.1° − 3000 cos 20° 362.70

β = 49.7° 

or

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PROBLEM 11.173 A particle moves along the spiral shown; determine the magnitude of the velocity of the particle in terms of b, θ , and θ.

SOLUTION Hyperbolic spiral.

r=

b

θ

dr b dθ b =− 2 = − 2 θ dt θ dt θ b  b vr = r = − 2 θ vθ = rθ = θ r =

θ

θ

2

 1  1 v = vr2 + vθ2 = bθ  − 2  +    θ  θ  bθ = 2 1+θ 2

2

θ

v=

b

θ2

1 + θ 2 θ 

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PROBLEM 11.174 A particle moves along the spiral shown; determine the magnitude of the velocity of the particle in terms of b, θ , and θ.

SOLUTION Logarithmic spiral.

r = ebθ r =

dr dθ = bebθ = bebθ θ dt dt

vr = r = bebθ θ vθ = rθ = ebθ θ v=

vr2

+ vθ = e θ b 2 + 1 bθ

2

v = ebθ 1 + b 2 θ 

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PROBLEM 11.175 A particle moves along the spiral shown. Knowing that θ is constant and denoting this constant by ω , determine the magnitude of the acceleration of the particle in terms of b, θ , and ω.

SOLUTION b

Hyperbolic spiral.

r=

From Problem 11.173

r = −

θ

 r =−

b

θ2

θ

b  2b  2 θ + 3θ 2

θ

θ

b 2b b ar =  r − rθ 2 = − 2 θ + 3 θ 2 − θ 2

θ

θ

θ

b b b  b  aθ = rθ + 2rθ = θ + 2  − 2 θ θ = θ − 2 2 θ 2 θ θ θ  θ 

Since θ = ω = constant, θ = 0, and we write: b 2 bω 2 2 − ω ω = 3 (2 − θ 2 ) θ θ3 θ 2 b bω aθ = −2 2 ω 2 = − 3 (2θ ) θ θ ar = +

2b

a = ar2 + aθ2 =

bω 2

θ

3

(2 − θ 2 ) 2 + (2θ ) 2 =

bω 2

θ

3

4 − 4θ 2 + θ 4 + 4θ 2 a=

bω 2

θ3

4 +θ 4 

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PROBLEM 11.176 A particle moves along the spiral shown. Knowing that θ is constant and denoting this constant by ω , determine the magnitude of the acceleration of the particle in terms of b, θ , and ω.

SOLUTION Logarithmic spiral.

r = ebθ

dr = bebθ θ dt r = bebθ θ + b 2 ebθ θ 2 = bebθ (θ + bθ 2 ) r =

ar = r − rθ 2 = bebθ (θ + bθ 2 ) − ebθ θ 2 a = rθ + 2rθ = ebθ θ + 2(bebθ θ)θ θ

Since θ = ω = constant, θ = θ , and we write ar = bebθ (bω 2 ) − ebθ ω 2 = ebθ (b 2 − 1)ω 2 aθ = 2bebθ ω 2 a = ar2 + aθ2 = ebθ ω 2 (b2 − 1)2 + (2b) 2 = ebθ ω 2 b4 − 2b 2 + 1 + 4b 2 = ebθ ω 2 b 4 + 2b 2 + 1 = ebθ ω 2 (b2 + 1) 2 = ebθ ω 2 (b2 + 1) a = (1 + b 2 )ω 2 ebθ 

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PROBLEM 11.177 The motion of a particle on the surface of a right circular cylinder is defined by the relations R = A, θ = 2π t , and z = B sin 2π nt , where A and B are constants and n is an integer. Determine the magnitudes of the velocity and acceleration of the particle at any time t.

SOLUTION R=A R = 0

θ = 2π t

z = B sin 2π nt

θ = 2π

z = 2π n B cos 2π nt

 = 0 R

θ = 0

 z = −4π 2 n 2 B sin 2π nt

Velocity (Eq. 11.49) v = R e R + Rθeθ + zk v=

+ A(2π )eθ + 2π n B cos 2π nt k v = 2π A2 + n 2 B 2 cos 2 2π nt 

Acceleration (Eq. 11.50)  - Rθ 2 )e + ( Rθ + 2 Rθ)e +  a = (R zk R θ

a = −4π 2 Aek − 4π 2 n 2 B sin 2π nt k a = 4π 2 A2 + n4 B 2 sin 2 2π nt 

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PROBLEM 11.178 Show that r = hφ sin θ knowing that at the instant shown, step AB of the step exerciser is rotating counterclockwise at a constant rate φ.

SOLUTION From the diagram r 2 = d 2 + h 2 − 2dh cos φ

Then Now or

2rr = 2dhφ sin φ r d = sin φ sin θ r=

d sin φ sin θ

Substituting for r in the expression for r  d sin φ    sin θ  r = dhφ sin φ  

or

r = hφ sin θ

Q.E.D.



Alternative solution. First note

α = 180° − (φ + θ )

Now

v = v r + vθ = re r + rθeθ

With B as the origin vP = dφ

( d = constant  d = 0)

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PROBLEM 11.178 (Continued)

With O as the origin

(vP )r = r

where

(vP )r = vP sin α

Then Now or substituting

r = dφ sin α h d = sin α sin θ d sin α = h sin θ r = hφ sin θ

Q.E.D.

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PROBLEM 11.179 The three-dimensional motion of a particle is defined by the relations R = A(1 − e −t ), θ = 2π t , and z = B(1 − e− t ). Determine the magnitudes of the velocity and acceleration when (a) t = 0, (b) t = ∞.

SOLUTION R = A(1 − e −t ) R = Ae −t  = − Ae−t R

θ = 2π t

z = B (1 − e−t )

θ = 2π

z = Be−t

θ = 0

 z = − Be−t

Velocity (Eq. 11.49) v = R e R + Rθeθ + zk v = Ae −t e R + 2π A(1 − e −t )eθ + Be−t k

(a)

When

t = 0: e−t = e0 = 1;

(b)

When

t = ∞ : e− t = e −∞ = 0

v = A2 + B 2 

v = Ae R + Bk v = 2π Aeθ

v = 2π A 

Acceleration (Eq. 11.50)  − Rθ1 )e + ( Rθ + 2 Rθ)e +  a = (R zk R θ = [ − Ae−t − A(1 − e −t )4π 2 ]e R + [0 + 2 Ae−t (2π )]eθ − Be −t k

(a)

When

t = 0: e −t = e0 = 1

a = − Ae R + 4π Aeθ − B k a = A2 + (4π A) 2 + B 2

(b)

When

a = (1 + 16π 2 ) A2 + B 2 

t = ∞ : e −t = e−∞ = 0

a = −4π 2 Ae R

a = 4π 2 A 

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PROBLEM 11.180* For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis. PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

SOLUTION First note that the vectors v and a lie in the osculating plane. Now

r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k

Then

v=

dr = R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k dt

and

a=

dv dt

(

)

= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i +R

(

ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t

)k

= ωn R[−(2sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t )k ]

It then follows that the vector ( v × a) is perpendicular to the osculating plane.

i j k ( v × a) = ωn R R(cos ωn t − ωn t sin ωn t ) c R(sin ωn t + ωn t cos ωn t ) −(2sin ωn t + ωn t cos ωn t ) 0 (2cos ωn t − ωn t sin ωn t ) = ωn R{c(2 cos ωn t − ωn t sin ωn t )i + R[ −(sin ωn t + ωn t cos ωn t )(2sin ωn t + ωn t cos ωn t ) − (cos ωn t − ωn t sin ωn t )(2 cos ωn t − ωn t sin ωn t )] j + c(2sin ωn t + ωn t cos ωn t )k

(

)

= ωn R c(2 cos ωn t − ωn t sin ωn t )i − R 2 + ωn2 t 2 j + c(2sin ωn t + ωn t cos ωn t )k   

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 260

PROBLEM 11.180* (Continued)

The angle α formed by the vector ( v × a) and the y axis is found from cos α =

Where

( v × a) ⋅ j | ( v × a) || j |

|j| =1

(

( v × a) ⋅ j = −ωn R 2 2 + ωn2 t 2

)

(

|( v × a) | = ωn R c 2 (2 cos ωn t − ωn t sin ωn t ) 2 + R 2 2 + ωn2 t 2 

)

2

1/ 2

+ c 2 (2sin ωn t + ωn t cos ωn t )2 

(

)

2 1/2

(

) 

= ωn R c 2 4 + ωn2 t 2 + R 2 2 + ωn2t 2 

Then

( ) ω R c ( 4 + ω t ) + R ( 2 + ω t )    −R ( 2 + ω t ) = c 4 + ω t + R 2 + ω t  ) ( )   ( −ωn R 2 2 + ωn2t 2

cos α =

2

n

2 2 n

2

2 2 n

2

1/2

2 2 n

2

2 2 n

2

2 2 n

2 1/2

The angle β that the osculating plane forms with y axis (see the above diagram) is equal to

β = α − 90° Then

cos α = cos ( β + 90°) = −sin β − sin β =

Then

tan β =

( ) ) + R ( 2 + ω t ) 

− R 2 + ωn2t 2

(

c 2 4 + ω 2t 2 n 

(

R 2 + ωn2 t 2 c

2

2 2 n

2

1/ 2

)

4 + ωn2 t 2

(

 R 2 + ωn2t 2 β = tan   c 4 + ω 2t 2 n  −1

or

)    

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PROBLEM 11.181* Determine the direction of the binormal of the path described by the particle of Problem 11.96 when (a) t = 0, (b) t = π /2 s.

SOLUTION Given:

)

(

r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k r − ft, t − s;

A = 3, B − 1

First note that eb is given by eb =

v×a |v ×a |

)

(

Now

r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k

Then

v=

dr dt 3t

= 3(cos t − t sin t )i +

j + (sin t + t cos t )k

(

t

)

2 dv t 2 +1 a= = 3( − sin t − sin t − t cos t )i + 3 t + 12− t j dt t +1 + (cos t + cos t − t sin t )k 3 = −3(2sin t + t cos t )i + 2 j + (2 cos t − t sin t )k (t + 1)3/2

and

(a)

t2 +1

At t = 0:

v = (3 ft/s)i a = (3 ft/s 2 ) j + (2 ft/s 2 )k

Then

and Then

v × a = 3i × (3j + 2k ) = 3(−2 j + 3k ) | v × a | = 3 (−2) 2 + (3) 2 = 3 13 eb =

3( −2 j + 3k ) 3 13

cos θ x = 0

or

θ x = 90°

=

cos θ y = −

1 13 2

(−2 j + 3k )

13

θ y = 123.7°

cos θ z =

3 13

θ z = 33.7°



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PROBLEM 11.181* (Continued)

(b)

At t =

Then

π 2

s:

  3π   3π v = − ft/s  i +  ft/s  j + (1 ft/s)k    2   π2 +4    π 24  a = −(6 ft/s 2 )i +  2 ft/s 2  j −  ft/s 2  k 3/2   (π + 4)  2 i 3π v×a = − 2 −6

j 3π 2 (π + 4)1/ 2 24 (π 2 + 4)3/ 2

k 1 −

π 2

   3π 2 24 3π 2 6 i = − + − +   2 1/2 4 (π 2 + 4)3/ 2    2(π + 4)   36π 18π k + − 2 + 2 3/2 1/2  (π + 4)   (π + 4) = −4.43984i − 13.40220 j + 12.99459k

and

Then

or

  j 

| v × a | = [(−4.43984) 2 + ( −13.40220)2 + (12.99459) 2 ]1/ 2 = 19.18829 1 (−4.43984i − 13.40220 j + 12.99459k ) 19.1829 4.43984 13.40220 12.99459 cos θ x = − cos θ y = − cos θ z = 19.18829 19.18829 19.18829

eb =

θ x = 103.4°

θ y = 134.3°

θ z = 47.4°



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PROBLEM 11.182 The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION x = 2t 3 − 15t 2 + 24t + 4

dx = 6t 2 − 30t + 24 dt dv a= = 12t − 30 dt v=

so

(a)

0 = 6t 2 − 30t + 24 = 6 (t 2 − 5t + 4)

Times when v = 0.

(t − 4)(t − 1) = 0

(b)

t = 1.00 s, t = 4.00 s 

Position and distance traveled when a = 0. a = 12t − 30 = 0

t = 2.5 s

x2 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4

so

x = 1.50 m 

Final position For

0 ≤ t ≤ 1 s, v > 0.

For

1 s ≤ t ≤ 2.5 s, v ≤ 0.

At

t = 0,

At

t = 1 s, x1 = (2)(1)3 − (15)(1) 2 + (24)(1) + 4 = 15 m

x0 = 4 m.

Distance traveled over interval: x1 − x0 = 11 m For

1 s ≤ t ≤ 2.5 s,

v≤0

Distance traveled over interval | x2 − x1 | = |1.5 − 15 | = 13.5 m

Total distance:

d = 11 + 13.5

d = 24.5 m 

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PROBLEM 11.183 A particle starting from rest at x = 1 m is accelerated so that its velocity doubles in magnitude between x = 2 m and x = 8 m. Knowing that the acceleration of the particle is defined by the relation a = k[ x − (A/x)], determine the values of the constants A and k if the particle has a velocity of 29 m/s when x = 16 m.

SOLUTION We have

v



When x = 1 ft, v = 0:

v 0

dv A  = a = kx−  dx x 

vdv =



x

1

A  k  x −  dx x 

1 2 1 v = k  x 2 − A ln 2 2

or

x

 x 1

1 1 = k  x 2 − A ln x −  2 2

1 2 1 1 3  v2 = k  (2) 2 − A ln 2 −  = k  − A ln 2  2 2 2 2    

At x = 2 ft:

1 2 1 1 v8 = k  (8) 2 − A ln 8 −  = k (31.5 − A ln 8) 2 2 2

x = 8 ft:

Now

1 2 v 2 8 1 2 v 2 2

v8 = 2: v2

= (2) 2 =

k (31.5 − A ln 8) k ( 32 − A ln 2 )

6 − 4 A ln 2 = 31.5 − A ln 8

1 25.5 = A(ln 8 − 4 ln 2) = A(ln 8 − ln 24 ) = A ln   2 A=

When x = 16 m, v = 29 m/s:

25.5 ln 12

A = −36.8 m 2 

1 1 25.5 1 (29) 2 = k  (16) 2 − ln(16) −  1 2 2 2 ln ( 2 )   1  420.5k = k 128 + 102 −  = 230.5k 2  = 230.5k

k = 1.832 s −2 

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PROBLEM 11.184 A particle moves in a straight line with the acceleration shown in the figure. Knowing that the particle starts from the origin with v0 = −2 m/s, (a) construct the v −t and x −t curves for 0 < t < 18 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 18 s.

SOLUTION Compute areas under a − t curve. A1 = (−0.75)(8) = −6 m/s A2 = (2)(4) = 8 m/s A3 = (6)(6) = 36 m/s v0 = −2 m/s v8 = v0 + A1 = −8 m/s v12 = v8 + A2 = 0 v18 = v12 + A3

v18 = 36 m/s 

Sketch v −t curve using straight line portions over the constant acceleration periods. Compute areas under the v −t curve. 1 (−2 − 8)(8) = −40 m 2 1 A5 = (−8)(4) = −16 m 2 1 A6 = (36)(6) = 108 m 2

A4 =

x0 = 0 x8 = x0 + A4 = −40 m x12 = x8 + A5 = −56 m x18 = x12 + A6

Total distance traveled = 56 + 108

x18 = 52 m  d = 164 m 

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PROBLEM 11.185 The velocities of commuter trains A and B are as shown. Knowing that the speed of each train is constant and that B reaches the crossing 10 min after A passed through the same crossing, determine (a) the relative velocity of B with respect to A, (b) the distance between the fronts of the engines 3 min after A passed through the crossing.

SOLUTION (a)

We have

v B = v A + v B/A

The graphical representation of this equation is then as shown. Then

vB2 /A = 662 + 482 − 2(66)(48) cos 155°

or

vB/A = 111.366 km/h

and or

48 111.366 = sin α sin 155°

α = 10.50° v B/A = 111.4 km/h

(b)

10.50° 

First note that at t = 3 min, A is (66 km/h) ( 603 ) = 3.3 km west of the crossing. 7 at t = 3 min, B is (48 km/h) ( 60 ) = 5.6 km southwest of the crossing.

Now

rB = rA + rB/A

Then at t = 3 min, we have rB2/A = 3.32 + 5.62 − 2(3.3)(5.6) cos 25° rB/A = 2.96 km 

or

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PROBLEM 11.186 Slider block B starts from rest and moves to the right with a constant acceleration of 1 ft/s 2. Determine (a) the relative acceleration of portion C of the cable with respect to slider block A, (b) the velocity of portion C of the cable after 2 s.

SOLUTION Let d be the distance between the left and right supports. Constraint of entire cable: xB + ( xB − x A ) + 2(d − x A ) = constant 2vB − 3v A = 0 aA =

and

2 2 aB = (1) = 0.667 ft/s 2 3 3

Constraint of Point C:

2aB − 3a A = 0 a A = 0.667 ft/s 2

or

2(d − x A ) + yC/ A = constant

−2v A + vC/ A = 0

− 2a A + aC/ A = 0

and

aC/ A = 2a A = 2(0.667) = 1.333 ft/s 2

(a)

aC/ A = 1.333 ft/s 2



Velocity vectors after 2s: v A = (0.667)(2) = 1.333 ft/s vC/ A = (1.333)(2) = 2.666 ft/s v C = v A + v C/ A

Sketch the vector addition. vC2 = v A2 + vC2 / A = (1.333) 2 + (2.666) 2 = 8.8889(ft/s) 2 vC = 2.981 ft/s tan θ =

vC/ A vA

=

2.666 = 2, 1.333

θ = 63.4°

vC = 2.98 ft/s

(b)

63.4° 

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PROBLEM 11.187 Collar A starts from rest at t = 0 and moves downward with a constant acceleration of 7 in./s 2 . Collar B moves upward with a constant acceleration, and its initial velocity is 8 in./s. Knowing that collar B moves through 20 in. between t = 0 and t = 2 s, determine (a) the accelerations of collar B and block C, (b) the time at which the velocity of block C is zero, (c) the distance through which block C will have moved at that time.

SOLUTION 

From the diagram



− y A + ( yC − y A ) + 2 yC + ( yC − yB ) = constant

Then

−2v A − vB + 4vC = 0

(1)

and

−2a A − aB + 4aC = 0

(2)

(v A )0 = 0

Given:

(a A ) = 7 in./s 2 ( v B )0 = 8 in./s

a B = constant

 

At t = 2 s (a)

y − (yB )0 = 20 in. y B = ( y B ) 0 + ( vB ) 0 t +

We have At t = 2 s:

−20 in. = (−8 in./s)(2 s) + aB = −4 in./s 2

1 aB t 2 2 1 aB (2 s) 2 2

or

a B = 2 in./s 2 

Then, substituting into Eq. (2) −2(7 in./s 2 ) − (−2 in./s 2 ) + 4aC = 0 aC = 3 in./s 2

or

aC = 3 in./s 2 

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PROBLEM 11.187 (Continued)

(b)

Substituting into Eq. (1) at t = 0 −2(0) − (−8 in./s) + 4(vC )0 = 0 or (vC )0 = −2 in./s  vC = (vC )0 + aC t

Now

(c)

When vC = 0:

0 = (−2 in./s) + (3 in./s 2 )t

or

t=

2 3

t = 0.667 s 

yC = ( yC )0 + (vC )0 t +

We have At t =

2 s 3

s:

2 yC − ( yC )0 = ( −2 in./s)  3 = −0.667 in.

1 aC t 2 2

 1 2 s  + (3 in./s 2 )   2 3 or

 s 

2

y C − (y C )0 = 0.667 in. 

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PROBLEM 11.188 A golfer hits a ball with an initial velocity of magnitude v0 at an angle α with the horizontal. Knowing that the ball must clear the tops of two trees and land as close as possible to the flag, determine v0 and the distance d when the golfer uses (a) a six-iron with α = 31°, (b) a five-iron with α = 27°.

SOLUTION The horizontal and vertical motions are x t cos α 1 2 1 y = (v0 sin α )t − gt = x tan α − gt 2 2 2 x = (v0 cos α )t

v0 =

or

(1)

2( x tan α − y) g

or

t2 =

At the landing Point C:

yC = 0,

And

xC = (v0 cos α )t =

(2) t =

2v0 sin α g

2v02 sin α cos α g

(3)

(a) α = 31° To clear tree A:

x A = 30 m, y A = 12 m

From (2),

t A2 =

From (1),

(v0 ) A =

To clear tree B:

2(30 tan 31° − 12) = 1.22851 s 2 , 9.81

t A = 1.1084 s

30 = 31.58 m/s 1.1084cos 31°

xB = 100 m,

yB = 14 m

From (2),

(t B ) 2 =

2(100 tan 31° − 14) = 9.3957 s 2 , 9.81

From (1),

(v0 ) B =

100 = 38.06 m/s 3.0652 cos 31°

The larger value governs,

v0 = 38.06 m/s

From (3),

xC =

t B = 3.0652 s

v0 = 38.1 m/s 

(2)(38.06) 2 sin 31° cos 31° = 130.38 m 9.81

d = xC − 110

d = 20.4 m 

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PROBLEM 11.188 (Continued)

(b)

α = 27° By a similar calculation,

t A = 0.81846 s, t B = 2.7447 s,

(v0 ) A = 41.138 m/s, (v0 ) B = 40.890 m/s,

v0 = 41.138 m/s

v0 = 41.1 m/s 

xC = 139.56 m,

d = 29.6 m 

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PROBLEM 11.189 As the truck shown begins to back up with a constant acceleration of 4 ft/s 2 , the outer section B of its boom starts to retract with a constant acceleration of 1.6 ft/s 2 relative to the truck. Determine (a) the acceleration of section B, (b) the velocity of section B when t = 2 s.

SOLUTION For the truck,

a A = 4 ft/s 2

For the boom,

a B/ A = 1.6 ft/s 2

50°

(a) a B = a A + a B/ A Sketch the vector addition. By law of cosines: aB2 = a A2 + aB2/ A − 2a AaB/ A cos 50° = 42 + 1.62 − 2(4)(1.6) cos 50° aB = 3.214 ft/s 2

Law of sines:

sin α =

aB/ A sin 50° aB

α = 22.4°,

=

1.6 sin 50° = 0.38131 3.214

a B = 3.21 ft/s 2

22.4° 

v B = (vB )0 + aBt = 0 + (3.214)(2)

(b)

v B = 6.43 ft/s 2

22.4° 

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PROBLEM 11.190 A motorist traveling along a straight portion of a highway is decreasing the speed of his automobile at a constant rate before exiting from the highway onto a circular exit ramp with a radius of 560-ft. He continues to decelerate at the same constant rate so that 10 s after entering the ramp, his speed has decreased to 20 mi/h, a speed which he then maintains. Knowing that at this constant speed the total acceleration of the automobile is equal to one-quarter of its value prior to entering the ramp, determine the maximum value of the total acceleration of the automobile.

SOLUTION First note

v10 = 20 mi/h =

88 ft/s 3

While the car is on the straight portion of the highway. a = astraight = at

and for the circular exit ramp a = at2 + an2

where

an =

v2

ρ

By observation, amax occurs when v is maximum, which is at t = 0 when the car first enters the ramp. For uniformly decelerated motion v = v0 + at t

and at t = 10 s:

v = constant  a = an = a=

Then or

astraight

2 v10

ρ

1 ast. 4

88 v 2 ( 3 ft/s ) 1 = at  at = 10 = ρ 4 560 ft

2

at = −6.1460 ft/s 2

(The car is decelerating; hence the minus sign.)

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PROBLEM 11.190 (Continued)

Then at t = 10 s: or Then at t = 0:

88 ft/s = v0 + (−6.1460 ft/s 2 )(10 s) 3 v0 = 90.793 ft/s amax =

at2

 v2 + 0 ρ 

  

2

1/2

2   (90.793 ft/s) 2    2 2 = (−6.1460 ft/s ) +    560 ft    

amax = 15.95 ft/s 2 

or

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PROBLEM 11.191 Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle α = 25° with the horizontal, determine (a) the speed v0 of the belt, (b) the radius of curvature of the trajectory described by the sand at Point B.

SOLUTION The motion is projectile motion. Place the origin at Point A. Then x0 = 0 and y0 = 0. The coordinates of Point B are xB = 30 ft and yB = −18 ft. Horizontal motion:

Vertical motion:

vx = v0 cos 25°

(1)

x = v0 t cos 25°

(2)

v y = v0 sin 25° − gt y = v0t sin 25° −

(3)

1 2 gt 2

(4)

At Point B, Eq. (2) gives v0 t B =

xB 30 = = 33.101 ft cos 25° cos 25°

Substituting into Eq. (4), 1 −18 = (33.101)(sin 25°) − (32.2)t B2 2 tB = 1.40958 s

(a)

Speed of the belt.

v0 =

v0t B 33.101 = = 23.483 1.40958 tB





v0 = 23.4 ft/s 

Eqs. (1) and (3) give vx = 23.483cos 25° = 21.283 ft/s v y = (23.483) sin 25° − (32.2)(1.40958) = −35.464 ft/s tan θ

−v y vx

= 1.66632

θ = 59.03°

v = 41.36 ft/s

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PROBLEM 11.191 (Continued)

Components of acceleration. a = 32.2 ft/s 2

at = 32.2sin θ

an = 32.2cos θ = 32.2 cos 59.03° = 16.57 ft/s 2

(b)

Radius of curvature at B.

an =

ρ=

v2

ρ v 2 (41.36) 2 = an 16.57

ρ = 103.2 ft 

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PROBLEM 11.192 The end Point B of a boom is originally 5 m from fixed Point A when the driver starts to retract the boom with a constant radial acceleration of  r = −1.0 m/s 2 and lower it with a constant angular acceleration θ = −0.5 rad/s 2 . At t = 2 s, determine (a) the velocity of Point B, (b) the acceleration of Point B, (c) the radius of curvature of the path.

SOLUTION r0 = 5 m, r0 = 0,  r = −1.0 m/s 2

Radial motion.

1 2  rt = 5 + 0 − 0.5t 2 2 r = r0 +  rt = 0 − 1.0t r = r0 + r0 t +

r = 5 − (0.5)(2)2 = 3 m r = (−1.0)(2) = −2 m/s

At t = 2 s,

θ 0 = 60° =

Angular motion.

π 3

rad, θ0 = 0, θ = −0.5 rad/s 2

1 π + 0 − 0.25t 2 2 3 θ = θ0 + θt = 0 − 0.5t

θ = θ0 + θ0 + θt 2 =

θ=

At t = 2 s,

π

+ 0 − (0.25)(2) 2 = 0.047198 rad = 2.70°

3  θ = −(0.5)(2) = −1.0 rad/s

Unit vectors er and eθ .

(a)

Velocity of Point B at t = 2 s. v B = re r + rθeθ = −(2 m/s)e r + (3 m)(−1.0 rad/s)eθ



 v B = (−2.00 m/s)er + ( −3.00 m/s)eθ 

 tan α = v=

vθ −3.0 = = 1.5 vr −2.0 vr2

2

α = 56.31° 2



2

+ vθ = (−2) + (−3) = 3.6055 m/s

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PROBLEM 11.192 (Continued)

Direction of velocity. et =

v −2er − 3eθ = = −0.55470er − 0.83205eθ 3.6055 v

θ + α = 2.70 + 56.31° = 59.01°  (b)

v B = 3.61 m/s



59.0° 

Acceleration of Point B at t = 2 s. a B = ( r − rθ 2 )er + ( rθ + 2rθ)eθ

= [−1.0 − (3)(−1) 2 ]e r + [(3)(−0.5) + (2)(−1.0)(−0.5)]eθ a B = ( −4.00 m/s 2 )er + (2.50 m/s 2 )eθ 

tan β =

aθ 2.50 = = −0.625 ar −4.00

β = −32.00°

a = ar2 + aθ2 = ( −4) 2 + (2.5) 2 = 4.7170 m/s 2

θ + β = 2.70° − 32.00° = −29.30° a B = 4.72 m/s 2

Tangential component:

29.3° 

at = (a ⋅ et )et at = (−4er + 2.5eθ ) ⋅ (−0.55470er − 0.83205eθ )et

= [(−4)(−0.55470) + (2.5)( −0.83205)]et = (0.138675 m/s 2 )et = 0.1389 m/s 2

Normal component:

59.0°

a n = a − at

a n = −4er + 2.5eθ − (0.138675)(−0.55470e r − 0.83205eθ )

= (−3.9231 m/s 2 )er + (2.6154 m/s 2 )eθ an = (3.9231) 2 + (2.6154) 2 = 4.7149 m/s 2

(c)

Radius of curvature of the path. an =

ρ=

v2

ρ v 2 (3.6055 m/s)2 = an 4.7149 m/s

ρ = 2.76 m 

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PROBLEM 11.193 A telemetry system is used to quantify kinematic values of a ski jumper immediately before she leaves the ramp. According to the system r = 500 ft, r = −105 ft/s,  r = −10 ft/s 2 , θ = 25°, θ = 0.07 rad/s, θ = 0.06 rad/s 2 . Determine (a) the velocity of the skier immediately before she leaves the jump, (b) the acceleration of the skier at this instant, (c) the distance of the jump d neglecting lift and air resistance.

SOLUTION (a)

Velocity of the skier.

(r = 500 ft, θ = 25°) v = vr e r + vθ eθ = re r + rθeθ

= ( −105 ft/s)e r + (500 ft)(0.07 rad/s)eθ v = ( −105 ft/s)e r + (35 ft/s)eθ 

Direction of velocity:

v = ( −105cos 25° − 35cos 65°)i + (35sin 65° − 105sin 25°) j = ( −109.95 ft/s)i + ( −12.654 ft/s) j v y −12.654 α = 6.565° tan α = = vx −109.95

v = (105) 2 + (35)2 = 110.68 ft/s v = 110.7 ft/s

6.57° 

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PROBLEM 11.193 (Continued)

(b)

Acceleration of the skier. a = ar er + aθ eθ = ( r − rθ 2 )er + (rθ + 2rθ)eθ

ar = −10 − (500)(0.07)2 = −12.45 ft/s 2 aθ = (500)(0.06) + (2)(−105)(0.07) = 15.30 ft/s 2 a = (−12.45 ft/s 2 )er + (15.30 ft/s 2 )eθ 

a = (−12.45)(i cos 25° + j sin 25°) + (15.30)(−i cos 65° + j sin 65°)

= (−17.750 ft/s 2 )i + (8.6049 ft/s 2 ) j ay 8.6049 tan β = = β = −25.9° ax −17.750 a = (12.45) 2 + (15.30)2 = 19.725 ft/s 2 a = 19.73 ft/s 2

(c)

25.9° 

Distance of the jump d. Projectile motion. Place the origin of the xy-coordinate system at the end of the ramp with the x-coordinate horizontal and positive to the left and the y-coordinate vertical and positive downward. Horizontal motion: (Uniform motion)







 Vertical motion:





x0 = 0 x0 = 109.95 ft/s





x = x0 + x0 t = 109.95t 

(Uniformly accelerated motion) y0 = 0 y 0 = 12.654 ft/s  y = 32.2 ft/s



(from Part a )

(from Part a) 

2

y = y0 + y 0 t +

1 2  yt = 12.654t − 16.1t 2  2

At the landing point, x = d cos 30° y = 10 + d sin 30° or

(1) y − 10 = d sin 30°

(2)

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PROBLEM 11.193 (Continued)

Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract x sin 30° − ( y − 10) cos 30° = 0 (109.95t )sin 30° − (12.654t + 16.1t 2 − 10) cos 30° = 0 −13.943t 2 + 44.016t + 8.6603 = 0 t = −0.1858 s and 3.3427 s

Reject the negative root. x = (109.95 ft/s)(3.3427 s) = 367.53 ft d=

x cos 30°

d = 424 ft 

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CHAPTER 12

PROBLEM 12.CQ1 A 1000 lb boulder B is resting on a 200 lb platform A when truck C accelerates to the left with a constant acceleration. Which of the following statements are true (more than one may be true)? (a) The tension in the cord connected to the truck is 200 lb (b) The tension in the cord connected to the truck is 1200 lb (c) The tension in the cord connected to the truck is greater than 1200 lb (d ) The normal force between A and B is 1000 lb (e) The normal force between A and B is 1200 lb ( f ) None of the above

SOLUTION Answer: (c) The tension will be greater than 1200 lb and the normal force will be greater than 1000 lb.

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PROBLEM 12.CQ2 Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed from the top, which of the following choices best describes the path of the marble after leaving the tube? (a) 1 (b) 2 (c) 3 (d ) 4 (e) 5

SOLUTION Answer: (d ) The particle will have velocity components along the tube and perpendicular to the tube. After it leaves, it will travel in a straight line.

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PROBLEM 12.CQ3 The two systems shown start from rest. On the left, two 40 lb weights are connected by an inextensible cord, and on the right, a constant 40 lb force pulls on the cord. Neglecting all frictional forces, which of the following statements is true? (a) Blocks A and C will have the same acceleration (b) Block C will have a larger acceleration than block A (c) Block A will have a larger acceleration than block C (d ) Block A will not move (e) None of the above

SOLUTION Answer: (b) If you draw a FBD of B, you will see that since it is accelerating downward, the tension in the cable will be less than 40 lb, so the acceleration of A will be less than the acceleration of C. Also, the system on the left has more inertia, so it is harder to accelerate than the system on the right. 

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PROBLEM 12.CQ4 The system shown is released from rest in the position shown. Neglecting friction, the normal force between block A and the ground is (a) less than the weight of A plus the weight of B (b) equal to the weight of A plus the weight of B (c) greater than the weight of A plus the weight of B

SOLUTION Answer: (a) Since B has an acceleration component downward the normal force between A and the ground will be less than the sum of the weights.

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PROBLEM 12.CQ5 People sit on a Ferris wheel at Points A, B, C and D. The Ferris wheel travels at a constant angular velocity. At the instant shown, which person experiences the largest force from his or her chair (back and seat)? Assume you can neglect the size of the chairs, that is, the people are located the same distance from the axis of rotation. (a) A (b) B (c) C (d ) D (e) The force is the same for all the passengers.

SOLUTION Answer: (c) Draw a FBD and KD at each location and it will be clear that the maximum force will be experiences by the person at Point C.

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PROBLEM 12.F1 Crate A is gently placed with zero initial velocity onto a moving conveyor belt. The coefficient of kinetic friction between the crate and the belt is μk. Draw the FBD and KD for A immediately after it contacts the belt.

SOLUTION Answer:

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PROBLEM 12.F2 Two blocks weighing WA and WB are at rest on a conveyor that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Assuming the coefficient of friction between the boxes and the belt is μk, draw the FBDs and KDs for blocks A and B. How would you determine if A and B remain in contact?

SOLUTION Answer: Block A

Block B

To see if they remain in contact assume aA = aB and then check to see if NAB is greater than zero.

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PROBLEM 12.F3 Objects A, B, and C have masses mA, mB, and mC respectively. The coefficient of kinetic friction between A and B is μk, and the friction between A and the ground is negligible and the pulleys are massless and frictionless. Assuming B slides on A draw the FBD and KD for each of the three masses A, B and C.

SOLUTION Answer: Block A

Block B

Block C

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PROBLEM 12.F4 Blocks A and B have masses mA and mB respectively. Neglecting friction between all surfaces, draw the FBD and KD for each mass.

SOLUTION Block A

Block B

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PROBLEM 12.F5 Blocks A and B have masses mA and mB respectively. Neglecting friction between all surfaces, draw the FBD and KD for the two systems shown.

SOLUTION System 1

System 2

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PROBLEM 12.F6 A pilot of mass m flies a jet in a half vertical loop of radius R so that the speed of the jet, v, remains constant. Draw a FBD and KD of the pilot at Points A, B and C.

SOLUTION

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PROBLEM 12.F7 Wires AC and BC are attached to a sphere which revolves at a constant speed v in the horizontal circle of radius r as shown. Draw a FBD and KD of C.

SOLUTION

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PROBLEM 12.F8 A collar of mass m is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant k. Knowing that the collar has a speed v at Point B, draw the FBD and KD of the collar at this point.

SOLUTION

where x = 7/12 ft and r = 5/12 ft.

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PROBLEM 12.1 Astronauts who landed on the moon during the Apollo 15, 16 and 17 missions brought back a large collection of rocks to the earth. Knowing the rocks weighed 139 lb when they were on the moon, determine (a) the weight of the rocks on the earth, (b) the mass of the rocks in slugs. The acceleration due to gravity on the moon is 5.30 ft/s2.

SOLUTION Since the rocks weighed 139 lb on the moon, their mass is m=

(a)

Wmoon 139 lb = = 26.226 lb ⋅ s 2 /ft 2 g moon 5.30 ft/s

On the earth, Wearth = mg earth w = (26.226 lb ⋅ s 2 /ft)(32.2 ft/s 2 )

(b)

Since 1 slug = 1 lb ⋅ s 2 /ft,

w = 844 lb  m = 26.2 slugs 

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PROBLEM 12.2 The value of g at any latitude φ may be obtained from the formula g = 32.09(1 + 0.0053 sin 2φ ) ft/s 2

which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly spherical. Determine to four significant figures (a) the weight in pounds, (b) the mass in pounds, (c) the mass in lb ⋅ s 2 /ft, at the latitudes of 0°, 45°, and 60°, of a silver bar, the mass of which has been officially designated as 5 lb.

SOLUTION g = 32.09(1 + 0.0053 sin 2φ ) ft/s 2

(a)

Weight:

(b)

Mass: At all latitudes:

(c)

or

φ = 0° :

g = 32.09 ft/s 2

φ = 45°:

g = 32.175 ft/s 2

φ = 90°:

g = 32.26 ft/s 2

W = mg

φ = 0° :

W = (0.1554 lb ⋅ s 2 /ft)(32.09 ft/s 2 ) = 4.987 lb



φ = 45°:

W = (0.1554 lb ⋅ s 2 /ft)(32.175 ft/s 2 ) = 5.000 lb



φ = 90°:

W = (0.1554 lb ⋅ s 2 /ft)(32.26 ft/s 2 ) = 5.013 lb

 m = 5.000 lb 

m=

5.00 lb 32.175 ft/s 2

m = 0.1554 lb ⋅ s 2 /ft 

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PROBLEM 12.3 A 400-kg satellite has been placed in a circular orbit 1500 km above the surface of the earth. The acceleration of gravity at this elevation is 6.43 m/s2. Determine the linear momentum of the satellite, knowing that its orbital speed is 25.6 × 103 km/h.

SOLUTION Mass of satellite is independent of gravity:

m = 400 kg

v = 25.6 × 103 km/h  1h  3 = (25.6 × 106 m/h)   = 7.111 × 10 m/s  3600 s  L = mv = (400 kg)(7.111 × 103 m/s)

L = 2.84 × 106 kg ⋅ m/s 

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PROBLEM 12.4 A spring scale A and a lever scale B having equal lever arms are fastened to the roof of an elevator, and identical packages are attached to the scales as shown. Knowing that when the elevator moves downward with an acceleration of 1 m/s 2 the spring scale indicates a load of 60 N, determine (a) the weight of the packages, (b) the load indicated by the spring scale and the mass needed to balance the lever scale when the elevator moves upward with an acceleration of 1 m/s2.

SOLUTION Assume

g = 9.81 m/s 2 m=

W g

ΣF = ma : Fs − W = −

W a g

 a W 1 −  = Fs g  

or

W=

Fs a 1− g

=

60 1 1− 9.81 W = 66.8 N 

(b) ΣF = ma : Fs − W =

W a g

 a Fs = W 1 +  g  1   = 66.811 +  9.81  

Fs = 73.6 N 

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PROBLEM 12.4 (Continued)

But

a  Fw = Ww 1 +  g 

and

 a Fp = W p 1 +  g 

so that

Ww = W p

and

mw =

Wp g

=

66.81 9.81

mw = 6.81 kg 

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PROBLEM 12.5 In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3 ft/s 2 while still on a level section of the highway. Knowing that the speed of the bus is 60 mi/h as it begins to climb the grade and that the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus up the grade when its speed has decreased to 50 mi/h.

SOLUTION First consider when the bus is on the level section of the highway. alevel = 3 ft/s 2

We have

ΣFx = ma: P =

W alevel g

Now consider when the bus is on the upgrade. We have Substituting for P

ΣFx = ma: P − W sin 7° =

W a′ g

W W alevel − W sin 7° = a′ g g a′ = alevel − g sin 7°

or

= (3 − 32.2 sin 7°) ft/s 2 = −0.92419 ft/s 2

For the uniformly decelerated motion 2 v 2 = (v0 ) upgrade + 2a′( xupgrade − 0)

 5  Noting that 60 mi/h = 88 ft/s, then when v = 50 mi/h  = v0  , we have  6  2

5  2 2  6 × 88 ft/s  = (88 ft/s) + 2(−0.92419 ft/s ) xupgrade  

or

xupgrade = 1280.16 ft xupgrade = 0.242 mi 

or

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PROBLEM 12.6 A hockey player hits a puck so that it comes to rest 10 s after sliding 100 ft on the ice. Determine (a) the initial velocity of the puck, (b) the coefficient of friction between the puck and the ice.

SOLUTION (a)

Assume uniformly decelerated motion. Then

v = v0 + at

At t = 10 s:

0 = v0 + a(10) a=−

v0 10

v 2 = v02 + 2a( x − 0)

Also At t = 10 s:

0 = v02 + 2a(100)

Substituting for a

 v  0 = v02 + 2  − 0  (100) = 0  10  v0 = 20.0 ft/s a=−

and Alternate solution to part (a)

or v0 = 20.0 ft/s 

20 = −2 ft/s 2 10

1 2 at 2 1 v  d = v0 t +  − 0  t 2 2 t  d = d0 + v0 t +

1 v0 t 2 2d v0 = t d=

(b)

We have

+ ΣFy = 0: N − W = 0

Sliding:

N = W = mg

F = μ k N = μ k mg − μk mg = ma

ΣFx = ma : −F = ma

μk = −

a −2.0 ft/s 2 =− g 32.2 ft/s 2

μk = 0.0621 

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PROBLEM 12.7 The acceleration of a package sliding at Point A is 3 m/s2. Assuming that the coefficient of kinetic friction is the same for each section, determine the acceleration of the package at Point B.

SOLUTION For any angle θ . Use x and y coordinates as shown. ay = 0 ΣFy = ma y : N − mg cos θ = 0

N = mg cos θ ΣFx = max : mg sin θ − μ k N = max

ax = g (sin θ − μ k cos θ )

At Point A.

θ = 30°, ax = 3 m/s 2 μk =

g sin 30° − ax g cos 30°

9.81 sin 30° − 3 9.81 cos 30° = 0.22423 =

At Point B.

θ = 15°, μk = 0.22423 ax = 9.81(sin 15° − 0.22423 cos 15°) = 0.414 m/s

a = 0.414 m/s 2

15° 

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PROBLEM 12.8 Determine the maximum theoretical speed that may be achieved over a distance of 60 m by a car starting from rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) four-wheel drive, (b) front-wheel drive, (c) rear-wheel drive.

SOLUTION (a)

Four-wheel drive ΣFy = 0: N1 + N 2 − W = 0

F1 + F2 = μ N1 + μ N 2 = μ ( N1 + N 2 ) = μ w ΣFx = ma : F1 + F2 = ma

μ w = ma

μW

mg = μ g = 0.80(9.81) m m a = 7.848 m/s 2 a=



v 2 = 2ax = 2(7.848 m/s 2 )(60 m) = 941.76 m 2 /s 2 v = 30.69 m/s

(b)

v = 110.5 km/h 

Front-wheel drive F2 = ma

μ (0.6 W ) = ma 0.6μW 0.6μ mg = m m = 0.6 μ g = 0.6(0.80)(9.81)

a=

a = 4.709 m/s 2 v 2 = 2ax = 2(4.709 m/s 2 )(60 m) = 565.1 m 2 /s 2 v = 23.77 m/s

v = 85.6 km/h 

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PROBLEM 12.8 (Continued) (c)

Rear-wheel drive F1 = ma

μ (0.4 W ) = ma 0.4μW 0.4μ mg = m m = 0.4 μ g = 0.4(0.80)(9.81)

a=

a = 3.139 m/s 2 v 2 = 2ax = 2(3.139 m/s 2 )(60 m) = 376.7 m 2 /s 2 v = 19.41 m/s

v = 69.9 km/h 

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PROBLEM 12.9 If an automobile’s braking distance from 90 km/h is 45 m on level pavement, determine the automobile’s braking distance from 90 km/h when it is (a) going up a 5° incline, (b) going down a 3-percent incline. Assume the braking force is independent of grade.

SOLUTION Assume uniformly decelerated motion in all cases. For braking on the level surface, v0 = 90 km/h = 25 m/s, v f = 0 x f − x0 = 45 m v 2f = v02 + 2a( x f − x0 ) a= =

v 2f − v02 2( x f − x0 ) 0 − (25) 2 (2)(45)

= −6.9444 m/s 2

Braking force. Fb = ma W = a g 6.944 W =− 9.81 = −0.70789W

(a)

Going up a 5° incline. ΣF = ma W − Fb − W sin 5° = a g F + W sin 5° a=− b g W = −(0.70789 + sin 5°)(9.81) = −7.79944 m/s 2 x f − x0 =

v 2f − v02

2a 0 − (25) 2 = (2)(−7.79944)

x f − x0 = 40.1 m 

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PROBLEM 12.9 (Continued)

(b)

Going down a 3 percent incline. 3 β = 1.71835° 100 W − Fb + W sin β = a g a = −(0.70789 − sin β )(9.81) = −6.65028 m/s 0 − (25) 2 x f = x0 =  (2)(−6.65028) tan β =



x f − x0 = 47.0 m 

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PROBLEM 12.10 A mother and her child are skiing together, and the mother is holding the end of a rope tied to the child’s waist. They are moving at a speed of 7.2 km/h on a gently sloping portion of the ski slope when the mother observes that they are approaching a steep descent. She pulls on the rope with an average force of 7 N. Knowing the coefficient of friction between the child and the ground is 0.1 and the angle of the rope does not change, determine (a) the time required for the child’s speed to be cut in half, (b) the distance traveled in this time.

SOLUTION Draw free body diagram of child. ΣF = ma :

x-direction:

mg sin 5° − μ k N − T cos15° = ma

y-direction:

N − mg cos5° + T sin15° = 0

From y-direction,

N = mg cos 5° − T sin15° = (20 kg)(9.81 m/s 2 ) cos 5° − (7 N) sin15° = 193.64 N

From x-direction,

μk N

T cos15° m m (0.1)(193.64 N) (7 N) cos 15° = (9.81 m/s 2 )sin 5° − − 20 kg 20 kg

a = g sin 5° −



= −0.45128 m/s 2

(in x-direction.)

v0 = 7.2 km/h = 2 m/s vf =

x0 = 0

1 v0 = 1 m/s 2

v f = v0 + at

t=

v f − v0 a

=

−1 m/s = 2.2159 s −0.45128 m/s 2

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PROBLEM 12.10 (Continued) t = 2.22 s 

(a)

Time elapsed.

(b)

Corresponding distance. x = x0 + v0t +

1 2 at 2

= 0 + (2 m/s)(2.2159 s) +

1 (−0.45128 m/s 2 )(2.2159 s) 2 2 x = 3.32 m 

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PROBLEM 12.11 The coefficients of friction between the load and the flat-bed trailer shown are μs = 0.40 and μk = 0.30. Knowing that the speed of the rig is 72 km/h, determine the shortest distance in which the rig can be brought to a stop if the load is not to shift.

SOLUTION Load: We assume that sliding of load relative to trailer is impending: F = Fm = μs N

Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max . ΣFy = 0: N − W = 0 N = W Fm = μ s N = 0.40 W ΣFx = ma : Fm = mamax 0.40 W =

W amax g

amax = 3.924 m/s 2 a max = 3.92 m/s 2

Uniformly accelerated motion. v 2 = v02 + 2ax with v = 0

v0 = 72 km/h = 20 m/s

a = − amax = 3.924 m/s 2 0 = (20)2 + 2( −3.924) x

x = 51.0 m 

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PROBLEM 12.12 A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Knowing that car A has a mass of 25 Mg and car B a mass of 20 Mg, and that the braking force is 30 kN on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is showing down.

SOLUTION v0 = 90 km/h = 90/3.6 = 25 m/s

(a)

Both cars:

ΣFx = Σma: 60 × 103 N = (45 × 103 kg)a

a = 1.333 m/s 2

v 2 = v01 + 2ax: 0 = (25) 2 + 2(−1.333) x 



x = 234 m 

Stopping distance: (b)

Car A:





ΣFx = ma: 30 × 103 + P = (25 × 103 )a P = (25 × 103 )(1.333) − 30 × 103

 

Coupling force:

P = +3332 N

P = 3.33 kN (tension) 

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PROBLEM 12.13 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable.

SOLUTION (a)

We note that aB =

1 aA. 2

Block A

ΣFx = m A a A : T − (200 lb) sin 30° =

200 aA 32.2

(1)

Block B

350  1  aA   32.2  2 

ΣFy = mB aB : 350 lb − 2T =

(a)

(2)

Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T: −2(200) sin 30° + 350 = 2 150 =





(b)

From Eq. (1), T − (200) sin 30° =

aB =

200 350  1  aA + aA 32.2 32.2  2 

575 aA 32.2 1 1 a A = (8.40 ft/s 2 ), 2 2

200 (8.40) 32.2

a A = 8.40 ft/s 2

30° 

a B = 4.20 ft/s 2 

T = 152.2 lb 

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PROBLEM 12.14 Solve Problem 12.13, assuming that the coefficients of friction between block A and the incline are μs = 0.25 and μk = 0.20. PROBLEM 12.13 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable.

SOLUTION We first determine whether the blocks move by computing the friction force required to maintain block A in equilibrium. T = 175 lb. When B in equilibrium, ΣFx = 0: 175 − 200sin 30° − Freq = 0 Freq = 75.0 lb ΣFy = 0: N − 200 cos 30° = 0 N = 173.2 lb FM = μs N = 0.25(173.2 lb) = 43.3 lb

Since Freq > Fm , blocks will move (A up and B down). We note that aB =

1 aA. 2

Block A

F = μk N = (0.20)(173.2) = 34.64 lb. ΣFx = m A 0 A : − 200sin 30° − 34.64 + T =

200 aA 32.2

(1)

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PROBLEM 12.14 (Continued)

Block B

ΣFy = mB aB : 350 lb − 2T =

(a)

(2)

Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T: −2(200) sin 30° − 2(34.64) + 350 = 2 81.32 = aB =

(b)

350  1  aA 32.2  2 

200 350  1  aA + aA 32.2 32.2  2 

575 aA 32.2

a A = 4.55 ft/s 2

1 1 a A = (4.52 ft/s 2 ), 2 2

From Eq. (1), T − (200) sin 30° − 34.64 =

30° 

aB = 2.28 ft/s 2  200 (4.52) 32.2

T = 162.9 lb 

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PROBLEM 12.15 Each of the systems shown is initially at rest. Neglecting axle friction and the masses of the pulleys, determine for each system (a) the acceleration of block A, (b) the velocity of block A after it has moved through 10 ft, (c) the time required for block A to reach a velocity of 20 ft/s.

SOLUTION Let y be positive downward for both blocks. Constraint of cable: y A + yB = constant a A + aB = 0

For blocks A and B,

aB = − a A

or

ΣF = ma : WA aA g

Block A:

WA − T =

Block B:

P + WB − T =

WB W aB = − B a A g g

P + WB − WA +

WA W aA = − B aA g g

Solving for aA,

T = WA −

or

aA =

WA aA g

WA − WB − P g WA + WB

(1)

v A2 − (v A )02 = 2a A [ y A − ( y A )0 ] with (v A )0 = 0 v A = 2a A [ y A − ( y A ) 0 ] v A − (v A )0 = a A t t=

(2)

with (v A )0 = 0

vA aA

(3)

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PROBLEM 12.15 (Continued)

(a)

Acceleration of block A. System (1): By formula (1), System (2): By formula (1), System (3): By formula (1),

(b)

(c)

WA = 200 lb, WB = 100 lb, P = 0 (a A )1 =

200 − 100 (32.2) 200 + 100

(a A )1 = 10.73 ft/s 2 

WA = 200 lb, WB = 0, P = 50 lb (a A ) 2 =

200 − 100 (32.2) 200

(a A ) 2 = 16.10 ft/s 2 

WA = 2200 lb, WB = 2100 lb, P = 0 ( a A )3 =

2200 − 2100 (32.2) 2200 + 2100

(a A )3 = 0.749 ft/s 2 

v A at y A − ( y A )0 = 10 ft. Use formula (2).

System (1):

(v A )1 = (2)(10.73)(10)

(v A )1 = 14.65 ft/s 

System (2):

(v A ) 2 = (2)(16.10)(10)

(v A ) 2 = 17.94 ft/s 

System (3):

(v A )3 = (2)(0.749)(10)

(v A )3 = 3.87 ft/s 

Time at v A = 20 ft/s. Use formula (3). System (1):

t1 =

20 10.73

t1 = 1.864 s 

System (2):

t2 =

20 16.10

t2 = 1.242 s 

System (3):

t3 =

20 0.749

t3 = 26.7 s 

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PROBLEM 12.16 Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are ( μk ) A = 0.30 and ( μk ) B = 0.32, determine the initial acceleration of each box.

SOLUTION Assume that aB > a A so that the normal force NAB between the boxes is zero. A:

ΣFy = 0: NA − WA cos 15° = 0

or

NA = WA cos 15°

Slipping:

FA = ( μk ) A NA

A:

= 0.3WA cos 15° ΣFx = m A a A : FA − WA sin 15° = m A a A 0.3WA cos 15° − WA sin 15° =

or

WA aA g

a A = (32.2 ft/s 2 )(0.3 cos 15° − sin 15°)

or

= 0.997 ft/s 2

B:

B:

ΣFy = 0: N B − WB cos 15° = 0

or

N B = WB cos 15°

Slipping:

FB = ( μk ) B N B = 0.32WB cos 15° ΣFx = mB aB : FB − WB sin 15° = mB aB

or or

0.32WB cos 15° − WB sin 15° =

WB aB g

aB = (32.2 ft/s 2 )(0.32 cos 15° − sin 15°) = 1.619 ft/s 2 aB > a A  assumption is correct

a A = 0.997 ft/s 2

15° 

a B = 1.619 ft/s 2

15° 

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PROBLEM 12.16 (Continued)

Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then assuming NAB to be compression, a A = aB

and find (ΣFx = ma) for each box.

A:

0.3WA cos 15° − WA sin 15° − N AB =

WA a g

B:

0.32WB cos 15° − WB sin 15° + N AB =

WB a g

Solving yields a = 1.273 ft/s 2 and NAB = −0.859 lb, which contradicts the assumption.

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PROBLEM 12.17 A 5000-lb truck is being used to lift a 1000 lb boulder B that is on a 200 lb pallet A. Knowing the acceleration of the truck is 1 ft/s2, determine (a) the horizontal force between the tires and the ground, (b) the force between the boulder and the pallet.

SOLUTION aT = 1 m/s 2

Kinematics:

a A = a B = 0.5 m/s 2 5000 = 155.28 slugs 32.2 200 mA = = 6.211 slugs 32.2 1000 mB = = 31.056 slugs 32.2

mT =

Masses:

Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder.

Vertical components

: 2T − (m A + mB ) g = (m A + mB ) a A 2T − (37.267)(32.2) = (37.267)(0.5) T = 609.32 lb

Apply Newton’s second law to the truck.

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PROBLEM 12.17 (Continued) Horizontal components (a)

: F − T = mT aT

Horizontal force between lines and ground. F = T + mT aT = 609.32 + (155.28)(1.0)

F = 765 lb 

Apply Newton’s second law to the boulder.

Vertical components + :

FAB − mB g = mB aB

FAB = mB ( g + a) = 31.056(32.2 + 0.5)

(b)

FAB = 1016 lb 

Contact force:

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 322

PROBLEM 12.18 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μs = 0.20 and μk = 0.15. If P = 0, determine (a) the acceleration of block B, (b) the tension in the cord.

SOLUTION From the constraint of the cord: 2 x A + xB/A = constant

Then

2v A + vB/A = 0

and

2a A + aB/A = 0

Now

a B = a A + a B/A

Then

aB = a A + ( −2a A )

or

aB = − a A

(1)

First we determine if the blocks will move for the given value of θ . Thus, we seek the value of θ for which the blocks are in impending motion, with the impending motion of A down the incline. B:

ΣFy = 0: N AB − WB cos θ = 0

or

N AB = mB g cos θ

Now

FAB = μ s N AB

B:

= 0.2mB g cos θ ΣFx = 0: − T + FAB + WB sin θ = 0 T = mB g (0.2cos θ + sin θ )

or A:

A:

ΣFy = 0: N A − N AB − WA cos θ = 0

or

N A = (m A + mB ) g cos θ

Now

FA = μ s N A = 0.2( mA + mB ) g cos θ ΣFx = 0: − T − FA − FAB + WA sin θ = 0

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PROBLEM 12.18 (Continued) T = m A g sin θ − 0.2(m A + mB ) g cos θ − 0.2mB g cos θ

or

= g[ mA sin θ − 0.2(m A + 2mB ) cos θ ]

Equating the two expressions for T mB g (0.2cos θ + sin θ ) = g[m A sin θ − 0.2(mA + 2mB ) cos θ ] 8(0.2 + tan θ ) = [40 tan θ − 0.2(40 + 2 × 8)]

or

tan θ = 0.4

or

or θ = 21.8° for impending motion. Since θ < 25°, the blocks will move. Now consider the motion of the blocks. ΣFy = 0: N AB − WB cos 25° = 0

(a)

B:

or

N AB = mB g cos 25°

Sliding:

FAB = μk N AB = 0.15mB g cos 25° ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB

or

T = mB [ g (0.15cos 25° + sin 25°) − aB ] = 8[9.81(0.15cos 25° + sin 25°) − aB ] = 8(5.47952 − aB )

A:

(N)

ΣFy = 0: N A − N AB − WA cos 25° = 0

or

N A = (m A + mB ) g cos 25°

Sliding:

FA = μk N A = 0.15(m A + mB ) g cos 25° ΣFx = m A a A : − T − FA − FAB + WA sin 25° = m A a A

Substituting and using Eq. (1) T = m A g sin 25° − 0.15( mA + mB ) g cos 25° − 0.15mB g cos 25° − m A (− aB ) = g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°] + m A aB = 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40aB = 91.15202 + 40aB

(N)

Equating the two expressions for T 8(5.47952 − aB ) = 91.15202 + 40aB

or

aB = −0.98575 m/s 2

a B = 0.986 m/s 2

(b)

We have or

25° 

T = 8[5.47952 − (−0.98575)] T = 51.7 N 

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PROBLEM 12.19 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μ s = 0.20 and μk = 0.15. If P = 40 N , determine (a) the acceleration of block B, (b) the tension in the cord.

SOLUTION From the constraint of the cord. 2 x A + xB/A = constant

Then

2v A + vB/A = 0

and

2a A + aB/A = 0

Now

a B = a A + a B/A

Then

aB = a A + ( −2a A )

or

aB = − a A

(1)

First we determine if the blocks will move for the given value of P. Thus, we seek the value of P for which the blocks are in impending motion, with the impending motion of a down the incline. B:

ΣFy = 0: N AB − WB cos 25° = 0

or

N AB = mB g cos 25°

Now

FAB = μ s N AB

B:

= 0.2 mB g cos 25° ΣFx = 0: − T + FAB + WB sin 25° = 0

A:

T = 0.2 mB g cos 25° + mB g sin 25°

or

= (8 kg)(9.81 m/s 2 ) (0.2 cos 25° + sin 25°) = 47.39249 N

A: or

ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0 N A = (m A + mB ) g cos 25° − P sin 25°

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PROBLEM 12.19 (Continued)

Now

FA = μ s N A

or

FA = 0.2[(mA + mB ) g cos 25° − P sin 25°] ΣFx = 0: − T − FA − FAB + WA sin 25° + P cos 25° = 0 −T − 0.2[(m A + mB ) g cos 25° − P sin 25°] − 0.2mB g cos 25° + mA g sin 25° + P cos 25° = 0

or or

P(0.2 sin 25° + cos 25°) = T + 0.2[(m A + 2mB ) g cos 25°] − m A g sin 25°

Then

P(0.2 sin 25° + cos 25°) = 47.39249 N + 9.81 m/s 2 {0.2[(40 + 2 × 8) cos 25° − 40 sin 25°] kg} P = −19.04 N for impending motion.

or

Since P, < 40 N, the blocks will move. Now consider the motion of the blocks. ΣFy = 0: N AB − WB cos 25° = 0

(a)

B:

or

N AB = mB g cos 25°

Sliding:

FAB = μk N AB

= 0.15 mB g cos 25° ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB T = mB [ g (0.15 cos 25° + sin 25°) − aB ]

or

= 8[9.81(0.15 cos 25° + sin 25°) − aB ] = 8(5.47952 − aB )

(N)

A:

ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0

or

N A = (m A + mB ) g cos 25° − P sin 25°

Sliding:

FA = μk N A

= 0.15[(m A + mB ) g cos 25° − P sin 25°] ΣFx = m A a A : − T − FA − FAB + WA sin 25° + P cos 25° = m A a A

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PROBLEM 12.19 (Continued)

Substituting and using Eq. (1) T = m A g sin 25° − 0.15[( mA + mB ) g cos 25° − P sin 25°] − 0.15 mB g cos 25° + P cos 25° − m A (− aB ) = g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°] + P(0.15 sin 25° + cos 25°) + m A aB = 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40(0.15 sin 25° + cos 25°) + 40aB = 129.94004 + 40aB

Equating the two expressions for T 8(5.47952 − aB ) = 129.94004 + 40aB

or

aB = −1.79383 m/s 2

a B = 1.794 m/s 2

(b)

We have

25° 

T = 8[5.47952 − (−1.79383)] T = 58.2 N 

or

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PROBLEM 12.20 A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.3 s with a constant acceleration of 2 m/s2. The belt then moves with a constant deceleration a2 and comes to a stop after a total displacement of 2.2 m. Knowing that the coefficients of friction between the package and the belt are μ s = 0.35 and μk = 0.25, determine (a) the deceleration a2 of the belt, (b) the displacement of the package relative to the belt as the belt comes to a stop.

SOLUTION (a)

Kinematics of the belt. vo = 0 1. Acceleration phase with a1 = 2 m/s 2 v1 = vo + a1t1 = 0 + (2)(1.3) = 2.6 m/s x1 = xo + vo t1 +

1 2 1 a1t1 = 0 + 0 + (2)(1.3) 2 = 1.69 m 2 2

2. Deceleration phase: v2 = 0 since the belt stops. v22 − v12 = 2a2 ( x2 − x1 ) a2 = t2 − t1 =

(b)

v22 − v12 0 − (2.6)2 = = −6.63 2( x2 − x1 ) 2(2.2 − 1.69)

a 2 = 6.63 m/s 2



v2 − v1 0 − 2.6 = = 0.3923 s −6.63 a2

Motion of the package. 1. Acceleration phase. Assume no slip. (a p )1 = 2 m/s 2 ΣFy = 0: N − W = 0 or N = W = mg ΣFx = ma : F f = m(a p )1

The required friction force is Ff. The available friction force is μ s N = 0.35W = 0.35mg Ff m

= (a p )1 , <

μs N = μ s g = (0.35)(9.81) = 3.43 m/s 2 m

Since 2.0 m/s 2 < 3.43 m/s 2, the package does not slip. (v p )1 = v1 = 2.6 m/s and (x p )1 = 1.69 m.

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PROBLEM 12.20 (Continued)

2. Deceleration phase. Assume no slip. (a p ) 2 = −11.52 m/s 2 ΣFx = ma : − F f = m( a p )2 Ff m

μs N m

=

μs mg m

= (a p ) 2 = −6.63 m/s 2

= μ s g = 3.43 m/s 2 < 6.63 m/s 2

Since the available friction force μ s N is less than the required friction force Ff for no slip, the package does slip. (a p ) 2 < 6.63 m/s 2 ,

F f = μk N

ΣFx = m( a p )2 : − μk N = m(a p ) 2

μk N

= − μk g m = −(0.25)(9.81)

(a p ) 2 = −

= −2.4525 m/s 2 (v p ) 2 = (v p )1 + (a p ) 2 (t2 − t1 ) = 2.6 + (−2.4525)(0.3923) = 1.638 m/s 2 1 ( a p )2 (t2 − t1 )2 2 1 = 1.69 + (2.6)(0.3923) + ( −2.4525)(0.3923) 2 2

( x p ) 2 = ( x p )1 + (v p )1 (t2 − t1 ) +

= 2.521 m

Position of package relative to the belt ( x p )2 − x2 = 2.521 − 2.2 = 0.321 x p/belt = 0.321 m



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PROBLEM 12.21 A baggage conveyor is used to unload luggage from an airplane. The 10-kg duffel bag A is sitting on top of the 20-kg suitcase B. The conveyor is moving the bags down at a constant speed of 0.5 m/s when the belt suddenly stops. Knowing that the coefficient of friction between the belt and B is 0.3 and that bag A does not slip on suitcase B, determine the smallest allowable coefficient of static friction between the bags.

SOLUTION Since bag A does not slide on suitcase B, both have the same acceleration. a=a

20°

Apply Newton’s second law to the bag A – suitcase B combination treated as a single particle.

ΣFy = ma y : − (mB + m A ) g cos 20° + N = 0 N = (m A + mB ) g cos 30° = (30)(9.81) cos 20° = 276.55 N

μ B N = (0.3)(276.55) = 82.965 N ΣFx = max : μ B N + ( mA + mB ) g sin 20° = (m A + mB )a a = g sin 20° +

μB N m A mB

= 9.81sin 20° −

a = 0.58972 m/s 2

82.965 30 a = 0.58972 m/s 2

20°

Apply Newton’s second law to bag A alone.

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PROBLEM 12.21 (Continued)

ΣFy = ma y : N AB − mA g cos 20° = 0 N AB = ma g sin 20° = (10)(9.81) cos 20° = 92.184 N

ZFx = max : M A g sin 20° − FAB = m A a FAB = m A ( g sin 20° − a) = (10)(9.81sin 20° − 0.58972) = 27.655 N

Since bag A does not slide on suitcase B,

μs >

FAB 27.655 = = 0.300 N AB 92.184

μs > 0.300 

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PROBLEM 12.22 To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are μs = 0.40 and μk = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s.

SOLUTION Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and a P/T be the acceleration of the plywood relative to the truck. (a)

Find the value of aT so that the relative motion of the plywood with respect to the truck is impending. aP = aT and F1 = μs N1 = 0.40 N1 ΣFy = mP a y : N1 − WP cos 20° = − mP aT sin 20° N1 = mP ( g cos 20° − aT sin 20°) ΣFx = max : F1 − WP sin 20° = mP aT cos 20° F1 = mP ( g sin 20° + aT cos 20°) mP ( g sin 20° + aT cos 20°) = 0.40 mP ( g cos 20° − aT sin 20°) (0.40 cos 20° − sin 20°) g cos 20° + 0.40sin 20° = (0.03145)(9.81) = 0.309

aT =

aT = 0.309 m/s 2

(b)

xP/T = ( xP/T )o + (vP /T )t + aP /T =

2 xP /T t

2

=



1 1 aP / T t 2 = 0 + 0 + aP / T t 2 2 2

(2)(2) = 4.94 m/s 2 (0.9) 2

a P / T = 4.94 m/s 2

20°

a P = aT + a P / T = (aT →) + (4.94 m/s 2

20°)

Fy = mP a y : N 2 − WP cos 20° = −mP aT sin 20° N 2 = mP ( g cos 20° − aT sin 20°)

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PROBLEM 12.22 (Continued)

ΣFx = Σmax : F2 − WP sin 20° = mP aT cos 20° − mP aP / T F2 = mP ( g sin 20° + aT cos 20° − aP / T )

For sliding with friction

F2 = μk N 2 = 0.30 N 2

mP ( g sin 20° + aT cos 20° − aP /T ) = 0.30mP ( g cos 20° + aT sin 20°) aT =

(0.30cos 20° − sin 20°) g + aP /T cos 20° + 0.30sin 20°

= (−0.05767)(9.81) + (0.9594)(4.94) = 4.17

aT = 4.17 m/s 2



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PROBLEM 12.23 To transport a series of bundles of shingles A to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform BC which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration a1 as shown. The lift then decelerates at a constant rate a2 and comes to rest at D, near the top of the ladder. Knowing that the coefficient of static friction between a bundle of shingles and the horizontal platform is 0.30, determine the largest allowable acceleration a1 and the largest allowable deceleration a2 if the bundle is not to slide on the platform.

SOLUTION Acceleration a1: Impending slip. ΣFy = m A a y :

F1 = μ s N1 = 0.30 N1

N1 − WA = mA a1 sin 65°

N1 = WA + mA a1 sin 65° = m A ( g + a1 sin 65°) ΣFx = m A ax : F1 = m A a1 cos 65° F1 = μ s N

or

m A a1 cos 65° = 0.30m A ( g + a1 sin 65°) a1 =

0.30 g cos 65° − 0.30 sin 65°

= (1.990)(9.81) = 19.53 m/s 2

Deceleration a 2 : Impending slip.

a1 = 19.53 m/s 2

65° 

F2 = μ s N 2 = 0.30 N 2

ΣFy = ma y : N1 − WA = − mA a2 sin 65°

N1 = WA − mA a2 sin 65° ΣFx = max :

F2 = mA a2 cos 65° F2 = μ s N 2

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PROBLEM 12.23 (Continued)

or

m A a2 cos 65° = 0.30 m A ( g − a2 cos 65°) 0.30 g cos 65° + 0.30 sin 65° = (0.432)(9.81)

a2 =

= 4.24 m/s 2

a 2 = 4.24 m/s 2

65° 

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PROBLEM 12.24 An airplane has a mass of 25 Mg and its engines develop a total thrust of 40 kN during take-off. If the drag D exerted on the plane has a magnitude D = 2.25v 2 , where v is expressed in meters per second and D in newtons, and if the plane becomes airborne at a speed of 240 km/h, determine the length of runway required for the plane to take off.

SOLUTION F = ma: 40 × 103 N − 2.25v 2 = (25 × 103 kg)a a=v

Substituting



x1 0

dv dv : 40 × 103 − 2.25 v 2 = (25 × 103 ) v dx dx

(25 × 103 )vdv 0 40 × 103 − 2.25v 2 25 × 103 x1 = − [ln(40 × 103 − 2.25v 2 )]v01 2(2.25)

dx =

=



v1

25 × 103 40 × 103 ln 4.5 40 × 103 − 2.25v12

For v1 = 240 km/h = 66.67 m/s x1 =

25 × 103 40 × 103 ln = 5.556ln1.333 3 4.5 40 × 10 − 2.25(66.67) 2

= 1.5982 × 103 m

x1 = 1.598 km 

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PROBLEM 12.25 The propellers of a ship of weight W can produce a propulsive force F0; they produce a force of the same magnitude but of opposite direction when the engines are reversed. Knowing that the ship was proceeding forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship travels before coming to a stop. Assume that the frictional resistance of the water varies directly with the square of the velocity.

SOLUTION

At maximum speed a = 0.

F0 = kv02 = 0

k=

F0 v02

When the propellers are reversed, F0 is reversed.

ΣFx = ma : − F0 − kv 2 = ma − F0 − F0

v2 = ma v02 dx =



x 0

a−

(v

F0

2 0

mv02

mv02 vdv vdv = a F0 v02 + v 2

dx = −

x=− =−

(

mv02 F0



)

v02

mv02 1 ln v02 + v 2 F0 2

(

mv02

)

vdv + v2

0

v0

+ v2

)

0 v0

2 ln v02 − ln 2v02  = mv0 ln 2  2 F0 2 F0 

( )

x = 0.347

m0 v02  F0

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PROBLEM 12.26 A constant force P is applied to a piston and rod of total mass m to make them move in a cylinder filled with oil. As the piston moves, the oil is forced through orifices in the piston and exerts on the piston a force of magnitude kv in a direction opposite to the motion of the piston. Knowing that the piston starts from rest at t = 0 and x = 0, show that the equation relating x, v, and t, where x is the distance traveled by the piston and v is the speed of the piston, is linear in each of these variables.

SOLUTION ΣF = ma : P − kv = ma dv P − kv =a= dt m t v m dv dt = 0 0 P − kv m v = − ln ( P − kv) 0 k m = − [ln ( P − kv) − ln P ] k m P − kv P − kv kt t = − ln =− or ln k P m m P − kv P v = (1 − e− kt/m ) = e− kt/m or m k





x= = x=



t 0

v dt =

Pt k

t

t

− 0

P  k − kt/m  − e  k  m 0

Pt P − kt/m Pt P + (e − 1) = − (1 − e− kt/m ) k m k m Pt kv − , which is linear. k m



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PROBLEM 12.27 A spring AB of constant k is attached to a support at A and to a collar of mass m. The unstretched length of the spring is . Knowing that the collar is released from rest at x = x0 and neglecting friction between the collar and the horizontal rod, determine the magnitude of the velocity of the collar as it passes through Point C.

SOLUTION Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle L = 2 + x 2

The elongation of the spring is e = L − , and the magnitude of the force exerted by the spring is Fs = ke = k (  2 + x 2 − ) x

cos θ =

By geometry,

 + x2 2

ΣFx = max : − Fs cos θ = ma − k (  2 + x 2 − )

a=−



v 0

v dv =

k x x− 2 m   + x2

x

 + x2 2

= ma

   

0

 0 a dx x

v

1 2 k v =− m 2 0



 x x− 2 x0   + x2  0

0  k 1 2 2 2  = − − + dx x x      m  2 x  0

k 1 2 1  v = −  0 −  2 − x02 +   2 + x02  m 2 2  k v2 = 2 2 + x02 − 2  2 + x02 m k =   2 + x02 − 2  2 + x02 +  2   m 

)

(

(

)

answer: v =

k m

(

)

2 + x02 −  

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PROBLEM 12.28 Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.

SOLUTION Let the position coordinate y be positive downward. y A + yD = constant

Constraint of cord AD:

v A + vD = 0,

a A + aD = 0

( yB − yD ) + ( yC − yD ) = constant

Constraint of cord BC:

vB + vC − 2vD = 0,

aB + aC − 2aD = 0

2a A + aB + aC = 0

Eliminate aD .

(1)

We have uniformly accelerated motion because all of the forces are constant. y B = ( y B ) 0 + ( vB ) 0 t + aB =

2[ yB − ( yB )0 ] t

2

=

1 a B t 2 , ( vB ) 0 = 0 2 (2)(3) = 1.5 m/s 2 2 (2)

ΣFy = 0: 2TBC − TAD = 0

Pulley D:

TAD = 2TBC ΣFy = ma y : WA − TAD = m A a A

Block A:

aA =

or

WA − TAD WA − 2TBC = mA mA

(2)

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PROBLEM 12.28 (Continued)

ΣFy = ma y : WC − TBC = mC aC

Block C:

aC =

or

WC − TBC mC

(3)

Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving for TBC ,

 WC − TBC   W − 2TBC  2 A =0  + aB +  mA    mC   m g − 2TBC 2 A mA 

 mC g − TBC   + aB +  mC  

 =0 

 4 1  +   TBC = 3g + aB  mA mC   4 1  10 + 5  TBC = 3(9.81) + 1.5 or TBC = 51.55 N  

ΣFy = ma y : P + WB − TBC = mB aB

Block B: (a)

Magnitude of P. P = TBC − WB + mB aB = 51.55 − 5(9.81) + 5(1.5)

(b)

P = 10.00 N 

Tension in cord AD. TAD = 2TBC = (2)(51.55)

TAD = 103.1 N 

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PROBLEM 12.29 A 40-lb sliding panel is supported by rollers at B and C. A 25-lb counterweight A is attached to a cable as shown and, in cases a and c, is initially in contact with a vertical edge of the panel. Neglecting friction, determine in each case shown the acceleration of the panel and the tension in the cord immediately after the system is released from rest.

SOLUTION (a)

F = Force exerted by counterweight

Panel:

T −F =

ΣFx = ma :

40 a g

(1)

Counterweight A: Its acceleration has two components a A = a P + a A /P = a → + a

ΣFx = max : F =

25 a g

ΣFg = mag : 25 − T =

(2) 25 a g

(3)

Adding (1), (2), and (3): 40 + 25 + 25 a g 25 25 a= g= (32.2) 90 90

T − F + F + 25 − T =

a = 8.94 ft/s 2



Substituting for a into (3): 25 − T =

25  25  g g  90 

T = 25 −

625 90

T = 18.06 lb 

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PROBLEM 12.29 (Continued)

(b)

Panel:

ΣFy = ma :

T=

40 a g

(1)

25 − T =

25 a g

(2)

Counterweight A: ΣFy = ma :

Adding (1) and (2):



40 + 25 a g 25 a= g 65

T + 25 − T =



Substituting for a into (1): T=

(c)

a = 12.38 ft/s 2

40  25  1000 g = g  65  65

T = 15.38 lb 

Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice versa. Panel:

ΣFx = ma :

T=

40 a g

(1)

Counterweight A: Same free body as in Part (b): ΣFy = ma :

25 − T =

25 a g

(2)

Since Eqs. (1) and (2) are the same as in (b), we get the same answers: a = 12.38 ft/s 2

; T = 15.38 lb 

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PROBLEM 12.30 The coefficients of friction between blocks A and C and the horizontal surfaces are μ s = 0.24 and μk = 0.20. Knowing that m A = 5 kg, mB = 10 kg, and mC = 10 kg, determine (a) the tension in the cord, (b) the acceleration of each block.

SOLUTION We first check that static equilibrium is not maintained: ( FA ) m + ( FC )m = μs ( mA + mC ) g = 0.24(5 + 10) g = 3.6 g

Since WB = mB g = 10g > 3.6g, equilibrium is not maintained. ΣFy : N A = m A g

Block A:

FA = μk N A = 0.2m A g ΣFλ = mA a A : T − 0.2mA g = mA a A

(1)

ΣFy : NC = mC g

Block C:

FC = μk NC = 0.2mC g Σ Fx = mC aC : T − 0.2mC g = mC aC ΣFy = mB aB

Block B:

mB g − 2T = mB aB

From kinematics: (a)

(2)

aB =

(3) 1 (a A + aC ) 2

Tension in cord. Given data:

(4)

m A = 5 kg mB = mC = 10 kg

Eq. (1): T − 0.2(5) g = 5a A

a A = 0.2T − 0.2 g

(5)

Eq. (2): T − 0.2(10) g = 10aC

aC = 0.1T − 0.2 g

(6)

Eq. (3): 10 g − 2T = 10aB

aB = g − 0.2T

(7)

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PROBLEM 12.30 (Continued)

Substitute into (4): 1 (0.2T − 0.2 g + 0.1T − 0.2 g ) 2 24 24 1.2 g = 0.35T T= g= (9.81 m/s 2 ) 7 7

g − 0.2T =

(b)

T = 33.6 N 

Substitute for T into (5), (7), and (6):  24  a A = 0.2  g  − 0.2 g = 0.4857(9.81 m/s 2 )  7   24  aB = g − 0.2  g  = 0.3143(9.81 m/s 2 )  7   24  aC = 0.1 g  − 0.2 g = 0.14286(9.81 m/s 2 )  7 

a A = 4.76 m/s 2



a B = 3.08 m/s 2  aC = 1.401 m/s 2



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PROBLEM 12.31 A 10-lb block B rests as shown on a 20-lb bracket A. The coefficients of friction are μs = 0.30 and μk = 0.25 between block B and bracket A, and there is no friction in the pulley or between the bracket and the horizontal surface. (a) Determine the maximum weight of block C if block B is not to slide on bracket A. (b) If the weight of block C is 10% larger than the answer found in a determine the accelerations of A, B and C.

SOLUTION Kinematics. Let x A and xB be horizontal coordinates of A and B measured from a fixed vertical line to the left of A and B. Let yC be the distance that block C is below the pulley. Note that yC increases when C moves downward. See figure.

The cable length L is fixed. L = ( xB − x A ) + ( xP − x A ) + yC + constant

Differentiating and noting that x P = 0, vB − 2v A + vC = 0 −2a A + aB + aC = 0

(1)

Here, a A and aB are positive to the right, and aC is positive downward. Kinetics. Let T be the tension in the cable and FAB be the friction force between blocks A and B. The free body diagrams are: Bracket A:

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PROBLEM 12.31 (Continued)

Block B:

Block C:

Bracket A:

ΣFx = max : 2T − FAB =

Block B:

ΣFx = max : FAB − T =

WA aA g

WB aB g

(2) (3)

+ ΣFy = ma y : N AB − WB = 0

N AB = WB

or Block C:

ΣFy = ma y : mC − T =

WC aC g

(4)

Adding Eqs. (2), (3), and (4), and transposing, W WA W a A + B aB + C aC = WC g g g

(5)

Subtracting Eq. (4) from Eq. (3) and transposing, W WB aB − C aC = FAB − WC g g

(a)

No slip between A and B.

aB = a A

From Eq. (1),

a A = aB = aC = a

From Eq. (5), For impending slip,

a=

(6)

WC g WA + WB + WC

FAB = μs N AB = μsWB

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PROBLEM 12.31 (Continued) Substituting into Eq. (6), (WB − WC )(WC g ) = μ sWB − WC WA + WB + WC

Solving for WC , WC = =

μsWB (WA + WB ) WA + 2WB − μ sWB (0.30)(10)(20 + 10) 20 + (2)(10) − (0.30)(10) WC = 2.43 lbs 

(b)

WC increased by 10%.

WC = 2.6757 lbs

Since slip is occurring,

FAB = μk N AB = μkWB W WB aB − C aC = μ kWB − WC g g

Eq. (6) becomes or

10aB − 2.6757 aC = [(0.25)(10) − 2.6757](32.2)

(7)

With numerical data, Eq. (5) becomes 20a A + 10aB + 2.6757aC = (2.6757)(32.2)

(8)

Solving Eqs. (1), (7), and (8) gives a A = 3.144 ft/s 2 , aB = 0.881 ft/s 2 , aC = 5.407 ft/s 2

a A = 3.14 ft/s 2



a B = 0.881 ft/s 2



aC = 5.41 ft/s 2 

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PROBLEM 12.32 The masses of blocks A, B, C and D are 9 kg, 9 kg, 6 kg and 7 kg, respectively. Knowing that a downward force of magnitude 120 N is applied to block D, determine (a) the acceleration of each block, (b) the tension in cord ABC. Neglect the weights of the pulleys and the effect of friction.

SOLUTION Note: As shown, the system is in equilibrium. From the diagram: 2 y A + 2 yB + yC = constant

Cord 1: Then

2v A + 2vB + vC = 0

and

2a A + 2aB + aC = 0 ( yD − y A ) + ( yD − yB ) = constant

Cord 2:

A:

(1)

Then

2 vD − v A − vB = 0

and

2aD − a A − aB = 0

(2)

ΣFy = mA a A : mA g − 2T1 + T2 = m A a A

(a)

9(9.81) − 2T1 + T2 = 9a A

or

(3)

ΣFy = mB aB : mB g − 2T1 + T2 = mB aB

B:

9(9.81) − 2T1 + T2 = 9aB

or

(4)

Note: Eqs. (3) and (4)  a A = a B Then

Eq. (1)  aC = −4a A Eq. (2)  aD = a A

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PROBLEM 12.32 (Continued)

ΣFy = mC aC : mC g − T1 = mC aC

C:

T1 = mC ( g − aC ) = 6( g + 4a A )

or

(5)

ΣFy = mD aD : mD g − 2T2 + ( FD )ext = mD aD 1 1 T2 = [mD ( g − aD ) + 120] = 94.335 − (7a A ) 2 2

or

(6)

Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3) D:

1 9(9.81) − 2 × 6( g + 4a A ) + 94.335 − (7a A ) = 9a A 2 aA =

or

9(9.81) − 2 × 6(9.81) + 94.335 = 1.0728 m/s 2 48 + 3.5 + 9

a A = a B = a D = 1.073 m/s 2  aC = −4(1.0728 m/s 2 )

and (b)

or aC = 4.29 m/s 2 

Substituting into Eq. (5) T1 = 6 ( 9.81 + 4(1.0728) )

or T1 = 84.6 N 

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PROBLEM 12.33 The masses of blocks A, B, C and D are 9 kg, 9 kg, 6 kg and 7 kg, respectively. Knowing that a downward force of magnitude 50 N is applied to block B and that the system starts from rest, determine at t = 3 s the velocity (a) of D relative to A, (b) of C relative to D. Neglect the weights of the pulleys and the effect of friction.

SOLUTION Note: As shown, the system is in equilibrium. From the constraint of the two cords, 2 y A + 2 yB + yC = constant

Cord 1: Then

2v A + 2vB + vC = 0

and

2a A + 2aB + aC = 0

(1)

( yD − y A ) + ( yD − yB ) = constant

Cord 2: Then

2 vD − v A − vB = 0

and

2aD − a A − aB = 0

(2)

We determine the accelerations of blocks A, C, and D using the blocks as free bodies. WA aA g m A g − 2T1 + T2 = m A a A

ΣFy = m A a A : WA − 2T1 + T2 =

A: or

WB aB g mB g − 2T1 + T2 + ( FB )ext = mB aB

(3)

ΣFy = mB aB : WB − 2T1 + T2 + ( FB )ext =

B: or

(4)

(3) − (4)  −( FB )ext = 9( a A − aB ) (F ) aB = a A + B ext mB

Forming or

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PROBLEM 12.33 (Continued) Then

 (F ) 2a A + 2  a A + B ext  mg 

Eq. (1):

  + aC = 0   aC = −4a A −

or

 (F ) 2aD − a A −  a A + B ext mB 

Eq. (2):

 =0 

aD = a A +

or

2( FB )ext mB

C:

( FB )ext 2mB

 2( FB )ext  ΣFy = mC aC : WC − T1 = mC aC = mC  −4a A −  mB   T1 = mC g + 4mC a A +

or D:

2mC ( FB )ext mB

(5)

ΣFy = mD aD : WD − 2T2 = mD aD T2 =

or

 (F )  1 × mD  g − a A − B ext  2 2mB  

(6)

Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)   2mC ( FB )ext  1 ( FB )ext  m A g − 2  mC g + 4mC a A +  + × mD  g − a A −  = mAa A mB 2mB    2  m A g − 2mC g −

4 mC ( FB )ext mB

or

aA =

Then

aC = −4( −2.2835 m/s 2 ) −

mA + 8mC +

aD = −2.2835 m/s 2 +



mD ( FB )ext 4 mB

+

mD 2

mD g 2

= −2.2835 m/s 2

2(50) = −1.9771 m/s 2 9

(50) = 0.4943 m/s 2 2(9)

Note: We have uniformly accelerated motion, so that v = 0 + at

(a)

We have

v D/ A = v D − v A

or

v D /A = aD t − a A t = [0.4943 − (−2.2835)] m/s 2 × 3 s v D /A = 8.33 m/s 

or (b)

And

v C/ D = v C = v D

or

vC /D = aC t − aD t = (−1.9771 − 0.4943) m/s 2 × 3 s v C /D = 7.41m/s 

or

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PROBLEM 12.34 The 15-kg block B is supported by the 25-kg block A and is attached to a cord to which a 225-N horizontal force is applied as shown. Neglecting friction, determine (a) the acceleration of block A, (b) the acceleration of block B relative to A.

SOLUTION (a)

First we note a B = a A + a B/A , where a B/A is directed along the inclined surface of A. ΣFx = mB ax : P − WB sin 25° = mB a A cos 25° + mB aB/A

B: or

225 − 15 g sin 25° = 15( a A cos 25° + aB/A )

or

15 − g sin 25° = a A cos 25° + aB/A

B:

(1)

ΣFy = mB a y : N AB − WB cos 25° = −mB a A sin 25° N AB = 15( g cos 25° − a A sin 25°)

or A:

ΣFx′ = mA a A : P − P cos 25° + N AB sin 25° = m A a A

or

N AB = [25a A − 225(1 − cos 25°)] / sin 25°

A:

Equating the two expressions for N AB 15( g cos 25° − a A sin 25°) =

or

25a A − 225(1 − cos 25°) sin 25°

3(9.81) cos 25° sin 25° + 45(1 − cos 25°) 5 + 3sin 2 25° = 2.7979 m/s 2

aA =

a A = 2.80 m/s 2

(b)



From Eq. (1) aB/A = 15 − (9.81)sin 25° − 2.7979cos 25°

a B/A = 8.32 m/s 2

or

25° 

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PROBLEM 12.35 Block B of mass 10-kg rests as shown on the upper surface of a 22-kg wedge A. Knowing that the system is released from rest and neglecting friction, determine (a) the acceleration of B, (b) the velocity of B relative to A at t = 0.5 s.

SOLUTION A:

ΣFx = m A a A : WA sin 30° + N AB cos 40° = m A a A

(a)

NAB =

or

22 ( a A − 12 g ) cos 40°

Now we note: a B = a A + a B /A , where a B/A is directed along the top surface of A. B:

ΣFy ′ = mB a y′ : NAB − WB cos 20° = −mB a A sin 50° NAB = 10 ( g cos 20° − a A sin 50°)

or Equating the two expressions for NAB

1   22  a A − g  2   = 10( g cos 20° − a A sin 50°) cos 40°

or

aA =

(9.81)(1.1 + cos 20° cos 40°) = 6.4061 m/s 2 2.2 + cos 40° sin 50°

ΣFx′ = mB ax′ : WB sin 20° = mB aB/A − mB a A cos 50°

or

aB/A = g sin 20° + a A cos 50° = (9.81sin 20° + 6.4061cos 50°) m/s 2 = 7.4730 m/s 2

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PROBLEM 12.35 (Continued)

Finally

a B = a A + a B/ A

We have

aB2 = 6.40612 + 7.47302 − 2(6.4061 × 7.4730) cos 50°

or

aB = 5.9447 m/s 2

and or

7.4730 5.9447 = sin α sin 50°

α = 74.4° a B = 5.94 m/s 2

(b)

75.6° 

Note: We have uniformly accelerated motion, so that v = 0 + at

Now

v B/A = v B − v A = a B t − a At = a B/At

At t = 0.5 s:

vB/A = 7.4730 m/s 2 × 0.5 s

v B/A = 3.74 m/s

or

20° 

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PROBLEM 12.36 A 450-g tetherball A is moving along a horizontal circular path at a constant speed of 4 m/s. Determine (a) the angle θ that the cord forms with pole BC, (b) the tension in the cord.

SOLUTION a A = an =

First we note

v 2A

ρ

ρ = l AB sin θ

where ΣFy = 0: TAB cos θ − WA = 0

(a)

TAB =

or

mA g cos θ

ΣFx = m A a A : TAB sin θ = m A

v A2

ρ

Substituting for TAB and ρ mA g v A2 sin θ = m A cos θ l AB sin θ 1 − cos 2 θ =

or

sin 2 θ = 1 − cos 2 θ

(4 m/s) 2 cos θ 1.8 m × 9.81 m/s 2

cos 2 θ + 0.906105cos θ − 1 = 0 cos θ = 0.64479

Solving

θ = 49.9° 

or (b)

From above

TAB =

m A g 0.450 kg × 9.81 m/s 2 = cos θ 0.64479 TAB = 6.85 N 

or

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 356

PROBLEM 12.37 During a hammer thrower’s practice swings, the 7.1-kg head A of the hammer revolves at a constant speed v in a horizontal circle as shown. If ρ = 0.93 m and θ = 60°, determine (a) the tension in wire BC, (b) the speed of the hammer’s head.

SOLUTION First we note

a A = an =

v 2A

ρ

ΣFy = 0: TBC sin 60° − WA = 0

(a) or

7.1 kg × 9.81 m/s 2 sin 60° = 80.426 N

TBC =

TBC = 80.4 N  ΣFx = m A a A : TBC cos 60° = mA

(b)

or

v A2 =

v A2

ρ

(80.426 N) cos 60° × 0.93 m 7.1 kg v A = 2.30 m/s 

or

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 357

PROBLEM 12.38 A single wire ACB passes through a ring at C attached to a sphere which revolves at a constant speed v in the horizontal circle shown. Knowing that the tension is the same in both portions of the wire, determine the speed v.

SOLUTION ΣFx = ma: T (sin 30° + sin 45°) =

mv 2

ρ

(1)

ΣFy = 0: T (cos 30° + cos 45°) − mg = 0 T (cos 30° + cos 45°) = mg

Divide Eq. (1) by Eq. (2):

(2)

sin 30° + sin 45° v 2 = cos 30° + cos 45° ρ g

v 2 = 0.76733ρ g = 0.76733 (1.6 m)(9.81 m/s 2 ) = 12.044 m 2 /s 2

v = 3.47 m/s 

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PROBLEM 12.39 Two wires AC and BC are tied at C to a sphere which revolves at a constant speed v in the horizontal circle shown. Determine the range of values of v for which both wires remain taut.

SOLUTION

ΣFx = ma: TAC sin 30° + TBC sin 45° =

mv 2

(1)

ρ

ΣFy = 0: TAC cos 30° + TBC cos 45° − mg = 0

Divide Eq (1) by Eq. (2):

TAC cos 30° + TBC cos 45° = mg

(2)

TAC sin 30° + TBC sin 45° v 2 = TAC cos 30° + TBC cos 45° ρ g

(3)

When AC is slack, TAC = 0. Eq. (3) yields

v12 = ρ g tan 45° = (1.6 m) (9.81 m/s2 ) tan 45° = 15.696 m 2 /s2

Wire AC will remain taut if v ≤ v1 , that is, if

v1 = 3.96 m/s

v ≤ 3.96 m/s 

When BC is slack, TBC = 0. Eq. (3) yields

v22 = ρ g tan 30° = (1.6 m)(9.81 m/s 2 ) tan 30° = 9.0621 m 2 /s 2

Wire BC will remain taut if v ≥ v2 , that is, if

v2 = 3.01 m/s

v ≥ 3.01 m/s 

Combining the results obtained, we conclude that both wires remain taut for 3.01 m/s ≤ v ≤ 3.96 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 359

PROBLEM 12.40 Two wires AC and BC are tied at C to a sphere which revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 60 N.

SOLUTION From the solution of Problem 12.39, we find that both wires remain taut for

3.01 m/s ≤ v ≤ 3.96 m/s 

To determine the values of v for which the tension in either wire will not exceed 60 N, we recall Eqs. (1) and (2) from Problem 12.39: TAC sin 30° + TBC sin 45° =

mv 2

ρ

(1)

TAC cos 30° + TBC cos 45° = mg

(2)

Subtract Eq. (1) from Eq. (2). Since sin 45° = cos 45°, we obtain TAC (cos 30° − sin 30°) = mg −

mv 2

(3)

ρ

Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract: TBC (sin 45° cos 30° − cos 45° sin 30°) = TBC sin15° =

mv 2

ρ

mv 2

ρ

cos 30° − mg sin 30°

cos 30° − mg sin 30°

(4)

Making TAC = 60 N, m = 5 kg, ρ = 1.6 m, g = 9.81 m/s 2 in Eq. (3), we find the value v1 of v for which TAC = 60 N:

60(cos 30° − sin 30°) = 5(9.81) − 21.962 = 49.05 −

5v12 1.6

v12 0.32

v12 = 8.668,

We have TAC ≤ 60 N for v ≥ v1 , that is, for

v1 = 2.94 m/s v ≥ 2.94 m/s 

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PROBLEM 12.40 (Continued)

Making TBC = 60 N, m = 5 kg, ρ = 1.6 m, g = 9.81 m/s 2 in Eq. (4), we find the value v2 of v for which TBC = 60 N:

60sin 15° =

5v22 cos 30° − 5(9.81) sin 30° 1.6

15.529 = 2.7063v22 − 24.523

v22 = 14.80,

v2 = 3.85 m/s

We have TBC ≤ 60 N for v ≤ v2 , that is, for

v ≤ 3.85 m/s 

Combining the results obtained, we conclude that the range of allowable value is

3.01 m/s ≤ v ≤ 3.85 m/s 

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PROBLEM 12.41 A 100-g sphere D is at rest relative to drum ABC which rotates at a constant rate. Neglecting friction, determine the range of the allowable values of the velocity v of the sphere if neither of the normal forces exerted by the sphere on the inclined surfaces of the drum is to exceed 1.1 N.

SOLUTION aD = an =

First we note

vD2

ρ

ρ = 0.2 m

where

ΣFx = mD aD : N1 cos 60° + N 2 cos 20° = mD

vD2

ρ

(1)

ΣFy = 0: N1 sin 60° + N 2 sin 20° − WD = 0 N1 sin 60° + N 2 sin 20° = mD g

or

(2)

Case 1: N1 is maximum. N1 = 1.1 N

Let Eq. (2)

(1.1 N) sin 60° + N 2 sin 20° = (0.1 kg) (9.81 m/s 2 ) N 2 = 0.082954 N

or

( N 2 )( N1 )max < 1.1 N

OK

0.2 m (1.1cos 60° + 0.082954 cos 20°) N 0.1 kg

Eq. (1)

(vD2 )( N1 )max =

or

(vD )( N1 )max = 1.121 m/s (sin 20°) × [Eq. (1)] − (cos 20°) × [Eq. (2)]

Now we form

N1 cos 60° sin 20° − N1 sin 60° cos 20° = mD

or

− N1 sin 40° = mD

vD2

ρ vD2

ρ

sin 20° − mD g cos 20° sin 20° − mD g cos 20°

(vD )min occurs when N1 = ( N1 ) max (vD )min = 1.121 m/s PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 362

PROBLEM 12.41 (Continued)

Case 2: N 2 is maximum. N 2 = 1.1 N

Let Eq. (2)

N1 sin 60° + (1.1 N)sin 20° = (0.1 kg)(9.81 m/s 2 ) N1 = 0.69834 N

or

( N1 )( N 2 )max ≤ 1.1 N

OK

0.2 m (0.69834 cos 60° + 1.1cos 20°) N 0.1 kg

Eq. (1)

(vD2 )( N 2 )max =

or

(vD )( N 2 )max = 1.663 m/s (sin 60°) × [Eq. (1)] − (cos 60°) × [Eq. (2)]

Now we form

N 2 cos 20° sin 60° − N 2 sin 20° cos 60° = mD

or

N 2 cos 40° = mD

vD2

ρ vD2

ρ

sin 60° − mD g cos 60° sin 60° − mD g cos 60°

(vD ) max occurs when N 2 = ( N 2 )max (vD ) max = 1.663 m/s

For N1 ≤ N 2 < 1.1 N 1.121 m/s < vD < 1.663 m/s 

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PROBLEM 12.42* As part of an outdoor display, a 12-lb model C of the earth is attached to wires AC and BC and revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 26 lb.

SOLUTION aC = an =

First note

vC2

ρ

ρ = 3 ft

where

ΣFx = mC aC : TCA sin 40° + TCB sin15° =

WC vC2 g ρ

ΣFy = 0: TCA cos 40° − TCB cos15° − WC = 0

(1) (2)

Note that Eq. (2) implies that (a)

when

TCB = (TCB ) max , TCA = (TCA ) max

(b)

when

TCB = (TCB ) min ,

TCA = (TCA )min

Case 1: TCA is maximum.

Let Eq. (2) or

TCA = 26 lb (26 lb) cos 40° − TCB cos15° − (12 lb) = 0

TCB = 8.1964 lb (TCB )(TCA )max < 26 lb

OK

[(TCB ) max = 8.1964 lb]

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PROBLEM 12.42* (Continued)

Eq. (1) (vC2 )(TCA )max =

(32.2 ft/s 2 )(3 ft) (26sin 40° + 8.1964 sin15°) lb 12 lb

(vC )(TCA )max = 12.31 ft/s

or

(cos15°)(Eq. 1) + (sin15°)(Eq. 2)

Now we form

TCA sin 40° cos15° + TCA cos 40° sin15° =

WC vC2 cos15° + WC sin15° g ρ

TCA sin 55° =

WC vC2 cos15° + WC sin15° g ρ

or

(3)

(vc ) max occurs when TCA = (TCA ) max (vC )max = 12.31 ft/s Case 2: TCA is minimum.

Because (TCA ) min occurs when TCB = (TCB ) min , let TCB = 0 (note that wire BC will not be taut). Eq. (2)

TCA cos 40° − (12 lb) = 0 TCA = 15.6649 lb, 26 lb OK

or

Note: Eq. (3) implies that when TCA = (TCA )min , vC = (vC ) min . Then (32.2 ft/s 2 )(3 ft) (15.6649 lb) sin 40° 12 lb

Eq. (1)

(vC2 )min =

or

(vC )min = 9.00 ft/s

0 < TCA ≤ TCB < 6 lb when

9.00 ft/s < vC < 12.31 ft/s 

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PROBLEM 12.43* The 1.2-lb flyballs of a centrifugal governor revolve at a constant speed v in the horizontal circle of 6-in. radius shown. Neglecting the weights of links AB, BC, AD, and DE and requiring that the links support only tensile forces, determine the range of the allowable values of v so that the magnitudes of the forces in the links do not exceed 17 lb.

SOLUTION v2

First note

a = an =

where

ρ = 0.5 ft

ρ

W v2 g ρ

(1)

ΣFy = 0: TDA cos 20° − TDE cos 30° − W = 0

(2)

ΣFx = ma : TDA sin 20° + TDE sin 30° =

Note that Eq. (2) implies that (a)

when

TDE = (TDE ) max ,

TDA = (TDA ) max

(b)

when

TDE = (TDE ) min ,

TDA = (TDA ) min

Case 1: TDA is maximum.

Let Eq. (2)

TDA = 17 lb (17 lb) cos 20° − TDE cos 30° − (1.2 lb) = 0

or

TDE = 17.06 lb unacceptable ( > 17 lb)

Now let

TDE = 17 lb

Eq. (2) or

TDA cos 20° − (17 lb) cos 30° − (1.2 lb) = 0 TDA = 16.9443 lb OK ( ≤ 17 lb)

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PROBLEM 12.43* (Continued)

(TDA ) max = 16.9443 lb (TDE ) max = 17 lb (v 2 )(TDA )max =

Eq. (1)

(32.2 ft/s 2 )(0.5 ft) (16.9443sin 20° + 17sin 30°) lb 1.2 lb

v(TDA )max = 13.85 ft/s

or

(cos 30°) × [Eq. (1)] + (sin 30°) × [Eq. (2)]

Now form

TDA sin 20° cos 30° + TDA cos 20° sin 30° =

W v2 cos 30° + W sin 30° g ρ

TDA sin 50° =

W v2 cos 30° + W sin 30° g ρ

or

(3)

vmax occurs when TDA = (TDA ) max vmax = 13.85 ft/s Case 2: TDA is minimum.

Because (TDA )min occurs when TDE = (TDE )min , let TDE = 0. Eq. (2)

TDA cos 20° − (1.2 lb) = 0

or

TDA = 1.27701 lb, 17 lb

OK

Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then Eq. (1)

(v 2 )min =

or

(32.2 ft/s 2 ) (0.5 ft) (1.27701 lb) sin 20° 1.2 lb

vmin = 2.42 ft/s

0 < TAB , TBC , TAD , TDE < 17 lb 2.42 ft/s < v < 13.85 ft/s 

when

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PROBLEM 12.44 A 130-lb wrecking ball B is attached to a 45-ft-long steel cable AB and swings in the vertical arc shown. Determine the tension in the cable (a) at the top C of the swing, (b) at the bottom D of the swing, where the speed of B is 13.2 ft/s.

SOLUTION (a)

At C, the top of the swing, vB = 0; thus an =

vB2 =0 LAB

ΣFn = 0: TBA − WB cos 20° = 0 TBA = (130 lb) × cos 20°

or

TBA = 122.2 lb 

or ΣFn = man : TBA − WB = mB

(b)

or

(vB ) 2D LAB

 130 lb TBA = (130 lb) +  2  32.2 ft/s

  (13.2 ft/s)     45 ft

2

    

TBA = 145.6 lb 

or

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PROBLEM 12.45 During a high-speed chase, a 2400-lb sports car traveling at a speed of 100 mi/h just loses contact with the road as it reaches the crest A of a hill. (a) Determine the radius of curvature ρ of the vertical profile of the road at A. (b) Using the value of ρ found in part a, determine the force exerted on a 160-lb driver by the seat of his 3100-lb car as the car, traveling at a constant speed of 50 mi/h, passes through A.

SOLUTION (a)

Note:

100 mi/h = 146.667 ft/s

ΣFn = man : Wcar =

or

Wcar v A2 g ρ

(146.667 ft/s) 2 32.2 ft/s 2 = 668.05 ft

ρ=

ρ = 668 ft 

or (b)

Note: v is constant  at = 0; 50 mi/h = 73.333 ft/s ΣFn = man : W − N =

or

W v 2A g ρ

  (73.333 ft/s) 2 N = (160 lb) 1 −  2  (32.2 ft/s )(668.05 ft)  N = 120.0 lb 

or

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PROBLEM 12.46 A child having a mass of 22 kg sits on a swing and is held in the position shown by a second child. Neglecting the mass of the swing, determine the tension in rope AB (a) while the second child holds the swing with his arms outstretched horizontally, (b) immediately after the swing is released.

SOLUTION Note: The factors of “ 12 ” are included in the following free-body diagrams because there are two ropes and only one is considered. (a)

For the swing at rest 1 ΣFy = 0: TBA cos 35° − W = 0 2 TBA =

or

22 kg × 9.81 m/s 2 2 cos 35° TBA = 131.7 N 

or (b)

At t = 0, v = 0, so that

an =

v2

ρ

=0

1 ΣFn = 0: TBA − W cos 35° = 0 2

or

TBA =

1 (22 kg)(9.81 m/s 2 ) cos 35° 2 TBA = 88.4 N 

or

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PROBLEM 12.47 The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 km/h when the brakes are suddenly applied, causing the wheels of the car to slide on the track ( μk = 0.20). Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B.

SOLUTION ΣFn = man : N − mg = m

(a)

v2

ρ

 v2 N = m  g + ρ 

  

 v2 F = μk N = μk m  g + ρ 

  

ΣFt = mat : F = mat at =

Given data:

 F v2 = μ k  g + ρ m 

  

μk = 0.20, v = 72 km/h = 20 m/s ρ = 30 m

g = 9.81 m/s 2 ,

 (20)2  at = 0.20 9.81 +  30  

(b)

an = 0

at = 4.63 m/s 2 

ΣFn = man = 0: N − mg = 0 N = mg F = μ k N = μk mg

ΣFt = mat : F = mat

at =

F = μ k g = 0.20(9.81) m

at = 1.962 m/s 2 

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PROBLEM 12.47 (Continued)

(c)

ΣFn = man : mg − N =

mv 2

ρ

 v2 N = m  g − ρ 

  

 v2 F = μk N = μk m  g − ρ 

  

ΣFt = mat : F = mat at =

 F v2 = μk  g − m ρ 

  (20) 2   = 0.20 9.81 −  45    at = 0.1842 m/s 2 

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PROBLEM 12.48 A 250-g block fits inside a small cavity cut in arm OA, which rotates in the vertical plane at a constant rate such that v = 3 m/s. Knowing that the spring exerts on block B a force of magnitude P = 1.5 N and neglecting the effect of friction, determine the range of values of θ for which block B is in contact with the face of the cavity closest to the axis of rotation O.

SOLUTION ΣFn = man : P + mg sin θ − Q = m

v2

ρ

To have contact with the specified surface, we need Q ≥ 0, or

Q = P + mg sin θ − sin θ >

Data:

mv 2

ρ

>0

1  v2 P   −  g  ρ m 

(1)

m = 0.250 kg, v = 3 m/s, P = 1.5 N, ρ = 0.9 m

Substituting into (1): 1  (3) 2 1.5  −   9.81  0.9 0.25  sin θ > 0.40775 sin θ >

24.1° < θ < 155.9° 

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PROBLEM 12.49 A series of small packages, each with a mass of 0.5 kg, are discharged from a conveyor belt as shown. Knowing that the coefficient of static friction between each package and the conveyor belt is 0.4, determine (a) the force exerted by the belt on a package just after it has passed Point A, (b) the angle θ defining the Point B where the packages first slip relative to the belt.

SOLUTION Assume package does not slip. at = 0, F f ≤ μ s N

On the curved portion of the belt an =

v2

ρ

=

(1 m/s) 2 = 4 m/s 2 0.250 m

For any angle θ

ΣFy = ma y : N − mg cos θ = − man = −

mv 2

N = mg cos θ −

ρ mv 2

ρ

(1)

ΣFx = max : −F f + mg sin θ = mat = 0

F f = mg sin θ

(a)

At Point A,

θ = 0° N = (0.5)(9.81)(1.000) − (0.5)(4)

(b)

At Point B,

(2)

N = 2.905 N 

F f = μs N mg sin θ = μs ( mg cos θ − man )  a  4   sin θ = μs  cos θ − n  = 0.40 cos θ − g  9.81   

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PROBLEM 12.49 (Continued)

Squaring and using trigonometic identities, 1 − cos 2 θ = 0.16 cos 2 θ − 0.130479cos θ + 0.026601 1.16 cos 2 θ − 0.130479cos θ − 0.97340 = 0 cos θ = 0.97402

θ = 13.09° 

Check that package does not separate from the belt. N=

Ff

μs

=

mg sin θ

μs

N > 0. 

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PROBLEM 12.50 A 54-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a constant rate. Knowing that the pilot’s apparent weights at Points A and C are 1680 N and 350 N, respectively, determine the force exerted on her by the seat of the trainer when the trainer is at Point B.

SOLUTION First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on the seat of the jet trainer. At A:

ΣFn = man : N A − W = m

v A2

ρ

 1680 N  v A2 = (1200 m)  − 9.81 m/s 2   54 kg 

or

= 25,561.3 m 2 /s 2

At C:

ΣFn = man : N C + W = m

or

vC2

ρ

 350 N  vC2 = (1200 m)  + 9.81 m/s 2  54 kg   = 19,549.8 m 2 /s 2

Since at = constant, we have from A to C vC2 = v 2A + 2at Δs AC

or or

19,549.8 m 2/s 2 = 25,561.3 m 2 /s 2 + 2at (π × 1200 m) at = −0.79730 m/s 2

Then from A to B vB2 = v 2A + 2at Δs AB

π  = 25,561.3 m 2/s 2 + 2(−0.79730 m/s 2 )  × 1200 m  2  = 22,555 m 2/s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 376

PROBLEM 12.50 (Continued)

At B:

ΣFn = man : N B = m

vB2

ρ 22,555 m 2 /s 2 1200 m

or

N B = 54 kg

or

N B = 1014.98 N

ΣFt = mat : W + PB = m | at |

or

PB = (54 kg)(0.79730 − 9.81) m/s 2

or

PB = 486.69 N

Finally,

( Fpilot ) B = N B2 + PB2 = (1014.98) 2 + (486.69) 2 = 1126 N (Fpilot ) B = 1126 N

or

25.6° 

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PROBLEM 12.51 A carnival ride is designed to allow the general public to experience high acceleration motion. The ride rotates about Point O in a horizontal circle such that the rider has a speed v0. The rider reclines on a platform A which rides on rollers such that friction is negligible. A mechanical stop prevents the platform from rolling down the incline. Determine (a) the speed v0 at which the platform A begins to roll upwards, (b) the normal force experienced by an 80-kg rider at this speed.

SOLUTION Radius of circle:

R = 5 + 1.5cos 70° = 5.513 m ΣF = ma:

Components up the incline,

70°:

− m A g cos 20° = −

mv02 sin 20° R

1

(a)

Components normal to the incline, N − mg sin 20° =

(b)

1

 g R  2  (9.81 m/s) (5.513 m  2 Speed v0 : v0 =   =  = 12.1898 m/s tan 20°  tan 20°   

Normal force:

v0 = 12.19 m/s 

20°. mv02 cos 20°. R

N = (80)(9.81)sin 20° +

80(12.1898) 2 cos 20° = 2294 N 5.513

N = 2290 N 

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PROBLEM 12.52 A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (See Sample Problem 12.6 for the definition of rated speed). Knowing that a racing car starts skidding on the curve when traveling at a speed of 180 mi/h, determine (a) the banking angle θ , (b) the coefficient of static friction between the tires and the track under the prevailing conditions, (c) the minimum speed at which the same car could negotiate that curve.

SOLUTION W = mg

Weight

a=

Acceleration

v2

ρ

ΣFx = max : F + W sin θ = ma cos θ F=

mv 2

ρ

cos θ − mg sin θ

(1)

ΣFy = ma y : N − W cos θ = ma sin θ N=

(a)

mv 2

ρ

sin θ + mg cos θ

(2)

Banking angle. Rated speed v = 120 mi/h = 176 ft/s. F = 0 at rated speed. 0=

mv 2

ρ

cos θ − mg sin θ

(176)2 v2 = = 0.96199 ρ g (1000) (32.2) θ = 43.89°

tan θ =

(b)

Slipping outward.

θ = 43.9° 

v = 180 mi/h = 264 ft/s

F = μN

μ=

F v 2 cos θ − ρ g sin θ = N v 2 sin θ + ρ g cos θ

(264) 2 cos 43.89° − (1000) (32.2)sin 43.89° (264) 2 sin 43.89° + (1000) (32.2) cos 43.89° = 0.39009

μ=

μ = 0.390 

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PROBLEM 12.52 (Continued)

(c)

Minimum speed.

F = −μ N v 2 cos θ − ρ g sin θ v 2 sin θ + ρ g cos θ ρ g (sin θ − μ cos θ ) v2 = cos θ + μ sin θ

−μ =

=

(1000) (32.2) (sin 43.89° − 0.39009 cos 43.89°) cos 43.89° + 0.39009 sin 43.89°

= 13.369 ft 2 /s 2 v = 115.62 ft/s

v = 78.8 mi/h 

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PROBLEM 12.53 Tilting trains, such as the American Flyer which will run from Washington to New York and Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force Fs (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 100 mi/h on a curved section of track banked through an angle θ = 6° and with a rated speed of 60 mi/h, determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt (φ = 0), (b) the required angle of tilt φ if the passenger is to feel no side force. (See Sample Problem 12.6 for the definition of rated speed.)

SOLUTION Rated speed:

vR = 60 mi/h = 88 ft/s, 100 mi/h = 146.67 ft/s

From Sample Problem 12.6, vR2 = g ρ tan θ

ρ=

or

vR2 (88) 2 = = 2288 ft g tan θ 32.2 tan 6°

Let the x-axis be parallel to the floor of the car. ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) =

(a)

mv 2

ρ

cos (θ + φ )

φ = 0.  v2  cos (θ + φ ) − sin (θ + φ )  Fs = W   gρ   (146.67) 2  =W  cos 6° − sin 6°   (32.2)(2288) 

= 0.1858W

Fs = 0.1858W 

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PROBLEM 12.53 (Continued)

(b)

For Fs = 0, v2 cos (θ + φ ) − sin (θ + φ ) = 0 gρ v2 (146.67)2 = = 0.29199 g ρ (32.2)(2288) θ + φ = 16.28° φ = 16.28° − 6°

tan (θ + φ ) =

φ = 10.28° 

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PROBLEM 12.54 Tests carried out with the tilting trains described in Problem 12.53 revealed that passengers feel queasy when they see through the car windows that the train is rounding a curve at high speed, yet do not feel any side force. Designers, therefore, prefer to reduce, but not eliminate, that force. For the train of Problem 12.53, determine the required angle of tilt φ if passengers are to feel side forces equal to 10% of their weights.

SOLUTION vR = 60 mi/h = 88 ft/s, 100 mi/h = 146.67 ft/s

Rated speed: From Sample Problem 12.6,

vR2 = g ρ tan θ

or

ρ=

vR2 (88) 2 = = 2288 ft g tan θ 32.2 tan 6°

Let the x-axis be parallel to the floor of the car. ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) =

Solving for Fs,

Now So that Let Then

mv 2

ρ

cos (θ + φ )

 v2  cos (θ + φ ) − sin (θ + φ )  Fs = W   gρ 

v2 (146.67) 2 = = 0.29199 and Fs = 0.10W g ρ (32.2)(2288) 0.10W = W [0.29199 cos (θ + φ ) − sin (θ + φ )] u = sin (θ + φ ) cos (θ + φ ) = 1 − u 2 0.10 = 0.29199 1 − u 2 − u or 0.29199 1 − u 2 = 0.10 + u

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PROBLEM 12.54 (Continued)

Squaring both sides,

0.08526(1 − u 2 ) = 0.01 + 0.2u + u 2

or

1.08526u 2 + 0.2u − 0.07526 = 0

The positive root of the quadratic equation is u = 0.18685 Then,

θ + φ = sin −1 u = 10.77° φ = 10.77° − 6°

φ = 4.77° 

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PROBLEM 12.55 A 3-kg block is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Knowing that the coefficient of static friction is 0.5 and that r = 2 m, determine the maximum allowable velocity v of the block.

SOLUTION Let β be the slope angle of the dish. tan β = At r = 2 m, tan β = 1

or

dy 1 = r dr 2

β = 45°

Draw free body sketches of the sphere. ΣFy = 0: N cos β − μS N sin β − mg = 0 N =

mg cos β − μ S sin β

ΣFn = man: N sin β + μ S N cos β =

mv 2

ρ

mg (sin β + μS N cos β ) mv 2 = cos β − μ S sin β ρ v2 = ρ g

sin β + μS cos β sin 45° + 0.5cos 45° = (2)(9.81) = 58.86 m 2 /s 2 cos β − μ S sin β cos 45° − 0.5sin 45° v = 7.67 m/s 

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PROBLEM 12.56 Three seconds after a polisher is started from rest, small tufts of fleece from along the circumference of the 225-mm-diameter polishing pad are observed to fly free of the pad. If the polisher is started so that the fleece along the circumference undergoes a constant tangential acceleration of 4 m/s 2 , determine (a) the speed v of a tuft as it leaves the pad, (b) the magnitude of the force required to free a tuft if the average mass of a tuft is 1.6 mg.

SOLUTION (a)

at = constant  uniformly acceleration motion

Then

v = 0 + at t

At t = 3 s:

v = (4 m/s 2 )(3 s) v = 12.00 m/s 

or (b)

ΣFt = mat : Ft = mat

or

Ft = (1.6 × 10−6 kg)(4 m/s 2 ) = 6.4 × 10−6 N ΣFn = man : Fn = m

At t = 3 s:

Fn = (1.6 × 10−6 kg)

v2

ρ (12 m/s) 2 m) ( 0.225 2

= 2.048 × 10−3 N

Finally,

Ftuft = Ft 2 + Fn2 = (6.4 × 10−6 N) 2 + (2.048 × 10−3 N)2 Ftuft = 2.05 × 10−3 N 

or

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PROBLEM 12.57 A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.24 m/s 2 , determine the coefficient of static friction between the trunk and the turntable.

SOLUTION First we note that (aB )t = constant implies uniformly accelerated motion. vB = 0 + ( a B ) t t

At t = 10 s:

vB = (0.24 m/s 2 )(10 s) = 2.4 m/s

In the plane of the turntable ΣF = mB a B : F = mB (a B )t + mB (a B ) n

Then

F = mB ( aB )t2 + (aB ) n2 = mB (aB )t2 +

( ) vB2

2

ρ

+ ΣFy = 0: N − W = 0

or

N = mB g

At t = 10 s:

F = μ s N = μ s mB g

Then

 

2

 

2

μs mB g = mB ( aB )t2 +  vρB  1/2

or

1 μs = 9.81 m/s 2

2   (2.4 m/s)2    2 2 (0.24 m/s ) +     2.5 m   

μs = 0.236 

or

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PROBLEM 12.58 A small, 300-g collar D can slide on portion AB of a rod which is bent as shown. Knowing that α = 40° and that the rod rotates about the vertical AC at a constant rate of 5 rad/s, determine the value of r for which the collar will not slide on the rod if the effect of friction between the rod and the collar is neglected.

SOLUTION First note

vD = rθABC + ΣFy = 0: N sin 40° − W = 0

or

N=

mg sin 40°

ΣFn = man : N cos 40° = m

vD2 r

or

( rθABC ) 2 mg cos 40° = m sin 40° r

or

r=

g 2  θ

ABC

1 tan 40°

9.81 m/s 2 1 = 2 (5 rad/s) tan 40° = 0.468 m r = 468 mm 

or

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PROBLEM 12.59 A small, 200-g collar D can slide on portion AB of a rod which is bent as shown. Knowing that the rod rotates about the vertical AC at a constant rate and that α = 30° and r = 600 mm, determine the range of values of the speed v for which the collar will not slide on the rod if the coefficient of static friction between the rod and the collar is 0.30.

SOLUTION Case 1: v = vmin , impending motion downward ΣFx = max : N − W sin 30° = m

or

v2 cos 30° r

  v2 N = m  g sin 30° + cos 30°  r  

ΣFy = ma y : F − W cos 30° = −m

v2 sin 30° r

or

  v2 F = m  g cos 30° − sin 30°  r  

Now

F = μs N

Then

or

    v2 v2 m  g cos 30° − sin 30°  = μs × m  g sin 30° + cos 30°  r r    

v 2 = gr

1 − μ s tan 30° μ s + tan 30°

= (9.81 m/s 2 )(0.6 m)

or

1 − 0.3tan 30° 0.3 + tan 30°

vmin = 2.36 m/s

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PROBLEM 12.59 (Continued)

Case 2: v = vmax , impending motion upward ΣFx = max : N − W sin 30° = m

v2 cos 30° r

  v2 N = m  g sin 30° + cos 30°  r  

or

ΣFy = ma y : F + W cos 30° = m

v2 sin 30° r

or

  v2 F = m  − g cos 30° + sin 30°  r  

Now

F = μs N

Then

    v2 v2 m  − g cos 30° + sin 30°  = μs × m  g sin 30° + cos30°  r r    

or

v 2 = gr

1 + μs tan 30° tan 30° − μ s

= (9.81 m/s 2 )(0.6 m)

or

1 + 0.3tan 30° tan 30° − 0.3

vmax = 4.99 m/s

For the collar not to slide 2.36 m/s < v < 4.99 m/s 

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PROBLEM 12.60 A semicircular slot of 10-in. radius is cut in a flat plate which rotates about the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are μs = 0.35 and μk = 0.25, determine whether the block will slide in the slot if it is released in the position corresponding to (a) θ = 80°, (b) θ = 40°. Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released.

SOLUTION First note

Then

1 (26 − 10 sin θ ) ft 12 vE = ρφABCD

ρ=

an =

vE2

ρ

= ρ (φABCD ) 2

1  =  (26 − 10 sin θ ) ft  (14 rad/s) 2 12  98 = (13 − 5 sin θ ) ft/s 2 3

Assume that the block is at rest with respect to the plate. ΣFx = max : N + W cos θ = m

or

vE2

ρ

  v2 N = W  − cos θ + E sin θ    gρ   ΣFy = ma y : −F + W sin θ = − m

or

sin θ

vE2

ρ

cos θ

  v2 F = W  sin θ + E cos θ    gρ  

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PROBLEM 12.60 (Continued)

(a)

We have

θ = 80°

Then 1 98   N = (0.8 lb)  − cos80° + × (13 − 5sin 80°) ft/s 2 × sin 80° 2 3 32.2 ft/s   = 6.3159 lb 1 98   F = (0.8 lb) sin 80° + × (13 − 5sin 80°) ft/s 2 × cos80°  2 3 32.2 ft/s   = 1.92601 lb Fmax = μs N = 0.35(6.3159 lb) = 2.2106 lb

Now

The block does not slide in the slot, and F = 1.926 lb

(b)

We have

80° 

θ = 40°

Then 1 98   N = (0.8 lb)  − cos 40° + × (13 − 5sin 40°) ft/s 2 × sin 40° 2 3 32.2 ft/s   = 4.4924 lb 1 98   F = (0.8 lb) sin 40° + × (13 − 5sin 40°) ft/s 2 × cos 40° 2 3 32.2 ft/s   = 6.5984 lb

Now

Fmax = μs N ,

from which it follows that

F > Fmax

Block E will slide in the slot and

a E = a n + a E/plate = a n + (a E/plate )t + (a E/plate ) n

At t = 0, the block is at rest relative to the plate, thus (a E/plate )n = 0 at t = 0, so that a E/plate must be directed tangentially to the slot. ΣFx = max : N + W cos 40° = m

or

vE2

ρ

sin 40°

  v2 N = W  − cos 40° + E sin 40°  (as above) gρ   = 4.4924 lb

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PROBLEM 12.60 (Continued)

Sliding:

F = μk N

= 0.25(4.4924 lb) = 1.123 lb

Noting that F and a E/plane must be directed as shown (if their directions are reversed, then ΣFx is while ma x is ), we have the block slides downward in the slot and

F = 1.123 lb

40° 

Alternative solutions. (a)

Assume that the block is at rest with respect to the plate. ΣF = ma : W + R = ma n

Then

tan (φ − 10°) =

=

or and Now so that

W W g = = 2  man W vE ρ (φ ABCD ) 2 g ρ 32.2 ft/s 2 98 (13 − 5sin 80°) ft/s 2 3

(from above)

φ − 10° = 6.9588° φ = 16.9588° tan φs = μs

μ s = 0.35

φs = 19.29°

0 < φ < φs  Block does not slide and R is directed as shown. Now

F = R sin φ

Then

F = (0.8 lb)

and R =

W sin (φ − 10°)

sin16.9588° sin 6.9588° = 1.926 lb

F = 1.926 lb

The block does not slide in the slot and

80° 

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PROBLEM 12.60 (Continued)

(b)

Assume that the block is at rest with respect to the plate. ΣF = ma : W + R = ma n

From Part a (above), it then follows that tan (φ − 50°) =

g

ρ (φABCD )2

=

32.2 ft/s 2 98 (13 − 5sin 40°) ft/s 2 3

φ − 50° = 5.752°

or

φ = 55.752°

and

φs = 19.29°

Now

φ > φs

so that

The block will slide in the slot and then

φ = φk , where φk = 14.0362°

or

tan φk = μ k

μk = 0.25

To determine in which direction the block will slide, consider the free-body diagrams for the two possible cases.

ΣF = ma : W + R = ma n + ma E/plate

Now

From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding downward. Then ΣFx = max : W cos 40° + R cos φk = m

or

ρ

sin 40°

F = R sin φk

Now Then

vE2

W cos 40° +

F W vE2 = sin 40° tan φk g ρ   v2 F = μ kW  − cos 40° + E sin 40°    gρ   = 1.123 lb (see the first solution)

F = 1.123 lb

The block slides downward in the slot and

40° 

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PROBLEM 12.61 A small block B fits inside a lot cut in arm OA which rotates in a vertical plane at a constant rate. The block remains in contact with the end of the slot closest to A and its speed is 1.4 m/s for 0 ≤ θ ≤ 150°. Knowing that the block begins to slide when θ = 150°, determine the coefficient of static friction between the block and the slot.

SOLUTION Draw the free body diagrams of the block B when the arm is at θ = 150°. v = at = 0,

g = 9.81 m/s 2

ΣFt = mat : − mg sin 30° + N = 0 N = mg sin 30° ΣFn = man : mg cos 30° − F = m F = mg cos30° −

Form the ratio

μs =

v2

ρ

mv 2

ρ

F , and set it equal to μ s for impending slip. N F g cos 30° − v 2 /ρ 9.81 cos 30° − (1.4)2 /0.3 = = N g sin 30° 9.81 sin 30°

μ s = 0.400 

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PROBLEM 12.62 The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at stations E, F, and G by picking it up at a station when θ = 0 and depositing it at the next station when θ = 180°. Knowing that member BC remains horizontal throughout its motion and that links AB and CD rotate at a constant rate in a vertical plane in such a way that vB = 2.2 ft/s, determine (a) the minimum value of the coefficient of static friction between the component and BC if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending.

SOLUTION ΣFx = max : F =

W vB2 cos θ g ρ

+ ΣFy = ma y : N − W = −

or

Now

W vB2 sin θ g ρ

  v2 N = W 1 − B sin θ    gρ     v2 Fmax = μ s N = μ s W 1 − B sin θ    gρ  

and for the component not to slide F < Fmax

or or

  v2 W vB2 cos θ < μ s W 1 − B sin θ    g ρ gρ  

μs >

cos θ gρ vB2

− sin θ

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PROBLEM 12.62 (Continued)

We must determine the values of θ which maximize the above expression. Thus d dθ

   

 − sin θ = − sin θ  

cos θ gρ vB2

(

gρ vB2

)

− sin θ − (cos θ )(− cos θ )

(

gρ vB2

− sin θ

)

2

or

sin θ =

vB2 gρ

Now

sin θ =

(2.2 ft/s) 2 = 0.180373 (32.2 ft/s 2 ) ( 10 ft ) 12

μs = (μ s )min

for

θ = 10.3915° and θ = 169.609°

or (a)

=0

From above, ( μs ) min = ( μs ) min =

cos θ gρ vB2

where sin θ =

− sin θ

vB2 gρ

cos θ cos θ sin θ = = tan θ − sin θ 1 − sin 2 θ

1 sin θ

= tan10.3915° ( μ s )min = 0.1834 

or (b)

We have impending motion to the left for

θ = 10.39° 

to the right for

θ = 169.6° 

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PROBLEM 12.63 Knowing that the coefficients of friction between the component I and member BC of the mechanism of Problem 12.62 are μs = 0.35 and μk = 0.25, determine (a) the maximum allowable constant speed vB if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending.

SOLUTION ΣFx = max : F =

W vB2 cos θ g ρ

+ ΣFy = ma y : N − W = −

W vB2 sin θ g ρ

  v2 N = W 1 − B sin θ    gρ  

or

Fmax = μ s N

Now

  v2 = μ s W 1 − B sin θ    gρ  

and for the component not to slide F < Fmax

or

or

  v2 W vB2 cos θ < μ s W 1 − B sin θ    g ρ gρ   vB2 < μs

gρ cos θ + μs sin θ

(1)

( )

To ensure that this inequality is satisfied, vB2 must be less than or equal to the minimum value max of μs g ρ /(cos θ + μs sin θ ), which occurs when (cos θ + μ s sin θ ) is maximum. Thus d (cos θ + μs sin θ ) = − sin θ + μ s cos θ = 0 dθ

or

tan θ = μ s

μ s = 0.35 or

θ = 19.2900°

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PROBLEM 12.63 (Continued)

(a)

The maximum allowed value of vB is then

(v ) 2 B

max

= μs

gρ cos θ + μs sin θ

= gρ

tan θ = g ρ sin θ cos θ + (tan θ ) sin θ

where tan θ = μs

 10  = (32.2 ft/s 2 )  ft  sin 19.2900°  12  (vB ) max = 2.98 ft/s 

or (b)

First note that for 90° < θ < 180°, Eq. (1) becomes vB2 < μs

gρ cos α + μs sin α

where α = 180° − θ . It then follows that the second value of θ for which motion is impending is

θ = 180° − 19.2900° = 160.7100°

we have impending motion to the left for

θ = 19.29° 

to the right for

θ = 160.7° 

Alternative solution. ΣF = ma : W + R = man

For impending motion, φ = φs . Also, as shown above, the values of θ for which motion is impending

(

minimize the value of vB, and thus the value of an is an =

vB2

ρ

). From the above diagram, it can be concluded

that an is minimum when ma n and R are perpendicular.

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PROBLEM 12.63 (Continued)

Therefore, from the diagram

θ = φs = tan −1 μs and or

(as above)

man = W sin φs m

vB2

or

ρ

= mg sin θ

vB2 = g ρ sin θ

(as above)

α = 180° − θ

(as above)

For 90° ≤ θ ≤ 180°, we have from the diagram

α = φs and or

man = W sin φs vB2 = g ρ sin θ

(as above)

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PROBLEM 12.64 In the cathode-ray tube shown, electrons emitted by the cathode and attracted by the anode pass through a small hole in the anode and then travel in a straight line with a speed v0 until they strike the screen at A. However, if a difference of potential V is established between the two parallel plates, the electrons will be subjected to a force F perpendicular to the plates while they travel between the plates and will strike the screen at Point B, which is at a distance δ from A. The magnitude of the force F is F = eV /d , where −e is the charge of an electron and d is the distance between the plates. Derive an expression for the deflection d in terms of V, v0 , the charge −e and the mass m of an electron, and the dimensions d, , and L.

SOLUTION Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x =  at the right edge of the plate. At the screen, x=

 2

+L

Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0. Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when it impacts on the screen. For uniform horizontal motion, x = v0t ,

so that

t1 =

and

t2 =

 v0

 2v0

+

L . v0

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between plates (0 < t < t1 ), the vertical force on the electron is Fy = eV /d . After it passes the plates (t1 < t < t2 ), it is zero. For 0 < t < t1 ,

ΣFy = ma y : a y =

Fy m

=

eV md

v y = (v y ) 0 + a y t = 0 + y = y0 + ( v y ) 0 t +

eVt md

1 eVt 2 ayt 2 = 0 + 0 + 2 2md

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PROBLEM 12.64 (Continued)

At t = t1 ,

(v y )1 =

eVt1 md

and

y1 =

eVt12 2md

For t1 < t < t2 , a y = 0 y = y1 + (v y )1 (t − t1 )

At t = t2 ,

y2 = δ = y1 + (v y )1 (t2 − t1 )

δ= =

eVt12 eVt1 eVt 1 + ( t2 − t1 ) = 1  t2 − t1  md  2md md 2  eV    L 1  + −   mdv0  2v0 v0 2 v0 

or

δ=

eV L  mdv02

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PROBLEM 12.65 In Problem 12.64, determine the smallest allowable value of the ratio d / in terms of e, m, v0, and V if at x =  the minimum permissible distance between the path of the electrons and the positive plate is 0.05d . Problem 12.64 In the cathode-ray tube shown, electrons emitted by the cathode and attracted by the anode pass through a small hole in the anode and then travel in a straight line with a speed v0 until they strike the screen at A. However, if a difference of potential V is established between the two parallel plates, the electrons will be subjected to a force F perpendicular to the plates while they travel between the plates and will strike the screen at point B, which is at a distance δ from A. The magnitude of the force F is F = eV /d , where −e is the charge of an electron and d is the distance between the plates. Derive an expression for the deflection d in terms of V, v0 , the charge −e and the mass m of an electron, and the dimensions d, , and L.

SOLUTION Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x =  at the right edge of the plate. At the screen, x=

 2

+L

Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0. Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when it impacts on the screen. For uniform horizontal motion, x = v0t ,

so that

t1 =

and

t2 =

 v0

 2v0

+

L . v0

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between the plates (0 < t < t1 ), the vertical force on the electron is Fy = eV/d . After it passes the plates (t1 < t < t2 ), it is zero. For 0 < t < t1 ,

ΣFy = ma y : a y =

Fy m

=

eV md

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PROBLEM 12.65 (Continued)

v y = (v y ) 0 + a y t = 0 + y = y0 + ( v y ) 0 t +

At t = t1 , But

so that

 v0

, y= y<

eVt md

eVt 2 1 ayt 2 = 0 + 0 + 2 2md

eV  2 2mdv02 d − 0.05d = 0.450 d 2

eV  2 < 0.450 d 2mdv02 d2 1 eV eV > = 1.111 2 2 2 0.450  mv0 2mv0

d eV > 1.054   mv02

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PROBLEM 12.F9 Four pins slide in four separate slots cut in a horizontal circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. Each pin has a mass m and maintains the same velocity relative to the plate when the plate rotates about O with a constant counterclockwise angular velocity ω. Draw the FBDs and KDs to determine the forces on pins P1 and P2.

SOLUTION

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PROBLEM 12.F10 At the instant shown, the length of the boom AB is being decreased at the constant rate of 0.2 m/s, and the boom is being lowered at the constant rate of 0.08 rad/s. If the mass of the men and lift connected to the boom at Point B is m, draw the FBD and KD that could be used to determine the horizontal and vertical forces at B.

SOLUTION

Where r = 6 m, r = −0.2 m/s, r = 0, θ = −0.08 rad/s, θ = 0

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PROBLEM 12.F11 Disk A rotates in a horizontal plane about a vertical axis at the constant rate θ0 . Slider B has a mass m and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when r = 0. Knowing that the slider is released with no radial velocity in the position r = r0, draw a FBD and KD at an arbitrary distance r from O.

SOLUTION

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PROBLEM 12.F12 Pin B has a mass m and slides along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate θ0 , draw a FBD and KD that can be used to determine the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively.

SOLUTION

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PROBLEM 12.66 Rod OA rotates about O in a horizontal plane. The motion of the 0.5-lb collar B is defined by the relations r = 10 + 6 cosπ t and θ = π (4t 2 − 8t ), where r is expressed in inches, t in seconds, and θ in radians. Determine the radial and transverse components of the force exerted on the collar when (a) t = 0, (b) t = 0.5 s.

SOLUTION Use polar coordinates and calculate the derivatives of the coordinates r and θ with respect to time.

Mass of collar:

(a)

r = 10 + 6 cos π t in.

θ = π (4t 2 − 8t ) rad

r = −6π sin π t in./s

θ = π (8t − 8) rad/s

 r = −6π 2 cos π t in./s 2

θ = 8π rad/s 2

m=

0.5 lb = 0.015528 lb s 2 /ft = 1.294 × 10−3 lb ⋅ s 2 /in. 2 32.2 ft/s

t = 0: r = 16 in.

θ =0

r = 0

θ = −8π = −25.1327 rad/s

 r = −6π 2 = −59.218 in./s 2

θ = 8π = −25.1327 rad/s 2

ar = r − rθ 2 = −59.218 − (16)( −25.1327)2 = −10165.6 in./s 2 aθ = rθ + 2rθ = (16)(25.1327) + 0 = 402.12 in./s 2 Fr = mar = (1.294 × 10−3 lb ⋅ s 2 /in.)( − 10165.6 in./s 2 ) Fθ = maθ = (1.294 × 10−3 lb ⋅ s 2 /in.)(402.12 in./s 2 )

Fr = −13.15 lb  Fθ = 0.520 lb 

θ =0 

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PROBLEM 12.66 (Continued)

(b)

t = 0.5 s: r = 10 + 6 cos(0.5π ) = 10 in.

θ = π [(4)(0.25) − (8)(0.5)] = −9.4248 rad = −540° = 180°

r = −6π sin(0.5π ) = −18.8496 in./s

θ = π [(8)(0.5) − 8] = −12.5664 rad/s

 r = −6π 2 cos(0.5π ) = 0

θ = 8π = 25.1327 rad/s 2

ar = r − rθ 2 = 0 − (10)(−12.5664) 2 = −1579.14 in./s 2 aθ = rθ + 2rθ = (10)(25.1327) + (2)(−18.8496)(−12.5664) = 725.07 in./s 2 Fr = mar = (1.294 × 10−3 lb ⋅ s 2 /in.)( −1579.14 in./s 2 )

Fr = −2.04 lb 

Fθ = maθ = (1.294 × 10−3 lb ⋅ s 2 /in.)(725.07 in./s 2 )

Fθ = 0.938 lb 

θ = 180° 

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PROBLEM 12.67 Rod OA oscillates about O in a horizontal plane. The motion of the 2-lb collar B is defined by the relations r = 6(1 − e−2t ) and θ = (3/π )(sin π t ), where r is expressed in inches, t in seconds, and θ in radians. Determine the radial and transverse components of the force exerted on the collar when (a) t = 1 s, (b) t = 1.5 s.

SOLUTION Use polar coordinates and calculate the derivatives of the coordinates r and θ with respect to time. r = 6(1 − e −2t ) in. r = 12e

−2 t

in./s

r = −24e−2t in./s 2

Mass of collar:

(a)

t = 1 s:

m=

θ = (3/π )sin π t radians θ = 3cos π t rad/s θ = −3π sin π t rad/s 2

2 lb = 0.06211 lb ⋅ s 2 /ft = 5.176 × 10−3 lb ⋅ s 2 /in. 2 32.2 ft/s

e−2t = 0.13534,

sin π t = 0,

cos π t = −1

r = 6(1 − 0.13534) = 5.188 in.

θ =0

r = (12)(0.13534) = 1.62402 in./s

θ = −3.0 rad/s

 r = ( −24) (0.13534) = −3.2480 in./s 2

θ = 0

ar =  r − rθ 2 = −3.2480 − (5.188)(−3.0)2 = −49.94 in./s 2 aθ = rθ + 2rθ = 0 + (2)(1.62402) (−3) = −9.744 in./s 2 Fr = mar = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 49.94 in./s 2 )

Fr = −0.258 lb 

Fθ = maθ = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 9.744 in./s 2 )

Fθ = −0.0504 lb 

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PROBLEM 12.67 (Continued)

(b)

t = 1.5 s:

e −2t = 0.049787,

sin π t = −1,

cos π t = 0

r = 6 (1 − 0.049787) = 5.7013 in.

θ = (3/π )(−1) = −0.9549 rad = −54.7°

r = (12)(0.049787) = 0.59744 in./s 2

θ = 0

 r = −(24)(0.049787) = −1.19489 in./s 2 θ = −(3π )(−1) = 9.4248 rad/s 2 ar =  r − rθ 2 = −1.19489 − 0 = −1.19489 in./s 2 aθ = rθ + 2rθ = (5.7013)(9.4248) + 0 = 53.733 in./s 2 Fr = mar = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 1.19489 in./s 2 ) Fθ = maθ = (5.176 × 10−3 lb ⋅ s 2 /in.)(53.733 in./s 2 )

Fr = −0.00618 lb  Fθ = 0.278 lb 

θ = −54.7° 

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PROBLEM 12.68 The 3-kg collar B slides on the frictionless arm AA′. The arm is attached to drum D and rotates about O in a horizontal plane at the rate θ = 0.75t , where θ and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA′.

SOLUTION Kinematics dr = r = 0.5 m/s dt

We have At t = 0, r = 0:



r 0

dr =



t

0.5 dt

0

or

r = (0.5t ) m

Also,

 r =0

θ = (0.75t ) rad/s θ = 0.75 rad/s 2

Now

ar = r − rθ 2 = 0 − [(0.5t ) m][(0.75t ) rad/s]2 = −(0.28125t 3 ) m/s 2

and

aθ = rθ + 2rθ = [(0.5t ) m][0.75 rad/s 2 ] + 2(0.5 m/s)[(0.75t ) rad/s] = (1.125t ) m/s 2

Kinetics ΣFr = mar : − T = (3 kg)( −0.28125t 3 ) m/s 2

or

T = (0.84375t 3 ) N ΣFθ = mB aθ : Q = (3 kg)(1.125t ) m/s 2

or

Q = (3.375t ) N

Now require that

T =Q

or or

(0.84375t 3 ) N = (3.375t ) N t 2 = 4.000 t = 2.00 s 

or

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PROBLEM 12.69 The horizontal rod OA rotates about a vertical shaft according to the relation θ = 10t , where θ and t are expressed in rad/s and seconds, respectively. A 250-g collar B is held by a cord with a breaking strength of 18 N. Neglecting friction, determine, immediately after the cord breaks, (a) the relative acceleration of the collar with respect to the rod, (b) the magnitude of the horizontal force exerted on the collar by the rod.

SOLUTION

θ = 10t rad/s, θ = 10 rad/s 2 m = 250 g = 0.250 kg

Before cable breaks: Fr = −T and  r = 0. Fr = mar : − T = m( r − rθ 2 ) mrθ 2 = mr + T

or

θ 2 =

mr + T 0 − 18 = = 144 rad 2 /s 2 mr (0.25)(0.5)

θ = 12 rad/s Immediately after the cable breaks: Fr = 0, r = 0 (a)

Acceleration of B relative to the rod. m( r − rθ 2 ) = 0

or

 r = rθ 2 = (0.5)(12)2 = 72 m/s 2 a B /rod = 72 m/s 2 radially outward 

(b)

Transverse component of the force. Fθ = maθ : Fθ = m (rθ + 2rθ) Fθ = (0.250)[(0.5)(10) + (2)(0)(12)] = 1.25

Fθ = 1.25 N 

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PROBLEM 12.70 Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that θ = 15 rad/s and θ = 250 rad/s 2 for the position θ = 20°, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE.

SOLUTION Kinematics. From the geometry of the system, we have r = 2b cos θ

Then

r = − (2b sin θ )θ

 r = −2b(θsin θ + θ 2 cos θ )

and

ar = r − rθ 2 = −2b(θsin θ + θ 2 cos θ ) − (2b cos θ )θ 2 = − 2b(θsin θ + 2θ 2 cos θ )

Now

 20  = − 2 ft  [(250 rad/s 2 )sin 20° + 2(15 rad/s) 2 cos 20°] = −1694.56 ft/s 2 12  

aθ = rθ + 2rθ = (2b cos θ )θ + 2(−2bθ sin θ )θ = 2b(θcos θ − 2θ 2 sin θ )

and

 20  = 2 ft  [(250 rad/s 2 ) cos 20° − 2(15 rad/s) 2 sin 20°] = 270.05 ft/s 2  12 

Kinetics. (a)

We have

Fr = mar =

and

Fθ = maθ =

1 4

lb

32.2 ft/s 1 4

2

lb

32.2 ft/s 2

× (−1694.56 ft/s 2 ) = −13.1565 lb × (270.05 ft/s 2 ) = 2.0967 lb

Fr = −13.16 lb  Fθ = 2.10 lb 

ΣFr : −Fr = −Q cos 20°

(b) or

Q=

1 (13.1565 lb) = 14.0009 lb cos 20°

ΣFθ : Fθ = P − Q sin 20°

or

P = (2.0967 + 14.0009sin 20°) lb = 6.89 lb

P = 6.89 lb

70° 

Q = 14.00 lb

40° 

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PROBLEM 12.71 The two blocks are released from rest when r = 0.8 m and θ = 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine (a) the initial tension in the cable, (b) the initial acceleration of block A, (c) the initial acceleration of block B.

SOLUTION Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Constraint of cable:

r + yB = constant, r + vB = 0,

For block A,

 r + aB = 0

r = −aB

(1)

ΣFx = m Aa A : T cos θ = m Aa A or T = m Aa A sec θ

(2)

or

+ ΣFy = mB aB : mB g − T = mB aB

For block B,

Adding Eq. (1) to Eq. (2) to eliminate T, mB g = m Aa A secθ + mB aB

(3) (4)

Radial and transverse components of a A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.  r − rθ 2 = ar = a A ⋅ er = −a A cos θ

(5)

Noting that initially θ = 0, using Eq. (1) to eliminate r, and changing signs gives aB = a A cosθ

(6)

Substituting Eq. (6) into Eq. (4) and solving for a A , aA =

mB g (25) (9.81) = = 5.48 m/s 2 m A secθ + mB cos θ 20sec 30° + 25cos 30°

From Eq. (6), aB = 5.48cos30° = 4.75 m/s 2 (a)

From Eq. (2), T = (20)(5.48)sec30° = 126.6

(b)

Acceleration of block A.

(c)

Acceleration of block B.

T = 126.6 N  a A = 5.48 m/s 2



a B = 4.75 m/s 2 

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PROBLEM 12.72 The velocity of block A is 2 m/s to the right at the instant when r = 0.8 m and θ = 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine, at this instant, (a) the tension in the cable, (b) the acceleration of block A, (c) the acceleration of block B.

SOLUTION Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Radial and transverse components of v A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components. r = vr = v A ⋅ er = −v A cos30° = −2cos30° = −1.73205 m/s

rθ = vθ = v A ⋅ eθ = −v A sin 30° = 2sin 30° = 1.000 m/s 2

θ =

vθ 1.000 = = 1.25 rad/s r 0.8

Constraint of cable: r + yB = constant, r + vB = 0,

For block A,

 r + aB = 0

or

r = −aB

ΣFx = m Aa A : T cosθ = m Aa A or T = m Aa A secθ + ΣFy = mB aB : mB g − T = mB aB

For block B,

Adding Eq. (1) to Eq. (2) to eliminate T, mB g = m Aa A secθ + mB aB

(1) (2) (3) (4)

Radial and transverse components of a A. Use a method similar to that used for the components of velocity. r − rθ 2 = ar = a A ⋅ er = −a A cos θ

(5)

Using Eq. (1) to eliminate r and changing signs gives aB = a A cosθ − rθ 2

(6)

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PROBLEM 12.72 (Continued)

Substituting Eq. (6) into Eq. (4) and solving for a A , aA =

(

mB g + rθ 2

)

m A secθ + mB cos θ

=

(25)[9.81 + (0.8)(1.25) 2 ] = 6.18 m/s 2 20sec30° + 25cos 30°

From Eq. (6), aB = 6.18cos 30° − (0.8)(1.25) 2 = 4.10 m/s 2 (a)

From Eq. (2), T = (20)(6.18) sec30° = 142.7

(b)

Acceleration of block A.

(c)

Acceleration of block B.

T = 142.7 N 

a A = 6.18 m/s 2



a B = 4.10 m/s 2 

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PROBLEM 12.73* Slider C has a weight of 0.5 lb and may move in a slot cut in arm AB, which rotates at the constant rate θ0 = 10 rad/s in a horizontal plane. The slider is attached to a spring of constant k = 2.5 lb/ft, which is unstretched when r = 0. Knowing that the slider is released from rest with no radial velocity in the position r = 18 in. and neglecting friction, determine for the position r = 12 in. (a) the radial and transverse components of the velocity of the slider, (b) the radial and transverse components of its acceleration, (c) the horizontal force exerted on the slider by arm AB.

SOLUTION Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring. Fr = − k (r − l0 ) Σ Fr = mar : − k (r − l0 ) = m(r − rθ 2 ) kl k   r =  θ 2 −  r + 0 m m 

(1)

But d dr dr dr ( r) = = r dt dr dt dr  kl  k   = rdr  =  θ 2 −  r + 0  dr rdr m m   r=

Integrate using the condition r = r0 when r = r0 . 1 2 r  1   2 k  2 kl0  r r r =  θ −  r + m m  r0 2 r0  2  kl 1 2 1 2 1  2 k  2 r − r0 =  θ −  r − r02 + 0 (r − r0 ) 2 2 2 m m 2kl0 k  r 2 = r02 +  θ 2 −  r 2 − r02 + ( r − r0 ) m m 

(

)

(

Data:

m=

)

W 0.5 lb = = 0.01553 lb ⋅ s 2 /ft g 32.2 ft/s 2

θ = 10 rad/s, k = 2.5 lb/ft, l0 = 0 r0 = (vr )0 = 0, r0 = 18 in. = 1.5 ft, r = 12 in. = 1.0 ft

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PROBLEM 12.73* (Continued)

(a)

Components of velocity when r = 12 in. 2.5   2 2 r 2 = 0 + 102 −  (1.0 − 1.5 ) + 0 0.01553   = 76.223 ft 2 /s 2 vr = r = ±8.7306 ft/s

Since r is decreasing, vr is negative

(b)

r = −8.7306 ft/s

vr = −8.73 ft/s 

vθ = rθ = (1.0)(10)

vθ = 10.00 ft/s 

Components of acceleration. Fr = −kr + kl0 = −(2.5)(1.0) + 0 = −2.5 lb ar =

Fr 2.5 =− m 0.01553

ar = 161.0 ft/s 2 

aθ = rθ + 2rθ = 0 + (2)( −8.7306)(10) aθ = −174.6 ft/s 2 

(c)

Transverse component of force. Fθ = maθ = (0.01553)( −174.6)

Fθ = −2.71 lb 

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PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.27), express the radial and transverse components of the velocity v of the particle as functions of θ .

SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27),

h = r 2θ = h0 = r0 v0

or

θ =

r0 v0 r

2

=

r0 v0 cos 2θ r0

2

=

v0 cos 2θ r0

Radial component of velocity. vr = r = = r0

r0 dr  d  θ=  dθ dθ  cos 2θ

sin 2θ v0 cos 2θ (cos 2θ )3/ 2 r

 sin 2θ θ θ = r0 3/2 (cos 2 ) θ  vr = v0

sin 2θ cos 2θ



Transverse component of velocity. vθ =

h r0 v0 = cos 2θ r r0

vθ = v0 cos 2θ 

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PROBLEM 12.75 For the particle of Problem 12.74, show (a) that the velocity of the particle and the central force F are proportional to the distance r from the particle to the center of force O, (b) that the radius of curvature of the path is proportional to r3. PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.27), express the radial and transverse components of the velocity v of the particle as functions of θ.

SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27), h = r 2θ = h0 = r0 v0

or

θ =

r0 v0 r

2

=

r0 v0 cos 2θ r0

2

=

v0 cos 2θ r0

Differentiating the expression for r with respect to time, r =

r0 dr  d  θ=  dθ dθ  cos 2θ

 sin 2θ sin 2θ v0 sin 2θ θ = r0 cos 2θ = v0 θ = r0 3/2 3/ 2 r (cos 2θ ) (cos 2θ ) cos 2θ  0

Differentiating again, r =

(a)

dr  d  sin 2θ   2 cos 2 2θ + sin 2 2θ  v0 2 2 cos 2 2θ + sin 2 2θ θ= θ=  v0 θ = v0 dθ dθ  r0 (cos 2θ )3/2 cos 2θ  cos 2θ

vr = r = v0

sin 2θ cos 2θ

=

v0 r sin 2θ r0

vr vθ = rθ = 0 cos 2θ r0

v = (vr ) 2 + (vθ ) 2 =

v0 r sin 2 2θ + cos 2 2θ r0

v=

v0 r  r0

v 2 2cos 2 2θ + sin 2 2θ r0 v0 2 ar =  r − rθ 2 = 0 − cos 2 2θ r0 cos 2θ cos 2θ r0 2 =

v0 2 cos 2 2θ + sin 2 2θ v0 v 2r = = 02 r0 r0 cos 2θ r0 cos 2θ

Fr = mar =

mv02 r r02

Fr =

:

mv02 r r02



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PROBLEM 12.75 (Continued)

Since the particle moves under a central force, aθ = 0. Magnitude of acceleration. a = ar 2 + aθ 2 =

v0 2 r r0 2

Tangential component of acceleration. at =

v0 2 r dv d  v0 r  v0  r =  = = sin 2θ  dt dt  r0  r0 r0 2

Normal component of acceleration. at = a 2 − at 2 = r  cos 2θ =  0  r 

But

an =

Hence, (b)

But an =

v2

ρ

or

v0 2 r r0 2

1 − sin 2 2θ =

v0 2 r cos 2θ r0 2

2

v0 2 r

ρ=

v 2 v0 2 r 2 r = 2 ⋅ 2 an r0 v0

ρ=

r3  r02

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PROBLEM 12.76 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F along a semicircular path of diameter OA. Observing that r = r0 cos θ and using Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ .

SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27),

h = r 2θ = h0 = r0 v0

or

θ =

r0 v0 r

2

=

r0 v0 r0 cos θ 2

2

=

v0 r0 cos 2 θ

Radial component of velocity. vr = r =

d (r0 cos θ ) = −( r0 sin θ )θ dt

Transverse component of velocity. vθ = rθ = ( r0 cos θ )θ

Speed.

v = vr 2 + vθ 2 = r0θ =

r0 v0 r0 cos θ 2

v=

v0 cos 2 θ



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PROBLEM 12.77 For the particle of Problem 12.76, determine the tangential component Ft of the central force F along the tangent to the path of the particle for (a) θ = 0, (b) θ = 45°. PROBLEM 12.76 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F along a semicircular path of diameter OA. Observing that r = r0 cos θ and using Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ .

SOLUTION Since the particle moves under a central force, h = constant Using Eq. (12.27), h = r 2θ = h0 = r0 v0

θ =

r0 v0 r

2

=

r0 v0 r0 cos θ 2

2

=

v0 r0 cos 2 θ

Radial component of velocity. vr = r =

d (r0 cos θ ) = −( r0 sin θ )θ dt

Transverse component of velocity. vθ = rθ = ( r0 cos θ )θ

Speed. v = vr 2 + vθ 2 = r0θ =

r0 v0 r0 cos θ 2

=

v0 cos 2 θ

Tangential component of acceleration. at =

v0 dv (−2)( − sin θ )θ 2v0 sin θ = v0 = ⋅ 3 3 dt cos θ cos θ r0 cos 2 θ =

2v0 2 sin θ r0 cos5 θ

Tangential component of force. Ft = mat : Ft =

(a)

θ = 0,

(b)

θ = 45°,

2mv0 2 sin θ r0 cos5 θ

Ft = 0 Ft =

Ft = 0  2mv0 sin 45°

Ft =

cos 45° 5

8mv0 2  r0

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PROBLEM 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth.

SOLUTION Mm [Eq. (12.28)] r2

We have

F =G

and

F = Fn = man = m

Then

or

Now

G

v2 r

Mm v2 = m r r2 M= v=

r 2 v G 2π r

τ 2

so that

2

r  2π r  1  2π  3 M=  r =   G τ  G  τ 

Noting that

τ = 27.32 days = 2.3604 × 106 s

and

r = 238,910 mi = 1.26144 × 109 ft

we have

2

  2π 9 3 M=   (1.26144 × 10 ft) −9 4 4  6  34.4 × 10 ft /lb ⋅ s  2.3604 × 10 s  1

M = 413 × 1021 lb ⋅ s 2 /ft 

or

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PROBLEM 12.79 Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration of gravity g at the surface of the earth, and the time τ required for the moon to complete one full revolution about the earth. Compute r knowing that τ = 27.3 days, giving the answer in both SI and U.S. customary units.

SOLUTION Mm r2

We have

F =G

and

F = Fn = man = m

Then or Now so that

G

[Eq. (12.28)] v2 r

Mm v2 m = r r2 v2 =

GM r

GM = gR 2 v2 =

[Eq. (12.30)]

gR 2 g or v = R r r 2π r 2π r = v R gr

For one orbit,

τ=

or

 gτ 2 R 2 r =  2  4π

Now

τ = 27.3 days = 2.35872 × 106 s

1/ 3

  



Q.E.D.

R = 3960 mi = 20.9088 × 106 ft 1/3

SI:

 9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m) 2  r=  4π 2  

= 382.81 × 106 m r = 383 × 103 km 

or U.S. customary units: 1/3

 32.2 ft/s2 × (2.35872 × 106 s) 2 × (20.9088 × 106 ft) 2  r=  4π 2  

= 1256.52 × 106 ft r = 238 × 103 mi 

or

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PROBLEM 12.80 Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with which they describe their orbit. Give the answers in both SI and U.S. customary units.

SOLUTION For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

or

v=

GM r

Let τ be the period time to complete one orbit. But

vτ = 2π r

Then

GM τ 2 r = 4π 2 3

or

GM τ 2 = 4π 2 r 2 r

 GM τ 2 r =  2  4π

1/3

  

τ = 23.934 h = 86.1624 × 103 s

Data: (a)

v 2τ 2 =

or

In SI units:

g = 9.81 m/s 2 , R = 6.37 × 106 m GM = gR 2 = (9.81)(6.37 × 106 ) 2 = 398.06 × 1012 m3 /s 2 1/3

 (398.06 × 1012 )(86.1624 × 103 )2  6 r=  = 42.145 × 10 m 2 4 π  

altitude h = r − R = 35.775 × 106 m In U.S. units:

h = 35,800 km 

g = 32.2 ft/s 2 , R = 3960 mi = 20.909 × 106 ft GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 1/3

 (14.077 × 1015 )(86.1624 × 103 ) 2  6 r=  = 138.334 × 10 ft 2 4 π  

altitude h = r − R = 117.425 × 106 ft

h = 22, 200 mi 

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PROBLEM 12.80 (Continued)

(b)

In SI units: v=

GM 398.06 × 1012 = = 3.07 × 103 m/s 6 r 42.145 × 10

v=

GM 14.077 × 1015 = = 10.09 × 103 ft/s r 138.334 × 106

v = 3.07 km/s 

In U.S. units: v = 10.09 × 103 ft/s 

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PROBLEM 12.81 Show that the radius r of the orbit of a moon of a given planet can be determined from the radius R of the planet, the acceleration of gravity at the surface of the planet, and the time τ required by the moon to complete one full revolution about the planet. Determine the acceleration of gravity at the surface of the planet Jupiter knowing that R = 71,492 km and that τ = 3.551 days and r = 670.9 × 103 km for its moon Europa.

SOLUTION Mm r2

We have

F =G

and

F = Fn = man = m

Then or Now so that

G

[Eq. (12.28)] v2 r

Mm v2 m = r r2 v2 =

GM r

GM = gR 2 v2 =

[Eq. (12.30)]

gR 2 r

or

v=R

g r

2π r 2π r = v R gr

For one orbit,

τ=

or

 gτ 2 R 2 r =  2  4π

Solving for g,

g = 4π 2

1/3

  

Q.E.D.



r3 τ 2 R2

and noting that τ = 3.551 days = 306,806 s, then g Jupiter = 4π 2 = 4π 2

3 rEur 2 RJup τ Eur

(670.9 × 106 m)3 (306,806 s)2 (71.492 × 106 m) 2

g Jupiter = 24.8 m/s 2 

or Note:

g Jupiter ≈ 2.53g Earth 

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PROBLEM 12.82 The orbit of the planet Venus is nearly circular with an orbital velocity of 126.5 × 103 km/h. Knowing that the mean distance from the center of the sun to the center of Venus is 108 × 106 km and that the radius of the sun is 695 × 103 km, determine (a) the mass of the sun, (b) the acceleration of gravity at the surface of the sun.

SOLUTION Let M be the mass of the sun and m the mass of Venus. For the circular orbit of Venus, GMm mv 2 = ma = n r r2

GM = rv 2

where r is radius of the orbit. r = 108 × 106 km = 108 × 109 m

Data:

v = 126.5 × 103 km/hr = 35.139 × 103 m/s GM = (108 × 109 )(35.139 × 103 ) 2 = 1.3335 × 1020 m3 /s 2

(a) Mass of sun. (b) At the surface of the sun,

M =

GM 1.3335 × 1020 m3 /s 2 = G 66.73 × 10−12

M = 1.998 × 1030 kg 

R = 695.5 × 103 km = 695.5 × 106 m GMm = mg R2 g =

GM 1.3335 × 1020 = R2 (695.5 × 106 ) 2

g = 276 m/s 2 

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PROBLEM 12.83 A satellite is placed into a circular orbit about the planet Saturn at an altitude of 2100 mi. The satellite describes its orbit with a velocity of 54.7 × 103 mi/h. Knowing that the radius of the orbit about Saturn and the periodic time of Atlas, one of Saturn’s moons, are 85.54 × 103 mi and 0.6017 days, respectively, determine (a) the radius of Saturn, (b) the mass of Saturn. (The periodic time of a satellite is the time it requires to complete one full revolution about the planet.)

SOLUTION 2π rA

Velocity of Atlas.

vA =

where

v A = 85.54 × 103 mi = 451.651 × 106 ft

and

τ A = 0.6017 days = 51,987 s

Gravitational force.

τA

vA =

(2π ) (451.651 × 106 ) = 54.587 × 103 ft/s 51,987

F=

GMm mv 2 = r r2

from which

GM = rv 2 = constant

For the satellite,

rs vs2 = rAv 2A rA v A2



rs =

where

vs = 54.7 × 103 mi/h = 80.227 × 103 ft/s

vs 2



(451.651 × 106 )(54.587 × 103 ) 2 = 209.09 × 106 ft (80.227 × 103 ) 2 rs = 39, 600 mi rs =

(a)

Radius of Saturn. R = rs − (altitude) = 39, 600 − 2100

(b)

R = 37,500 mi 

Mass of Saturn. M=

rAv 2A (451.651 × 106 )(54.587 × 103 ) 2 = G 34.4 × 10−9 M = 39.1 × 1024 lb ⋅ s 2 /ft 

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PROBLEM 12.84 The periodic times (see Problem 12.83) of the planet Uranus’s moons Juliet and Titania have been observed to be 0.4931 days and 8.706 days, respectively. Knowing that the radius of Juliet’s orbit is 40,000 mi, determine (a) the mass of Uranus, (b) the radius of Titania’s orbit.

SOLUTION 2π rJ

Velocity of Juliet.

vJ =

where

rJ = 40,000 mi = 2.112 × 108 ft

and

τ J = 0.4931 days = 42, 604 s

Gravitational force.

vJ =

(2π )(2.112 × 108 ft) = 3.11476 × 10 4 ft/s 42, 604 s

F=

GMm mv 2 = r r2

GM = rv 2 = constant

from which (a)

τJ

M=

Mass of Uranus.

rJ vJ2 G

(2.112 × 108 )(3.11476 × 104 ) 2 34.4 × 10−9 = 5.95642 × 10 24 lb ⋅ s 2 /ft

M=

M = 5.96 × 1024 lb ⋅ s 2 /ft 

(b)

Radius of Titania’s orbit. GM = rT vT2 =

rT3

=

τ rJ3  T τJ

4π 2 rT3

τT 2

=

4π 2 rJ3

τJ2

2

2  8 3  8.706  27 3  = (2.112 × 10 )   = 2.93663 × 10 ft 0.4931   

rT = 1.43202 × 109 ft = 2.71216 × 105 mi

rT = 2.71 × 105 mi 

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PROBLEM 12.85 A 500 kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass of the earth and that the radius of the moon is 1737 km, determine (a) the gravitational force exerted on the spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the surface of the moon.

SOLUTION First note that

rE = RE + hE = (6.37 × 106 + 4.5 × 106 ) m

Then

(a)

RE = 6.37 × 106 m

= 10.87 × 106 m

We have and

F=

GMm r2

[Eq. (12.28)]

GM = gR 2

[Eq. (12.29)] m R =W   2 r r

2

Then

F = gR 2

For the earth orbit,

 6.37 × 106 m  F = (500 kg)(9.81 m/s 2 )   6  10.87 × 10 m 

2

F = 1684 N 

or (b)

From the solution to Problem 12.78, we have 2

1  2π  3 M=  r G  τ 

Then

Now

τ=

2π r 3/ 2 GM

τE =τM 

2π rE3/ 2 GM E

=

2π rM3/2 GM M

(1)

1/3

or

M  rM =  M  rE = (0.01230)1/3 (10.87 × 106 m)  ME 

or

rM = 2.509 × 106 m

rM = 2510 km 

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PROBLEM 12.85 (Continued)

(c)

GM = gR 2

We have

[Eq.(12.29)]

Substituting into Eq. (1) 2π rE3/2 RE g E

=

2π rM3/ 2 RM g M 2

or

3

2

gM

 R  r   R  M  =  E   M  gE =  E   M  gE  RM   rE   RM   M E 

gM

 6370 km  2 =  (0.01230)(9.81 m/s ) 1737 km  

using the results of Part (b). Then 2

g moon = 1.62 m/s 2 

or Note:

g moon ≈

1 g earth 6

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PROBLEM 12.86 A space vehicle is in a circular orbit of 2200-km radius around the moon. To transfer it to a smaller circular orbit of 2080-km radius, the vehicle is first placed on an elliptic path AB by reducing its speed by 26.3 m/s as it passes through A. Knowing that the mass of the moon is 73.49 × 1021 kg, determine (a) the speed of the vehicle as it approaches B on the elliptic path, (b) the amount by which its speed should be reduced as it approaches B to insert it into the smaller circular orbit.

SOLUTION For a circular orbit,

Σ Fn = man : F = m F =G

Eq. (12.28): Then

G

or

v2 r Mm r2

Mm v2 = m r r2 GM v2 = r 66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg 2200 × 103 m

Then

2 (v A )circ =

or

(v A )circ = 1493.0 m/s

and

2 (vB )circ =

or

(vB )circ = 1535.5 m/s

(a)

We have

66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg 2080 × 103 m

(v A )TR = (v A )circ + Δv A = (1493.0 − 26.3) m/s = 1466.7 m/s

Conservation of angular momentum requires that rA m(v A )TR = rB m(vB )TR

or

2200 km × 1466.7 m/s 2080 km = 1551.3 m/s

(vB )TR =

(vB )TR = 1551 m/s 

or (b)

Now or

(v B )circ = (vB )TR + ΔvB ΔvB = (1535.5 − 1551.3) m/s ΔvB = −15.8 m/s 

or

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PROBLEM 12.87 Plans for an unmanned landing mission on the planet Mars called for the earth-return vehicle to first describe a circular orbit at an altitude dA = 2200 km above the surface of the planet with a velocity of 2771 m/s. As it passed through Point A, the vehicle was to be inserted into an elliptic transfer orbit by firing its engine and increasing its speed by Δv A = 1046 m/s. As it passed through Point B, at an altitude dB = 100,000 km, the vehicle was to be inserted into a second transfer orbit located in a slightly different plane, by changing the direction of its velocity and reducing its speed by ΔvB = −22.0 m/s. Finally, as the vehicle passed through Point C, at an altitude dC = 1000 km, its speed was to be increased by ΔvC = 660 m/s to insert it into its return trajectory. Knowing that the radius of the planet Mars is R = 3400 km, determine the velocity of the vehicle after completion of the last maneuver.

SOLUTION rA = 3400 + 2200 = 5600 km = 5.60 × 106 m rB = 3400 + 100, 000 = 103, 400 km = 103.4 × 106 m rC = 3400 + 1000 = 4400 km = 4.40 × 106 m

First transfer orbit. v A = 2771 m/s + 1046 m/s = 3817 m/s

Conservation of angular momentum: rA m v A = rB m vB (5.60 × 10 )(3817) = (103.4 × 106 )vB 6

vB = 206.7 m/s

Second transfer orbit.

vB′ = vB + ΔvB = 206.7 − 22.0 = 184.7 m/s

Conservation of angular momentum: rB mvB′ = rC mvC (103.4 × 106 )(184.7) = (4.40 × 106 )vC vC = 4340 m/s

After last maneuver. v = vC + ΔvC = 4340 + 660

v = 5000 m/s 

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PROBLEM 12.88 To place a communications satellite into a geosynchronous orbit (see Problem 12.80) at an altitude of 22,240 mi above the surface of the earth, the satellite first is released from a space shuttle, which is in a circular orbit at an altitude of 185 mi, and then is propelled by an upper-stage booster to its final altitude. As the satellite passes through A, the booster’s motor is fired to insert the satellite into an elliptic transfer orbit. The booster is again fired at B to insert the satellite into a geosynchronous orbit. Knowing that the second firing increases the speed of the satellite by 4810 ft/s, determine (a) the speed of the satellite as it approaches B on the elliptic transfer orbit, (b) the increase in speed resulting from the first firing at A.

SOLUTION For earth,

R = 3960 mi = 20.909 × 106 ft GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 rA = 3960 + 185 = 4145 mi = 21.8856 × 106 ft rB = 3960 + 22, 240 = 26, 200 mi = 138.336 × 106 ft

Speed on circular orbit through A. (v A )circ = =

GM rA 14.077 × 1015 21.8856 × 106

= 25.362 × 103 ft/s

Speed on circular orbit through B. (vB )circ = =

GM rB 14.077 × 1015 138.336 × 106

= 10.088 × 103 ft/s

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PROBLEM 12.88 (Continued)

(a)

Speed on transfer trajectory at B. (vB ) tr = 10.088 × 103 − 4810 = 5.278 × 103 Conservation of angular momentum for transfer trajectory.

5280 ft/s 

rA (v A ) tr = rB (vB ) tr (v A ) tr =

rB (vB ) tr rA

(138.336 × 106 )(5278) 21.8856 × 106 = 33.362 × 103 ft/s

=

(b)

Change in speed at A. Δv A = (v A ) tr − (v A )circ = 33.362 × 103 − 25.362 × 103 = 8.000 × 103 Δv A = 8000 ft/s 

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PROBLEM 12.89 A space shuttle S and a satellite A are in the circular orbits shown. In order for the shuttle to recover the satellite, the shuttle is first placed in an elliptic path BC by increasing its speed by ΔvB = 280 ft/s as it passes through B. As the shuttle approaches C, its speed is increased by ΔvC = 260 ft/s to insert it into a second elliptic transfer orbit CD. Knowing that the distance from O to C is 4289 mi, determine the amount by which the speed of the shuttle should be increased as it approaches D to insert it into the circular orbit of the satellite.

SOLUTION R = 3960 mi = 20.9088 × 106 ft

First note

rA = (3960 + 380) mi = 4340 mi = 22.9152 × 106 ft rB = (3960 + 180) mi = 4140 mi = 21.8592 × 106 ft ΣFn = man :

For a circular orbit,

F =G

Eq. (12.28): Then or

G

F =m

Mm r2

Mm v2 =m 2 r r v2 =

GM gR 2 = r r

using Eq. (12.29).

32.2 ft/s 2 × (20.9088 × 106 ft) 2 22.9152 × 106 ft

Then

2 (v A )circ =

or

(v A )circ = 24, 785 ft/s

and

2 (vB )circ =

or

(vB )circ = 25,377 ft/s

We have

v2 r

32.2 ft/s 2 × (20.9088 × 106 ft) 2 21.8592 × 106 ft

(vB )TRBC = (vB )circ + ΔvB = (25,377 + 280) ft/s = 25, 657 ft/s

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PROBLEM 12.89 (Continued)

Conservation of angular momentum requires that

From Eq. (1)

BC : rB m (vB )TRBC = rC m (vC )TRBC

(1)

CD : rC m (vC )TRCD = rA m (vD )TRCD

(2)

(vC )TRBC =

rB 4140 mi (vB )TRBC = × 25, 657 ft/s 4289 mi rC

= 24, 766 ft/s

Now

(vC )TRCD = (vC )TRBC + ΔvC = (24, 766 + 260) ft/s = 25, 026 ft/s

From Eq. (2)

(vD )TRCD =

rC 4289 mi (vC )TRCD = × 25,026 ft/s 4340 mi rA

= 24, 732 ft/s

Finally, or

(v A )circ = (vD )TRCD + ΔvD ΔvD = (24,785 − 24, 732) ft/s ΔvD = 53 ft/s 

or

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PROBLEM 12.90 A 1 kg collar can slide on a horizontal rod, which is free to rotate about a vertical shaft. The collar is initially held at A by a cord attached to the shaft. A spring of constant 30 N/m is attached to the collar and to the shaft and is undeformed when the collar is at A. As the rod rotates at the rate θ = 16 rad/s, the cord is cut and the collar moves out along the rod. Neglecting friction and the mass of the rod, determine (a) the radial and transverse components of the acceleration of the collar at A, (b) the acceleration of the collar relative to the rod at A, (c) the transverse component of the velocity of the collar at B.

SOLUTION Fsp = k ( r − rA )

First note (a)

Fθ = 0 and at A,

Fr = − Fsp = 0 (a A ) r = 0  (a A )θ = 0 

ΣFr = mar :

(b)

Noting that

−Fsp = m( r − rθ 2 ) acollar/rod =  r , we have at A 0 = m[acollar/rod − (150 mm)(16 rad/s) 2 ] acollar/rod = 38400 mm/s 2 (acollar/rod ) A = 38.4 m/s 2 

or (c)

After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular momentum about the shaft is conserved. rA m (v A )θ = rB m (vB )θ

Then

(vB )θ =

where (v A )θ = rA θ0

150 mm [(150 mm)(16 rad/s)] = 800 mm/s 450 mm (vB )θ = 0.800 m/s 

or

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PROBLEM 12.91 A 1-lb ball A and a 2-lb ball B are mounted on a horizontal rod which rotates freely about a vertical shaft. The balls are held in the positions shown by pins. The pin holding B is suddenly removed and the ball moves to position C as the rod rotates. Neglecting friction and the mass of the rod and knowing that the initial speed of A is v A = 8 ft/s, determine (a) the radial and transverse components of the acceleration of ball B immediately after the pin is removed, (b) the acceleration of ball B relative to the rod at that instant, (c) the speed of ball A after ball B has reached the stop at C.

SOLUTION Let r and θ be polar coordinates with the origin lying at the shaft. Constraint of rod: θ B = θ A + π radians; θB = θA = θ; θB = θA = θ. (a) Components of acceleration Sketch the free body diagrams of the balls showing the radial and transverse components of the forces acting on them. Owing to frictionless sliding of B along the rod, ( FB ) r = 0. Radial component of acceleration of B. Fr = mB (aB ) r :

(a B ) r = 0 

Transverse components of acceleration. (a A )θ = rAθ + 2rAθ = raθ (aB )θ = rBθ + 2rBθ

(1)

Since the rod is massless, it must be in equilibrium. Draw its free body diagram, applying Newton’s 3rd Law. ΣM 0 = 0: rA ( FA )θ + rB ( FB )θ = rAm A (a A )θ + rB mB (aB )θ = 0 rAmArAθ + rB mB (rBθ + 2rBθ) = 0

θ = At t = 0,

rB = 0

so that

−2mB rBθ m ArA2 + mB rB 2

θ = 0. (aB )θ = 0 

From Eq. (1),

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PROBLEM 12.91 (Continued)

(b)

Acceleration of B relative to the rod.

(v ) 96 At t = 0, (v A )θ = 8 ft/s = 96 in./s, θ = A θ = = 9.6 rad/s rA 10 rB − rBθ 2 = (aB ) r = 0

 rB = rBθ 2 = (8) (9.6)2 = 737.28 in./s 2  rB = 61.4 ft/s 2 

(c)

Speed of A.

Substituting

d (mr 2θ) for rFθ in each term of the moment equation dt

gives

(

)

(

)

d d m ArA2θ + mB rB 2θ = 0 dt dt

Integrating with respect to time,

(

2 2 2 m ArA θ + mB rB θ = m ArA θ

) + ( m r θ ) 2

0

B B

0

Applying to the final state with ball B moved to the stop at C,  WA 2 WB 2   W  W rA + rC θ f =  A rA2 + B (rB )02  θ0  g g  g   g 

θ f =

WArA2 + WB (rB )02  (1)(10) 2 + (2)(8) 2 = (9.6) = 3.5765 rad/s θ 0 WArA 2 + WB rC 2 (1)(10) 2 + (2)(16) 2

(v A ) f = rAθ f = (10)(3.5765) = 35.765 in./s

(v A ) f = 2.98 ft/s 

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PROBLEM 12.92 Two 2.6-lb collars A and B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O and a stop prevents collar B from moving. The frame is rotating at the rate θ = 12 rad/s and r = 0.6 ft when the stop is removed allowing collar A to move out along rod OE. Neglecting friction and the mass of the frame, determine, for the position r = 1.2 ft, (a) the transverse component of the velocity of collar A, (b) the tension in the cord and the acceleration of collar A relative to the rod OE.

SOLUTION Masses: m A = mB =

2.6 = 0.08075 lb ⋅ s 2 /ft 32.2

(a) Conservation of angular momentum of collar A: ( H 0 ) 2 = ( H 0 )1 m Ar1(vθ )1 = m Ar2 (vθ ) 2 (vθ ) 2 =

r1(vθ )1 r12θ1 (0.6) 2 (12) = = = 3.6 r2 r2 1.2 (vθ ) 2 = 3.60 ft/s 

θ2 =

(vθ ) 2 3.6 = = 3.00 rad/s rA 1.2

(b) Let y be the position coordinate of B, positive upward with origin at O. Constraint of the cord: r − y = constant

or

 y =  r

Kinematics: (aB ) y =  y = r

Collar B: Collar A:

and

(a A ) r = r − rθ 2

ΣFy = mB aB : T − WB = mB  y = mB r

(1)

ΣFr = mA (a A )r : − T = m A ( r − rθ 2 )

(2)

Adding (1) and (2) to eliminate T, −WB = (m A + mB ) r + m Arθ 2 a A/rod =  r=

m Arθ 2 − WB (0.08075)(1.2)(3.00) 2 − (2.6) = = −10.70 ft/s 2 m A + mB 0.08075 + 0.08075

T = mB ( r + g ) = (0.08075)(−10.70 + 32.2)

T = 1.736 lb 

a A / rod = 10.70 ft/s 2 radially inward. 

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PROBLEM 12.93 A small ball swings in a horizontal circle at the end of a cord of length l1 , which forms an angle θ1 with the vertical. The cord is then slowly drawn through the support at O until the length of the free end is l2 . (a) Derive a relation among l1 , l2 , θ1 , and θ 2 . (b) If the ball is set in motion so that initially l1 = 0.8 m and θ1 = 35°, determine the angle θ 2 when l2 = 0.6 m.

SOLUTION (a)

For state 1 or 2, neglecting the vertical component of acceleration, ΣFy = 0: T cos θ − W = 0 T = W cos θ ΣFx = man : T sin θ = W sin θ cos θ =

But ρ =  sin θ

mv 2

ρ

so that v2 =

ρW m

sin 2 θ cos θ =  g sin θ tan θ

v1 = 1 g sin θ1 tan θ1

and

v2 =  2 g sin θ 2 tan θ 2 ΣM y = 0: H y = constant r1mv1 = r2 mv2

v11 sin θ1 = v2  2 sin θ 2

or

3/2 3/2 1 g sin θ1 sin θ1 tan θ1 =  2 sin θ 2 sin θ 2 tan θ 2

31 sin 3 θ1 tan θ1 = 32 sin 3 θ 2 tan θ 2  (b)

With θ1 = 35°, 1 = 0.8 m, and  2 = 0.6 m (0.8)3 sin 3 35° tan 35° = (0.6)3 sin 3 θ 2 tan θ 2 sin 3 θ 2 tan θ 2 − 0.31320 = 0

θ 2 = 43.6° 

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PROBLEM 12.CQ6 A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about Point D, that is, the upper left corner of the crate? (a) 0 (b) mv1a (c) mv1b (d ) mv1 a 2 + b 2

SOLUTION Answers: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum, mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point D.

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PROBLEM 12.CQ7 A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about Point A, that is, the point of contact between the front tire of the forklift and the ground? (a) 0 (b) mv1d (c) 3mv1 (d ) mv1 32 + d 2

SOLUTION Answer: (b)

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PROBLEM 12.94 A particle of mass m is projected from Point A with an initial velocity v 0 perpendicular to OA and moves under a central force F along an elliptic path defined by the equation r = r0 /(2 − cos θ ). Using Eq. (12.37), show that F is inversely proportional to the square of the distance r from the particle to the center of force O.

SOLUTION u=

1 2 − cos θ = , r r0

d 2u 2 F +u = = 2 r0 mh 2u 2 dθ

Solving for F,

F=

du sin θ = , dθ r0

d 2 u cos θ = r0 dθ 2

by Eq. (12.37).

2mh 2 u 2 2mh 2 = r0 r0 r 2

Since m, h, and r0 are constants, F is proportional to

1 r2

, or inversely proportional to r 2 . 

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PROBLEM 12.95 A particle of mass m describes the logarithmic spiral r = r0 ebθ under a central force F directed toward the center of force O. Using Eq. (12.37) show that F is inversely proportional to the cube of the distance r from the particle to O.

SOLUTION u=

1 1 −bθ = e r r0

du b = − e −bθ dθ r0 d 2u b 2 −bθ e = r0 dθ 2 d 2u b 2 + 1 −bθ F +u = e = 2 r dθ mh 2u 2 0 F = =

(b 2 + 1)mh 2u 2 −bθ e r0 (b 2 + 1)mh 2u 2 (b 2 + 1)mh 2 = r r3

Since b, m, and h are constants, F is proportional to

1 r3

, or inversely proportional to r 3.

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PROBLEM 12.96 For the particle of Problem 12.74, and using Eq. (12.37), show that the central force F is proportional to the distance r from the particle to the center of force O. PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v 0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.27), express the radial and transverse components of the velocity v of the particle as functions of θ .

SOLUTION u =

1 = r

cos 2θ , r0

sin 2θ du =− dθ r0 cos 2θ

d 2u cos 2θ (2 cos 2θ ) − sin 2θ (− sin 2θ / cos 2θ ) =− dθ r0 cos 2θ =−

Eq. (12.37):

2 cos 2 2θ + sin 2 2θ (1 + cos 2 2θ ) = − r0 (cos 2θ )3/2 r0 (cos 2θ )3/2

d 2u F +u = 2 dθ mh 2u 2

Solving for F,  d 2u  F = mh 2u 2  + u   dθ  = mh 2

cos 2θ r0 2

 1 + cos 2 2θ + − 3/2  r0 (cos 2θ )

cos 2θ   r0 

= mh 2

cos 2θ r0 2

 1 − − 3/ 2  r0 (cos 2θ )

cos 2θ + r0

=−

mh 2 mh 2 = − 3 4 r0 cos 2θ r0

r0 cos 2θ

cos 2θ r0

   F =−

mh 2r  r0 4

The force F is proportional to r. The minus sign indicates that it is repulsive.

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PROBLEM 12.97 A particle of mass m describes the path defined by the equation r = r0 sin θ under a central force F directed toward the center of force O. Using Eq. (12.37), show that F is inversely proportional to the fifth power of the distance r from the particle to O.

SOLUTION We have

d 2u F +u = 2 dθ mh 2u 2

u=

where

1 r

Eq. (12.37)

and mh 2 = constant

 d 2u  F × u 2  2 + u   dθ  u=

Now

Then

and

du 1  1  1 cos θ =  =− dθ dθ  r0 sin θ  r0 sin 2 θ d 2u 1  − sin θ (sin 2 θ ) − cos θ (2 sin θ cos θ )  = −   r0  dθ 2 sin 4 θ  =

Then

1 1 = r r0 sin θ

 1 F × 2 r

1 1 + cos 2 θ r0 sin 3 θ

2 1   1 1 + cos θ +   3 r r sin θ sin θ  0 0

= mh 2

1 1  1 + cos 2 θ sin 2 θ +  r0 r 2  sin 3 θ sin 3 θ

2 1 1 = mh r0 r 2 sin 3 θ 2

= mh 2

  

  

r sin θ =    r0 

3

3

2r02 r5

F is proportional to

1 1 F× 5 5 r r

Q.E.D. 

Note: F > 0 implies that F is attractive.

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PROBLEM 12.98 It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 14.1 km/s as it reached its minimum altitude of 303 km above the surface of the earth. Determine the eccentricity of the trajectory of the spacecraft during this portion of its flight.

SOLUTION For earth, R = 6.37 × 106 m r0 = 6.37 × 106 + 303. × 103 = 6.673 × 106 m h = r0v0 = (6.673 × 106 )(14.1 × 103 ) = 94.09 × 109 m 2 /s GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 1 GM = 2 (1 + ε ) r0 h 1+ε =

h2 (94.09 × 109 )2 = = 3.33 r0GM (6.673 × 106 )(398.06 × 1012 )

ε = 3.33 − 1

ε = 2.33 

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PROBLEM 12.99 It was observed that during the Galileo spacecraft’s first flyby of the earth, its maximum altitude was 600 mi above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the maximum velocity of Galileo during its first flyby of the earth.

SOLUTION For the earth: R = 3960 mi = 20.909 × 106 ft GM = gR 2 = (32.2) (20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2

For a parabolic trajectory, ε = 1. Eq. (12.39′) : At θ = 0,

1 GM = 2 (1 + cos θ ) r h 1 2GM 2GM = = 2 2 2 r0 h r0 v0

or

v0 =

2GM r0

At r0 = 3960 + 600 = 4560 mi = 24.077 × 106 ft, v0 =

(2)(14.077 × 1015 ) = 34.196 × 103 ft/s 6 24.077 × 10 v0 = 6.48 mi/s 

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PROBLEM 12.100 As a space probe approaching the planet Venus on a parabolic trajectory reaches Point A closest to the planet, its velocity is decreased to insert it into a circular orbit. Knowing that the mass and the radius of Venus are 4.87 × 1024 kg and 6052 km, respectively, determine (a) the velocity of the probe as it approaches A, (b) the decrease in velocity required to insert it into the circular orbit.

SOLUTION First note (a)

rA = (6052 + 280) km = 6332 km

From the textbook, the velocity at the point of closest approach on a parabolic trajectory is given by v0 =

2GM r0 1/2

Thus,

(v A )par

 2 × 66.73 × 10−12 m3 /kg ⋅ s 2 × 4.87 × 1024 kg  =  6332 × 103 m   = 10,131.4 m/s

(v A ) par = 10.13 km/s 

or (b)

We have

(v A )circ = (v A ) par + Δv A

Now

(v A )circ =

Then

GM r0

=

1

Δv A =

1

2

2

Eq. (12.44)

(v A )par

(v A ) par − (v A )par

 1  = − 1 (10.1314 km/s)  2  = −2.97 km/s | Δv A | = 2.97 km/s 

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PROBLEM 12.101 It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn, it was at a distance of 185 × 103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295 × 103 km at a speed of 11.35 km/s, determine the eccentricity of the trajectory of Voyager I on its approach to Saturn.

SOLUTION For a circular orbit, Eq. (12.44)

v=

GM r

For the orbit of Tethys, GM = rT vT2

For Voyager’s trajectory, we have 1 GM = 2 (1 + ε cos θ ) r h

where h = r0 v0 At O,

r = r0 , θ = 0

Then

1 GM (1 + ε ) = r0 (r0 v0 ) 2

or

ε=

r0 v02 r v2 − 1 = 0 02 − 1 GM rT vT 2

=

185 × 103 km  21.0 km/s  ×  −1 295 × 103 km  11.35 km/s 

ε = 1.147 

or

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PROBLEM 12.102 A satellite describes an elliptic orbit about a planet of mass M. Denoting by r0 and r1 , respectively, the minimum and maximum values of the distance r from the satellite to the center of the planet, derive the relation 1 1 2GM + = 2 r0 r1 h

where h is the angular momentum per unit mass of the satellite.

SOLUTION Using Eq. (12.39),

1 GM = 2 + C cos θ A rA h

and

1 GM = 2 + C cos θ B . rB h

But

θ B = θ A + 180°,

so that Adding,

cos θ A = − cos θ B . 1 1 1 1 2GM + = + = 2  rA rB r0 r1 h

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PROBLEM 12.103 A space probe is describing a circular orbit about a planet of radius R. The altitude of the probe above the surface of the planet is α R and its speed is v0. To place the probe in an elliptic orbit which will bring it closer to the planet, its speed is reduced from v0 to β v0 , where β < 1, by firing its engine for a short interval of time. Determine the smallest permissible value of β if the probe is not to crash on the surface of the planet.

SOLUTION GM rA

For the circular orbit,

v0 =

where

rA = R + α R = R(1 + α )

Eq. (12.44),

GM = v02 R(1 + α )

Then

From the solution to Problem 12.102, we have for the elliptic orbit, 1 1 2GM + = 2 rA rB h h = hA = rA (v A ) AB

Now

= [ R(1 + α )]( β v0 )

Then

2v02 R(1 + α ) 1 1 + = R(1 + α ) rB [ R(1 + α ) β v0 ]2 =

2 β R(1 + α ) 2

Now β min corresponds to rB → R. Then

1 1 2 + = 2 R(1 + α ) R β min R(1 + α )

β min =

or

2  2 +α

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PROBLEM 12.104 At main engine cutoff of its thirteenth flight, the space shuttle Discovery was in an elliptic orbit of minimum altitude 60 km and maximum altitude 500 km above the surface of the earth. Knowing that at Point A the shuttle had a velocity v0 parallel to the surface of the earth and that the shuttle was transferred to a circular orbit as it passed through Point B, determine (a) the speed v0 of the shuttle at A, (b) the increase in speed required at B to insert the shuttle into the circular orbit.

SOLUTION For earth, R = 6370 km = 6370 × 103 m GM = gR 2 = (9.81)(6370 × 103 ) 2 = 3.9806 × 1014 m3 /s 2 rA = 6370 + 60 = 6430 km = 6430 × 103 m rB = 6370 + 500 = 6870 km = 6870 × 103 m

Elliptic trajectory. Using Eq. (12.39),

1 GM = 2 + C cos θ A rA h

But

θ B = θ A + 180°, so that cos θ A = − cos θ B

Adding,

1 GM = 2 + C cos θ B . rB h

1 1 rA + rB 2GM + = = 2 rA rB rA rB h h=

(a)

and

2GMrA rB (2)(3.9806 × 1014 )(6430 × 103 )(6870 × 103 ) = = 51.422 × 109 m 2 /s rA + rB 6430 × 103 + 6870 × 103

Speed v0 at A. v0 = v A = (vB )1 =

h 51.422 × 109 = rA 6430 × 103

v0 = 8.00 × 103 m/s 

h 51.422 × 109 = = 7.48497 × 103 m/s 3 rA 6870 × 10

For a circular orbit through Point B, (vB )circ =

(b)

GM 3.9806 × 1014 = = 7.6119 × 103 m/s rB 6870 × 103

Increase in speed at Point B. ΔvB = (vB )circ − (vB )1 = 126.97 m/s

ΔvB = 127 m/s 

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PROBLEM 12.105 A space probe is to be placed in a circular orbit of 5600 mi radius about the planet Venus in a specified plane. As the probe reaches A, the point of its original trajectory closest to Venus, it is inserted in a first elliptic transfer orbit by reducing its speed by Δv A . This orbit brings it to Point B with a much reduced velocity. There the probe is inserted in a second transfer orbit located in the specified plane by changing the direction of its velocity and further reducing its speed by ΔvB . Finally, as the probe reaches Point C, it is inserted in the desired circular orbit by reducing its speed by ΔvC . Knowing that the mass of Venus is 0.82 times the mass of the earth, that rA = 9.3 × 103 mi and rB = 190 × 103 mi, and that the probe approaches A on a parabolic trajectory, determine by how much the velocity of the probe should be reduced (a) at A, (b) at B, (c) at C.

SOLUTION For Earth,

R = 3690 mi = 20.9088 × 106 ft, g = 32.2 ft/s 2 GM earth = gR 2 = (32.2)(20.9088 × 106 )2 = 14.077 × 1015 ft 3 /s 2

For Venus, For a parabolic trajectory with

GM = 0.82GM earth = 11.543 × 1015 ft 3 /s 2 rA = 9.3 × 103 mi = 49.104 × 106 ft (v A )1 = vesc =

First transfer orbit AB.

2GM (2)(11.543 × 1015 ) = = 21.683 × 103 ft/s 6 rA 49.104 × 10

rB = 190 × 103 mi = 1003.2 × 106 ft

At Point A, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rA hAB hAB

(1)

1 GM GM = 2 + C cos 0 = 2 + C rB hAB hAB

(2)

At Point B, where θ = 0°

Adding,

r + rA 2GM 1 1 + = B = 2 rA rB rA rB hAB

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PROBLEM 12.105 (Continued)

Solving for hAB , hAB =

2GMrA rB (2)(11.543 × 1015 )(49.104 × 106 )(1003.2 × 106 ) = = 1.039575 × 1012 ft 2 /s rB + rA 1052.3 × 106

(v A ) 2 =

hAB 1.039575 × 1012 = = 21.174 × 103 ft/s 6 rA 49.104 × 10

(vB )1 =

hAB 1.039575 × 1012 = = 1.03626 × 103 ft/s 6 rB 1003.2 × 10 rC = 5600 mi = 29.568 × 106 ft

Second transfer orbit BC. At Point B, where θ = 0

1 GM GM = 2 + C cos 0 = 2 + C rB hBC hBC

At Point C, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rC hBC hBC

Adding,

1 1 rB + rC 2GM + = = 2 rB rC rB rC hBC hBC =

2GMrB rC (2)(11.543 × 1015 )(1003.2 × 106 )(29.568 × 106 ) = = 814.278 × 109 ft 2 /s rB + rC 1032.768 × 106

( vB ) 2 =

hBC 814.278 × 109 = = 811.69 ft/s rB 1003.2 × 106

(vC )1 =

hBC 814.278 × 109 = = 27.539 × 103 ft/s rC 29.568 × 106

Final circular orbit.

rC = 29.568 × 106 ft (vC )2 =

GM 11.543 × 1015 = = 19.758 × 103 ft/s rC 29.568 × 106

Speed reductions. (a)

At A:

(v A )1 − (v A ) 2 = 21.683 × 103 − 21.174 × 103

Δv A = 509 ft/s 

(b)

At B:

(vB )1 − (vB ) 2 = 1.036 × 103 − 811.69

ΔvB = 224 ft/s 

(c)

At C:

(vC )1 − (vC )2 = 27.539 × 103 − 19.758 × 103 

ΔvC = 7.78 × 103 ft/s 

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PROBLEM 12.106 For the space probe of Problem 12.105, it is known that rA = 9.3 × 103 mi and that the velocity of the probe is reduced to 20, 000 ft/s as it passes through A. Determine (a) the distance from the center of Venus to Point B, (b) the amounts by which the velocity of the probe should be reduced at B and C, respectively.

SOLUTION Data from Problem 12.105:

rC = 29.568 × 106 ft, M = 0.82 M earth

For Earth,

R = 3960 mi = 20.9088 × 106 ft, g = 32.2 ft/s 2 GM earth = gR 2 = (32.2)(20.9088 × 106 )2 = 14.077 × 1015 m3 /s 2 GM = 0.82GM earth = 11.543 × 1015 ft 3 /s 2

For Venus,

v A = 20, 000 ft/s, rA = 9.3 × 103 mi = 49.104 × 106 ft

Transfer orbit AB:

hAB = rAv A = (49.104 × 106 )(20, 000) = 982.08 × 109 ft 2 /s

At Point A, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rA hAB hAB

At Point B, where θ = 0° 1 GM GM = 2 + C cos 0 = 2 + C rB hAB hAB

Adding,

1 1 2GM + = 2 rA rB hAB 1 2GM 1 = 2 − rB rA hAB =

(2)(11.543 × 1015 ) 1 − 9 2 (982.08 × 10 ) 49.104 × 106

= 3.57125 × 10 −9 ft −1

(a)

Radial coordinate rB .

rB = 280.01 × 106 ft (vB )1 =

rB = 53.0 × 103 mi 

hAB 982.08 × 109 = = 3.5073 × 103 ft/s 6 rB 280.01 × 10

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PROBLEM 12.106 (Continued)

rC = 5600 mi = 29.568 × 106 ft

Second transfer orbit BC. At Point B, where θ = 0

1 GM GM = 2 + C cos 0 = 2 + C rB hBC hBC

At Point C, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rC hBC hBC

Adding,

r + rC 1 1 2GM + = B = 2 rB rC rB rC hBC hBC =

2GMrB rC = rB + rC

(2)(11.543 × 1015 )(280.01 × 106 )(29.568 × 106 ) 309.578 × 106

= 785.755 × 109 ft 2 /s ( vB ) 2 =

hBC 785.755 × 109 = = 2.8062 × 103 ft/s 6 rB 280.01 × 10

(vC )1 =

hBC 785.755 × 109 = = 26.575 × 103 ft/s 6 rC 29.568 × 10

rC = 29.568 × 106 ft

Circular orbit with

(vC )2 =

(b)

GM 11.543 × 1015 = = 19.758 × 103 ft/s 6 rC 29.568 × 10

Speed reductions at B and C. At B:

(vB )1 − (vB ) 2 = 3.5073 × 103 − 2.8062 × 103 ΔvB = 701 ft/s 



At C:

(vC )1 − (vC )2 = 26.575 × 103 − 19.758 × 103 ΔvC = 6.82 × 103 ft/s 

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PROBLEM 12.107 As it describes an elliptic orbit about the sun, a spacecraft reaches a maximum distance of 202 × 106 mi from the center of the sun at Point A (called the aphelion) and a minimum distance of 92 × 106 mi at Point B (called the perihelion). To place the spacecraft in a smaller elliptic orbit with aphelion at A′ and perihelion at B ′, where A′ and B ′ are located 164.5 × 106 mi and 85.5 × 106 mi, respectively, from the center of the sun, the speed of the spacecraft is first reduced as it passes through A and then is further reduced as it passes through B ′. Knowing that the mass of the sun is 332.8 × 103 times the mass of the earth, determine (a) the speed of the spacecraft at A, (b) the amounts by which the speed of the spacecraft should be reduced at A and B ′ to insert it into the desired elliptic orbit.

SOLUTION First note

Rearth = 3960 mi = 20.9088 × 106 ft rA = 202 × 106 mi = 1066.56 × 109 ft rB = 92 × 106 mi = 485.76 × 109 ft

From the solution to Problem 12.102, we have for any elliptic orbit about the sun 1 1 2GM sun + = r1 r2 h2

(a)

For the elliptic orbit AB, we have r1 = rA , r2 = rB , h = hA = rA v A

Also,

GM sun = G[(332.8 × 103 ) M earth ] 2 = gRearth (332.8 × 103 )

Then

using Eq. (12.30).

2 (332.8 × 103 ) 1 1 2 gRearth + = rA rB (rAv A )2

1/2

or

R v A = earth rA

 665.6 g × 103    1   + r1 r A B  

3960 mi  665.6 × 103 × 32.2 ft/s 2 = 202 × 106 mi  1066.561× 109 ft + 485.761× 109 ft  = 52, 431 ft/s

1/ 2

   

v A = 52.4 × 103 ft/s 

or

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PROBLEM 12.107 (Continued)

(b)

From Part (a), we have 1 1 2GM sun = (rA v A ) 2  +   rA rB 

Then, for any other elliptic orbit about the sun, we have

(

2 1 1 1 (rA v A ) rA + + = r1 r2 h2

1 rB

)

For the elliptic transfer orbit AB′, we have r1 = rA , r2 = rB′ , h = htr = rA (v A ) tr

Then

or

(

2 1 1 1 1 ( rAv A ) rA + rB + = rA rB′ [rA (v A ) tr ]2

 (v A ) tr = v A   

1 rA

+

1 rB

1 rA

+

1 rB′

1/2

   

)  1 + rA rB = vA   1 + rA rB′ 

1/ 2

   

1/2

 1 + 202  92 = (52, 431 ft/s)   1 + 202  85.5   = 51,113 ft/s

htr = ( hA ) tr = (hB′ ) tr : rA (v A ) tr = rB′ (vB′ ) tr

Now Then

(vB′ ) tr =

202 × 106 mi × 51,113 ft/s = 120,758 ft/s 85.5 × 106 mi

For the elliptic orbit A′B′, we have r1 = rA′ , r2 = rB′ , h = rB′ vB′

Then

or

(

2 1 1 1 (rA v A ) rA + + = rA′ rB′ (rB′vB′ )2

r  vB ′ = v A A  rB′  

1 rB

1 rA

+

1 rB

1 rA′

+

1 rB′

) 1/ 2

   

1 1 202 × 106 mi  202 × 106 + 92 × 106 = (52, 431 ft/s) 85.5 × 106 mi  164.51× 106 + 85.51× 106  = 116,862 ft/s

1/ 2

   

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PROBLEM 12.107 (Continued)

Finally, or

(v A ) tr = v A + Δv A Δv A = (51,113 − 52, 431) ft/s |Δv A | = 1318 ft/s 

or and or

vB′ = (vB′ ) tr + ΔvB ΔvB′ = (116,862 − 120, 758) ft/s = −3896 ft/s |ΔvB | = 3900 ft/s 

or

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PROBLEM 12.108 Halley’s comet travels in an elongated elliptic orbit for which the minimum distance from the sun is approximately 12 rE , where rE = 150 × 106 km is the mean distance from the sun to the earth. Knowing that the periodic time of Halley’s comet is about 76 years, determine the maximum distance from the sun reached by the comet.

SOLUTION We apply Kepler’s Third Law to the orbits and periodic times of earth and Halley’s comet: 2

τH   aH    =   τE   aE 

Thus

3

τ  aH = aE  H   τE 

2/3

 76 years  = rE    1 year  = 17.94rE

But

2/3

1 ( rmin + rmax ) 2 11  17.94rE =  rE + rmax  22  aH =

1 rE 2 = (35.88 − 0.5) rE

rmax = 2(17.94 rE ) − = 35.38 rE

rmax = (35.38)(150 × 106 km) rmax = 5.31 × 109 km 

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PROBLEM 12.109 Based on observations made during the 1996 sighting of comet Hyakutake, it was concluded that the trajectory of the comet is a highly elongated ellipse for which the eccentricity is approximately ε = 0.999887. Knowing that for the 1996 sighting the minimum distance between the comet and the sun was 0.230 RE , where RE is the mean distance from the sun to the earth, determine the periodic time of the comet.

SOLUTION For Earth’s orbit about the sun, v0 =

2π RE 2π RE 3/ 2 GM = , τ0 = RE v0 GM

or

GM =

2π RE3/ 2

τ0

(1)

For the comet Hyakutake, 1 GM = 2 = (1 + ε ), r0 h a=

1 GM 1+ ε = 2 (1 + ε ), r1 = r0 r1 1− ε h

r 1 1+ ε r0 (r0 + r1 ) = 0 , b = r0 r1 = 2 1− ε 1−ε

h = GMr0 (1 + ε )

τ= =

2π r02 (1 + ε )1/ 2 2π ab = h (1 − ε )3/ 2 GMr0 (1 + ε ) 2π r03/ 2 GM (1 − ε )3/2

 r  = 0   RE 

3/2

1 (1 − ε )3/2

= (0.230)3/2

Since

=

2π r03/ 2τ 0 2π RE3 (1 − ε )3/2

τ0

1 τ 0 = 91.8 × 103τ 0 (1 − 0.999887)3/ 2

τ 0 = 1 yr, τ = (91.8 × 103 )(1.000)

τ = 91.8 × 103 yr 

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PROBLEM 12.110 A space probe is to be placed in a circular orbit of radius 4000 km about the planet Mars. As the probe reaches A, the point of its original trajectory closest to Mars, it is inserted into a first elliptic transfer orbit by reducing its speed. This orbit brings it to Point B with a much reduced velocity. There the probe is inserted into a second transfer orbit by further reducing its speed. Knowing that the mass of Mars is 0.1074 times the mass of the earth, that rA = 9000 km and rB = 180,000 km, and that the probe approaches A on a parabolic trajectory, determine the time needed for the space probe to travel from A to B on its first transfer orbit.

SOLUTION For earth, R = 6373 km = 6.373 × 106 m GM = gR 2 = (9.81) (6.373 × 106 ) 2 = 398.43 × 1012 m3 /s 2

For Mars, GM = (0.1074)(398.43 × 1012 ) = 42.792 × 1012 m3 /s 2 rA = 9000 km = 9.0 × 106 m

rB = 180000 km = 180 × 106 m

For the parabolic approach trajectory at A, (v A )1 =

2GM = rA

(2)(42.792 × 1012 ) = 3.0837 × 103 m/s 9.0 × 106

First elliptic transfer orbit AB. Using Eq. (12.39), But

1 GM = 2 + C cosθ A rA hAB

θ B = θ A + 180°,

Adding,

so that

and

1 GM = 2 + C cosθ B . rB hAB

cosθ A = − cosθ B .

1 1 r + rB 2GM + = A = 2 rA rB rArB hAB hAB =

2GMrArB = rA + rB

(2)(42.792 × 1012 )(9.0 × 106 )(180 × 106 ) 189.0 × 106

hAB = 27.085 × 109 m 2 /s a = b=

1 1 (rA + rB ) = (9.0 × 106 + 180 × 106 ) = 94.5 × 106 m 2 2 rArB =

(9.0 × 106 )(180 × 106 ) = 40.249 × 106 m

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PROBLEM 12.110 (Continued)

Periodic time for full ellipse: For half ellipse AB,

τ AB = τ AB =

τ =

2π ab h

1 π ab τ = 2 h

π (94.5 × 106 )(40.249 × 106 ) 27.085 × 10

9

= 444.81 × 103 s

τ AB = 122.6 h 

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PROBLEM 12.111 A space shuttle is in an elliptic orbit of eccentricity 0.0356 and a minimum altitude of 300 km above the surface of the earth. Knowing that the radius of the earth is 6370 km, determine the periodic time for the orbit.

SOLUTION For earth,

g = 9.81 m/s 2 ,

R = 6370 km = 6.370 × 106 m

GM = gR 2 = (9.81)(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2

For the orbit,

r0 = 6370 + 300 = 6670 km = 6.670 × 106 m 1 GM = 2 (1 − ε ) r1 h

1 GM = 2 (1 + ε ) r0 h r1 = r0 a =

1+ε 1.0356 − (6.670 × 106 ) = 7.1624 × 106 m 1−ε 0.9644

1 (r0 + r1) = 6.9162 × 106 m 2

b=

r0r1 = 6.9118 × 106 m

h=

(1 + ε )GMr0 =

(1.0356)(398.06 × 1012 )(6.670 × 106 )

= 52.436400 × 109 m 2 /s

τ =

2π ab 2π (6.9118 × 106 )(6.9162 × 106 ) = h 52.436400 × 109 m 2 /s

τ = 95.5 min 

= 5.7281 × 103 s

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PROBLEM 12.112 The Clementine spacecraft described an elliptic orbit of minimum altitude hA = 400 km and a maximum altitude of hB = 2940 km above the surface of the moon. Knowing that the radius of the moon is 1737 km and that the mass of the moon is 0.01230 times the mass of the earth, determine the periodic time of the spacecraft.

SOLUTION For earth,

R = 6370 km = 6.370 × 106 m GM = gR 2 = (9.81)(6.370 × 106 )2 = 398.06 × 1012 m3 /s 2

For moon,

GM = (0.01230)(398.06 × 1012 ) = 4.896 × 1012 m3 /s 2 rA = 1737 + 400 = 2137 km = 2.137 × 106 m rB = 1737 + 2940 = 4677 km = 4.677 × 106 m

Using Eq. (12.39),

1 GM = 2 + C cos θ A rA h

But

θ B = θ A + 180°, so that cos θ A = − cos θ B .

Adding,

and

1 GM = 2 + C cos θ B . rB h

1 1 r +r 2GM + = A B = 2 rA rB rA rB hAB hAB =

2GMrA rB (2)(4.896 × 1012 )(2.137 × 106 )(4.677 × 106 ) = rA + rB 6.814 × 106

= 3.78983 × 109 m 2 /s 1 a = ( rA + rB ) = 3.402 × 106 m 2 b = rA rB = 3.16145 × 106 m

Periodic time.

τ=

2π ab 2π (3.402 × 106 )(3.16145 × 106 ) = = 17.831 × 103 s hAB 3.78983 × 109

τ = 4.95 h 



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PROBLEM 12.113 Determine the time needed for the space probe of Problem 12.100 to travel from B to C.

SOLUTION From the solution to Problem 12.100, we have (v A )par = 10,131.4 m/s (v A )circ =

and

1 2

(v A ) par = 7164.0 m/s

rA = (6052 + 280) km = 6332 km

Also, For the parabolic trajectory BA, we have

1 GM v = 2 (1 + ε cos θ ) r hBA

[Eq. (12.39′)]

where ε = 1. Now at A, θ = 0:

1 GM v = 2 (1 + 1) rA hBA

or

rA =

at B, θ = −90°:

1 GM v = 2 (1 + 0) rB hBA

or

rB =

2 hBA 2GM v

2 hBA GM v

rB = 2rA

As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and Point B. Thus, (Area swept out) BA = ABA =

Now

2 4 rA rB = rA2 3 3

dA 1 = h dt 2

where h = constant

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PROBLEM 12.113 (Continued)

Then

A=

2A 1 ht or tBA = BA 2 hBA tBA = =

2 × 43 rA2 rAv A

hBA = rAv A =

8 rA 3 vA

8 6332 × 103 m 3 10,131.4 m/s

= 1666.63 s

For the circular trajectory AC, t AC =

Finally,

π

r 2 A

(v A )circ

=

π 6332 × 103 m 2 7164.0 m/s

= 1388.37 s

t BC = t BA + t AC = (1666.63 + 1388.37) s =3055.0 s t BC = 50 min 55 s 

or

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PROBLEM 12.114 A space probe is describing a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O. As the probe passes through Point A, its velocity is reduced from v0 to β v0, where β < 1, to place the probe on a crash trajectory. Express in terms of n and β the angle AOB, where B denotes the point of impact of the probe on the planet.

SOLUTION r0 = rA = nR

For the circular orbit,

v0 =

GM GM = r0 nR

The crash trajectory is elliptic. v A = β v0 =

β 2 GM nR

h = rAv A = nRv A = β 2 nGMR GM 1 = 2 h2 β nR 1 GM 1 + ε cos θ = 2 (1 + ε cos θ ) = r β 2 nR h

At Point A, θ = 180° 1 1 1− ε = = 2 rA nR β nR

or β 2 = 1 − ε

or ε = 1 − β 2

At impact Point B, θ = π − φ 1 1 = rB R 1 1 + ε cos (π − φ ) 1 − ε cos φ = = R β 2 nR β 2 nR

ε cos φ = 1 − nβ 2 or cos φ =

1 − nβ 2

ε

=

1 − nβ 2 1− β 2

φ = cos −1[(1 − nβ 2 ) /(1 − β 2 )] 

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PROBLEM 12.115 A long-range ballistic trajectory between Points A and B on the earth’s surface consists of a portion of an ellipse with the apogee at Point C. Knowing that Point C is 1500 km above the surface of the earth and the range Rφ of the trajectory is 6000 km, determine (a) the velocity of the projectile at C, (b) the eccentricity ε of the trajectory.

SOLUTION For earth, R = 6370 km = 6.37 × 106 m GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2

For the trajectory, rC = 6370 + 1500 = 7870 km = 7.87 × 106 m rA = rB = R = 6.37 × 106 m,

rC 7870 = = 1.23548 rA 6370

Range A to B: s AB = 6000 km = 6.00 × 106 m s AB 6.00 × 106 = = 0.94192 rad = 53.968° R 6.37 × 106

ϕ = For an elliptic trajectory, At A,

θ = 180° −

At C,

θ = 180° ,

ϕ 2

1 GM = 2 (1 + ε cos θ ) r h 1 GM = 2 (1 + ε cos153.016°) rA h

= 153.016°,

1 GM = 2 (1 − ε ) rC h

(1) (2)

Dividing Eq. (1) by Eq. (2), rC 1 + ε cos153.016° = = 1.23548 rA 1−ε

ε =

1.23548 − 1 = 0.68384 1.23548 + cos153.016°

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PROBLEM 12.115 (Continued)

From Eq. (2), h = h=

GM (1 − ε )rC (398.06 × 1012 )(0.31616)(7.87 × 106 ) = 31.471 × 109 m 2 /s

(a) Velocity at C. (b)

vC =

h 31.471 × 109 = = 4.00 × 103 m/s 6 rC 7.87 × 10

vC = 4 km/s 

ε = 0.684 

Eccentricity of trajectory.

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PROBLEM 12.116 A space shuttle is describing a circular orbit at an altitude of 563 km above the surface of the earth. As it passes through Point A, it fires its engine for a short interval of time to reduce its speed by 152 m/s and begin its descent toward the earth. Determine the angle AOB so that the altitude of the shuttle at Point B is 121 km. (Hint: Point A is the apogee of the elliptic descent trajectory.)

SOLUTION GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 rA = 6370 + 563 = 6933 km = 6.933 × 106 m rB = 6370 + 121 = 6491 km = 6.491 × 106 m

For the circular orbit through Point A, vcirc =

GM = rA

398.06 × 1012 = 7.5773 × 103 m/s 6.933 × 106

For the descent trajectory, v A = vcirc + Δv = 7.5773 × 103 − 152 = 7.4253 × 103 m/s h = rAv A = (6.933 × 106 )(7.4253 × 103 ) = 51.4795 × 109 m 2 /s 1 GM = 2 (1 + ε cos θ ) r h

At Point A,

θ = 180°,

r = rA 1 GM = 2 (1 − ε ) rA h 1−ε =

h2 (51.4795 × 109 )2 = = 0.96028 GM rA (398.06 × 1012 )(6.933 × 106 )

ε = 0.03972 1 GM = 2 (1 + ε cos θ B ) rB h 1 + ε cos θ B =

h2 (51.4795 × 109 ) 2 = = 1.02567 GM rB (398.06 × 1012 )(6.491 × 106 ) cosθ B =

θ B = 49.7°

1.02567 − 1

ε

= 0.6463

AOB = 180° − θ B = 130.3°

AOB = 130.3° 

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PROBLEM 12.117 As a spacecraft approaches the planet Jupiter, it releases a probe which is to enter the planet’s atmosphere at Point B at an altitude of 280 mi above the surface of the planet. The trajectory of the probe is a hyperbola of eccentricity ε = 1.031. Knowing that the radius and the mass of Jupiter are 44423 mi and 1.30 × 1026 slug, respectively, and that the velocity vB of the probe at B forms an angle of 82.9° with the direction of OA, determine (a) the angle AOB, (b) the speed vB of the probe at B.

SOLUTION First we note (a)

rB = (44.423 × 103 + 280) mi = 44.703 × 103 mi

We have

1 GM j = 2 (1 + ε cos θ ) r h

At A, θ = 0:

1 GM j = 2 (1 + ε ) rA h

or

At B, θ = θ B =  AOB :

or Then or

[Eq. (12.39′)]

h2 = rA (1 + ε ) GM j 1 GM j = 2 (1 + ε cos θ B ) rB h h2 = rB (1 + ε cos θ B ) GM j rA (1 + ε ) = rB (1 + ε cos θ B ) cos θ B = =

 1  rA  (1 + ε ) − 1 ε  rB   1  44.0 × 103 mi (1 + 1.031) − 1  3 1.031  44.703 × 10 mi 

= 0.96902

or

θ B = 14.2988°

 AOB = 14.30° 

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PROBLEM 12.117 (Continued)

(b)

From above where

Then

h 2 = GM j rB (1 + ε cos θ B ) 1 | rB × mv B | = rB vB sin φ m φ = (θ B + 82.9°) = 97.1988° h=

(rB vB sin φ )2 = GM j rB (1 + ε cos θ B )

or 1/2

vB =

 1  GM j (1 + ε cos θ B )   sin φ  rB 

1/2

−9 4 4 26 1  34.4 × 10 ft /lb ⋅ s × (1.30 × 10 slug)  = × [1 + (1.031)(0.96902)]  6 sin 97.1988°  236.03 × 10 ft 

vB = 196.2 ft/s 

or

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PROBLEM 12.118 A satellite describes an elliptic orbit about a planet. Denoting by r0 and r1 the distances corresponding, respectively, to the perigee and apogee of the orbit, show that the curvature of the orbit at each of these two points can be expressed as 1

ρ

=

1 1 1   +  2  r0 r1 

SOLUTION Using Eq. (12.39),

1 GM = 2 + C cos θ A rA h

and

1 GM = 2 + C cos θ B . rB h

But

θ B = θ A + 180°,

so that Adding,

cos θ A = − cos θ B 1 1 2GM + = 2 rA rB h

At Points A and B, the radial direction is normal to the path. an =

But

Fn = 1

ρ

=

v2

ρ

=

h2 r 2ρ

GMm mh 2 ma = = n r2 r 2ρ GM 1  1 1  =  +  2  rA rB  h2

1

ρ

=

1 1 1   +   2  r0 r1 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 481

PROBLEM 12.119 (a) Express the eccentricity ε of the elliptic orbit described by a satellite about a planet in terms of the distances r0 and r1 corresponding, respectively, to the perigee and apogee of the orbit. (b) Use the result obtained in Part a and the data given in Problem 12.109, where RE = 149.6 × 106 km, to determine the approximate maximum distance from the sun reached by comet Hyakutake.

SOLUTION (a)

We have

1 GM = 2 (1 + ε cos θ ) r h

At A, θ = 0:

1 GM = 2 (1 + ε ) r0 h

or At B, θ = 180°:

or Then

Eq. (12.39′)

h2 = r0 (1 + ε ) GM 1 GM = 2 (1 − ε ) r1 h h2 = r1 (1 − ε ) GM r0 (1 + ε ) = r1 (1 − ε )

ε=

or

(b)

1+ ε r0 1−ε

From above,

r1 =

where

r0 = 0.230 RE

Then

r1 =

1 + 0.999887 × 0.230(149.6 × 109 m) 1 − 0.999887 r1 = 609 × 1012 m 

or Note: r1 = 4070 RE

r1 − r0  r1 + r0

or r1 = 0.064 lightyears.

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PROBLEM 12.120 Derive Kepler’s third law of planetary motion from Eqs. (12.39) and (12.45).

SOLUTION For an ellipse,

2a = rA + rB

Using Eq. (12.39),

1 GM = 2 + C cos θ A rA h

and

1 GM = 2 + C cos θ B . rB h

But

θ B = θ A + 180°,

so that Adding,

and b = rA rB

cos θ A = − cos θ B . 1 1 rA + rB 2a 2GM + = = 2 = 2 rA rB rA rB b h h=b

GM a

By Eq. (12.45),

τ= τ2 =

2π ab 2π ab a 2π a3/ 2 = = h b GM GM 4π 2 a3 GM

For Orbits 1 and 2 about the same large mass,

and

τ12 =

4π 2 a13 GM

τ 22 =

4π 2 a23 GM 2

3

 τ1   a1    =    τ 2   a2 

Forming the ratio,

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PROBLEM 12.121 Show that the angular momentum per unit mass h of a satellite describing an elliptic orbit of semimajor axis a and eccentricity ε about a planet of mass M can be expressed as h = GMa(1 − ε 2 )

SOLUTION By Eq. (12.39′),

1 GM = 2 (1 + ε cos θ ) r h

At A, θ = 0°:

1 GM = 2 = (1 + ε ) rA h

or

rA =

h2 GM (1 + ε )

At B, θ = 180°:

1 GM = 2 = (1 − ε ) rB h

or

rB =

h2 GM (1 − ε )

h2 1  2h 2  1 = + =  GM  1 + ε 1 − ε  GM (1 − ε 2 )

Adding,

rA + rB =

But for an ellipse,

rA + rB = 2a 2a =

2h2 GM (1 − ε 2 )

h = GMa(1 − ε 2 ) 

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PROBLEM 12.122 In the braking test of a sports car its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if the car skids. Ignore air resistance and rolling resistance.

SOLUTION (a)

Coefficient of static friction. ΣFy = 0:

N −W = 0

N =W

v0 = 70 mi/h = 102.667 ft/s v 2 v02 − = at (s − s0 ) 2 2 at =

v 2 − v02 0 − (102.667) 2 = = − 31.001 ft/s 2 2(s − s0 ) (2)(170)

For braking without skidding μ = μ s , so that μ s N = m | at | ΣFt = mat : − μ s N = mat

μs = − (b)

mat a 31.001 = − t = W g 32.2

μ s = 0.963 

Stopping distance with skidding. Use μ = μk = (0.80)(0.963) = 0.770 ΣF = mat : μk N = −mat

at = −

μk N m

= − μ k g = − 24.801 ft/s 2

Since acceleration is constant, (s − s0 ) =

v 2 − v02 0 − (102.667) 2 = 2at (2)(− 24.801) s − s0 = 212 ft 

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PROBLEM 12.123 A bucket is attached to a rope of length L = 1.2 m and is made to revolve in a horizontal circle. Drops of water leaking from the bucket fall and strike the floor along the perimeter of a circle of radius a. Determine the radius a when θ = 30°.

SOLUTION Initial velocity of drop = velocity of bucket ΣFy = 0:

T cos 30° = mg

ΣFx = max :

Divide (2) by (1):

(1)

T sin 30° = man tan 30° =

(2)

an v2 = g pg

Thus

v 2 = ρ g tan 30°

But

ρ = L sin 30° = (1.2 m) sin 30° = 0.6 m

Thus

v 2 = 0.6(9.81) tan 30° = 3.398 m 2 /s 2

v = 1.843 m/s

Assuming the bucket to rotate clockwise (when viewed from above), and using the axes shown, we find that the components of the initial velocity of the drop are (v0 ) x = 0, (v0 ) y = 0, (v0 ) 2 = 1.843 m/s

Free fall of drop y = y0 + (v0 ) y t −

1 2 gt 2

y = y0 −

1 2 gt 2

When drop strikes floor: y =0

y0 −

1 2 gt = 0 2

But

y0 = 2L − L cos 30° = 2(1.2) − 1.2 cos 30° = 1.361 m

Thus

1.361 −

1 (9.81) t 2 = 0 2

t = 0.5275

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PROBLEM 12.123 (Continued)

Projection on horizontal floor (uniform motion) x = x0 + (v0 ) z t = L sin 30° + 0,

x = 0.6 m

z = z0 + (v0 ) z t = 0 + 1.843(0.527) = 0.971 m

Radius of circle: a = a =

x2 + z 2 (0.6)2 + (0.971)2

a = 1.141 m 

Note: The drop travels in a vertical plane parallel to the yz plane.

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PROBLEM 12.124 A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of A, (b) the acceleration of B relative to A.

SOLUTION Acceleration vectors: a A = aA

30°, a B/A = aB/A

a B = a A + a B/ A

Block B:

ΣFx = max : mB aB /A − mB a A cos 30° = 0 aB/A = a A cos 30°

(1)

ΣFy = ma y : N AB − WB = − mB a A sin 30° N AB = WB − (WB sin 30°)

Block A:

aA g

(2)

ΣF = ma : WA sin 30° + N AB sin 30° = WA WA sin 30° + WB sin 30° − (WB sin 2 30°)

aA =

aA g

aA a = WA A g g

(WA + WB ) sin 30° (30 + 12)sin 30° g= (32.2) = 20.49 ft/s 2 2 2 WA + WB sin 30° 30 + 12sin 30°

a A = 20.49 ft/s 2

(a)

30° 

aB/ A = (20.49) cos 30° = 17.75 ft/s 2 a B/A = 17.75 ft/s 2

(b)



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PROBLEM 12.125 A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A which rides on an inclined I-beam as shown. Knowing that at the instant shown the trolley has an acceleration of 1.2 ft/s2 up and to the right, determine (a) the acceleration of B relative to A, (b) the tension in cable CD.

SOLUTION (a)

First we note: a B = a A + a B/A , where a B/A is directed perpendicular to cable AB. ΣFx = mB ax : 0 = −mB ax + mB a A cos 25°

B:

aB/A = (1.2 ft/s 2 ) cos 25°

or

a B/A = 1.088 ft/s 2

or (b)



For crate B ΣFy = mB a y : TAB − WB =

or

WB a A sin 25° g

A:

 (1.2 ft/s 2 )sin 25°  TAB = (500 lb) 1 +  32.2 ft/s 2   = 507.87 lb

For trolley A ΣFx = m A a A : TCD − TAB sin 25° − WA sin 25° =

or

WA aA g

 1.2 ft/s 2  TCD = (507.87 lb)sin 25° + (40 lb)  sin 25° +  32.2 ft/s 2   TCD = 233 lb 

or

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PROBLEM 12.126 The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 km/h when the brakes are suddenly applied, causing the wheels of the car to slide on the track (μ k = 0.25). Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B.

SOLUTION v = 72 km/h = 20 m/s

(a)

Almost reached Point A.

ρ = 30 m an =

v2

ρ

=

(20) 2 = 13.333 m/s 2 30

ΣFy = ma y : N R + N F − mg = man N R + N F = m( g + an ) F = μk ( N R + N F ) = μ k m( g + an ) ΣFx = max : − F = mat at = −

F = − μ k ( g + an ) m

| at | = μk ( g + an ) = 0.25(9.81 + 13.33)

(b)

Between A and B.

| at | = 5.79 m/s 2 

ρ =∞ an = 0 | at | = μk g = (0.25)(9.81)

(c)

Just passed Point B.

| at | = 2.45 m/s 2 

ρ = 45 m an =

v2

ρ

=

(20)2 = 8.8889 m/s 2 45

ΣFy = ma y : N R + N F − mg = − man

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PROBLEM 12.126 (Continued)

or

N R + N F = m ( g − an ) F = μ k ( N R + N F ) = μ k m ( g − an ) ΣFx = max : − F = mat at = −

F = − μ k ( g − an ) m

| at | = μk ( g − an ) = (0.25)(9.81 − 8.8889) | at | = 0.230 m/s 2 

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PROBLEM 12.127 The 100-g pin B slides along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate θ0 = 12 rad/s, determine for any given value of θ (a) the radial and transverse components of the resultant force F exerted on pin B, (b) the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively.

SOLUTION Kinematics From the drawing of the system, we have r=

Then

0.2 m cos θ

sin θ    r =  0.2 θ  m/s 2 cos θ  

θ = 12 rad/s θ = 0

and

cos θ (cos θ ) − sin θ (−2cos θ sin θ )  2 θ cos 4 θ  1 + sin 2θ  2  =  0.2 θ  m/s 2 3 cos θ  

 r = 0.2

2

Substituting for θ  sin θ sin θ  (12) =  2.4  m/s 2 cos θ cos 2 θ    1 + sin 2 θ 1 + sin 2 θ (12)2 =  28.8 r = 0.2 3 cos θ cos3 θ  r = 0.2

Now

 2  m/s 

ar =  r − rθ 2  1 + sin 2 θ   0.2 =  28.8 − cos3 θ   cos θ   sin 2 θ  2 =  57.6 3   m/s θ cos  

 2  (12) 

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PROBLEM 12.127 (Continued)

aθ = rθ + 2rθ

and

 sin θ  = 0 + 2  2.4  (12) cos 2 θ    sin θ  2 =  57.6 2   m/s θ cos  

Kinetics (a)

We have

 sin 2 θ Fr = mB ar = (0.1 kg)  57.6 cos3 θ 

  2  m/s    Fr = (5.76 N) tan 2 θ sec θ 

or and

 sin θ Fθ = mB aθ = (0.1 kg)  57.6 cos 2 θ 

  2  m/s    Fθ = (5.76 N) tan θ sec θ 

or

(b)

Now or

ΣFy : Fθ cos θ + Fr sin θ = P cos θ P = 5.76 tan θ sec θ + (5.76 tan 2 θ sec θ ) tan θ

P = (5.76 N) tan θ sec2 θ

or

θ 

ΣFr : Fr = Q cos θ

or

Q = (5.76 tan 2 θ sec θ )

1 cos θ

Q = (5.76 N) tan 2 θ sec2 θ

or



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PROBLEM 12.128 A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) θ = 90°, (b) θ = 75°, (c) θ = 45°. Indicate in each case the direction of the impending motion.

SOLUTION vC = (r sin θ )φAB

First note

= (0.6 m)(6 rad/s)sin θ = (3.6 m/s)sin θ

(a)

With θ = 90°,

vC = 3.6 m/s ΣFy = 0: F − WC = 0

or

F = mC g

Now

F = μs N

or

N=

1

μs

ΣFn = mC an : N = mC 1

or

or

μs

mC g = mC

μs =

mC g vC2 r vC2 r

gr (9.81 m/s 2 )(0.6 m) = vC2 (3.6 m/s) 2 ( μ s ) min = 0.454 

or

The direction of the impending motion is downward. 

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PROBLEM 12.128 (Continued)

(b) and (c) First observe that for an arbitrary value of θ, it is not known whether the impending motion will be upward or downward. To consider both possibilities for each value of θ, let Fdown correspond to impending motion downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown, we have ΣFy = 0: N cos θ ± F sin θ − WC = 0 F = μs N

Now

N cos θ ± μ s N sin θ − mC g = 0

Then or

N=

mC g cos θ ± μs sin θ

and

F=

μ s mC g cos θ ± μ s sin θ

ΣFn = mC an : N sin θ  F cos θ = mC

vC2

ρ

ρ = r sin θ

Substituting for N and F mC g μs mC g v2 sin θ  cos θ = mC C cos θ ± μs sin θ cos θ ± μ s sin θ r sin θ

μs vC2 tan θ  = 1 ± μs tan θ 1 ± μs tan θ gr sin θ

or

μs = ±

or

1+

vC2 gr sin θ

vC2 gr sin θ

tan θ

vC2 [(3.6 m/s) sin θ ]2 = = 2.2018sin θ gr sin θ (9.81 m/s 2 )(0.6 m)sin θ

Now

Then (b)

tan θ −

μs = ±

tan θ − 2.2018 sin θ 1 + 2.2018sin θ tan θ

μs = ±

tan 75° − 2.2018sin 75° = ± 0.1796 1 + 2.2018sin 75° tan 75°

θ = 75°

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PROBLEM 12.128 (Continued)

Then downward:

μs = + 0.1796

upward:

μs < 0

not possible ( μ s )min = 0.1796 

The direction of the impending motion is downward.  (c)

θ = 45° μs = ±

tan 45° − 2.2018sin 45° = ± (− 0.218) 1 + 2.2018sin 45° tan 45°

Then downward:

μs < 0

upward:

μs = 0.218

not possible ( μ s ) min = 0.218 

The direction of the impending motion is upward.  Note: When or

tan θ − 2.2018sin θ = 0

θ = 62.988°,

μs = 0. Thus, for this value of θ , friction is not necessary to prevent the collar from sliding on the rod.

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PROBLEM 12.129 Telemetry technology is used to quantify kinematic values of a 200-kg roller coaster cart as it passes overhead. According to the system, r = 25 m, r = −10 m/s, r = −2 m/s 2 , θ = 90°, θ = −0.4 rad/s, θ = −0.32 rad/s 2. At this instant, determine (a) the normal force between the cart and the track, (b) the radius of curvature of the track.

SOLUTION Find the acceleration and velocity using polar coordinates. vr = r = −10 m/s vθ = rθ = (25 m)( − 0.4 rad/s) = −10 m/s

So the tangential direction is

45° and v = 10 2 m/s.

ar =  r − rθ 2 = −2 m/s − (25 m)( − 0.4 rad/s)2 = −6 m/s 2

aθ = rθ + 2rθ = (25 m)(−0.32) rad/s 2 | +2(−10 m/s)( − 0.4 rad/s) =0

So the acceleration is vertical and downward. (a)

To find the normal force use Newton’s second law. y-direction N − mg sin 45° = −ma cos 45° N = m( g sin 45° − a cos 45°) = (200 kg)(9.81) m/s 2 − 6 m/s 2 )(0.70711) = 538.815 N N = 539 N 

(b)

Radius l curvature of the track. an =

ρ =

v2

ρ v2 (10 2)2 = an 6 cos 45°

ρ = 47.1 m 

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PROBLEM 12.130 The radius of the orbit of a moon of a given planet is equal to twice the radius of that planet. Denoting by ρ the mean density of the planet, show that the time required by the moon to complete one full revolution about the planet is (24 π /G ρ )1/2 , where G is the constant of gravitation.

SOLUTION For gravitational force and a circular orbit, Fr =

GMm mv 2 = r r2

or

v=

GM r

Let τ be the periodic time to complete one orbit. vτ = 2π r

Solving for τ,

But

τ=

τ=

GM = 2π r r

hence,

GM = 2

2π r 3/2 GM

4 M = π R3 ρ , 3

Then

τ

or

3π  r  G ρ  R 

π 3

G ρ R3/2

3/2

Using r = 2R as a given leads to

τ = 23/2

3π = Gρ

24π Gρ

τ = (24π /G ρ )1/2 

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PROBLEM 12.131 At engine burnout on a mission, a shuttle had reached Point A at an altitude of 40 mi above the surface of the earth and had a horizontal velocity v0. Knowing that its first orbit was elliptic and that the shuttle was transferred to a circular orbit as it passed through Point B at an altitude of 170 mi, determine (a) the time needed for the shuttle to travel from A to B on its original elliptic orbit, (b) the periodic time of the shuttle on its final circular orbit.

SOLUTION For Earth,

R = 3960 mi = 20.909 × 106 ft, g = 32.2 ft/s 2 GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2

(a)

For the elliptic orbit,

rA = 3960 + 40 = 4000 mi = 21.12 × 106 ft rB = 3960 + 170 = 4130 mi = 21.8064 × 106 ft

1 (rA + rB ) = 21.5032 × 106 ft 2 b = rA rB = 21.4605 × 106 ft

a=

Using Eq. 12.39,

1 GM = 2 + C cos θ A rA h

and

1 GM = 2 + C cos θ B rB rB

But θ B = θ A + 180°, so that cos θ A = − cos θ B Adding,

1 1 rA + rB 2a 2GM + = = 2 = 2 rA rB rA rB b h

or

h=

Periodic time.

τ= τ=

GMb 2 a 2π ab 2π ab a 2π a3/ 2 = = h GM GMb2 2π (21.5032 × 106 )3/ 2 14.077 × 1015

= 5280.6 s = 1.4668 h

The time to travel from A to B is one half the periodic time

τ AB = 0.7334 h

τ AB = 44.0 min 

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PROBLEM 12.131 (Continued)

(b)

For the circular orbit,

a = b = rB = 21.8064 × 106 ft

τ circ =

2π a3/2 GM

=

2π (21.8064 × 106 )3/2 14.077 × 1015

τ circ = 1.498 h

= 5393 s

τ circ = 89.9 min 

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PROBLEM 12.132 It was observed that as the Galileo spacecraft reached the point on its trajectory closest to Io, a moon of the planet Jupiter, it was at a distance of 1750 mi from the center of Io and had a velocity of 49.4 × 103 ft/s. Knowing that the mass of Io is 0.01496 times the mass of the earth, determine the eccentricity of the trajectory of the spacecraft as it approached Io.

SOLUTION First note

r0 = 1750 mi = 9.24 × 106 ft Rearth = 3960 mi = 20.9088 × 106 ft

We have

1 GM = 2 (1 + ε cos θ ) r h

At Point O,

r = r0 , θ = 0, h = h0 = r0 v0

Also,

Eq. (12.39)

GM Io = G (0.01496 M earth ) 2 = 0.01496gRearth

using Eq. (12.30).

Then

2 1 0.01496 gRearth = (1 + ε ) r0 (r0 v0 ) 2

or

ε= =

r0 v02 2 0.01496 gRearth

−1

(9.24 × 106 ft)(49.4 × 103 ft/s) 2 −1 0.01496(32.2 ft/s 2 )(20.9088 × 106 ft) 2

ε = 106.1 

or

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PROBLEM 12.133* Disk A rotates in a horizontal plane about a vertical axis at the constant rate θ0 = 10 rad/s. Slider B has mass 1 kg and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when r = 0. Knowing that the slider is released with no radial velocity in the position r = 500 mm, determine the position of the slider and the horizontal force exerted on it by the disk at t = 0.1 s for (a) k = 100 N/m, (b) k = 200 N/m.

SOLUTION First we note

r = 0, xsp = 0  Fsp = kr

when

r0 = 500 mm = 0.5 m

and

θ = θ0 = 12 rad/s

then

θ = 0 ΣFr = mB ar : − Fsp = mB ( r − rθ02 )  k   − θ02  r = 0 r +  mB 

ΣFθ = mB aθ : FA = mB (0 + 2rθ0 )

(a)

(1) (2)

k = 100 N/m

Substituting the given values into Eq. (1) 100 N/m  r +  − (10 rad/s)2  r = 0  1 kg   r =0

Then

dr r = 0 and at t = 0, r = 0 : =  dt



r 0

dr =



0.1

(0) dt

0

r = 0

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PROBLEM 12.133* (Continued)

and

dr = r = 0 and at t = 0, r0 = 0.5 m dt



r r0

dr =



0.1

(0) dt

0

r = r0 r = 0.5 m 

Note: r = 0 implies that the slider remains at its initial radial position. With r = 0, Eq. (2) implies FH = 0 

(b)

k = 200 N/m

Substituting the given values into Eq. (1)  200   r+ − (10 rad/s)2  r = 0 1 kg   r + 100 r = 0

d (r) dt

Now

 r=

Then

r = vr

so that

d dr d d = = vr dt dt dr dr

dvr dr

dvr + 100r = 0 dr

vr



At t = 0, vr = 0, r = r0 :

r = vr

vr 0

vr d vr = −100



r

r dr

r0

vr2 = −100(r 2 − r02 ) vr = 10 r02 − r 2 vr =

Now At t = 0, r = r0 :



r r0

dr r02

−r

2

=

dr = 10 r02 − r 2 dt t

 10 dt = 10t 0

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PROBLEM 12.133* (Continued)

r = r0 sin φ ,

Let

sin −1 ( r/r0 )

π

Then

/2

dr = r0 cos φ dφ

r0 cos φ dφ r02 − r02 sin 2 φ sin −1 ( r/r0 )

π

/2

r sin −1   r0

= 10t

dφ = 10t

 π  − = 10 t  2

π  r = r0 sin 10 t +  = r0 cos10 t = (0.5 ft) cos10 t 2  r = −(5 m/s) sin10 t

Then Finally,

at t = 0.1 s: r = (0.5 ft) cos (10 × 0.1) r = 0.270 m 

Eq. (2)

FH = 1 kg × 2 × [ −(5 ft/s)sin (10 × 0.1)] (10 rad/s) FH = −84.1 N 

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CHAPTER 13

PROBLEM 13.CQ1 Block A is traveling with a speed v0 on a smooth surface when the surface suddenly becomes rough with a coefficient of friction of μ causing the block to stop after a distance d. If block A were traveling twice as fast, that is, at a speed 2v0, how far will it travel on the rough surface before stopping? (a) d/2 (b) d (c)

2d

(d) 2d (e) 4d

SOLUTION Answer: (e)

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PROBLEM 13.1 A 400-kg satellite was placed in a circular orbit 1500 km above the surface of the earth. At this elevation the acceleration of gravity is 6.43 m/s2. Determine the kinetic energy of the satellite, knowing that its orbital speed is 25.6 × 103 km/h.

SOLUTION Mass of satellite:

m = 400 kg

Velocity:

v = 25.6 × 103 km/h = 7.111 × 103 m/s

Kinetic energy:

T=

1 2 1 mv = (400 kg)(7.111 × 103 m/s) 2 2 2

T = 10.113 × 109 J

T = 10.11 GJ 

Note: Acceleration of gravity has no effect on the mass of the satellite.

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PROBLEM 13.2 A 1-lb stone is dropped down the “bottomless pit” at Carlsbad Caverns and strikes the ground with a speed of 95ft/s. Neglecting air resistance, determine (a) the kinetic energy of the stone as it strikes the ground and the height h from which it was dropped, (b) Solve Part a assuming that the same stone is dropped down a hole on the moon. (Acceleration of gravity on the moon = 5.31 ft/s2.)

SOLUTION W lb 1 lb = = 0.031056 lb ⋅ s 2 /ft g 32.2 ft/s 2

Mass of stone:

m=

Initial kinetic energy:

T1 = 0

(a)

(rest)

Kinetic energy at ground strike: T2 =

1 2 1 mv2 = (0.031056)(95)2 = 140.14 ft ⋅ lb 2 2 T2 = 140.1 ft ⋅ lb 

Use work and energy: where

T1 + U1→ 2 = T2 U1→2 = wh = mgh 0 + mgh = h=

(b)

On the moon:

1 2 mv2 2

v22 (95) 2 = 2 g (2)(32.2)

h = 140.1 ft 

g = 5.31 ft/s 2 T2 = 140.1 ft ⋅ lb 

T1 and T2 will be the same, hence h=

v22 (95) 2 = 2 g (2)(5.31)

h = 850 ft 

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PROBLEM 13.3 A baseball player hits a 5.1-oz baseball with an initial velocity of 140 ft/s at an angle of 40° with the horizontal as shown. Determine (a) the kinetic energy of the ball immediately after it is hit, (b) the kinetic energy of the ball when it reaches its maximum height, (c) the maximum height above the ground reached by the ball.

SOLUTION  1 lb  W = (5.1 oz)   = 0.31875 lb  16 oz 

Mass of baseball:

m=

(a)

W 0.31875 lb = = 0.009899 lb ⋅ s 2 /ft 2 g 32.2 ft/s

Kinetic energy immediately after hit. v = v0 = 140 ft/s T1 =

(b)

1 2 1 mv = (0.009899)(140) 2 2 2

T1 = 97.0 ft ⋅ lb 

Kinetic energy at maximum height: v = v0 cos 40° = 140cos 40° = 107.246 ft/s T2 =

Principle of work and energy:

1 2 1 mv = (0.009899)(107.246) 2 2 2

T2 = 56.9 ft ⋅ lb 

T1 + U1→ 2 = T2 U1→2 = T2 − T1 = −40.082 ft ⋅ lb

Work of weight:

U1→2 = −Wd

Maximum height above impact point. d=

(c)

T2 − T1 −40.082 ft ⋅ lb = = 125.7 ft −W −0.31875 lb

125.7 ft 

Maximum height above ground: h = 125.7 ft + 2 ft

h = 127.7 ft 

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PROBLEM 13.4 A 500-kg communications satellite is in a circular geosynchronous orbit and completes one revolution about the earth in 23 h and 56 min at an altitude of 35800 km above the surface of the earth. Knowing that the radius of the earth is 6370 km, determine the kinetic energy of the satellite.

SOLUTION Radius of earth:

R = 6370 km

Radius of orbit:

r = R + h = 6370 + 35800 = 42170 km = 42.170 × 106 m

Time one revolution:

t = 23 h + 56 min t = (23 h)(3600 s/h) + (56 min)(60 s/min) = 86.160 × 103 s

Speed:

v=

2π r 2π (42.170 × 106 ) = = 3075.2 m/s t 86.160 × 103

Kinetic energy:

T=

1 2 mv 2

T=

1 (500 kg)(3075.2 m/s) 2 = 2.3643 × 109 J 2 T = 2.36 GJ 

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PROBLEM 13.5 In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The bucket is to swing no more than 10 ft horizontally when the crane is brought to a sudden stop. Determine the maximum allowable speed v of the crane.

SOLUTION Let position  be the position with bucket B directly below A, and position  be that of maximum swing where d = 10 ft. Let L be the length AB. T1 =

Kinetic energies:

1 2 mv , T2 = 0 2

U1→2 = −Wh = − mgh

Work of the weight:

where h is the vertical projection of position  above position  From geometry (see figure),

y = L2 − d 2 h= L− y = L − L2 − d 2 = 30 − (30) 2 − (10)2 = 1.7157 ft

Principle of work and energy:

T1 + U1→ 2 = T2 1 2 mv − mgh = 0 2

v 2 = 2 gh = (2)(32.2 ft/s 2 )(1.7157 ft) = 110.49 ft 2 /s 2

v = 10.51 ft/s 

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PROBLEM 13.6 In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The crane is traveling at a speed of 10 ft/s when it is brought to a sudden stop. Determine the maximum horizontal distance through which the bucket will swing.

SOLUTION Let position  be the position with bucket B directly below A, and position  be that of maximum swing where the horizontal distance is d. Let L be the length AB. T1 =

Kinetic energies:

1 2 mv , T2 = 0 2

U1→2 = −Wh = − mgh

Work of the weight:

where h is the vertical projection of position  above position . Principle of work and energy:

T1 + U1→ 2 = T2

1 2 mv − mgh = 0 2 v2 (10 ft/s) 2 h= = = 1.5528 ft 2 g (2)(32.2 ft/s 2 )

From geometry (see figure),

d = L2 − y 2 = L2 − ( L − h) 2 = (30) 2 − (30 − 1.5528) 2 = 9.53 ft

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PROBLEM 13.7 Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive.

SOLUTION Let W be the weight and m the mass.

W = mg

(a)

N = 0.60W = 0.60 mg μs = 0.75

Front wheel drive:

Maximum friction force without slipping: F = μ s N = (0.75)(0.60W ) = 0.45mg U1→2 = Fd = 0.45 mgd T1 = 0,

Principle of work and energy:

T2 =

1 2 mv2 2

T1 + U1→ 2 = T2

1 2 mv2 2 v22 = (2)(0.45 gd ) = (2)(0.45)(9.81 m/s 2 )(110 m) = 971.19 m 2 /s 2

0 + 0.45mgd =

v2 = 31.164 m/s

(b)

v2 = 112.2 km/h 

N = 0.40W = 0.40mg μs = 0.75

Rear wheel drive:

Maximum friction force without slipping: F = μ s N = (0.75)(0.40W ) = 0.30mg U1→2 = Fd = 0.30mgd T1 = 0,

Principle of work and energy:

T2 =

1 2 mv2 2

T1 + U1→ 2 = T2 1 mv22 2 v22 = (2)(0.30) gd = (2)(0.30)(9.81 m/s 2 )(110 m) = 647.46 m 2 /s 2

0 + 0.30 mgd =

v2 = 25.445 m/s

v2 = 91.6 km/h 

Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for static and dynamic conditions. Compare with sample Problem 16.1 where the vehicle is treated as a rigid body. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514

PROBLEM 13.8 Skid marks on a drag racetrack indicate that the rear (drive) wheels of a car slip for the first 20 m of the 400-m track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the speed of the car at the end of the first 20-m portion of the track if it starts from rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed for the car at the finish line if, after skidding for 20 m, it is driven without the wheels slipping for the remainder of the race? Assume that while the car is rolling without slipping, 60 percent of the weight of the car is on the rear wheels and the coefficient of static friction is 0.75. Ignore air resistance and rolling resistance.

SOLUTION (a)

For the first 20 m, the normal force at the real wheels is equal to the weight of the car. Since the wheels are skidding, the friction force is F = μ k N = μk W = μ k mg

Principle of work and energy:

T1 + U1→ 2 = T2

1 2 mv2 2 1 0 + μk mgd = mv22 2 2 v2 = 2 μk gd = (2)(0.6)(9.81 m/s2 )(20 m) = 235.44 m 2 /s 2 0 + Fd =

 (b)

v2 = 15.34 m/s 

 Assume that for the remainder of the race, sliding is impending and N = 0.6 W F = μ s N = μ s (0.6W ) = (0.75)(0.6 mg ) = 0.45 mg

Principle of work and energy:

T2 + U 2→3 = T3 1 2 1 mv2 + (0.45mg )d ′ = mv32 2 2

v32 = v22 (2)(0.45) gd ′ = 235.44 m 2 /s 2 + (2)(0.45)(9.81 m/s 2 )(400 m − 20 m) = 3590.5 m 2 /s 2 v3 = 59.9 m/s 

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PROBLEM 13.9 A package is projected up a 15° incline at A with an initial velocity of 8 m/s. Knowing that the coefficient of kinetic friction between the package and the incline is 0.12, determine (a) the maximum distance d that the package will move up the incline, (b) the velocity of the package as it returns to its original position.

SOLUTION (a)

Up the plane from A to B: 1 2 1W W mv A = (8 m/s) 2 = 32 TB = 0 2 2 g g = ( −W sin15° − F )d F = μk N = 0.12 N

TA = U A−B

ΣF = 0 N − W cos15° = 0 N = W cos15° U A−B = −W (sin15° + 0.12 cos15°) d = −Wd (0.3747) TA + U A−B = TB : 32

W − Wd (0.3743) = 0 g d=

(b)

32 (9.81)(0.3747)

d = 8.71 m 

Down the plane from B to A: (F reverses direction)

U B−A

1W 2 vA TB = 0 2 g = (W sin15° − F )d

U B−A

= W (sin15° − 0.12cos15°)(8.70 m/s) = 1.245W

TA =

TB + U B−A = TA

0 + 1.245W =

d = 8.71 m/s

1W 2 vA 2 g

v A2 = (2)(9.81)(1.245) = 24.43 v A = 4.94 m/s

vA = 4.94 m/s

15° 

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PROBLEM 13.10 A 1.4 kg model rocket is launched vertically from rest with a constant thrust of 25 N until the rocket reaches an altitude of 15 m and the thrust ends. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, (c) the speed of the rocket when it returns to the ground.

SOLUTION Weight:

W = mg = (1.4)(9.81) = 13.734 N

(a)

T1 = 0

First stage:

U1→2 = (25 − 13.734)(15) = 169.0 N ⋅ m T1 + U1→ 2 = T2

(b)

T2 =

1 2 mv = U1→2 = 169.0 N ⋅ m 2

v2 =

2U1→2 (2) (169.0) = m 1.4

v2 = 15.54 m/s 

Unpowered flight to maximum height h: T2 = 169.0 N ⋅ m

T3 = 0

U 2→3 = −W (h − 15) T2 + U 2→3 = T3 W (h − 15) = T2 h − 15 =

(c)

T2 169.0 =  W 13.734

h = 27.3 m 

Falling from maximum height: T3 = 0

T4 =

1 2 mv4 2

U 3→4 = Wh = mgh T3 = U 3→4 = T4 :

0 + mgh =

1 2 mv4 2

v42 = 2 gh = (2)(9.81 m/s 2 )(27.3 m) = 535.6 m 2 /s 2

v4 = 23.1 m/s 

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PROBLEM 13.11 Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that μk = 0.25 between the packages and the surface ABC, determine the distance d if the packages are to arrive at C with a velocity of 2 m/s.

SOLUTION N AB = mg cos 30°

On incline AB:

FAB = μk N AB = 0.25 mg cos 30° U A→ B = mgd sin 30° − FAB d = mgd (sin 30° − μk cos 30°) N BC = mg

On level surface BC:

xBC = 7 m

FBC = μk mg U B →C = − μk mg xBC

At A,

TA =

1 2 mv A 2

and v A = 1 m/s

At C,

TC =

1 2 mvC 2

and vC = 2 m/s

Assume that no energy is lost at the corner B. TA + U A→ B + U B →C = TC

Work and energy.

1 2 1 mv A + mgd (sin 30° − μ k cos 30°) − μ k mg xBC = mv02 2 2

Dividing by m and solving for d, vC2 /2 g + μ k xBC − v A2 /2 g   d= (sin 30° − μk cos 30°) =

(2) 2/(2)(9.81) + (0.25)(7) − (1)2/(2)(9.81) sin 30° − 0.25cos 30° d = 6.71 m 

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PROBLEM 13.12 Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that d = 7.5 m and μk = 0.25 between the packages and all surfaces, determine (a) the speed of the package at C, (b) the distance a package will slide on the conveyor belt before it comes to rest relative to the belt.

SOLUTION (a)

On incline AB:

N AB = mg cos 30° FAB = μk N AB = 0.25 mg cos 30° UA→ B = mgd sin 30° − FAB d = mgd (sin 30° − μ k cos 30°)

On level surface BC:

N BC = mg

xBC = 7 m

FBC = μ k mg U B →C = − μk mg xBC

At A,

TA =

1 2 mv A 2

and v A = 1 m/s

At C,

TC =

1 2 mvC 2

and vC = 2 m/s

Assume that no energy is lost at the corner B. Work and energy.

TA + U A→ B + U B →C = TC 1 2 1 mv A + mgd (sin 30° − μ k cos 30°) − μ k mg xBC = mv02 2 2

Solving for vC2 , vC2 = v 2A + 2 gd (sin 30° − μk cos 30°) − 2 μk g xBC = (1) 2 + (2)(9.81)(7.5)(sin 30° − 0.25cos 30°) − (2)(0.25)(9.81)(7) = 8.3811 m 2/s 2

(b)

vC = 2.90 m/s 

Box on belt: Let xbelt be the distance moves by a package as it slides on the belt. ΣFy = ma y

N − mg = 0 N = mg

Fx = μk N = μk mg

At the end of sliding,

v = vbelt = 2 m/s

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PROBLEM 13.12 (Continued)

Principle of work and energy: 1 2 1 2 mvC − μ k mg xbelt = mvbelt 2 2 2 vC2 − vbelt xbelt = 2 μk g =

8.3811 − (2) 2 (2)(0.25)(9.81)

xbelt = 0.893 m 

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PROBLEM 13.13 Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A where they slide and eventually fall off at B. Knowing that μk = 0.40, determine the velocity of the conveyor belt if the boxes leave the incline at B with a velocity of 8 ft/s.

SOLUTION Forces when box is on AB. ΣFy = 0: N − W cos15° = 0 N = W cos15°

Box is sliding on AB. F f = μk N = μkW cos15°

Distance

AB = d = 20 ft

Work of gravity force:

(U A− B ) g = Wd sin15° − F f d = − μkWd cos15°

Work of friction force: Total work

U A→ B = Wd (sin15° − μ k sin15°)

Kinetic energy:

TA =

1W 2 v0 2 g 1W 2 TB = vB 2 g

Principle of work and energy:

TA + U A→ B = TB

1W 2 1W 2 v0 + Wd (sin15° − μk cos15°) = vB 2 g 2 g v02 = vB2 − 2 gd (sin15° − μk cos15°) = (8) 2 − (2)(32.2)(20)[sin15° − (0.40)(cos15°)] = 228.29 ft 2 /s 2

v 0 = 15.11 ft/s

15° 

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PROBLEM 13.14 Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A where they slide and eventually fall off at B. Knowing that μk = 0.40, determine the velocity of the conveyor belt if the boxes are to have zero velocity at B.

SOLUTION Forces when box is on AB. ΣFy = 0: N − W cos15° = 0 N = W cos15°

Box is sliding on AB. F f = μk N = μkW cos15°

Distance

AB = d = 20 ft

Work of gravity force:

(U A− B ) g = Wd sin15° − F f d = − μkWd cos15°

Work of friction force: Total work

U A− B = Wd (sin15° − μk cos15°)

Kinetic energy:

TA =

1W 2 v0 2 g 1W 2 TB = vB 2 g

Principle of work and energy:

TA + U A− B = TB

1W 2 1W 2 v0 + Wd (sin15° − μk cos15°) = vB 2 g 2 g v02 = vB2 − 2 gd (sin15° − μk cos15°) = 0 − (2)(32.2)(20)[sin15° − (0.40)(cos15°)] = 164.29 ft 2 /s 2 v 0 = 12.81 ft/s

15° 

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PROBLEM 13.15 A 1200-kg trailer is hitched to a 1400-kg car. The car and trailer are traveling at 72 km/h when the driver applies the brakes on both the car and the trailer. Knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N, respectively, determine (a) the distance traveled by the car and trailer before they come to a stop, (b) the horizontal component of the force exerted by the trailer hitch on the car.

SOLUTION Let position 1 be the initial state at velocity v1 = 72 km/h = 20 m/s and position 2 be at the end of braking (v2 = 0). The braking forces and FC = 5000 N for the car and 4000 N for the trailer.

(a)

Car and trailer system.

(d = braking distance) 1 (mC + mT )v12 2 = −( FC + FT )d

T1 = U1→ 2

T2 = 0

T1 + U1→ 2 = T2 1 (mC + mT )v12 − ( FC + FT )d = 0 2 d=

(b)

(mC + mT )v12 (2600)(20) 2 = = 57.778 2( FC + FT ) (2)(9000)

d = 57.8 m 

Car considered separately. Let H be the horizontal pushing force that the trailer exerts on the car through the hitch. 1 mC v12 2 = ( H − FC )d

T1 = U1→ 2

T2 = 0

T1 + U1→ 2 = T2 1 mC v12 + ( H − FC ) d = 0 2 H = FC −

mC v12 (1400)(20) 2 = 5000 − 2d (2)(57.778) H = 154 N

Trailer hitch force on car:



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PROBLEM 13.16 A trailer truck enters a 2 percent uphill grade traveling at 72 km/h and reaches a speed of 108 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average force at the wheels of the cab, (b) the average force in the coupling between the cab and the trailer.

SOLUTION Initial speed:

v1 = 72 km/h = 20 m/s

Final speed:

v2 = 108 km/h = 30 m/s

Vertical rise:

h = (0.02)(300) = 6.00 m

Distance traveled:

d = 300 m

(a)

Traction force. Use cab and trailer as a free body. m = 1800 + 5400 = 7200 kg T1 + U1→ 2 = T2

Work and energy: Ft =

1 d

W = mg = (7200)(9.81) = 70.632 × 103 N 1 2 1 mv1 − Wh + Ft d = mv22 2 2

1 1 1 1 2 2 2 3 2  2 mv1 + Wh − mv1  = 300  2 (7200)(30) + (70.632 × 10 )(6.00) − 2 (7200)(20)   

= 7.4126 × 103 N

(b)

Ft = 7.41 kN 

Coupling force Fc . Use the trailer alone as a free body. m = 5400 kg

W = mg = (5400)(9.81) = 52.974 × 103 N

Assume that the tangential force at the trailer wheels is zero. Work and energy:

T1 + U1→ 2 = T2

1 2 1 mv1 − Wh + Fc d = mv22 2 2

The plus sign before Fc means that we have assumed that the coupling is in tension. Fc =

1 1 2 1 1 1 1   mv2 + Wh − mv12  = (5400)(30) 2 + (52.974 × 103 )(6.00) − (5400)(20) 2    2 d 2 2   300  2

= 5.5595 × 103 N

Fc = 5.56 kN (tension) 

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PROBLEM 13.17 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION

μk = 0.35 FB = (0.35)(100 kips) = 35 kips FC = (0.35)(80 kips) = 28 kips v1 = 30 mi/h = 44 ft/s

(a)

Entire train:

v2 = 0 T2 = 0

T1 + U1− 2 = T2 1 (80 kips + 100 kips + 80 kips) (44 ft/s) 2 − (28 kips +35 kips) x = 0 2 32.2 ft/s 2 x = 124.07 ft 





(b)

Force in each coupling: Recall that x = 124.07 ft

x = 124.1 ft 

Car A: Assume FAB to be in tension T1 + V1− 2 = T2 1 80 kips (44) 2 − FAB (124.07 ft) = 0 2 32.2 FAB = +19.38 kips FAB = 19.38 kips (tension) 

Car C:

T1 + U1− 2 = T2

1 80 kips (44)2 + ( FBC − 28 kips)(124.07 ft) = 0 2 32.2 FBC − 28 kips = −19.38 kips FBC = +8.62 kips

FBC = 8.62 kips (tension) 

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PROBLEM 13.18 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars A, causing it to slide on the track, but are not applied on the wheels of cars A or B. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION (a)

Entire train:

FA = μ N A = (0.35)(80 kips) = 28 kips v2 = 0 T2 = 0

v1 = 30 mi/h = 44 ft/s ← T1 + V1− 2 = T2 1 (80 kips + 100 kips + 80 kips) (44 ft/s) 2 − (28 kips) x = 0 2 2 32.2 ft/s x = 279.1 ft

(b)

x = 279 ft 

Force in each coupling: Car A: Assume FAB to be in tension T1 + V1− 2 = T2 1 80 kips (44 ft/s) 2 − (28 kips + FAB )(279.1 ft) = 0 2 32.2 ft/s 2 28 kips + FAB = +8.62 kips FAB = −19.38 kips

Car C:

FAB = 19.38 kips (compression) 

T1 + V1− 2 = T2 1 80 kips (44 ft/s) 2 + FBC (279.1 ft) = 0 2 32.2 ft/s 2 FBC = −8.617 kips

FBC = 8.62 kips (compression) 

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PROBLEM 13.19 Blocks A and B weigh 25 lbs and 10 lbs, respectively, and they are both at a height 6 ft above the ground when the system is released from rest. Just before hitting the ground block A is moving at a speed of 9 ft/s. Determine (a) the amount of energy dissipated in friction by the pulley, (b) the tension in each portion of the cord during the motion.

SOLUTION By constraint of the cable block B moves up a distance h when block A moves down a distance h. (h = 6 ft) Their speeds are equal. Let FA and FB be the tensions on the A and B sides, respectively, of the pulley. Masses:

WA 25 = = 0.7764 lb ⋅ s 2 /ft 32.2 g W 10 MB = B = = 0.31056 lb ⋅ s 2 /ft 32.2 g MA =

Let position 1 be the initial position with both blocks a distance h above the ground and position 2 be just before block A hits the ground. Kinetic energies:

(T1 ) A = 0,

(T1 ) B = 0

1 1 mA v 2 = (0.7764)(9)2 = 31.444 ft ⋅ lb 2 2 1 1 (T2 ) B = mB v 2 = (0.31056)(9) 2 = 12.578 ft ⋅ lb 2 2 (T2 ) A =

Principle of work and energy:

T1 + U1→ 2 = T2

Block A:

U1→2 = (WA − FA ) h 0 + (25 − FA )(6) = 31.444

Block B:

FA = 19.759 lb

U1→2 = ( FB − WB ) h 0 + ( FB − 10)(6) = 12.578

FB = 12.096 lb

At the pulley FA moves a distance h down, and FB moves a distance h up. The work done is U1→2 = ( FA − FB ) h = (19.759 − 12.096)(6) = 46.0 ft ⋅ lb

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PROBLEM 13.19 (Continued)

Since the pulley is assumed to be massless, it cannot acquire kinetic energy; hence, (a)

Energy dissipated by the pulley:

(b)

Tension in each portion of the cord:

E p = 46.0 ft ⋅ lb  A :19.76 lb  B :12.10 lb 

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PROBLEM 13.20 The system shown is at rest when a constant 30 lb force is applied to collar B. (a) If the force acts through the entire motion, determine the speed of collar B as it strikes the support at C. (b) After what distance d should the 30 lb force be removed if the collar is to reach support C with zero velocity?

SOLUTION Let F be the cable tension and vB be the velocity of collar B when it strikes the support. Consider the collar B. Its movement is horizontal so only horizontal forces acting on B do work. Let d be the distance through which the 30 lb applied force moves. (T1 ) B + (U1→ 2 ) B = (T2 ) B 0 + 30d − (2 F )(2) =

1 18 2 vB 2 32.2

30d − 4 F = 0.27950vB2

(1)

Now consider the weight A. When the collar moves 2 ft to the left, the weight moves 4 ft up, since the cable length is constant. Also, v A = 2vB . (T1 ) A + (U1− 2 ) A = (T2 ) B 1 WA 2 vA 2 g 1 6 4 F − (6)(4) = (2vB ) 2 2 32.2

0 + ( F − WA )(4) =

4 F − 24 = 0.37267 vB2

(2)

30d − 24 = 0.65217vB2

(3)

Add Eqs. (1) and (2) to eliminate F.

(a)

Case a:

d = 2 ft, vB = ? (30)(2) − (24) = 0.65217vB2 vB2 = 55.2 ft 2 /s 2

(b)

Case b:

vB = 7.43 ft/s 

d = ?, vB = 0. 30d − 24 = 0

d = 0.800 ft 

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PROBLEM 13.21 Car B is towing car A at a constant speed of 10 m/s on an uphill grade when the brakes of car A are fully applied causing all four wheels to skid. The driver of car B does not change the throttle setting or change gears. The masses of the cars A and B are 1400 kg and 1200 kg, respectively, and the coefficient of kinetic friction is 0.8. Neglecting air resistance and rolling resistance, determine (a) the distance traveled by the cars before they come to a stop, (b) the tension in the cable.

SOLUTION Given: Car B tows car A at 10 m/s uphill.

μk = 0.8

Car A brakes so 4 wheels skid.

Car B continues in same gear and throttle setting. Find: (a) Distance d, traveled to stop (b) Tension in cable (a)

F1 = traction force (from equilibrium) F1 = (1400 g ) sin 5° + (1200 g )sin 5°

F = 0.8 N A

= 2600(9.81)sin 5°

For system: A + B U1− 2 = [( F1 − 1400 g sin 5° − 1200 g sin 5°) − F ]d = T2 − T1 = 0 −

Since

1 1 mA + Bv 2 = − (2600)(10) 2 2 2

( F1 − 1400 g sin 5° − 1200 g sin 5°) = 0 − Fd = −0.8[1400(9.81) cos 5°]d = −130, 000 N ⋅ m d = 11.88 m 

(b)

Cable tension, T U1− 2 = [T − 0.8N A ](11.88) = T2 − T1 (T − 0.8(1400)(9.81) cos 5°)11.88 = −

1400 (10) 2 2

(T − 10945) = −5892 = 5.053 kN T = 5.05 kN 

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PROBLEM 13.22 The system shown is at rest when a constant 250-N force is applied to block A. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the horizontal surface, determine (a) the velocity of block B after block A has moved 2 m, (b) the tension in the cable.

SOLUTION Constraint of cable: x A + 3 yB = constant Δx A + 3ΔyB = 0 v A + 3vB = 0

Let F be the tension in the cable. Block A:

m A = 30 kg, P = 250 N, (T1 ) A = 0 (T1 ) A + (U1→2 ) A = (T2 ) A 1 mA v A2 2 1 0 + (250 − F )(2) = (30)(3vB ) 2 2

0 + ( P − F )(Δx A ) =

500 − 2 F = 135vB2

Block B:

(1)

mB = 25 kg, WB = mB g = 245.25 N (T1 ) B + (U1→ 2 ) B = (T2 ) B 1 mB vB2 2 2 1 (3F ) − 245.25)   = (25) vB2 3 2

0 + (3F − WB )(−ΔyB ) =

2 F − 163.5 = 12.5 vB2

(2)

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PROBLEM 13.22 (Continued)

Add Eqs. (1) and (2) to eliminate F. 500 − 163.5 = 147.5vB2 vB2 = 2.2814 m 2 /s 2

(a)

Velocity of B.

(b)

Tension in the cable. From Eq. (2),

v B = 1.510 m/s

2 F − 163.5 = (12.5)(2.2814)



F = 96.0 N 

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PROBLEM 13.23 The system shown is at rest when a constant 250-N force is applied to block A. Neglecting the masses of the pulleys and the effect of friction in the pulleys and assuming that the coefficients of friction between block A and the horizontal surface are μ s = 0.25 and μ k = 0.20, determine (a) the velocity of block B after block A has moved 2 m, (b) the tension in the cable.

SOLUTION Check the equilibrium position to see if the blocks move. Let F be the tension in the cable. Block B:

3F − mB g = 0 F=

Block A:

mB g (25)(9.81) = = 81.75 N 3 3

ΣFy = 0:

N A − mA g = 0 N A = m A g = (30)(9.81) = 294.3 N

ΣFx = 0: 250 − FA − F = 0 FA = 250 − 81.75 = 168.25 N

μs N A = (0.25)(294.3) = 73.57 N

Available static friction force: Since FA > μs N A , the blocks move. The friction force, FA, during sliding is

FA = μk N A = (0.20)(294.3) = 58.86 N

Constraint of cable: x A + 3 yB = constant Δ x A + 3ΔyB = 0 v A + 3vB = 0

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PROBLEM 13.23 (Continued)

m A = 30 kg, P = 250 N, (T1 ) A = 0.

Block A:

(T1 ) A + (U1→2 ) A = (T2 ) A 1 mA v 2A 2 1 0 + (250 − 58.86 − F )(2) = (30)(3vB ) 2 2 0 + ( P − FA − F )(Δx A ) =

382.28 − 2 F = 135vB2

(1)

M B = 25 kg, WB = mB g = 245.25 N

Block B:

(T1 ) B + (U1→2 ) B = (T2 ) B 1 mB vB2 2 2 1 (3F − 245.25)   = (25)vB2 3 2

0 + (3F − WB )(−ΔyB ) =

2 F − 163.5 = 12.5vB2

(2)

Add Eqs. (1) and (2) to eliminate F. 382.28 − 163.5 = 147.5vB2 vB2 = 1.48325 m 2 /s 2 v B = 1.218 m/s

(a)

Velocity of B:

(b)

Tension in the cable: From Eq. (2),

2 F − 163.5 = (12.5)(1.48325)



F = 91.0 N 

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PROBLEM 13.24 Two blocks A and B, of mass 4 kg and 5 kg, respectively, are connected by a cord which passes over pulleys as shown. A 3 kg collar C is placed on block A and the system is released from rest. After the blocks have moved 0.9 m, collar C is removed and blocks A and B continue to move. Determine the speed of block A just before it strikes the ground.

SOLUTION Position  to Position .

v1 = 0

T1 = 0

At  before C is removed from the system 1 1 ( mA + mB + mC )v22 = (12 kg)v22 = 6v22 2 2 = (m A + mC − mB ) g (0.9 m)

T2 = U1− 2

U1− 2 = (4 + 3 − 5)( g )(0.9 m) = (2 kg)(9.81 m/s 2 )(0.9 m) U1− 2 = 17.658 J T1 + U1− 2 = T2 : 0 + 17.658 = 6v22

v22 = 2.943

At Position , collar C is removed from the system. Position  to Position . T3 =

T2′ =

1 9  (m A + mB )v22 =  kg  (2.943) = 13.244 J 2 2 

1 9 ( mA + mB )(v3 ) 2 = v32 2 2 U 2′−3 = ( mA − mB )( g )(0.7 m) = (−1 kg)(9.81 m/s 2 )(0.7 m) = −6.867 J T2′ + U 2 −3 = T3

13.244 − 6.867 = 4.5v32

v32 = 1.417 v A = v3 = 1.190 m/s

v A = 1.190 m/s 

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PROBLEM 13.25 Four packages, each weighing 6 lb, are held in place by friction on a conveyor which is disengaged from its drive motor. When the system is released from rest, package 1 leaves the belt at A just as package 4 comes onto the inclined portion of the belt at B. Determine (a) the speed of package 2 as it leaves the belt at A, (b) the speed of package 3 as it leaves the belt at A. Neglect the mass of the belt and rollers.

SOLUTION Slope angle: (a)

sin β =

6 ft 15 ft

β = 23.6°

Package falls off the belt and 2, 3, 4 move down 6 = 2 ft. 3 1  3  6 lb T2 = 3  mv22  =  2 2  2  32.2 ft/s

 2 2  v2 = 0.2795v2 

U1− 2 = (3)(W )( R) = (3)(6 lb)(2 ft) = 36 lb ⋅ ft T1 + U1− 2 = T2 0 + 36 = 0.2795 v22

v22 = 128.8 v 2 = 11.35 ft/s

(b)



 

23.6° 

Package 2 falls off the belt and its energy is lost to the system and 3 and 4 move down 2 ft. 1   6 lb  (128.8) T2′ = (2)  m v22  =  2  2   32 ft/s   T2′ = 24 lb ⋅ ft 1   6 lb  2 (v3 ) = 0.18634v32 T3 = (2)  m v32  =  2   2   32.2 ft/s  U 2−3 = (2)(W )(2) = (2)(6 lb)(2 ft) = 24 lb ⋅ ft T2 + U 2−3 = T3 24 + 24 = 0.18634v32

v32 = 257.6 

 v3 = 16.05 ft/s

23.6° 

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PROBLEM 13.26 A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

SOLUTION Call blocks A and B. (a)

m A = 2 kg, mB = 3 kg

Position 1: Block B has just been removed. Spring force:

FS = −(m A + mB ) g = −k x

Spring stretch:

x1 = −

(m A + mB ) g (5 kg)(9.81 m/s 2 ) =− = −1.22625 m k 40 N/m

Let position 2 be a later position while the spring still contacts block A. Work of the force exerted by the spring:

(U1→2 )e = −



x2

k x dx

x1

1 = − k x2 2 =

Work of the gravitational force:

x2 x1

=

1 1 k x12 − k x22 2 2

1 1 (40)(−1.22625) 2 − (40) x22 = 30.074 − 20 x22 2 2

(U1→2 ) g = − mA g ( x2 − x1 ) = −(2)(9.81)( x2 + 1.22625) = −19.62 x2 − 24.059

Total work: Kinetic energies:

U1→2 = −20 x22 + 19.62 x2 + 6.015 T1 = 0 T2 =

Principle of work and energy:

1 1 m Av22 = (2)v22 = v22 2 2

T1 + U1→ 2 = T2 0 + 20 x22 − 19.62 x2 + 6.015 = v22

Speed squared:

v22 = −20 x22 − 19.62 x2 + 6.015

At maximum speed,

dv2 =0 dx2

(1)

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PROBLEM 13.26 (Continued)

Differentiating Eq. (1), and setting equal to zero, 2v2

Substituting into Eq. (1),

dv2 = −40 x2 = −19.62 = 0 dx 19.62 x2 = − = −0.4905 m 40

v22 = −(20)(−0.4905) 2 − (19.62)( −0.4905) + 6.015 = 10.827 m 2 /s 2 v 2 = 3.29 m/s 

Maximum speed: (b)

Position 3: Block A reaches maximum height. Assume that the block has separated from the spring. Spring force is zero at separation. Work of the force exerted by the spring: (U1→3 )e = −



0 x1

kxdx =

1 2 1 kx1 = (40)(1.22625)2 = 30.074 J 2 2

Work of the gravitational force: (U1→3 ) g = − m A gh = −(2)(9.81) h = −19.62 h

Total work:

U1→3 = 30.074 − 19.62 h

At maximum height,

v3 = 0, T3 = 0

Principle of work and energy:

T1 + U1→3 = T3 0 + 30.074 − 19.62 h = 0 h = 1.533 m 

Maximum height:

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PROBLEM 13.27 Solve Problem 13.26, assuming that the 2-kg block is attached to the spring. PROBLEM 13.26 A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

SOLUTION Call blocks A and B. (a)

m A = 2 kg, mB = 3 kg

Position 1: Block B has just been removed. Spring force:

FS = −(m A + mB ) g = − kx1

Spring stretch:

x1 = −

(m A + mB ) g (5 kg)(9.81 m/s 2 ) =− = −1.22625 m k 40 N/m

Let position 2 be a later position. Note that the spring remains attached to block A. Work of the force exerted by the spring:

(U1→2 )e = −



x2

kxdx

x1

1 = − kx 2 2 =

Work of the gravitational force:

x2 x1

=

1 2 1 2 kx1 − kx2 2 2

1 1 (40)(−1.22625) 2 − (40) x22 = 30.074 − 20 x22 2 2

(U1→2 ) g = − mA g ( x2 − x1 ) = −(2)(9.81)( x2 + 1.22625) = −19.62 x2 − 24.059

Total work: Kinetic energies:

U1→2 = −20 x22 − 19.62 x2 + 6.015 T1 = 0 T2 =

Principle of work and energy:

1 1 m Av22 = (2)v22 = v22 2 2

T1 + U1→ 2 = T2 0 + 20 x22 − 19.62 x2 + 6.015 = v22

Speed squared:

v22 = −20 x22 − 19.62 x2 + 6.015

At maximum speed,

dv2 =0 dx2

(1)

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PROBLEM 13.27 (Continued)

Differentiating Eq. (1) and setting equal to zero, 2v2

dv2 = −40 x2 = −19.62 = 0 dx2 x2 = −

Substituting into Eq. (1),

19.62 = −0.4905 m 40

v22 = −(20)( −0.4905)2 − (19.62)(−0.4905) + 6.015 = 10.827 m1 /s 2 v2 = 3.29 m/s 

Maximum speed: (b)

Maximum height occurs when v2 = 0. Substituting into Eq. (1),

0 = −20 x22 − 19.62 x2 + 6.015

Solving the quadratic equation x2 = −1.22625 m and 0.24525 m

Using the larger value,

x2 = 0.24525 m

Maximum height:

h = x2 − x1 = 0.24525 + 1.22625

h = 1.472 m 

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PROBLEM 13.28 An 8-lb collar C slides on a horizontal rod between springs A and B. If the collar is pushed to the right until spring B is compressed 2 in. and released, determine the distance through which the collar will travel assuming (a) no friction between the collar and the rod, (b) a coefficient of friction μk = 0.35.

SOLUTION k B = 144 lb/ft

(a)

k A = 216 lb/ft

Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A. TA = 0 U A− B =



TB = 0 2/12 0

k B xdx −



y 0

k A xdx 2

U A− B

 144 lb/ft  2   216 lb/ft  2 =  ft  −   ( y) 2 2   12   

TA + U A− B = TB :

0 + 2 − 108 y 2 = 0

y = 0.1361 ft = 1.633 in.

Total distance

d = 2 + 16 − (6 − 1.633) d = 13.63 in. 

(b)

Assume that C does not reach the spring at B because of friction. N = W = 6 lb F f = (0.35)(8 lb) = 2.80 lb TA = TD = 0 U A− D =



2/12 0

TA + U A− D = TD

144 × dx − F f ( y ) = 2 − 2.80 y 0 + 2 − 2.80 y = 0

y = 0.714 ft = 8.57 in.

The collar must travel 16 − 6 + 2 = 12 in. before it engages the spring at B. Since y = 8.57 in., it stops before engaging the spring at B. d = 8.57 in. 

Total distance

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PROBLEM 13.29 A 6-lb block is attached to a cable and to a spring as shown. The constant of the spring is k = 8 lb/in. and the tension in the cable is 3 lb. If the cable is cut, determine (a) the maximum displacement of the block, (b) the maximum speed of the block.

SOLUTION k = 8 lb/in. = 96 lb/ft

ΣFy = 0: ( Fs )1 = 6 − 3 = 3 lb C v1 = 0 T1 = 0: T2 =

For weight:

U1− 2 = (6 lb)x = 6 x

For spring:

U1− 2 = −



x 0

1  6 lb  2 2   v2 = 0.09317v2 2  32.2 

(3 + 96 x)dx = −3x − 48 x 2

T1 + U1− 2 = T2 : 0 + 6 x − 3x − 48 x 2 = 0.09317v22 3x − 48 x 2 = 0.09317v22

(a)

For xm , v2 = 0:

(1)

3x − 48 x 2 = 0 x = 0,

xm =

3 1 = ft 48 16

x m = 0.75 in. 

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PROBLEM 13.29 (Continued)

(b)

For vm we see maximum of

U1− 2 = 3x − 48 x 2 dU1− 2 = 3 − 96 x = 0 dx

x=

3 1 ft = ft 96 32

2

Eq. (1):

 1   1  3 ft  − 48  ft  = 0.09317vm2 32 32    

vm2 = 0.5031

vm = 0.7093 ft/s

vm = 8.51 in./s  

Note: U1− 2 for the spring may be computed using F6 − x curve U1− 2 = area 1 = 3x + 96x 2 2

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PROBLEM 13.30 A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs unstretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block.

SOLUTION

(a)

W = weight of the block = 10 (9.81) = 98.1 N xB =

1 xA 2 1 1 k A ( x A ) 2 − k B ( xB ) 2 2 2 (Gravity) (Spring A) (Spring B)

U1− 2 = W ( x A ) −

U1 − 2 = (98.1 N)(0.05 m) − −

1 (2000 N/m)(0.05 m) 2 2

1 (2000 N/m)(0.025 m) 2 2

U1− 2 =

1 1 (m)v 2 = (10 kg) v 2 2 2

4.905 − 2.5 − 0.625 =

1 (10)v 2 2

v = 0.597 m/s 

(b)

Let x = distance moved down by the 10 kg block 2

U1− 2

1 1  x 1 = W ( x) − k A ( x) 2 − k B   = (m)v 2 2 2 2 2

d 1 k  (m)v 2  = 0 = W − k A ( x) − B (2 x)  dx  2 8 

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PROBLEM 13.30 (Continued)

0 = 98.1 − 2000 ( x) −

2000 (2 x) = 98.1 − (2000 + 250) x 8

x = 0.0436 m (43.6 mm)

For

x = 0.0436, U = 4.2772 − 1.9010 − 0.4752 =

1 (10)v 2 2

vmax = 0.6166 m/s vmax = 0.617 m/s 

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PROBLEM 13.31 A 5-kg collar A is at rest on top of, but not attached to, a spring with stiffness k1 = 400 N/m; when a constant 150-N force is applied to the cable. Knowing A has a speed of 1 m/s when the upper spring is compressed 75 mm, determine the spring stiffness k2. Ignore friction and the mass of the pulley.

SOLUTION Use the method of work and energy applied to the collar A. T1 + U1→ 2 = T2

Since collar is initially at rest,

T1 = 0.

In position 2, where the upper spring is compressed 75 mm and v2 = 1.00 m/s, the kinetic energy is T2 =

1 2 1 mv2 = (5 kg)(1.00 m/s) 2 = 2.5 J 2 2

As the collar is raised from level A to level B, the work of the weight force is (U1→2 ) g = − mgh

where m = 5 kg, g = 9.81 m/s 2 and h = 450 mm = 0.450 m Thus, (U1→2 ) g = −(5)(9.81)(0.450) = −22.0725 J In position 1, the force exerted by the lower spring is equal to the weight of collar A. F1 = mg = −(5 kg)(9.81 m/s) = −49.05 N

As the collar moves up a distance x1, the spring force is F = F1 − k1 x2

until the collar separates from the spring at xf =

F1 49.05 N = = 0.122625 m = 122.625 mm k1 400 N/m

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PROBLEM 13.31 (Continued)

Work of the force exerted by the lower spring: (U1→2 )1 =



xf

0

( F1 − k1 x)dx

= F1 x f − =

1 2 1 1 kx f = k1 x 2f − k1 x 2f = k1 x 2f 2 2 2

1 (400 N/m)(0.122625)2 = 3.0074 J 2

In position 2, the upper spring is compressed by y = 75 mm = 0.075 m. The work of the force exerted by this spring is 1 1 (U1→2 ) 2 = − k2 y 2 = − k2 (0.075) 2 = −0.0028125 k 2 2 2

Finally, we must calculate the work of the 150 N force applied to the cable. In position 1, the length AB is (l AB )1 = (450) 2 + (400) 2 = 602.08 mm

In position 2, the length AB is (l AB ) 2 = 400 mm. The displacement d of the 150 N force is d = (l AB )1 − (l AB )2 = 202.08 mm = 0.20208 m

The work of the 150 N force P is (U1→2 ) P = Pd = (150 N)(0.20208 m) = 30.312 J

Total work:

U1→2 = −22.0725 + 3.0074 − 0.0028125k2 + 30.312 = 11.247 − 0.0028125k2

Principle of work and energy:

T1 + U1→ 2 = T2 0 + 11.247 − 0.0028125k2 = 2.5 k2 = 3110 N/m

k2 = 3110 N/m 

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PROBLEM 13.32 A piston of mass m and cross-sectional area A is equilibrium under the pressure p at the center of a cylinder closed at both ends. Assuming that the piston is moved to the left a distance a/2 and released, and knowing that the pressure on each side of the piston varies inversely with the volume, determine the velocity of the piston as it again reaches the center of the cylinder. Neglect friction between the piston and the cylinder and express your answer in terms of m, a, p, and A.

SOLUTION Pressures vary inversely as the volume pL Aa = P Ax pR Aa = P A(2a − x)

Initially at ,

v=0 x=

pL =

pa x

pR =

pa (2a − x)

a 2

T1 = 0

At ,

x = a, T2 = U1− 2 =



a a/2

1 2 mv 2

( pL − pR ) Adx =



1  1 paA  −  dx a/2  x 2a − x  a

U1− 2 = paA[ln x + ln (2a − x)]aa/2  a  3a   U1− 2 = paA ln a + ln a − ln   − ln    2  2    3a 2  4 U1− 2 = paA ln a 2 − ln  = paA ln   4  3  4 1 T1 + U1− 2 = T2 0 + paA ln   = mv 2 3 2 v2 =

2 paA ln ( 43 ) m

= 0.5754

paA m

v = 0.759

paA  m

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PROBLEM 13.33 An uncontrolled automobile traveling at 65 mph strikes squarely a highway crash cushion of the type shown in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F of the force required to crush the barrels is shown as a function of the distance x the automobile has moved into the cushion. Knowing that the weight of the automobile is 2250 lb and neglecting the effect of friction, determine (a) the distance the automobile will move into the cushion before it comes to rest, (b) the maximum deceleration of the automobile.

SOLUTION (a)

65 mi/h = 95.3 ft/s Assume auto stops in 5 ≤ d ≤ 14 ft.

v1 = 95.33 ft/s 1 2 1  2250 lb mv1  2 2  32.2 ft/s 2 T1 = 317,530 lb ⋅ ft T1 =

 2  (95.3 ft/s) 

= 317.63 k ⋅ ft v2 = 0 T2 = 0 U1− 2 = (18 k)(5 ft) + (27 k)( d − 5)

T1 + U1− 2

= 90 + 27d − 135 = 27d − 45 k ⋅ ft = T2

317.53 = 27d − 45 d = 13.43 ft 

Assumption that d ≤ 14 ft is ok. (b)

Maximum deceleration occurs when F is largest. For d = 13.43 ft, F = 27 k. Thus, F = maD  2250 lb (27, 000 lb) =  2  32.2 ft/s

  ( aD ) 

aD = 386 ft/s 2 

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PROBLEM 13.34 Two types of energy-absorbing fenders designed to be used on a pier are statically loaded. The force-deflection curve for each type of fender is given in the graph. Determine the maximum deflection of each fender when a 90-ton ship moving at 1 mi/h strikes the fender and is brought to rest.

SOLUTION W1 = (90 ton)(2000 lb/ton) = 180 × 103 lb

Weight:

W 180 × 103 = = 5590 lb ⋅ s 2 /ft g 32.2

Mass:

m=

Speed:

v1 = 1 mi/h =

Kinetic energy:

T1 =

5280 ft = 1.4667 ft/s 3600 s

1 2 1 mv1 = (5590)(1.4667) 2 2 2 = 6012 ft ⋅ lb

T2 = 0

Principle of work and energy:

(rest)

T1 + U1→ 2 = T2 6012 + U1→2 = 0 U1→2 = −6012 ft ⋅ lb = −72.15 kip ⋅ in.

The area under the force-deflection curve up to the maximum deflection is equal to 72.15 kip ⋅ in. Fender A: From the force-deflection curve F = kx Area =



x

0

fdx =

k=



x

0

Fmax 60 = = 5 kip/in. xmax 12

kx dx =

1 2 kx 2

1 (5) x 2 = 72.51 2 x 2 = 28.86 in.2 Fender B: We divide area under curve B into trapezoids Partial area

x = 5.37 in. 

Total Area

From x = 0

to x = 2 in.:

1 (2 in.)(4 kips) = 4 kip ⋅ in. 2

4 kip ⋅ in.

From x = 2 in.

to x = 4 in.:

1 (2 in.)(4 + 10) = 14 kip ⋅ in. 2

18 kip ⋅ in.

From x = 4 in.

to x = 6 in.:

1 (2 in.)(10 + 18) = 28 kip ⋅ in. 2

46 kip ⋅ in.

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PROBLEM 13.34 (Continued) We still need ΔU = 72.15 − 46 = 26.15 kip ⋅ in. Equation of straight line approximating curve B from x = 6 in. to x = 8 in. is

Δx F − 18 = F = 18 + 6Δx 2 30 − 18

1 ΔU = 18Δx + (6Δx)Δx = 26.15 kip ⋅ in. 2 2 (Δx) + 6Δx − 8.716 = 0 Δx = 1.209 in. Thus:

x = 6 in. + 1.209 in. = 7.209 in.

x = 7.21 in. 

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PROBLEM 13.35 Nonlinear springs are classified as hard or soft, depending upon the curvature of their force-deflection curve (see figure). If a delicate instrument having a mass of 5 kg is placed on a spring of length l so that its base is just touching the undeformed spring and then inadvertently released from that position, determine the maximum deflection xm of the spring and the maximum force Fm exerted by the spring, assuming (a) a linear spring of constant k = 3 kN/m, (b) a hard, nonlinear spring, for which F = (3 kN/m)( x + 160 x 2 ) .

SOLUTION W = mg = (5 kg) g W = 49.05 N

T1 = T2 = 0, T1 + U1− 2 = T2 yields U1− 2 = 0

Since

U1− 2 = Wxm −

(a)



xm

0

Fdx = 49.05 xm −



xm

0

Fdx = 0

For F = kx = (300 N/m) x Eq. (1):

49.05 xm −



xm

0

3000 x dx = 0 xm = 32.7 × 10−3 m = 32.7 mm 

49.05 xm − 1500 xm2 = 0 Fm = 3000 xm = 3000(32.7 × 10−3 )

(b)

(1)

For Eq. (1)

F = (3000 N/m) x(1 + 160 x 2 ) 49.05 xm −



xm

0

3000( x + 160 x3 )dx = 0

1  49.05 xm − 3000  xm2 + 40 xm4  = 0 2  

Solve by trial:

Fm = 98.1 N 

(2)

xm = 30.44 × 10−3 m

xm = 30.4 mm 

Fm = (3000)(30.44 × 10−3 )[1 + 160(30.44 × 10−3 ) 2 ]

Fm = 104.9 N 

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PROBLEM 13.36 A rocket is fired vertically from the surface of the moon with a speed v0 . Derive a formula for the ratio hn /hu of heights reached with a speed v, if Newton’s law of gravitation is used to calculate hn and a uniform gravitational field is used to calculate hu . Express your answer in terms of the acceleration of gravity g m on the surface of the moon, the radius Rm of the moon, and the speeds v and v0 .

SOLUTION Newton’s law of gravitation T1 = U1− 2 =

1 2 mv0 2



Rm + hn Rm

T2 =

1 2 mv 2

( − Fn )dr

U1− 2 = − mg m Rm2



Rm + hn Rm

Fn =

mg m Rm2 r2

dr r2

 1  1 U1− 2 = mg m Rm2  −   Rm Rm + hn  T1 + U1− 2 = T2  Rm  1 2 1 2 mv0 + mg m  Rm −  = mv Rm + hn  2 2  hn =

(v

)

Rm  − v2   2 2 (v −v )  2 g m  Rm − 0 2 gm  

2 0

(1)

Uniform gravitational field T1 = U1− 2 =

1 2 mv0 T2 = mv 2 2



Rm + hn Rm

1 2 1 mv0 − mg m hu = mv 2 2 2

T1 + U1− 2 = T2

hu

( − Fu )dr = −mg m ( Rm + hu − Rm ) = −mghu

(v =

2 0

− v2

)

(2)

2 gm hn = hu

Dividing (1) by (2)

1

( v −v ) 1 − (2 g R ) 2 0

2



m m

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PROBLEM 13.37 Express the acceleration of gravity g h at an altitude h above the surface of the earth in terms of the acceleration of gravity g0 at the surface of the earth, the altitude h and the radius R of the earth. Determine the percent error if the weight that an object has on the surface of earth is used as its weight at an altitude of (a) 1 km, (b) 1000 km.

SOLUTION F=

At earth’s surface, (h = 0)

=

(h + R ) 2

GM E m / R 2

( Rh + 1)

2

mg h

GM E m = mg0 R2 GM E R

2

GM E

= g0

gh =

Thus,

GM E m

gh =

R2 h   +1  R 

2

g0 h   +1 R 

2

R = 6370 km

At altitude h, “true” weight Assumed weight

F = mg h = WT W0 = mg0 Error = E =

W0 − WT mg0 − mg h g0 − g h = = W0 mg0 g0 g0 −

gh =

g0

( Rh + 1)

2

E=

g0

(1 + ) h R

2

g0

1   = 1 − 2 h  (1 + R ) 

(a)

h = 1 km:

1   P = 100E = 100 1 − 2 1  (1 + 6370 ) 

P = 0.0314% 

(b)

h = 1000 km:

1   P = 100E = 100 1 −  1000 2  (1 + 6370 ) 

P = 25.3% 

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PROBLEM 13.38 A golf ball struck on earth rises to a maximum height of 60 m and hits the ground 230 m away. How high will the same golf ball travel on the moon if the magnitude and direction of its velocity are the same as they were on earth immediately after the ball was hit? Assume that the ball is hit and lands at the same elevation in both cases and that the effect of the atmosphere on the earth is neglected, so that the trajectory in both cases is a parabola. The acceleration of gravity on the moon is 0.165 times that on earth.

SOLUTION

Solve for hm .

At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the same in both cases. 1 2 mv 2 = −mge he

T1 = U1− 2

T2 = Earth

U1− 2 = −mg m hm

Earth

1 2 1 mv − mge he = mvH2 2 2

Moon

1 2 1 mv − mg m hm = mvH2 2 2

Moon

hm g e = he g m

− ge he + g m hm = 0

Subtracting

1 2 mvH 2

 ge  hm = (60 m)    0.165 ge 

hm = 364 m 

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PROBLEM 13.39 The sphere at A is given a downward velocity v 0 of magnitude 5 m/s and swings in a vertical plane at the end of a rope of length l = 2 m attached to a support at O. Determine the angle θ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

SOLUTION 1 2 1 mv0 = m (5) 2 2 2 T1 = 12.5 m T1 =

1 2 mv 2 = mg (l ) sin θ

T2 = U1− 2

T1 + U1− 2 = T2

12.5m + 2mg sin θ =

25 + 4 g sin θ = v 2

1 2 mv 2

(1)

Newton’s law at . v2 v2 =m  2 2 v = 4 g − 2 g sin θ

2mg − mg sin θ = m

(2)

Substitute for v 2 from Eq. (2) into Eq. (1) 25 + 4 g sin θ = 4 g − 2 g sin θ (4)(9.81) − 25 sin θ = = 0.2419 (6)(9.81)

θ = 14.00° 

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PROBLEM 13.40 The sphere at A is given a downward velocity v 0 and swings in a vertical circle of radius l and center O. Determine the smallest velocity v 0 for which the sphere will reach Point B as it swings about Point O (a) if AO is a rope, (b) if AO is a slender rod of negligible mass.

SOLUTION 1 2 mv0 2 1 T2 = mv 2 2 U1− 2 = − mgl T1 =

1 2 1 mv0 − mgl = mv 2 2 2 v02 = v 2 + 2 gl

T1 + U1− 2 = T2

Newton’s law at  (a)

For minimum v, tension in the cord must be zero. Thus, v 2 = gl v02 = v 2 + 2 gl = 3gl

(b)

v0 = 3gl 

Force in the rod can support the weight so that v can be zero. v02 = 0 + 2 gl

Thus,

v0 = 2 gl 

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PROBLEM 13.41 A small sphere B of weight W is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord (a) just before the sphere comes in contact with the peg, (b) just after it comes in contact with the peg.

SOLUTION Velocity of the sphere as the cord contacts A vB = 0

TB = 0

1 2 mvC 2 = ( mg )(1)

TC = U B −C

TB + U B −C = TC 0 + 1mg =

1 2 mvC 2

vC2 = (2)( g )

Newton’s law (a)

Cord rotates about Point O ( R = L)

T − mg (cos 60°) = m

vC2 L

T = mg (cos 60°) + T=

3 mg 2

m(2) g 2 T = 1.5 W 

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PROBLEM 13.41 (Continued)

(b)

L  Cord rotates about A  R =  2 

T − mg (cos 60°) = T=

mvC2 L 2

mg m(2)( g ) + 2 1

5 1  T =  + 2  mg = mg 2 2 

T = 2.5W 

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PROBLEM 13.42 A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 40-ft diameter, and moves up and down past Point E. Knowing that h = 60 ft and assuming no energy loss due to friction, determine (a) the force exerted by his seat on a 160-lb rider at B and D, (b) the minimum value of the radius of curvature at E if the roller coaster is not to leave the track at that point.

SOLUTION Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position 2 be at P. Apply the principle of work and energy. T1 = 0

T2 =

1 2 mvP 2

U1→2 = mgyP T1 + U1→ 2 = T2 : 0 + mgyP =

1 2 mvP 2

vP2 = 2 gyP

Magnitude of normal acceleration at P: ( aP ) n =

(a)

vP2

ρP

=

2 gyP

ρP

Rider at Point B. yB = h = 60 ft

ρ B = r = 20 ft an =

(2 g )(60) = 6g 20

ΣF = ma: N B − mg = m(6 g ) N B = 7 mg = 7W = (7)(160 lb) N B = 1120 lb 

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PROBLEM 13.42 (Continued)

Rider at Point D. yD = h − 2r = 20 ft

ρ D = 20 ft an =

(2 g )(20) = 2g 20

ΣF = ma: N D + mg = m(2 g ) N D = mg = W = 160 lb N D = 160 lb 

(b)

Car at Point E.

yE = h − r = 40 ft NE = 0 ΣF = man : mg = m ⋅

2 gyE

ρE

ρ E = 2 yE

ρ = 80.0 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 561

PROBLEM 13.43 In Problem 13.42, determine the range of values of h for which the roller coaster will not leave the track at D or E, knowing that the radius of curvature at E is ρ = 75 ft. Assume no energy loss due to friction. PROBLEM 13.42 A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 40-ft diameter, and moves up and down past Point E. Knowing that h = 60 ft and assuming no energy loss due to friction, determine (a) the force exerted by his seat on a 160-lb rider at B and D, (b) the minimum value of the radius of curvature at E if the roller coaster is not to leave the track at that point.

SOLUTION Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position 2 be at P. Apply the principle of work and energy. T2 =

T1 = 0

1 2 mvP 2

U1→2 = mg yP

T1 + U1→ 2 = T2 : 0 + mgy p =

1 2 mvP 2

vP2 = 2 g yP

Magnitude of normal acceleration of P: ( aP ) n =

vP2

ρP

=

2 g yP

ρP

The condition of loss of contact with the track at P is that the curvature of the path is equal to ρp and the normal contact force N P = 0. Car at Point D.

ρ D = r = 20 ft y D = h − 2r ( aD ) n =

2 g (h − 2r ) r

ΣF = ma

2 g ( h − 2r ) r 2 h − 5r N D = mg r

N D + mg = m

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PROBLEM 13.43 (Continued)

For N D > 0 2 h − 5r > 0 5 h > r = 50 ft 2

Car at Point E.

ρ E = ρ = 75 ft yE = h − r = h − 20 ft ( aE ) n =

2 g (h − 20) 75

ΣF = ma 2mg ( h − 20) 75 115 − 2h N E = mg 75

N E − mg = −

For

N E > 0,

115 − 2h > 0

h < 57.5 ft 50.0 ft ≤ h ≤ 57.5 ft 

Range of values for h:

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 563

PROBLEM 13.44 A small block slides at a speed v on a horizontal surface. Knowing that h = 0.9 m, determine the required speed of the block if it is to leave the cylindrical surface BCD when θ = 30°.

SOLUTION At Point C where the block leaves the surface BCD the contact force is reduced to zero. Apply Newton’s second law at Point C.

n-direction:

N − mg cos θ = −man = −

mvC2 h vc2 = gh cos θ

With N = 0, we get

Apply the work-energy principle to the block sliding over the path BC. Let position 1 correspond to Point B and position 2 to C. T1 =

1 2 mvB 2

T2 =

1 1 mvC2 = mgh cos θ 2 2

U1→2 = weight × change in vertical distance = mgh (1 − cos θ ) 1 2 1 mvB + mg (1 − cos θ ) = mgh cos θ 2 2 2 vB = gh cos θ − 2 gh(1 − cos θ ) = gh(3cos θ − 2)

T1 + U1→ 2 = T2 :

Data:

g = 9.81 m/s 2 , h = 0.9 m, θ = 30°. vB2 = (9.81)(0.9)(3cos 30° − 2) = 5.2804 m 2 /s 2 vB = 2.30 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 564

PROBLEM 13.45 A small block slides at a speed v = 8 ft/s on a horizontal surface at a height h = 3 ft above the ground. Determine (a) the angle θ at which it will leave the cylindrical surface BCD, (b) the distance x at which it will hit the ground. Neglect friction and air resistance.

SOLUTION

Block leaves surface at C when the normal force N = 0. mg cos θ = man g cos θ =

vC2 h

vC2 = gh cos θ = gy

(1)

Work-energy principle. TB =

(a)

1 2 1 mv = m(8) 2 = 32m 2 2

1 2 mvC 2 = TC

TC = TB + U B −C

Use Eq. (1)

32m + mg (h − y ) =

1 2 mvC 2

32 + g (h − yC ) =

1 g yC 2

U B −C = W (h − g ) = mg ( h − yC )

(2)

3 g yC 2 (32 + gh) yC = ( 32 g )

32 + gh =

yC =

(32 + (32.2)(3)) 3 (32.2) 2

yC = 2.6625 ft yC = h cos θ

cos θ =

yC 2.6625 = = 0.8875 3 h

(3)

θ = 27.4° 

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PROBLEM 13.45 (Continued)

(b)

From (1) and (3) vC = gy vC = (32.2)(2.6625) vC = 9.259 ft/s

At C:

(vC ) x = vC cos θ = (9.259)(cos 27.4°) = 8.220 ft/s (vC ) y = −vC sin θ = −(9.259)(sin 27.4°) = 4.261 ft/s y = yC + (vC ) y t −

At E:

yE = 0:

1 2 gt = 2.6625 − 4.261t − 16.1t 2 2

t 2 + 0.2647t − 0.1654 = 0

t = 0.2953 s

At E:

x = h(sin θ ) + (vC ) x t = (3)(sin 27.4°) + (8.220)(0.2953) x = 1.381 + 2.427 = 3.808 ft x = 3.81 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 566

PROBLEM 13.46 A chair-lift is designed to transport 1000 skiers per hour from the base A to the summit B. The average mass of a skier is 70 kg and the average speed of the lift is 75 m/min. Determine (a) the average power required, (b) the required capacity of the motor if the mechanical efficiency is 85 percent and if a 300 percent overload is to be allowed.

SOLUTION Note: Solution is independent of speed. (a)

Average power =

ΔU (1000)(70 kg)(9.81 m/s 2 )(300 m) N⋅m = = 57, 225 Δt 3600 s s

Average power = 57.2 kW  (b)

Maximum power required with 300% over load =

100 + 300 (57.225 kW) = 229 kW 100

Required motor capacity (85% efficient) Motor capacity =

229 kW = 269 kW  0.85

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 567

PROBLEM 13.47 It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic car-lift platform to a height of 2.8 m. Determine (a) the average output power delivered by the hydraulic pump to lift the system, (b) the average power electric required, knowing that the overall conversion efficiency from electric to mechanical power for the system is 82 percent.

SOLUTION (a)

( PP ) A = ( F )(v A ) = (mC + mL )( g )(v A ) v A = s/t = (2.8 m)/(15 s) = 0.18667 m/s

( PP ) A = [(1200 kg) + (300 kg)](9.81 m/s 2 )(0.18667 m/s)3

(b)

( PP ) A = 2.747 kJ/s

( PP ) A = 2.75 kW 

( PE ) A = ( PP )/η = (2.75 kW)/(0.82)

( PE ) A = 3.35 kW 

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PROBLEM 13.48 The velocity of the lift of Problem 13.47 increases uniformly from zero to its maximum value at mid-height 7.5 s and then decreases uniformly to zero in 7.5 s. Knowing that the peak power output of the hydraulic pump is 6 kW when the velocity is maximum, determine the maximum life force provided by the pump. PROBLEM 13.47 It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic car-lift platform to a height of 2.8 m. Determine (a) the average output power delivered by the hydraulic pump to lift the system, (b) the average power electric required, knowing that the overall conversion efficiency from electric to mechanical power for the system is 82 percent.

SOLUTION Newton’s law Mg = ( M C + M L ) g = (1200 + 300) g Mg = 1500 g

ΣF = F − 1500 g = 1500a

(1)

Since motion is uniformly accelerated, a = constant Thus, from (1), F is constant and peak power occurs when the velocity is a maximum at 7.5 s. a=

vmax 7.5 s

P = (6000 W) = ( F )(vmax ) vmax = (6000)/F a = (6000)/(7.5)( F )

Thus,

(2)

Substitute (2) into (1) F − 1500 g = (1500)(6000)/(7.5)( F ) F 2 − (1500 kg)(9.81 m/s 2 ) F −

(1500 kg)(6000 N ⋅ m/s) =0 (7.5 s)

F 2 − 14, 715F − 1.2 × 106 = 0 F = 14,800 N

F = 14.8 kN 

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PROBLEM 13.49 (a) A 120-lb woman rides a 15-lb bicycle up a 3-percent slope at a constant speed of 5 ft/s. How much power must be developed by the woman? (b) A 180-lb man on an 18-lb bicycle starts down the same slope and maintains a constant speed of 20 ft/s by braking. How much power is dissipated by the brakes? Ignore air resistance and rolling resistance.

SOLUTION

tan θ =

3 100

θ = 1.718°

W = WB + WW = 15 + 120

(a)

W = 135 lb PW = W ⋅ v = (W sin θ ) (v) PW = (135)(sin 1.718°)(5)

(a)

PW = 20.24 ft ⋅ lb/s PW = 20.2 ft ⋅ lb/s  W = WB + Wm = 18 + 180

(b)

W = 198 lb

Brakes must dissipate the power generated by the bike and the man going down the slope at 20 ft/s. PB = W ⋅ v = (W sin θ )(v) PB = (198)(sin 1.718)(20) PB = 118.7 ft ⋅ lb/s 

    (b)

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PROBLEM 13.50 A power specification formula is to be derived for electric motors which drive conveyor belts moving solid material at different rates to different heights and distances. Denoting the efficiency of the motors by η and neglecting the power needed to drive the belt itself, derive a formula (a) in the SI system of units for the power P in kW, in terms of the mass flow rate m in kg/h, the height b and horizontal distance l in meters, and (b) in U.S. customary units, for the power in hp, in terms of the material flow rate w in tons/h, and the height b and horizontal distance l in feet.

SOLUTION (a)

Material is lifted to a height b at a rate, (m kg/h)( g m/s 2 ) = [mg (N/h)] Thus, ΔU [mg (N/h)][b(m)]  mgb  = =  N ⋅ m/s Δt (3600 s/h)  3600  1000 N ⋅ m/s = 1 kW

Thus, including motor efficiency, η P (kw) =

mgb (N ⋅ m/s)  1000 N ⋅ m/s  (3600)   (η ) kW   P(kW) = 0.278 × 10−6

mgb

η



ΔU [W (tons/h)(2000 lb/ton)][b(ft)] = Δt 3600 s/h

(b)

=

With η ,

Wb ft ⋅ lb/s; 1hp = 550 ft ⋅ lb/s 1.8

 Wb   1 hp   1  hp =  (ft ⋅ lb/s)      1.8   550 ft ⋅ lb/s  η  hp =

1.010 × 10−3Wb

η



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PROBLEM 13.51 In an automobile drag race, the rear (drive) wheels of a 1000 kg car skid for the first 20 m and roll with sliding impending during the remaining 380 m. The front wheels of the car are just off the ground for the first 20 m, and for the remainder of the race 80 percent of the weight is on the rear wheels. Knowing that the coefficients of friction are μs = 0.90 and μk = 0.68, determine the power developed by the car at the drive wheels (a) at the end of the 20-m portion of the race, (b) at the end of the race: Give your answer in kW and in hp. Ignore the effect of air resistance and rolling friction.

SOLUTION (a)

First 20 m.

(Calculate velocity at 20 m.) Force generated by rear wheels = μkW , since car skids.

Thus,

Fs = (0.68)(1000)( g )

Fs = (0.68)(1000 kg)(9.81 m/s 2 ) = 6670.8 N

Work and energy.

T1 = 0, T2 =

1 2 2 mv20 = 500v20 2

T1 + U1− 2 = T2 U1− 2 = (20 m)(Fs ) = (20 m)(6670.8 N) U1− 2 = 133, 420 J 2 0 + 133,420 = 500v20 2 v20 =

133,420 = 266.83 500

v20 = 16.335 m/s

Power = ( Fs )(v20 ) = (6670.8 N)(16.335 m/s) Power = 108,970 J/s =108.97 kJ/s 1 kJ/s = 1 kW 1 hp = 0.7457 kW

Power = 109.0 kJ/s = 109.0 kW Power =



(109.0 kW) = 146.2 hp  (0.7457 kW/hp)

 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 572

PROBLEM 13.51 (Continued)

(b)

End of race. (Calculate velocity at 400 m.) For remaining 380 m, with 80% of weight on rear wheels, the force generated at impending sliding is ( μ s )(0.80)(mg ) FI = (0.90)(0.80)(1000 kg)(9.81 m/s 2 ) FI = 7063.2 N

Work and energy, from 20 m  to 28 m . v2 = 16.335 m/s [from part (a)] T2 =

1 (1000 kg)(16.335 m/s) 2 2

T2 = 133, 420 J T3 =

1 2 2 mv380 = 500v380 2

U 2 −3 = ( FI )(380 m) = (7063.2 N)(380 m) U 2 −3 = 2, 684,000 J T2 + U 2 −3 = T3 2 (133, 420 J) + (2, 684, 000 J) = 500v30

v30 = 75.066 m/s

Power = ( FI )(v30 ) = (7063.2 N)(75.066 m/s) = 530, 200 J/s

kW Power = 530, 200 J = 530 kW hp Power =



530 kW = 711 hp  (0.7457 kW/hp)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 573

PROBLEM 13.52 The frictional resistance of a ship is known to vary directly as the 1.75 power of the speed v of the ship. A single tugboat at full power can tow the ship at a constant speed of 4.5 km/h by exerting a constant force of 300 kN. Determine (a) the power developed by the tugboat, (b) the maximum speed at which two tugboats, capable of delivering the same power, can tow the ship.

SOLUTION (a)

Power developed by tugboat at 4.5 km/h. v0 = 4.5 km/h = 1.25 m/s F0 = 300 kN P0 = F0 v0 = (300 kN)(1.25 m/s)

(b)

P0 = 375 kW 

Maximum speed. Power required to tow ship at speed v: 1.75

 v  F = F0    v0 

1.75

 v  P = Fv = F0 v    v0 

 v  = F0 v0    v0 

2.75

(1)

Since we have two tugboats, the available power is twice maximum power F0 v0 developed by one tugboat.  v  2 F0 v0 = F0 v0    v0   v     v0 

Recalling that

2.75

2.75

=2

v = v0 (2)1/ 2.75 = v0 (1.2867)

v0 = 4.5 km/h v = (4.5 km/h)(1.2867) = 5.7902 km/h

v = 5.79 km/h 

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PROBLEM 13.53 A train of total mass equal to 500 Mg starts from rest and accelerates uniformly to a speed of 90 km/h in 50 s. After reaching this speed, the train travels with a constant velocity. The track is horizontal and axle friction and rolling resistance result in a total force of 15 kN in a direction opposite to the direction of motion. Determine the power required as a function of time.

SOLUTION Let FP be the driving force and FR be the resisting force due to axle friction and rolling resistance. Uniformly accelerated motion. (t < 50 s): v = v0 + at

v0 = 0

t = 50 s,

At

v = 90 km/h = 25 m/s

25 m/s = 0 + a(50) a = 0.5 m/s 2 v = (0.5 m/s 2 )t FP − FR = ma

Newton’s second law:

FR = 15 kN = 15 × 103 N

where

m = 500 Mg = 500 × 103 kg a = 0.5 m/s 2 FP = FR + ma = 15 × 103 + (500 × 103 )(0.5)

= 265 × 103 N = 265 kN FP v = (265 × 103 )(0.5t )

Power:

(0 < 50s)

Uniform motion.

(t > 50 s):

Power = (132.5 kW/s)t 

a=0 FP = FR = 15 × 103 N; v = 25 m/s

Power:

FP v = (15 × 103 )(25 m/s) = 375 × 103W (t > 50 s)

Power = 375 kW 

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PROBLEM 13.54 The elevator E has a weight of 6600 lbs when fully loaded and is connected as shown to a counterweight W of weight of 2200 lb. Determine the power in hp delivered by the motor (a) when the elevator is moving down at a constant speed of 1 ft/s, (b) when it has an upward velocity of 1 ft/s and a deceleration of 0.18 ft/s 2 .

SOLUTION (a) Acceleration = 0 Elevator

Counterweight

Motor

ΣFy = 0: TW − WW = 0

ΣF = 0: 2TC + TW − 6600 = 0

TW = 2200 lb

Kinematics:

TC = 2200 lb 2 xE = xC , 2 x E = xC , vC = 2vE = 2 ft/s

P = TC ⋅ vC = (2200 lb)(2 ft/s) = 4400 lb ⋅ ft/s = 8.00 hp P = 8.00 hp  aE = 0.18 ft/s 2 , vE = 1 ft/s

(b) Counterweight

Elevator

Counterweight:

ΣF = Ma : TW − W =

W (aW ) g

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PROBLEM 13.54 (Continued)

TW = (2200 lb) +

(2200 lb)(0.18 ft/s 2 ) (32.2 ft/s 2 )

TW = 2212 lb

Elevator ΣF = ma

2TC + TW − WE =

2TC = (−2212 lb) + (6600 lb) −

−WE (a E ) g

(6600 lb)(0.18 ft/s 2 ) (32.2 ft/s 2 )

2TC = 4351 lb TC = 2175.6 lb vC = 2 ft/s (see part(a)) P = TC ⋅ vC = (2175.6 lb)(2 ft/s) = 4351.2 lb ⋅ ft/s

= 7.911 hp







P = 7.91 hp 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 577

PROBLEM 13.CQ2 Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air resistance, which of the following statements are true when the two balls hit the ground? (a)

The kinetic energy of A is the same as the kinetic energy of B.

(b)

The kinetic energy of A is half the kinetic energy of B.

(c)

The kinetic energy of A is twice the kinetic energy of B.

(d)

The kinetic energy of A is four times the kinetic energy of B.

SOLUTION Answer: (c)

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PROBLEM 13.CQ3 Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as the initial height of the block. Will A make it completely around the loop without losing contact with the track? (a) Yes (b) No (c) need more information

SOLUTION Answer: (b) In order for A to not maintain contact with the track, the normal force must remain greater than zero, which requires a non-zero speed at the top of the loop.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 579

PROBLEM 13.55 A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0 . In each of the two cases shown, derive an expression for the constant ke , in terms of k1 and k2 , of the single spring equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when subjected to the same force P.

SOLUTION System is in equilibrium in deflected x0 position. Case (a)

Force in both springs is the same = P x0 = x1 + x2

Thus,

x0 =

P ke

x1 =

P k1

x2 =

P k2

P P P = + ke k1 k2 1 1 1 = + ke k1 k2

Case (b)

ke =

k1k2  k1 + k2

Deflection in both springs is the same = x0 P = k1 x0 + k2 x0 P = (k1 + k2 ) x0 P = ke x0

Equating the two expressions for P = (k1 + k2 ) x0 = ke x0

ke = k1 + k2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 580

PROBLEM 13.56 A loaded railroad car of mass m is rolling at a constant velocity v0 when it couples with a massless bumper system. Determine the maximum deflection of the bumper assuming the two springs are (a) in series (as shown), (b) in parallel.

SOLUTION Let position A be at the beginning of contact and position B be at maximum deflection. 1 2 mv0 2 VA = 0

(zero force in springs)

TB = 0

(v = 0 at maximum deflection)

TA =

1 2 1 k1 x1 + k2 x22 2 2 where x1 is deflection of spring k1 and x2 is that of spring k2. VB =

Conservation of energy:

TA + VA = TB + VB 1 2 1 1 mv0 + 0 = 0 + k1 x12 + k2 x22 2 2 2 k1 x12 + k2 x22 = mv02

(a)

(1)

Springs are in series. Let F be the force carried by the two springs. F k1

Then,

x1 =

Eq. (1) becomes

1 1  F 2  +  = mv02  k1 k2 

so that

The maximum deflection is

and x2 =

F k2

1 1  F = v0 m /  +   k1 k2 

1 1  + F  k1 k2 

δ = x1 + x2 = 

1 1 1  1  =  +  v0 m /  +   k1 k2   k1 k2  1 1  = v0 m  +   k1 k2 

δ = v0 m(k1 + k2 )/k1k2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 581

PROBLEM 13.56 (Continued)

(b)

Springs are in parallel. x1 = x2 = δ

Eq. (1) becomes

(k1 + k2 )δ 2 = mv02

δ = v0

m  k1 + k2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 582

PROBLEM 13.57 A 600-g collar C may slide along a horizontal, semicircular rod ABD. The spring CE has an undeformed length of 250 mm and a spring constant of 135 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at D.

SOLUTION First calculate the lengths of the spring when the collar is at positions A, B, and D. l A = 4402 + 3002 + 1802 = 562.14 mm lB = 2402 + 3002 + 202 = 384.71 mm lD = 402 + 3002 + 1802 = 352.14 mm

The elongations of springs are given by e = l − l0 . eA = 562.14 − 250 = 312.14 mm = 0.31214 m eB = 384.71 − 250 = 134.71 mm = 0.13471 m eD = 352.14 − 250 = 102.14 mm = 0.10214 m V=

Potential energies:

1 2 ke 2

1 (135 N/m)(0.31214 m) 2 = 6.5767 J 2 1 VB = (135 N/m)(0.13471 m) 2 = 1.2249 J 2 1 VD = (135 N/m)(0.10214 m)2 = 0.7042 J 2 VA =

Since the semicircular rod ABC is horizontal, there is no change in gravitational potential energy. m = 600 g = 0.600 kg

Mass of collar: Kinetic energies:

1 2 mv A = 0.300 v A2 = 0 2 1 TB = mvB2 = 0.300 vB2 2 1 2 TD = mvD = 0.300 vD2 2 TA =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 583

PROBLEM 13.57 (Continued)

(a)

Speed of collar at B. Conservation of energy:

TA + VA = TB + VB 0 + 6.5767 = 0.300vB2 + 1.2249 vB2 = 17.839 m 2 /s 2

(b)

vB = 4.22 m/s 

Speed of collar at D. Conservation of energy:

TA + VA = TD + VD 0 + 6.5767 = 0.300 v02 + 0.7042 vD2 = 19.575 m 2 /s 2

vD = 4.42 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 584

PROBLEM 13.58 A 3-lb collar is attached to a spring and slides without friction along a circular rod in a horizontal plane. The spring has an undeformed length of 7 in. and a constant k = 1.5 lb/in. Knowing that the collar is in equilibrium at A and is given a slight push to get it moving, determine the velocity of the collar (a) as it passes through B, (b) as it passes through C.

SOLUTION

L0 = 7 in., LDA = 20 in. LDB =

(8 + 6)2 + 62 = 15.23 in. LDC = 8 in.

Δ LDA = 20 − 7 = 13 in. Δ LDB = 15.23 − 7 = 8.23 in.  Δ LDC = 8 − 7 = 1 in. 

(a)

TA = 0, VA =

1 1 k (ΔLDA )2 = (1.5)(13) 2 = 126.75 lb ⋅ in. 2 2 = 10.5625 lb ⋅ ft

TB = VB =

1 2 1.5 2 mvB = vB 2 g

1 (1.5)(8.23) 2 = 50.8 lb ⋅ in. = 4.233 lb ⋅ ft 2

TA + VA = TB + VB : 0 + 10.5625 =

1.5vB2 + 4.233 32.2

vB = 11.66 ft/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 585

PROBLEM 13.58 (Continued)

(b)

TA = 0, VA = 10.5625 lb ⋅ ft, TC = VC =

1.5 2 vC 32.2

1 (1.5)(1) 2 = 0.75 lb ⋅ in. = 0.0625 lb ⋅ ft 2

TA + VA = TC + VC : 0 + 10.5625 =

1.5 2 vc + 0.0625  32.2

vC = 15.01 ft/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 586

PROBLEM 13.59 A 3-lb collar C may slide without friction along a horizontal rod. It is attached to three springs, each of constant k = 2 lb/in. and 6 in. undeformed length. Knowing that the collar is released from rest in the position shown, determine the maximum speed it will reach in the ensuing motion.

SOLUTION Maximum velocity occurs at E where collar is passing through position of equilibrium. Position  T1 = 0

Note: Undeformed length of springs is 6 in. = 0.5 ft. Spring AC:

L = (1 ft) 2 + (0.5 ft) 2 = 1.1180 ft Δ = 1.1180 − 0.50 = 0.6180 ft

Spring CD:

L = (0.5 ft)2 + (0.5 ft)2 = 0.70711 ft Δ = 0.70711 − 0.50 = 0.20711 ft

Spring BD:

L = 0.50 ft, Δ = 0

Potential energy.

(k = 2 lb/in. = 24 lb/ft for each spring)

1 1 1 V1 = Σ k Δ 2 = k ΣΔ 2 = (24 lb/ft)[(0.6180 ft) 2 + (0.20711 ft) 2 + 0] 2 2 2 V1 = 5.0983 lb ⋅ ft Position  3.0 lb 1 1 m= = 0.093168 slug; T2 = mv22 = (0.093168 slug) v22 2 2 2 32.2 ft/s

Spring AC:

L = (0.5 ft)2 + (0.5 ft)2 = 0.7071067 ft Δ = 0.70711 − 0.50 = 0.20711 ft

Spring CD:

L = 0.50 ft Δ=0

Spring BC:

L = (0.5 ft) 2 + (0.5 ft)2 = 0.7071067 ft Δ = 0.70711 − 0.50 = 0.20711 ft

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 587

PROBLEM 13.59 (Continued)

Potential energy.

1 1 V2 = Σ k Δ 2 = k ΣΔ 2 2 2 1 V2 = (24 lb/ft)[(0.20711 ft) 2 + 0 + (0.20711 ft) 2 ] = 1.0294 lb ⋅ ft 2

Conservation of energy. T1 + V1 = T2 + V2 : 0 + 5.0983 lb ⋅ ft =

1 (0.093168 slug)v22 + 1.0294 lb ⋅ ft 2

v22 = 87.345

v 2 = 9.35 ft/s ↔ 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 588

PROBLEM 13.60 A 500-g collar can slide without friction on the curved rod BC in a horizontal plane. Knowing that the undeformed length of the spring is 80 mm and that k = 400 kN/m, determine (a) the velocity that the collar should be given at A to reach B with zero velocity, (b) the velocity of the collar when it eventually reaches C.

SOLUTION (a)

Velocity at A: TA =

1 2  0.5  mv A =  kg  v A2 2 2  

TA = (0.25)v A2 ΔLA = 0.150 m − 0.080 m ΔLA = 0.070 m 1 k ( ΔLA )2 2 1 VA = (400 × 103 N/m)(0.070 m)2 2 VA = 980 J VA =

vB = 0

TB = 0

ΔLB = 0.200 m − 0.080 m = 0.120 m 1 1 k ( ΔLB )2 = (400 × 103 N/m)(0.120 m)2 2 2 VB = 2880 J VB =

Substitute into conservation of energy. TA + VA = TB + VB v A2 =

0.25v 2A + 980 = 0 + 2880

(2880 − 980) (0.25)

v A2 = 7600 m 2 /s 2 vA = 87.2 m/s  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 589

PROBLEM 13.60 (Continued)

(b)

Velocity at C: Since slope at B is positive, the component of the spring force FP , parallel to the rod, causes the block to move back toward A. TB = 0, VB = 2880 J [from part (a)] 1 2 (0.5 kg) 2 mvC = vC = 0.25vC2 2 2 ΔLC = 0.100 m − 0.080 m = 0.020 m TC =

VC =

1 1 k (Δ LC ) 2 = (400 × 103 N/m)(0.020 m) 2 = 80.0 J 2 2

Substitute into conservation of energy. TB + VB = TC + VC

0 + 2880 = 0.25vC2 + 80.0

vC2 = 11, 200 m 2 /s 2 vC = 105.8 m/s 

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PROBLEM 13.61 An elastic cord is stretched between two Points A and B, located 800 mm apart in the same horizontal plane. When stretched directly between A and B, the tension is 40 N. The cord is then stretched as shown until its midpoint C has moved through 300 mm to C ′; a force of 240 N is required to hold the cord at C ′. A 0.1 kg pellet is placed at C ′, and the cord is released. Determine the speed of the pellet as it passes through C.

SOLUTION Let  = undeformed length of cord. Position 1.

Length AC ′B = 1.0 m; Elongation = x1 = 1.0 − 

3  ΣFx = 0: 2  F1  − 240 N = 0 F1 = 200 N 5 

Position 2.

Length ACB = 0.8 m; Elongation = x2 = 0.8 −  Given

F2 = 40 N F1 = kx, F2 = k x2 F1 − F2 = k ( x1 − x2 ) 200 − 40 = k [(1.0 − ) − (0.8 − )] = 0.2k k=

160 = 800 N/m 0.2

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PROBLEM 13.61 (Continued)

Position : Position:

T1 = 0

x1 =

F1 200 N = = 0.25 m k 800 N/m

x2 =

F2 40 N = = 0.05 m k 800 N/m

V1 =

1 2 1 kx1 = (800 N/m)(0.25 m) 2 = 25.0 N ⋅ m 2 2

m = 0.10 kg 1 2 1 mv2 = (0.1 kg) v22 = 0.05 v22 2 2 1 1 V2 = k x22 = (800 N/m)(0.05 m) 2 = 1 N ⋅ m 2 2 T2 =

Conservation of energy: T1 + V1 = T2 + V2 0 + 25.0 N ⋅ m = 0.05v22 + 1.0 N ⋅ m 24.0 = 0.05v22

v2 = 21.909 m/s

v2 = 21.9 m/s 

Note: The horizontal force applied at the midpoint of the cord is not proportional to the horizontal distance C ′C. A solution based on the work of the horizontal force would be rather involved.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 592

PROBLEM 13.62 An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 ft long when unstretched, and to stretch to a total length of 100 ft when a 600-lb weight is attached to it and dropped from the tower. Determine (a) the required spring constant k of the cable, (b) how close to the ground a 186-lb man will come if he uses this cable to jump from the tower.

SOLUTION (a)

Conservation of energy: V1 = 0 T1 = 0 V1 = 100 W

V1 = (100 ft)(600 lb)

Datum at :

= 6 × 104 ft ⋅ lb V2 = 0 T2 = 0

V2 = Vg + Ve = 0 +

1 k (15 ft) 2 2

T1 + V1 = T2 + V2 0 + 6 × 104 = 0 + (112.5)k k = 533 lb/ft 

 (b)

From (a),

k = 533 lb/ft T1 = 0 W = 186 lb V1 = (186)(130 − d ) T2 = 0

Datum:

1 (533)(130 − 85 − d ) 2 2 V2 = (266.67)(45 − d )2 V2 = Vg + Ve = 0 +

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PROBLEM 13.62 (Continued)

d = distance from the ground

T1 + V1 = T2 + V2

0 + (186)(130 − d ) = 0 + (266.67)(45 − d ) 2 266.7 d 2 − 23815d + 515827 = 0

d=

23815  (23815) 2 − 4(266.7)(515827) 36.99 ft = 52.3 ft (2)(266.7)

Discard 52.3 ft (since the cord acts in compression when rebound occurs). d = 37.0 ft 

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PROBLEM 13.63 It is shown in mechanics of materials that the stiffness of an elastic cable is k = AE/L where A is the cross sectional area of the cable, E is the modulus of elasticity and L is the length of the cable. A winch is lowering a 4000-lb piece of machinery using at a constant speed of 3ft/s when the winch suddenly stops. Knowing that the steel cable has a diameter of 0.4 in., E = 29 × 106 lb/in2, and when the winch stops L = 30 ft, determine the maximum downward displacement of the piece of machinery from the point it was when the winch stopped.

SOLUTION m=

Mass of machinery:

W 4000 = = 124.22 lb ⋅ s 2 /ft g 32.2

Let position 1 be the state just before the winch stops and the gravitational potential Vg be equal to zero at this state. For the cable,

A=

π 4

(diameter)2 =

π 4

(0.4 in.)2 = 0.12566 in 2

AE = (0.12556 in.2 )(29 × 106 lb/in 2 ) = 3.6442 × 106 lb AE 3.6442 × 106 lb = = 121.47 × 103 lb/ft L 30 ft

For

L = 30 ft,

Initial force in cable (equilibrium):

F1 = W = 4000 lb.

Elongation in position 1:

x1 =

F1 4000 = = 0.03293 ft k 121.47 × 103

Potential energy:

V1 =

1 2 F12 kx1 = 2 2k

V1 =

(4000 lb) 2 = 65.860 ft ⋅ lb (2)(121.47 × 103 lb/ft)

T1 =

1 2 mv1 2

T1 =

1 (124.22 lb ⋅ s 2 /ft)(3 ft/s2 ) = 558.99 ft ⋅ lb 2

Kinetic energy:

k=

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PROBLEM 13.63 (Continued)

Let position 2 be the position of maximum downward displacement. Let x2 be the elongation in this position. Potential energy:

V2 =

1 2 kx2 − W ( x2 − x1 ) 2

1 (121.47 × 103 ) x22 − (4000)( x2 − 0.03293) 2 = 60.735 × 103 x 2 − 4000 x2 + 131.72

V2 =

Kinetic energy:

T2 = 0

(since v2 = 0)

Principle of work and energy:

T1 + V1 = T2 + V2 558.99 + 65.860 = 60.735 × 103 x22 − 4000 x2 + 131.72 60.735 × 103 x22 − 4000 x2 − 493.13 = 0 x2 = 0.12887 ft

Maximum displacement:

δ = x2 − x1 = 0.09594 ft

δ = 1.151 in.

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PROBLEM 13.64 A 2-kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when the collar is at C and its constant is 600 N/m. If the collar is released at A with no initial velocity, determine its velocity (a) as it passes through B, (b) as it reaches C.

SOLUTION Spring elongations: At A,

x A = 250 mm − 150 mm = 100 mm = 0.100 m

At B,

xB = 200 mm − 150 mm = 50 mm = 0.050 m

At C ,

xC = 0

Potential energies for springs. 1 2 1 kx A = (600)(0.100) 2 = 3.00 J 2 2 1 2 1 (VB )e = kxB = (600)(0.050) 2 = 0.75 J 2 2 (VC )e = 0 (VA )e =

Gravitational potential energies: Choose the datum at level AOC. (VA ) g = (VC ) g = 0 (VB ) g = −mg y = −(2)(9.81)(0.200) = −3.924 J

Kinetic energies:

TA = 0 1 2 mvB = 1.00 vB2 2 1 TC = mvC2 = 1.00 vC2 2 TB =

(a)

Velocity as the collar passes through B. Conservation of energy:

TA + VA = TB + VB 0 + 3.00 + 0 = 1.00 vB2 + 0.75 − 3.924 vB2 = 6.174 m 2 /s 2

v B = 2.48 m/s



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PROBLEM 13.64 (Continued)

(b)

Velocity as the collar reaches C. Conservation of energy:

TA + VA = TC + VC

0 + 3.00 + 0 = 1.00vC2 + 0 + 0 vC2 = 3.00 m 2 /s 2

vC = 1.732 m/s 

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PROBLEM 13.65 A 1-kg collar can slide along the rod shown. It is attached to an elastic cord anchored at F, which has an undeformed length of 250 mm and a spring constant of 75 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E.

SOLUTION LAF = (0.5) 2 + (0.4) 2 + (0.3) 2 LAF = 0.70711 m LBF = (0.4) 2 + (0.3) 2 LBF = 0.5 m LFE = (0.5) 2 + (0.3) 2 LFE = 0.58309 m V = Ve + Vg

(a)

Speed at B:

v A = 0,

TA = 0

Point A: (VA )e =

1 k (Δ LAF ) 2 2

Δ LAF = LAF − L0 = 0.70711 − 0.25 Δ LAF = 0.45711 m

1 (75 N/m)(0.45711 m) 2 2 (VA )e = 7.8355 N ⋅ m (VA )e =

(VA ) g = ( mg )(0.4) = (1.0 kg)(9.81 m/s 2 )(0.4 m) = 3.9240 N ⋅ m VA = (VA )e + (VA ) g = 7.8355 + 3.9240 = 11.7595 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 599

PROBLEM 13.65 (Continued)

Point B:

1 2 1 mvB = (1.0 kg)vB2 2 2 2 TB = 0.5 vB TB =

(VB )e =

1 k (Δ LBF ) 2 2

Δ LBF = LBF − L0 = 0.5 − 0.25 Δ LBF = 0.25 m

1 (75 N/m)(0.25 m) 2 = 2.3438 N ⋅ m 2 (VB ) g = ( mg )(0.4) = (1.0 kg)(9.81 m/s2 )(0.4 m) = 3.9240 N ⋅ m (VB )e =

VB = (vB )e + (VB ) g = 2.3438 + 3.9240 = 6.2678 N ⋅ m TA + VA = TB + VB 0 + 11.7595 = 0.5 vB2 + 6.2678 vB2 = (5.49169)/(0.5) vB2 = 10.983 m 2 /s 2

(b)

vB = 3.31 m/s 

Speed at E: Point A:

TA = 0

VA = 11.7595 N ⋅ m (from part (a))

Point E: 1 2 1 mvE = (1.0 kg)vE2 = 0.5vE2 2 2 1 (VE )e = k ( Δ LFE ) 2 Δ LFE = LFE − L0 = 0.5831 − 0.25 2 TE =

ΔLFE = 0.3331 m 1 (75 N/m)(0.3331 m)2 = 4.1607 N ⋅ m 2 (VE ) g = 0 VE = 4.1607 N ⋅ m (VE )e =

TA + VA = TE + VE

0 + 11.7595 = 0.5 vE2 + 4.1607

vE2 = 7.5988/0.5 vE2 = 15.1976 m 2 /s 2

vE = 3.90 m/s  

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PROBLEM 13.66 A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 3 lb/ft and undeformed length equal to the arc of circle AB. An 8-oz collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released from rest at an angle θ with the vertical, determine (a) the smallest value of θ for which the collar will pass through D and reach Point A, (b) the velocity of the collar as it reaches Point A.

SOLUTION (a)

Smallest angle θ occurs when the velocity at D is close to zero. vC = 0

vD = 0

TC = 0

TD = 0

V = Ve + Vg

Point C: Δ LBC = (1 ft)(θ ) = θ ft 1 k (ΔLBC ) 2 2 3 (VC )e = θ 2 2 (VC )e =

R = 12 in. = 1 ft

(VC ) g = WR(1 − cos θ )  8 oz  (VC ) g =   (1 ft)(1 − cosθ )  16 oz/lb  (VC ) g =

1 (1 − cos θ ) 2

3 1 VC = (VC )e + (VC ) g = θ 2 + (1 − cos θ ) 2 2

Point D: (VD )e = 0 (spring is unattached) (VD ) g = W (2 R ) = (2)(0.5 lb)(1 ft) = 1 lb ⋅ ft TC + VC = TD + VD

By trial,

3 1 0 + θ 2 + (1 − cos θ ) = 1 2 2 2 (1.5)θ − (0.5) cos θ = 0.5

θ = 0.7592 rad

θ = 43.5° 

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PROBLEM 13.66 (Continued)

(b)

Velocity at A: Point D: VD = 0 TD = 0 VD = 1 lb ⋅ ft[see Part (a)] 

Point A: TA =

1 2 1 (0.5 lb) 2 mv A = vA 2 2 (32.2 ft/s 2 )

TA = 0.0077640v A2 VA = (VA ) g = W ( R) = (0.5 lb)(1 ft) = 0.5 lb ⋅ ft TA + VA = TD + VD 0.0077640v A2 + 0.5 = 0 + 1 v A2 = 64.4 ft 2 /s 2

vA = 8.02 ft/s  

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PROBLEM 13.67 The system shown is in equilibrium when φ = 0. Knowing that initially φ = 90° and that block C is given a slight nudge when the system is in that position, determine the velocity of the block as it passes through the equilibrium position φ = 0. Neglect the weight of the rod.

SOLUTION Find the unstretched length of the spring. 1.1 0.3 = 1.3045 rad θ = 74.745°

θ = tan −1

LBD = (1.1) 2 + .32 LBD = 1.140 ft

Equilibrium

ΣM A = (0.3)( Fs sin θ ) − (25)(2.1) = 0 (25 lb)(2.1 ft) (0.3 ft)(sin 74.745°) = 181.39 lb Fs = k ΔLBD Fs =

181.39 lb = (600 lb/ft)(ΔLBD ) ΔLBD = 0.30232 ft

Unstretched length

L0 = LBD − ΔLBD L0 = 1.140 − 0.3023 = 0.83768 ft

′ , when φ = 90°. Spring elongation, ΔLBD ′ = (1.1 ft + 0.3 ft) − L0 ΔLBD ′ = 1.4 ft − 0.8377 ft ΔLBD = 0.56232 ft

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PROBLEM 13.67 (Continued)

At , (φ = 90°)

v1 = 0, T1 = 0 V1 = (V1 )e + (V1 ) g

1 ′ )2 k (ΔLBD 2 1 (V1 )e = (600 lb/ft)(0.5623 ft) 2 2 (V1 )e = 94.86 lb ⋅ ft

(V1 )e =

(V1 ) g = −(25 lb)(2.1 ft) = −52.5 ft ⋅ lb V1 = 94.86 − 52.5 = 42.36 ft ⋅ lb

At , (φ = 0°) 1 1 k ( ΔLBD ) 2 = (600 lb/ft)(0.3023 ft) 2 2 2 (V2 )e = 27.42 lb ⋅ ft (V2 )e =

(V2 ) g = 0

V2 = 27.42 ft ⋅ lb

1 2 1  25 lb mv2 =  2 2  32.2 ft/s 2 T1 + V1 = T2 + V2 T2 =

 2 2  v2 = 0.3882 v2 

0 + 42.36 = 0.3882 v22 + 27.42 v22 = (14.941)/(0.3882) v22 = 38.48 ft 2 /s 2

v2 = 6.20 ft/s 

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PROBLEM 13.68 A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

SOLUTION Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the spring. Use the principle of conservation of energy. T1 + V1 = T2 + V2 . Position 1:

1 2 1 mv1 = (50)(2)2 = 100 J 2 2 V1g = mgh1 = (50)(9.81)(8 sin 20°) = 1342.09 J T1 =

V1e =

1 2 1 ke1 = (30 × 103 )(0.050) 2 = 37.5 J 2 2

1 2 mv2 = 0 since v2 = 0. 2 = mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x

T2 =

Position 2:

V2 g

V2e =

1 2 1 ke2 = (30 × 103 )(0.05 + x)2 = 37.5 + 1500 x + 15000 x 2 2 2

Principle of conservation of energy: 100 + 1342.09 + 37.5 = −167.61x + 37.5 + 1500 x + 15000 x 2 15, 000 x 2 + 1332.24 x − 1442.09 = 0

Solving for x, x = 0.26882 and − 0.357 64

x = 0.269 m  

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PROBLEM 13.69 Solve Problem 13.68 assuming the kinetic coefficient of friction between the package and the incline is 0.2. PROBLEM 13.68 A spring is used to stop a 50-kg package which is moving down a 20° incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

SOLUTION Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the spring. Use the principle of work and energy. T1 + V1 + U1→2 = T2 + V2 Position 1.

1 2 1 mv1 = (50)(2) 2 = 100 J 2 2 V1g = mgh1 = (50)(9.81)(8sin 20°) = 1342.09 J T1 =

V1e =

Position 2.

1 2 1 ke1 = (30 × 103 )(0.05) 2 = 37.5 J 2 2

1 2 mv2 = 0 since v2 = 0. 2 = mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x

T2 = V2 g

V2e =

1 2 1 ke2 = (30 × 103 )(0.05 + x) 2 = 37.5 + 1500 x + 15,000 x 2 2 2

Work of the friction force. ΣFn = 0 N − mg cos 20° = 0 N = mg cos 20° = (50)(9.81) cos 20° = 460.92 N F f = μk N

U1→2

= (0.2)(460.92) = 92.184 = − Ff d = −92.184(8 + x) = −737.47 − 92.184 x

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PROBLEM 13.69 (Continued)

Principle of work and energy:

T1 + V1 + U1− 2 = T2 + V2 100 + 1342.09 + 37.5 − 737.47 − 92.184 x

= −167.76 x + 37.5 + 1500 x + 15, 000 x 2 15, 000 x 2 + 1424.42 x − 704.62 = 0

Solving for x, x = 0.17440 and −0.26936

x = 0.1744 m 

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PROBLEM 13.70 A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach Point A with practically no velocity and then drop freely along the track. Determine the normal force exerted by the track on the car as the car reaches point B. Ignore air resistance and rolling resistance.

SOLUTION Calculate the speed of the car as it reaches Point B using the principle of conservation of energy as the car travels from position A to position B. Position A:

v A = 0,

TA =

Position B:

VB = −mgh

1 2 mv A = 0, VA = 0 (datum) 2

where h is the decrease in elevation between A and B. TB =

Conservation of energy:

1 2 mvB 2

TA + VA = TB + VB : 1 2 mvB − mgh 2 vB2 = 2 gh

0+0=

= (2)(9.81 m/s 2 )(27 m)(1 − cos 40°) = 123.94 m 2 /s 2

Normal acceleration at B: ( aB ) n =

vB2

ρ

=

123.94 m 2 /s 2 = 4.59 m/s 2 27 m

(a B )n = 4.59 m/s 2

50°

Apply Newton’s second law to the car at B.

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PROBLEM 13.70 (Continued)

50°ΣFn = man : N − mg cos 40° = − man N = mg cos 40° − man = m( g cos 40° − an ) = (250 kg)[(9.81 m/s 2 ) cos 40° − 4.59 m/s 2 ] = 1878.7 − 1147.5

N = 731 N 

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PROBLEM 13.71 A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach Point A with practically no velocity and then drop freely along the track. Determine the maximum and minimum values of the normal force exerted by the track on the car as the car travels from A to D. Ignore air resistance and rolling resistance.

SOLUTION Calculate the speed of the car as it reaches Point P, any point on the roller coaster track. Apply the principle of conservation of energy. Position A:

v A = 0,

TA =

Position P:

VP = −mgh

1 2 mv A = 0, VA = 0 (datum) 2

where h is the decrease in elevation along the track. TP =

Conservation of energy:

1 mv 2 2

TA + VA = TP + VP 0+0=

1 2 mv − mgh 2

v 2 = 2 gh

(1)

Calculate the normal force using Newton’s second law. Let θ be the slope angle of the track.

ΣFn = man : N − mg cos θ = man N = mg cos θ + man

Over portion AB of the track, and

(2)

h = ρ (1 − cos θ ) an = −

mv 2

ρ

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PROBLEM 13.71 (Continued)

where ρ is the radius of curvature. ( ρ = 27 m) N = mg cos θ −

2mg ρ (1 − cos θ )

ρ

= mg (3cos θ − 2)

At Point A (θ = 0)

N A = mg = (250)(9.81) = 2452.5 N

At Point B (θ = 40°)

N B = (2452.5)(3cos 40° − 2) N B = 731 N

Over portion BC,

θ = 40°,

an = 0

(straight track)

N BC = mg cos 40° = 2452.5cos 40° N BC = 1879 N

Over portion CD,

h = hmax − r (1 − cos θ ) an =

and

mv 2 r

where r is the radius of curvature. (r = 72 m) 2mgh r h  = mg cos θ + 2mg  max − 1 − cos θ   r 

N = mg cos θ +

2h   = mg  3cos θ − 2 + max  r  

which is maximum at Point D, where 2h   N D = mg 1 + max  r  

Data:

hmax = 27 + 18 = 45 m,

r = 72 m

 (2) (45)  N D = (2452.5) 1 + = 5520 N 72  

Summary: 

minimum (just above B):

731 N  5520 N 

maximum (at D):

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PROBLEM 13.72 A 1-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant k = 10 lb/ft. Knowing that the collar is released from being held at A determine the speed of the collar and the normal force between the collar and the rod as the collar passes through B.

SOLUTION W 1 = = 0.031056 lb ⋅ s 2 /ft g 32.2

For the collar,

m=

For the spring,

k = 10 lb/ft l0 = 5 in.

 A = 7 + 5 + 5 = 17 in.

At A:

 Δ −  0 = 12 in. = 1 ft  B = (7 + 5) 2 + 52 = 13 in.

At B:

 B −  0 = 1.8 in. =

2 ft 3

Velocity of the collar at B. Use the principle of conservation of energy. TA + VA = TB + VB

Where

TA =

1 2 mv A = 0 2

1 k ( A −  0 ) 2 + W (0) 2 1 = (10)(1) 2 + 0 = 5 ft ⋅ lb 2 1 2 1 TB = mvB = (0.031056)vB2 = 0.015528vB2 2 2 1 VB = k ( B −  0 ) 2 + Wh 2 VA =

2

1 2  5 (10)   + (1)  −  2 3  12  = 1.80556 ft ⋅ lb =

0 + 5 = 0.015528vB2 = 1.80556 vB = 14.34 ft/s 

vB2 = 205.72 ft 2 /s 2

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PROBLEM 13.72 (Continued)

Forces at B. 2 Fs = k ( B −  0 ) = (10)   = 6.6667 lb. 3 5 sinα = 13 5 ρ = 5 in. = ft 12 mvB2 man =

ρ

(0.031056)(205.72) 5/12 = 15.3332 lb =

ΣFy = ma y : Fs sin α − W + N = man N = man + W − Fs sin α  5 = 15.3332 + 1 − (6.6667)    13  N = 13.769 lb

N = 13.77 lb 

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PROBLEM 13.73 A 10-lb collar is attached to a spring and slides without friction along a fixed rod in a vertical plane. The spring has an undeformed length of 14 in. and a constant k = 4 lb/in. Knowing that the collar is released from rest in the position shown, determine the force exerted by the rod on the collar at (a) Point A, (b) Point B. Both these points are on the curved portion of the rod.

SOLUTION Mass of collar:

m=

W 10 = = 0.31056 lb ⋅ s 2 /ft g 32.2

Let position 1 be the initial position shown, and calculate the potential energies of the spring for positions 1, A, and B. l0 = 14 in. l1 = (14 + 14)2 + (14) 2 = 31.305 in. x1 = l1 − l0 = 31.305 − 14 = 17.305 in. (V1 )e =

1 2 1 kx1 = (4)(17.305) 2 = 598.92 in ⋅ lb = 49.910 ft ⋅ lb 2 2

l A = (14) 2 + (14) 2 = 19.799 in. x A = l A − l0 = 19.799 − 14 = 5.799 in. (VA )e =

1 2 1 kx A = (4)(5.799)2 = 67.257 in ⋅ lb = 5.605 ft ⋅ lb 2 2

lB = 14 + 14 = 28 in. xB = lB − l0 = 28 − 14 = 14 in. (VB )e =

Gravitational potential energies:

1 2 1 kxB = (4)(14) 2 = 392 in ⋅ lb = 32.667 ft ⋅ lb 2 2

Datum at level A. (V1 ) g = 0 (VA ) g = 0 (VB ) g = Wy = (10 lb)(−14 in.) = −140 in ⋅ lb = −11.667 ft ⋅ lb

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 614

PROBLEM 13.73 (Continued)

Total potential energies:

V = Ve + Vg V1 = 49.910 ft ⋅ lb, VA = 5.605 ft ⋅ lb, VB = 21.0 ft ⋅ lb

Kinetic energies:

T1 = 0 1 2 1 mv A = (0.31056)v A2 = 0.15528v A2 2 2 1 2 1 TB = mvB = (0.31056)vB2 = 0.15528vB2 2 2 TA =

Conservation of energy:

T1 + V1 = TA + VA : 0 + 49.910 = 0.15528v 2A + 5.605

Conservation of energy:

T1 + V1 = TB + VB 0 + 49.910 = 0.15528vB2 + 21.0

Normal accelerations at A and B.

v 2A = 285.32 ft 2 /s 2

vB2 = 186.18 ft 2 /s2

an = v 2 /ρ

ρ = 14 in. = 1.16667 ft

Spring forces at A and B:

(a A )n =

285.32 ft 2 /s 2 1.16667 ft

(a A )n = 244.56 ft/s 2

( aB ) n =

189.10 ft 2 /s 2 1.16667 ft

(a B ) n = 159.58 ft/s 2

F = kx FA = (4 lb/in.)(5.799 in.) FB = (4 lb/in.)(14 in.)

FA = 23.196 lb

45°

FB = 56.0 lb

To determine the forces (N A and N B ) exerted by the rod on the collar, apply Newton’s second law. (a)

At Point A: ΣF = m ( a A ) n : W + N A − FA sin 45° = m( a A ) n 10 + N A − 23.196sin 45° = (0.31056)(244.56) N A = 82.4 lb 

(b) At Point B: ΣF = m ( a B ) n : N B = (0.31056)(159.58)

N B = 49.6 lb



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PROBLEM 13.74 An 8-oz package is projected upward with a velocity v0 by a spring at A; it moves around a frictionless loop and is deposited at C. For each of the two loops shown, determine (a) the smallest velocity v0 for which the package will reach C, (b) the corresponding force exerted by the package on the loop just before the package leaves the loop at C.

SOLUTION Loop 1

(a) The smallest velocity at B will occur when the force exerted by the tube on the package is zero.

ΣF = 0 + mg =

mvB2 r

vB2 = rg = 1.5 ft(32.2 ft/s 2 ) vB2 = 48.30 TA =

At A

1 2 mv0 2

0.5   VA = 0  8 oz = 0.5 lb  = = 0.01553  32.2   TB =

At B

1 2 1 mvB = m (48.30) = 24.15 m 2 2

VB = mg (7.5 + 1.5) = 9 mg = 9(0.5) = 4.5 lb ⋅ ft TA + VA = TB + VB :

1 (0.01553)v02 = 24.15(0.01553) + 4.5 2

v02 = 627.82



v0 = 25.056 

v0 = 25.1 ft/s 

At C TC =

1 2 mvC = 0.007765vC2 2

VC = 7.5 mg = 7.5(0.5) = 3.75

TA + VA = TC + VC : 0.007765v02 = 0.007765vC2 + 3.75 0.007765(25.056) 2 − 3.75 = 0.007765vC2 vC2 = 144.87 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 616

PROBLEM 13.74 (Continued)

(b)

ΣF = man : N = 0.01553

(144.87) 1.5

N = 1.49989

Loop 2

{Package in tube} NC = 1.500 lb



(a) At B, tube supports the package so, vB ≈ 0 vB = 0, TB = 0

VB = mg (7.5 + 1.5) = 4.5 lb ⋅ ft

TA + VA = TB + VB 1 (0.01553)v A2 = 4.5  v A = 24.073 2 v A = 24.1 ft/s  TC = 0.007765vC2 , VC = 7.5 mg = 3.75

(b) At C

TA + VA = TC + VC : 0.007765(24.073) 2 = 0.007765vC2 + 3.75 vC2 = 96.573

 96.573  NC = 0.01553   = 0.99985  1.5 

{Package on tube} NC = 1.000 lb 

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PROBLEM 13.75 If the package of Problem 13.74 is not to hit the horizontal surface at C with a speed greater than 10 ft/s, (a) show that this requirement can be satisfied only by the second loop, (b) determine the largest allowable initial velocity v 0 when the second loop is used.

SOLUTION (a)

Loop 1 From Problem 13.74, at B

vB2 = gr = 48.3 ft 2 /s 2  vB = 6.9498 ft/s TB =

1 2 1 mvB = (0.01553)(48.3) = 0.37505 2 2

VB = mg (7.5 + 1.5) = (0.5)(9) = 4.5 lb ⋅ ft TC =

1 2 1 mvC = (0.01553)vC2 = 0.007765vC2 2 2

VC = 7.5(0.5) = 3.75 lb ⋅ ft TB + VB = TC + VC : 0.37505 + 4.5 = 0.007765vC2 + 3.75 vC2 = 144.887  vC = 12.039 ft/s 12.04 ft/s > 10 ft/s  Loop (1) does not work  1 TA = mv02 = 0.007765v02 2 VA = 0

(b) Loop 2 at A

vC = 10 ft/s

At C assume TC =

1 2 mvC = 0.007765(10) 2 = 0.7765 2 vC = 7.5(0.5) = 3.75

TA + VA = TC + VC : 0.007765v02 = 0.7765 + 3.75

 v0 = 24.144 



v0 = 24.1 ft/s 

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PROBLEM 13.76 A small package of weight W is projected into a vertical return loop at A with a velocity v0. The package travels without friction along a circle of radius r and is deposited on a horizontal surface at C. For each of the two loops shown, determine (a) the smallest velocity v0 for which the package will reach the horizontal surface at C, (b) the corresponding force exerted by the loop on the package as it passes Point B.

SOLUTION Loop 1: (a)

Newton’s second law at position C: ΣF = ma: mg = m

vc2 r

vc2 = gr

Conservation of energy between position A and B. 1 2 mv0 2 VA = 0 TA =

1 2 1 mvC = mgr 2 2 VC = mg (2r ) = 2mgr TC =

TA + VA = TC + VC :

1 2 1 mv0 + 0 = mgr + 2mgr 2 2 v02 = 5gr v0 =

Smallest velocity v 0:

5 gr



(b) Conservation of energy between positions A and B. (b) TB =

1 2 mvB ; 2

VB = mg (r )

TA + VA = TB + VB:

1 2 1 mv A + 0 = mvB2 + mgr 2 2

1 1 m(5 gr ) + 0 = mvB2 + mgr  2 2

vB2 = 3gr

 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 619

PROBLEM 13.76 (Continued) Newton’s second law at position B. man = m

vB2 3gr =m = 3 mg r r

ΣF = ΣFeff :

N B = 3 mg N B = 3W

Force exerted by the loop:



Loop 2: (a)

At point C, vc = 0 Conservation of energy between positions A and C. 1 2 mvC = 0 2 VC = mg (2r ) = 2mgr TC =

TA + VA = TC + VC : 1 2 mv0 + 0 = 0 + 2mgr 2 v02 = 4 gr v0 =

Smallest velocity v 0: (b)

4 gr



Conservation of energy between positions A and B. TA + VA − TB + VB :

1 2 1 mv0 + 0 = mvB2 + mgr 2 2 1 1 m(4 gr ) = mvB2 + mgr  vB2 = 2 gr 2 2

Newton’s second law at position B. man = m

vB2 2 gr =m = 2 mg r r

ΣF = Σeff : N = 2 mg

Force exerted by loop:

N = 2W



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PROBLEM 13.77 The 1 kg ball at A is suspended by an inextensible cord and given an initial horizontal velocity of 5 m/s. If l = 0.6 m and xB = 0, determine yB so that the ball will enter the basket.

SOLUTION Let position 1 be at A.

v1 = v0

Let position 2 be the point described by the angle where the path of the ball changes from circular to parabolic. At position 2, the tension Q in the cord is zero. Relationship between v2 and θ based on Q = 0. Draw the free body diagram.

ΣF = 0: Q + mg sin θ = man =

With

Q = 0, v22 = gl sin θ

mv22 l

or v2 = gl sin θ

(1)

Relationship among v0 , v2 and θ based on conservation of energy. T1 + V1 = T2 + V2 1 2 1 mv0 − mgl = mv22 + mgl sin θ 2 2 v02 − v22 = 2 gl (1 + sin θ )

(2)

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PROBLEM 13.77 (Continued)

Eliminating v2 from Eqs. (1) and (2), v02 − gl sin θ = 2 gl (1 + sin θ )  1  (5) 2  1  v02 − 2  = 0.74912  − 2 =  3  gl  3  (9.81)(0.6)  θ = 48.514°

sin θ =

From Eq. (1),

v22 = (9.81)(0.6)sin 48.514° = 4.4093 m 2/s 2 v2 = 2.0998 m/s

x and y coordinates at position 2. x2 = l cos θ = 0.6cos 48.514° = 0.39746 m y2 = l sin θ = 0.6sin 48.514° = 0.44947 m

Let t2 be the time when the ball is a position 2. Motion on the parabolic path. The horizontal motion is x = −v2 sin θ = −2.0998 sin 48.514° = −1.5730 m/s x = x2 − 1.5730(t − t2 )

At Point B,

xB = 0 0 = 0.39746 − 1.5730 (t B − t2 ) t B − t2 = 0.25267 s

The vertical motion is y = y2 + v2 cos θ (t − t2 ) −

At Point B,

1 g (t − t2 ) 2 2

yB = y2 + v2 cos θ (t B − t2 ) −

1 g (t B − t2 ) 2 2

yB = 0.44947 + (2.0998 cos 48.514°)(0.25267) 1 − (9.81)(0.25267) 2 2 = 0.48779 m

yB = 0.448 m 

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PROBLEM 13.78* Packages are moved from Point A on the upper floor of a warehouse to Point B on the lower floor, 12 ft directly below A, by means of a chute, the centerline of which is in the shape of a helix of vertical axis y and radius R = 8 ft. The cross section of the chute is to be banked in such a way that each package, after being released at A with no velocity, will slide along the centerline of the chute without ever touching its edges. Neglecting friction, (a) express as a function of the elevation y of a given Point P of the centerline the angle φ formed by the normal to the surface of the chute at P and the principal normal of the centerline at that point, (b) determine the magnitude and direction of the force exerted by the chute on a 20-lb package as it reaches Point B. Hint: The principal normal to the helix at any Point P is horizontal and directed toward the y axis, and the radius of curvature of the helix is ρ = R[1 – (h/2πR)2].

SOLUTION (a)

v A = 0 TA = 0

At Point A:

VA = mgh TP =

At any Point P:

1 2 mv 2

VP = Wy = mgy TA + VA = TP + VP 1 2 mv + mgy 2 v 2 = 2 g (h − y )

0 + mgh =

en along principal normal, horizontal and directed toward y axis et tangent to centerline of the chute eD along binormal

β = tan −1

h (12 ft) = tan −1 = 13.427° 2π R 2π (8 ft)

mab = 0

since

ab = 0

Note: Friction is zero, ΣFt = mat : mg sin β = mat

at = g sin β

ΣFb = mab : N b − W cos β = 0 N B = W cos β ΣFn = man : N n =

mv 2 m 2 g (h − y ) (h − y) = = 2W e e e

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 623

PROBLEM 13.78* (Continued)

The total normal force is the resultant of Nb and Nm, it lies in the b–m plane, and forms angle φ with m axis. tan φ = Nb /N n 2( w(h − y ) e tan φ = (e /2(h − y )) cos β tan φ = W cos β

  h 2  R e = R 1 +   = R(1 + tan 2 β ) =  cos 2 β   2π R  

Given:

e R cos β = 2( h − y ) 2( h − y ) cos β 8 ft 4.112 tan φ = = 2(12 − y ) cos13.427° 12 − y tan φ =

Thus,

cot φ = 0.243(12 − y ) 

or (b)

At Point B:

y=0

for x, y, z axes, we write, with W = 20 lb, N x = Nb sin β = W cos β sin β = (20 lb) cos 13.427° sin13.427° N x = 4.517 lb N y = Nb cos β = W cos 2 β = (20 lb) cos 2 13.427° N y = 18.922 lb N z = − N n = −2w N z = −2(20 lb)

h− y h− y = −2W e R/cos 2 β

(12 ft-0) cos 2 13.427° N z = −56.765 lb 8 ft

N = (4.517) 2 + (18.922) 2 + (−56.765) 2

N = 60.0 lb 

cos θ x =

N x 4.517 = N 60

θ x = 85.7° 

cos θ y =

Ny

θ y = 71.6° 

cos θ z =

Nz 56.742 =− N 60

N

=

18.922 60

θ z = 161.1° 

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PROBLEM 13.79* Prove that a force F(x, y, z) is conservative if, and only if, the following relations are satisfied:

∂ Fx ∂ Fy = ∂y ∂x

∂ Fy ∂z

∂ Fz ∂y

=

∂ Fz ∂ Fx = ∂x ∂z

SOLUTION For a conservative force, Equation (13.22) must be satisfied. Fx = −

We now write

Since

∂V ∂x

∂ Fx ∂ 2V =− ∂y ∂ x∂ y

Fy = −

∂ Fy ∂x

∂V ∂y =−

Fz = −

∂V ∂z

∂ 2V ∂ y∂ x ∂ Fx ∂ Fy =  ∂y ∂x

∂ 2V ∂ 2V = : ∂ x∂ y ∂ y∂ x

We obtain in a similar way

∂ Fy ∂z

=

∂ Fz ∂y

∂ Fz ∂ Fx =  ∂x ∂z

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PROBLEM 13.80 The force F = ( yzi + zxj + xyk )/xyz acts on the particle P( x, y, z ) which moves in space. (a) Using the relation derived in Problem 13.79, show that this force is a conservative force. (b) Determine the potential function associated with F.

SOLUTION Fx =

(a)

yz xyz

Fy =

zx xyz

Fz =

xy xyz

()

∂ Fy ∂ 1y ∂ Fx ∂ ( 1x ) = =0 = =0 ∂y ∂y ∂x ∂x Thus,

∂ Fx ∂ Fy = ∂y ∂x

The other two equations derived in Problem 13.79 are checked in a similar way. (b)

Recall that

Fx = −

∂V , ∂x

Fy = −

∂V , ∂y

Fz = −

∂V ∂z

Fx =

1 ∂V =− x ∂x

V = − ln x + f ( y, z )

(1)

Fy =

1 ∂V =− y ∂y

V = − ln y + g ( z , x)

(2)

Fz =

1 ∂V =− z ∂z

V = − ln z + h( x, y )

(3)

Equating (1) and (2) − ln x + f ( y , z ) = − ln y + g ( z , x)

Thus,

f ( y, z ) = − ln y + k ( z )

(4)

g ( z , x) = − ln x + k ( z )

(5)

Equating (2) and (3) − ln z + h( x, y ) = − ln y + g ( z , x) g ( z , x) = − ln z + l ( x)

From (5), g ( z , x) = − ln x + k ( z )

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PROBLEM 13.80 (Continued)

Thus, k ( z ) = − ln z l ( x) = − ln x

From (4), f ( y, z ) = − ln y − ln z

Substitute for f ( y, z ) in (1) V = − ln x − ln y − ln z V = − ln xyz 



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PROBLEM 13.81* A force F acts on a particle P(x, y) which moves in the xy plane. Determine whether F is a conservative force and compute the work of F when P describes in a clockwise sense the path A, B, C, A including the quarter circle x 2 + y 2 = a 2, if (a) F = kyi, (b) F = k ( yi + xj).

SOLUTION (a)

Fx = ky

∂ Fy

∂ Fx =k ∂y

Fy = 0

=0

∂x

∂ Fx ∂ Fy ≠ ∂y ∂x

Thus,

F is not conservative.



UABCA =

F ⋅ dr =

ABCA

 

C B

B A



B A

kyi ⋅ dyj +



C B

kyi ⋅ (dxi + dyj) +



A B

kyi ⋅ dxj

= 0, F is perpendicular to the path.

kyi ⋅ (dxi + dyj) =



C

ky dx

B

From B to C, the path is a quarter circle with origin at A. x2 + y2 = a2

Thus,

y = a2 − x2



Along BC,



A C

C B

kydx =

Fx = k y

Fy = kx

0

k a 2 − x 2 dx =

π ka 2 4

kyi ⋅ dx j = 0 ( y = 0 on CA) UABCA =

(b)



a

B

C

   A

+

B

+

A C

=0+

π ka 2 4

+0

U ABCA =

π ka 2 4



∂ Fy ∂ Fx =k =k ∂y ∂x ∂ Fx ∂ Fy = , F is conservative. ∂y ∂x

Since ABCA is a closed loop and F is conservative,

U ABCA = 0 

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PROBLEM 13.82* The potential function associated with a force P in space is known to be V ( x, y, z ) = −( x 2 + y 2 + z 2 )1/2. (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D.

SOLUTION (a)

(b)

Px = −

∂V ∂ [− ( x 2 + y 2 + z 2 )1/2 ] =− = x( x 2 + y 2 + z 2 ) −1/ 2 ∂x ∂x



Py = −

∂V ∂ [− ( x 2 + y 2 + z 2 )1/ 2 ] =− = y ( x 2 + y 2 + z 2 ) −1/2 ∂y ∂y



Pz = −

∂V ∂ [−( x 2 + y 2 + z 2 )1/2 ] =− = z ( x 2 + y 2 + z 2 ) −1/ 2 ∂z ∂z



U OABD = U OA + U AB + U BD

O–A: Py and Px are perpendicular to O–A and do no work. x = y = 0 and Pz = 1

Also, on O–A

UO− A =

Thus,



a

0

Pz dz =



a

0

dz = a

A–B: Pz and Py are perpendicular to A–B and do no work. y = 0, z = a and Px =

Also, on A–B

U A− B =

Thus,



a

0

x (x + a 2 )1/ 2 2

xdx ( x + a 2 )1/2 2

= a ( 2 − 1)

B–D: Px and Pz are perpendicular to B–D and do no work. On

B−D,

k =a z=a Py =

y ( y + 2a 2 )1/ 2 2

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PROBLEM 13.82* (Continued)

Thus,

U BD =



a

0

a y dy = ( y 2 + 2a 2 )1/ 2 2 1/ 2 0 ( y + 2a )

2

U BD = (a 2 + 2a 2 )1/ 2 − (2a 2 )1/ 2 = a ( 3 − 2 ) U OABD = U O − A + U A− B + U B − D = a + a ( 2 − 1) + a ( 3 − 2)

U OABD = a 3 

ΔVOD = V (a, a, a ) − V (0, 0, 0) = −(a 2 + a 2 + a 2 )1/ 2 − 0

Thus,

ΔVOD = − a 3 

U OABD = −ΔVOD

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PROBLEM 13.83* (a) Calculate the work done from D to O by the force P of Problem 13.82 by integrating along the diagonal of the cube. (b) Using the result obtained and the answer to part b of Problem 13.82, verify that the work done by a conservative force around the closed path OABDO is zero. PROBLEM 13.82 The potential function associated with a force P in space is known to be V(x, y, z) = −( x 2 + y 2 + z 2 )1/ 2. (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D.

SOLUTION From solution to (a) of Problem 13.82 P=

(a)

U OD =



D O

xi + yj + zk ( x + y 2 + z 2 )1/ 2 2

P ⋅ dr

r = xi + yj + zk dr = dxi + dyj + dzk xi + yj + zk P= 2 ( x + y 2 + z 2 )1/ 2

Along the diagonal. Thus,

x= y=z P ⋅ dr =

UO−D =

3x = 3 (3x 2 )1/2



a

0

3 dx = 3a

U OD = 3a 

U OABDO = U OABD + U DO

(b) From Problem 13.82

U OABD = 3a

at left

The work done from D to O along the diagonal is the negative of the work done from O to D. U DO = −U OD = − 3a

[see part (a)]

Thus, U OABDO = 3a − 3a = 0



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PROBLEM 13.84* The force F = ( xi + yj + zk )/(x 2 + y 2 + z 2 )3/ 2 acts on the particle P( x, y , z ) which moves in space. (a) Using the relations derived in Problem 13.79, prove that F is a conservative force. (b) Determine the potential function V(x, y, z) associated with F.

SOLUTION Fx =

(a)

x

( x + y + z 2 )3/2 y Fy = 2 2 ( x + y + z 2 )1/2 2

2

x ( − 32 ) (2 y ) ∂ Fx = 2 ∂ y ( x + y 2 + z 2 )5/2

y ( − 32 ) 2 y ∂ Fy = 2 ∂ x ( x + y 2 + z 2 )5/2

Thus,

∂ Fx ∂ Fy = ∂y ∂x

The other two equations derived in Problem 13.79 are checked in a similar fashion. (b)

Recalling that

∂V ∂V ∂V , Fy = − , Fz = − ∂x ∂y ∂z x ∂V Fx = − V =− dx 2 2 ∂x ( x + y + z 2 )3/2 Fx = −



V = ( x 2 + y 2 + z 2 )−1/2 + f ( y , z )

Similarly integrating ∂ V/∂ y and ∂ V/∂ z shows that the unknown function f ( x, y ) is a constant. V=



1 ( x + y + z 2 )1/2 2

2



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PROBLEM 13.85 (a) Determine the kinetic energy per unit mass which a missile must have after being fired from the surface of the earth if it is to to reach an infinite distance from the earth. (b) What is the initial velocity of the missile (called the escape velocity)? Give your answers in SI units and show that the answer to part b is independent of the firing angle.

SOLUTION g = 9.81 m/s 2

At the surface of the earth,

r1 = R = 6370 km = 6.37 × 106 m

Centric force at the surface of the earth, GMm R2 GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 F = mg =

Let position 1 be on the surface of the earth (r1 = R) and position 2 be at r2 = OD. Apply the conservation of energy principle. T1 + V1 = T2 + V2 1 2 GMm 1 2 GMm = mv2 + mv1 − r1 r2 2 2 GMm GMm − R ∞ T1 T2 GM T2 = + = + gR m m R m

T1 = T2 +

For the escape condition set

T2 =0 m T1 = gR = (9.81 m/s 2 )(6.37 × 106 m) = 62.49 × 106 m 2 /s 2 m T1 = 62.5 MJ/kg  m

(a) 1 2 mvesc = gr 2 vesc = 2 gR

(b)

vesc = (2)(9.81)(6.37 × 106 ) = 11.18 × 103 m/s

vesc = 11.18 km/s 

Note that the escape condition depends only on the speed in position 1 and is independent of the direction of the velocity (firing angle). PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 633

PROBLEM 13.86 A satellite describes an elliptic orbit of minimum altitude 606 km above the surface of the earth. The semimajor and semiminor axes are 17,440 km and 13,950 km, respectively. Knowing that the speed of the satellite at Point C is 4.78 km/s, determine (a) the speed at Point A, the perigee, (b) the speed at Point B, the apogee.

SOLUTION rA = 6370 + 606 = 6976 km = 6.976 × 106 m rC = (17440 − 6976) 2 + (13950) 2 = 17438.4 km = 17.4384 × 106 m rB = (2)(17440) − 6976 = 27904 km = 27.904 × 106 m

For earth,

R = 6370 km = 6.37 × 106 m GM = gR 2 = (9.81 m/s 2 )(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2 vC = 4.78 km/s = 4780 m/s

(a)

Speed at Point A: Use conservation of energy. TA + VA = TC + VC 1 2 GMm 1 2 GMm mv A − = mvC − rA rC 2 2

1 1  v A2 = vC2 + 2GM  −   rA rC  1 1   = (4780) 2 + (2)(398.06 × 1012 )  − 6 6  6.976 × 10 17.4384 × 10  = 91.318 × 106 m 2 /s 2 VA = 9.556 × 103 m/s

v A = 9.56 km/s 

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PROBLEM 13.86 (Continued)

(b)

Speed at Point B: Use conservation of energy. TB + VB = TC + VC 1 2 GMm 1 2 GMm mvB − = mvC − rB rC 2 2

1 1  vB2 = vC2 + 2GM  −   rB rC  1 1   = (4780) 2 + (2)(398.06 × 1012 )  − 6 6 × × 27.904 10 17.4384 10   = 5.7258 × 106 m 2 /s 2 vB = 2.39 × 103 m/s

vB = 2.39 km/s 

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PROBLEM 13.87 While describing a circular orbit 200 mi above the earth a space vehicle launches a 6000-lb communications satellite. Determine (a) the additional energy required to place the satellite in a geosynchronous orbit at an altitude of 22,000 mi above the surface of the earth, (b) the energy required to place the satellite in the same orbit by launching it from the surface of the earth, excluding the energy needed to overcome air resistance. (A geosynchronous orbit is a circular orbit in which the satellite appears stationary with respect to the ground).

SOLUTION Geosynchronous orbit r1 = 3960 + 200 = 4160 mi = 21.965 × 106 ft r2 = 3960 + 22,000 = 25,960 mi = 137.07 × 106 ft E = T +V =

Total energy

1 2 GMm mv − 2 r

M = mass of earth m = mass of satellite

Newton’s second law T =

F = man :

1 2 GM mv = m 2 2r

E = T +V = GM = gRE2 E =−

GMm mv 2 GM =  v2 = 2 r r r

V =−

GMm r

1 GMm GMm 1 GMm − =− 2 r r 2 r E =−

1 gRE2 m 1 RE2W =− where (W = mg ) 2 r 2 r

1 (6000)(20.9088 × 106 ft) 2 1.3115 × 1018 =− ft ⋅ lb 2 r r

Geosynchronous orbit at r2 = 137.07 × 106 ft EGs =

−1.3115 × 1018 = −9.5681 × 109 ft ⋅ lb 6 137.07 × 10

(a) At 200 mi, E200 = −

r1 = 21.965 × 106 ft 1.3115 × 1018 = −5.9709 × 1010 21.965 × 106

ΔE300 = EGs − E200 = 5.0141 × 1010 Δ E300 = 50.1 × 109 ft ⋅ lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 636

PROBLEM 13.87 (Continued)

(b)

Launch from earth At launch pad

EE = −

− gRE2 m GMm = = −WRE RE RE

EE = −6000(3960 × 5280) = −1.25453 × 1011 Δ EE = EGs − EE = −9.5681 × 109 + 125.453 × 109

Δ E E = 115.9 × 109 ft ⋅ lb 

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PROBLEM 13.88 A lunar excursion module (LEM) was used in the Apollo moon-landing missions to save fuel by making it unnecessary to launch the entire Apollo spacecraft from the moon’s surface on its return trip to earth. Check the effectiveness of this approach by computing the energy per pound required for a spacecraft (as weighed on the earth) to escape the moon’s gravitational field if the spacecraft starts from (a) the moon’s surface, (b) a circular orbit 50 mi above the moon’s surface. Neglect the effect of the earth’s gravitational field. (The radius of the moon is 1081 mi and its mass is 0.0123 times the mass of the earth.)

SOLUTION Note:

GM moon = 0.0123 GM earth

By Eq. 12.30,

GM moon = 0.0123 gRE2

At ∞ distance from moon: r2 = ∞, Assume v2 = 0 E2 = T2 + V2 GM M m ∞ =0−0 =0− =0

(a)

On surface of moon:

RM = 1081 mi = 5.7077 × 106 ft

v1 = 0 T1 = 0

RE = 3960 mi = 20.909 × 106 ft

V1 = −

GM M m RM

E1 = T1 + V1 = 0 − E1 = −

0.0123 gRE2 m RM

(0.0123)(32.2 ft/s2 )(20.909 × 106 ft)2 m (5.7077 × 106 ft)

WE = Weight of LEM on the earth E1 = ( −30.336 × 106 ft 2 /s 2 )m

m=

WE g

 30.336 × 106 2 2  E1 =  − ft /s WE 2  32.2 ft/s  ΔE = E2 − E1 = 0 + (942.1 × 103 ft ⋅ lb/lb)WE ΔE = 942 × 103 ft ⋅ lb/lb  WE

Energy per pound:

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PROBLEM 13.88 (Continued)

(b)

r1 = RM + 50 mi r1 = (1081 mi + 50 mi) = 1131 mi = 5.9717 × 106 ft

Newton’s second law: F = man :

GM M m r12

=m

v12 =

v12 r1

GMm r1

V1 = −

1 2 1 GM M mv1 = m r1 2 2

GM M m r1

E1 = T1 + V1 = E1 = −

T1 =

1 GM M m GM M m − r1 r1 2

1 GM M m 1 0.0123 gRE2 m =− r1 r1 2 2

1 (0.0123)(32.2 ft/s 2 )(20.909 × 106 ft)2 m 2 5.9717 × 106 ft 6 2 2 (14.498 × 10 ft /s )WE E1 = = 450.2 × 103 ft ⋅ lb/lb WE 2 (32.2 ft/s ) E1 = −

ΔE = E2 − E1 = 0 + 450.2 × 103 ft ⋅ lb/lb WE ΔE = 450 × 103 ft ⋅ lb/lb  WE

Energy per pound:

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PROBLEM 13.89 Knowing that the velocity of an experimental space probe fired from the earth has a magnitude v A = 32.5 Mm/h at Point A, determine the speed of the probe as it passes through Point B.

SOLUTION rA = R + hA = 6370 + 4300 = 10740 km = 10.670 × 106 m rB = 6370 + 12700 = 19070 km = 19.070 × 106 m GM = gR 2 = (9.81)(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2 v A = 32.5 Mm/h = 9.0278 × 103 m/s

Use conservation of energy. TB + VB = TA + VA 1 2 GMm 1 2 GMm mvB − = mv A − 2 rB 2 rA 1 1 vB2 = v 2A + 2GM  −   rB rA  1 1   = (9.0278 × 103 ) 2 + (2)(398.06 × 1012 )  − 6 6 19.070 × 10 10.670 × 10  = 48.635 × 106 m 2 /s 2 vB = 6.97 × 103 m/s

vB = 25.1 Mm/h 

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PROBLEM 13.90 A spacecraft is describing a circular orbit at an altitude of 1500 km above the surface of the earth. As it passes through Point A, its speed is reduced by 40 percent and it enters an elliptic crash trajectory with the apogee at Point A. Neglecting air resistance, determine the speed of the spacecraft when it reaches the earth’s surface at Point B.

SOLUTION Circular orbit velocity vC2 GM = 2 , GM = gR 2 r r vC2 =

GM gR 2 (9.81 m/s 2 )(6.370 × 106 m) 2 = = r r (6.370 × 106 m + 1.500 × 106 m)

vC2 = 50.579 × 106 m 2 /s 2 vC = 7112 m/s

Velocity reduced to 60% of vC gives v A = 4267 m/s. Conservation of energy: TA + VA = TB + VB 1 GM m 1 GM m = m vB2 − m v A2 − rB rA 2 2 1 9.81(6.370 × 106 )2 vB2 9.81(6.370 × 106 ) 2 (4.267 × 103 ) 2 − = − 2 2 (7.870 × 106 ) (6.370 × 106 ) vB = 6.48 × 103 m/s

vB = 6.48 km/s 

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PROBLEM 13.91 Observations show that a celestial body traveling at 1.2 × 106 mi/h appears to be describing about Point B a circle of radius equal to 60 light years. Point B is suspected of being a very dense concentration of mass called a black hole. Determine the ratio MB/MS of the mass at B to the mass of the sun. (The mass of the sun is 330,000 times the mass of the earth, and a light year is the distance traveled by light in one year at a velocity of 186,300 mi/s.)

SOLUTION

One light year is the distance traveled by light in one year. Speed of light = 186,300 mi/s r = (60 yr)(186,300 mi/s)(5280 ft/mi)(365 days/yr)(24 h/day)(3600 s/h) r = 1.8612 × 1018 ft

Newton’s second law GM B m v2 = m r r2 2 rv MB = G 2 GM earth = gRearth F=

= (32.2 ft/s 2 )(3960 mi × 5280 ft/mi) 2 = 14.077 × 1015 (ft 3 /s 2 ) M sun = 330, 000 M E : GM sun = 330, 000 GM earth

GM sun = (330,000)(14.077 × 1015 ) = 4.645 × 1021 ft 3 /s 2 G= MB =

4.645 × 1021 M sun rv 2 M sun rv 2 = G 4.645 × 1021

MB (1.8612 × 1018 )(1.76 × 106 ) 2 = M sun 4.645 × 1021

MB = 1.241 × 109  M sun

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PROBLEM 13.92 (a) Show that, by setting r = R + y in the right-hand member of Eq. (13.17′) and expanding that member in a power series in y/R, the expression in Eq. (13.16) for the potential energy Vg due to gravity is a first-order approximation for the expression given in Eq. (13.17′). (b) Using the same expansion, derive a second-order approximation for Vg.

SOLUTION Vg = −

WR 2 WR 2 WR setting r = R + y : Vg = − =− r R+ y 1 + Ry

y  Vg = −WR 1 +  R 

−1

 (−1) y (−1)(−2)  y 2  = −WR 1 + + +    1 R 1⋅ 2  R   

We add the constant WR, which is equivalent to changing the datum from r = ∞ to r = R :  y  y 2  Vg = WR  −   +   R  R  

(a)

First order approximation:  y Vg = WR   = Wy  R

[Equation 13.16]

(b)

Second order approximation:

 y  y 2  Vg = WR  −     R  R  

Vg = Wy −

Wy 2  R

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PROBLEM 13.93 Collar A has a mass of 3 kg and is attached to a spring of constant 1200 N/m and of undeformed length equal to 0.5 m. The system is set in motion with r = 0.3 m, vθ = 2 m/s, and vr = 0. Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when r = 0.6 m.

SOLUTION Let position 1 be the initial position. r1 = 0.3 m (vr )1 = 0,

(vθ )1 = 2 m/s,

v1 = 2 m/s

x1 = r1 − l0 = (0.3 − 0.5) = −0.2 m

Let position 2 be when r = 0.6 m. r2 = 0.6 m (vr )2 = ?,

(vθ ) 2 = ?,

v2 = ?

x2 = r2 − l0 = (0.6 − 0.5) = 0.1 m

Conservation of angular momentum:

r1m(vθ )1 = v2 m(vθ ) 2 (vθ ) 2 =

Conservation of energy:

r1 (vθ )1 (0.3)(2) = = 1.000 m/s 0.6 r2

T1 + V1 = T2 + V2 1 2 1 2 1 2 1 2 mv1 + kx1 = mv2 + kx2 2 2 2 2 k 2 2 2 2 v2 = v1 + x1 − x2 m 1200 (0.2)2 − (0.1)2  = 16 m 2 /s 2 = (2) 2 +  3  2 2 2 2 2 (vr ) 2 = v2 − (vθ ) = 16 − 1 = 15 m /s

(

)

vr = ±3.87 m/s

vr = ±3.87 m/s  vθ = 1.000 m/s 

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PROBLEM 13.94 Collar A has a mass of 3 kg and is attached to a spring of constant 1200 N/m and of undeformed length equal to 0.5 m. The system is set in motion with r = 0.3 m, vθ = 2 m/s, and vr = 0. Neglecting the mass of the rod and the effect of friction, determine (a) the maximum distance between the origin and the collar, (b) the corresponding speed. (Hint: Solve the equation obtained for r by trial and error.)

SOLUTION Let position 1 be the initial position. r1 = 0.3 m (vr )1 = 0,

(vθ )1 = 2 m/s,

v1 = 2 m/s

x1 = r1 − l0 = 0.3 − 0.5 = −0.2 m 1 2 1 mv = (3)(2) 2 = 6 J 2 2 1 2 1 V1 = kx1 = (1200)(−0.2) 2 = 24 J 2 2 T1 =

Let position 2 be when r is maximum. (vr ) 2 = 0 r2 = rm x2 = (rm − 0.5) 1 2 1 mv2 = (3)(vθ ) 22 = 1.5(vθ ) 22 2 2 1 1 V2 = kx22 = (1200)(rm − 0.5) 2 2 2 T2 =

= 600(rm − 0.5)2

Conservation of angular momentum:

r1m(vθ )1 = v2 m(vθ ) 2 (vθ ) 2 =

Conservation of energy:

r1 (0.3) 0.6 (vθ ), = (2) = r2 rm rm

T1 + V1 = T2 + V2 6 + 24 = 1.5(vθ ) 22 + 600( rm − 0.5)2 2

 0.6  2 30 = (1.5)   + 600( rm − 0.5) r  m  0.54 f (rm ) = 2 + 600( rm − 0.5) 2 − 30 = 0 rm

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PROBLEM 13.94 (Continued)

Solve for rm by trial and error. rm (m)

0.5

1.0

0.8

0.7

0.72

0.71

f (rm )

–27.8

120.5

24.8

–4.9

0.080

–2.469

rm = 0.72 −

(a)

Maximum distance.

(b)

Corresponding speed.

(0.01)(0.08) = 0.7197 m 2.467 + 0.08 rm = 0.720 m 

(vθ ) 2 =

0.6 = 0.8337 m/s 0.7197 (vr ) 2 = 0

v2 = 0.834 m/s 

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PROBLEM 13.95 A 4-lb collar A and a 1.5-lb collar B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O. At the instant shown, the velocity vA of collar A has a magnitude of 6 ft/s and a stop prevents collar B from moving. If the stop is suddenly removed, determine (a) the velocity of collar A when it is 8 in. from O, (b) the velocity of collar A when collar B comes to rest. (Assume that collar B does not hit O, that collar A does not come off rod OE, and that the mass of the frame is negligible.)

SOLUTION 4 = 0.12422 lb ⋅ s 2 /ft 32.2 1.5 mB = = 0.04658 lb ⋅ s 2 /ft 32.2 mA =

Masses:

Constraint of the cord. Let r be the radial distance to the center of collar A and y be the distance that collar B moves up from its initial level. y = Δr ; y = vr (a)

Let position 1 be the initial position just after the stop at B is removed and position 2 be when the collar is 8 in. (0.66667 ft) from O. r1 = 4 in. = 0.33333 ft

(vr )1 = 0

r2 = 8 in. = 0.66667 ft Δr = y2 = 8 − 4 = 4 in. = 0.33333 ft V1 = 0,

Potential energy:

V2 = WB y2 = (1.5)(0.33333) = 0.5 ft ⋅ lb

Conservation of angular momentum of collar A: m A r1 (vθ )1 = m A r2 (vθ ) 2 (vθ ) 2 =

Conservation of energy:

r1 (vθ )1 (0.33333)(6) = = 3 ft/s 0.66667 r2

T1 + V1 = T2 + V2

1 1 1 1 m A [(vr )12 + (vθ )12 ] + mB y12 = m A [(vr ) 22 + (vθ ) 22 ] + mB y 2 + V2 2 2 2 2 1 1 1 m A [0 + (vθ )12 ] + 0 + 0 = m A [(vr ) 22 + (vθ ) 22 ] + mB (vr ) 22 + 0.5 2 2 2

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PROBLEM 13.95 (Continued) 1 1 1 (0.12422)(6) 2 = (0.12422)[(vr ) 22 + (3) 2 ] + (0.04658)(vr ) 22 + 0.5 2 2 2 2.236 = 0.06211(vr )22 + 0.559 + 0.02329(vr ) 22 + 0.5 0.0854(vr )22 = 1.177 (vr ) 2 = 13.78 ft 2 /s 2

(vr ) 2 = 3.71 ft/s  (vθ ) 2 = 3.00 ft/s  v = 4.77 ft/s 

(b)

Let position 3 be when collar B comes to rest. y3 = r3 − 0.33333,

(vr )3 = 0,

y3 = 0

Conservation of angular momentum of collar A. m A r1 (vθ )1 = m A r3 (vθ )3 (vθ )3 =

r1 (vθ )1 (0.33333)(6) 2 = = r3 r3 r3 T1 + V1 = T3 + V3

Conservation of energy:

1 1 1 1 m A [(vr )12 + (vθ )12 ] + mB y12 = m A [(vr )32 + (vθ )32 ] + mB y32 + wB y3 2 2 2 2

  2 2  1 1 (0.12422)[0 + (6)2 ] + 0 = (0.12422) 0 +    + 0 + (1.5)(r3 − 0.33333) 2 2   r3     2.236 =

0.24844 + 1.5r3 − 0.5 r32

1.5r33 − 2.736r32 + 0.24844 = 0

Solving the cubic equation for r3, r3 = 1.7712 ft, − 0.2805 ft, 0.33333 ft

Since r3 > r1 = 0.33333 ft, the required root is r3 = 1.7712 ft

Corresponding velocity of collar A: (vr )3 = 0  (vθ )3 =





2 2 = r3 1.7712

(vθ )3 = 1.129 ft/s  v3 = 1.129 ft/s 



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PROBLEM 13.96 A 1.5-lb ball that can slide on a horizontal frictionless surface is attached to a fixed Point O by means of an elastic cord of constant k = 1 lb/in. and undeformed length 2 ft. The ball is placed at Point A, 3 ft from O, and given an initial velocity v 0 perpendicular to OA. Determine (a) the smallest allowable value of the initial speed v0 if the cord is not to become slack, (b) the closest distance d that the ball will come to Point O if it is given half the initial speed found in part a.

SOLUTION Let L1 be the initial stretched length of the cord and L2 the length of the closest approach to Point O if the cord does not become slack. Let position 1 be the initial state and position 2 be that of closest approach to Point O. The only horizontal force acting on the ball is the conservative central force due to the elastic cord. At the point of closest approach the velocity of the ball is perpendicular to the cord. Conservation of angular momentum:

r1m v1 = r2m v2 L1m v0 = L2m v2 or v2 =

L1v0 L2

T1 + V1 = T2 + V2

Conservation of energy:

1 2 1 1 1 mv1 + k ( L1 − L0 ) 2 = mv22 + k ( L2 − L0 ) 2 2 2 2 2 k v12 − v22 = − [( L1 − L0 ) 2 + ( L2 − L0 ) 2 ] m v02 −

L12 2 k v = − [( L1 − L0 ) 2 + ( L2 − L0 ) 2 ] 2 0 m L2 L0 = 2 ft, L1 = 3 ft

Data:

L2 = L0 = 2 ft for zero tension in the cord at the point of closest approach. k = 1 lb/in. = 12 lb/ft m = W /g = 1.5/32.2 = 0.04658 lb ⋅ s 2 /ft v02 −

(3) 2 12 v0 = − [(3 − 2) 2 + (2 − 2) 2 ] 0.04658 (2) 2

−1.25 v02 = −257.6

(a)

v02 = 206.1 ft 2 /s 2

v0 = 14.36 ft/s 

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PROBLEM 13.96 (Continued)

(b)

Let v0 =

1 2

(14.36 ft/s) = 7.18 ft/s 2 so that the cord is slack in the position of closest approach to Point O.

Let position 1 be the initial position and position 2 be position of closest approach with the cord being slack. Conservation of energy:

T1 + V1 = T2 + V2 1 2 1 1 mv0 + k ( L1 − L0 )2 = mv22 2 2 2 k ( L − L0 ) 2 m 12 (3 − 2) 2 = 309.17 ft 2 /s 2 = (7.18) 2 + 0.04658 v2 = 17.583 ft/s

v22 = v02 +

Conservation of angular momentum: r1m v1 = r2m v2 sin φ r2 sin φ = d = d =

r1v1 Lv = 10 v2 v2

(3)(7.18) 17.583

d = 1.225 ft 

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PROBLEM 13.97 A 1.5-lb ball that can slide on a horizontal frictionless surface is attached to a fixed Point O by means of an elastic cord of constant k = 1 lb/in. and undeformed length 2 ft. The ball is placed at Point A, 3 ft from O, and given an initial velocity v 0 perpendicular to OA, allowing the ball to come within a distance d = 9 in. of Point O after the cord has become slack. Determine (a) the initial speed v0 of the ball, (b) its maximum speed.

SOLUTION Let L1 be the initial stretched length of the cord. Let position 1 be the initial position. Let position 2 be the position of closest approach to point after the cord has become slack. While the cord is slack there are no horizontal forces acting on the ball, so the velocity remains constant. While the cord is stretched, the only horizontal force acting on the ball is the conservative central force due to the elastic cord. At the point of closest approach the velocity of the ball is perpendicular to the radius vector. Conservation of angular momentum:

r1m v1 = r2 mv2 L1v0 = d v2

Conservation of energy:

v2 =

or

L1 v0 d

T1 + V1 = T2 + V2 1 2 1 1 mv0 + k ( L1 − L0 )2 = mv22 + 0 2 2 2 k v02 − v22 = − ( L1 − L0 ) 2 m 2

k L  v02 −  12 v0  = − ( L1 − L0 ) m d  L0 = 2 ft, L1 = 3 ft, d = 9 in. = 0.75 ft

Data:

k = 1 lb/in. = 12 lb/ft m = W /g = 1.5/32.2 = 0.04658 lb ⋅ s 2 /ft 2

12 3v  v02 −  0  = − (3 − 2) 2 0.04658  0.75  − 15 v02 = −257.6 v02 = 17.17 ft 2 /s 2

(a) (b)

Maximum speed.

vm = v2 =

3v0 0.75

v0 = 4.14 ft/s  vm = 16.58 ft/s 

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PROBLEM 13.98 Using the principles of conservation of energy and conservation of angular momentum, solve part a of Sample Problem 12.9.

SOLUTION R = 6370 km r0 = 500 km + 6370 km r0 = 6870 km = 6.87 × 106 m v0 = 36,900 km/h =

36.9 × 106 m 3.6 × 103 s

= 10.25 × 103 m/s

Conservation of angular momentum: r0 mv0 = r1mv A,

r0 = rmin , r1 = rmax

 6.870 × 106 r  v A′ =  0  v0 =  r1  r1   v A′ =

 3  (10.25 × 10 ) 

70.418 × 109 r1

(1)

Conservation of energy: Point A: v0 = 10.25 × 103 m/s 1 2 1 mv0 = m(10.25 × 103 ) 2 2 2 TA = (m)(52.53 × 106 )(J) TA =

VA = −

GMm r0

GM = gR 2 = (9.81 m/s 2 )(6.37 × 106 m) 2 GM = 398 × 1012 m3 /s 2 r0 = 6.87 × 106 m VA = −

(398 × 1012 m3 /s 2 )m (6.87 × 106 m)

= −57.93 × 106 m (J)

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PROBLEM 13.98 (Continued)

Point A′: 1 2 mv A′ 2 GMm VA′ = − r1 TA′ =

=−

398 × 1012 m (J) r1

TA + VA = TA′ + VA′ 52.53 × 106 m − 57.93 × 106 m =

1 398 × 1012 m m v A2 ′ − r1 2

Substituting for v A′ from (1) −5.402 × 106 =

(70.418 × 109 ) 2 398 × 1012 − r1 (2)(r1 ) 2

−5.402 × 106 =

(2.4793 × 1021 ) 398 × 1012 − r1 r12

(5.402 × 106 )r12 − (398 × 1012 )r1 + 2.4793 × 1021 = 0 r1 = 66.7 × 106 m, 6.87 × 106 m

rmax = 66,700 km 

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PROBLEM 13.99 Solve sample Problem 13.8, assuming that the elastic cord is replaced by a central force F of magnitude (80/r2) N directed toward O. PROBLEM 13.8 Skid marks on a drag racetrack indicate that the rear (drive) wheels of a car slip for the first 20 m of the 400-m track.(a) Knowing that the coefficient of kinetic friction is 0.60, determine the speed of the car at the end of the first 20-m portion of the track if it starts from rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed for the car at the finish line if, after skidding for 20 m, it is driven without the wheels slipping for the remainder of the race? Assume that while the car is rolling without slipping, 60 percent of the weight of the car is on the rear wheels and the coefficient of static friction is 0.75. Ignore air resistance and rolling resistance.

SOLUTION (a)

The force exerted on the sphere passes through O. Angular momentum about O is conserved. Minimum velocity is at B, where the distance from O is maximum. Maximum velocity is at C, where distance from O is minimum. rA mv A sin 60° = rm mvm (0.5 m)(0.6 kg)(20 m/s)sin 60° = rm (0.6 kg)vm vm =

8.66 rm

(1)

Conservation of energy: At Point A,

1 2 1 mv A = (0.6 kg)(20 m/s)2 = 120 J 2 2 80 −80 , V = Fdr = 2 dr = r r −80 VA = = −160 J 0.5 TA =



At Point B,

TB =



1 2 1 mvm = (0.6 kg)vm2 = 0.3vm2 2 2

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PROBLEM 13.99 (Continued)

VB =

and Point C :

−80 rm

TA + VA = TB + VB 120 − 160 = 0.3vm2 −

80 rm

(2)

Substitute (1) into (2) 2

 8.66  80 −40 = (0.3)   − rm  rm  rm2 − 2 rm + 0.5625 = 0 rm′ = 0.339 m and rm = 1.661 m rmax = 1.661 m  rmin = 0.339 m 

(b)

Substitute rm′ and rm from results of part (a) into (1) to get corresponding maximum and minimum values of the speed. vm′ =

8.66 = 25.6 m/s 0.339

vmax = 25.6 m/s 

vm =

8.66 = 5.21 m/s 1.661

vmin = 5.21 m/s 

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PROBLEM 13.100 A spacecraft is describing an elliptic orbit of minimum altitude hA = 2400 km and maximum altitude hB = 9600 km above the surface of the earth. Determine the speed of the spacecraft at A.

SOLUTION rA = 6370 km + 2400 km rA = 8770 km rB = 6370 km + 9600 km = 15,970 km rA mv A = rB mvB

Conservation of momentum:

vB =

Conservation of energy:

TA =

1 2 mv A 2

rA 8770 vA = v A = 0.5492v A 15,970 rB

VA =

−GMm rA

TB =

1 2 mvB 2

(1) VB =

−GMm rB

GM = gR 2 = (9.81 m/s 2 )(6370 × 103 m)2 = 398.1 × 1012 m3/s 2 −(398.1 × 1012 ) m = −45.39 × 106 m 3 8770 × 10 −(398.1 × 1012 ) m VB = = −24.93 m (15,970 × 103 ) VA =

TA + VA = TB + VB : 1 1 m v 2A − 45.39 × 106 m = m vB2 − 24.93 × 106 m 2 2

(2)

Substituting for vB in (2) from (1) v A2 [1 − (0.5492) 2 ] = 40.92 × 106 v A2 = 58.59 × 106 m 2/s 2 v A = 7.65 × 103 m/s

v A = 27.6 × 103 km/h 

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PROBLEM 13.101 While describing a circular orbit, 185 mi above the surface of the earth, a space shuttle ejects at Point A an inertial upper stage (IUS) carrying a communication satellite to be placed in a geosynchronous orbit (see Problem 13.87) at an altitude of 22,230 mi above the surface of the earth. Determine (a) the velocity of the IUS relative to the shuttle after its engine has been fired at A, (b) the increase in velocity required at B to plae the satellite in its final orbit.

SOLUTION For earth,

R = 3960 mi = 20.909 × 106 ft g = 32.2 ft 2 GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2

Speed on a circular orbits of radius r, rA, and rB. F = man GMm mv 2 = r r2 GM v2 = r

v=

GM r

rA = 3960 + 185 = 4145 mi = 21.886 × 106 ft (v A )circ =

14.077 × 1015 = 25.362 × 103 ft/s 21.886 × 106

rB = 3960 + 22230 = 26190 mi = 138.283 × 106 ft (vB )circ =

14.077 × 1015 = 10.089 × 103 ft/s 138.283 × 106

Calculate speeds at A and B for path AB. Conservation of angular momentum: mrA v A sin φ A = mrB vB sin φ A vB =

rA v A sin 90° 21.886 × 106 v A = = 0.15816 v A rB sin 90° 138.283 × 106

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PROBLEM 13.101 (Continued)

TA + VA = TB + VB

Conservation of energy:

1 2 GMm 1 2 GMm mv A + = mvB − 2 rA 2 rB  1 1  2GM (rB − rA ) v A2 − vB2 = 2GM  −  = rA rB  rA rB  v A2 − (0.15816v A ) 2 =

(2)(14.077 × 1015 )(116.397 × 106 ) (21.886 × 106 )(138.283 × 106 )

0.97499v A2 = 1.082796 × 109 v A = 33.325 × 103 ft/s vB = (0.15816)(33.325 × 106 ) = 5.271 × 103 ft/s

(a)

(b)

Increase in speed at A: Δv A = 33.325 × 103 − 25.362 × 103 = 7.963 × 103 ft/s

Δv A = 7960 ft/s 

ΔvB = 10.089 × 103 − 5.271 × 103 = 4.818 × 103 ft/s

ΔvB = 4820 ft/s 

Increase in speed at B:

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PROBLEM 13.102 A spacecraft approaching the planet Saturn reaches Point A with a velocity v A of magnitude 68.8 × 103 ft/s. It is to be placed in an elliptic orbit about Saturn so that it will be able to periodically examine Tethys, one of Saturn’s moons. Tethys is in a circular orbit of radius 183 × 103 mi about the center of Saturn, traveling at a speed of 37.2 × 103 ft/s. Determine (a) the decrease in speed required by the spacecraft at A to achieve the desired orbit, (b) the speed of the spacecraft when it reaches the orbit of Tethys at B.

SOLUTION 

(a) rA = 607.2 × 106 ft rB = 966.2 × 106 ft v′A = speed of spacecraft in the elliptical orbit after its speed has been decreased.

Elliptical orbit between A and B. Conservation of energy 1 mv′A2 2 −GM sat m VA = rA TA =

Point A:   

M sa = Mass of Saturn, determine GM sa from the speed of Tethys in its circular orbit. vcirc =

(Eq. 12.44)

GM sat r

2 GM sat = rB vcirc

GM sat = (966.2 × 106 ft 2 )(37.2 × 103 ft/s) 2 = 1.337 × 1018 ft 3/s 2 VA = −

(1.337 × 1018 ft 3/s 2 ) m (607.2 × 106 ft)

= −2.202 × 109 m



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PROBLEM 13.102 (Continued)

TB =

Point B:

1 2 mvB 2

VB =

−GM sat m (1.337 × 1018 ft 3 /s 2 ) m =− rB (966.2 × 106 ft)

VB = 1.384 × 109 TA + VA = TB + VB ; 1 1 m v′A2 − 2.202 × 109 m = m vB2 − 1.384 × 109 m 2 2 v′A2 − vB2 = 1.636 × 109

Conservation of angular momentum: rA mv′A = rB mvB

vB =

rA 607.2 × 106 v′A = v′A = 0.6284v′A rB 966.2 × 106

v′A2 [1 − (0.6284) 2 ] = 1.636 × 109 v′A = 52, 005 ft/s

(a)

Δv A = v A − v′A = 68,800 − 52, 005

(b)

vB =

Δv A = 16,795 ft/s 

rA v′A = (0.6284)(52, 005) rB

vB = 32, 700 ft/s 

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PROBLEM 13.103 A spacecraft traveling along a parabolic path toward the planet Jupiter is expected to reach Point A with a velocity vA of magnitude 26.9 km/s. Its engines will then be fired to slow it down, placing it into an elliptic orbit which will bring it to within 100 × 103 km of Jupiter. Determine the decrease in speed Δv at Point A which will place the spacecraft into the required orbit. The mass of Jupiter is 319 times the mass of the earth.

SOLUTION Conservation of energy. Point A: 1 m(v A − Δv A ) 2 2 −GM J m VA = rA TA =

GM J = 319GM E = 319 gRE2 RE = 6.37 × 106 m GM J = (319)(9.81 m/s 2 )(6.37 × 106 m) 2 GM J = 126.98 × 1015 m3 /s 2 rA = 350 × 106 m VA =

−(126.98 × 1015 m3 /s 2 )m (350 × 106 m)

VA = −(362.8 × 106 )m

Point B: 1 2 mvB 2 −GM J m −(126.98 × 1015 m3 /s 2 )m VB = = rB (100 × 106 m) TB =

VB = −(1269.8 × 106 ) m TA + VA = TB + VB 1 1 m(v A − Δv A ) 2 − 362.8 × 106 m = mvB2 − 1269.8 × 106 m 2 2 (v A − Δv A ) 2 − vB2 = −1814 × 106

(1)

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PROBLEM 13.103 (Continued)

Conservation of angular momentum. rA = 350 × 106 m rB = 100 × 106 m rA m(v A − Δv A ) = rB mvB

r  vB =  A  (v A − Δv A )  rB   350  =  (v A − Δv A )  100 

(2)

Substitute vB in (2) into (1) (v A − Δv A ) 2 [1 − (3.5) 2 ] = −1814 × 106 (v A − Δv A ) 2 = 161.24 × 106 (v A − Δv A ) = 12.698 × 103 m/s

(Take positive root; negative root reverses flight direction.) v A = 26.9 × 103 m/s

(given)

Δv A = (26.9 × 103 m/s − 12.698 × 103 m/s) Δv A = 14.20 km/s 

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PROBLEM 13.104 As a first approximation to the analysis of a space flight from the earth to Mars, it is assumed that the orbits of the earth and Mars are circular and coplanar. The mean distances from the sun to the earth and to Mars are 149.6 × 106 km and 227.8 × 106 km, respectively. To place the spacecraft into an elliptical transfer orbit at Point A, its speed is increased over a short interval of time to v A which is faster than the earth’s orbital speed. When the spacecraft reaches Point B on the elliptical transfer orbit, its speed vB is increased to the orbital speed of Mars. Knowing that the mass of the sun is 332.8 × 103 times the mass of the earth, determine the increase in velocity required (a) at A, (b) at B.

SOLUTION M = mass of the sun

GM = 332.8(10)3 (9.81 m/s 2 )(6.37 × 106 m)2 = 1.3247(10)20 m3 /s 2

Circular orbits

Earth vE =

GM = 29.758 m/s 149.6(10)9

Mars vM =

GM = 24.115 m/s 227.8(10)9

Conservation of angular momentum Elliptical orbit

v A (149.6) = vB (227.8)

Conservation of energy 1 2 GM 1 GM vA − = vB2 − 9 2 2 149.6(10) 227.8(10)9 v A = vB

(227.8) = 1.52273 vB (149.6)

1 1.3247(10) 20 1 1.3247(10)20 = vB2 − (1.52273) 2 vB2 − 9 2 2 149.6(10) 227.8(10)9 0.65935vB2 = 3.0398(10)8 vB2 = 4.6102(10)8 vB = 21, 471 m/s, v A = 32, 695 m/s

(a)

Increase at A,

v A − vE = 32.695 − 29.758 = 2.94 km/s 

(b)

Increase at B,

vB − vM = 24.115 − 21.471 = 2.64 km/s 

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PROBLEM 13.105 The optimal way of transferring a space vehicle from an inner circular orbit to an outer coplanar circular orbit is to fire its engines as it passes through A to increase its speed and place it in an elliptic transfer orbit. Another increase in speed as it passes through B will place it in the desired circular orbit. For a vehicle in a circular orbit about the earth at an altitude h1 = 200 mi, which is to be transferred to a circular orbit at an altitude h2 = 500 mi, determine (a) the required increases in speed at A and at B, (b) the total energy per unit mass required to execute the transfer.

SOLUTION Elliptical orbit between A and B Conservation of angular momentum mrAv A = mrBvB vA =

rB 7.170 vB = vB 6.690 rA

rA = 6370 km + 320 km = 6690 km,

rA = 6.690 × 106 m

v A = 1.0718vB rB = 6370 km + 800 km = 7170 km,

(1) rB = 7.170 × 106 m

R = (6370 km) = 6.37 × 106 m

Conservation of energy GM = gR 2 = (9.81 m/s 2 )(6.37 × 106 m)2 = 398.060 × 1012 m3 /s 2

Point A: TA =

1 2 mv A 2

VA = −

GMm (398.060 × 1012 )m =− rA (6.690 × 106 )

VA = 59.501 × 106 m

Point B: TB =

1 2 mvB 2

VB = −

GMm (398.060 × 1012 )m =− rB (7.170 × 106 )

VB = 55.5 × 106 m

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PROBLEM 13.105 (Continued)

TA + VA = TB + VB 1 2 1 mv A − 59.501 × 106 m = mvB2 − 55.5 × 106 m 2 2 v A2 − vB2 = 8.002 × 106

From (1)

v A = 1.0718vB

vB2 [(1.0718) 2 − 1] = 8.002 × 106

vB2 = 53.79 × 106 m 2 /s 2 ,

vB = 7334 m/s

v A = (1.0718)(7334 m/s) = 7861 m/s

Circular orbit at A and B (Equation 12.44) (v A )C =

GM = rA

398.060 × 1012 = 7714 m/s 6.690 × 106

(vB )C =

GM = rB

398.060 × 1012 = 7451 m/s 7.170 × 106

(a) Increases in speed at A and B Δv A = v A − (v A )C = 7861 − 7714 = 147 m/s  ΔvB = (vB )C − vB = 7451 − 7334 = 117 m/s 

(b) Total energy per unit mass E/m = E/m =

1 [(v A ) 2 − (v A )C2 + (vB )C2 − (vB ) 2 ] 2

1 [(7861) 2 − (7714) 2 + (7451) 2 − (7334)2 ] 2 E/m = 2.01 × 106 J/kg 

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PROBLEM 13.106 During a flyby of the earth, the velocity of a spacecraft is 10.4 km/s as it reaches its minimum altitude of 990 km above the surface at Point A. At Point B the spacecraft is observed to have an altitude of 8350 km. Determine (a) the magnitude of the velocity at Point B, (b) the angle φB .

SOLUTION At A:

hA = vr = [1.04(10) 4 m/s][6.37(10)6 m + 0.990(10)6 m] hA = 76.544(10)9 m 2 /s 1 1 GM (TA + VA ) = v 2 − m 2 r =

1 (9.81)[6.37(10)6 ]2 ≅0 [1.04(10) 4 ]2 − 2 [6.37(10)6 + 0.990(10)6 ]

(Parabolic orbit) At B:

1 1 GM (TB + VB ) = vB2 − =0 m 2 rB 1 2 (9.81)[6.37(106 )]2 vB = 2 [6.37(10)6 + 8.35(10)6 ] vB2 = 54.084(10)6 vB = 7.35 km/s 

(a) hB = vB sin φB rB = 76.544(10)9 sin φB =

76.544(10)9 7.35(106 )[6.37(10)6 + 8.35(10)6 ]

= 0.707483

φ B = 45.0° 

(b)

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PROBLEM 13.107 A space platform is in a circular orbit about the earth at an altitude of 300 km. As the platform passes through A, a rocket carrying a communications satellite is launched from the platform with a relative velocity of magnitude 3.44 km/s in a direction tangent to the orbit of the platform. This was intended to place the rocket in an elliptic transfer orbit bringing it to Point B, where the rocket would again be fired to place the satellite in a geosynchronous orbit of radius 42,140 km. After launching, it was discovered that the relative velocity imparted to the rocket was too large. Determine the angle γ at which the rocket will cross the intended orbit at Point C.

SOLUTION R = 6370 km rA = 6370 km + 300 km rA = 6.67 × 106 m rC = 42.14 × 106 m GM = gR 2 GM = (9.81 m/s)(6.37 × 106 m) 2 GM = 398.1 × 1012 m3 /s 2

For any circular orbit: Fn = man =

2 mvcirc

r 2

mv GMm = m circ 2 r r GM = r

Fn = vcirc

Velocity at A: (v A )circ =

GM (398.1 × 1012 m3/s3 ) = = 7.726 × 103 m/s rA (6.67 × 106 m)

v A = (v A )circ + (v A ) R = 7.726 × 103 + 3.44 × 103 = 11.165 × 103 m/s

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PROBLEM 13.107 (Continued)

Velocity at C: Conservation of energy:

TA + VA = TC + VC 1 GM m 1 GM m m v A2 − = m vC2 − 2 2 rA rC

 1 1 vC2 = v 2A + 2GM  −   rC rA  1 1   = (11.165 × 103 ) 2 + 2(398.1 × 1012 )  − 6 6  42.14 10 6.67 10 × ×   vC2 = 124.67 × 106 − 100.48 × 106 = 24.19 × 106 m 2 /s 2 vC = 4.919 × 103 m/s

Conservation of angular momentum: rA mv A = rC mvC cos γ cos γ =

rA v A rC vC

(6.67 × 106 )(11.165 × 103 ) (42.14 × 106 )(4.919 × 103 ) cos γ = 0.35926 =

γ = 68.9° 

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PROBLEM 13.108 A satellite is projected into space with a velocity v0 at a distance r0 from the center of the earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle α with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite.

SOLUTION For circular orbit of radius r0 v02 GMm = m r0 r02

F = man v02 =

GM r0

But v0 forms an angle α with the intended circular path. For elliptic orbit. Conservation of angular momentum: r0 mv0 cos α = rA mv A r  v A =  0 cos α  v0  rA 

(1)

Conservation of energy: 1 2 GMm 1 2 GMm mv0 − = mv A − 2 r0 2 rA r0  2GM  v02 − v 2A = 1 −  r0  rA 

Substitute for v A from (1)   r 2  2GM v02 1 −  0  cos 2 α  = r0   rA    

But

v02 =

GM , r0

 r0  1 −   rA  2

thus

r   r  1 −  0  cos 2 α = 2 1 − 0   rA   rA 

2

r  r  cos 2 α  0  − 2  0  + 1 = 0  rA   rA  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 669

PROBLEM 13.108 (Continued)

Solving for

r0 rA r0 + 2 ± 4 − 4cos 2 α 1 ± sin α = = 2 rA 2cos α 1 − sin 2 α (1 + sin α )(1 − sin α ) rA = r0 = (1  sin α )r0 1 ± sin α

also valid for Point A′ Thus, rmax = (1 + sin α )r0

rmin = (1 − sin α )r0 

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PROBLEM 13.109 Upon the LEM’s return to the command module, the Apollo spacecraft of Problem 13.88 was turned around so that the LEM faced to the rear. The LEM was then cast adrift with a velocity of 200 m/s relative to the command module. Determine the magnitude and direction (angle φ formed with the vertical OC) of the velocity vC of the LEM just before it crashed at C on the moon’s surface.

SOLUTION Command module in circular orbit rB = 1740 + 140 = 1880 km = 1.88 × 106 m GM moon = 0.0123 GM earth = 0.0123 gR 2 = 0.0123(9.81)(6.37 × 106 )2 = 4.896 × 1012 m3 /s 2 R = 1740 km ΣF = man v0 =

GM m m

=

rB2

GMm = rB

mv02 rB

4.896 × 1012 1.88 × 106

v0 = 1614 m/s

vB = 1614 − 200 = 1414 m/s

Conservation of energy between B and C: 1 2 GM m m 1 2 GM m m mvB − = mvC − rB rC 2 2 vC2 = vB2 +

rC = R

2GMm  rB  − 1  rB  R 

vC2 = (1414 m/s)2 + 2

 (4.896 × 1012 m3/s 2 )  1.88 × 106 − 1  6 6  (1.88 × 10 m)  1.74 × 10 

vC2 = 1.999 × 106 + 0.4191 × 106 = 2.418 × 106 m 2 /s 2

vC = 1555 m/s 

Conservation of angular momentum: rB mvB = RmvC sin φ rB vB (1.88 × 106 m)(1414 m/s) = = 0.98249 rC vC (1.74 × 106 m)(1555 m/s) φ = 79.26°

sin φ =

φ = 79.3° 

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PROBLEM 13.110 A space vehicle is in a circular orbit at an altitude of 225 mi above the earth. To return to earth, it decreases its speed as it passes through A by firing its engine for a short interval of time in a direction opposite to the direction of its motion. Knowing that the velocity of the space vehicle should form an angle φB = 60° with the vertical as it reaches Point B at an altitude of 40 mi, determine (a) the required speed of the vehicle as it leaves its circular orbit at A, (b) its speed at Point B.

SOLUTION (a)

rA = 3960 mi + 225 mi = 4185 mi rA = 4185 mi × 5280 ft/mi = 22,097 × 103 ft rB = 3960 mi + 40 mi = 4000 mi rB = 4000 × 5280 = 21,120 × 103 ft R = 3960 mi = 20,909 × 103 ft GM = gR 2 = (32.2 ft/s 2 )(20,909 × 103 ft) 2 GM = 14.077 × 1015 ft 3 /s 2

Conservation of energy: 1 2 mv A 2 −GMm VA = rA TA =

=

−14.077 × 1015 m 22, 097 × 103

= −637.1 × 106 m 1 2 mvB 2 −GMm VB = rB TB =

=

−14.077 × 1015 m 21,120 × 103

= −666.5 × 106 m TA + VA = TB + VB 1 1 m v A2 − 637.1 × 106 m = m vB2 − 666.5 × 106 m 2 2 2 v A = vB2 − 58.94 × 106

(1)

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PROBLEM 13.110 (Continued)

Conservation of angular momentum: rA mv A = rB mvB sin φB vB =

(rA )v A 4185  1  = vA (rB )(sin φB ) 4000  sin 60° 

vB = 1.208 v A

(2)

Substitute vB from (2) in (1) v 2A = (1.208v A ) 2 − 58.94 × 106 v A2 [(1.208) 2 − 1] = 58.94 × 106 v 2A = 128.27 × 106 ft 2 /s 2 v A = 11.32 × 103 ft/s 

(a) (b)

From (2) vB = 1.208v A = 1.208(11.32 × 106 ) = 13.68 × 103 ft/s vB = 13.68 × 103 ft/s 

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PROBLEM 13.111* In Problem 13.110, the speed of the space vehicle was decreased as it passed through A by firing its engine in a direction opposite to the direction of motion. An alternative strategy for taking the space vehicle out of its circular orbit would be to turn it around so that its engine would point away from the earth and then give it an incremental velocity Δv A toward the center O of the earth. This would likely require a smaller expenditure of energy when firing the engine at A, but might result in too fast a descent at B. Assuming this strategy is used with only 50 percent of the energy expenditure used in Problem 13.109, determine the resulting values of φB and vB .

SOLUTION rA = 3960 mi + 225 mi rA = 4185 mi = 22.097 × 106 ft rB = 3960 mi + 40 mi = 4000 mi rB = 21.120 × 106 ft GM = gR 2 = (32.2 ft/s 2 )[(3960)(5280) ft 2 ] GM = 14.077 × 1015 ft 3 /s 2

Velocity in circular orbit at 225 m altitude:

Newton’s second law F = man : (v A )circ =

2 GMm m(v A )circ = rA rA2

GM 14.077 × 1015 = rA 22.097 × 106

= 25.24 × 103 ft/s

Energy expenditure: From Problem 13.110, Energy,

v A = 11.32 × 103 ft/s 1 1 2 m(v A )circ − mv 2A 2 2 1 1 = m(25.24 × 103 )2 − m(11.32 × 103 ) 2 2 2 6 = 254.46 × 10 m ft ⋅ lb

ΔE109 = ΔE109 ΔE109

ΔE110 = (0.50)ΔE109 =

(254.46 × 106 m) ft ⋅ lb 2

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PROBLEM 13.111* (Continued)

Thus, additional kinetic energy at A is 1 (254.46 × 106 m) m( Δv A ) 2 = ΔE110 = ft ⋅ lb 2 2

(1)

Conservation of energy between A and B: TA =

1 2 m[(v A )circ + ( Δv A ) 2 ] 2

TB =

1 2 mvB 2

VB =

VA =

−GMm rA

−GMm rA

TA + VA = TB + VB 1 254.46 × 106 m 14.077 × 1015 m 1 2 14.077 × 1015 m m(25.24 × 103 ) 2 + − = mvB − 2 2 2 22.097 × 106 21.120 × 106 vB2 = 637.06 × 106 + 254.46 × 106 − 1274.1 × 106 + 1333 × 106 vB2 = 950.4 × 103 vB = 30.88 × 103 ft/s 

Conservation of angular momentum between A and B: rA m(v A )circ = rB mvB sin φB

 r  (v ) (4185) (25.24 × 103 ) = 0.8565 sin φB =  A  A circ = (4000) (30.88 × 103 )  rB  (vB )

φB = 58.9° 

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PROBLEM 13.112 Show that the values vA and vP of the speed of an earth satellite at the apogee A and the perigee P of an elliptic orbit are defined by the relations v A2 =

2GM rP rA + rP rA

vP2 =

2GM rA rA + rP rP

where M is the mass of the earth, and rA and rP represent, respectively, the maximum and minimum distances of the orbit to the center of the earth.

SOLUTION Conservation of angular momentum: rA mv A = rP mvP vA =

rP vP rA

(1)

Conservation of energy: 1 2 GMm 1 2 GMm mvP − = mv A − 2 rP 2 rA

(2)

Substituting for v A from (1) into (2) 2

vP2

2GM  rP  2 2GM − =   vP − rP rA  rA 

  r 2    1 −  P   vP2 = 2GM  1 − 1    rA    rP rA    rA2 − rP2 rA2

with

vP2 = 2GM

rA − rP rA rP

rA2 − rP2 = (rA − rP )( rA + rP ) vP2 =

2GM rA + rP

 rA    (3)   rP 

Exchanging subscripts P and A v A2 =

2GM rA + rP

 rP     rA 

Q.E.D.



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 676

PROBLEM 13.113 Show that the total energy E of an earth satellite of mass m describing an elliptic orbit is E = −GMm /(rA + rP ), where M is the mass of the earth, and rA and rP represent, respectively, the maximum and minimum distances of the orbit to the center of the earth. (Recall that the gravitational potential energy of a satellite was defined as being zero at an infinite distance from the earth.)

SOLUTION See solution to Problem 13.112 (above) for derivation of Equation (3). vP2 =

2GM rA (rA + rP ) rP

Total energy at Point P is E = TP + VP = =

1 2 GMm mvP − 2 rP

1  2GMm rA  GMm  − rP 2  ( rA + r0 ) rP 

 rA 1 = GMm  −   rP ( rA + rP ) rP  (r − r − r ) = GMm A A P rP (rA + rP ) E=−

GMm  rA + rP

Note: Recall that gravitational potential of a satellite is defined as being zero at an infinite distance from the earth.

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PROBLEM 13.114* A space probe describes a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O. Show that (a) in order for the probe to leave its orbit and hit the planet at an angle θ with the vertical, its velocity must be reduced to α v0, where

α = sin θ

2(n − 1) n − sin 2 θ

(b) the probe will not hit the planet if α is larger than

2 /(1 + n).

2

SOLUTION (a)

Conservation of energy: 1 m (α v0 ) 2 2 GMm VA = − nR TA =

At A:

1 mv 2 2 GMm VB = − R TB =

At B:

M = mass of planet m = mass of probe TA + VA = TB + VB 1 GMm 1 GMm = mv 2 − m (α v0 ) 2 − 2 nR 2 R

(1)

Conservation of angular momentum: nR mα v0 = Rmv sin θ v=

nα v0 sin θ

(2)

Replacing v in (1) by (2) 2

(α v0 )2 −

2GM  nα v0  2GM = −  nR R  sin θ 

(3)

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PROBLEM 13.114* (Continued)

For any circular orbit. an =

v2 r

Newton’s second law 2 −GMm m(v) circ = r r2 GM vcirc = r

v0 = vcirc =

For r = nR,

GM nR

Substitute for v0 in (3)

α2

GM 2GM n 2α 2  GM  2GM − =  − nR nR R sin 2 θ  nR 



α 2 1 − 

n2   = 2(1 − n) sin 2 θ 

α2 =

2(1 − n)(sin 2 θ ) 2(n − 1)sin 2 θ = 2 (sin 2 θ − n 2 ) (n − sin 2 θ )

α = sin θ (b)

2(n − 1) n − sin 2 θ 2

Q.E.D.



Probe will just miss the planet if θ > 90°,

α = sin 90° Note:

2( n − 1) = n − sin 2 90° 2

2 n +1



n 2 − 1 = ( n − 1)(n + 1) 

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PROBLEM 13.115 A missile is fired from the ground with an initial velocity v 0 forming an angle φ 0 with the vertical. If the missile is to reach a maximum altitude equal to α R, where R is the radius of the earth, (a) show that the required angle φ 0 is defined by the relation sin φ0 = (1 + α ) 1 −

α  vesc    1 + α  v0 

2

where vesc is the escape velocity, (b) determine the range of allowable values of v0 .

SOLUTION rA = R

(a)

Conservation of angular momentum: Rmv0 sin φ0 = rB mvB rB = R + α R = (1 + α ) R vB =

Rv0 sin φ0 v0 sin φ0 = (1 + α ) R (1 + α )

(1)

Conservation of energy: 1 2 GMm 1 2 GMm = mvB − mv0 − R 2 2 (1 + α ) R

TA + VA = TB + VB v02 − vB2 =

2 GMm  1 1−  R  1+ α

 2 GMm  α   = R 1+α    

Substitute for vB from (1)  sin 2 φ0 v02 1 −  (1 + α ) 2 

From Equation (12.43):

 2 GMm  α   =   R 1+α  

2 = vesc

 sin 2 φ0 v02 1 −  (1 + α ) 2 

2GM R

 2  α   = vesc   1+α   2

v  α = 1 −  esc  2 (1 + α )  v0  1 + α sin 2 φ0

α  vesc  sin φ0 = (1 + α ) 1 −   1 + α  v0 

(2) 2

Q.E.D.

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PROBLEM 13.115 (Continued)

(b)

Allowable values of v0 (for which maximum altitude = α R) 0 < sin 2 φ0 < 1

For sin φ0 = 0, from (2) 2

v  α 0 = 1 −  esc   v0  1 + α v0 = vesc

α 1+ α

For sin φ0 = 1, from (2) 2

v  α 1 = 1 −  esc  2 (1 + α )  v0  1 + α 2

 vesc  1 1   = 1 + α − α 1+ α  v0  v0 = vesc

2  1 + 2α + α − 1 2 + α = =  α (1 + α ) 1+ α 

1+α 2 +α vesc

α 1+α < v0 < vesc  1+α 2 +α

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PROBLEM 13.116 A spacecraft of mass m describes a circular orbit of radius r1 around the earth. (a) Show that the additional energy ΔE which must be imparted to the spacecraft to transfer it to a circular orbit of larger radius r2 is ΔE =

GMm(r2 − r1 ) 2r1r2

where M is the mass of the earth. (b) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path AB, the amounts of energy Δ E A and Δ EB which must be imparted at A and B are, respectively, proportional to r2 and r1 : Δ EA =

r2 r1 + r2

ΔE

Δ EB =

r1 r1 + r2

ΔE

SOLUTION (a)

For a circular orbit of radius r F = man :

GMm v2 = m r r2 GM v2 = r E = T +V =

1 2 GMm 1 GMm mv − =− r 2 2 r

(1)

Thus ΔE required to pass from circular orbit of radius r1 to circular orbit of radius r2 is 1 GMm 1 GMm + 2 r1 2 r2 GMm(r2 − r1 ) ΔE = Q.E.D. 2r1r2 Δ E = E1 − E2 = −

(b)

(2)

For an elliptic orbit, we recall Equation (3) derived in Problem 13.113 (with vP = v1 ) v12 =

2Gm r2 (r1 + r2 ) r1

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PROBLEM 13.116 (Continued)

At Point A: Initially spacecraft is in a circular orbit of radius r1 . GM 2 = vcirc r1 1 2 1 GM Tcirc = mvcirc = m r1 2 2 After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall

and

v12 =

2GM r2 ⋅ (r1 + r2 ) r1

T1 =

2GMr2 1 2 1 mv1 = m 2 2 r1 (r1 + r2 )

At Point A, the increase in energy is ΔE A = T1 − Tcirc =

Recall Equation (2):

2GMr2 1 1 GM − m m 2 r1 (r1 + r2 ) 2 r1

ΔE A =

GMm(2r2 − r1 − r2 ) GMm( r2 − r1 ) = 2r1 (r1 + r2 ) 2r1 (r1 + r2 )

ΔE A =

r2 r1 + r2

ΔE A =

r2 ΔE (r1 + r2 )

Q.E.D.

r1 ΔE (r1 + r2 )

Q.E.D.

 GMm(r2 − r1 )    2r1r2  

A similar derivation at Point B yields, ΔEB =

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PROBLEM 13.117* Using the answers obtained in Problem 13.108, show that the intended circular orbit and the resulting elliptic orbit intersect at the ends of the minor axis of the elliptic orbit. PROBLEM 13.108 A satellite is projected into space with a velocity v0 at a distance r0 from the center of the earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle α with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite.

SOLUTION If the point of intersection P0 of the circular and elliptic orbits is at an end of the minor axis, then v0 is parallel to the major axis. This will be the case only if α + 90° = θ0 , that is if cos θ0 = − sin α . We must therefore prove that cos θ0 = − sin α

(1)

We recall from Equation (12.39): 1 GM = 2 + C cos θ r h

When θ = 0,

r = rmin

(2)

and rmin = r0 (1 − sin α )

1 GM = 2 +C r0 (1 − sin α ) h

For θ = 180°, 

(3)

r = rmax = r0 (1 + sin α ) 1 GM = 2 −C r0 (1 + sin α ) h

(4)

Adding (3) and (4) and dividing by 2: GM 1  1 1  = +  2 2r0  1 − sin α 1 + sin α  h 1 = r0 cos 2 α

Subtracting (4) from (3) and dividing by 2: 1  1 1 −  2r0  1 − sin α 1 + sin α sin α C= r0 cos 2 α C=

  1 =   2r0

 2sin α  2  1 − sin α 

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PROBLEM 13.117* (Continued)

Substituting for

GM h2

and C into Equation (2) 1 1 (1 + sin α cos θ ) = r r0 cos 2 α

(5)

Letting r = r0 and θ = θ0 in Equation (5), we have cos 2 α = 1 + sin α cos θ0 cos 2 α − 1 sin α sin 2 α =− sin α = − sin α

cos θ0 =

This proves the validity of Equation (1) and thus P0 is an end of the minor axis of the elliptic orbit.

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PROBLEM 13.118* (a) Express in terms of rmin and vmax the angular momentum per unit mass, h, and the total energy per unit mass, E/m, of a space vehicle moving under the gravitational attraction of a planet of mass M (Figure 13.15). (b) Eliminating vmax between the equations obtained, derive the formula 1 rmin

GM = 2 h

2  1 + 1 + 2 E  h    m  GM    

(c) Show that the eccentricity ε of the trajectory of the vehicle can be expressed as

ε = 1+

2E  h  m  GM 

2

(d ) Further show that the trajectory of the vehicle is a hyperbola, an ellipse, or a parabola, depending on whether E is positive, negative, or zero.

SOLUTION (a)

Point A: Angular momentum per unit mass.

H0 m r mv = min max m

h=

h = rmin vmax

(1) 

Energy per unit mass E 1 = (T + V ) m m E 1 1 2 GMm  1 2 GM =  mvmax −  = vmax − m m2 rmin  2 rmin

(b)

(2) 

From Eq. (1): vmax = h/rmin substituting into (2) E 1 h 2 GM = − 2 m 2 rmin rmin E 2 2   1  2GM 1 m −  2  =0   − 2 ⋅ rmin h h  rmin 

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PROBLEM 13.118* (Continued)

Solving the quadratic:

1 rmin

=

2 2 ( mE ) GM  GM  + +  2  h2 h2  h 

=

GM h2

Rearranging 1 rmin

(c)

2  1 + 1 + 2 E  h   m  GM  

(3) 

Eccentricity of the trajectory: Eq. (12.39′)

1 GM = 2 (1 + ε cos θ ) r h

When θ = 0,

cos θ = 1 and r = rmin

Thus, 1 rmin

Comparing (3) and (4), (d )

=

GM (1 + ε ) h2

ε = 1+

(4)

2E  h  m  GM 

2

(5)

Recalling discussion in section 12.12 and in view of Eq. (5)



1. Hyperbola if ε > 1, that is, if E > 0



2. Parabola if ε = 1, that is, if E = 0



3. Ellipse if ε < 1, that is, if E < 0



Note: For circular orbit ε = 0 and 2

2E  h  1+ =0 m  GM  GM r

2

 GM  m , E = −   h  2

or

but for circular orbit

v2 =

thus

1 (GM ) 2 1 GMm E=− m =−  2 GMr 2 r

and

h2 = v 2 r 2 = GMr ,

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PROBLEM 13.CQ4 A large insect impacts the front windshield of a sports car traveling down a road. Which of the following statements is true during the collision? (a) The car exerts a greater force on the insect than the insect exerts on the car. (b) The insect exerts a greater force on the car than the car exerts on the insect. (c) The car exerts a force on the insect, but the insect does not exert a force on the car. (d) The car exerts the same force on the insect as the insect exerts on the car. (e) Neither exerts a force on the other; the insect gets smashed simply because it gets in the way of the car.

SOLUTION Answer: (d) This is Newton’s 3rd Law.

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PROBLEM 13.CQ5 The expected damages associated with two types of perfectly plastic collisions are to be compared. In the first case, two identical cars traveling at the same speed impact each other head on. In the second case, the car impacts a massive concrete wall. In which case would you expect the car to be more damaged? (a) Case 1 (b) Case 2 (c) The same damage in each case

SOLUTION Answer: (c) In both cases the car will come to a complete stop, so the applied impulse will be the same.

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PROBLEM 13.F1 The initial velocity of the block in position A is 30 ft/s. The coefficient of kinetic friction between the block and the plane is μk = 0.30. Draw impulse-momentum diagrams that could be used to determine the time it takes for the block to reach B with zero velocity, if θ = 20°.

SOLUTION Answer:

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PROBLEM 13.F2 A 4-lb collar which can slide on a frictionless vertical rod is acted upon by a force P which varies in magnitude as shown. Knowing that the collar is initially at rest, draw impulse-momentum diagrams that could be used to determine its velocity at t = 3 s.

SOLUTION Answer:

Where

ò

t2 = 3 t1 = 0

Pdt is the area under the curve.

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PROBLEM 13.F3 The 15-kg suitcase A has been propped up against one end of a 40-kg luggage carrier B and is prevented from sliding down by other luggage. When the luggage is unloaded and the last heavy trunk is removed from the carrier, the suitcase is free to slide down, causing the 40-kg carrier to move to the left with a velocity vB of magnitude 0.8 m/s. Neglecting friction, draw impulse-momentum diagrams that could be used to determine (a) the velocity of A as it rolls on the carrier and (b) the velocity of the carrier after the suitcase hits the right side of the carrier without bouncing back.

SOLUTION Answer: (a)

(b)

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PROBLEM 13.F4 Car A was traveling west at a speed of 15 m/s and car B was traveling north at an unknown speed when they slammed into each other at an intersection. Upon investigation it was found that after the crash the two cars got stuck and skidded off at an angle of 50° north of east. Knowing the masses of A and B are mA and mB respectively, draw impulse-momentum diagrams that could be used to determine the velocity of B before impact.

SOLUTION Answer:

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PROBLEM 13.F5 Two identical spheres A and B, each of mass m, are attached to an inextensible inelastic cord of length L and are resting at a distance a from each other on a frictionless horizontal surface. Sphere B is given a velocity v0 in a direction perpendicular to line AB and moves it without friction until it reaches B' where the cord becomes taut. Draw impulse-momentum diagrams that could be used to determine the magnitude of the velocity of each sphere immediately after the cord has become taut.

SOLUTION Answer:

Where v A¢ y = vB¢ y since the cord is inextensible.

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PROBLEM 13.119 A 35,000 Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 150 kN.

SOLUTION m = 35, 000 Mg = 35 × 106 kg F = 150 × 103 N v1 = 4 km/hr = 1.1111 m/s

mv1 − Ft = 0 (35 × 10 kg)(1.1111 m/s) − (150 × 103 N)t = 0 t = 259.26 s 6

t = 4 min19 s 

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PROBLEM 13.120 A 2500-lb automobile is moving at a speed of 60 mi/h when the brakes are fully applied, causing all four wheels to skid. Determine the time required to stop the automobile (a) on dry pavement ( μ k = 0.75), (b) on an icy road ( μ k = 0.10).

SOLUTION

v1 = 60 mph = 88 ft/s mv1 − μkWt = 0 t=

(a)

(b)

mv1 mv1 v = = 1 μkW μk mg μk g

For μk = 0.75 t=

88 ft/s (0.75)(32.2 ft/s 2 )

t = 3.64 s 

t=

88 ft/s (0.10)(32.2 ft/s 2 )

t = 27.3 s 

For μk = 0.10

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PROBLEM 13.121 A sailboat weighing 980 lb with its occupants is running down wind at 8 mi/h when its spinnaker is raised to increase its speed. Determine the net force provided by the spinnaker over the 10-s interval that it takes for the boat to reach a speed of 12 mi/h.

SOLUTION v1 = 8 mi/h = 11.73 ft/s t1− 2 = 10 sec v2 = 12 mi/h = 17.60 ft/s

m ⋅ v1 + imp1− 2 = mv2 m(11.73 ft/s) + Fn (10 s) = m(17.60 ft/s) Fn =

(980 lb)(17.60 ft/s − 11.73 ft/s) (32.2 ft/s 2 )(10 s)

Fn = 178.6 lb 

Note: Fn is the net force provided by the sails. The force on the sails is actually greater and includes the force needed to overcome the water resistance on the hull.

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PROBLEM 13.122 A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2500 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5m/s.

SOLUTION Apply the principle of impulse and momentum to the log.

mv1 + ΣImp1 → 2 = mv 2

Components in y-direction: 0 + Nt − mgt cos 20° = 0 N = mg cos 20°

Components in x-direction: 0 + Tt − mgt sin 20° − μk Nt = mv2 (T − mg sin 20° − μk mg cos 20°)t = mv2 [T − mg (sin 20° + μk cos 20°)]t = mv2

Data:

T = 2500 N, m = 300 kg, g = 9.81 m/s 2 ,

μk = 0.45,

v2 = 0.5 m/s

[2500 − (300)(9.81)(sin 20° + 0.45cos 20°)] t = (300)(0.5) 248.95 t = 150

t = 0.603 s 

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PROBLEM 13.123 A truck is traveling down a road with a 3-percent grade at a speed of 55 mi/h when the brakes are applied. Knowing the coefficients of friction between the load and the flatbed trailer shown are μs = 0.40 and μk = 0.35, determine the shortest time in which the rig can be brought to a stop if the load is not to shift.

SOLUTION Apply the principle impulse-momentum to the crate, knowing that, if the crate does not shift, the velocity of the crate matches that of the truck. For impending slip the friction and normal components of the contact force between the crate and the flatbed trailer satisfy the following equation: F f = μs N

mv1 + ΣImp1→2 = mv 2

Components in y-direction:

0 + Nt − mgt cos θ = 0 N = mg cos θ mv1 + mgt sin θ − μ s Nt = mv2

Components in x-direction:

mv1 + mgt (sin θ − μ s cos θ ) = 0 t=

Data:

v1 g ( μ s cos θ − sin θ )

v1 = 55 mi/h = 80.667 ft/s, v2 = 0, g = 32.2 ft/s 2 , μ s = 0.40,

tan θ = 3/100

θ = 1.71835° μs cos θ − sin θ = 0.36983 t=

80.667 (32.2)(0.36983)

t = 6.77 s 

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PROBLEM 13.124 Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop. A 10-ton truck enters a 15° ramp at a high speed v0 = 108 ft/s and travels for 6 s before its speed is reduced to 36 ft/s. Assuming constant deceleration, determine (a) the magnitude of the braking force, (b) the additional time required for the truck to stop. Neglect air resistance and rolling resistance.

SOLUTION W = 20, 000 lb m=

20,000 = 621.118 lb ⋅ s 2 /ft 32.2

Momentum in the x direction x : mv0 − ( F + mg sin15°)t = mv1 621.118(108) − ( F + mg sin15°)6 = (621.118)(36) F + mg sin15° = 7453.4

(a)

F = 7453.4 − 20, 000 sin15° = 2277 lb F = 2280 lb 

(b)

mv0 − ( F + mg sin15°)t = 0

t = total time

621.118(108) − 7453.4 t = 0;

t = 9.00 s

Additional time = 9 – 6

t = 3.00 s 

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PROBLEM 13.125 Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. The train is travelling down a 5 percent grade when it decreases its speed at a constant rate from 120 mi/h to 60 mi/h in a time interval of 12 s. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the baggage car if the trunk is not to slide.

SOLUTION

v1 = 120 mi/h = 176 ft/s v2 = 60 mi/h = 88 ft/s t1− 2 = 12 s Nt1− 2 = Wt1− 2 cos θ

m v1 − μs m gt1− 2 cos θ + m gt1− 2 sin θ = m v2 (176 ft/s) − μs (32.2 ft/s 2 )(12 s)(cos 2.86°) + (32.2 ft/s 2 )(12 s)(sin 2.86°) = 88 ft/s

μs =

176 − 88 + (32.2)(12)(sin 2.86°) (32.2)(12)(cos 2.86°)

μs = 0.278 

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PROBLEM 13.126 A 2-kg particle is acted upon by the force, expressed in newtons, F = (8 − 6t )i + (4 − t 2 ) j + (4 + t )k. Knowing that the velocity of the particle is v = (150 m/s)i + (100 m/s)j − (250 m/s)k at t = 0, determine (a) the time at which the velocity of the particle is parallel to the yz plane, (b) the corresponding velocity of the particle.

SOLUTION



mv 0 + Fdt = mv



Where

Fdt =

(1)

t

 [(8 − 6t )i + (4 − t 0

2

) j + (4 + t )k ]dt

1   1   = (8t − 3t 2 ) i +  4t − t 3  j +  4t + t 2  k 3   2  

Substituting m = 2 kg,

v0 = 150i + 100 j − 250k into (1):

1   1   (2 kg)(150i + 100 j − 250k ) + (8t − 3t 2 )i +  4t − t 3  j +  4t + t 2  k = (2 kg) v 3   2   3   1   1   v = 150 + 4t − t 2  i + 100 + 2t − t 3  j +  −250 + 2t + t 2  k 2   6   4  

(a)

v is parallel to yz plane when vx = 0, that is, when 3 150 + 4t − t 2 = 0 t = 11.422 s 2

(b)

t = 11.42 s 

1   v = 100 + 2(11.422) − (11.422)3  j 6   1   +  −250 + 2(11.422) + (11.422) 2  k 4   v = −(125.5 m/s) j − (194.5 m/s)k 

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PROBLEM 13.127 A truck is traveling down a road with a 4-percent grade at a speed of 60 mi/h when its brakes are applied to slow it down to 20 mi/h. An antiskid braking system limits the braking force to a value at which the wheels of the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels is 0.60, determine the shortest time needed for the truck to slow down.

SOLUTION

θ = tan −1

4 = 2.29° 100

mv1 + Σ imp1− 2 = mv 2 mv1 + Wt sin θ − Ft = mv2 v1 = 60 mi/h = 88 ft/s

N = W cos θ W = mg

v2 = 20 mi/h = 29.33 ft/s

F = μ s N = μ sW cos θ

( m )(88 ft/s) + ( m )(32.2 ft/s 2 )(t )(sin 2.29°) − (0.60)( m )(32.2 ft/s 2 )(cos 2.29°)(t ) = ( m )(29.33 ft/s)

t=

88 − 29.33 32.2[(0.60) cos 2.29° − sin 2.29°]

t = 3.26 s 

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PROBLEM 13.128 Skid marks on a drag race track indicate that the rear (drive) wheels of a car slip for the first 20 m of the 400-m track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the shortest possible time for the car to travel the initial 20-m portion of the track if it starts from rest with its front wheels just off the ground. (b) Determine the minimum time for the car to run the whole race if, after skidding for 20 m, the wheels roll without sliding for the remainder of the race. Assume for the rolling portion of the race that 65 percent of the weight is on the rear wheels and that the coefficient of static friction is 0.85. Ignore air resistance and rolling resistance.

SOLUTION (a)

First 20 m Velocity at 20 m. Rear wheels skid to generate the maximum force resulting in maximum velocity and minimum time since all the weight is on the rear wheel: This force is F = μ k N = 0.60W . T0 + U 0− 20 = T20

Work and energy. T0 = 0

U 0 − 20 = ( F )(20)

T20 =

1 2 mv20 2

1 2 m v20 2 = (2)(0.60)(20 m)(9.81 m/s 2 )

0 + μk mg (20) = 2 v20

v20 = 15.344 m/s

Impulse-momentum.

0 + μk mgt0 − 20 = mv20 t0− 20 =

v20 = 15.344 m/s

15.344 m/s (0.60)(9.81 m/s 2 )

t0− 20 = 2.61 s 

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PROBLEM 13.128 (Continued)

(b)

For the whole race: The maximum force on the wheels for the first 20 m is F = μk mg = 0.60 mg. For remaining 360 m, the maximum force, if there is no sliding and 65 percent of the weight is on the rear (drive) wheels, is F = μ s (0.65) mg = (0.85)(0.65) mg = 0.5525 mg

Velocity at 400 m. T0 + U 0− 20 + U 20 − 400 = T400

Work and energy. T0 = 0 T400 =

U 0 − 20 = (0.60mg )(20 m),

U 60 − 400 = (0.5525mg )(380 m)

1 2 mv400 2 0 + 12 mg + (0.5525)(380)mg =

1 2 mv400 2

v400 = 65.990 m/s

Impulse–momentum. From 20 m to 400 m

F = μs N = 0.510mg v20 = 15.344 m/s v400 = 65.990 m/s m (15.344) + 0.5525 mgt20 − 400 = m (65.990); t20 − 400 = 9.3442 s t0− 400 = t0 − 20 + t20 − 400 = 2.61 + 9.34

t0− 400 = 11.95 s 

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PROBLEM 13.129 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the time required to bring the train to a stop, (b) the force in each coupling.

SOLUTION Weights of cars:

WA = WC = 80,000 lb, WB = 100, 000 lb

Masses of cars:

m A = mC = 2484 lb ⋅ s 2 /ft, mB = 3106 lb ⋅ s 2 /ft

For each car the normal force (upward) is equal in magnitude to the weight of the car. N A = NC = 80,000 lb

Friction forces:

FA = 0

N B = 100, 000 lb

(brakes not applied)

FB = (0.35)(100, 000) = 35000 lb FC = (0.35)(80, 000) = 28, 000 lb

Stopping data: (a)

v1 = 30 mi/h = 44 ft/s, v2 = 0.

Apply the principle of impulse-momentum to the entire train. m = m A + mB + mC = 8074 lb ⋅ s 2 /ft F = FA + FB + FC = 63, 000 lb − mv1 + Ft = mv2 t =

(b)

m(v1 − v2 ) (8074)(44) = = 5.639 s F 63, 000

t = 5.64 s 

Coupling force FAB: Apply the principle of impulse-momentum to car A alone.

−m Av1 + FAt + FABt = 0 −(2484)(44) + 0 + FAB (5.639) = 0 FAB = 19,390 lb

FAB = 19,390 lb (tension) 

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PROBLEM 13.129 (Continued)

Coupling force FBC : Apply the principle of impulse-momentum to car C alone.

−mC v1 + FC t − FBC t = 0 −(2484)(44) + (28000)(5.639) − FBC (5.639) = 0 FBC = 8620 lb

FBC = 8620 lb (tension) 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 707

PROBLEM 13.130 Solve Problem 13.129 assuming that the brakes are applied only on the wheels of car A. PROBLEM 13.129 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the time required to bring the train to a stop, (b) the force in each coupling.

SOLUTION Weights of cars:

WA = WC = 80,000 lb, WB = 100, 000 lb

Masses of cars:

m A = mC = 2484 lb ⋅ s 2 /ft, mB = 3106 lb ⋅ s 2 /ft

For each car the normal force (upward) is equal in magnitude to the weight of the car. N A = N C = 80,000 lb

N B = 100, 000 lb

FA = (0.35)(80, 000) = 28, 000 lb

Friction forces:

FB = 0   (brakes not applied) FC = 0 

Stopping data: (a)

v1 = 30 mi/h = 44 ft/s, v2 = 0.

Apply the principle of impulse-momentum to the entire train. m = m A + mB + mC = 8074 lb ⋅ s 2 /ft F = FA + FB + FC = 28, 000 lb − mv1 + Ft = mv2 t =

(b)

m(v1 − v2 ) (8074)(44) = = 12.688 s F 28, 000

t = 12.69 s 

Coupling force FAB: Apply the principle of impulse-momentum to car A alone.

−mAv1 + FAt + FABt = 0 −(2484)(44) + (28, 000)(12.688) + FAB (12.688) = 0 FAB = −19,390 lb

FAB = 19,390 lb (compression) 

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PROBLEM 13.130 (Continued)

Coupling force FBC : Apply the principle of impulse-momentum to car C alone.

−mC v1 + FC t − FBC t = 0 −(2484)(44) + (0) − FBC (12.688) = 0 FBC = −8620 lb

FBC = 8620 lb (compression) 

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PROBLEM 13.131 A trailer truck with a 2000-kg cab and an 8000-kg trailer is traveling on a level road at 90 km/h. The brakes on the trailer fail and the antiskid system of the cab provides the largest possible force which will not cause the wheels of the cab to slide. Knowing that the coefficient of static friction is 0.65, determine (a) the shortest time for the rig to come to a stop, (b) the force in the coupling during that time.

SOLUTION v = 90 km/h = 25 m/s

(a)

The shortest time for the rig to come to a stop will be when the friction force on the wheels is maximum. The downward force exerted by the trailer on the cab is assumed to be zero. Since the trailer brakes fail, all of the braking force is supplied by the wheels of the cab, which is maximum when the wheels of the cab are at impending sliding.

Ft1− 2 = μs N C t1− 2

N C = mC g = (2000) g

Ft1− 2 = (0.65)(2000) gt [( mC + mT )v]2 = − Ft + [(mC + mT )v]1 0 = −(0.65)(2000 kg)(9.81 m/s 2 )(t1− 2 ) = 10, 000 kg(25 m/s) t1− 2 = 19.60 s 

(b)

For the trailer:

[mT v]2 = −Qt1− 2 + [mT v]1

From (a),

t1− 2 = 19.60 s 0 = −Q(19.60 s) + (8000 kg)(25 m/s) Q = 10, 204 N Q = 10.20 kN (compression) 

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PROBLEM 13.132 The system shown is at rest when a constant 150-N force is applied to collar B. Neglecting the effect of friction, determine (a) the time at which the velocity of collar B will be 2.5 m/s to the left, (b) the corresponding tension in the cable.

SOLUTION

Constraint of cord. When the collar B moves 1 unit to the left, the weight A moves up 2 units. Thus v A = 2 vB

Masses and weights:

vB =

m A = 3 kg

1 vA 2

WA = 29.43 N

mB = 8 kg

Let T be the tension in the cable. Principle of impulse and momentum applied to collar B.

: 0 + 150t − 2Tt = mB (vB ) 2 For (vB ) 2 = 2.5 m/s

150t − 2Tt = (8 kg)(2.5 m/s) 150t − 2Tt = 20

(1)

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PROBLEM 13.132 (Continued)

Principle of impulse and momentum applied to weight A.

: 0 + Tt − WAt = mA (v A ) 2 Tt + WAt = m A (2VB 2 ) Tt − 29.43t = (3 kg)(2)(2.5 m/s) Tt − 29.43t = 15

(2)

To eliminate T multiply Eq. (2) by 2 and add to Eq. (1). (a)

(b)

Time:

91.14t = 50

From Eq. (2),

T=

t = 0.549 s 

15 + 29.43 t T = 56.8 N 

Tension in the cable.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 712

PROBLEM 13.133 An 8-kg cylinder C rests on a 4-kg platform A supported by a cord which passes over the pulleys D and E and is attached to a 4-kg block B. Knowing that the system is released from rest, determine (a) the velocity of block B after 0.8 s, (b) the force exerted by the cylinder on the platform.

 

SOLUTION

(a)

Blocks A and C: [( mA + mC )v]1 − T (t1− 2 ) + ( mA + mC ) gt1− 2 = [(m A + mC )v]2 0 + (12 g − T )(0.8) = 12v

(1)

Block B: [mB v]1 + (T )t1− 2 − mB gt1− 2 = (mB v) 2



0 + (T − 4 g )(0.8) = 4v



(2)

Adding (1) and (2), (eliminating T) (12 g − 4 g )(0.8) = (12 + 4)



v=

(b)

(8 kg)(9.81 m/s 2 )(0.8 s) 16 kg

v = 3.92 m/s 

Collar A: (m Av)1 = 0 0 + ( FC + m A g )

(3)

From Eq. (2) with v = 3.92 m/s 4v T= + 4g 0.8 (4 kg)(3.92 m/s) T= + (4 kg)(9.81 m/s 2 ) (0.8 s) T = 58.84 N

 

Solving for FC in (3) FC =

(4 kg)(3.92 m/s) − (4 kg)(9.81 m/s 2 ) + 58.84 N (0.8 s)



FC = 39.2 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 713

PROBLEM 13.134 An estimate of the expected load on over-the-shoulder seat belts is to be made before designing prototype belts that will be evaluated in automobile crash tests. Assuming that an automobile traveling at 45 mi/h is brought to a stop in 110 ms, determine (a) the average impulsive force exerted by a 200-lb man on the belt, (b) the maximum force Fm exerted on the belt if the force-time diagram has the shape shown.

SOLUTION (a)

Force on the belt is opposite to the direction shown. v1 = 45 mi/h = 66 ft/s, W = 200 lb



mv1 − Fdt = mv 2

 Fdt = F

ave Δt

Δt = 0.110 s (200 lb)(66 ft/s) − Fave (0.110 s) = 0 (32.2 ft/s 2 ) Fave =

(b)

(200)(66) = 3727 lb (32.2)(0.110)

Fave = 3730 lb 

1 Fm (0.110 s) 2 From (a), impulse = Fave Δt = (3727 lb)(0.110 s)

Impulse = area under F −t diagram =

1 Fm (0.110) = (3727)(0.110) 2 Fm = 7450 lb 

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PROBLEM 13.135 A 60-g model rocket is fired vertically. The engine applies a thrust P which varies in magnitude as shown. Neglecting air resistance and the change in mass of the rocket, determine (a) the maximum speed of the rocket as it goes up, (b) the time for the rocket to reach its maximum elevation.

SOLUTION m = 0.060 kg

Mass:

mg = (0.060)(9.81) = 0.5886 N

Weight: Forces acting on the model rocket: Thrust:

P(t )(given function of t )

Weight:

W (constant)

Support:

S (acts until P > W )

Over

0 < t < 0.2 s:

13 t = 65t 0.2 W = 0.5886 N P=

S = W − P = 0.5886 − 65t

Before the rocket lifts off, S become zero when t = t1.

0 = 0.5886 − 65t1

t1 = 0.009055 s.

Impulse due to S: (t > t1 )



t 0

Sdt =



t1

Sdt

0

1 mgt1 2 = (0.5)(0.5886)(0.009055) =

= 0.00266 N ⋅ s

The maximum speed occurs when

dv = a = 0. dt

At this time, W − P = 0, which occurs at t2 = 0.8 s.

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PROBLEM 13.135 (Continued)

(a)

Maximum speed (upward motion): Apply the principle of impulse-momentum to the rocket over 0 ≤ t ≤ t2 .



0.8 0

Pdt = area under the given thrust-time plot. 1 1 (0.2)(13) + (0.1)(13 + 5) + (0.8 − 0.3)(5) 2 2 = 4.7 N ⋅ s =



0.8 0

m1v1 +



Wdt = (0.5886)(0.8) = 0.47088 N ⋅ S 0.8

0

Pdt +



0.8 0

Sdt −



0.8 0

Wdt = mv2

0 + 4.7 + 0.00266 − 0.47088 = 0.060 v2 v2 = 70.5 m/s 

(b)

Time t3 to reach maximum height: (v3 = 0) mv1 +



t3

0

Pdt +



t3

0

Sdt − Wt3 = mv3

0 + 4.7 + 0.00266 − 0.5886t3 = 0

t3 = 7.99 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 716

PROBLEM 13.136 A simplified model consisting of a single straight line is to be obtained for the variation of pressure inside the 10-mm-diameter barrel of a rifle as a 20-g bullet is fired. Knowing that it takes 1.6 ms for the bullet to travel the length of the barrel and that the velocity of the bullet upon exit is 700 m/s, determine the value of p0.

SOLUTION

At

t = 0,

p = p0 = c1 − c2t c1 = p0

At

t = 1.6 × 10−3 s,

p=0 0 = c1 − c2 (1.6 × 10−3 s) c2 =

p0 1.6 × 10−3 s

m = 20 × 10−3 kg 0+ A



1.6 × 10 −3 s 0

pdt = mv2

π (10 × 10−3 )2

A=

4 A = 78.54 × 10−6 m 2 0+ A



1.6 ×10−3 s 0

(c1 − c2t ) dt =

20 × 10−3 g

 (c )(1.6 × 10−3 s) 2  −3 (78.54 × 10−6 m 2 ) (c1 )(1.6 × 10−3 s) − 2  = (20 × 10 kg)(700 m/s) 2   1.6 × 10−3 c1 − 1.280 × 10−6 c2 = 178.25 × 103 (1.6 × 10−3 m 2 ⋅ s) p0 −

(1.280 × 10−6 m 2 ⋅ s 2 ) p0 = 178.25 × 103 kg ⋅ m/s −3 (1.6 × 10 s) p0 = 222.8 × 106 N/m 2

p0 = 223 MPa  

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PROBLEM 13.137 A 125-lb block initially at rest is acted upon by a force P which varies as shown. Knowing that the coefficients of friction between the block and the horizontal surface are μs = 0.50 and μk = 0.40, determine (a) the time at which the block will start moving, (b) the maximum speed reached by the block, (c) the time at which the block will stop moving.

SOLUTION





0 + Pdt − Fdt = mv v=

At any time: (a)

1 Pdt − Fdt   m 





(1)

Block starts moving at t.

P = Fs = μsW = (0.50)(125 lb) = 62.5 lb t1 t1 8s 8s ; = = Fs 100 lb 62.5 lb 100 lb

(b)

t1 = 5.00 s 

Maximum velocity: At t = tm P = Fk = μkW = 0.4(125) = 50 lb

where Block moves at t = 5 s.

Shaded area is maximum net impulse

 Pdt −  F dt R

when t = tm1 v = vm Eq. (1):

vm = vm =

1 shaded  1  1 1  1 (12.5 + 50)(3) + (50)(4)  = (193.75)  = 2 m  area  m  2  m 1 125 lb 32.2

[193.75] = 49.91 ft/s

v m = 49.9 ft/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 718

PROBLEM 13.137 (Continued)

(c)

Block stops moving when  Pdt − Fdt  = 0; or  





 Qdt =  Fdt

Assume tm > 16 s.

1

 Pdt = 2 (100)(16) = 800 lb ⋅ s 1

 Fdt = 2 (62.5)(5) + (50)(t

m

 Pdt −  Fdt = 800 − [156.25 + 50(t tm > 16 s OK

m

− 5) − 5)] = 0

tm = 17.875 s tm = 17.88 s 

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PROBLEM 13.138 Solve Problem 13.137, assuming that the weight of the block is 175 lb. PROBLEM 13.137 A 125-lb block initially at rest is acted upon by a force P which varies as shown. Knowing that the coefficients of friction between the block and the horizontal surface are μs = 0.50 and μk = 0.40, determine (a) the time at which the block will start moving, (b) the maximum speed reached by the block, (c) the time at which the block will stop moving.

SOLUTION See solution of Problem 13.137. W = 175 lb

(a)

Block starts moving:

v=

1 Pdt − Fdt   m 





(1)

P = Fs = μsW = (0.50)(175) = 87.5 lb

See first figure of Problem 13.137. t1 t1 8s 8s = = ; Fs 100 lb 87.5 lb 100 s

(b)

Maximum velocity:

t1 = 7.00 s 

P = Fk = μkW = 0.4(175) = 70 lb 16 − tm 8s = 70 lb 100 lb  8  16 − tm = 70   = 5.6  100  tm = 10.40 s

Eq. (1):

vm =

1 shaded    m  area 

1 1 1  (17.5 + 30)(1.0) + (30)(10.4 − 8)   m 2 2  1 = (59.75) m =

vm =

1 175 lb 32.2

[59.75]

= 10.994 ft/s

v m = 10.99 ft/s



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PROBLEM 13.138 (Continued)

(c)

Block stops moving when net impulse

 ( P − F )dt  = 0  



Assume ts , < 16 s.



ts

0

Pdt = =



ts

0

Fdt =

(16 − t s )  1 1 (100)(8) + 100 + 100  (t s − 8) 2 2 8  1 1  100  (100)(16) −  (16 − t s ) 2 2 2  8  1 (87.5)(7) + (70)(t s − 7) 2 100

 Pdt −  Fdt = 800 − 16 (16 − t )

2

s

Solving for ts ,

ts = 13.492 s

− 306.25 − 70(ts − 7) = 0 ts = 13.49 s 

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PROBLEM 13.139 A baseball player catching a ball can soften the impact by pulling his hand back. Assuming that a 5-oz ball reaches his glove at 90 mi/h and that the player pulls his hand back during the impact at an average speed of 30 ft/s over a distance of 6 in., bringing the ball to a stop, determine the average impulsive force exerted on the player’s hand.

SOLUTION

v = 90 mi/h = 132 ft/s m=

5 /g 16

( ) d = 12 = 1 s 60 vav 30 6

t= 0 = Fav t + mv Fav =

( )

Wv gt

Fav = =

mv t

( 165 lb ) (132 ft/s) 1 s (32.2 ft/s 2 ) ( 60 )

Fav = 76.9 lb 

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PROBLEM 13.140 A 1.62 ounce golf ball is hit with a golf club and leaves it with a velocity of 100 mi/h. We assume that for 0 ≤ t ≤ t0, where t0 is the duration of the impact, the magnitude F of the force exerted on the ball can be expressed as F = Fm sin (π t/t0 ). Knowing that t0 = 0.5 ms, determine the maximum value Fm of the force exerted on the ball.

SOLUTION W = 1.62 ounces = 0.10125 lb m = 3.1444 × 10−3 slug t = 0.5 ms = 0.5 × 10−3 s v = 100 mi/h = 146.67 ft/s

The impulse applied to the ball is



t0 0

Fdt =



t0 0

=−

Fm sin

Fm t0

π

πt t0

dt = −

Fm t0

π

(cos π − cos 0) =

cos

πt t0

t0

0

2 Fm t0

π

Principle of impulse and momentum. mv1 +



t0 0

F dt = mv 2

with v1 = 0, 0+

Solving for Fm,

Fm =

2 Fm t0

π π mv2 2t0

= mv2 =

π (3.1444 × 10−3 )(146.67) (2)(0.5 × 10−3 )

= 1.4488 × 103 lb

Fm = 1.45 kip  

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PROBLEM 13.141 The triple jump is a track-and-field event in which an athlete gets a running start and tries to leap as far as he can with a hop, step, and jump. Shown in the figure is the initial hop of the athlete. Assuming that he approaches the takeoff line from the left with a horizontal velocity of 10 m/s, remains in contact with the ground for 0.18 s, and takes off at a 50° angle with a velocity of 12 m/s, determine the vertical component of the average impulsive force exerted by the ground on his foot. Give your answer in terms of the weight W of the athlete.

SOLUTION

mv1 + (P − W)Δt = mv 2

Δt = 0.18 s

Vertical components W (12)(sin 50°) g (12)(sin 50°) Pv = W + W (9.81)(0.18)

0 + ( Pv − W )(0.18) =

Pv = 6.21W 

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PROBLEM 13.142 The last segment of the triple jump track-and-field event is the jump, in which the athlete makes a final leap, landing in a sand-filled pit. Assuming that the velocity of a 80-kg athlete just before landing is 9 m/s at an angle of 35° with the horizontal and that the athlete comes to a complete stop in 0.22 s after landing, determine the horizontal component of the average impulsive force exerted on his feet during landing.

SOLUTION

m = 80 kg Δt = 0.22 s mv1 + (P − W)Δt = mv 2

Horizontal components m(9)(cos 35°) − PH (0.22) = 0 PH =

(80 kg)(9 m/s)(cos 35°) = 2.6809 kN (0.22 s)

PH = 2.68 kN 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 725

PROBLEM 13.143 The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur. Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200 g implant determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest.

SOLUTION m = 200 g = 0.200 kg Fave = 2 kN = 2000 N Δt = 2 ms = 0.002 s

(a)

Velocity immediately after impact: Use principle of impulse and momentum: v1 = 0

v2 = ?

Imp1→2 = Fave (Δt )

mv1 + Imp1→2 = mv 2 0 + Fave (Δt ) = mv2 v2 =

(b)

Fave (Δt ) (2000)(0.002) = m 0.200

v2 = 20.0 m/s 

Average resistance to penetration: Δx = 1 mm = 0.001 m v2 = 20.0 ft/s v3 = 0

Use principle of work and energy. T2 + U 2→3 = T3 Rave =

or

1 2 mv2 − Rave ( Δx) = 0 2

mv22 (0.200)(20.0) 2 = = 40 × 103 N 2(Δx) (2)(0.001)

Rave = 40.0 kN 

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PROBLEM 13.144 A 25-g steel-jacketed bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10-mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate.

SOLUTION Impulse and momentum. Bullet alone:

mv1 + F Δ t = mv 2

t direction:

mv1 cos15° − Ft Δt = mv2 cos 20° Ft Δt = (0.025 kg)[600 m/s cos 15° − 400 m/s cos 20°] = 5.092 kg ⋅ m/s

Δt =

S BC 0.010 m = = 20 × 10−6 s v AV 500 m/s

Ft = (5.092 kg ⋅ m/s)/(20 × 10−6 s) = 254.6 × 103 kg ⋅ m/s 2 = 254.6 kN

n direction:

− mv1 sin15° + Fn Δt = mv2 sin 20° Fn Δt = (0.025 kg)[600 m/s sin 15° + 400 m/s sin 20°] = 7.3025 kg ⋅ m/s Fn = (43025 kg ⋅ m/s)/(20 × 10−6 ) = 365.1 × 103 kg ⋅ m/s 2 = 365.1 kN

Force on bullet:

F = Fn2 + Ft 2 = 365.12 + 254.62 = 445 kN tanθ =

Fn 365.1 = Ft 254.6

θ = 55.1°

θ − 15° = 40.1° F = 445 kN

Force on plate:

40.1°

F ′ = −F

F ′ = 445 kN

40.1° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 727

PROBLEM 13.145 A 25-ton railroad car moving at 2.5 mi/h is to be coupled to a 50 ton car which is at rest with locked wheels ( μk = 0.30). Determine (a) the velocity of both cars after the coupling is completed, (b) the time it takes for both cars to come to rest.

SOLUTION Weight and mass: (Label cars A and B.) Car A : WA = 50 tons = 100, 000 lb, mA = 3106 lb ⋅ s 2 /ft Car B: WB = 25 tons = 50,000 lb, mB = 1553 lb ⋅ s 2 /ft

Initital velocities:

vA = 0

vB = 2.5 mi/h = 3.6667 ft/s

(a)

v B = 3.6667 ft/s

The momentum of the system consisting of the two cars is conserved immediately before and after coupling.

Let v′ be the common velocity of that cars immediately after coupling. Apply conservation of momentum. : mB vB = m A v′ + mB v′ v′ =

(b)

mB vB (3106)(3.6667) = = 2.444 ft/s 4569 mA + mB

v′ = 1.667 mi/h



After coupling: The friction force acts only on car A. + ΣF = 0 A : N A − W A = 0

N A = WA

FA = μk N A = μkWA

(sliding)

FB = 0 (Car B is rolling.)

Apply impuslse-momentum to the coupled cars.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 728

PROBLEM 13.145 (Continued) : −(m A + mB )v′ + FAt = 0 t=

(m A + mB )v11 mB vB = μk W A FA

t=

(1553)(3.6667) = 0.1898 (0.30)(100, 000) t = 0.190 s 

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PROBLEM 13.146 At an intersection car B was traveling south and car A was traveling 30° north of east when they slammed into each other. Upon investigation it was found that after the crash the two cars got stuck and skidded off at an angle of 10° north of east. Each driver claimed that he was going at the speed limit of 50 km/h and that he tried to slow down but couldn’t avoid the crash because the other driver was going a lot faster. Knowing that the masses of cars A and B were 1500 kg and 1200 kg, respectively, determine (a) which car was going faster, (b) the speed of the faster of the two cars if the slower car was traveling at the speed limit.

SOLUTION (a)

Total momentum of the two cars is conserved.

Σmv, x :

m A v A cos 30° = (m A + mB )v cos 10°

(1)

Σmv, y :

m A v A sin 30° − mB vB = (mA + mB )v sin 10°

(2)

Dividing (1) into (2), mB vB sin 30° sin 10° − = cos 30° m Av A cos 30° cos 10° vB (tan 30° − tan 10°)( mA cos 30°) = vA mB vB (1500) = (0.4010) cos 30° vA (1200) vB = 0.434 vA

v A = 2.30 vB

A was going faster. 

Thus, (b)

Since vB was the slower car. vB = 50 km/h v A = (2.30)(50)

v A = 115.2 km/h 

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PROBLEM 13.147 The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0), determine the average resistance of the ground to penetration.

SOLUTION Velocity of the hammer at impact: Conservation of energy. T1 = 0 VH = mg (1.2 m) VH = (650 kg)(9.81 m/s 2 )(1.2 m) V1 = 7652 J 1 m 2 650 2 VH2 = v = 325 vH2 2 V2 = 0 T2 =

T1 + V1 = T2 + V2 0 + 7652 = 325 v 2 v 2 = 23.54 m 2 /s 2 v = 4.852 m/s

Velocity of pile after impact: Since the impact is plastic (e = 0), the velocity of the pile and hammer are the same after impact. Conservation of momentum:

The ground reaction and the weights are non-impulsive.

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PROBLEM 13.147 (Continued)

Thus,

mH vH = (mH + m p )v′ v′ =

Work and energy:

mH vH (650) = (4.852 m/s) = 3.992 m/s (mH + m p ) (650 + 140) d = 0.110 m T2 + U 2 −3 = T3

1 (mH + mH )(v′)2 2 T3 = 0 T2 =

1 (650 + 140)(3.992) 2 2 T2 = 6.295 × 103 J T2 =

U 2 −3 = ( mH + m p ) gd − FAV d = (650 + 140)(9.81)(0.110) − FAV (0.110) U 2 −3 = 852.49 − (0.110) FAV T2 + U 2 −3 = T3 6.295 × 103 + 852.49 − (0.110) FAV = 0 FAV = (7147.5)/(0.110) = 64.98 × 103 N

FAV = 65.0 kN 

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PROBLEM 13.148 A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet and the energy absorbed by the rivet under each blow, knowing that the head of the hammer has a weight of 1.5 lbs and that it strikes the rivet with a velocity of 20 ft/s. Assume that the hammer does not rebound and that the anvil is supported by springs and (a) has an infinite mass (rigid support), (b) has a weight of 9 lb.

SOLUTION Weight and mass: Hammer: WH = 1.5 lb

mH = 0.04658 lb ⋅ s 2 /ft

Anvil: Part a: WA = ∞

mA = ∞

Part b: WA = 9 lb

mA = 0.2795 lb ⋅ s 2 /ft

Kinetic energy before impact: 1 1 mH vH2 = (0.04658)(20) 2 = 9.316 ft ⋅ lb 2 2 Let v2 be the velocity common to the hammer and anvil immediately after impact. Apply the principle of conservation of momentum to the hammer and anvil over the duration of the impact. T1 =

: Σmv1 = Σ mv2 mH vH = ( mH + mA )v2 v2 =

mH vH mH + m A

TA =

1 1 mH2 vH2 (mH + m A )v22 = 2 2 mH + mA

T2 =

mH T1 mH + m A

(1)

Kinetic energy after impact:

(2)

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PROBLEM 13.148 (Continued) Impulse exerted on the hammer:

: mH vH − F (Δt ) = mH v2 F Δt = mH (vH − v2 )

(a)

(3)

WA = ∞ : By Eq. (1),

v2 = 0

By Eq. (2),

T2 = 0 T1 − T2 = 9.32 ft ⋅ lb 

Energy absorbed: By Eq. (3),

F (Δt ) = (0.04658)(20 − 0) = 0.932 lb ⋅ s

The impulse exerted on the rivet the same magnitude but opposite to direction. F Δt = 0.932 lb ⋅ s 

(b)

WA = 9 lb:

By Eq. (1),

v2 =

(0.04658)(20) = 2.857 ft/s 0.32608

By Eq. (2),

T2 =

(0.04658)(9.316) = 1.331 ft ⋅ lb 0.32608 T1 − T2 = 7.99 ft ⋅ lb 

Energy absorbed: By Eq. (3),

F (Δt ) = (0.04658)(20 − 2.857)

F (Δt ) = 0.799 lb ⋅ s 

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PROBLEM 13.149 Bullet B weighs 0.5 oz and blocks A and C both weigh 3 lb. The coefficient of friction between the blocks and the plane is μk = 0.25. Initially the bullet is moving at v0 and blocks A and C are at rest (Figure 1). After the bullet passes through A it becomes embedded in block C and all three objects come to stop in the positions shown (Figure 2). Determine the initial speed of the bullet v0.

SOLUTION Masses: 0.5 = 970.5 × 10−6 lb ⋅ s 2 /ft (16)(32.2)

Bullet:

mB =

Blocks A and C:

m A = mC =

Block C + bullet:

3 = 93.168 × 10−3 lb ⋅ s 2 /ft 32.2

mC + mB = 94.138 × 10−3 lb ⋅ s 2 /ft

Normal forces for sliding blocks from N − mg = 0 Block A:

N A = m A g = 3.00 lb.

Block C + bullet:

N C = (mC + mB ) g = 3.03125 lb.

Let v0 be the initial speed of the bullet; v1 be the speed of the bullet after it passes through block A; vA be the speed of block A immediately after the bullet passes through it; vC be the speed block C immediately after the bullet becomes embedded in it. Four separate processes and their governing equations are described below. 1.

The bullet hits block A and passes through it. Use the principle of conservation of momentum. (v A )0 = 0 mB v0 + mA (v A )0 = mB v1 + m Av A v0 = v1 +

2.

mAv A mB

(1)

The bullet hits block C and becomes embedded in it. Use the principle of conservation of momentum. (vC )0 = 0 mB v1 + mC (vC )0 = (mB + mC )vC v1 =

(mB + mC )vC mB

(2)

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PROBLEM 13.149 (Continued)

3.

Block A slides on the plane. Use principle of work and energy. T1 + U1→2 = T2 1 m Av A2 − μ k N A d A = 0 or v A = 2

4.

2μk N A d A mA

(3)

Block C with embedded bullet slides on the plane. Use principle of work and energy. dC = 4 in. = 0.33333 ft T1 + U1→ 2 = T2 1 (mC + mB )vC2 − μ k N C dC = 0 or vC = 2

2 μk NC dC mC + mB

(4)

Applying the numerical data: (2)(0.25)(3.03125)(0.33333) 94.138 × 10−3 = 2.3166 ft/s

From Eq. (4),

vC =

From Eq. (3),

vA =

From Eq. (2),

v1 =

From Eq. (1),

v0 = 224.71 +

(2)(0.25)(3.00)(0.5) 93.168 × 10−3 = 2.8372 ft/s (94.138 × 10−3 )(2.3166) 970.5 × 10−6 = 224.71 ft/s (93.138 × 10−3 )(2.8372) 970.5 × 10−6

v0 = 497 ft/s 

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PROBLEM 13.150 A 180-lb man and a 120-lb woman stand at opposite ends of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first.

SOLUTION (a)

Woman dives first: Conservation of momentum: −

120 300 + 180 (16 − v1 ) + v1 = 0 g g v1 =

(120)(16) = 3.20 ft/s 600

Man dives next. Conservation of momentum:

300 + 180 300 180 v1 = − v2 + (16 − v2 ) g g g v2 =

(b)

480v1 − (180)(16) = 2.80 ft/s 480

v 2 = 2.80 ft/s



Man dives first: Conservation of momentum: 180 300 + 120 (16 − v1′ ) − v1′ = 0 g g v1′ =

(180)(16) = 4.80 ft/s 600

Woman dives next. Conservation of momentum: −

300 + 120 300 120 v1′ = v2′ + (16 − v2′ ) g g g −420v1′ + (120)(16) v2′ = = −0.229 ft/s 420 v′2 = 0.229 ft/s



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PROBLEM 13.151 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supported by springs. Knowing that the height of the rebound is 0.6 m, determine (a) the velocity of the plate immediately after the impact, (b) the energy lost due to the impact.

SOLUTION Just before impact

vy =

Just after impact

vy =

2 g (1.6) = 5.603 m/s

2 g (0.6) = 3.431 m/s

(a) Conservation of momentum: (+ y ) mballv y + 0 = −mballv′y + mplatev′plate (0.075)(5.603) + 0 = −0.075(3.431) + 0.4v′plate v′plate = 1.694 m/s 

(b)

Energy loss Initial energy Final energy

(T + V )1 = (T + V ) 2 =

1 (0.075)(2) 2 + 0.075 g (1.6) 2

1 1 (0.075)(2)2 + 0.075 g (0.6) + (0.4)(1.694) 2 2 2

Energy lost = (1.3272 − 1.1653) J = 0.1619 J 

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PROBLEM 13.152 A 2-kg sphere A is connected to a fixed Point O by an inextensible cord of length 1.2 m. The sphere is resting on a frictionless horizontal surface at a distance of 1.5 ft from O when it is given a velocity v 0 in a direction perpendicular to line OA. It moves freely until it reaches position A′, when the cord becomes taut. Determine the maximum allowable velocity v 0 if the impulse of the force exerted on the cord is not to exceed 3 N ⋅ s.

SOLUTION For the sphere at A′ immediately before and after the cord becomes taut

mv0 + F Δt = mv A′ mv0 sin θ − F Δt = 0

F Δt = 3 N ⋅ s

m = 2 kg 2(sin 65.38°)v0 = 3 v0 = 1.650 m/s 

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PROBLEM 13.153 A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

SOLUTION Weight and mass.

Bullet: w = 1 oz = Block: W = 5 lb

(a)

1 lb 16

m = 0.001941 lb ⋅ s 2 /ft. M = 0.15528 lb ⋅ s 2 /ft.

Use the principle of impulse and momentum applied to the bullet and the block together.

Σm v1 + ΣImp1→ 2 = mv 2 mv0 cos 30° + 0 = (m + M )v′

Components :

mv0 cos 30° (0.001941)(1400)cos 30° = 0.157221 m+M v′ = 14.968 ft/s v′ =

v′ = 14.97 ft/s 

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PROBLEM 13.153 (Continued)

(b)

Use the principle of impulse and momentum applied to the bullet alone.

x-components:

−mv0 sin 30° + Rx Δt = 0 Rx Δt = mv0 sin 30° = (0.001941)(1400) sin 30° = 1.3587 lb ⋅ s

y-components:

Rx Δt = 1.359 lb ⋅ s 

−mv0 cos 30° + Ry Δt = −mv′ Ry Δt = m(v0 cos 30° − v′) = (0.001941)(1400 cos 30° − 14.968)

Ry Δt = 2.32 lb ⋅ s 

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PROBLEM 13.154 In order to test the resistance of a chain to impact, the chain is suspended from a 240-lb rigid beam supported by two columns. A rod attached to the last link is then hit by a 60-lb block dropped from a 5-ft height. Determine the initial impulse exerted on the chain and the energy absorbed by the chain, assuming that the block does not rebound from the rod and that the columns supporting the beam are (a) perfectly rigid, (b) equivalent to two perfectly elastic springs.

SOLUTION Velocity of the block just before impact: T1 = 0

V1 = Wh = (60 lb)(5 ft) = 300 lb ⋅ ft

1 2 mv V2 = 0 2 T1 + V1 = T2 + V2 T2 =

0 + 300 =

1  60  2  v 2 g 

(600)(32.2) 60 = 17.94 ft/s

v=

(a)

Rigid columns:

− mv + F Δt = 0

 60    (17.94) = F Δt  g 

F Δt = 33.43 lb ⋅ s on the block.

F Δt = 33.4 lb ⋅ s 

All of the kinetic energy of the block is absorbed by the chain. T=

1  60  2   (17.94) 2 g 

= 300 ft ⋅ lb

E = 300 ft ⋅ lb 

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PROBLEM 13.154 (Continued)

(b)

Elastic columns: Momentum of system of block and beam is conserved.

mv = ( M + m)v′ m 60 v′ = v= (17.94 ft/s) (m + M ) 300

v′ = 3.59 ft/s

Referring to figure in part (a), − mv + F Δt = − mv′ F Δt = m(v − v′)  60  =   (17.94 − 3.59)  g  1 2 1 1 mv − mv′2 − Mv′2 2 2 2 60 240 = [(17.94)2 − (3.59)2 ] − (3.59)2 2g 2g

F Δt = 26.7 lb ⋅ s 

E=

E = 240 ft ⋅ lb 

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PROBLEM 13.CQ6 A 5 kg ball A strikes a 1 kg ball B that is initially at rest. Is it possible that after the impact A is not moving and B has a speed of 5v? (a) Yes (b) No Explain your answer.

SOLUTION Answer: (b) No. Conservation of momentum is satisfied, but the coefficient of restitution equation is not. The coefficient of restitution must be less than 1.

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PROBLEM 13.F6 A sphere with a speed v0 rebounds after striking a frictionless inclined plane as shown. Draw impulse-momentum diagrams that could be used to find the velocity of the sphere after the impact.

SOLUTION Answer:

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PROBLEM 13.F7 An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B which can slide along the floor of the car (μk = 0.25). The flatcar was at rest with its brakes released. Instead of A and C coupling as expected, it is observed that A rebounds with a speed of 2 km/h after the impact. Draw impulsemomentum diagrams that could be used to determine (a) the coefficient of restitution and the speed of the flatcar immediately after impact, and (b) the time it takes the load to slide to a stop relative to the car.

SOLUTION Answer: (a)

Look at A and C (the friction force between B and C is not impulsive) to find the velocity after impact.

(b)

Consider just B and C to find their final velocity.

Consider just B to find the time.

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PROBLEM 13.F8 Two frictionless balls strike each other as shown. The coefficient of restitution between the balls is e. Draw the impulse-momentum diagrams that could be used to find the velocities of A and B after the impact.

SOLUTION Answer:

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PROBLEM 13.F9 A 10-kg ball A moving horizontally at 12 m/s strikes a 10-kg block B. The coefficient of restitution of the impact is 0.4 and the coefficient of kinetic friction between the block and the inclined surface is 0.5. Draw impulsemomentum diagrams that could be used to determine the speeds of A and B after the impact.

SOLUTION Answer:

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PROBLEM 13.F10 Block A of mass mA strikes ball B of mass mB with a speed of vA as shown. Draw impulse-momentum diagrams that could be used to determine the speeds of A and B after the impact and the impulse during the impact.

SOLUTION Answer:

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PROBLEM 13.155 The coefficient of restitution between the two collars is known to be 0.70. Determine (a) their velocities after impact, (b) the energy loss during impact.

SOLUTION Impulse-momentum principle (collars A and B): Σmv1 + ΣImp1→2 = Σmv 2

Horizontal components Using data,

: m A v A + mB vB = mA v′A + mB vB′ (5)(1) + (3)( −1.5) = 5v′A + 3vB′ 5v′A + 3vB′ = 0.5

or

(1)

Apply coefficient of restitution. vB′ − v′A = e(v A − vB ) vB′ − v′A = 0.70[1 − ( −0.5)] vB′ − v′A = 1.75

(a)

(2)

Solving Eqs. (1) and (2) simultaneously for the velocities, v′A = −0.59375 m/s

v A = 0.594 m/s



vB′ = 1.15625 m/s

v B = 1.156 m/s



1 1 1 1 m Av 2A + mB vB2 = (5)(1)2 + (3)(−1.5)2 = 5.875 J 2 2 2 2 1 1 1 1 T2 = m A (v′A )2 + mB (vB′ )2 = (5)( −0.59375) 2 + (3)(1.15625) 2 = 2.8867 J 2 2 2 2

Kinetic energies: T1 =

(b)

T1 − T2 = 2.99 J 

Energy loss:

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PROBLEM 13.156 Collars A and B, of the same mass m, are moving toward each other with identical speeds as shown. Knowing that the coefficient of restitution between the collars is e, determine the energy lost in the impact as a function of m, e and v.

SOLUTION Impulse-momentum principle (collars A and B): Σmv1 + ΣImp1→2 = Σmv 2

Horizontal components

: m A v A + mB vB = mA v′A + mB vB′ mv + m(−v) = mv′A + mvB′

Using data,

v′A + vB′ = 0

or

(1)

Apply coefficient of restitution. vB′ − v′A = e(v A − vB ) vB′ − v′A = e [v − (−v)] vB′ − v′A = 2ev

Subtracting Eq. (1) from Eq. (2),

−2v A = 2ev v A = −ev

Adding Eqs. (1) and (2),

(2)

v A = ev

2vB = 2ev vB = ev

v B = ev

1 1 1 1 m A v 2A + mB vB2 = mv 2 + m(−v) 2 = mv 2 2 2 2 2 1 1 1 1 T2 = m A (v′A ) 2 + mB (vB′ )2 = m(ev)2 + m(ev)2 = e 2 mv 2 2 2 2 2

Kinetic energies: T1 =

T1 − T2 = (1 − e2 ) mv 2 

Energy loss:

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PROBLEM 13.157 One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 100 in., the height of the first bounce of the ball must be in the range 53 in. ≤ h ≤ 58 in. Determine the range of the coefficient of restitution of the tennis balls satisfying this requirement.

SOLUTION Uniform accelerated motion: v = 2 gh v′ = 2 gh′

Coefficient of restitution:

e= e=

Height of drop Height of bounce Thus,

v′ v h′ h

h = 100 in. 53 in. ≤ h′ ≤ 58 in. 53 58 ≤e≤ 100 100

0.728 ≤ e ≤ 0.762 

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PROBLEM 13.158 Two disks sliding on a frictionless horizontal plane with opposite velocities of the same magnitude v0 hit each other squarely. Disk A is known to have a weight of 6-lb and is observed to have zero velocity after impact. Determine (a) the weight of disk B, knowing that the coefficient of restitution between the two disks is 0.5, (b) the range of possible values of the weight of disk B if the coefficient of restitution between the two disks is unknown.

SOLUTION Total momentum conserved: m A v A + mB vB = m A v′A + mB v′ (m A )v0 + mB (−v0 ) = 0 + mB v′

m  v′ =  A − 1 v0  mB 

(1)

Relative velocities: vB′ − v′A = e(v A − vB ) v′ = 2ev0

(2)

Subtracting Eq. (2) from Eq. (1) and dividing by v0 , mA − 1 − 2e = 0 mB

mA = 1 + 2e mB

mB =

mA 1 + 2e WB =

Since weight is proportional to mass, (a)

(b)

WA (3) 1 + 2e

With WA = 6 lb and e = 0.5, WB =

6 1 + (2)(0.5)

WB =

6 = 2 lb 1 + (2)(1)

WB =

6 = 6 lb 1 + (2)(0)

WB = 3.00 lb 

With WA = 6 lb and e = 1,

With WA = 6 lb and e = 0,



2.00 lb ≤ WB ≤ 6.00 lb 

Range:

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PROBLEM 13.159 To apply shock loading to an artillery shell, a 20-kg pendulum A is released from a known height and strikes impactor B at a known velocity v0. Impactor B then strikes the 1-kg artillery shell C. Knowing the coefficient of restitution between all objects is e, determine the mass of B to maximize the impulse applied to the artillery shell C.

SOLUTION m A = 20 kg, mB = ?

First impact: A impacts B.

Σmv + ΣImp1→2 = Σmv 2

Impulse-momentum:

m A v0 = m Av′A + mB vB′

Components directed left:

20v0 = 20v′A + mB vB′

(1)

vB′ − v′A = e(v A − vB )

Coefficient of restitution:

vB′ − v′A = ev0 v′A = vB′ − ev0

(2)

Substituting Eq. (2) into Eq. (1) yields 20v0 = 20(vB′ − ev0 ) + mB vB′ 20v0 (1 + e) = (+ mB )vB′ vB′ =

Second impact: B impacts C. Impulse-momentum:

20v0 (1 + e) 20 + mB

(3)

mB = ?, mC = 1 kg Σmv 2 + ΣImp 2→3 = Σmv3

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PROBLEM 13.159 (Continued)

Components directed left: mB vB′ = mB vB′′ + mC vC′′ mB vB′ = mB vB′′ + vC′′

(4)

vC′′ − vB′′ = e(vB′ − vC′ )

Coefficient of restitution:

vC′′ − vB′′ = evB′ vB′′ − vC′′ = evC′

(5)

Substituting Eq. (4) into Eq. (5) yields mB vB′ = mB (vC′′ − evB′ ) + mC vC′′ mB vB′ (1 + e) = (1 + mB )vC′′ vC′′ =

mB vB′ (1 + e) 1 + mB

vC′′ =

20mB v0 (1 + e) 2 (20 + mB )(1 + mB )

(6)

Substituting Eq. (3) for vB′ in Eq. (6) yields

The impulse applied to the shell C is mC vC′′ =

(1)(20) mB v0 (1 + e) 2 (20 + mB )(1 + mB )

To maximize this impulse choose mB such that Z=

mB (20 + mB )(1 + mB )

is maximum. Set dZ /dmB equal to zero. (20 + mB )(1 + mB ) − mB [(20 + mB ) + (1 + mB )] dZ = =0 dmB (20 + mB ) 2 (1 + mB )2 20 + 21mB + mB2 − mB (21 + 2mB ) = 0 20 − mB2 = 0 mB = 4.47 kg 

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PROBLEM 13.160 Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the same weight, has been pushed by dockworkers and hits car B with a velocity of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between B and C and 0.5 between A and B, determine the velocity of each car after all collisions have taken place.

SOLUTION m A = mB = mC = m

Collision between B and C: The total momentum is conserved:

mvB′ + mvC′ = mvB + mvC vB′ + vC′ = 0 + 1.5

(1)

Relative velocities: (vB − vC )(eBC ) = (vC′ − vB′ ) (−1.5)(0.8) = (vC′ − vB′ ) −1.2 = vC′ − vB′

(2)

Solving (1) and (2) simultaneously, vB′ = 1.35 m/s vC′ = 0.15 m/s

v′C = 0.150 m/s



Since vB′ > vC′ , car B collides with car A. Collision between A and B:

mv′A + mvB′′ = mv A + mvB′ v′A + vB′′ = 0 + 1.35

(3)

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PROBLEM 13.160 (Continued)

Relative velocities: (v A − vB′ )eAB = (vB′′ − v′A ) (0 − 1.35)(0.5) = vB′′ − v′A v′A − vB′′ = 0.675

(4)

Solving (3) and (4) simultaneously, 2v′A = 1.35 + 0.675



v′A = 1.013 m/s



v′′B = 0.338 m/s



Since vC′ < vB′′ < v′A , there are no further collisions.

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PROBLEM 13.161 Three steel spheres of equal weight are suspended from the ceiling by cords of equal length which are spaced at a distance slightly greater than the diameter of the spheres. After being pulled back and released, sphere A hits sphere B, which then hits sphere C. Denoting by e the coefficient of restitution between the spheres and by v 0 the velocity of A just before it hits B, determine (a) the velocities of A and B immediately after the first collision, (b) the velocities of B and C immediately after the second collision. (c) Assuming now that n spheres are suspended from the ceiling and that the first sphere is pulled back and released as described above, determine the velocity of the last sphere after it is hit for the first time. (d ) Use the result of Part c to obtain the velocity of the last sphere when n = 5 and e = 0.9.

SOLUTION (a)

First collision (between A and B): The total momentum is conserved: mv A + mvB = mv′A + mvB′ v0 = v′A + vB′

(1)

Relative velocities: (v A − vB )e = (vB′ − v′A ) v0 e = vB′ − v′A

(2)

Solving Equations (1) and (2) simultaneously,

(b)

v′A =

v0 (1 − e)  2

vB′ =

v0 (1 + e)  2

Second collision (between B and C): The total momentum is conserved. mvB′ + mvC = mvB′′ + mvC′

Using the result from (a) for vB′ v0 (1 + e) + 0 = vB′′ + vC′ 2

(3)

Relative velocities: (vB′ − 0)e = vC′ − vB′′

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PROBLEM 13.161 (Continued)

Substituting again for vB′ from (a) v0

(1 + e) (e) = vC′ − vB′′ 2

(4)

Solving equations (3) and (4) simultaneously, vC′ =

(c)

1  v0 (1 + e) (e)  + v0 (1 + e)   2 2 2  vC′ =

v0 (1 + e) 2  4

vB′′ =

v0 (1 − e2 )  4

For n spheres n balls (n − 1)th collision,

we note from the answer to part (b) with n = 3 vn′ = v3′ = vC′ = v3′ =

or

v0 (1 + e)2 4

v0 (1 + e)(3−1) 2(3−1)

Thus, for n balls vn′ =

(d )

v0 (1 + e)( n −1) 2( n −1)



For n = 5, e = 0.90,

from the answer to part (c) with n = 5 vB′ = =

v0 (1 + 0.9)(5−1) 2(5−1) v0 (1.9)4 (2) 4 vB′ = 0.815 v0 

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PROBLEM 13.162 At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vB = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same time, (b) car A hits car B before car C does.

SOLUTION Assume that each car with its rider may be treated as a particle. The masses are: m A = 200 + 40 = 240 kg, mB = 200 + 60 = 260 kg, mC = 200 + 35 = 235 kg.

Assume velocities are positive to the right. The initial velocities are: v A = 2 m/s vB = 0 vC = −1.5 m/s

Let v′A , vB′ , and vC′ be the final velocities. (a)

Cars A and C hit B at the same time. Conservation of momentum for all three cars. m Av A + mB vB + mC vC = m Av′A + mB vB′ + mC vC′ (240)(2) + 0 + (235)(−1.5) = 240v′A + 260vB′ + 235vC′

(1)

Coefficient of restition for cars A and B. vB′ − v′A = e(v A − vB ) = (0.8)(2 − 0) = 1.6

(2)

Coefficient of restitution for cars B and C. vC′ − vB′ = e(vB − vC ) = (0.8)[0 − ( −1.5)] = 1.2

(3)

Solving Eqs. (1), (2), and (3) simultaneously, v′A = −1.288 m/s vB′ = 0.312 m/s vC′ = 1.512 m/s vA′ = 1.288 m/s



vB′ = 0.312 m/s



vC′ = 1.512 m/s



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PROBLEM 13.162 (Continued)

(b)

Car A hits car B before C does. First impact. Car A hits car B. Let v′A and vB′ be the velocities after this impact. Conservation of momentum for cars A and B. m A v A + mB vB = m A v′A + mB vB′ (240)(2) + 0 = 240v′A + 260vB′

(4)

Coefficient of restitution for cars A and B. vB′ − v′A = e(v A − vB ) = (0.8)(2 − 0) = 1.6

(5)

Solving Eqs. (4) and (5) simultaneously, v′A = 0.128 m/s, vB′ = 1.728 m/s v′A = 0.128 m/s v′B = 1.728 m/s

Second impact. Cars B and C hit. Let vB′′ and vC′′ be the velocities after this impact. Conservation of momentum for cars B and C. mB vB′ + mC vC = mB vB′′ + mC vC′′ (260)(1.728) + (235)(−1.5) = 260vB′′ + 235vC′′

(6)

Coefficient of restitution for cars B and C. vC′′ − vB′′ = e(vB′ − vC) = (0.8)[1.728 − (−1.5)] = 2.5824

(7)

Solving Eqs. (6) and (7) simultaneously, vB′′ = −1.03047 m/s vC′′ = 1.55193 m/s v′′B = 1.03047 m/s v′′C = 1.55193 m/s

Third impact. Cars A and B hit again. Let v′′′A and vB′′′ be the velocities after this impact. Conservation of momentum for cars A and B. mA v′A + mB vB′′ = mA v′′′A + mB vB′′′ (240)(0.128) + (260)(−1.03047) = 240vA′′′ + 260vB′′′

(8)

Coefficient of restitution for cars A and B. vB′′′ − v′′′A = e(v′A − vB′′ ) = (0.8)[0.128 − (−1.03047)] = 0.926776

(9)

Solving Eqs. (8) and (9) simultaneously, v′′′A = −0.95633 m/s vB′′′ = −0.02955 m/s v′′′A = 0.95633 m/s v′′′B = 0.02955 m/s PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 761

PROBLEM 13.162 (Continued)

There are no more impacts. The final velocities are: v′′′A = 0.956 m/s



v′′′B = 0.0296 m/s



v′′C = 1.552 m/s



We may check our results by considering conservation of momentum of all three cars over all three impacts. m A v A + mB vB + mC vC = (240)(2) + 0 + (235)(−1.5) = 127.5 kg ⋅ m/s m A v′′′A + mB vB′′′ + mC vC′′ = (240)( −0.95633) + (260)(−0.02955) + (235)(1.55193) = 127.50 kg ⋅ m/s.

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PROBLEM 13.163 At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s when it hits stationary car B. The coefficient of restitution between each car is 0.8. Determine the velocity of car C so that after car B collides with car C the velocity of car B is zero.

SOLUTION Assume that each car with its rider may be treated as a particle. The masses are: m A = 200 + 40 = 240 kg mB = 200 + 60 = 260 kg mC = 200 + 35 = 235 kg

Assume velocities are positive to the right. The initial velocities are: v A = 2 m/s, vB = 0, vC = ?

First impact. Car A hits car B. Let v′A and vB′ be the velocities after this impact. Conservation of momentum for cars A and B. m A v A + mB vB = m A v′A + mB vB′ (240)(2) + 0 = 240v′A + 260vB′

(1)

Coefficient of restitution for cars A and B. vB′ − v′A = e(v A − vB ) = (0.8)(2 − 0) = 1.6

(2)

Solving Eqs. (1) and (2) simultaneously, v′A = 0.128 m/s vB′ = 1.728 m/s v′A = 0.128 m/s v′B = 1.728 m/s

Second impact. Cars B and C hit. Let vB′′ and vC′′ be the velocities after this impact. vB′′ = 0. Coefficient of restitution for cars B and C. vC′′ − vB′′ = e(vB′ − vC ) = (0.8)(1.728 − vC ) vC′′ = 1.3824 − 0.8vC

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 763

PROBLEM 13.163 (Continued)

Conservation of momentum for cars B and C. mB vB′ + mC vC = mB vB′′ + mC vC′′ (260)(1.728) + 235vC = (260)(0) + (235)(1.3824 − 0.8vC ) (235)(1.8)vC = (235)(1.3824) − (260)(1.728) vC = −0.294 m/s

vC = 0.294 m/s



Note: There will be another impact between cars A and B.

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PROBLEM 13.164 Two identical billiard balls can move freely on a horizontal table. Ball A has a velocity v0 as shown and hits ball B, which is at rest, at a Point C defined by θ = 45°. Knowing that the coefficient of restitution between the two balls is e = 0.8 and assuming no friction, determine the velocity of each ball after impact.

SOLUTION

Ball A: t-dir

mv0 sin θ = mv′At  v′At = v0 sin θ

Ball B: t-dir 0 = mB v′Bt  v′Bt = 0

Balls A + B: n-dir mv0 cosθ + 0 = m v′An + m v′Bn

(1)

Coefficient of restitution vBn ′ − v′An = e (v An − vBn ) v′Bn − v′An = e (v0 cosθ − 0)

(2)

Solve (1) and (2) 1 − e  1 + e  v′An = v0  cosθ  ; v′Bn = v0   cosθ  2   2 

With numbers e = 0.8; θ = 45° v′At = v0 sin 45° = 0.707 v0  1 − 0.8  cos 45°  = 0.0707 v0 v′An = v0   2 

v′Bt = 0  1 + 0.8  v′Bn = v0   cos 45° = 0.6364 v0  2  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 765

PROBLEM 13.164 (Continued) (A)

1

v" A = [(0.707 v0 ) 2 + (0.0707v0 )2 ] 2 = 0.711v0



= 0.711v0   0.0707 

β = tan −1   = 5.7106°  0.707  So

θ = 45 − 5.7106 = 39.3°

(B)

v′A = 0.711v0

39.3° 

v′B = 0.636 v0

45° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 766

PROBLEM 13.165 The coefficient of restitution is 0.9 between the two 2.37-in. diameter billiard balls A and B. Ball A is moving in the direction shown with a velocity of 3 ft/s when it strikes ball B, which is at rest. Knowing that after impact B is moving in the x direction, determine (a) the angle θ , (b) the velocity of B after impact.

SOLUTION (a)

Since vB′ is in the x-direction and (assuming no friction), the common tangent between A and B at impact must be parallel to the y-axis,

tan θ =

10 6−D

10 6 − 2.37 = 70.04°

θ = tan −1

θ = 70.0°  (b)

Conservation of momentum in x(n) direction: mv A cos θ + m(vB ) n = m(v′A ) n + mvB′ (3)(cos 70.04) + 0 = (v′A )n + vB′ 1.0241 = (v′A )n + (vB′ )

(1)

Relative velocities in the n direction: e = 0.9

(v A cos θ − (vB ) n )e = vB′ − (v′A ) n (1.0241 − 0)(0.9) = vB′ − (v′A ) n

(1) + (2)

2vB′ = 1.0241(1.9)

(2) v′B = 0.972 ft/s



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PROBLEM 13.166 A 600-g ball A is moving with a velocity of magnitude 6 m/s when it is hit as shown by a 1-kg ball B which has a velocity of magnitude 4 m/s. Knowing that the coefficient of restitution is 0.8 and assuming no friction, determine the velocity of each ball after impact.

SOLUTION

Before

After v A = 6 m/s

(v A ) n = (6)(cos 40°) = 4.596 m/s (v A )t = −6(sin 40°) = −3.857 m/s vB = (vB ) n = −4 m/s (vB )t = 0

t-direction: Total momentum conserved: m A(v A )t + mB (vB )t = mA (vB′ )t + mB (vB′ )t (0.6 kg)( −3.857 m/s) + 0 = (0.6 kg)(v′A )t + (1 kg)(vB′ )t −2.314 m/s = 0.6 (v′A )t + (vB′ )t

(1)

Ball A alone: Momentum conserved: m A (v A )t = mA (v′A )t −3.857 = (v′A )t (v′A )t = −3.857 m/s

(2)

Replacing (v′A )t in (2) in Eq. (1) −2.314 = (0.6)(−3.857) + (vB′ )t −2.314 = −2.314 + (vB′ )t (vB′ )t = 0

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PROBLEM 13.166 (Continued)

n-direction: Relative velocities: [(v A ) n − (vB ) n ]e = (vB′ )n − (v′A )n [(4.596) − (−4)](0.8) = (vB′ )n − (v′A )n 6.877 = (vB′ )n − (v′A )n

(3)

Total momentum conserved: m A (v A ) n + mB (vB ) n = m A (v′A ) n + mB (vB′ ) n (0.6 kg)(4.596 m/s) + (1 kg)( − 4 m/s) = (1 kg)(vB′ ) n + (0.6 kg)(v′A ) n −1.2424 = (vB′ ) n + 0.6(v′A ) n

(4)

Solving Eqs. (4) and (3) simultaneously, (v′A )n = 5.075 m/s (vB′ )n = 1.802 m/s

Velocity of A: tan β =

|(v A )t | |(v A ) n |

3.857 5.075 β = 37.2° =

β + 40° = 77.2°

v′A = (3.857) 2 + (5.075) 2 = 6.37 m/s v′A = 6.37 m/s v′B = 1.802 m/s

Velocity of B:

77.2°  40° 

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PROBLEM 13.167 Two identical hockey pucks are moving on a hockey rink at the same speed of 3 m/s and in perpendicular directions when they strike each other as shown. Assuming a coefficient of restitution e = 0.9, determine the magnitude and direction of the velocity of each puck after impact.

SOLUTION Use principle of impulse-momentum:

Σmv1 + ΣImp1→2 = Σmv 2

t-direction for puck A: − mv A sin 20° + 0 = m(v′A )t (v′A )t = v A sin 20° = 3sin 20° = 1.0261 m/s

t-direction for puck B: − mvB cos 20° + 0 = m(vB′ )t (vB′ )t = vB cos 20° = −3cos 20° = −2.8191 m/s

n-direction for both pucks: mv A cos 20° − mvB sin 20° = m(v′A ) n + m(vB′ ) n (v′A )n + (vB′ ) n = v A cos 20° − vB sin 20° = 3cos 20° − 3sin 20°

(1)

e = 0.9

Coefficient of restitution:

(vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ] = 0.9[3cos 20° − (−3)sin 20°]

(2)

Solving Eqs. (1) and (2) simultaneously, (v′A )n = −0.8338 m/s

(vB′ )n = 2.6268 m/s

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PROBLEM 13.167 (Continued)

Summary:

( v′A )n = 0.8338 m/s

20°

( v′A )t = 1.0261 m/s

70°

( v′B )n = 2.6268 m/s

20°

( v′B )t = 2.8191 m/s

70°

v A = (0.8338) 2 + (1.0261) 2 = 1.322 m/s tan α =

1.0261 0.8338

α = 50.9°

α + 20° = 70.9° v′A = 1.322 m/s

70.9° 

vB′ = (2.6268) 2 + (2.8191) 2 = 3.85 m/s tan β =

2.8191 2.6268

β = 47.0°

β − 20° = 27.0° v′B = 3.85 m/s

27.0° 

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PROBLEM 13.168 Two identical pool balls of 57.15-mm diameter, may move freely on a pool table. Ball B is at rest and ball A has an initial velocity v = v0 i. (a) Knowing that b = 50 mm and e = 0.7, determine the velocity of each ball after impact. (b) Show that if e = 1, the final velocities of the balls from a right angle for all values of b.

SOLUTION Geometry at instant of impact: b 50 = d 57.15 θ = 61.032°

sin θ =

Directions n and t are shown in the figure. Principle of impulse and momentum:

Ball B:

Ball A:

Ball A, t-direction:

mv0 sin θ + 0 = m(v A )t

Ball B, t-direction:

0 + 0 = m(v B )t

Balls A and B, n-direction:

Coefficient of restitution: (a)

(v A )t = v0 sin θ

(1)

(v B )t = 0

(2)

mv0 cos θ + 0 + m(v A ) n + m(vB ) n (v A )n + (vB ) n = v0 cos θ

(3)

(vB ) n − (v A ) n = e[v0 cos θ ]

(4)

e = 0.7. From Eqs. (1) and (2), (v A )t = 0.87489v0

(1)′

( vB ) t = 0

(2)′

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PROBLEM 13.168 (Continued)

From Eqs. (3) and (4), (v A ) n + (vB )n = 0.48432v0

(3)′

(vB ) n − (v A ) n = (0.7)(0.48432v0 )

(4)′

Solving Eqs. (5) and (6) simultaneously, (v A )n = 0.072648v0

(vB ) n = 0.41167v0

v A = (v A )n2 + (v A )t2 = (0.072648v0 ) 2 + (0.87489v02 = 0.87790v0 tan β =

(v A ) n 0.072648v0 = = 0.083037 (v A )t 0.87489v0

β = 4.7468° ϕ = 90° − θ − β = 90° − 61.032° − 4.7468° = 24.221°

(b)

v A = 0.878v0

24.2° 

v B = 0.412v0

61.0° 

e = 1. Eqs. (3) and (4) become (v A )n + (vB )n = v0 cos θ

(3)′′

(vB )n − (v A ) n = v0 cosθ

(4)′′

Solving Eqs. (3)′′ and (4)′′ simultaneously, (v A )n = 0, (vB )t = v0 cos θ

But

(v A )t = v0 sin θ , and (vB )t = 0

vA is in the t-direction and vB is in the n-direction; therefore, the velocity vectors form a right angle.

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PROBLEM 13.169 A boy located at Point A halfway between the center O of a semicircular wall and the wall itself throws a ball at the wall in a direction forming an angle of 45° with OA. Knowing that after hitting the wall the ball rebounds in a direction parallel to OA, determine the coefficient of restitution between the ball and the wall.

SOLUTION Law of sines: sin θ R 2

=

sin135° R

θ = 20.705° α = 45° − 20.705° = 24.295°

Conservation of momentum for ball in t-direction: −v sin θ = −v′ sin α

Coefficient of restitution in n: Dividing,

v(cos θ )e = v′ cos α tan θ = tan α e e=

tan 20.705° tan 24.295°

e = 0.837 

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PROBLEM 13.170 The Mars Pathfinder spacecraft used large airbags to cushion its impact with the planet’s surface when landing. Assuming the spacecraft had an impact velocity of 18 m/s at an angle of 45° with respect to the horizontal, the coefficient of restitution is 0.85 and neglecting friction, determine (a) the height of the first bounce, (b) the length of the first bounce. (Acceleration of gravity on the Mars = 3.73 m/s2.)

SOLUTION Use impulse-momentum principle. Σmv1 + ΣImp1→2 = Σmv 2

The horizontal direction (x to the right) is the tangential direction and the vertical direction ( y upward) is the normal direction. Horizontal components:

mv0 sin 45° = 0 = mvx vx = v0 sin 45°.

v x = 12.728 m/s

Vertical components, using coefficient of restitution e = 0.85 v y − 0 = e[0 − (−v0 cos 45°)] v y = (0.85)(18cos 45°)

v y = 10.819 m/s

The motion during the first bounce is projectile motion. Vertical motion:

Horizontal motion:

1 2 gt 2 v y = (v y )0 − gt y = (v y ) 0 t −

x = vx t

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PROBLEM 13.170 (Continued)

(a)

Height of first bounce: v y = 0:

0 = (v y )0 = gt

(v y ) 0

10.819 m/s = 2.901 s g 3.73 m/s 2 1 y = (10.819)(2.901) − (3.73)(2.901) 2 2 t=

=

y = 15.69 m 

(b)

Length of first bounce: 1 10.819t − (3.73) t 2 = 0 2

y = 0: t = 5.801 s

x = (12.728)(5.801)

x = 73.8 m 

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PROBLEM 13.171 A girl throws a ball at an inclined wall from a height of 3 ft, hitting the wall at A with a horizontal velocity v 0 of magnitude 25 ft/s. Knowing that the coefficient of restitution between the ball and the wall is 0.9 and neglecting friction, determine the distance d from the foot of the wall to the Point B where the ball will hit the ground after bouncing off the wall.

SOLUTION

Momentum in t direction is conserved mv sin 30° = mv′t (25)(sin 30°) = v′t v′t = 12.5 ft/s

Coefficient of restitution in n-direction (v cos 30°)e = v′n (25)(cos 30°)(0.9) = v′n

v′n = 19.49 ft/s

Write v′ in terms of x and y components (v′x )0 = v′n (cos 30°) − vt′ (sin 30°) = 19.49(cos 30°) − 12.5(sin 30°) = 10.63 ft/s (v′y )0 = vn′ (sin 30°) + vt′ (cos 30°) = 19.49(sin 30°) + 12.5(cos 30°) = 20.57 ft/s PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 777

PROBLEM 13.171 (Continued) Projectile motion y = y0 + (v′y )0 t −

At B,

1 2 t2 gt = 3 ft + (20.57 ft/s)t − (32.2 ft/s 2 ) 2 2

y = 0 = 3 + 20.57t B − 16.1tB2 ;

t B = 1.4098 s

xB = x0 + (v′x )0 tB = 0 + 10.63(1.4098);

xB = 14.986 ft

d = xB − 3cot 60° = (14.986 ft) − (3 ft) cot 60° = 13.254 ft d = 13.25 ft 

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PROBLEM 13.172 A sphere rebounds as shown after striking an inclined plane with a vertical velocity v0 of magnitude v0 = 5 m/s. Knowing that α = 30° and e = 0.8 between the sphere and the plane, determine the height h reached by the sphere.

SOLUTION Rebound at A Conservation of momentum in the t-direction: mv0 sin 30° = m(v′A )t (v′A )t = (5 m/s)(sin 30°) = 2.5 m/s

Relative velocities in the n-direction: (−v0 cos 30° − 0)e = 0 − (v′A ) n (v′A ) n = (0.8)(5 m/s)(cos 30°) = 3.4641 m/s

Projectile motion between A and B: After rebound (vx )0 = (v′A )t cos 30° + (v′A )n sin 30° (vx )0 = (2.5)(cos 30°) + (3.4641) sin 30 = 3.8971 m/s (v y )0 = −(v′A )t sin 30° + (v′A ) n cos30° (v y )0 = −(2.5)(sin 30°) + (3.4641) cos 30° = 1.750 m/s

x-direction:

x = (v x ) 0 t

v x = (v x ) 0

x = 3.8971t vx = 3.8971 m/s = vB

y-direction:

At A:

1 2 gt 2 v y = (v y )0 − gt y = (v y ) 0 t −

v y = 0 = (v y )0 − gt AB t AB = (v y )0 /g =

1.75 m/s 9.81 m/s 2

t A− B = 0.17839 s

At B: gt A2 − B 2 9.81 (0.17839) 2 h = (1.75)(0.17839) − 2 y = h = (v y ) 0 t A − B −

h = 0.156 m 

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PROBLEM 13.173 A sphere rebounds as shown after striking an inclined plane with a vertical velocity v0 of magnitude v0 = 6 m/s. Determine the value of α that will maximize the horizontal distance the ball travels before reaching its maximum height h assuming the coefficient of restitution between the ball and the ground is (a) e = 1, (b) e = 0.8.

SOLUTION Directions x, y, n, and t are shown in the sketch. Analysis of the impact: Use the principle of impulse and momentum for components in the t-direction. mv0 sin α + 0 = m(vt′ )1 (vt )1 = v0 sin α

Coefficient of restitution:

(1)

( v n )1 = −e( v n )0 (vn )1 = ev0 cos α

(2)

x and y components of velocity immediately after impact: (vx )1 = (vn )1 sin α + (vt )1 cos α = v0 (1 + e)sin α cos α =

1 v0 (1 + e) sin 2 α 2

(3)

(v y )1 = (vn )1 cos α − (vt )1 sin α = v0 (e cos 2 α − sin 2 α ) 1 v0 [e(1 + cos 2α ) − (1 − cos 2α )] 2 1 = v0 [(1 + e) cos 2α − (1 − e)] 2 =

(4)

Projectile motion: Use coordinates x and y with the origin at the point of impact. x0 = 0 y0 = 0

Vertical motion:

v y = (v y )1 − gt vy =

1 v0 [(1 + e) cos 2α − (1 − e)] − gt 2

v y = 0 at the position of maximum height where t2 =

(v y )1 g

=

v0 [(1 + e) cos α − (1 − e)] 2g

(5)

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PROBLEM 13.173 (Continued)

vx = (vx )1 =

Horizontal motion:

1 v0 (1 + e)sin 2α 2

x = (vx )1 t

At the point of maximum height, x2 = (vx )1 t2 =

v02 (1 + e) sin 2α [(1 + e) cos 2α − (1 − e)] 4g

Let θ = 2α and Z = 4 gx2 /v02 (1 + e). To determine the value of θ that maximizes x2 (or Z ), differentiate Z with respect to θ and set the derivative equal to zero. Z = sin θ [(1 + e) cos θ − (1 − e)] dZ = cos θ [(1 + e) cos θ − (1 − e)] − (1 + e) sin 2 θ dθ = (1 + e) cos 2 θ − (1 − e) cos θ − (1 + e)(1 − cos 2 θ ) = 0 2(1 + e) cos 2 θ − (1 − e) cos θ − (1 + e) = 0

This is a quadratic equation for cos θ . (a)

e =1

4 cos 2 θ − 2 = 0 cos 2 θ =

1 2

cos θ = ±

2 2

θ = ±45°

and

± 135°

α = 22.5°

and

67.5°

Reject the negative values of θ which make x2 negative. Reject α = 67.5° since it makes a smaller maximum height.

α = 22.5°  (b)

e = 0.8

3.6 cos 2 θ − 0.2cos θ − 1.8 = 0

cos θ = 0.73543 and −0.67987 θ = ±42.656° and ±132.833° α = ±21.328° and ±66.417°

α = 21.3° 

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PROBLEM 13.174 Two cars of the same mass run head-on into each other at C. After the collision, the cars skid with their brakes locked and come to a stop in the positions shown in the lower part of the figure. Knowing that the speed of car A just before impact was 5 mi/h and that the coefficient of kinetic friction between the pavement and the tires of both cars is 0.30, determine (a) the speed of car B just before impact, (b) the effective coefficient of restitution between the two cars.

SOLUTION (a)

At C:

Conservation of total momentum: mA = mB = m

5 mi/h = 7.333 ft/s m A v A + mB vB = mA v′A + mB vB′ −7.333 + vB = v′A + vB′

(1)

Work and energy. Care A (after impact): 1 m A (v′A )2 2 T2 = 0 T1 =

U1− 2 = F f (12) U1− 2 = μk mA g (12 ft) T1 + U1− 2 = T2

1 m A (v′A ) 2 − mk m A g (12) = 0 2 (v′A ) 2 = (2)(12 ft)(0.3)(32.2 ft/s 2 ) = 231.84 ft/s 2 v′A = 15.226 ft/s

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PROBLEM 13.174 (Continued)

Car B (after impact): 1 mB (vB′ )2 2 T2 = 0 T1 =

U1− 2 = μk mB g (3) T1 + U1− 2 = T2

1 mB (vB′ ) 2 − μk mB g (3) 2 vB′2 = (2)(3 ft)(0.3)(32.2 ft/s 2 ) (vB′ )2 = 57.96 ft/s 2 vB′ = 7.613 ft/s

From (1)

vB = 7.333 + v′A + vB′ = 7.333 + 15.226 + 7.613 vB = 30.2 ft/s = 20.6 mi/h 

(b)

Relative velocities: (−v A − vB ) e = vB′ − v′A (−7.333 − 30.2) e = 7.613 − 15.226 −(7.613) = 0.2028 e= −(37.53)

e = 0.203 

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PROBLEM 13.175 A 1-kg block B is moving with a velocity v0 of magnitude v0 = 2 m/s as it hits the 0.5-kg sphere A, which is at rest and hanging from a cord attached at O. Knowing that μk = 0.6 between the block and the horizontal surface and e = 0.8 between the block and the sphere, determine after impact (a) the maximum height h reached by the sphere, (b) the distance x traveled by the block.

SOLUTION Velocities just after impact

Total momentum in the horizontal direction is conserved: m A v A + mB vB = m A v′A + mB vB′ 0 + (1 kg)(2 m/s) = (0.5 kg)(v′A ) + (1 kg)(vB′ ) 4 = v′A + 2vB′

(1)

Relative velocities: (v A − vB ) e = (vB′ − v′A ) (0 − 2)(0.8) = vB′ − v′A −1.6 = vB′ − v′A

(2)

Solving Eqs. (1) and (2) simultaneously: vB′ = 0.8 m/s v′A = 2.4 m/s

(a)

Conservation of energy: 1 m Av12 V1 = 0 2 1 T1 = m A (2.4 m/s) 2 = 2.88 m A 2 T1 =

T2 = 0 V2 = m A gh

T1 + V1 = T2 + V2

2.88 m A + 0 = 0 + m A (9.81) h h = 0.294 m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 784

PROBLEM 13.175 (Continued)

(b)

Work and energy:

1 1 mB v12 = mB (0.8 m/s) 2 = 0.32mB T2 = 0 2 2 = − F f x = − μk Nx = − μ x mB gx = −(0.6)(mB )(9.81) x

T1 = U1− 2

U1− 2 = −5.886mB x T1 + U1− 2 = T2 :

0.32mB − 5.886mB x = 0 x = 0.0544 m

x = 54.4 mm 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 785

PROBLEM 13.176 A 0.25-lb ball thrown with a horizontal velocity v0 strikes a 1.5-lb plate attached to a vertical wall at a height of 36 in. above the ground. It is observed that after rebounding, the ball hits the ground at a distance of 24 in. from the wall when the plate is rigidly attached to the wall (Figure 1) and at a distance of 10 in. when a foamrubber mat is placed between the plate and the wall (Figure 2). Determine (a) the coefficient of restitution e between the ball and the plate, (b) the initial velocity v0 of the ball.

SOLUTION

(a)

Figure (1), ball alone relative velocities v0e = (v′B )1

Projectile motion t = time for the ball to hit ground 2 ft = v0et

(1)

Figure (2), ball and plate relative velocities (vB − v A )e = v′P + (v′B ) 2 vB = v0 ,

vP = 0

v0e = v′P + (v′B ) 2

(2)

Conservation of momentum mBvB + mPvP = mBv′B + mP v′P 0.25 0.25 1.5 (−v′B ) 2 + v0 + 0 = v′P g g g

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 786

PROBLEM 13.176 (Continued)

0.25v0 = −0.25(v′B ) 2 + 1.5v′p  v0 = −(v′B ) 2 − 6v′p (v′B ) 2 =

Solving (2) and (3) for (v′B ) 2 ,

(3)

(6e − 1) v0 7

Projectile motion 0.8333 =

(6e − 1) v0t 7

(4)

Dividing Equation (4) by Equation (1) 0.8333 6e − 1 = ; 2.91655e = 6e − 1 2 7e e = 0.324 

(b)

From Figure (1) h=

Projectile motion,

1 2 1 gt ; 3 = (32.2)t 2 2 2

6 = 32.2t 2

(5)

From Equation (1), 2 = v0et  t =

2 6.1728 = 0.324v0 v0

Using Equation (5) 2

 6.1728  2 6 = 32.2    6v0 = 1226.947  v0  v02 = 204.49 v0 = 14.30 ft/s 

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PROBLEM 13.177 After having been pushed by an airline employee, an empty 40-kg luggage carrier A hits with a velocity of 5 m/s an identical carrier B containing a 15-kg suitcase equipped with rollers. The impact causes the suitcase to roll into the left wall of carrier B. Knowing that the coefficient of restitution between the two carriers is 0.80 and that the coefficient of restitution between the suitcase and the wall of carrier B is 0.30, determine (a) the velocity of carrier B after the suitcase hits its wall for the first time, (b) the total energy lost in that impact.

SOLUTION (a)

Impact between A and B: Total momentum conserved:

m A v A + mB vB = m A v′A + mB vB′

mA = mB = 40 kg

5 m/s + 0 = v′A + vB′

(1)

Relative velocities: (v A − vB )eAB = vB′ − v′A (5 − 0)(0.80) = vB′ − v′A

(2)

Adding Eqs. (1) and (2) (5 m/s)(1 + 0.80) = 2vB′ vB′ = 4.5 m/s

Impact between B and C (after A hits B) Total momentum conserved: mB vB′ + mC vC′ = mB vB′′ + mC vC′′ (40 kg)(4.5 m/s) + 0 = (40 kg) vB′′ + (15 kg) vC′′ 4.5 = vB′′ + 0.375 vC′′

(3)

Relative velocities: (vB′ − vC′ ) eBC = vC′′ − vB′′ (4.5 − 0)(0.30) = vC′′ − vB′′

(4)

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PROBLEM 13.177 (Continued)

Adding Eqs. (4) and (3) (4.5)(1 + 0.3) = (1.375)vC′′ vC′′ = 4.2545 m/s vB′′ = 4.5 − 0.375(4.2545) vB′′ = 2.90 m/s

(b)

vB′ = 2.90 m/s 

ΔTL = (TB′ + TC′ ) − (TB′′ + TC′′) 1  40  kg  (4.5 m/s) 2 = 405 J mB (vB′ )2 =  2  2  1  40  kg  (2.90) 2 = 168.72 J TC′ = 0 TB′′ = mB (vB′′ ) 2 =  2  2  TB′ =

TC′′ =

1  15  kg  (4.2545 m/s) 2 = 135.76 J mC (vC′′ )2 =  2  2 

Δ TL = (405 + 0) − (168.72 + 135.76) = 100.5 J

ΔTL = 100.5 J 

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PROBLEM 13.178 Blocks A and B each weigh 0.8 lb and block C weighs 2.4 lb. The coefficient of friction between the blocks and the plane is μk = 0.30. Initially block A is moving at a speed v0 = 15 ft/s and blocks B and C are at rest (Fig. 1). After A strikes B and B strikes C, all three blocks come to a stop in the positions shown (Fig. 2). Determine (a) the coefficients of restitution between A and B and between B and C, (b) the displacement x of block C.

SOLUTION (a)

Work and energy Velocity of A just before impact with B: T1 =

1 WA 2 v0 2 g

T2 =

( )

1 WA 2 vA 2 g

2

U1− 2 = − μk WA (1 ft) T1 + U1− 2 = T2

( )

1 WA 2 1 WA 2 v0 − μk WA (1) = vA 2 g 2 g

2

(v A2 )2 = v02 − 2μk g = (15 ft/s)2 − 2(0.3)(32.2 ft/s 2 )(1 ft) (v A2 )2 = 205.68 ft/s 2 ,

(v A ) 2 = 14.342 ft/s

Velocity of A after impact with B:

(v′A ) 2

1 WA (v′A ) 22 T3 = 0 2 g = − μk WA (3/12)

T2′ = U 2−3

T2′ + U 2−3 = T3 ,

1 WA (v′A ) 22 − ( μk )(WA /4) = 0 2 g

1  (v′A )22 = 2(0.3)(32.2 ft/s 2 )  ft  = 4.83 ft 2 /s 2 4  ′ = (v A )2 2.198 ft/s

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PROBLEM 13.178 (Continued)

Conservation of momentum as A hits B: (v A )2 = 14.342 ft/s (v′A )2 = 2.198 ft/s m A (v A ) 2 + mB vB = mB (v′A ) 2 + mB vB′ mA = mB vB′ = 12.144 ft/s 14.342 + 0 = 2.198 + vB′

Relative velocities A and B: [(v A )2 − vB ]eAB = vB′ − (v′A )2 (14.342 − 0)eAB = 12.144 − 2.198

eAB = 0.694 

Work and energy. Velocity of B just before impact with C: W 1 WB (vB′ ) 22 = B (12.144) 2 2 g 2g W 1 WB (vB′ ) 42 = B (vB′ ) 42 T4 = 2 g 2g (vB′ ) = 12.144 ft/s U 2− 4 = − μk WB (1 ft) = (0.3) WB T2 =

F f = μk WB

(v ′ ) 2 (12.144)2 − 0.3 = B 4 2g 2g

T2 + U 2 − 4 = T4 ,

(vB′ )4 = 11.321 ft/s

Conservation of momentum as B hits C: 0.8 g 2.4 mC = g mB =

(vB′ )4 = 11.321 ft/s mB (vB′ ) 4 + mC vC = mB (vB′′ ) 4 + mC vC′ 0.8 0.8 (2.4) (11.321) + 0 = (vB′′ )4 + (vC′ ) g g g 11.321 = (vB′′ ) 4 + 3vC′

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PROBLEM 13.178 (Continued)

Velocity of B after B hits C ,(vB′′ ) 4 = 0.

(Compare Figures (1) and (2).) vC′ = 3.774 ft/s

Relative velocities B and C: ((vB′ ) 4 − vC )eBC = vC′ − (vB′′ ) 4 (11.321 − 0)eBC = 3.774 − 0 eBC = 0.333 

(b)

Work and energy, Block C: 1 WC (vC )2 T5 = 0 U 4−5 = − μkWC ( x) 2 g 1 WC (3.774) 2 − (0.3) WC ( x) = 0 = T5 2 g

T4 = T4 + U 4−5

x=

(3.774) 2 = 0.737 ft 2(32.2)(0.3) x = 8.84 in. 

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PROBLEM 13.179 A 0.5-kg sphere A is dropped from a height of 0.6 m onto a 1.0 kg plate B, which is supported by a nested set of springs and is initially at rest. Knowing that the coefficient of restitution between the sphere and the plate is e = 0.8, determine (a) the height h reached by the sphere after rebound, (b) the constant k of the single spring equivalent to the given set if the maximum deflection of the plate is observed to be equal to 3h.

SOLUTION Velocity of A and B after impact. m A = 0.5 kg mB = 1.0 kg

Sphere A falls. Use conservation of energy to find v A , the speed just before impact. Use the plate surface as the datum. 1 m Av A2 , V2 = 0 2 1 0 + m A gh0 = m Av A2 + 0 2

T1 = 0, V1 = m A gh0 , T2 = T1 + V1 = T2 + V2

With

h0 = 0.6 m, v A = 2 gh0 = (2)(9.81)(0.6)

v A = 3.4310 m/s ↓

Analysis of the impact. Conservation of momentum. m A v A + mB v B = m A v′A + mB v′B

Dividing by m A and using y-components

with

vB = 0

with (mB /mA = 2)

−3.4310 + 0 = (v′A ) y + 2(vB′ ) y

Coefficient of restitution.

(1)

(vB′ ) y − (v′A ) y = e [(v A ) y − (vB ) y ] (vB′ ) y − (v′A ) y = e (v A ) y = −3.4310e

(2)

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PROBLEM 13.179 (Continued)

Solving Eqs. (1) and (2) simultaneously with e = 0.8 gives (v′A ) y = 0.68621 m/s (vB′ ) y = −2.0586 m/s

v′A = 0.68621 m/s v′B = 2.0586 m/s

(a)

Sphere A rises. Use conservation of energy to find h. 1 m A (v′A )2 , V1 = 0, T2 = 0, V2 = m A gh 2 1 T1 + V1 = T2 + V2: m A (v′A )2 + 0 = 0 + m A gh 2 T1 =

h=

(b)

(v′A ) 2 (0.68621) 2 = 2g (2)(9.81)

h = 0.0240 m 

Plate B falls and compresses the spring. Use conservation of energy. Let δ 0 be the initial compression of the spring and Δ be the additional compression of the spring after impact. In the initial equilibrium state, ΣFy = 0: kδ 0 − WB = 0 or kδ 0 = WB T1 =

Just after impact:

1 1 mB (vB′ )2 , V1 = kδ 02 2 2 T2 = 0

At maximum deflection of the plate,

V2 = (V2 ) g + (V2 )e = −WB Δ +

Conservation of energy:

(3)

1 k (δ 0 + Δ) 2 2

T1 + V1 = T2 + V2

1 1 1 1 mB (vB′ ) 2 + kδ 02 = 0 − WB Δ + kδ 02 + kδ 0 Δ + k Δ 2 2 2 2 2

Invoking the result of Eq. (3) gives 1 1 mB (vB′ )2 = k Δ 2 2 2

Data:

(4)

mB = 1.0 kg, vB′ = 2.0586 m/s Δ = 3h = (3)(0.024) = 0.072 m k=

mB (vB′ ) 2 Δ2

=

(1.0)(2.0586) 2 (0.072) 2

k = 817 N/m 

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PROBLEM 13.180 A 0.5-kg sphere A is dropped from a height of 0.6 m onto a 1.0-kg plate B, which is supported by a nested set of springs and is initially at rest. Knowing that the set of springs is equivalent to a single spring of constant k = 900 N/m , determine (a) the value of the coefficient of restitution between the sphere and the plate for which the height h reached by the sphere after rebound is maximum, (b) the corresponding value of h, (c) the corresponding value of the maximum deflection of the plate.

SOLUTION m A = 0.5 kg mB = 1.0 kg k = 900 N/m

Sphere A falls. Use conservation of energy to find v A , the speed just before impact. Use the plate surface as the datum. T1 = 0 T2 =

With h0 = 0.6 m,

V1 = m A gh0

1 m Av A2 , 2

V2 = 0

v A = 2 gh0 = (2)(9.81)(0.6)

v A = 3.4310 m/s

Analysis of impact. Conservation of momentum. m A v A + mB v B = mA v′A + mB vB′ with v B = 0

Dividing by m A and using y components

with (mB /mA = 2)

−3.4310 = (v′A ) y + 2(vB′ ) y

Coefficient of restitution.

(1)

(vB′ ) y − (v′A ) y = e [(v A ) y − (vB ) y ] (vB′ ) y − (v′A ) y = e(v A ) y = −3.4310e (vB′ ) y = −3.4310 + (v′A ) y

(2)

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PROBLEM 13.180 (Continued)

Substituting into Eq. (1), −3.4310 = (v′A ) y + (2)[−3.4310e + (v′A ) y ]

From Eq. (2), (a)

(v′A ) y = 1.1437 (2e − 1)

(3)

(vB′ ) y = −1.1437(1 + e)

(4)

Sphere A rises. Use conservation of energy to find h. 1 m A (v′A )2 , V1 = 0 2 T2 = 0, V2 = mA gh T1 =

1 m A (v′A )2 + 0 = 0 + m A gh 2 (v′ ) 2 (1.1437) 2 (2e − 1)2 h= A = 2g (2)(9.81)

T1 + V1 = T2 + V2 :

Since h is to be maximum, e must be as large as possible. e = 1.000 

Coefficient of restitution for maximum h: (b)

Corresponding value of h.

(v′A ) = 1.1437[(2)(1) − 1] = 1.1437 m/s h=

(c)

(v′A ) 2 (1.1437)2 = 2g (2)(9.81)

h = 0.0667 m 

Plate B falls and compresses the spring. Use conservation of energy. Let δ 0 be the initial compression of the spring and Δ be the additional compression of the spring after impact. In the initial equilibrium state, ΣFy = 0 kδ 0 − WB = 0 or kδ 0 = WB

Just after impact:

T1 =

1 mB (vB′ )2 , 2

At maximum deflection of the plate,

V1 =

1 2 kδ 0 2

T2 = 0

V2 = (V2 ) g + (V2 )e = −WB Δ +

Conservation of energy:

(3)

1 k (δ 0 + Δ) 2 2

T1 + V1 = T2 + V2

1 1 1 1 mB (vB′ ) 2 + kδ 02 = 0 − WB Δ + kδ 02 + kδ 0 Δ + k Δ 2 2 2 2 2

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PROBLEM 13.180 (Continued)

Invoking the result of Eq. (3) gives 1 1 mB (vB′ )2 = k Δ 2 2 2

Data:

mB = 1.0 kg, (vB′ ) y = −1.1437(1 + 1) = −2.2874 m/s.

v′B = 2.2874 m/s , k = 900 N/m Δ2 =

mB (vB′ ) 2 (1.0)(2.2874) 2 = = 0.0058133 m 2 k 900

Δ = 0.0762 m 

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PROBLEM 13.181 The three blocks shown are identical. Blocks B and C are at rest when block B is hit by block A, which is moving with a velocity v A of 3 ft/s. After the impact, which is assumed to be perfectly plastic (e = 0), the velocity of blocks A and B decreases due to friction, while block C picks up speed, until all three blocks are moving with the same velocity v. Knowing that the coefficient of kinetic friction between all surfaces is μk = 0.20, determine (a) the time required for the three blocks to reach the same velocity, (b) the total distance traveled by each block during that time.

SOLUTION (a)

Impact between A and B, conservation of momentum mv A + mvB + mvC = mv′A + mvB′ + mvC′ 3 + 0 = v′A + vB′ + 0

Relative velocities (e = 0) (v A − vB )e = vB′ − v′A 0 = vB′ − v′A v′A = vB′

3 = 2vB′ vB′ = 1.5 ft/s

v = Final (common) velocity

Block C: Impulse and momentum

WC vC + F f t = 0 + (0.2)t =

WC v g

F f = μkWC

v g

v = (0.2) gt

(1)

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PROBLEM 13.181 (Continued)

Blocks A and B: Impulse and momentum

2

W W (1.5) − 4(0.2)Wt = 2 v g g 1.5 − 0.4 gt = v

(2)

Substitute v from Eq. (1) into Eq. (2) 1.5 − 0.4 gt = 0.2 gt

t=

(b)

(1.5 ft/s) 0.6(32.2 ft/s 2 )

t = 0.0776 s 

Work and energy: v = (0.2)(32.2)(0.0776) = 0.5 ft/s

From Eq. (1) Block C:

T1 = 0

T2 =

1W W (v ) 2 = (0.5) 2 2 g 2g

U1− 2 = F f xC = μk WxC = 0.2WxC T1 + U1− 2 = T2 xC =

0 + (0.2)(W ) xC =

1W 2 v 2 g

(0.5 ft/s) 2 = 0.01941 ft 0.2(2)(32.2 ft/s 2 )

xC = 0.01941 ft 

 Blocks A and B:

T1 =

1 W 2 2 g

 2  (1.5) = 2.25W 

T2 =

1 W 2 2 g

 2  (0.5) = 0.25W 

U1− 2 = −4 μkWgx A = −0.8Wgx A T1 + U1− 2 = T2 2.25W − 4(0.2)W (32.2) x A = 0.25W



x A = 0.07764 ft

x A = 0.0776 ft 

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PROBLEM 13.182 Block A is released from rest and slides down the frictionless surface of B until it hits a bumper on the right end of B. Block A has a mass of 10 kg and object B has a mass of 30 kg and B can roll freely on the ground. Determine the velocities of A and B immediately after impact when (a) e = 0, (b) e = 0.7.

SOLUTION Let the x-direction be positive to the right and the y-direction vertically upward. Let (v A ) x , (v A ) y , (v A ) x and (vB ) y be velocity components just before impact and (v′A ) x ,(v′A )′y , (vB′ ) x , and (vB′ ) y those just after impact. By inspection, (v A ) y = (vB ) y = (v′A ) y = (vB′ ) y = 0

Conservation of momentum for x-direction: While block is sliding down:

0 + 0 = m A (v A ) x + mB (vB ) x

(vB ) x = − β (v A ) x

(1)

Impact:

0 + 0 = m A (v′A ) x + mB (vB′ ) x

(vB′ ) x = − β (v′A ) x

(2)

β = mA /mB

where

Conservation of energy during frictionless sliding: Initial potential energies: m A gh for A,

0 for B.

Potential energy just before impact:

V1 = 0

Initial kinetic energy:

T0 = 0 (rest)

Kinetic energy just before impact:

T1 =

1 1 m Av 2A + mB vB2 2 2

T0 + V0 = T1 + V1 1 1 1 m Av A2 + mB vB2 = (m A + mB β 2 )v A2 2 2 2 1 = m A (1 + β ) v A2 2

m A gh =

v A2 = (v A ) 2x =

2 gh 1+ β

vA =

2 gh 1+ β

(3)

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PROBLEM 13.182 (Continued)

Velocities just before impact:

2 gh 1+ β

vA = vB = β

2 gh 1+ β

Analysis of impact. Use Eq. (2) together with coefficient of restitution. (vB′ ) x − (v′A ) x = e[(v A ) x − (vB ) x ] − β (v′A ) x − (v′A ) x = e[(v A ) x + β (v A ) x ] (v′A ) x = −e(v A ) x

(4)

m A = 10 kg

Data:

mB = 30 kg h = 0.2 m g = 9.81 m/s 2

β= From Eq. (3),

(a)

e = 0:

10 kg = 0.33333 30 kg

(2)(9.81)(0.2) 1.33333 = 1.71552 m/s

vA =

(v′A ) x = 0

(vB′ ) x = 0

v′A = 0  v′B = 0 

(b)

e = 0.7:

(v′A ) x = −(0.7)(1.71552)

= −1.20086 m/s ′ (vB ) x = −(0.33333)(1.20086) = 0.40029 m/s v′A = 1.201 m/s



v′B = 0.400 m/s



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PROBLEM 13.183 A 20-g bullet fired into a 4-kg wooden block suspended from cords AC and BD penetrates the block at Point E, halfway between C and D, without hitting cord BD. Determine (a) the maximum height h to which the block and the embedded bullet will swing after impact, (b) the total impulse exerted on the block by the two cords during the impact.

SOLUTION

Total momentum in x is conserved: ′ mbl vbl + mbu vbu cos 20° = mbl vbl′ + mbu vbu

′ ) (vbl′ = vbu

0 + (0.02 kg)(−600 m/s)( cos 20) = (4.02 kg)(vbl′ )

vbl′ = −2.805 m/s

Conservation of energy: 1 (mbl + mbu )(vbl′ ) 2 2  4.02 kg  2 T1 =   (2.805 m/s) 2   T1 = 15.815 J

T1 =

V1 = 0 T2 = 0

V2 = (mbl + mbu ) gh

V2 = (4.02 kg)(9.81 m/s 2 )(h) = 39.44h T1 + V1 = T2 + V2 15.815 + 0 = 0 + 39.44h

h = 0.401 m

(b)

h = 401 mm 

Refer to figure in part (a). Impulse-momentum in y-direction: mbu vbu sin 20 + F Δt = (mbl + mbu )(vbl′ ) y (vbl ) y = 0 (0.02 kg) (−600 m/s) ( sin 20°) + FΔt = 0

FΔt = 4.10 N ⋅ s 

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PROBLEM 13.184 A 2-lb ball A is suspended from a spring of constant 10 lb/in and is initially at rest when it is struck by 1-lb ball B as shown. Neglecting friction and knowing the coefficient of restitution between the balls is 0.6, determine (a) the velocities of A and B after the impact, (b) the maximum height reached by A.

SOLUTION 2 lb = 0.062112 lb ⋅ s 2 /ft 2 32.2 ft/s

Masses:

mA =

mB =

Other data:

k = (10 lb/in.)(12 in./ft.) = 120 lb/ft, e = 0.6

1 lb = 0.031056 lb ⋅ s 2 /ft 2 32.2 ft/s

v A = 0, vB = 2 ft/s

For analysis of the impact use the principle of impulse and momentum. Σmv1 + ΣImp1 2 = Σmv 2

t-direction for ball A: 0 + 0 = mA (v′A )t

t-direction for ball B:

(v′A )t = 0

mB vB sin 20° + 0 = mB (vB′ )t (vB′ )t = vB sin 20° = (2)(sin 20°) = 0.6840 ft/s

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PROBLEM 13.184 (Continued)

n-direction for balls A and B: mB vB cos 20° + 0 = mB (vB′ ) n + m A (v′A ) n (vB′ )n +

mA (v′A )n = vB cos 20° mB

(vB′ )n + 2(v′A ) n = (2) cos 20°

Coefficient of restitution:

(1)

(vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ] = e[0 − (vB cos 20°)] = −(0.6)(2) cos 20°

(2)

Solving Eqs. (1) and (2) simultaneously, (v′A )n = 1.00234 ft/s

(a)

(vB′ ) n = −0.12529 ft/s

Velocities after the impact: v′A = 1.00234 ft/s

v′A = 1.002 ft/s 

v′B = (0.6840 ft/s →) + (0.12529 ft/s )

vB = (0.6840) 2 + (0.12529)2 = 0.695 ft/s tan β =

0.12529 0.6840

β = 10.4° v′B = 0.695 ft/s

(b)

10.4° 

Maximum height reached by A: Use conservation of energy for ball A after the impact. Position 1: Just after impact. T1 =

1 1 m A (v′A ) 2 = (0.062112)(1.00234)2 = 0.0312021 ft ⋅ lb 2 2

Force in spring = weight of A x1 = −

W 2 lb F =− A =− = −0.016667 ft 120 lb/ft k k 2

(V1 )e =

1 2 1  WB  WA2 kx1 = k  = 2 2  k  2k

(2 lb) 2 = 0.016667 ft ⋅ lb (2)(120) (V1 ) g = 0 (datum) =

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PROBLEM 13.184 (Continued) Position 2: Maximum height h. V2 = 0 T2 = 0 1 1 k (h + x1 ) 2 = (120)(h − 0.016667) 2 2 2 2 = 60h − 2h + 0.016667

(V2 )e =

(V2 ) g = WA h = (2 lb)h = 2h

Conservation of energy: T1 + V1 = T2 + V2 0.031202 + 0.016667 = 0 + 60h 2 − 2h + 0.016667 + 2h 60h 2 = 0.031202

h = ±0.022804 ft h = 0.274 in. 

Using the positive root,

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PROBLEM 13.185 Ball B is hanging from an inextensible cord. An identical ball A is released from rest when it is just touching the cord and drops through the vertical distance hA = 8 in. before striking ball B. Assuming perfectly elastic impact (e = 0.9) and no friction, determine the resulting maximum vertical displacement hB of ball B.

SOLUTION Ball A falls T1 = 0 V2 = 0 T1 + V1 = T2 + V2

(Put datum at 2)

h = 8 in. = 0.66667 ft 1 mgh = mv 2A 2 v A = 2 gh = (2)(32.2)(0.66667) = 6.5524 ft/s

Impact

θ = sin −1

r = 30° 2r

Impulse-Momentum

Unknowns:

vB′ , v′At , v′An

x-dir 0 + 0 = mB vB′ + m Av′An sin 30° + mA v′At cos 30°

(1)

Noting that m A = mB and dividing by mA vB′ + v′An sin 30° + v′At cos 30° = 0

(1)

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PROBLEM 13.185 (Continued)

Ball A alone: Momentum in t-direction: − m Av A sin 30° + 0 = mA v At v′At = −v A sin 30° = −6.5524 sin 30° = −3.2762 ft/s

(2)

Coefficient of restitution: ′ − v′An = e(v An − ven ) vBn vB′ sin 30° − v′An = 0.9(v A cos 30° − 0)

(3)

With known value for vAt, Eqs. (1) and (3) become vB′ + v′An sin 30° = 3.2762 cos 30° vB′ sin 30° − v′An = (0.9)(6.5524) cos 30°

Solving the two equations simultaneously, vB′ = 4.31265 ft/s v′An = −2.9508 ft/s

After the impact, ball B swings upward. Using B as a free body T ′ + V ′ = TB + VB

where

1 mB (vB′ )2 , 2 V ′ = 0, T′ =

TB = 0

and

VB = mB ghB 1 mB (vB′ ) 2 = mB ghB 2 hB =

1 (vB′ )2 2 g

1 (4.31265) 2 2 32.2 = 0.2888 ft =

hB = 3.47 in. 

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PROBLEM 13.186 A 70 g ball B dropped from a height h0 = 1.5 m reaches a height h2 = 0.25 m after bouncing twice from identical 210-g plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine (a) the coefficient of restitution between the ball and the plates, (b) the height h1 of the ball’s first bounce.

SOLUTION (a)

Plate on hard ground (first rebound): Conservation of energy: 1 1 1 mB v 2y + mB v02 = mB gh0 + mB vx2 2 2 2 v0 = 2 gh0

Relative velocities., n-direction: v0 e = v1

v1 = e 2 gh0

′ = vBx vBx

t-direction Plate on foam rubber support at C.

Conservation of energy: V1 = V3 = 0

Points  and :

1 1 1 1 ′ )2 ′ ) 2 × mB v12 = mB (v3 )2B + mB (vBx mB (vBx 2 2 2 2 (v3 ) B = e 2 gh0

Conservation of momentum: At :

mB ( −v3 ) B + mP vP = mB (v3′ ) B − mP vP′ mP 210 = =3 mB 70

− e 2 gh0 = (v3′ ) B − 3vP′

(1)

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PROBLEM 13.186 (Continued)

Relative velocities: [( −v3 ) B − (vP )]e = −vP′ − (v3′ ) B e2 2 gh0 + 0 = vP′ + (v3′ ) B

(2)

Multiplying (2) by 3 and adding to (1) 4(v3′ ) B = 2 gh0 (3e2 − e)

Conservation of energy at , Thus,

(v3′ ) B = 2 gh2 4 2 gh2 = 2 gh0 (3e2 − e) 3e 2 − e = 4

h2 0.25 =4 = 1.63299 h0 1.5

3e2 − e − 1.633 = 0

(b)

e = 0.923 

Points  and : Conservation of energy. 1 1 1 ′ )2 + mB v12 = mB (vBx ′ )2 ; mB (vBx 2 2 2

1 2 e (2 gh0 ) = gh1 2

h1 = e2 h0 = (0.923) 2 (1.5) h1 = 1.278 m 

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PROBLEM 13.187 A 700-g sphere A moving with a velocity v0 parallel to the ground strikes the inclined face of a 2.1-kg wedge B which can roll freely on the ground and is initially at rest. After impact the sphere is observed from the ground to be moving straight up. Knowing that the coefficient of restitution between the sphere and the wedge is e = 0.6, determine (a) the angle θ that the inclined face of the wedge makes with the horizontal, (b) the energy lost due to the impact.

SOLUTION

(a)

Momentum of sphere A alone is conserved in the t-direction: m Av0 cos θ = m Av′A sin θ v0 = v′A tan θ

(1)

Total momentum is conserved in the x-direction: mB vB + m Av0 = mB vB′ + (v′A ) x vB = 0, (v′A ) x = 0 0 + 0.700 v0 = 2.1vB′ + 0 vB′ = v0 /3

(2)

Relative velocities in the n-direction: (−v0 sin θ − 0)e = −vB′ sin θ − v′A cos θ (v0 )(0.6) = vB′ + v′A cot θ

(3)

Substituting vB′ from Eq. (2) into Eq. (3) 0.6v0 = 0.333 v0 + v′A cot θ 0.267v0 = v′A cot θ

(4)

Divide (4) into (1) 1 tan θ = = tan 2 θ 0.267 cot θ tan θ = 1.935

(b)

From (1)

θ = 62.7° 

v0 = v′A tan θ = v′A (1.935) v′A = 0.5168v0 , vB′ = v0 /3 Tlost =

(

1 1 mA v 2A − m A (v′A ) 2 + mB vB2 2 2

(2)

)

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PROBLEM 13.187 (Continued)

1 1 (0.7)(v0 )2 − [(0.7)(0.5168v0 )2 + (2.1)(v0 /3) 2 ] 2 2 1 = [0.7 − 0.1870 − 0.2333]v02 2 = 0.1400v02 J

Tlost = Tlost Tlost

Tlost = 0.1400v02 

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PROBLEM 13.188 When the rope is at an angle of α = 30° the 1-lb sphere A has a speed v0 = 4 ft/s. The coefficient of restitution between A and the 2-lb wedge B is 0.7 and the length of rope l = 2.6 ft. The spring constant has a value of 2 lb/in. and θ = 20°. Determine (a) the velocities of A and B immediately after the impact, (b) the maximum deflection of the spring assuming A does not strike B again before this point.

SOLUTION Masses:

m A = (1/32.2) lb ⋅ s 2 /ft

mB = (2/32.2) lb ⋅ s 2 /ft

Analysis of sphere A as it swings down: Initial state:

α = 30°, h0 = l (1 − cos α ) = (2.6)(1 − cos 30°) = 0.34833 ft V0 = m A gh0 = (1)(0.34833) = 0.34833 lb ⋅ ft T0 =

Just before impact:

α = 0, h1 = 0, V1 = 0 T1 =

Conservation of energy:

1 1  1.0  2 mA v02 =   (4) = 0.24845 lb ⋅ ft 2 2  32.2 

1 1  1.0  2  1.0  2 m A v 2A =  vA =   vA 2 2  32.2   64.4  T0 + V0 = T1 + V1 1 2 vA + 0 64.4 v 2A = 38.433 ft 2 /s 2

0.24845 + 0.34833 =

v A = 6.1994 ft/s

Analysis of the impact. Use conservation of momentum together with the coefficient of restitution. e = 0.7.

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PROBLEM 13.188 (Continued)

Sphere A: Momentum in t-direction: m A v A sin θ + 0 = m A (v′A )t (v′A )t = v A sin θ = 6.1994 sin 20° = 2.1203 m/s ( v A )t = 2.1203 m/s 70°

Both A and B: Momentum in x-direction: m A v A + 0 = m A (v′A ) n cos θ + mA (v′A )t sin θ + mB vB′ (1/32.2)(6.1994) = (1/32.2)(v A ) n cos 20° + (1/32.2)(2.120323) sin 20° + (2/32.2)vB′ (1/32.2)(v′A ) n cos 20° + (2/32.2)vB′ = 0.17001

(1)

Coefficient of restitution: (vB′ ) n − (v′A )n = e[(v A ) n − (vB ) n ] vB′ cos θ − (v′A )n = e[v A cos θ − 0] vB′ cos 20° − (v′A )n = (0.7)(6.1994) cos 20°

(2)

Solving Eqs. (1) and (2) simultaneously for (v′A ) n and vB′ , (v′A ) n = −1.0446 ft/s vB′ = 3.2279 ft/s

Resolve vA into horizontal and vertical components. tan β =

(v′A )t −(v′A )n

2.1203 1.0446 β = 63.77° =

β + 20° = 83.8°

v′A = (2.1203) 2 + (1.0446) 2 = 2.3637 ft/s

(a)

v′A = 2.36 ft/s

Velocities immediately after impact.

83.8° 

v′B = 3.23 ft/s

(b)



Maximum deflection of wedge B. Use conservation of energy:

TB1 + VB1 = TB 2 + VB 2 1 mB vB2 2 VB1 = 0 TB1 =

TB 2 = 0 VB 2 =

1 k (Δx) 2 2

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PROBLEM 13.188 (Continued)

The maximum deflection will occur when the block comes to rest (ie, no kinetic energy) 1 1 mB vB2 = k (Δx) 2 2 2 m v2 (Δ x) = B B = k 2

(

2 lb 32.2 ft/s 2

(Δ x) = 0.1642118 ft

) (3.2279 ft/s)

2

2 lb/in (12 in/ft) (Δ x) = 1.971 in. 

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PROBLEM 13.189 When the rope is at an angle of α = 30° the 1-kg sphere A has a speed v0 = 0.6 m/s. The coefficient of restitution between A and the 2-kg wedge B is 0.8 and the length of rope l = 0.9 m The spring constant has a value of 1500 N/m and θ = 20°. Determine, (a) the velocities of A and B immediately after the impact (b) the maximum deflection of the spring assuming A does not strike B again before this point.

SOLUTION m A = 1 kg

Masses:

mB = 2 kg

Analysis of sphere A as it swings down:

α = 30°, h0 = l (1 − cos α ) = (0.9)(1 − cos 30°) = 0.12058 m

Initial state:

V0 = m A gh0 = (1)(9.81)(0.12058) = 1.1829 N ⋅ m T0 =

1 2 1 mv0 = (1)(0.6)2 = 0.180 N ⋅ m 2 2

α = 0, h1 = 0, V1 = 0

Just before impact:

T1 =

Conservation of energy:

1 1 m Av A2 = (1)v A2 = 0.5v A2 2 2

T0 + V0 = T1 + V1

0.180 + 1.1829 = 0.5 v A2 + 0 v A2 = 2.7257 m 2 /s 2 v A = 1.6510 m/s

Analysis of the impact: Use conservation of momentum together with the coefficient of restitution. e = 0.8.

Note that the ball rebounds horizontally and that an impulse  Tdt is applied by the rope. Also, an impulse  Ndt is applied to B through its supports.

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PROBLEM 13.189 (Continued)

Both A and B: Momentum in x-direction: m A (v A ) x + 0 = m A (v′A ) x + mB (vB′ ) x (1)(1.6510) = (1)(v′A ) x + (2)(vB′ ) x

(1)

(v A )n = (v A ) x cos θ

Coefficient of restitution:

(vB )n = 0, (v′A )n = (v′A ) x cos θ , (vB′ ) x cos 30° (vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ] (vB′ ) x cos θ − (v′A ) x cos θ = e[(v A ) x cos θ ]

Dividing by cos θ and applying e = 0.8 gives (vB′ ) x − (v′A ) x = (0.8)(1.6510)

(2)

Solving Eqs. (1) and (2) simultaneously, (v′A ) x = −0.33020 m/s (vB′ ) x = 0.99059 m/s

(a)

Velocities immediately after impact.

(b)

Maximum deflection of wedge B. Use conservation of energy:

v′A = 0.330 m/s



v′B = 0.991 m/s



TB1 + VB1 = TB 2 + VB 2 1 mB vB2 2 VB1 = 0 TB1 =

TB 2 = 0 VB 2 =

1 k (Δx) 2 2

The maximum deflection will occur when the block comes to rest (ie, no kinetic energy) 1 1 mB vB2 = k (Δ x) 2 2 2 (Δ x) 2 =

mB vB2 (2)(0.99059 m/s)2 = 1500 N/m) k

(Δx) = 0.0362 m

Δ x = 36.2 mm 

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PROBLEM 13.190 A 32,000-lb airplane lands on an aircraft carrier and is caught by an arresting cable. The cable is inextensible and is paid out at A and B from mechanisms located below dock and consisting of pistons moving in long oil-filled cylinders. Knowing that the piston-cylinder system maintains a constant tension of 85 kips in the cable during the entire landing, determine the landing speed of the airplane if it travels a distance d = 95 ft after being caught by the cable.

SOLUTION Mass of airplaine:

m=

W 32000 lb = = 993.79 lb ⋅ s 2 /ft g 32.2 ft/s 2

Work of arresting cable force. Q = 85 kips = 85000 lb.

As the cable is pulled out, the cable tension acts paralled to the cable at the airplane hook. For a small displacement ΔU = −Q (Δl AC ) − Q (ΔlBC )

Since Q is constant, U1→2 = −Q  AC + BC − AB  d = 95 ft,

For

AC = BC = (35) 2 + (95) 2 = 101.24 ft

U1→2 = −(85000)(101.24 + 101.24 − 70) = −11.261 ft ⋅ lb

Principle of work and energy:

T1 + U1→ 2 = T2 1 2 1 mv1 + U1→2 = mv22 2 2

Since v2 = 0, we get v12 = −

Initial speed:

2U1→2 (2)(−11.261) =− = 22.663 × 103 ft 2 /s 2 m 993.79

v1 = 150.54 ft/s

v1 = 102.6 mi/h 

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PROBLEM 13.191 A 2-oz pellet shot vertically from a spring-loaded pistol on the surface of the earth rises to a height of 300 ft. The same pellet shot from the same pistol on the surface of the moon rises to a height of 1900 ft. Determine the energy dissipated by aerodynamic drag when the pellet is shot on the surface of the earth. (The acceleration of gravity on the surface of the moon is 0.165 times that on the surface of the earth.)

SOLUTION Since the pellet is shot from the same pistol the initial velocity v0 is the same on the moon and on the earth. Work and energy. 1 2 mv0 2 = − mg E (300 ft) − EL

T1 =

Earth:

U1− 2

( EL = Loss of energy due to drag) T1 =

Moon:

1 2 mv0 2

T2 = 0

U1− 2 = − mg M (1900)

T1 − 300mg E − EL = 0

(1)

T2 = 0 T1 − 1900mg M = 0

Subtracting (1) from (2)

(2)

−1900mg M + 300mg E + EL = 0 g M = 0.165 g E m=

(2/16) gE

EL = (1900)

(2/16) (2/16) (0.165 g E ) − 300 gE gE gE EL = 1.688 ft ⋅ lb 

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PROBLEM 13.192 A satellite describes an elliptic orbit about a planet of mass M. The minimum and maximum values of the distance r from the satellite to the center of the planet are, respectively, r0 and r1 . Use the principles of conservation of energy and conservation of angular momentum to derive the relation 1 1 2GM + = 2 r0 r1 h

where h is the angular momentum per unit mass of the satellite and G is the constant of gravitation.

SOLUTION Angular momentum: h = r0 , v0 = r1 v1 b = r0 v0 = r1v1 v0 =

h r0

v1 =

h r1

(1)

Conservation of energy: 1 2 mv0 2 GMm VA = − r0 TA =

1 2 mv1 2 GMm VB = − r1 TB =

TA + VA = TB + VB 1 2 GMm 1 2 GMm = mv1 − mv0 − r0 r1 2 2 1 1 r − r  v02 − v12 = 2GM  −  = 2GM  1 0   r0 r1   r1r0 

Substituting for v0 and v1 from Eq. (1) 1 r − r  1 h 2  2 − 2  = 2GM  1 0   r1r0   r0 r1   r2 − r2  r − r  h2 h 2  1 2 20  = 2 2 (r1 − r0 )( r1 + r0 ) = 2GM  1 0   r1r0   r1 r0  r1 r0

1 1 h 2  +  = 2GM  r0 r1 

 1 1  2GM  + = 2 h  r0 r1 

Q.E.D. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 819

PROBLEM 13.193 A 60-g steel sphere attached to a 200-mm cord can swing about Point O in a vertical plane. It is subjected to its own weight and to a force F exerted by a small magnet embedded in the ground. The magnitude of that force expressed in newtons is F = 3000/r 2 where r is the distance from the magnet to the sphere expressed in millimeters. Knowing that the sphere is released from rest at A, determine its speed as it passes through Point B.

SOLUTION m = 0.060 kg

Mass and weight:

W = mg = (0.060)(9.81) = 0.5886 N Vg = Wh

Gravitational potential energy: where h is the elevation above level at B. Potential energy of magnetic force:

3000 dV =− ( F , in newtons, r in mm) 2 dr r r 3000 3000 = N ⋅ mm Vm = − ∞ r2 r F=



Use conservation of energy:

T1 + V1 = T2 + V2

Position 1: (Rest at A.) v1 = 0

T1 = 0

h1 = 100 mm (Vg )1 = (0.5886 N)(100 mm) = 58.86 N ⋅ mm 2

From the figure, AD = 2002 − 1002 (mm 2 ) MD = 100 + 12 = 112 mm 2

r12 = AD + MD

2

= 2002 − 1002 + 1122 = 42544 mm 2 r1 = 206.26 mm (Vr )1 = −

3000 = −14.545 N ⋅ mm r1

V1 = 58.86 − 14.545 = 44.3015 N ⋅ mm = 44.315 × 10−3 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 820

PROBLEM 13.193 (Continued)

Position 2. (Sphere at Point B.) 1 2 1 mv2 = (0.060)v22 = 0.030 v22 2 2 (Vg )2 = 0 (since h2 = 0) T2 =

r2 = MB = 12 mm

(See figure.)

3000 = −250 N ⋅ mm = −250 × 10−3 N ⋅ mm 12 T1 + V1 = T2 + V2 (Vm )2 = −

0 + 44.315 × 10−3 = 0.030v22 − 250 × 10−3 v22 = 9.8105 m 2 /s 2

v2 = 3.13 m/s 

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PROBLEM 13.194 A shuttle is to rendezvous with a space station which is in a circular orbit at an altitude of 250 mi above the surface of the earth. The shuttle has reached an altitude of 40 mi when its engine is turned off at Point B. Knowing that at that time the velocity v 0 of the shuttle forms an angle φ0 = 55° with the vertical, determine the required magnitude of v 0 if the trajectory of the shuttle is to be tangent at A to the orbit of the space station.

SOLUTION Conservation of energy: TB =

1 2 mv0 2

VB = −

GMm rB

TA =

1 2 mv A 2

VA = −

GMm rA GM = gR 2 (Eq.12.30) TA + VA = TB + VB

1 gR 2 1 gR 2 m v02 − m = m v 2A m 2 rB 2 rA rA = 3960 + 250 = 4210 mi

 rB  1 −   rA  rB = 3960 + 40 = 4000 mi

v A2 = v02 −

2 gR 2 rB

v A2 = v02 −

2(32.2)(3960 × 528)3  4000  1 − 4210  (4000 × 5280)  

v A2 = v02 − 66.495 × 106

(1) rA v A = rB vB sin φ0 ;

Conservation of angular momomentum:

v A = (4000/4210)v0 sin 55° = 0.77829 v0

Eqs. (2) and (1)

[1 − (0.77829) 2 ] v02 = 66.495 × 106

(2) v0 = 12,990 ft/s 

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PROBLEM 13.195 A 300-g block is released from rest after a spring of constant k = 600 N/m has been compressed 160 mm. Determine the force exerted by the loop ABCD on the block as the block passes through (a) Point A, (b) Point B, (c) Point C. Assume no friction.

SOLUTION Conservation of energy to determine speeds at locations A, B, and C. Mass: m = 0.300 kg

Initial compression in spring: x1 = 0.160 m Place datum for gravitational potential energy at position 1. Position 1: v1 = 0 V1 =

T1 =

1 2 mv1 = 0 2

1 2 1 kx1 = (600 N/m)(0.160 m) 2 = 7.68 J 2 2

Position 2: T2 =

1 2 1 mv A = (0.3)v 2A = 0.15v A2 2 2

V2 = mgh2 = (0.3 kg)(9.81 m/s 2 )(0.800 m) = 2.3544 J T1 + V1 = T2 + V2 : 0 + 7.68 = 0.15v A2 + 2.3544 v A2 = 35.504 m 2 /s 2

Position 3:

T3 =

1 2 1 mvB = (0.3)vB2 = 0.15vB2 2 2

V3 = mgh3 = (0.3 kg)(9.81 m/s 2 )(1.600 m) = 4.7088 J T1 + V1 = T3 + V3 : 0 + 7.68 = 0.15vB2 + 4.7088 vB2 = 19.808 m 2 /s 2

Position 4: T2 =

1 2 1 mvC = (0.3)vC2 = 0.15vC2 2 2

V4 = mgh4 = (0.3 kg)(9.81 m/s)(0.800 m) = 2.3544 J T1 + V1 = T4 + V4 : 0 + 7.68 = 0.15vC2 + 2.3544 vC2 = 35.504 m 2 /s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 823

PROBLEM 13.195 (Continued)

(a)

Newton’s second law at A: an =

v 2A

ρ

=

35.504 m 2 /s 2 = 44.38 m/s 2 0.800 m

a n = 44.38 m/s 2 ΣF = man : N A = man N A = (0.3 kg)(44.38 m/s 2 )

(b)

N A = 13.31 N



Newton’s second law at B: an =

vB2

ρ

=

19.808 m 2 /s 2 = 24.76 m/s 2 0.800 m

a n = 24.76 m/s 2 ΣF = man : N B = mg = man N B = m( an − g ) = (0.3 kg)(24.76 m/s 2 − 9.81 m/s 2 )

(c)

N B = 4.49 N 

Newton’s second law at C: an =

vC2

ρ

=

35.504 m 2 /s 2 = 44.38 m/s 2 0.800 m

a n = 44.38 m/s 2 ΣF = man : NC = man N C = (0.3 kg)(44.38 m/s 2 )

N C = 13.31 N



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 824

PROBLEM 13.196 A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to Point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest Point C ′′ that the sphere will reach.

SOLUTION Velocity at Point C (before the cord is taut). Conservation of energy from B to C: TB = 0  2 VB = mg (2)  a = mga 2  2    1 TC = mvC2 VC = 0 2 TB + VB = TC + VC 0 + mga 2 =

1 2 mvC + 0 2

vC = 2 2 ga

Velocity at C (after the cord becomes taut). Linear momentum perpendicular to the cord is conserved:

θ = 45° − mvC sinθ = mvC′ vC′ =

(

vC′ =

)

 2 2 2  ga  2    1

2 ga = 2 4 ga

Note: The weight of the sphere is a non-impulsive force. Velocity at C: C to C ′ (conservation of energy):

1 m(vC′ )2 2 1 TC ′ = m(vC′ ′ ) 2 2 TC =

VC = 0 VC ′ = 0

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PROBLEM 13.196 (Continued)

Datum:

TC + VC = TC ′ + VC ′ 1 1 m(vC′ ) 2 + 0 = m(vC′ ) 2 + 0 2 2 vC′ = vC′ ′

C ′ to C ′′ (conservation of energy): 1 m(vC′ ′ ) 2 2 1 TC ′ = m 21/ 4 ga 2 2 TC ′ = mga 2 TC ′ + VC ′ = TC ′′ + VC ′′ TC ′ =

(

Datum:

)

2

VC ′ = 0 TC ′′ = 0 VC ′′ = mgh 2 mga + 0 = 0 + mgh 2 2 h= a 2

h = 0.707 a 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 826

PROBLEM 13.197 A 300-g collar A is released from rest, slides down a frictionless rod, and strikes a 900-g collar B which is at rest and supported by a spring of constant 500 N/m. Knowing that the coefficient of restitution between the two collars is 0.9, determine (a) the maximum distance collar A moves up the rod after impact, (b) the maximum distance collar B moves down the rod after impact.

SOLUTION After impact

Velocity of A just before impact, v0 v0 =

2 gh = =

2(9.81 m/s 2 )(1.2 m)sin 30° 2(9.81)(1.2)(0.5) = 3.431 m/s

Conservation of momentum m Av0 = mB vB − m Av A : 0.3v0 = 0.9vB − 0.3v A (1)

Restitution (v A + vB ) = e(v0 + 0) = 0.9v0 (2)

Substituting for vB from (2) in (1) 0.3v0 = 0.9(0.9v0 − v A ) − 0.3v A

1.2v A = 0.51v0

v A = 1.4582 m/s, vB = 1.6297 m/s

(a)

A moves up the distance d where: 1 m Av A2 = mA gd sin 30°; 2

1 (1.4582 m/s) 2 = (9.81 m/s 2 )d (0.5) 2 d A = 0.21675 m = 217 mm 

(b)

Static deflection = x0 , B moves down Conservation of energy (1) to (2) Position (1) – spring deflected, x0 k x0 = mB g sin 30° T1 + V1 = T2 + V2: T1 =

1 mBvB2 , T2 = 0 2

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PROBLEM 13.197 (Continued)

V1 = Ve + Vg = x + dB

V2 = Ve′ + Vg′ = 0 0

1 2 kx0 + mB gd B sin 30° 2

kxdx =

(

1 k d B2 + 2d B x0 + x02 2

(

)

)

1 2 1 1 kx0 + mgd B sin 30° + mB vB2 = k d B2 + 2d B x0 + x02 + 0 + 0 2 2 2 ∴ kd B2 = mBvB2 ;

500d B2 = 0.9 (1.6297) 2

d B = 0.0691 m d B = 69.1 mm 

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PROBLEM 13.198 Blocks A and B are connected by a cord which passes over pulleys and through a collar C. The system is released from rest when x = 1.7 m. As block A rises, it strikes collar C with perfectly plastic impact (e = 0). After impact, the two blocks and the collar keep moving until they come to a stop and reverse their motion. As A and C move down, C hits the ledge and blocks A and B keep moving until they come to another stop. Determine (a) the velocity of the blocks and collar immediately after A hits C, (b) the distance the blocks and collar move after the impact before coming to a stop, (c) the value of x at the end of one compete cycle.

SOLUTION (a)

Velocity of A just before it hits C: Conservations of energy: Datum at : Position : (v A )1 = (vB )1 = 0

T1 = 0 v1 = 0

Position :

1 1 m A (v A )2 + mB vB2 2 2 v A = vB (kinematics)

T2 =

1 11 (5 + 6)v A2 = v A2 2 2 V2 = m A g (1.7) − mB g (1.7) T2 =

= (5 − 6)( g )(1.7) V2 = −1.7 g T1 + V1 = T2 + V2 0+0=

11 2 v A − 1.7 g 2

 3.4  v A2 =   (9.81)  11  = 3.032 m 2 /s 2 v A = 1.741 m/s

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PROBLEM 13.198 (Continued)

Velocity of A and C after A hits C:

v′A = vC′ (plastic impact)

Impulse-momentum A and C: m A v A + T Δt = (m A + mC ) v′A (5)(1.741) + T Δt = 8v′A

(1)

vB = v A ; vB′ = v′A (cord remains taut)

B alone:

mB v A − T Δ t = mB v′A (6)(1.741) − T Δt = 6v′A

Adding Equations (1) and (2),

(2)

11(1.741) = 14v′A v′A = 1.3679 m/s v′A = vB′ = vC′ = 1.368 m/s 

(b)

Distance A and C move before stopping: Conservations of energy: Datum at : Position : 1 (m A + mB + mC )(v′A ) 2  14  T2 =   (1.3681) 2  2 T2 = 13.103 J T2 =

V2 = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 830

PROBLEM 13.198 (Continued)

Position : T3 = 0 V3 = (m A + mC ) gd − mB gd V3 = (8 − 6) gd = 2 gd T2 + V2 = T3 + V3 13.103 + 0 = 0 + 2gd d = (13.103)/(2)(9.81) = 0.6679 m

(c)

d = 0.668 m 

As the system returns to position  after stopping in position , energy is conserved, and the velocities of A, B, and C before the collar at C is removed are the same as they were in Part (a) above with the directions reversed. Thus, v′A = vC′ = vB′ = 1.3679 m/s. After the collar C is removed, the velocities of A and B remain the same since there is no impulsive force acting on either. Conversation of energy: Datum at : 1 (m A + mB )(v′A )2 2 1 T2 = (5 + 6)(1.3679) 2 2 T2 = 10.291 J T2 =

V2 = 0 T4 = 0

V4 = mB gx − m A gx V4 = (6 − 5) gx

T2 + V2 = T4 + V4 10.291 + 0 = (1)(9.81) x x = 1.049 m 

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PROBLEM 13.199 A 2-kg ball B is traveling horizontally at 10 m/s when it strikes 2-kg ball A. Ball A is initially at rest and is attached to a spring with constant 100 N/m and an unstretched length of 1.2 m. Knowing the coefficient of restitution between A and B is 0.8 and friction between all surfaces is negligible, determine the normal force between A and the ground when it is at the bottom of the hill.

SOLUTION Ball B impacts on ball A. Use the principle of impulse and momentum. Σmv1 + ΣImp1→2 = Σmv 2

v0 = 10 m/s

Velocity components: (v0 ) x = v0

(v0 )n = v0 cos 40° (v0 )t = v0 sin 40°

(v A ) x = v A

(v A )n = v A cos 40°

(vB ) x = (vB ) n cos 40° + (vB )t sin 40°

Impulse-momentum for ball B alone. t-direction: mB (v0 )t = mB (vB )t (vB )t = (v0 )t = 10sin 40° = 6.4279 m/s

(1)

Impulse-momentum for balls A and B. x-direction mB v0 + 0 = m Av A + mB (vB ) x + mB (vB )t (2)(10) + 0 = 2v A + 2[(vB )n cos 40° + 6.4279sin 40°] 2v A + 2(vB ) n cos 40° = 11.7365

(1)

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PROBLEM 13.199 (Continued)

(e = 0.8)

Coefficient of restitution.

(vB )n = (v A )n = e[0 − (v0 ) n ] (vB )n − v A cos 40° = −(0.8)(10) cos 40°

(2)

Solving Eqs. (1) and (2) simultaneously, v A = 6.6566 m/s

(vB ) n = −1.0291 m/s

As ball A moves from the impact location to the lowest point on the path, the spring compresses and the elevation decreases. Since friction is negligible, energy is conserved. T1 + V1 = T2 + V2 1 1 m Av A2 + (Ve )1 + (Vg )1 = m A v22 + (Ve ) 2 + (Vg ) 2 2 2

Position 1: (Just after impact.) 1 1 m Av A2 = (2)(6.6566) 2 = 44.3101 J 2 2 (Ve )1 = 0 (The spring is unstretched.) T1 =

(Vg )1 = 0 (Datum)

Position 2: (Lowest point on path.) T2 =

For the spring,

1 1 m Av22 = (2)v22 = v22 2 2

x2 = l2 − l0 = 0.4 m − 1.2 m = 0.8 m Fe = kx2 = (100)(0.8) = 80 N (V2 )e =

1 2 1 kx2 = (100)(0.8) 2 = 32 J 2 2 h2 = −0.4 m

Elevation above datum:

(V2 ) g = mA g h2 = (2)(9.81)(−0.4) = −7.848

Conservation of energy: 44.310 + 0 + 0 = v22 + 32 − 7.848 v22 = 20.158 m 2 /s 2

v2 = 4.489 m/s

Normal acceleration at lowest point on path: an =

v22

ρ

=

20.158 = 28.798 m/s 2 0.7

a n = 28.8 m/s 2

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PROBLEM 13.199 (Continued)

Apply Newton’s second law to the ball.

ΣF = man : N − mg − Fe = man N = mg + Fe + man = (2)(9.81) + 80 + (2)(28.798)

N = 157.2 N 

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PROBLEM 13.200 A 2-kg block A is pushed up against a spring compressing it a distance x = 0.1 m. The block is then released from rest and slides down the 20° incline until it strikes a 1-kg sphere B which is suspended from a 1 m inextensible rope. The spring constant k = 800 N/m, the coefficient of friction between A and the ground is 0.2, the distance A slides from the unstretched length of the spring d = 1.5 m and the coefficient of restitution between A and B is 0.8. When α = 40°, determine (a) the speed of B (b) the tension in the rope.

SOLUTION Data:

m A = 2 kg, mB = 1 kg, k = 800 N/m, x = 0.1 m, d = 1.5 m

μk = 0.2, e = 0.8, θ = 20°, α = 40°, l = 1.0 m Block slides down the incline: ΣFy = 0

N − m A g cos θ = 0 N = m A g cos θ = (2)(9.81) cos 20° = 18.4368 N F f = μk N = (0.2)(18.4368) = 3.6874 N

Use work and energy. Datum for Vg is the impact point near B. T1 = 0, (V1 )e =

1 2 1 k x1 = (800)(0.1) 2 = 4.00 J 2 2

(V1 ) g = m A gh1 = mA g ( x + d ) sin θ = (2)(9.81)(1.6)sin 20° = 10.7367 J U1→2 = − F f ( x + d ) = −(3.6874)(1.6) = −5.8998 J T2 =

1 1 m A v 2A = (1)(v A2 ) = 1.000 v A2 2 2

V2 = 0

T1 + V1 + U1→2 = T2 + V2 : 0 + 4.00 + 10.7367 − 5.8998 = 1.000 v 2A + 0 v A2 = 8.8369 m 2 /s 2 v A = 2.9727 m/s

20°

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PROBLEM 13.200 (Continued)

Impact: Conservation of momentum.

Both A and B, horizontal components

:

m A v A cos θ + 0 = m Av′A cos θ + mB vB

(2)(2.9727) cos 20° = 2v′A cos 20° + (1.00)vB

(1)

(vB′ )n − (v′A ) n = e[(vB ) n − (v A ) n ]

Relative velocities:

vB′ cos θ − v′A = e[v A − 0] vB′ cos 20° − v′A = (0.8)(2.9727)

(2)

Solving Eqs. (1) and (2) simultaneously, v′A = 1.0382 m/s vB′ = 3.6356 m/s

Sphere B rises: Use conservation of energy. 1 mB (vB′ ) 2 V1 = 0 2 1 T2 = mB v22 V2 = mB gh2 = mB gl (1 − cos α ) 2 1 1 T1 + V1 = T2 + V2 : mB (vB′ ) 2 + 0 = mB v22 + mB g (1 − cos) 2 2 v22 = (vB′ )2 − 2 gl (1 − cos α ) T1 =

= (3.6356)2 − (2)(9.81)(1 − cos 40°) = 8.6274 m 2 /s 2 v2 = 2.94 m/s 

(a) Speed of B: (b) Tension in the rope:

ρ = 1.00 m an =

v22

ρ

=

8.6274 = 8.6274 m/s 2 1.00

ΣFn = mB an : T − mB g cos α = mB an T = mB (an + g cos α ) = (1.0)(8.6274 + 9.81cos 40°) T = 16.14 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 836

PROBLEM 13.201* The 2-lb ball at A is suspended by an inextensible cord and given an initial horizontal velocity of v0. If l = 2 ft, xB = 0.3 ft and yB = 0.4 ft determine the initial velocity v so that the ball will enter in the basket. Hint: use a computer to solve the resulting set of equations.

SOLUTION v1 = v0

Let position 1 be at A.

Let position 2 be the point described by the angle θ where the path of the ball changes from circular to parabolic. At position 2 the tension Q in the cord is zero. Relationship between v2 and θ based on Q = 0. Draw the free body diagram.

ΣF = 0: Q + mg sin θ = man =

With Q = 0,

v22 = g  sin θ

mv22 

or v2 = g  sin θ

(1)

Relationship among v0 , v2 , and θ based on conservation of energy. T1 + V1 = T2 + V2

1 2 1 mv0 − mg  = mv22 + mg  sin θ 2 2 v02 = v22 + 2 g (1 + sin θ )

(2)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 837

PROBLEM 13.201* (Continued)

x and y coordinates at position 2: x2 =  cos θ

(3)

y2 =  sin θ

(4)

Let t2 be the time when the ball is in position 2. Motion on the parabolic path. The horizontal motion is x = −v2 sin θ x = x2 − (v2 sin θ )(t − t2 )

At Point B,

x = xB

(t B − t2 ) =

Vertical motion:

and t = t B .

(5) From Eq. (5),

 cos θ − xB vθ sin θ

(6)

y = v2 cos θ − g (t − t2 ) y = y2 + (v2 cos θ )(t − t2 ) −

1 g (t − t2 ) 2 2

At Point B, yB =  sin θ + (v2 cos θ )(t B − t2 ) −

Data:

 = 2 ft, xB = 0.3 ft,

1 g (t B − t2 )2 2

(7)

yB = 0.4 ft, g = 32.2 ft/s 2

With the numerical data, Eq. (1) becomes Eq. (6) becomes Eq. (7) becomes

v2 = 64.4sin θ t B − t2 =

(1)′

2 cos θ − 0.3 v2 sin θ

(6)′

yB = 2sin θ + (v2 cos θ )(t B − t2 ) − 16.1(t B − t2 ) 2

(7)′

Method of solution. From a trial value of θ, calculate v2 from Eq. (1)′, t B − t2 from Eq. (6)′, and yB from Eq. (7)′. Repeat until yB = 0.4 ft as required. Try θ = 30°.

v2 = 64.4sin 30° = 5.6745 ft/s

2 cos30° − 0.3 = 0.50473s 5.6745sin 30° yB = 2sin 30° + (5.6745cos 30°)(0.50473) − (16.1)(0.50473)2

t B − t2 =

= −0.62116 ft

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PROBLEM 13.201* (Continued)

Try θ = 45°.

v2 = 64.4sin 45° = 6.7482

2 cos 45° − 0.3 = 0.23351 s 6.7482sin 45° yB = 2sin 45° + (6.7482 cos 45°)(0.23351) − (16.1)(0.23351)2

t B − t2 =

= 1.65060 ft

Try θ = 37.5°.

v2 = 64.4sin 37.5° = 6.2613 ft/s t B − t2 =

2 cos 37.5° − 0.3 = 0.33757 s 6.2613 sin 37.5°

yB = 2sin 37.5° + (6.2613cos 37.5°)(0.33757) − (16.1)(0.33757) 2 = 1.05972 ft

Let u = θ − 30°.

The following sets of data points have be determined: (u, yB ) = (0°, −0.62114 ft), (7.5°, 1.05972 ft), (15°, 1.65060 ft)

The quadratic curve fit of this data gives yB = −0.62114 + 0.29678 u − 0.009688711 u 2

Setting yB = 0.4 ft gives the quadratic equation −0.009688711 u 2 + 0.29678 u − 1.02114 = 0

Solving for u,

u = 3.95° and 26.68°

Rejecting the second value gives

θ = 30° + u = 33.95°.

Try θ = 33.95°.

v2 = 64.4sin 33.95° = 5.997 ft/s t B − t2 =

2 cos 33.95° − 0.3 = 0.40578 s 5.9971 sin 33.95°

yB = 2sin 33.95° + (5.997 cos 33.95°)(0.40578) − (16.1)(0.40578) 2 = 0.48462 ft

The new quadratic curve-fit is based on the data points (u, yB ) = (0°, −0.62114 ft), (3.95°, 0.48462 ft), (7.5°, 1.05972 ft).

The quadratic curve fit of this data is yB = −0.62114 + 0.342053907 u − 0.015725232 u 2

Setting yB = 0.4 ft gives −0.015725232 u 2 + 0.342053907 u − 1.02114 = 0

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PROBLEM 13.201* (Continued)

Solving for u,

θ = 30° + 3.572° = 33.572°

u = 3.572°

Try θ = 33.572°.

v2 = 64.4 sin 33.572° = 5.9676 ft/s t B − t2 =

2 cos 33.572° − 0.3 = 0.41406 s 5.9676 sin 33.572°

yB = 2sin 33.572° + (5.9676 cos 33.572°)(0.41406) − (16.1)(0.41406) 2 = 0.40445 ft

which is close enough to 0.4 ft. Substituting θ = 33.572° and v2 = 5.9676 ft/s into Eq. (2) along with other data gives v02 = (5.9676) 2 + (2)(32.2)(2)(1 + sin 33.572°) = 235.64 ft 2 /s 2 v 0 = 15.35 ft/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 840

CHAPTER 14

PROBLEM 14.1 A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic and the coefficient of kinetic friction between B and C is 0.2, determine (a) the velocity of the bullet and B after the first impact, (b) the final velocity of the carrier.

SOLUTION For convenience, label the bullet as particle A of the system of three particles A, B, and C. (a)

Impact between A and B: Use conservation of linear momentum of A and B. Assume that the time period is so short that any impulse due to the friction force between B and C may be neglected. Σ mv1 + Σ Imp1 2 = Σ mv 2

Components

:

m A v0 + 0 = (m A + mB )v′

v′ =

m Av0 (30 × 10−3 kg)(450 m/s) = = 4.4554 m/s m A + mB (30 × 10−3 kg + 3 kg v′ = 4.46 m/s

(b)



Final velocity of the carrier: Particles A, B, and C have the same velocity v′′ to the left. Use conservation of linear momentum of all three particles. The friction forces between B and C are internal forces. Neglect friction at the wheels of the carrier. Σ mv 2 + Σ Imp 23 = Σ mv3

Components

:

(m A + mB )v′ + 0 = ( mA + mB + mC )v

v′′ = =

mA v0 (m A + mB )v′ = mA + mB + mC m A + mB + mC (30 × 10−3 kg)(450 m/s) = 0.4087 m/s 30 × 10−3 kg + 3 kg + 30 kg v′′ = 0.409 m/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 843

PROBLEM 14.2 A 30-g bullet is fired with a horizontal velocity of 450 m/s through 3-kg block B and becomes embedded in carrier C which has a mass of 30 kg. After the impact, block B slides 0.3 m on C before coming to rest relative to the carrier. Knowing the coefficient of kinetic friction between B and C is 0.2, determine (a) the velocity of the bullet immediately after passing through B, (b) the final velocity of the carrier.

SOLUTION For convenience, label the bullet as particle A of the system of three particles A, B, and C. (b)

Final velocity of carrier: Use conservation momentum for all three particles, since the impact forces and the friction force between B and C are internal forces of the system. Σ mv1 + Σ Imp1 2 = Σ mv 2

Components

:

m A v0 + 0 = (m A + mB + mC )v′′ v′′ =

m Av0 (0.030 kg)(450 m/s) = = 0.40872 m/s 33.03 kg mA + mB + mC v′′ = 0.409 m/s

(a)



Velocity v A of the bullet: The sequence of events described is broken into the following states and processes. The symbols for velocities of A, B, and C at the various states are given in the following table: Symbol for velocity State

A

B

C

Process

(1)

v0

0

0

Initial state

(2)

vA

vB

0

1

2: Bullet passes through block

(3)

v AC

vB

v AC

2

3: Bullet impacts end of carrier

(4)

v′′

v′′

v′′

3

4: Block slides to rest relative to carrier

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 844

PROBLEM 14.2 (Continued)

For process 1

2 apply conservation of momentum. m A v0 = m Av A + mB vB

For process 2

3 apply conservation of momentum to A and C. m A v A = (m A + mC )v AC

For process 3

(1)

(2)

4 apply conservation of momentum to A, B, and C. (m A + mC )v AC + mB vB = (m A + mB + mC )v′′

(3)

4 apply the principle of work and energy, since the work U 3→4 of the friction For process 3 force may be calculated. N = WB = mB g = (3 kg)(9.81 m/s) = 29.43 N

Normal force:

F f = μk N = (0.2)(29.43) = 5.886 N

Friction force:

U 3→4 = − F f d = −(5.886 N)(0.3 m) = −1.7658 J

Work: Principle of work and energy:

TAC + TB + U 3→ 4 = T ′′ TAC =

where

(4)

1 2 (mA + mC )v AC 2

TB =

1 mB vB2 2

T ′′ =

1 (m A + mB + mC )(v′′) 2 2

Applying the numerical data gives (0.030)(450) = 0.030v A + 3vB

(1)′

0.030v A = 30.03v AC

(2)′

30.03v AC + 3vB = (33.03)(0.40872)

(3)′

1 1 1 2 (30.03)v AC + (3)vB2 − 1.7658 = (33.03)(0.40872) 2 2 2 2 v AC =

From Eq. (3)′,

(4)′

(33.03)(0.40872) − 3vB = 0.44955 − 0.0999 vB 30.03

Substituting into Eq. (4)′ gives (15.015)(0.44955 − 0.0999vB )2 + 1.5vB2 − 1.7658 = 2.7586

which reduces to the quadratic equation 1.64985vB2 − 1.34865vB − 1.48995 = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 845

PROBLEM 14.2 (Continued)

Solving,

vB = 1.44319 and − 0.62575 vB = 1.44319 m/s

Using Eq. (1)′ with numerical data,

13.5 = 0.030v A + (3)(1.44319) v A = 306 m/s



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PROBLEM 14.3 Car A weighing 4000 lb and car B weighing 3700 lb are at rest on a 22-ton flatcar which is also at rest. Cars A and B then accelerate and quickly reach constant speeds relative to the flatcar of 7 ft/s and 3.5 ft/s, respectively, before decelerating to a stop at the opposite end of the flatcar. Neglecting friction and rolling resistance, determine the velocity of the flatcar when the cars are moving at constant speeds.

SOLUTION The masses are m A =

4000 3700 (22)(2000) = 124.2 slugs, mB = = 114.9 slugs, and mF = = 1366.5 slugs 32.2 32.2 32.2

Let v A , vB , and vF be the sought after velocities in ft/s, positive to the right. (v A )0 = (vB )0 = (vF )0 = 0.

Initial values:

m A (v A )0 + mB (vB )0 + mF (vF )0 = 0.

Initial momentum of system:

There are no horizontal external forces acting during the time period under consideration. Momentum is conserved. 0 = m Av A + mB vB + mF vF 124.2v A + 114.9vB + 1366.5vF = 0

(1)

The relative velocities are given as v A/F = v A − vF = − 7 ft/s

(2)

vB/F = vB − vF = − 3.5 ft/s

(3)

Solving (1), (2), and (3) simultaneously, v A = − 6.208 ft/s, vB = − 2.708 ft/s, vF = 0.7919 ft/s v F = 0.792 ft/s



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PROBLEM 14.4 A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B.

SOLUTION The masses are m for the bullet and m A and mB for the blocks. (a)

The bullet passes through block A and embeds in block B. Momentum is conserved. Initial momentum:

mv0 + mA (0) + mB (0) = mv0

Final momentum:

mvB + m Av A + mB vB

Equating,

mv0 = mvB + mA v A + mB vB m=

mA v A + mB vB (6)(5) + (4.95)(9) = = 0.0500 lb v0 − vB 1500 − 9 m = 0.800 oz 

(b)

The bullet passes through block A. Momentum is conserved. Initial momentum:

mv0 + mA (0) = mv0

Final momentum:

mv1 + m Av A

Equating,

mv0 = mv1 + mA v A v1 =

mv0 − mA v A (0.0500)(1500) − (6)(5) = = 900 ft/s 0.0500 m v1 = 900 ft/s



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PROBLEM 14.5 Two swimmers A and B, of weight 190 lb and 125 lb, respectively, are at diagonally opposite corners of a floating raft when they realize that the raft has broken away from its anchor. Swimmer A immediately starts walking toward B at a speed of 2 ft/s relative to the raft. Knowing that the raft weighs 300 lb, determine (a) the speed of the raft if B does not move, (b) the speed with which B must walk toward A if the raft is not to move.

SOLUTION (a)

The system consists of A and B and the raft R. Momentum is conserved. (Σmv)1 = (Σmv) 2 0 = mA vA + mB vB + mR vR

v A = v A/R + v R

v B − v B/R + v R

B

v A = 2 ft/s  + v R A

(1) vB/R = 0

vB = vR

B

0 = m A [2  + v R ] + mB v R + mR v e A

vR =

(b)

−2 m A −(2 ft/s)(190 lb) = (m A + mB + mR ) (190 lb + 125 lb + 300 lb)

vR = 0.618 ft/s 

From Eq. (1), 0 = m Av A + mB vB + 0

(vR = 0)

vB = −

mA v A mB

vB = −

(2 ft/s)(190 lb) = 3.04 ft/s (125 lb)

0

v A = v A/R + v R = 2 ft/s vB = 3.04 ft/s 

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PROBLEM 14.6 A 180-lb man and a 120-lb woman stand side by side at the same end of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first.

SOLUTION (a)

Woman dives first. Conservation of momentum: 120 300 + 180 (16 − v1 ) − v1 = 0 g g v1 =

(120)(16) = 3.20 ft/s 600

Man dives next. Conservation of momentum:



300 + 180 300 180 v1 = − v2 + (16 − v2 ) g g g v2 =

(b)

480v1 + (180)(16) = 9.20 ft/s 480

v 2 = 9.20 ft/s



v′2 = 9.37 ft/s



Man dives first. Conservation of momentum: 180 300 + 120 (16 − v1′ ) − v1′ = 0 g g v1′ =

(180)(16) = 4.80 ft/s 600

Woman dives next. Conservation of momentum: −

300 + 120 300 120 v1′ = − v2′ + (16 − v2′ ) g g g v2′ =

420v1′ + (120)(16) = 9.37 ft/s 420

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PROBLEM 14.7 A 40-Mg boxcar A is moving in a railroad switchyard with a velocity of 9 km/h toward cars B and C, which are both at rest with their brakes off at a short distance from each other. Car B is a 25-Mg flatcar supporting a 30-Mg container, and car C is a 35-Mg boxcar. As the cars hit each other they get automatically and tightly coupled. Determine the velocity of car A immediately after each of the two couplings, assuming that the container (a) does not slide on the flatcar, (b) slides after the first coupling but hits a stop before the second coupling occurs, (c) slides and hits the stop only after the second coupling has occurred.

SOLUTION Each term of the conservation of momentum equation is mass times velocity. As long as the same units are used in all terms, any unit may be used for mass and for velocity. We use Mg for mass and km/h for velocity and apply conservation of momentum. Note: Only moving masses are shown in the diagrams. Initial momentum: (a)

m A v0 = (40)(9) = 360

Container does not slide

360 = 95v1 = 130v2

(b)

v1 = 3.79 km/h



v 2 = 2.77 km/h



v1 = 5.54 km/h



v 2 = 2.77 km/h



Container slides after 1st coupling, stops before 2nd

360 = 65v1 = 130v2

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PROBLEM 14.7 (Continued)

(c)

Container slides and stops only after 2nd coupling

360 = 65v1 = 100v2

v1 = 5.54 km/h



v 2 = 3.60 km/h



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PROBLEM 14.8 Packages in an automobile parts supply house are transported to the loading dock by pushing them along on a roller track with very little friction. At the instant shown, packages B and C are at rest and package A has a velocity of 2 m/s. Knowing that the coefficient of restitution between the packages is 0.3, determine (a) the velocity of package C after A hits B and B hits C, (b) the velocity of A after it hits B for the second time.

SOLUTION (a)

Packages A and B:

Total momentum conserved: m Av A + mB vB = mA v′A + mB vB′ (8 kg)(2 m/s) + 0 = (8 kg)v′A + (4 kg)vB′ 4 = 2v′A + vB′

(1)

Relative velocities. (v A − vB )e = (vB′ − v′A ) (2)(0.3) = vB′ − v′A

(2)

Solving Equations (1) and (2) simultaneously, v′A = 1.133 m/s vB′ = 1.733 m/s

Packages B and C:

mB vB′ + mC vC = mB vB′′ + mC vC′ (4 kg)(1.733 m/s) + 0 = 4vB′′ + 6vc′ 6.932 = 4vB′′ + 6vC′

(3)

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PROBLEM 14.8 (Continued)

Relative velocities: (vB′ − vC )e = vC′ − vB′′ (1.733)(0.3) = 0.5199 = vC′ − vB′′

(4)

Solving equations (3) and (4) simultaneously, v′C = 0.901 m/s

(b)



Packages A and B (second time),

Total momentum conserved: (8)(1.133) + (4)(0.381) = 8v′′A + 4vB′′′ 10.588 = 8v′′A + 4vB′′

(5)

Relative velocities: (v′A − vB′′ )e = vB′′′ − v′′A (1.133 − 0.381)(0.3) = 0.2256 = vB′′′ − v′′A

(6)

Solving (5) and (6) simultaneously, v′′A = 0.807 m/s

v′′A = 0.807 m/s



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PROBLEM 14.9 A system consists of three particles A, B, and C. We know that m A = 3 kg, mB = 2 kg, and mC = 4 kg and that the velocities of the particles expressed in m/s are, respectively, v A = 4i + 2 j + 2k , v B = 4i + 3j, and vC = −2i + 4 j + 2k. Determine the angular momentum H O of the system about O.

SOLUTION Linear momentum of each particle expressed in kg ⋅ m/s. m A v A = 12i + 6 j + 6k mB v B = 8i + 6 j mC vC = −8i + 16 j + 8k rA = 3j,

Position vectors, (meters):

rB = 1.2i + 2.4 j + 3k ,

rC = 3.6i

Angular momentum about O, (kg ⋅ m 2/s). H O = rA × (mA v A ) + rB × (mB v B ) + rC × (mC vC ) i j k i j k i j k = 0 3 0 + 1.2 2.4 3 + 3.6 0 0 12 6 6 8 6 0 −8 16 8 = (18i − 36k ) + (−18i + 24 j − 12k ) + (−28.8j + 57.6k ) = 0i − 4.8j + 9.6k H O = −(4.80 kg ⋅ m 2 /s) j + (9.60 kg ⋅ m 2/s) k  

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PROBLEM 14.10 For the system of particles of Problem 14.9, determine (a) the position vector r of the mass center G of the system, (b) the linear momentum mv of the system, (c) the angular momentum H G of the system about G. Also verify that the answers to this problem and to problem 14.9 satisfy the equation given in Problem 14.27. PROBLEM 14.9 A system consists of three particles A, B, and C. We know that m A = 3 kg, mB = 2 kg, and mC = 4 kg and that the velocities of the particles expressed in m/s are, respectively, v A = 4i + 2 j + 2k , v B = 4i + 3j, and vC = −2i + 4 j + 2k. Determine the angular momentum H O of the system about O.

SOLUTION Position vectors, (meters):

(a)

Mass center:

rA = 3j,

rB = 1.2i + 2.4 j + 3k ,

rC = 3.6i

(m A + mB + mC ) r = m ArA + mBrB + mC rC 9r = (3)(3j) + (2)(1.2i + 2.4 j + 3k ) + (4)(3.6i) r = 1.86667i + 1.53333j + 0.66667k r = (1.867 m)i + (1.533 m) j + (0.667 m)k 

Linear momentum of each particle, (kg ⋅ m 2 /s). m A v A = 12i + 6 j + 6k mB v B = 8i + 6 j mC vC = −8i + 16 j + 8k

(b)

Linear momentum of the system, (kg ⋅ m/s.) mv = m A v A + mB v B + mC vC = 12i + 28 j + 14k mv = (12.00 kg ⋅ m/s)i + (28.0 kg ⋅ m/s) j + (14.00 kg ⋅ m/s)k 

Position vectors relative to the mass center, (meters). rA′ = rA − r = −1.86667i + 1.46667 j − 0.66667k rB′ = rB − r = −0.66667i + 0.86667 j + 2.33333k rC′ = rC − r = 1.73333i − 1.53333j − 0.66667k

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PROBLEM 14.10 (Continued)

(c)

Angular momentum about G, (kg ⋅ m 2 /s). H G = rA′ × m Av A + rB′ × mB v B + rC′ × mC vC i j k i j k = −1.86667 1.46667 −0.66667 + −0.66667 0.86667 2.33333 12 6 6 8 6 0 i j k + 1.73333 −1.53333 −0.66667 16 8 −8

= (12.8i + 3.2 j − 28.8k ) + (−14i + 18.6667 j − 10.9333k ) + (−1.6i − 8.5333j + 15.4667k ) = −2.8i + 13.3333j − 24.2667k H G = −(2.80 kg ⋅ m 2 /s)i + (13.33 kg ⋅ m 2 /s) j − (24.3 kg ⋅ m 2 /s)k 

i j k r × mv = 1.86667 1.53333 0.66667 12 28 14 = (2.8 kg ⋅ m 2 /s)i − (18.1333 kg ⋅ m 2 /s) j + (33.8667 kg ⋅ m 2 /s)k H G + r × mv = −(4.8 kg ⋅ m 2 /s) j + (9.6 kg ⋅ m 2/s)k

Angular momentum about O. H O = rA × (mA v A ) + rB × (mB v B ) + rC × (mC vC ) i j k i j k i j k = 0 3 0 + 1.2 2.4 3 + 3.6 0 0 12 6 6 8 6 0 −8 16 8 = (18i − 36k ) + (−18i + 24 j − 12k ) + (−28.8j + 57.6k ) = −(4.8 kg ⋅ m 2 /s) j + (9.6 kg ⋅ m 2 /s)k

Note that H O = H G + r × mv 

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PROBLEM 14.11 A system consists of three particles A, B, and C. We know that WA = 5lb, WB = 4 lb, and WC = 3 lb, and that the velocities of the particles expressed in ft/s are, respectively, v A = 2i + 3j − 2k , v B = vx i + v y j + vz k , and vC = −3i − 2 j + k. Determine (a) the components vx and vz of the velocity of particle B for which the angular momentum HO of the system about O is parallel to the x axis, (b) the value of HO.

SOLUTION i H O = Σri × mvi = Σmi xi (vi ) x i j k i 5 4 = 0 5 4 + 4 g g 2 3 −2 vx =

HO =

(a)

j yi (vi ) y

k zi (vi ) z

j k i j k 3 4 3 + 8 6 0 g 2 vz −3 −2 1

1 [5(−10 − 12) + 4(4vz − 6) + 3(6 − 0)]i g +

1 [5(8 − 0) + 4(3vx − 4vz ) + 3(0 − 8)]j g

+

1 [5(0 − 10) + 4(8 − 4vx ) + 3(−16 + 18)]k g

1 [(16vz − 116)i + (12vz − 16vz + 16) j + (−16vx − 12)k ] g

For H O to be parallel to the x axis, we must have H y = H z = 0: H z = 0: −16vx − 12 = 0

vx = −0.75 ft/s 

H y = 0: 12(−0.75) − 16vz + 16 = 0

(b)

(1)

vz = 0.4375 ft/s 

Substitute into Eq. (1): HO =

1 1 109.0 (16vz − 116)i = [16(0.4375) − 116]i = − i 32.2 g g H O = −(3.39 ft ⋅ lb ⋅ s)i 

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PROBLEM 14.12 For the system of particles of Problem 14.11, determine (a) the components vx and vz of the velocity of particle B for which the angular momentum HO of the system about O is parallel to the z axis, (b) the value of HO. PROBLEM 14.11 A system consists of three particles A, B, and C. We know that WA = 5lb, WB = 4 lb, and WC = 3 lb and that the velocities of the particles expressed in ft/s are, respectively, vA = 2i + 3j − 2k , vB = vx i + v y j + vz k , and vC = −3i − 2 j + k. Determine (a) the components vx and vz of the velocity of particle B for which the angular momentum HO of the system about O is parallel to the x axis, (b) the value of HO.

SOLUTION i H O = Σri × mvi = Σmi xi (vi ) x i j k i 5 4 = 0 5 4 + 4 g g 2 3 −2 vx =

HO =

(a)

(b)

j yi (vi ) y

k zi (vi ) z

j k i j k 3 4 3 + 8 6 0 g 2 vz −3 −2 1

1 [5(−10 − 12) + 4(4vz − 6) + 3(6 − 0)]i g +

1 [5(8 − 0) + 4(3vx − 4vz ) + 3(0 − 8)]j g

+

1 [5(0 − 10) + 4(8 − 4vx ) + 3(−16 + 18)]k g

1 [(16vz − 116)i + (12vz − 16vz + 16) j + (−16vx − 12)k ] g

(1)

For H O to be parallel to the z axis, we must have H x = H y = 0: H x = 0: 16vz − 116 = 0

vz = 7.25 ft/s 

H y = 0: 12vx − 16(7.25) + 16 = 0

vx = 8.33 ft/s 

Substituting into Eq. (1): HO =

1 [−16(8.33) − 12]k 32.2

H O = −(4.51 ft ⋅ lb ⋅ s)k 

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PROBLEM 14.13 A system consists of three particles A, B, and C. We know that mA = 3 kg, mB = 4 kg, and mC = 5 kg, and that the velocities of the particles expressed in m/s are, respectively, vA = −4i + 4 j + 6k, vB = −6i + 8 j + 4k , and vC = 2i − 6 j − 4k. Determine the angular momentum H O of the system about O.

SOLUTION Linear momentum of each particle, (kg ⋅ m/s): mA v A = −12i + 12 j + 18k mB v B = −24i + 32 j + 16k mC vC = 10i − 30 j − 20k

Position vectors, (meters): rA = 1.2 i + 1.5k , rB = 0.9i + 1.2 j + 1.2k , rC = 2.4 j + 1.8k

Angular momentum about O, (kg ⋅ m 2 /s): H O = rA × m A vA + rB × mB v B + rC × mC vC i j k i j k i j k = 1.2 0 1.5 + 0.9 1.2 1.2 + 0 2.4 1.8 10 −30 −20 −12 12 18 −24 32 16 = (−18i − 39.6 j + 14.4k ) + ( −19.2i − 43.2 j + 57.6k ) + (6i + 18 j − 24k ) = −31.2i − 64.8 j + 48.0k H O = − (31.2 kg ⋅ m 2 /s)i − (64.8 kg ⋅ m 2 /s) j + (48.0 kg ⋅ m 2/s)k 

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PROBLEM 14.14 For the system of particles of Problem 14.13, determine (a) the position vector r of the mass center G of the system, (b) the linear momentum mv of the system, (c) the angular momentum H G of the system about G. Also verify that the answers to this problem and to Problem 14.13 satisfy the equation given in Problem 14.27. PROBLEM 14.13 A system consists of three particles A, B, and C. We know that m A = 3 kg, mB = 4 kg, and mC = 5 kg and that the velocities of the particles expressed in m/s are, respectively, v A = −4i + 4 j + 6k , v B = −6i + 8 j + 4k , and vC = 2i − 6 j − 4k. Determine the angular momentum H O of the system about O.

SOLUTION Position vectors, (meters): rA = 1.2i + 1.5k , rB = 0.9i + 1.2 j + 1.2k , rC = 2.4 j + 1.8k

(a)

Mass center: (m A + mB + mC ) r = mArA + mB rB + mC rC 12 r = (3)(1.2i + 1.5k ) + (4)(0.9i + 1.2 j + 1.2k ) + (5)(2.4 j + 1.8k ) r = 0.6i + 1.4 j + 1.525k r = (0.600 m)i + (1.400 m) j + (1.525 m)k 

Linear momentum of each particle, (kg ⋅ m/s): mA v A = −12i + 12 j + 18k mB v B = −24i + 32 j + 16k mC vC = 10i − 30 j − 20k

(b)

Linear momentum of the system, (kg ⋅ m/s): mv = m A v A + mB v B + mC vC = −26i + 14 j + 14k mv = −(26.0 kg ⋅ m/s)i + (14.00 kg ⋅ m/s) j + (14.00 kg ⋅ m/s)k 

Position vectors relative to the mass center, (meters). rA′ = rA − r = 0.6i − 1.4 j − 0.025k rB′ = rB − r = 0.3i − 0.2 j − 0.325k rC′ = rC − r = −0.6i + 1.0 j + 0.275k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 861

PROBLEM 14.14 (Continued)

(c)

Angular momentum about G, (kg ⋅ m 2 /s): H G = rA′ × m A v A + rB′ × mB v B + rC′ × mC vC i j k i j k i j k = 0.6 −1.4 −0.025 + 0.3 −0.2 −0.325 + − 0.6 1.0 0.275 18 16 10 −30 −20 −12 12 −24 32 = (−24.9i − 10.5 j − 9.6k ) + (7.2i + 3.0 j + 4.8k ) + (−11.75i − 9.25 j + 8.0k ) = −29.45i − 16.75 j + 3.2k

H G = − (29.5 kg ⋅ m 2 /s)i − (16.75 kg ⋅ m 2 /s) j + (3.20 kg ⋅ m 2 /s)k  i j k r × mv = 0.6 1.4 1.525 = −1.75i − 48.05 j + 44.8k −26 14 14 H G + r × mv = − (31.2 kg ⋅ m 2 /s)i − (64.8 kg ⋅ m 2 /s) j + (48.0 kg ⋅ m 2 /s)k

Angular momentum about O, (kg ⋅ m 2/s): H O = rA × m A v A + rB × mB v B + rC × mC vC i j k i j k i j k = 1.2 0 1.5 + 0.9 1.2 1.2 + 0 2.4 1.8 10 −30 −20 −12 12 18 −24 32 16 = (−18i − 39.6 j + 14.4k ) + ( −19.2i − 43.2 j + 57.6k ) + (6i + 18 j − 24k ) = −(31.2 kg ⋅ m2 /s)i − (64.8 kg ⋅ m 2 /s) j + (48.0 kg ⋅ m 2 /s)k

Note that

H O = H G + r × mv.

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PROBLEM 14.15 A 13-kg projectile is passing through the origin O with a velocity v0 = (35 m/s)i when it explodes into two fragments A and B, of mass 5 kg and 8 kg, respectively. Knowing that 3 s later the position of fragment A is (90 m, 7 m, –14 m), determine the position of fragment B at the same instant. Assume a y = − g = −9.81 m/s2 and neglect air resistance.

SOLUTION Motion of mass center: It moves as if projectile had not exploded. r = v0 ti −

1 2 gt j 2

1 = (35 m/s)(3 s)i − (9.81 m/s 2 )(3 s) 2 j 2 = (105 m)i − (44.145 m) j

Equation (14.12): m r = Σmi ri : m r = m ArA + mB rB 13(105i − 44.145 j) = 5(90i + 7 j − 14k ) + 8rB 8rB = (13 × 105 − 5 × 90)i + (−13 × 44.145 − 5 × 7) j + (5 × 14)k = 915i − 608.89 j + 70k rB = (114.4 m)i − (76.1 m) j + (8.75 m)k 

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PROBLEM 14.16 A 300-kg space vehicle traveling with a velocity v0 = (360 m/s)i passes through the origin O at t = 0. Explosive charges then separate the vehicle into three parts A, B, and C, with mass, respectively, 150 kg, 100 kg, and 50 kg. Knowing that at t = 4 s, the positions of parts A and B are observed to be A (1170 m, –290 m, –585 m) and B (1975 m, 365 m, 800 m), determine the corresponding position of part C. Neglect the effect of gravity.

SOLUTION Motion of mass center: Since there is no external force, r = v 0t = (360 m/s) i (4 s) = (1440 m)i

Equation (14.12): m r = Σmi ri :

(300)(1440i ) = (150)(1170i − 290 j − 585k ) + (100)(1975i + 365 j + 800k ) + (50)rC 50rC = (300 × 1440 − 150 × 1170 − 100 × 1975)i + (150 × 290 − 100 × 365) j + (150 × 585 − 100 × 800)k = 59, 000i + 7, 000 j + 7,750k rC = (1180 m)i + (140 m) j + (155 m)k 

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PROBLEM 14.17 A 2-kg model rocket is launched vertically and reaches an altitude of 70 m with a speed of 30 m/s at the end of powered flight, time t = 0. As the rocket approaches its maximum altitude it explodes into two parts of masses m A = 0.7 kg and mB = 1.3 kg. Part A is observed to strike the ground 80 m west of the launch point at t = 6 s. Determine the position of part B at that time.

SOLUTION Choose a planar coordinate system having coordinates x and y with the origin at the launch point on the ground and the x-axis pointing east and the y-axis vertically upward. Let subscript E refer to the point where the explosion occurs, and A and B refer to the fragments A and B. Let t be the time elapsed after the explosition. Motion of the mass center: x = xE + (vx )t = 0 y = yE + (v y )t0 −

yE = 70 m and (v y )0 = 30 m/s

where At

1 2 gt 2

t = 6 s,

x =0 1 y = 70 + (30)(6) − (9.81)(6)2 = 73.42 m 2

Definition of mass center: m A x = m A x A + mB xB 0 = (0.7 kg)(−80 m) + (1.3 kg) xB xB = 43.1 m my = mA y A + mB yB (2 kg)(73.42 m) = (0.7 kg)(0) + (1.3 kg) yB yB = 113.0 m 43.1 m (east), 113.0 m (up) 

Position of part B:

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PROBLEM 14.18 An 18-kg cannonball and a 12-kg cannonball are chained together and fired horizontally with a velocity of 165 m/s from the top of a 15-m wall. The chain breaks during the flight of the cannonballs and the 12-kg cannonball strikes the ground at t = 1.5 s, at a distance of 240 m from the foot of the wall, and 7 m to the right of the line of fire. Determine the position of the other cannonball at that instant. Neglect the resistance of the air.

SOLUTION Let subscript A refer to the 12-kg cannonball and B to the 18-kg cannonball. The motion of the mass center of A and B is uniform in the x-direction, uniformly accelerated with acceleration − g = −9.81 m/s 2 in the y-direction, and zero in the z-direction. x = (v0 ) x t = (165 m/s)(1.5 s) = 247.5 m y = y0 + (v0 ) y t −

1 2 gt 2

1 = 15 m + 0 − (9.81 m/s 2 )(1.55)2 = 3.964 m 2 z =0 r = (247.5 m)i + (3.964 m)j

Definition of mass center: m r = m ArA + mB rB

Data:

m A = 12 kg,

mB = 18 kg,

m = mA + mB = 30 kg

t = 1.5 s, x A = 240 m, y A = 0,

zA = 7 m

(30)(247.5i + 3.964 j) = (12)(240i + 7k ) + (18)( xB i + yB j + z B k )



i: (30)(247.5) = (12)(240) + 18 xB

xB = 253 m 

j: (30)(3.964) = (12)(0) + 18 yB

yB = 6.61 m 

k: (30)(0) = (12)(7) + 18 z B

z B = −4.67 m 

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PROBLEM 14.19 Car A was traveling east at high speed when it collided at Point O with car B, which was traveling north at 45 mi/h. Car C, which was traveling west at 60 mi/h, was 32 ft east and 10 ft north of Point O at the time of the collision. Because the pavement was wet, the driver of car C could not prevent his car from sliding into the other two cars, and the three cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the weights of cars A, B, and C are, respectively, 3000 lb, 2600 lb, and 2400 lb, and neglecting the forces exerted on the cars by the wet pavement, solve the problems indicated. Knowing that the speed of car A was 75 mi/h and that the time elapsed from the first collision to the stop at P was 2.4 s, determine the coordinates of the utility pole P.

SOLUTION Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system consisting of cars A, B, and C during the impacts with one another. The mass center of the system moves at the velocity it had before the collision. Setting the origin at O, we can find the initial mass center r0 : at the moment of the first collision: (m A + mB + mC )( x0 i + y0 j) = m A (0) + mB (0) + mC ( xC i + yC j) x0 = 0.3 xC = (0.3)(32) = 9.6 ft,

y0 = 0.3 yC = (0.3)(10) = 3 ft

Given velocities: vA = (75 mi/h)i = (110 ft/s)i, v B = (45 mi/h)i = (66 ft/s)j,

vC = (60 mi/h)i = (88 ft/s)i

Velocity of mass center: (m A + mB + mC ) v = mA v A + mB v B + mC v C v = 0.375vA + 0.325vB + 0.3vC

Since the collided cars hit the pole at rP = xP i + yP j x P i + yP j = x0 i + y0 j + vt

Resolve into components.

x : xP = x0 + 0.375v At P − 0.3vC tP

(1)

y : yP = y0 + 0.325vB tP

(2)

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PROBLEM 14.19 (Continued)

Data:

t P = 2.4 s

From (1),

xP = 9.6 + (0.375)(110)(2.4) − (0.3)(88)(2.4) = 45.240

From (2),

yP = 3.0 + (0.325)(66)(2.4) = 54.480 ft xP = 45.2 ft  yP = 54.5 ft 

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PROBLEM 14.20 Car A was traveling east at high speed when it collided at Point O with car B, which was traveling north at 45 mi/h. Car C, which was traveling west at 60 mi/h, was 32 ft east and 10 ft north of Point O at the time of the collision. Because the pavement was wet, the driver of car C could not prevent his car from sliding into the other two cars, and the three cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the weights of cars A, B, and C are, respectively, 3000 lb, 2600 lb, and 2400 lb, and neglecting the forces exerted on the cars by the wet pavement, solve the problems indicated. Knowing that the coordinates of the utility pole are xP = 46 ft and yP = 59 ft, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A.

SOLUTION Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system consisting of cars A, B, and C during the impacts with one another. The mass center of the system moves at the velocity it had before the collision. Setting the origin at O, we can find the initial mass center r0 : at the moment of the first collision: (m A + mB + mC )( x0 i + y0 j) = m A (0) + mB (0) + mC ( xC i + yC j) x0 = 0.3 xC = (0.3)(32) = 9.6 ft,

y0 = 0.3 yC = (0.3)(10) = 3 ft

Given velocities: vA = vAi, v B = (45 mi/h)i = (66 ft/s)j,

vC = (60 mi/h)i = (88 ft/s)i

Velocity of mass center: (m A + mB + mC ) v = m A v A + mB v B + mC vC v = 0.375vA + 0.325vB + 0.3vC

Since the collided cars hit the pole at rP = xP i + yP j

xP i + yP j = x0 i + y0 j + vt

Resolve into components.

x : xP = x0 + 0.375v At P − 0.3vC tP

(1)

y : yP = y0 + 0.325vB tP

(2)

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PROBLEM 14.20 (Continued)

Data:

xP = 59 ft, yP = 46 ft

(a ) From (2),

46 = 3 + (0.325)(66)t P  t P = 2.0047 s

(b) From (1),

t P = 2.00 s 

59 = 9.6 + (0.375)v A (2.0047) − (0.3)(88)(2.0047) vA = 136.11 ft/s

vA = 92.8 mi/h 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 870

PROBLEM 14.21 An expert archer demonstrates his ability by hitting tennis balls thrown by an assistant. A 2-oz tennis ball has a velocity of (32 ft/s)i – (7 ft/s)j and is 33 ft above the ground when it is hit by a 1.2-oz arrow traveling with a velocity of (165 ft/s)j + (230 ft/s)k where j is directed upwards. Determine the position P where the ball and arrow will hit the ground, relative to Point O located directly under the point of impact.

SOLUTION Assume that the ball and arrow move together after the hit. Conservation of momentum of ball and arrow during the hit. mA =

1.2/16 = 2.3292 × 10−3 slug 32.2

mB =

2/16 = 3.8820 × 10 −3 slug 32.2 m A vA + mB v B = (m A + mB ) v

−3

(2.3292 × 10 )(165 j + 230k ) + (3.8820 × 10−3 )(32i − 7 j) = (2.3292 × 10−3 + 3.8820 × 10−3 )v v = (20.0 ft/s)i + (57.5 ft/s) j + (86.25 ft/s)k

After the hit, the ball and arrow move as a projectile. 1 2 gt 2 1 y = 33 + 57.5t − (32.2)t 2 2 y = y0 + (v y )0 t −

Vertical motion:

y = 0 at ground. −16.1t + 57.5t + 33 = 0 2

Solve for t.

After rejecting the negative root,

t = 4.0745 s

Horizontal motion:

x = x0 + (vx )0 t z = z0 + (v2 )0 t x = 0 + (20)(4.07448) = 81.490 ft z = 0 + (86.25)(4.07448) = 351.42 ft rP = (81.5 ft)i + (351 ft)k 



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PROBLEM 14.22 Two spheres, each of mass m, can slide freely on a frictionless, horizontal surface. Sphere A is moving at a speed v0 = 16 ft/s when it strikes sphere B which is at rest and the impact causes sphere B to break into two pieces, each of mass m/2. Knowing that 0.7 s after the collision one piece reaches Point C and 0.9 s after the collision the other piece reaches Point D, determine (a) the velocity of sphere A after the collision, (b) the angle θ and the speeds of the two pieces after the collision.

SOLUTION Velocities of pieces C and D after impact and fracture. (vC′ ) x =

xC 6.3 = = 9 ft/s, tC 0.7

(vC′ ) y = 9 tan 30° ft/s

(v′D ) x =

xD 6.3 = = 7 ft/s, tD 0.9

(v′D ) y = −7 tan θ ft/s

Assume that during the impact the impulse between spheres A and B is directed along the x-axis. Then, the y component of momentum of sphere A is conserved. 0 = m(v′A ) y

Conservation of momentum of system: : mAv0 + mB (0) = m Av′A + mC (vC′ ) x + mD (v′D ) x m (16) + 0 = mv′A +

m m (9) + (7) 2 2 v′A = 8.00 ft/s

(a )



: mA (0) + mB (0) = mA (v′A ) y + mC (vC′ ) y + mD (v′D ) y 0+0=0+ (b)

tan θ =

m m (9 tan 30°) − (7 tan θ ) 2 2

9 tan 30° = 0.7423 7

vC =

(vC ) 2x + (vC ) 2y =

(9)2 + (9 tan 30°) 2

vD =

(vD ) 2x + (vD )2y =

(7)2 + (7 tan 36.6°)2

θ = 36.6°  vC = 10.39 ft/s  vD = 8.72 ft/s 

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PROBLEM 14.23 In a game of pool, ball A is moving with a velocity v0 when it strikes balls B and C which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated, and that v0 = 12 ft/s and vC = 6.29 ft/s, determine the magnitude of the velocity of (a) ball A, (b) ball B.

SOLUTION Conservation of linear momentum. In x direction: m(12 ft/s) cos 45° = mv A sin 4.3° + mvB sin 37.4°

+ m(6.29) cos 30° 0.07498vA + 0.60738vB = 3.0380



(1)

In y direction: m(12 ft/s) sin 45° = mv A cos 4.3° − mvB cos 37.4°

+ m(6.29) sin 30° 0.99719vA − 0.79441vB = 5.3403

(a)

Multiply (1) by 0.79441, (2) by 0.60738, and add: 0.66524vA = 5.6570

(b)

(2)

vA = 8.50 ft/s 

Multiply (1) by 0.99719, (2) by –0.07498, and add: 0.66524vB = 2.6290 



vB = 3.95 ft/s 

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PROBLEM 14.24 A 6-kg shell moving with a velocity v0 = (12 m/s)i − (9 m/s) j − (360 m/s)k explodes at Point D into three fragments A, B, and C of mass, respectively, 3 kg, 2 kg, and 1 kg. Knowing that the fragments hit the vertical wall at the points indicated, determine the speed of each fragment immediately after the explosion. Assume that elevation changes due to gravity may be neglected.

SOLUTION rD = 4k

Position vectors (m): rA = −1.5i

rA/D = −1.5i − 4k

rA/D = 4.272

rB = 4i + 2 j

rB/D = 4i + 2 j − 4k

rB/D = 6

rC = −3j

rC/D = −3j − 4k

rC/D = 5

1 (−1.5i − 4k ) 4.272 1 Along rB/D , λ B = (4i + 2 j − 4k ) 6 1 Along rC/D , λ C = (−3j − 4k ) 5 Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explosion have the directions of the unit vectors.

Unit vectors:

Along rA/D ,

λA =

vA = vA λ A vB = vB λ B vC = vC λ C

Conservation of momentum:

mv 0 = m A vA + mB vB + mC vC  v  v 6(12i − 9 j − 360k ) = 3  A  (−1.5i − 4k ) + 2  B 4.272    6 v  + 1 C  (−3j − 4k )  5 

  (4i + 2 j − 4k ) 

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PROBLEM 14.24 (Continued) Resolve into components. 72 = −1.0534v A + 1.3333vB −54 = 0.66667vB − 0.60000vC −2160 = −2.8090v A − 1.3333vB − 0.80000vC

vA = 431 m/s 

Solving,

vB = 395 m/s  vC = 528 m/s 

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PROBLEM 14.25 A 6-kg shell moving with a velocity v0 = (12 m/s)i − (9 m/s) j − (360 m/s)k explodes at Point D into three fragments A, B, and C of mass, respectively, 2 kg, 1 kg, and 3 kg. Knowing that the fragments hit the vertical wall at the points indicated, determine the speed of each fragment immediately after the explosion. Assume that elevation changes due to gravity may be neglected.

SOLUTION rD = 4k

Position vectors (m): rA = −1.5i

rA/D = −1.5i − 4k

rA/D = 4.272

rB = 4i + 2 j

rB/D = 4i + 2 j − 4k

rB/D = 6

rC = −3j

rC/D = −3j − 4k

rC/D = 5

1 (−1.5i − 4k ) 4.272 1 Along rB/D , λ B = (4i + 2 j − 4k ) 6 1 Along rC/D , λ C = (−3j − 4k ) 5 Assume that elevation changes due to gravity may be neglected. Then the velocity vectors after the explosion have the directions of the unit vectors. vA = vAλ A

Unit vectors:

Along rA/D ,

λA =

v B = vB λ B vC = vC λ C

Conservation of momentum:

mv 0 = m A vA + mB vB + mC vC

Resolve into components.  v  v 6(12i − 9 j − 360k ) = 2  A  (−1.5i − 4k ) + 1 B  4.272   6 v  + 3  C  (3j − 4k )  5 

  (4i + 2 j − 4k ) 

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PROBLEM 14.25 (Continued)

72 = −0.70225v A + 0.66667vB −54 = 0.33333vB − 1.8000vC −2160 = −1.8727v A − 0.66667vB − 2.40000vC

vA = 646 m/s 

Solving,

vB = 789 m/s  vC = 176 m/s 

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PROBLEM 14.26 In a scattering experiment, an alpha particle A is projected with the velocity u0 = −(600 m/s)i + (750 m/s) j − (800 m/s)k into a stream of oxygen nuclei moving with a common velocity v0 = (600 m/s) j . After colliding successively with nuclei B and C, particle A is observed to move along the path defined by the Points A1(280, 240, 120)and A2(360, 320, 160), while nuclei B and C are observed to move along paths defined, respectively, by B1(147, 220, 130), B2(114, 290, 120),and by C1(240, 232, 90) and C2(240, 280, 75). All paths are along straight lines and all coordinates are expressed in millimeters. Knowing that the mass of an oxygen nucleus is four times that of an alpha particle, determine the speed of each of the three particles after the collisions.

SOLUTION Position vectors (mm):

Unit vectors:

 A1 A2 = 80i + 80 j + 40k  B1 B2 = −33i + 70 j − 10k  C1C2 = 48 j − 15k

( A1 A2 ) = 120 ( B1 B2 ) = 78.032 (C1C2 ) = 50.289

Along A1 A2 ,

λ A = 0.66667i + 0.66667 j + 0.33333k

Along B1 B2 ,

λB = −0.42290i + 0.89707 j − 0.12815k

Along C1C2 ,

λC = 0.95448 j − 0.29828k

Velocity vectors after the collisions: vA = vAλ A vB = vB λB vC = vC λC

Conservation of momentum: mu0 + 4mv 0 + 4mv 0 = mvA + 4mvB + 4mv C

Divide by m and substitute data. (−600i + 750 j − 800k ) + 2400 j + 2400 j = vAλ A + 4vB λB + 4vC λC

Resolving into components, i : − 600 = 0.66667vA − 1.69160vB j: 5550 = 0.66667vA + 3.58828vB + 3.81792vC k : − 800 = 0.33333vA − 0.51260vB − 1.19312vC

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PROBLEM 14.26 (Continued)

Solving the three equations simultaneously, vA = 919.26 m/s vB = 716.98 m/s vC = 619.30 m/s vA = 919 m/s  vB = 717 m/s  vC = 619 m/s 

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PROBLEM 14.27 Derive the relation H O = r × mv + H G

between the angular momenta H O and H G defined in Eqs. (14.7) and (14.24), respectively. The vectors r and v define, respectively, the position and velocity of the mass center G of the system of particles relative to the newtonian frame of reference Oxyz, and m represents the total mass of the system.

SOLUTION n

From Eq. (14.7),

HO =

 (r × m v ) i

i i

i =1 n

=

 ( r + r′) × m v  i

i i

i =1

n

=r×

n

 (m v ) +  ( r ′× m v ) i i

i =1

i

i i

i =1

= r × mv + H G

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PROBLEM 14.28 Show that Eq. (14.23) may be derived directly from Eq. (14.11) by substituting for H O the expression given in Problem 14.27.

SOLUTION n

HO =

From Eq. (14.7),

 (r × m v ) i

i i

i =1 n

=

 ( r + r′) × m v  i

i i

i =1

n

=r×



n

( mi vi ) +

i =1

 ( r ′× m v ) i

i i

i =1

= r × mv + H G

 = r × mv + r × mv + H  H O G

Differentiating, Using Eq. (14.11),

 ΣM O = r × mv + r × mv + H G  = v × mv + r × m a + H G  = 0 + r ×   n

But

 i =1

 F  + H i

 i =1

(1)

G

i =1

n

MO =



n

 r ×  i =1  n

MG +





n

 F  i

i =1

Subtracting r × ( Σin=1 Fi ) from each side of Eq. (1) gives  ΣM G = H G

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PROBLEM 14.29 Consider the frame of reference Ax′y ′z ′ in translation with respect to the newtonian frame of reference Oxyz. We define the angular momentum H′A of a system of n particles about A as the sum H ′A =

n

 r ′ × m v′ i

i i

(1)

i =1

of the moments about A of the momenta mi vi′ of the particles in their motion relative to the frame Ax′y ′z ′. Denoting by H A the sum n

HA =

 r′ × m v i

i i

(2)

i =1

of the moments about A of the momenta mi vi of the particles in their motion relative to the newtonian frame Oxyz, show that HA = H A′ at a given instant if, and only if, one of the following conditions is satisfied at that instant: (a) A has zero velocity with respect to the frame Oxyz, (b) A coincides with the mass center G of the system, (c) the velocity vA relative to Oxyz is directed along the line AG.

SOLUTION vi = vA + v′i n

HA =

 r′ × m v i

i i

i =1 n

=

 r′ × m ( v i

i

A

i =1 n

=

n

 (ri′ × mi vA ) +  ri′ × mi vi i =1 n

=

+ vi′ )

i =1

 ( m r ′) × v i i

A

+ H A′

i =1 n

=

 m (r − r ) × v i

i

A

A

+ H′A

i =1

= m( r − rA ) × vA + H′A H A = H ′A if, and only if,

m(r − rA ) × vA = 0

This condition is satisfied if (a ) vA = 0

Point A has zero velocity.

or

(b)

r = rA

Point A coincides with the mass center.

or

(c )

vA is parallel to r − rA.

Velocity vA is directed along line AG.

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PROBLEM 14.30  ′ , where H ′ is defined by Eq. (1) of Show that the relation ΣMA = H A A Problem 14.29 and where ΣM A represents the sum of the moments about A of the external forces acting on the system of particles, is valid if, and only if, one of the following conditions is satisfied: (a) the frame Ax′y ′z ′ is itself a newtonian frame of reference, (b) A coincides with the mass center G, (c) the acceleration aA of A relative to Oxyz is directed along the line AG.

SOLUTION From equation (1),

H ′A =

n

 ( r ′ × m v′ ) i

i i

i =1

H A′ =

n

[(r − r ) × m ( v i

A

i

i

− vA )]

i =1

Differentiate with respect to time. ′ = H A

n



n

[(ri − rA ) × mi ( vi − vA )] +

i =1

But

ri = vi v i = ai rA = vA

and

vA = aA

 ′ =0+ H A

Hence,

[(r − r ) × m ( v i

A

i

i

− v A )]

i =1

n

[(r − r ) × m (a i

A

i

i

− aA )]

i =1

n

=

[(r − r ) × (F − m a )] i

A

i

i A

i =1 n

=

n

[(r − r ) × F ] − [m (r − r )] × a i

A

i

i =1

i

i

A

A

i =1

= MA − m( r − rA ) × aA  H ′A = MA if, and only if,

m( r − rA ) × aA = 0

This condition is satisfied if (a ) aA = 0

The frame is newtonian.

or

(b)

r = rA

Point A coincides with the mass center.

or

(c )

aA is parallel to r − rA.

Acceleration aA is directed along line AG.



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PROBLEM 14.31 Determine the energy lost due to friction and the impacts for Problem 14.1. PROBLEM 14.1 A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic and the coefficient of kinetic friction between B and C is 0.2, determine (a) the velocity of the bullet and B after the first impact, (b) the final velocity of the carrier.

SOLUTION From the solution to Problem 4.1 the velocity of A and B after the first impact is v′ = 4.4554 m/s and the velocity common to A, B, and C after the sliding of block B and bullet A relative to the carrier C has ceased in v′′ = 0.4087 m/s. Friction loss due to sliding: N = WA + WB = (m A + mB ) g

Normal force:

= (0.030 kg + 3 kg)(9.81 m/s) = 29.724 N

Friction force:

F f = μk N = (0.2)(29.724) = 5.945 N

Relative sliding distance:

Assume d = 0.5 m.

Energy loss due to friction:

F f d = (5.945)(0.5)

F f d = 2.97 J 

Kinetic energy of block with embedded bullet immediately after first impact: ′ = TAB

1 1 (m A + mB )(v′) 2 = (3.03 kg)(4.4554 m/s)2 = 30.07 J 2 2

Final kinetic energy of A, B, and C together ′′ = TABC

1 1 (m A + mB + mC )(v′′)2 = (33.03 kg)(0.4087 m/s) 2 = 2.76 J 2 2

Loss due to friction and stopping impact:

′ − TABC ′′ = 30.07 − 2.76 = 27.31 J TAB

Since 27.31 J ≥ 2.97 J, the block slides 0.5 m relative to the carrier as assumed above.

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PROBLEM 14.31 (Continued)

Impact loss due to AB impacting the carrier: 27.31 − 2.97 = 24.34

Loss = 24.3 J 

Initial kinetic energy of system ABC. T0 =

1 1 m Av02 = (0.030 kg)(450 m/s) 2 = 3037.5 J 2 2

Impact loss at first impact: ′ = 3037.5 − 30.07 T0 − TAB

Loss = 3007 J 

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PROBLEM 14.32 In Problem 14.4, determine the energy lost as the bullet (a) passes through block A, (b) becomes embedded in block B.

SOLUTION The masses are m for the bullet and m A and mB for the blocks. The bullet passes through block A and embeds in block B. Momentum is conserved. Initial momentum: mv0 + mA (0) + mB (0) = mv0 Final momentum:

mvB + mA vA + mB vB

Equating,

mv0 = mvB + mA vA + mB vB m=

mA vA + mB vB (6)(5) + (4.95)(9) = = 0.0500 lb 1500 − 9 v0 − vB

The bullet passes through block A. Momentum is conserved. Initial momentum: mv0 + m A (0) = mv0 Final momentum:

mv1 + mA vA

Equating,

mv0 = mv1 + mA vA

mv0 − mA vA (0.0500)(1500) − (6)(5) = = 900 ft/s 0.0500 m 0.05 m= = 1.5528 × 10−3 lb ⋅ s 2 /ft The masses are: 32.2 6 mA = = 0.18633 lb ⋅ s 2/ft 32.2 4.95 mB = = 0.153727 lb ⋅ s2 /ft 32.2 (a) Bullet passes through block A. Kinetic energies: 1 1 Before: T0 = mv02 = (1.5528 × 10−3 )(1500) 2 = 1746.9 ft ⋅ lb 2 2 1 2 1 1 1 After: T1 = mv1 + mA v 2A = (1.5528 × 10−3 )(900) 2 + (0.18633)(5)2 = 631.2 ft ⋅ lb 2 2 2 2 T0 − T1 = 1746.9 − 631.2 = 1115.7 ft ⋅ lb energy lost = 1116 ft ⋅ lb  Lost: v1 =

(b)

Bullet becomes embedded in block B. Kinetic energies: 1 1 Before: T2 = mv12 = (1.5528 × 10−3 )(900) 2 = 628.9 ft ⋅ lb 2 2 1 1 T3 = ( m + mB )vB2 = (0.15528)(9)2 = 6.29 ft ⋅ lb After: 2 2 T2 − T3 = 628.9 − 6.29 = 622.6 ft ⋅ lb energy lost = 623 ft ⋅ lb  Lost:

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PROBLEM 14.33 In Problem 14.6, determine the work done by the woman and by the man as each dives from the boat, assuming that the woman dives first.

SOLUTION Woman dives first. Conservation of momentum: 120 300 + 180 (16 − v1 ) − v1 = 0 g g (120)(16) = 3.20 ft/s v1 = 600 16 − v1 = 12.80 ft/s

Kinetic energy before dive:

T0 = 0

Kinetic energy after dive:

T1 =

1 300 + 180 1 120 (3.20) 2 + (12.80) 2 2 32.2 2 32.2 = 381.61 ft ⋅ lb

T1 − T0 = 381.61 ft ⋅ lb

Work of woman:

T1 − T0 = 382 ft ⋅ lb 

Man dives next. Conservation of momentum:



300 + 180 300 180 v1 = − v2 + (16 − v2 ) g g g 480v1 + (180)(16) = 9.20 ft/s v2 = 480 16 − 9.20 = 6.80 ft/s

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PROBLEM 14.33 (Continued) 1 300 + 180 (3.20) 2 2 32.2 = 76.323 ft ⋅ lb

Kinetic energy before dive:

T1′ =

Kinetic energy after dive:

T2′ =

Work of man:

1 300 1 180 (9.20) 2 + (6.80) 2 2 32.2 2 32.2 = 523.53 ft ⋅ lb

T2′ − T1′ = 447.2 ft ⋅ lb

T2′ − T1′ = 447 ft ⋅ lb 

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PROBLEM 14.34 Determine the energy lost as a result of the series of collisions described in Problem 14.8. PROBLEM 14.8 Packages in an automobile parts supply house are transported to the loading dock by pushing them along on a roller track with very little friction. At the instant shown, packages B and C are at rest and package A has a velocity of 2 m/s. Knowing that the coefficient of restitution between the packages is 0.3, determine (a) the velocity of package C after A hits B and B hits C, (b) the velocity of A after it hits B for the second time.

SOLUTION From the solution to Problem 14.8 v A = 2 m/s, vB = vC = 0, v′A = 1.133 m/s, vB′ = 1.733 m/s, vB′′ = 0.382 m/s vB′ = 0.901 m/s v′′A = 0.807 m/s, vB′′ = 1.033 m/s m A = 8 kg,

A hits B:

mB = 4 kg,

1 1 m Av 2A = (8 kg)(2 m/s) 2 = 16 J 2 2 1 1 T2 = m A (v′A ) 2 + mB (vB′ )2 2 2 1 1 T = (8 kg)(1.133 m/s) 2 + (4 kg)(1.733 m/s) 2 = 11.14 J 2 2 T1 =

Loss = T1 − T2 :

B hits C:

Loss = 4.86 J 

1 1 mB (vB′ )2 = (4 kg)(1.733)2 = 6.007 J 2 2 1 1 T4 = mB (vB′′ )2 + mC (vC′ )2 2 2 1 1 = (4 kg)(0.382 m/s) 2 + (6 kg)(0.901 m/s) 2 = 2.727 J 2 2 T3 =

Loss = T3 − T4 :

A hits B again:

mC = 6 kg

Loss = 3.28 J 

1 1 m A (v′A )2 + mB (vB′′ ) 2 2 2 1 1 = (8 kg)(1.33 m/s) 2 + (4 kg)(0.382) 2 = 5.427 J 2 2 1 1 T6 = m A (v′′A )2 + mB (vB′′′ ) 2 2 2 1 1 = (8 kg)(0.807 m/s) 2 + (4 kg)(1.033 m/s) 2 = 4.739 J 2 2 T5 =

Loss = T5 − T6 :

Loss = 0.688 J

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PROBLEM 14.35 Two automobiles A and B, of mass mA and mB , respectively, are traveling in opposite directions when they collide head on. The impact is assumed perfectly plastic, and it is further assumed that the energy absorbed by each automobile is equal to its loss of kinetic energy with respect to a moving frame of reference attached to the mass center of the two-vehicle system. Denoting by EA and EB, respectively, the energy absorbed by automobile A and by automobile B, (a) show that EA /EB = mB /mA , that is, the amount of energy absorbed by each vehicle is inversely proportional to its mass, (b) compute EA and EB , knowing that mA = 1600 kg and mB = 900 kg and that the speeds of A and B are, respectively, 90 km/h and 60 km/h.

SOLUTION Velocity of mass center:

(mA + mB ) v = m A vA + mB vB v=

mA vA + mB vB mA + mB

Velocities relative to the mass center: v′A = vA − v = vA −

mA vA + mB vB mB ( vA + vB ) = mA + mB mA + mB

v′B = vB − v = vB −

mA vA + mB vB mA ( vA + vB ) = mA + mB mA + mB

Energies:

(a) (b)

EA =

m m 2 ( v + v B ) ⋅ ( vA + vB ) 1 mA vA′ ⋅ v′A = A B A 2 2(m A + mB ) 2

EB =

m 2 m ( v + vB ) ⋅ ( vA + vB ) 1 mB vB′ ⋅ vB′ = A B A 2 2(mA + mB ) 2 EA mB =  EB mA

Ratio: vA = 90 km/h = 25 m/s vB = 60 km/h = 16.667 m/s vA + vB = 41.667 m/s

EA =

(1600)(900) 2 (41.667) 2 = 180.0 × 103 J (2)(2500) 2

EB =

(1600) 2 (900)(41.667) 2 = 320 × 103 J 2 (2)(2500)

EA = 180.0 kJ  EB = 320 kJ 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 890

PROBLEM 14.36 It is assumed that each of the two automobiles involved in the collision described in Problem 14.35 had been designed to safely withstand a test in which it crashed into a solid, immovable wall at the speed v0. The severity of the collision of Problem 14.35 may then be measured for each vehicle by the ratio of the energy it absorbed in the collision to the energy it absorbed in the test. On that basis, show that the collision described in Problem 14.35 is (mA /mB ) 2 times more severe for automobile B than for automobile A.

SOLUTION Velocity of mass center:

(m A + mB ) v = mA vA + mB vB v=

mA vA + mB vB mA + mB

Velocities relative to the mass center: vA′ = vA − v = v A −

mA vA + mB vB mB ( vA + vB ) = m A + mB mA + mB

vB′ = vB − v = vB −

m A vA + mB vB m (v + v ) = A A B mA + mB mA + mB

Energies:

Energies from tests: Severities:

EA =

m m 2 ( v + vB ) ⋅ ( vA + vB ) 1 mA vA′ ⋅ vA′ = A B A 2 2( mA + mB ) 2

EB =

m 2 m ( v + vB ) ⋅ ( vA + vB ) 1 mB vB′ ⋅ vB′ = A B A 2 2( mA + mB ) 2

( E A )0 =

1 1 mA v02 , ( EB )0 = mB v02 2 2

SA =

EA m 2 ( v + vB ) ⋅ ( vA + vB ) = B A ( E A )0 ( mA + mB ) 2 v02

SB =

EB m 2 ( v + vB ) ⋅ ( vA + vB ) = A A ( E B )0 ( mA + mB ) 2 v02 SA mB2  = SB mA2

Ratio:

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PROBLEM 14.37 Solve Sample Problem 14.4, assuming that cart A is given an initial horizontal velocity v0 while ball B is at rest.

SOLUTION (a)

Velocity of B at maximum elevation: At maximum elevation, ball B is at rest relative to cart A. vB = vA Use impulse-momentum principle.

mAv0 + 0 = m AvA + mB vB

x components:

= (mA + mB )vB vB =

(b)

mA v0 mA + mB



Conservation of energy: 1 m A v02 , V1 = 0 2 1 1 T2 = m A v A2 + mB vB2 2 2 1 = (m A + mB )vB2 2 mA2 v02 = 2(m A + mB ) T1 =

V2 = mB gh T2 + V2 = T1 + V1 m A2 v02 2( mA + mB )

+ mB gh =

h=

1 m Av02 2

m A2 v02  1  2 m v −  A 0  2mB g  m A + mB 

h=

v02 mA  m A + mB 2 g

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 892

PROBLEM 14.38 Two hemispheres are held together by a cord which maintains a spring under compression (the spring is not attached to the hemispheres). The potential energy of the compressed spring is 120 J and the assembly has an initial velocity v 0 of magnitude v0 = 8 m/s. Knowing that the cord is severed when θ = 30°, causing the hemispheres to fly apart, determine the resulting velocity of each hemisphere.

SOLUTION Use a frame of reference moving with the mass center. Conservation of momentum: 0 = −m Av′A + mB vB′ v′A =

mB vB′ mA

V=

1 1 m A (v′A ) 2 + mB (vB′ ) 2 2 2

Conservation of energy:

2

m  1 1 = m A  B vB′  + mB (vB′ ) 2 2 2  mA  m (m + mB ) (vB′ ) 2 = B A 2m A vB′ =

Data:

2m AV mB ( mA + mB )

m A = 2.5 kg mB = 1.5 kg V = 120 J vB′ = v′A =

(2)(2.5)(120) = 10 (1.5)(4.0) 1.5 (10) = 6 2.5

vB′ = 10 m/s v′A = 6 m/s

30°

30°

Velocities of A and B. vA = [8 m/s

] + [6 m/s

vB = [8 m/s

] + [10 m/s

30°] 30°]

vA = 4.11 m/s vB = 17.39 m/s

46.9°  16.7° 

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PROBLEM 14.39 A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting friction, determine (a) the velocity of B relative to A after it has slid 3 ft down the inclined surface of the wedge, (b) the corresponding velocity of A.

SOLUTION Kinematics:

v B = v A + v B/A

Law of cosines:

vB2 = v A2 + vB2/A − 2v AvB/A cos 30°

(1)

Principle of impulse and momentum: Σmv 0 + ΣFt = Σmv

Components

:

0 + 0 = m Av A + mB (v A − vB/A cos 30°) vA =

mB vB /A cos 30° 15cos 30° = vB/A mA + mB 25 + 15

= 0.32476 vB/A

From Eq. (1)

vB2 = (0.32476) 2 vB2/A + vB2/A − (2)(0.32476) cos 30°vB2/A = 0.54297 vB2/A

Principle of conservation of energy:

T0 + V0 = T1 + V1 1 1 m A v A2 + mB vB2 − WB d sin 30° 2 2 1 WA 1 WB (0.32476vB/A ) 2 + (0.54297) vB2 /A = WB d sin 30° 2 g 2 g

0+0=

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PROBLEM 14.39 (Continued)

1 15  1 25  2 2  2 32.2 (0.32476) + 2 32.2 (0.54297)  vB/A = (15)(3)sin 30°   0.16741vB2/A = 22.5

v B/A = 11.59 ft/s

(a) (b)

v A = (0.32476)(11.59)

v A = 3.76 ft/s

30° 



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PROBLEM 14.40 A 40-lb block B is suspended from a 6-ft cord attached to a 60-lb cart A, which may roll freely on a frictionless, horizontal track. If the system is released from rest in the position shown, determine the velocities of A and B as B passes directly under A.

SOLUTION Conservation of linear momentum: Since block and cart are initially at rest, L0 = 0

Thus, as B passes under A, L = m A vA + mB vB = 0 m A v A + mB vB = 0 vA = −

mB vB mA

(1)

Conservation of energy: Initially, T0 = 0 V0 = mB gl (1 − cos θ )

As B passes under A,

Thus,

1 1 m Av A2 + mB vB2 2 2 V =0 T=

T0 + V0 = T + V : mB gl (1 − cos θ ) =

1 1 m Av A2 + mB vB2 2 2

Substituting for vA from (1) and multiplying by 2:  m2  2mB gl (1 − cos θ ) = mA  B2 vB2  + mB vB2 m   A   m2  m + mA 2 =  B + mB  vB2 = mB B vB m  mA  A 

vB =

2m A gl (1 − cos θ ) mA + mB

(2)

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PROBLEM 14.40 (Continued)

Given data:

wA = 60 lb wB = 40 lb, l = 6 ft g = 32.2 ft/s 2 θ = 25° 2mA 2 wA (2)(60) = = = 1.2 mA + mB wA + wB 60 + 40

From Eq. (2),

vB = (1.2)(32.2)(6)(1 − cos 25°)

v B = 4.66 ft/s



From Eq. (1),

vA = −

wB 40 vB = − (4.66) 60 wA

vA = 3.11 ft/s



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PROBLEM 14.41 In a game of pool, ball A is moving with a velocity v 0 of magnitude v0 = 15 ft/s when it strikes balls B and C, which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and assuming frictionless surfaces and perfectly elastic impact (i.e., conservation of energy), determine the magnitudes of the velocities vA, vB, and vC.

SOLUTION v 0 = v0 (cos 45° i + sin 45° j)

Velocity vectors:

v0 = 15 ft/s

vA = vA j vB = vB (sin 30° i − cos 30° j) vC = vC (cos 30° i + sin 30° j)

Conservation of momentum: mv 0 = mvA + mvB + mvC

Divide by m and resolve into components. i: v0 cos 45° = vB sin 30° + vC cos 30° j: v0 sin 45° = vA − vB cos 30° + vC sin 30° Solving for vB and vC , vB = −0.25882v0 + 0.86603vA vC = 0.96593v0 − 0.5vA

Conservation of energy: Divide by

1 2

1 2 1 2 1 2 1 2 mv0 = mvA + mvB + mvC 2 2 2 2

m and substitute for vB and vC . v02 = vA2 + (−0.25882 v0 + 0.86603vA )2 + (0.96593 v0 − 0.5vA )2 = 2v A2 + v02 − 1.41422 v0 vA vA = 0.70711v0 = 10.61 ft/s

vA = 10.61 ft/s 

vB = 0.35355 v0 = 5.30 ft/s

vB = 5.30 ft/s 

vC = 0.61237 v0 = 9.19 ft/s

vC = 9.19 ft/s 

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PROBLEM 14.42 In a game of pool, ball A is moving with a velocity v 0 of magnitude v0 = 15 ft/s when it strikes balls B and C, which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and assuming frictionless surfaces and perfectly elastic impact (i.e., conservation of energy), determine the magnitudes of the velocities vA, vB, and vC.

SOLUTION Velocity vectors:

v 0 = v0 (cos 30° i + sin 30° j)

v0 = 15 ft/s

vA = vA j vB = vB (sin 45° i − cos 45° j) vC = vC (cos 45° i + sin 45° j)

Conservation of momentum: mv 0 = mvA + mvB + mvC

Divide by m and resolve into components. i: v0 cos 30° = vB sin 45° + vC cos 45° j: v0 sin 30° = vA − vB cos 45° + vC sin 45° Solving for vB and vC , vB = 0.25882v0 + 0.70711v A vC = 0.96593v0 − 0.70711v A

Conservation of energy:

1 2 1 2 1 2 1 2 mv0 = mv A + mvB + mvC 2 2 2 2

Divide by m and substitute for vB and vC . v02 = v 2A + (0.25882v0 + 0.70711vA ) 2

+ (0.96593v0 − 0.70711vA ) 2 = v02 − v0 vA + 2v A2 vA = 0.5v0 = 7.500 ft/s

vA = 7.50 ft/s 

vB = 0.61237v0 = 9.1856 ft/s

vB = 9.19 ft/s 

vC = 0.61237v0 = 9.1856 ft/s

vC = 9.19 ft/s 

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PROBLEM 14.43 Three spheres, each of mass m, can slide freely on a frictionless, horizontal surface. Spheres A and B are attached to an inextensible, inelastic cord of length l and are at rest in the position shown when sphere B is struck squarely by sphere C, which is moving to the right with a velocity v 0 . Knowing that the cord is slack when sphere B is struck by sphere C and assuming perfectly elastic impact between B and C, determine (a) the velocity of each sphere immediately after the cord becomes taut, (b) the fraction of the initial kinetic energy of the system which is dissipated when the cord becomes taut.

SOLUTION (a)

Determination of velocities. Impact of C and B. Conservation of momentum: mv 0 = mv C + mv1 vC + v1 = v0

(1)

Conservation of energy (perfectly elastic impact): 1 2 1 2 1 2 mv0 = mvC + mv1 2 2 2

Square Eq. (1):

vC2 + v12 = v02

(2)

vC2 + 2vC v1 + v12 = v02

Subtract Eq. (2):

2vC v1 = 0

v1 = 0 corresponds to initial conditions and should be eliminated. Therefore,

From Eq. (1):

vC = 0 

v1 = v0

Cord AB becomes taut: Because cord is inextensible, component of vB along AB must be equal to vA. Conservation of momentum: mv 0 = 2mv A + mv B/A y comp:

0 = 2mv A sin 60° − mvB/A sin 30° vB/A = 2 3vA

x comp:

(3)

mv0 = 2mv A cos 60° + mvB/A cos 30°

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PROBLEM 14.43 (Continued)

Dividing by m and substituting for vB/A from Eq. (3): v0 = 2vA (0.5) + (2 3vA )( 3/2) v0 = 4vA

Carrying into Eq. (3):

vA = 0.250v0

vA = 0.250v0

60° 

vB/A = 2 3(0.250v0 ) = 0.866v0

vB = vA + vB/A

Thus,

= 0.250v0

60° + 0.866v0

30°

vB = (0.250v0 cos 60° + 0.866v0 cos 30°) i + (0.250v0 sin 60° − 0.866v0 sin 30°) j

vB = 0.875v0 i − 0.2165 j vB = 0.90139v0

(b)

vB = 0.901v0

13.90°

13.9° 

Fraction of kinetic energy lost: T0 =

1 2 mv0 2

1 2 1 2 1 2 mv A + mvB + mvC 2 2 2 1 1 1 = m(0.250v0 ) 2 + (0.90139v0 )2 + m(0) 2 2 2 1 2 = m(0.875)v0 2

Tfinal =

Kinetic energy lost = T0 − Tfinal =

1 1 1 m(1 − 0.875)v02 = ⋅ mv02 2 2 8

Fraction of kinetic energy lost =

1  8

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PROBLEM 14.44 In a game of pool, ball A is moving with the velocity v 0 = v0 i when it strikes balls B and C, which are at rest side by side. Assuming frictionless surfaces and perfectly elastic impact (i.e., conservation of energy), determine the final velocity of each ball, assuming that the path of A is (a) perfectly centered and that A strikes B and C simultaneously, (b) not perfectly centered and that A strikes B slightly before it strikes C.

SOLUTION (a)

A strikes B and C simultaneously: During the impact, the contact impulses make 30° angles with the velocity v0.

v B = vB (cos 30°i + sin 30° j)

Thus,

vC = vC (cos 30°i − sin 30° j) vA = vA i

By symmetry, Conservation of momentum:

m v 0 = mv A + mv B + mv C

y component:

0 = 0 + mvB sin 30° − mvC sin 30°

x component:

mv0 = mvA + mvB cos 30° + mvC cos 30°

vC = vB

v0 − vA 2 (v0 − vA ) = cos 30° 3 v −v vB = vC = 0 A 3

vB + vC =

Conservation of energy:

1 2 1 2 1 2 1 2 mv0 = mvA + mvB + mvC 2 2 2 2 2 v02 = v A2 + (v0 − vA ) 2 3 2 v02 − v 2A = (v0 − vA )(v0 + vA ) = (v0 − vA ) 2 3

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PROBLEM 14.44 (Continued)

2 1 5 v0 = − v A (v0 − vA ) 3 3 3 6 2 3 vB = vC = v0 = v0 5 5 3

v0 + vA =

1 vA = − v0 5

vA = 0.200v0

(b)



vB = 0.693v0

30° 

vC = 0.693v0

30° 

A strikes B before it strikes C: First impact: A strikes B. During the impact, the contact impulse makes a 30° angle with the velocity v 0 .

Thus, Conservation of momentum:

v B = vB (cos 30°i + sin 30° j) mv 0 = mvA + mvB

y component:

0 = m(vA′ ) y + mvB sin 30°

x component:

v0 = m(vA′ ) x + mvB cos 30°

(vA′ ) y = −vB sin 30° (vA′ ) x = v0 − vB cos 30°

Conservation of energy: 1 2 1 1 1 mv0 = m(vA′ ) 2x + m(vA′ ) 2y + mvB2 2 2 2 2 1 1 1 = m(v0 − vB cos 30°)2 + (vB sin 30°)2 + vB2 2 2 2 1 2 2 2 2 2 = m v0 − 2v0 vB + vB cos 30° + vB sin 30° + vB2 2

(

)

3 1 v0 , (v′A ) x = v0 sin 2 30° = v0 , 2 4 3 v0 (vA′ ) y = −v0 cos 30° sin 30° = − 4 vB = v0 cos 30° =

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PROBLEM 14.44 (Continued) Second impact: A strikes C. During the impact, the contact impulse makes a 30° angle with the velocity v 0 .

vC = vC (cos 30°i − sin 30° j)

Thus, Conservation of momentum: x component:

y component:

mv′A = mvA + mvC m(v′A ) x = m(vA ) x + mvC cos 30°, 1 (vA ) x = (v′A ) x − vC cos 30° = v0 − vC cos 30° 4 ′ m(v A ) y = m(vA ) y − mvC sin 30° (vA ) y = (vA′ ) y + vC sin 30° = −

3 v0 + vC sin 30° 4

Conservation of energy: 1 1 1 1 1 m(v′A )2x + m(v′A )2y = m(v A ) 2x + m(v A ) 2y + mvC2 2 2 2 2 2 2 2   1  1 2 3 2  1  1 3   m  v0 + v0  = m  v0 − vC cos 30°  +  − v0 + vC sin 30°  + vC2    2 16 16  2  4   4   1 1 2 1 = m  v0 − v0 vC cos 30° + vC2 cos 2 30° 2 16 2  3 3 + v02 − v0 vC sin 30° + vC2 sin 2 30° + vC2  16 2 

1  3 0 = −v0 vC  cos 30° + sin 30°  + 2vC2 2  2   1  3 3 vC = v0  cos 30° + v sin 30°  = 4  4 0 4   1 3 1 (v A ) x = v0 − v0 cos 30° = − v0 4 4 8 3 3 3 v0 + v0 sin 30° = − v0 (v A ) y = − 4 4 8

vA = 0.250v0

60° 

vB = 0.866v0

30° 

vC = 0.433v0

30° 

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PROBLEM 14.45 Two small spheres A and B, of mass 2.5 kg and 1 kg, respectively, are connected by a rigid rod of negligible weight. The two spheres are resting on a horizontal, frictionless surface when A is suddenly given the velocity v 0 = (3.5 m/s)i. Determine (a) the linear momentum of the system and its angular momentum about its mass center G, (b) the velocities of A and B after the rod AB has rotated through 180°.

SOLUTION Position of mass center: y=

(a)



mi yi 2.5(0) + 1(0.2) = = 0.057143 m 2.5 + 1 mi

Linear and angular momentum: L = mA v0 = 2.5 kg(3.5 m/s) i = (8.75 kg ⋅ m/s) i

L = (8.75 kg ⋅ m/s)i   H G = GA × m A v 0 = (0.05714285 m) j × (8.75 kg ⋅ m/s) i = −(0.50000 kg ⋅ m 2 /s) k

H G = −(0.500 kg ⋅ m 2 /s)k 

(b)

Velocities of A and B after 180° rotation Conservation of linear momentum: m Av0 = mA v′A + mB vB′ (2.5)(3.5) = (2.5)v′A + (1.0)vB′ 2.55v′A + vB′ = 8.75

(1)

Conservation of angular momentum about G ′: : rA m Av0 = −rA m Av′A + rB mB vB′ rB = 0.20 − rA = 0.14286 m (0.057143)(2.5)(3.5) = −(0.057143)(2.5)v′A + (0.14286)(1.0)vB′

Dividing by 0.057143: Add Eqs. (1) and (2): From Eq. (1):

−2.5v′A +

0.14286 vB′ = 8.75 0.057143

3.5vB′ = 17.5

vB′ = +5.00 m/s

2.5v′A + (5) = 8.75

v′A = +1.50 m/s

(2)

vA′ = (1.50 m/s)i; vB′ = (5.00 m/s)i 



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PROBLEM 14.46 A 900-lb space vehicle traveling with a velocity v 0 = (1500 ft/s) k passes through the origin O. Explosive charges then separate the vehicle into three parts A, B, and C, with masses of 150 lb, 300 lb, and 450 lb, respectively. Knowing that shortly thereafter the positions of the three parts are, respectively, A(250, 250, 2250), B (600, 1300, 3200), and C (–475, –950, 1900), where the coordinates are expressed in ft, that the velocity of B is vB = (500 ft/s)i + (1100 ft/s) j + (2100 ft/s) k , and that the x component of the velocity of C is −400 ft/s, determine the velocity of part A.

SOLUTION rA = 250i + 250 j + 2250 k

Position vectors (ft):

rB = 600i + 1300 j + 3200 k rC = −475i − 950 j + 1900 k

Since there are no external forces, linear momentum is conserved. (m A + mB + mC ) v 0 = m A v A + mB v B + mC vC

vA =

m A + mB + mC m m v0 − B vB − C vC = 6 v0 − 2 vB − 3vC mA mA mA

(1)

= (6)(1500k ) − (2)(500i + 1100 j + 2100k ) − (3)[ −400i + (vC ) y j + (vC ) z k ] = −3(vC ) y j − 3(vC ) z k + 200i − 2200 j + 4800 k (v A ) x = 200, (vC ) y = −3(vC ) y − 2200, (v A ) z = −3(vC ) z + 4800

Conservation of angular momentum about O: (H O ) 2 = ( H O )1

Since the vehicle passes through the origin, (H O )1 = 0. (H O ) 2 = rA × (m A vA ) + rB × ( mB vB ) + rC × ( mC vC ) = 0

Divide by m A . rA × vA +

m mB rB × vB + C rC × vC = rA × vA + 2rB × vB + 3rC × vC mA mA = rA × (6 v 0 − 2 v B − 3vC ) + 2rB × v B + 3rC × vC

= 3(rC − rA ) × vC + 6rA × v0 + 2(rB − rA ) × vB = (−2175 i − 3600 j − 1050 k ) × vC + (1500 i + 1500 j + 13500 k ) × v0 + (700i + 2100 j + 1900k ) × v B = 0

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PROBLEM 14.46 (Continued)

i j k i j k i j k −2175 −3600 −1050 + 1500 1500 13500 + 700 2100 1900 = 0 (vC ) x (vC ) y (vC ) z 0 0 1500 500 1100 2100

Resolve into components. i:

1050(vC ) y − 3600(vC ) z + 2, 250, 000 + 2,320,000 = 0

(2)

j:

2175(vC ) z − 1050(vC ) x − 2, 250, 000 − 520, 000 = 0

(3)

k:

3600(vC ) x − 2175(vC ) y + 0 − 280, 000 = 0

(4)

Set

(vC ) x = −400 ft/s

From Eq. (4),

(vC ) y = −790.80 ft/s

From Eq. (3),

(vC ) z = 1080.5 ft/s

From Eq. (1),

vA = (6)(1500k ) − (2)(500 i + 1100 j + 2100 k ) − (3)[ −400 i + (790.80) j + (1080.5) k ] 

= 200 i + 172.4 j + 1558.6 k vA = (200 ft/s) i + (172 ft/s)j + (1560 ft/s) k 



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PROBLEM 14.47 Four small disks A, B, C, and D can slide freely on a frictionless horizontal surface. Disks B, C, and D are connected by light rods and are at rest in the position shown when disk B is struck squarely by disk A, which is moving to the right with a velocity v 0 = (38.5 ft/s)i. The weights of the disks are WA = WB = WC = 15 lb, and WD = 30 lb. Knowing that the velocities of the disks immediately after the impact are v A = v B = (8.25 ft/s)i, vC = vC i, and v D = vD i, determine (a) the speeds vC and vD , (b) the fraction of the initial kinetic energy of the system which is dissipated during the collision.

SOLUTION There are no external forces. Momentum is conserved.

vC =

3m A 3(mA + mB ) v0 − vB = (0.5)(38.5) − (8.25) = 11 6mC 6mC

vC = 11.00 ft/s 

3mAv0 = 3(m A + mB )vB + 6mDvD

:

Moments about C vD =

3mAv0 = 6mC vC + 3(mA + mB )vB

:

(a) Moments about D

3mAv0 3(m A + mB ) − vB = (0.25)(38.5) − (0.5)(8.25) = 5.5 ft/s 6mD 6mD

vD = 5.50 ft/s 

(b) Initial kinetic energy: T1 =

1 WA 2 1 15 v0 = (38.5) 2 = 345.24 ft ⋅ lb 2 g 2 32.2

Final kinetic energy: T2 = =

1 WA + WB 2 1 WC 2 1 WD 2 vB + vC + vD 2 g 2 g 2 g 1 30 1 15 1 30 (8.25) 2 + (11.00) 2 + (5.50)2 = 73.98 ft ⋅ lb 2 32.2 2 32.2 2 32.2 345.24 − 73.98 = 271.26 ft ⋅ lb

Energy lost: Fraction of energy lost =

(T1 − T2 ) = 0.786  T1

271.26 = 0.786 345.24

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PROBLEM 14.48 In the scattering experiment of Problem 14.26, it is known that the alpha particle is projected from A0 (300, 0, 300) and that it collides with the oxygen nucleus C at Q(240, 200, 100), where all coordinates are expressed in millimeters. Determine the coordinates of Point B0 where the original path of nucleus B intersects the zx plane. (Hint: Express that the angular momentum of the three particles about Q is conserved.) PROBLEM 14.26 In a scattering experiment, an alpha particle A is projected with the velocity u0 = −(600 m/s)i + (750 m/s) j − (800 m/s)k into a stream of oxygen nuclei moving with a common velocity v0 = (600 m/s) j . After colliding successively with nuclei B and C, particle A is observed to move along the path defined by the Points A 1 (280, 240, 120) and A2 (360, 320, 160), while nuclei B and C are observed to move along paths defined, respectively, by B1 (147, 220, 130), B2 (114, 290, 120), and by C1(240, 232, 90) and C2 (240, 280, 75). All paths are along straight lines and all coordinates are expressed in millimeters. Knowing that the mass of an oxygen nucleus is four times that of an alpha particle, determine the speed of each of the three particles after the collisions.

SOLUTION Conservation of angular momentum about Q:       QA0 × ( mu 0 ) + QB 0 × (4mv0 ) + QC 0 × (4mv0 ) = QA1 × (mvA ) + QB1 × (4mvB ) + QC1 × (4mvC )    QA0 × ( mu 0 ) + QB 0 × (4mv 0) + 0 = 0 + QB1 × (4mvB ) + 0  where QA0 = rA0 − rQ = (300i + 300 k ) − (240 i + 200 j + 100 k )

(1)

= (60 mm)i − (200 mm) j + (200 mm)k  QB0 = (Δx)i + (Δy ) j + (Δz )k  QB1 = rB1 − rQ = (147i + 220 j + 130 k ) − (240i + 200 j + 100 k )

− (93 mm)i + (20 mm)j + (30 mm)k u0 = −(600 m/s)i + (750 m/s)j − (800 m/s)k v 0 = (600 m/s)j

and from the solution to Problem 14.26, vB = vB λ B = (716.98)(−0.42290i + 0.89707 j − 0.12815 k ) = −(303.21 m/s) i + (643.18 m/s) j − (91.88 m/s)k

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PROBLEM 14.48 (Continued)

Calculating each term and dividing by m, i j k  QA0 × u0 60 −200 200 = 10,000 i − 72, 000 j − 75,000 k −600 750 −800  QB0 × (4 v 0 ) = [( Δx)i + (Δy ) j + (Δz ) k ] × (2400 j)

= −2400(Δz )i + 2400( Δx) k  QB1 × (4 vB ) =

i j k 20 30 −93 = −84,532 i − 70,565 j − 215,006 k −1212.84 2572.72 −367.52

Collect terms and resolve into components.

Coordinates:

i:

10, 000 − 2400(Δz ) = −84,532

k:

−75, 000 + 2400( Δx) = −215, 006

Δz = 39.388 mm Δx = −58.336 mm

xB0 = xQ + Δx = 240 − 58.336

xB0 = 181.7 mm  yB0 = 0 

z B0 = zQ + Δz = 100 + 39.388

z B0 = 139.4 mm 

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PROBLEM 14.49 Three identical small spheres, each of weight 2 lb, can slide freely on a horizontal frictionless surface. Spheres B and C are connected by a light rod and are at rest in the position shown when sphere B is struck squarely by sphere A which is moving to the right with a velocity v0 = (8 ft/s)i. Knowing that θ = 45° and that the velocities of spheres A and B immediately after the impact are v A = 0 and v B = (6 ft/s)i + ( v B ) y j, determine (vB ) y and the velocity of C immediately after impact.

SOLUTION Let m be the mass of one ball. Conservation of linear momentum: (Σmv) = (Σmv)0 mv A + mv B + mvC = m( v A )0 + (mv B )0 + (mv C )0

Dividing by m and applying numerical data, 0 + [(6 ft/s)i + (vB ) y j] + [(vC ) x i + (vC ) y j] = (8 ft/s)i + 0 + 0

Components: x: 6 + (vC ) x = 8

(vC ) x = 2 ft/s 

y: (vB ) y + (vC ) y = 0 



(1)

Conservation of angular momentum about O: Σ[r × (mv)] = Σ[r × (mv 0 )]

where rA = 0,

rB = 0,

rC = (1.5 ft)(cos 45°i + sin 45° j)

(1.5)(cos 45°i + sin 45° j) × [m(vC ) x i + m(vC ) y j] = 0

Since their cross product is zero, the two vectors are parallel. (vC ) y = (vC ) x tan 45° = 2 tan 45° = 2 ft/s

From (1), (vB ) y = − 2 ft/s (vB ) y = − 2.00 ft/s 



vC = (2.00 ft/s)i + (2.00 ft/s) j 



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PROBLEM 14.50 Three small spheres A, B, and C, each of mass m, are connected to a small ring D of negligible mass by means of three inextensible, inelastic cords of length l. The spheres can slide freely on a frictionless horizontal surface and are rotating initially at a speed v0 about ring D which is at rest. Suddenly the cord CD breaks. After the other two cords have again become taut, determine (a) the speed of ring D, (b) the relative speed at which spheres A and B rotate about D, (c) the fraction of the original energy of spheres A and B which is dissipated when cords AD and BD again became taut.

SOLUTION Let the system consist of spheres A and B. State 1: Instant cord DC breaks.  3 1  m ( v A )1 = mv0  − i − j  2 2    3 1  i − j  m( v B )1 = mv0  2   2 L1 = m ( v A )1 + m( v B )1 = −mv0 j v =

L1 1 = − v0 j 2m 2

Mass center lies at Point G midway between balls A and B. 3 3 lj × (mv A )1 + − lj × (mv B )1 2 2 3 = lmv0k 2 1 1 T1 = mv02 + mv02 = mv02 2 2

(H G )1 =

State 2: The cord is taut. Conservation of linear momentum: 1 v D = v = − v0 j 2

(a) Let

( v A )2 = v + u A

vD = 0.500v0  and

vB = v + uB

L 2 = 2mv + mu A + mu B = L1 u B = −u A

uB = u A

(H G ) 2 = lmu Ak + lmuBk = 2lmu Ak

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PROBLEM 14.50 (Continued)

(b)

Conservation of angular momentum: (H G ) 2 = (H G )1 2lmu Ak =

3 lmv0k 2

u A = uB =

3 v0 4

u = 0.750v0 

1 1 1 (2m)v 2 + mu 2A + muB2 2 2 2 1 21 9 9  13 2 = mv0  + + = mv0 2  2 16 16  16

T2 =

(c)

Fraction of energy lost:

13 T1 − T2 1 − 16 3 = = T1 1 16

T1 − T2 = 0.1875  T1

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PROBLEM 14.51 In a game of billiards, ball A is given an initial velocity v 0 along the longitudinal axis of the table. It hits ball B and then ball C, which are both at rest. Balls A and C are observed to hit the sides of the table squarely at A′ and C′ , respectively, and ball B is observed to hit the side obliquely at B′. Knowing that v0 = 4 m/s, v A = 1.92 m/s, and a = 1.65 m, determine (a) the velocities v B and vC of balls B and C, (b) the Point C ′ where ball C hits the side of the table. Assume frictionless surfaces and perfectly elastic impacts (that is, conservation of energy).

SOLUTION Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball. ( v A )0 = v0i = 4i,

Before impacts:

v A = −1.92 j,

After impacts:

(v B )0 = ( vC )0 = 0

v B = (vB ) x i + (vB ) y j,

vC = vC i

v 0 = v A + v B + vC

Conservation of linear momentum: i: 4 = 0 + (vB ) x + vC j: 0 = −1.92 + (vB ) y + 0

(vB ) x = 4 − vC (vB ) y = 1.92

1 2 1 2 1 2 1 2 v0 = v A + vB + vC 2 2 2 2

Conservation of energy:

1 2 1 1 1 1 (4) = (1.92) 2 + (1.92)2 + (4 − vC )2 + vC2 2 2 2 2 2 vC2 − 4vC + 3.6864 = 0 vC =

4 ± (4) 2 − (4)(3.6864) = 2 ± 0.56 = 2.56 2

or

1.44

Conservation of angular momentum about B′: 0.75v0 = (1.8 − a)v A + cvC cvC = (0.75)(4) − (1.8 − 1.65)(1.92) = 2.712 c=

2.712 vC

If vC = 1.44,

c = 1.8833

If vC = 2.56,

c = 1.059

off the table. Reject.

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PROBLEM 14.51 (Continued)

(vB ) x = 4 − 2.56 = 1.44,

Then,

v B = 1.44i + 1.92 j

Summary. v B = 2.40 m/s

(a) 

53.1° 

vC = 2.56 m/s





c = 1.059 m 

(b)

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PROBLEM 14.52 For the game of billiards of Problem 14.51, it is now assumed that v0 = 5 m/s, vC = 3.2 m/s, and c = 1.22 m. Determine (a) the velocities v A and v B of balls A and B, (b) the Point A′ where ball A hits the side of the table. PROBLEM 14.51 In a game of billiards, ball A is given an initial velocity v 0 along the longitudinal axis of the table. It hits ball B and then ball C, which are both at rest. Balls A and C are observed to hit the sides of the table squarely at A′ and C′ , respectively, and ball B is observed to hit the side obliquely at B′. Knowing that v0 = 4 m/s, v A = 1.92 m/s, and a = 1.65 m, determine (a) the velocities v B and vC of balls B and C, (b) the Point C′ where ball C hits the side of the table. Assume frictionless surfaces and perfectly elastic impacts (that is, conservation of energy).

SOLUTION Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball. ( v A )0 = v0i = 5i,

Before impacts:

v A = −v A j,

After impacts:

(v B )0 = ( vC )0 = 0

v B = (vB ) x i + (vB ) y j,

vC = 3.2i

v 0 = v A + v B + vC

Conservation of linear momentum: i:

5 = 0 + (vB ) x + 3.2

(vB ) x = 1.8

j:

0 = −v A + (vB ) y + 0

(vB ) y = v A

1 2 1 2 1 2 1 2 v0 = v A + vB + vC 2 2 2 2

Conservation of energy:

1 2 1 1 1 1 (5) = (v A ) 2 + (1.8) 2 + (v A ) 2 + (3.2) 2 2 2 2 2 2

(a)

v A2 = 11.52 (vB ) y = 2.4

v A = 2.4

v A = 2.40 m/s 

v B = 1.8i + 2.4 j

v B = 3.00 m/s

53.1° 

Conservation of angular momentum about B′: 0.75v0 = (1.8 − a)v A + cvC av A = 1.8v A + cvC − 0.75v0 = (1.8)(2.4) + (1.22)(3.2) − (0.75)(5) = 4.474

(b)

a =

4.474 4.474 = vA 2.4

a = 1.864 m 

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PROBLEM 14.53 Two small disks A and B, of mass 3 kg and 1.5 kg, respectively, may slide on a horizontal, frictionless surface. They are connected by a cord, 600 mm, long, and spin counterclockwise about their mass center G at the rate of 10 rad/s. At t = 0, the coordinates of G are x0 = 0, y0 = 2 m, and its velocity is v 0 = (1.2 m/s)i + (0.96 m/s) j . Shortly thereafter, the cord breaks; disk A is then observed to move along a path parallel to the y axis and disk B along a path which intersects the x axis at a distance b = 7.5 m from O. Determine (a) the velocities of A and B after the cord breaks, (b) the distance a from the y-axis to the path of A.

SOLUTION Initial conditions. Location of G: AG BG AG + GB AB 0.6 m = = = = mB m A mB + mA m 4.5 kg

 0.6  AG = 1.5   = 0.2 m  4.5  BG = 0.4 m

Linear momentum: L 0 = m v0 = (4.5 kg)(1.2 i + 0.96 j)

= 5.4i + 4.32 j

Angular momentum: About G:

  (H G )0 = GA × m A v′A + GB × mB v′B

= (0.2 m)(3 kg)(0.2 m × 10 rad/s)k + (0.4 m)(1.5 kg)(0.4 m × 10 rad/s)k (H G )0 = (3.6 kg ⋅ m 2 /s)k

About O: Using formula derived in Problem 14.27, (H O )0 = r × mv0 + ( H G )0 = 2 j × (5.4i + 4.32 j) + 3.6k = −10.8k + 3.6k = −(7.2 kg ⋅ m 2 /s)k

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PROBLEM 14.53 (Continued)

Kinetic energy: Using Eq. (14.29), 1 1 1 1 1 mv02 + Σ m : vi′2 = mv02 + mA v′A2 + mB vB′2 2 2i 2 2 2 1 1 1 = (4.5)[(1.2)2 + (0.96)2 ] + (3)(0.2 × 10) 2 + (1.5)(0.4 × 10) 2 2 2 2 = (5.3136 + 6 + 12) = 23.314 J

T0 =

(a)

Conservation of linear momentum: L0 = L

5.4i + 4.32 j = m A v A + mB v B = 3(v A j) + 1.5[(vB ) z i + (vB ) y j]

Equating coefficients of i:

5.4 = 1.5(vB ) x (vB ) x = 3.6 m/s

(1)

Equating coefficients of j: 4.32 = 3v A + 1.5(vB ) y (vB ) y = 2.88 − 2vA

(2)

Conservation of energy: T0 = T : T0 = 23.314 J =

1 1 m Av A2 + mB vB2 2 2

1 1 (3) ( v 2A ) + (1.5)[(vB ) 2x + (vB ) 2y ] 2 2

Substituting from Eqs. (1) and (2): 23.314 = 1.5v 2A + 0.75(3.6) 2 + 0.75(2.88 − 2v A )2 4.5v 2A − 8.64vA − 7.373 = 0 v 2A − 1.92vA − 1.6389 = 0 vA = 0.96 + 1.60 = 2.56 m/s

vA = 2.56 m/s 

and vA = 0.96 − 1.60 = −0.64 m/s (rejected, since vA is shown directed up) From Eqs. (1) and (2):

(vB ) x = 3.6 m/s (vB ) y = 2.88 − 2(2.56) = −2.24 m/s vB = 3.6 i − 2.24 j

(b)

vB = 4.24 m/s

31.9° 

Conservation of angular momentum about O: (H O )0 = H O : − 7.2k = ai × mA vA + bi × mB vB − 7.2k = ai × 3(2.56 j) + 7.5i × 1.5(3.6i − 2.24 j) −7.2k = 7.68ak − 25.2k

a = 2.34 m 

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PROBLEM 14.54 Two small disks A and B, of mass 2 kg and 1 kg, respectively, may slide on a horizontal and frictionless surface. They are connected by a cord of negligible mass and spin about their mass center G. At t = 0, G is moving with the velocity v 0 and its coordinates are x0 = 0, y0 = 1.89 m. Shortly thereafter, the cord breaks and disk A is observed to move with a velocity v A = (5 m/s) j in a straight line and at a distance a = 2.56 m from the y-axis, while B moves with a velocity v B = (7.2 m/s)i − (4.6 m/s) j along a path intersecting the x-axis at a distance b = 7.48 m from the origin O. Determine (a) the initial velocity v 0 of the mass center G of the two disks, (b) the length of the cord initially connecting the two disks, (c) the rate in rad/s at which the disks were spinning about G.

SOLUTION Initial conditions. Location of G: AG BG AG + GB l = = = mB m A mB + m A m m 1 AG = B l = l 3 m mA 2 BG = l= l 3 m

Linear momentum: L 0 = mv 0 = 3 v 0

Angular momentum about G:

  (H G )0 = GA × m A v′A + GB × mB v ′B

1  1  2  2  =  l  (2 kg)  lω  k +  l  (1 kg)  lω  k 3  3  3  3  2 = l 2ω k 2 3

Kinetic energy: Using Eq. (14.29), T0 =

1 1 1 1 1 mv02 + Σ mi vi2 = mv02 + m Av′A2 + mB vB′2 2 2i 2 2 2 2

1 1 1  1 2  (3)v02 + (2)  lω  + (1)  lω  2 2 3  2 3  3 1 T0 = v02 + l 2ω 2 2 3

2

=

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PROBLEM 14.54 (Continued)

Conservation of linear momentum: mv 0 = m A v A + mB v B 3v 0 = (2)(5 j) + (1)(7.2i − 4.6 j) = 7.2i + 5.4 j v 0 = (2.4 m/s)i + (1.8 m/s) j 

(a) Conservation of angular momentum about O:

: (1.89 j) × mv 0 + ( H G )0 = (2.56i ) × m A vA + (7.48 i ) × mB vA Substituting for v 0 , (H G )0 , vA , vB and masses: 2 (1.89 j) × 3(2.4i + 1.8 j) + l 2ω k = (2.56i ) × 2(5 j) + (7.48i ) × (7.2i − 4.6 j) 3 2 −13.608k + l 2ω k = 25.6k − 34.408k 3 2 2 l ω = 4.80 3

l 2ω = 7.20

(1)

Conservation of energy: T0 = T :

3 2 1 2 2 1 1 v0 + l ω = m Av A2 + mB vB2 2 3 2 2

3 1 1 1 [(2.4)2 + (1.8) 2 ] + l 2ω 2 = (2)(5) 2 + (1)[(7.2) 2 + (4.6)2 ] 2 3 2 2 1 13.5 + l 2ω 2 = 25 + 36.5 3

l 2ω 2 = 144.0

(2)

Dividing Eq. (2) by Eq. (1), member by member:

ω=

144.0 = 20.0 rad/s 7.20

Original rate of spin = 20.0 rad/s 

(c) Substituting for ω into Eq. (1): l 2 (20.0) = 7.20

l 2 = 0.360

l = 0.600 m

Length of cord = 600 mm 

(b)

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PROBLEM 14.55 Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to three 9-in-long strings, which are tied to a ring G. Initially, the spheres rotate clockwise about the ring with a relative velocity of 2.6 ft/s and the ring moves along the x-axis with a velocity v0 = (1.3 ft/s)i. Suddenly, the ring breaks and the three spheres move freely in the xy plane with A and B, following paths parallel to the y-axis at a distance a = 1.0 ft from each other and C following a path parallel to the x-axis. Determine (a) the velocity of each sphere, (b) the distance d.

SOLUTION Conservation of linear momentum: L 0 = (3 m) v = 3m (1.3i) = m (3.9 ft/s)i

Before break:

L = mv A j − mvB j + mvC i

After break: L = L0 :

mvC i + m(v A − vB ) j = m(3.9 ft/s)i

Therefore,

vA = vB

(1)

vC = 3.9000 ft/s v C = 3.90 ft/s

(2)

Conservation of angular momentum: Before break:

( H O )0 = 3mlv′ = 3m (0.75ft)(2.6 ft/s) = 5.85m

After break:

H O = −mv A x A + mv A ( x A + 0.346) + mvC d H O = −m A x A + mv A ( x A + 1.0) + mvC d

H O = ( H O )0 :

1.0mvA + mvC d = 5.85m

Recalling Eq. (2): vA + 3.9d = 5.85 d = 1.5 − 0.25641vA

(3)

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PROBLEM 14.55 (Continued)

Conservation of energy. Before break: 1 1  (3 m)v 2 + 3  mv′2  2 2   3 3 = m v02 + v′2 = [(1.3)2 + (2.6)2 ]m = 12.675m 2 2

T0 =

(

)

After break: T=

1 2 1 2 1 2 mv A + mvB + mvC 2 2 2

T = T0 : Substituting for vB from Eq. (1) and vC from Eq. (2), 1 2 v A + v A2 + (3.900) 2  = 12.675  2 v A2 = 5.0700 v A = vB = 2.2517 ft/s

(a)

Velocities: vA = 2.25 ft/s ; v B = 2.25 ft/s ; vC = 3.9 ft/s

(b)



Distance d: From Eq. (3):

d = 1.5 − 0.25641(2.2517) = 0.92265 ft

d = 11.1 in. 

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PROBLEM 14.56 Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to three strings of length l which are tied to a ring G. Initially, the spheres rotate clockwise about the ring which moves along the x axis with a velocity v0. Suddenly the ring breaks and the three spheres move freely in the xy plane. Knowing that vA = (3.5 ft/s) j, vC = (6.0 ft/s)i, a = 16 in. and d = 9 in., determine (a) the initial velocity of the ring, (b) the length l of the strings, (c) the rate in rad/s at which the spheres were rotating about G.

SOLUTION Conservation of linear momentum: (3m) v = mvA + mvB + mvC 3mv0 i = m(3.5 ft/s) j − mvB j + m(6.0 ft/s)i

Equating coefficients of unit vectors: 3v0 = 6.00 ft/s 0 = 3.5 ft/s − vB

vB = 3.5 ft/s

(1) v0 = 2.00 ft/s

(a)



Conservation of angular momentum: Before break: After break:

( H O )0 = 3ml 2θ H O = −mv A xA + mv A ( x A + 16/12) + mvC (9/12) = m(3.5)(16/12) + m(6.0)(9/12) = m(9.1667)

( H O )0 = H 0 :

3ml 2θ = m(9.1667)

l 2θ = 3.0556

(2)

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PROBLEM 14.56 (Continued)

Conservation of energy: Before break: 1 3 1  3 (3m)v 2 + 3  mv′2  = mv02 + m(lθ) 2 2 2 2  2 3 3 = m(2.0) 2 + ml 2θ 2 2 2

T0 =

After break: 1 2 1 2 1 2 mv A + mvB + mvC 2 2 2 1 1 = m [(3.5)2 + (6.0)2 ] = m (60.5) 2 2

T=

T = T0 :

1 3 3 m(60.5) = m(2.0) 2 + ml 2θ 2 2 2 2 l 2θ 2 = 16.167

(3)

16.167 = 5.2909 Dividing Eq. (3) by Eq. (2): θ = 3.0556

(b)

From Eq. (2):

l2 =

3.0556 5.2909

l = 0.75994 ft l = 0.76 ft 

(c)

θ = 5.29 rad/s

Rate of rotation:



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PROBLEM 14.57 A stream of water of cross-sectional area A1 and velocity v1 strikes a circular plate which is held motionless by a force P. A hole in the circular plate of area A2 results in a discharge jet having a velocity v1. Determine the magnitude of P.

SOLUTION Mass flow rates. As the fluid ahead of the plate moves from section 1 to section 2 Δt , the mass Δm1 moved is Δm1 = ρ A1 ( Δt ) = ρ A1v1 (Δt )

so that

dm1 Δm1 = = ρ A1v1 dt Δt

Likewise, for the fluid that has passed through the hole Δm2 = ρ A2 (Δl ) = ρ A2 v1 (Δt )

so that

dm2 = ρ A2 v1 dt

Apply the impulse-momentum principle.



Σmv1 + F dt = Σmv 2

Components in the direction of the flow. (Δm1 )v1 − PΔt = (Δm2 )v1 P=

Δm1 Δm2 v1 − v1 = ρ A1v12 − ρ A2 v12 Δt Δt P = ρ ( A1 − A2 )v12 

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PROBLEM 14.58 A jet ski is placed in a channel and is tethered so that it is stationary. Water enters the jet ski with velocity v1 and exits with velocity v2. Knowing the inlet area is A1 and the exit area is A2, determine the tension in the tether.

SOLUTION Mass flow rates. Consider a cylindrical portion of the fluid lying in a section of pipe of cross sectional area A and length Δl. The volume and mass are Δm = ρ A(Δl ) Δm Δl = ρ A = ρ Av Δt Δt

Then At the pipe inlet and outlet, we get

Δm1 = ρ A1v1 , Δt

Δm2 = ρ A2 v2 Δt

Impulse and momentum principle:

Σmv1 + Imp1→2 = Σmv 2

Using horizontal components (+ →), (Δm1 )v1 cos θ + P (Δt ) = ( Δm2 )v2

Δm1 Δm1 v2 − v1 cos θ Δt Δt = ρ A2 v22 − ρ A1v12 cos θ

P=

P = ρ A2 v22 − ρ A1v12 cos θ 

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PROBLEM 14.59 A stream of water of cross-sectional area A and velocity v1 strikes a plate which is held motionless by a force P. Determine the magnitude of P, knowing that A = 0.75 in 2 , v1 = 80 ft/s, and V = 0.

SOLUTION Mass flow rate. As the fluid moves from section 1 to section 2 in time Δt, the mass Δm moved is Δm = ρ A(Δl ) dm Δm ρ A(Δl ) = = = ρ Av1 dt Δt Δt

Then Data:

γ = 62.4 lb/ft 3 ,

A = 0.75 in.2 = 0.0052083 ft 2 ,

v1 = 80 ft/s

dm (62.4) = (0.0052083)(80) = 0.80745 slug/s dt 32.2

Principle of impulse and momentum: :

(Δm)v1 − PΔt = 0 P=

Δm dm v= v Δt dt

P = (0.80745)(80) = 64.596 lb

P = 64.6 lb 

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PROBLEM 14.60 A stream of water of cross-sectional area A and velocity v1 strikes a plate which moves to the right with a velocity V. Determine the magnitude of V, knowing that A = 1 in2, v1 = 100 ft/s, and P = 90 lb.

SOLUTION

Consider velocities measured with respect to the plate, which is moving with velocity V. The velocity of the stream relative to the plate is u = v1 − V

(1)

Mass flow rate. As the fluid moves from section 1 to section 2 in time Δt, the mass Δm moved is Δm = ρ A(Δl ) dm Δm ρ A(Δl ) = = = ρ Au dt Δt Δt

Then

(2)

Principle of impulse and momentum: (Δm)u − P(Δt ) = 0 P=

Δm dm u= u = ρ Au 2 Δt dt u=

P ρA

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PROBLEM 14.60 (Continued)

From Eq. (1),

V = v1 − u = v1 −

Data:

P = 90 lb,

P

ρA

A = 1 in 2 = 0.0069444 ft 2

V1 = 100 ft/s, γ = 62.4 lb/ft 3 v = 100 −

90 (62.4/32.2)(0.0069444)

V = 18.2 ft/s 

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PROBLEM 14.61 A rotary power plow is used to remove snow from a level section of railroad track. The plow car is placed ahead of an engine which propels it at a constant speed of 20 km/h. The plow car clears 160 Mg of snow per minute, projecting it in the direction shown with a velocity of 12 m/s relative to the plow car. Neglecting friction, determine (a) the force exerted by the engine on the plow car, (b) the lateral force exerted by the track on the plow.

SOLUTION Velocity of the plow:

vP = 20 km/h = 5.5556 m/s

Velocity of thrown snow: v s = (12 m/s)(cos 30°i + sin 30° j) + (5.5556 m/s)k

Mass flow rate: dm (160000 kg/min) = = 2666.7 kg/s dt (60 s/min)

Let F be the force exerted on the plow and the snow. Apply impulse-momentum, noting that the snow is initially at rest and that the velocity of the plow is constant. Neglect gravity. F (Δt ) = ( Δm) v s

 dm  F=  vs = (2.666.7)(12 cos 30°i + 12sin 30° j + 5.5556k )  dt  = (27713 N)i + (16000 N) j + (14815 N)k (a )

Force exerted by engine.

Fz = 14.8 kN 

(b)

Lateral force exerted by track.

Fx = 27.7 kN 

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PROBLEM 14.62 Tree limbs and branches are being fed at A at the rate of 5 kg/s into a shredder which spews the resulting wood chips at C with a velocity of 20 m/s. Determine the horizontal component of the force exerted by the shredder on the truck hitch at D.

SOLUTION Eq. (14.38): (Δ m) vA + ΣFΔ t = (Δ m) vC ΣF =

Force exerted on chips = Σ F = 100 N

Δm vC = (5 kg/s)(20 m/s Δt

25°)

25°

Free body: shredder: ΣFx = 0: Fx − (100 N) cos 25° = 0 Fx = 90.6 N Fx = 90.6 N

On hitch:



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PROBLEM 14.63 Sand falls from three hoppers onto a conveyor belt at a rate of 90 lb/s for each hopper. The sand hits the belt with a vertical velocity v1 = 10 ft/s and is discharged at A with a horizontal velocity v2 = 13 ft/s. Knowing that the combined mass of the beam, belt system, and the sand it supports is 1300 lb with a mass center at G, determine the reaction at E.

SOLUTION Principle of impulse and momentum:

Moments about F: (Δm)v1(3a) + Δm v1 (2a) + (Δm) v1 a + (W Δt )c − ( RΔt ) L = 3(Δm)v2h R=

Data: L = 20 ft, R=

1 L

c = 13 ft,

Δm Δm   cW + 6av1 Δt − 3hv2 Δt     a = 5 ft,

h = 2.5 ft,

Δm ΔW /g 1 dW = = = 2.7950 slug/s Δt dt g dt

1 [13(1300) + (6)(5)(10)(2.7950) − (3)(2.5)(13)(2.7950)] 20 R = 873 lb 

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PROBLEM 14.64 The stream of water shown flows at a rate of 550 liters/min and moves with a velocity of magnitude 18 m/s at both A and B. The vane is supported by a pin and bracket at C and by a load cell at D which can exert only a horizontal force. Neglecting the weight of the vane, determine the components of the reactions at C and D.

SOLUTION dm (1000 kg/m3 )(550 liters/min)(1 min) = ρQ = dt (1000 liters/m3 )(60 sec) dm = 9.1667 kg/s dt

Mass flow rate:

vA = 18 m/s

Velocity vectors:

v B = 18 m/s

40°

Apply the impulse-momentum principle.

Moments about C : −0.040(Δm)v A + 0.150 D( Δt ) = 0.200( Δm)vB cos 40° + 0.165(Δm)vB sin 40°

1  Δm  [0.200vB cos 40° + 0.165vB sin 40° + 0.040v A ] 0.150  Δt  1 = (9.1667)[(0.200)(18) cos 40° + 0.165(18)sin 40° + 0.040(18)] 0.150 = 329.20 N Dx = 329 N 

D=

Dy = 0  x components:

C x (Δt ) + D (Δt ) = (Δm)vB cos 40°

 Δm  Cx =   vB cos 40° − D = (9.1667)(18cos 40°) − 329.20 = −202.79 N  Δt  y components:

C x = −203 N 

−(Δm)v A + C y (Δt ) = ( Δm)vB sin 40° Δm  Δm  Cy =  vA + vB sin 40° = (9.1667)(18 + 18sin 40°) = 271.06 N  Δt  Δt 

C y = 271 N 

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PROBLEM 14.65 The nozzle discharges water at the rate of 340 gal/min. Knowing the velocity of the water at both A and B has a magnitude of 65 ft/s and neglecting the weight of the vane, determine the components of the reactions at C and D. (1 ft3 = 7.48 gallons)

SOLUTION Volumetric flow rate:

Q = 340 gal/min × (1 ft 3 /7.48 gal) × (1 min/60 sec) = 0.75758 ft 3 /s

Mass density of water:

γ g

=

62.4 lb/ft 3 32.2 ft/s 2

62.4 dm γ = Q= (0.75758) = 1.4681 lb ⋅ s/ft 32.2 dt g

Mass flow rate:

Assume that the flow speed remains constant. Principle of impulse and momentum.

Moments about D: ((30/12) sin 50° − (23/12) cos 50° + (3/12) Δm V (30/12) Δt dm = 0.37324 V dt = (0.37324)(1.4681 lb ⋅ s/ft)(65 ft/s) = 35.617 lb

C=

C = 35.6 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 934

PROBLEM 14.65 (Continued)

Horizontal components: (Δm)V cos 50° + Dx (Δt ) = (Δm)V Dx = (1 − cos 50°)

Δm V Δt

dm V dt = (0.35721)(1.4681 lb ⋅ s/ft)(65 ft/s) = 34.087 lb = 0.35721

Vertical components:

Dx = 34.1 N



−(Δm)V sin 50° + C ( Δt ) + D y (Δt ) = 0 Δm V −C Δt dm = 0.76604 V −C dt = (0.76604)(1.4681 lb ⋅ s/ft)(65 ft/s) − 35.617 N = 37.484 lb

Dy = (sin 50°)

Dy = 37.5 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 935

PROBLEM 14.66 A high speed jet of air issues from the nozzle A with a velocity of v A and mass flow rate of 0.36 kg/s. The air impinges on a vane causing it to rotate to the position shown. The vane has a mass of 6-kg. Knowing that the magnitude of the air velocity is equal at A and B determine (a) the magnitude of the velocity at A, (b) the components of the reactions at O.

SOLUTION Assume that the speed of the air jet is the same at A and B. v A = vB = v

Apply the principle of impulse and momentum.

(a)

Moments about O : (0.190)(Δm)v − (0.250)W ( Δt ) = −(0.250)(Δm)v cos 50° − (0.500)( Δm)v sin 50° v=

0.250W Δt ⋅ Δm 0.150 cos50° + 0.500 sin 50° + 0.190

0.250W dm 0.66944 dt (0.250)(6)(9.81) = (0.66944)(0.36) = 61.058 m/s =

vA = 61.1 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 936

PROBLEM 14.66 (Continued)

(b)

x components: (Δm)v + Rx (Δt ) = −(Δm)v sin 50° Δm v (1 + sin 50°) Δt = −(0.36)(61.058)(1 + sin 50°) = −38.82 N

Rx = −

y components: 0 + Ry (Δt ) − W (Δt ) = −(Δm)v cos 50° Δm v cos 50° Δt = (6)(9.81) − (0.36)(61.058) cos 50° = 44.73 N

Ry = W +

R = (38.82)2 + (44.73)2 = 59.2 N tan α =

44.73 38.82

α = 49.0°

R = 59.2 N

49.0° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 937

PROBLEM 14.67 Coal is being discharged from a first conveyor belt at the rate of 120 kg/s. It is received at A by a second belt which discharges it again at B. Knowing that v1 = 3 m/s and v 2 = 4.25 m/s and that the second belt assembly and the coal it supports have a total mass of 472 kg, determine the components of the reactions at C and D.

SOLUTION Velocity before impact at A: (v A ) x = v1 = 3 m/s

(v A )2y = 2 g (Δy ) = (2)(9.81)(0.545) = 10.693 m 2 /s 2 tan θ =

Slope of belt:

2.4 − 1.2 , 2.25

(vA ) y = 3.270 m/s

θ = 28.07°

v 2 = 4.25 (cos θ i + sin θ j)

Velocity of coal leaving at B: Apply the impulse-momentum principle.

(Δm)(v A ) x + C x (Δt ) = (Δm)v2 cos θ

x components:

Cx =

Δm [v2 cos θ − (v A ) x ] = (120)(4.25cos 28.07° − 3) Δt C x = 90.0 N

 moments about C:



(Δm)[−1.2(v A ) x − 0.75(v A ) y ] + 3.00 D (Δt ) − 1.8W (Δt ) = (Δm)[ −2.4 v2 cos θ + 3v2 sin θ ]

Δm [−(1.2)(3) − (0.75)(3.270)] + 3D − (1.8)(472)(9.81) Δt =

Δm [−(2.4)(4.25cos θ ) + (3)(4.25sin θ )] Δt

D = 2775 + 1.0168

dm = 2775 + (1.0168)(120) = 2897 N dt

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 938

PROBLEM 14.67 (Continued)

y components:

(Δm)( −v A ) y + (C y + D − W )(Δt ) = (Δm)v2 sin θ Cy + D − W =

Δm (3.270 + 4.25sin θ ) Δt

= (120)(5.268) = 632.2 N C y = 4625.6 − 2897 + 632.2

C y = 2361 N  C x = 90.0 N, C y = 2360 N  Dx = 0, 



Dy = 2900 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 939

PROBLEM 14.68 A mass q of sand is discharged per unit time from a conveyor belt moving with a velocity v0. The sand is deflected by a plate at A so that it falls in a vertical stream. After falling a distance h the sand is again deflected by a curved plate at B. Neglecting the friction between the sand and the plates, determine the force required to hold in the position shown (a) plate A, (b) plate B.

SOLUTION (a)

When the sand impacts on plate A, it is momentarily brought to rest. Apply the principle of impulse and momentum to find the force on the sand.

(Δm)v0 + Ax (Δt ) = 0

x component:

Ax = −

Δm v0 = −qv0 Δt

0 + Ay (Δt ) = 0

y component:

Ay = 0

A = qv0



The sand falls vertically. Use conservation of energy for mass element Δm. Let v be the speed at the curved portion of plate B. T1 + V1 = T2 + V2 : 0 + (Δm) gh =

1 (Δm)v 2 + 0 2

v 2 = 2 gh v = 2 gh

Over the curved portion of plate B, there is negligible change of elevation. Hence, by conservation of energy, v is both the entrance speed and exit speed of the curved portion of plate B. (b)

Force exerted through plate B: Entrance velocity:

v = − 2gh j

Exit velocity:

v′ = 2 gh (− cos 30°i − sin 30° j)

Mass flow rate:

dm Δm = =q dt Δt

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 940

PROBLEM 14.68 (Continued)

Principle of impulse and momentum:

(Δm) v + B(Δt ) = (Δm) v′

 Δm  B=  ( v′ − v) = q 2 gh (− cos 30°i − sin 30° j + j)  Δt  3 2 gh 2 1 2 gh B y = 2 gh (1 − sin 30°) = 2 Bx = − 2 gh cos 30° = −

B = 2gh

30° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 941

PROBLEM 14.69 The total drag due to air friction on a jet airplane traveling at 900 km/h is 35 kN. Knowing that the exhaust velocity is 600 m/s relative to the airplane, determine the mass of air which must pass through the engine per second to maintain the speed of 900 km/h in level flight.

SOLUTION dm = mass flow rate dt u = exhaust relative to the airplane v = speed of airplane D = drag force

Symbols:

Principle of impulse and momentum:

(Δm)v + D(Δt ) = (Δm)u Δm dm D = = Δt dt u−v

Data: v = 900 km/h = 250 m/s u = 600 m/s D = 35 kN = 35000 N dm 35000 = dt 600 − 250

dm = 100 kg/s  dt

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 942

PROBLEM 14.70 While cruising in level flight at a speed of 600 mi/h, a jet plane scoops in air at the rate of 200 lb/s and discharges it with a velocity of 2100 ft/s relative to the airplane. Determine the total drag due to air friction on the airplane.

SOLUTION Flight speed: v = 600 mi/h = 880 ft/s Mass flow rate:

dm 200 lb/s = = 6.2112 slug/s dt 32.2 ft/s 2 ΣF = D =

dm (u − v) dt

or

D=

dm (v − u ) dt

where, for a frame of reference moving with the plane, v is the free stream velocity (equal to the air speed) and u is the relative exhaust velocity. D = (6.2112)(2100 − 880) = 7577.6 lb

D = 7580 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 943

PROBLEM 14.71 In order to shorten the distance required for landing, a jet airplane is equipped with movable vanes, which partially reverse the direction of the air discharged by each of its engines. Each engine scoops in the air at a rate of 120 kg/s and discharges it with a velocity of 600 m/s relative to the engine. At an instant when the speed of the airplane is 270 km/h, determine the reverse thrust provided by each of the engines.

SOLUTION Apply the impulse-momentum principle to the moving air. Use a frame of reference that is moving with the airplane. Let F be the force on the air. v = 270 km/h = 75 m/s u = 600 m/s

−(Δm)v + F(Δt ) = 2

(Δm) u sin 20° 2

dm Δm (v + u sin 20°) = (v + u sin 20°) dt Δt F = (120)(75 + 600sin 20°) = 33.6 × 103 N

F=

Force on airplane is − F.

F = 33.6 kN



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 944

PROBLEM 14.72 The helicopter shown can produce a maximum downward air speed of 80 ft/s in a 30-ft-diameter slipstream. Knowing that the weight of the helicopter and its crew is 3500 lb and assuming γ = 0.076 lb/ft 3 for air, determine the maximum load that the helicopter can lift while hovering in midair.

SOLUTION F=

The thrust is Calculation of

dm . dt

dm (v B − v A ) dt

mass = density × volume = density × area × length Δm = ρ AB ( Δl ) = ρ AB vB (Δt ) Δm γ dm = ρ AB vB = AB vB = Δt g dt

where AB is the area of the slipstream well below the helicopter and vB is the corresponding velocity in the slipstream. Well above the blade, v A ≈ 0. F=

Hence,

γ g

AB vB2

 0.076 lb/ft 3   π  2 2 =     (30 ft) (80 ft/s) 2  4 32.2 ft/s    = 10,678 lb F = 10,678 lb

The force on the helicopter is 10,678 lb . Weight of helicopter:

WH = 3500 lb

Weight of payload:

WP = WP

Statics:

ΣFy = F − WH − WP = 0 WP = F − WH = 10, 678 − 3500 = 7178 lb

W = 7180 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 945

PROBLEM 14.73 A floor fan designed to deliver air at a maximum velocity of 6 m/s in a 400-mm-diameter slipstream is supported by a 200-mm-diameter circular base plate. Knowing that the total weight of the assembly is 60 N and that its center of gravity is located directly above the center of the base plate, determine the maximum height h at which the fan may be operated if it is not to tip over. Assume ρ = 1.21 kg/m3 for air and neglect the approach velocity of the air.

SOLUTION Calculation of

dm dt

at a section in the airstream: mass = density × volume = density × area × length Δm = ρ A(Δl ) = ρ Av( Δt ) Δm dm = = ρ Av Δt dt

Thrust on the airstream: F=

dm ( vB − vA ) dt

where vB is the velocity just downstream of the fan and vA is the velocity for upstream. Assume that vA is negligible. π  F = ( ρ Av)v = ρ  D 2  v 2 4   π   F = (1.21 kg/m3 )   (0.400 m) 2 (6 m/s) 2 = 5.474 N 4 F = 5.474 N

Force on fan:

F ′ = −F = 5.474 N

Maximum height h:

1    e = 2 d = 100 mm = 0.1 m    ΣM E = 0 F ′h − We = 0 We (60)(0.1) h= = F′ 5.474

h = 1.096 m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 946

PROBLEM 14.74 The jet engine shown scoops in air at A at a rate of 200 lb/s and discharges it at B with a velocity of 2000 ft/s relative to the airplane. Determine the magnitude and line of action of the propulsive thrust developed by the engine when the speed of the airplane is (a) 300 mi/h, (b) 600 mi/h.

SOLUTION Use a frame of reference moving with the plane. Apply the impulse-momentum principle. Let F be the force that the plane exerts on the air.

x components:

(Δm)u A + F (Δt ) = (Δm) uB F=

moments about B:

Δm dm (uB − u A ) = (u B − u A ) Δt dt

(1)

−e(Δm)u A + M B ( Δt ) = 0 MB = e

dm uA dt

(2)

Let d be the distance that the line of action is below B. d=

Fd = M B

MB eu A = F uB − u A

(3)

dm 200 = 200 lb/s = = 6.2112 slugs/s, uB = 2000 ft/s, e = 12 ft dt 32.2

Data:

u A = 300 mi/h = 440 ft/s

(a) From Eq. (1),

F = (6.2112)(2000 − 440)

From Eq. (3),

d=

(12)(440) 2000 − 440

F = 9690 lb  d = 3.38 ft 

u A = 600 mi/h = 880 ft/s

(b) From Eq. (1),

F = (6.2112)(2000 − 880)

From Eq. (3),

d=

(12)(880) 2000 − 880

F = 6960 lb  d = 9.43 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 947

PROBLEM 14.75 A jet airliner is cruising at a speed of 900 km/h with each of its three engines discharging air with a velocity of 800 m/s relative to the plane. Determine the speed of the airliner after it has lost the use of (a) one of its engines, (b) two of its engines. Assume that the drag due to air friction is proportional to the square of the speed and that the remaining engines keep operating at the same rate.

SOLUTION Let v be the airliner speed and u be the discharge relative velocity. u = 800 m/s. dm (u − v) dt

Thrust formula for one engine:

F=

Drag formula:

D = kv 2

Three engines working.

Cruising speed = v0 = 900 km/h = 250 m/s 3F − D = 3

dm (u − v0 ) − k v02 = 0 dt

kv02 dm k (250) 2 = = = 37.879k dt 3(u − v0 ) 3(800 − 250)

(a)

One engine fails. Two engines working. Cruising speed = v1 2F − D = 2

dm (u − v1 ) − kv12 = 0 dt

(2)(37.879k )(800 − v1 ) − k v12 = 0 v12 + 75.758v1 − 60.606 × 103 = 0 v1 = 211.20 m/s

(b)

v1 = 760 km/h 

Two engines fail. One engine working. Cruising speed = v2 F−D=

dm (u − v2 ) − k v22 = 0 dt

(37.879k )(800 − v2 ) − k v22 = 0 v22 + 37.879v2 − 30.303 × 103 = 0 v2 = 156.17 m/s

v2 = 562 km/h 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 948

PROBLEM 14.76 A 16-Mg jet airplane maintains a constant speed of 774 km/h while climbing at an angle α = 18°. The airplane scoops in air at a rate of 300 kg/s and discharges it with a velocity of 665 m/s relative to the airplane. If the pilot changes to a horizontal flight while maintaining the same engine setting, determine (a) the initial acceleration of the plane, (b) the maximum horizontal speed that will be attained. Assume that the drag due to air friction is proportional to the square of the speed.

SOLUTION Calculate the propulsive force using velocities relative to the airplane. F=

dm (v B − v A ) dt

dm = 300 kg/s dt v A = 774 km/h

Data:

= 215 m/s vB = 665 m/s F = (300)(665 − 215) = 135, 000 N

Since there is no acceleration while the airplane is climbing, the forces are in equilibrium. +

18°ΣF = 0:

F − D − mg sin α = 0

F − D = mg sin α = (16, 000)(9.8)sin18° = 48, 454 N

(a)

Initial acceleration of airplane in horizontal flight: ma = F − D : 16,000a = 48.454 × 103



Corresponding drag force:

a = 3.03 m/s 2

18°

D = 135, 000 − 48, 454 = 86,546 N



Drag force factor:

D = k v A2

or

k= =

D v A2 86,546 (215) 2

= 1.87228 N ⋅ s 2 /m 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 949

PROBLEM 14.76 (Continued)

(b)

Maximum speed in horizontal flight: Since the acceleration is zero, the forces are in equilibrium. F −D=0 dm dm dm vA − vB = 0 (vB − v A ) − k v 2A = 0 k v A2 + dt dt dt 1.87228v A2 + 300v A − (300)(665) = 0 v A = 256.0 m/s

v A = 922 km/h 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 950

PROBLEM 14.77 The propeller of a small airplane has a 2-m-diameter slipstream and produces a thrust of 3600 N when the airplane is at rest on the ground. Assuming ρ = 1.225 kg/m3 for air, determine (a) the speed of the air in the slipstream, (b) the volume of air passing through the propeller per second, (c) the kinetic energy imparted per second to the air in the slipstream.

SOLUTION Calculation of

dm dt

at a section in the airstream: mass = density × volume = density × area × length Δm = ρ A( Δl ) = ρ Av Δt Δm dm = = ρ Av Δt dt

(a)

Thrust =

dm dt

( vB − vA ) where vB is the velocity just downstream of propeller and vA is the velocity far

upstream. Assume vA is negligible. π  Thrust = ( ρ Av)v = ρ  D 2  v 2 4   π  2 2 3600 = 1.225   (2) v 4 v 2 = 935.44 v = 30.585 m/s

(b) (c)

Q=

1 dm π π  = Av =  D 2  v = (2) 2 (30.585) = 96.086 4 ρ dt 4 

v = 30.6 m/s  Q = 96.1 m3 /s 

Kinetic energy of mass Δm : 1 1 1 (Δm)v 2 = ρ A( Δl )v 2 = ρ Av(Δt )v3 2 2 2 ΔT dT = dt Δt 1 = ρ Av3 2 1 π 2 3 = ρ  D v 2 4  1 π  = (1.225)   (2) 2 (30.585)3 2 4 = 55, 053 N ⋅ m/s ΔT =

dT = 55,100 N ⋅ m/s  dt PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 951

PROBLEM 14.78 The wind turbine-generator shown has an output-power rating of 1.5 MW for a wind speed of 36 km/h. For the given wind speed, determine (a) the kinetic energy of the air particles entering the 82.5-m-diameter circle per second, (b) the efficiency of this energy conversion system. Assume ρ = 1.21 kg/m3 for air.

SOLUTION (a)

Rate of kinetic energy in the slipstream. Let Δm be the mass moving through the slipstream of area A in the time Δt. Then, Δm = ρ A(Δl ) = ρ Av(Δt )

The kinetic energy carried by this mass is Δt =

1 1 (Δm)v 2 = ρ Av3 ( Δt ) 2 2

dT ΔT 1 = = ρ Av3 dt Δt 2

Data:

A=

π

d2 =

π

(82.5 m) 2 = 5345.6 m 2 4 4 v = 36 km/h = 10 m/s

dT 1 = (1.21 kg/m3 )(5345.6 m 2 )(10 m/s)3 dt 2 = 3.234 × 106 kg ⋅ m 2 /s3 dT = 3.234 MW  dt

(b)

Efficiency n:

η=

output power 1.5 MW = available input power 3.234 MW

η = 0.464 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 952

PROBLEM 14.79 A wind turbine-generator system having a diameter of 82.5 m produces 1.5 MW at a wind speed of 12 m/s. Determine the diameter of blade necessary to produce 10 MW of power assuming the efficiency is the same for both designs and ρ = 1.21 kg/m3 for air.

SOLUTION Rate of kinetic energy in the slipstream. Let Δm be the mass moving through the slipstream of area A in time Δt. Then Δm = ρ A(Δl ) = ρ Av(Δt )

The kinetic energy carried by this mass is 1 1 ( Δm) v 2 = ρ Av3 (Δt ) 2 2 dT ΔT 1 = = ρ Av3 Δt 2 dt

ΔT =

This is the available input power for the wind turbine. For a wind turbine of efficiency η , the output power P is P =η

dT η = ρ Av3 dt 2

We want to compare two turbines having P1 = 1.5 MW and P2 = 10 MW, respectively. Then P2 η2 ρ 2 A2 v23 = P1 η1 ρ1 A1v13

Since η 2 = η1 , ρ 2 = ρ1 , and v2 = v1 , we get P2 A2 d 22 10 = = = = 6.6667 P1 A1 d12 1.5 d 22 = 6.6667d12 = (6.6667)(82.5 in)2 d 2 = 213 m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 953

PROBLEM 14.80 While cruising in level flight at a speed of 570 mi/h, a jet airplane scoops in air at a rate of 240 lb/s and discharges it with a velocity of 2200 ft/s relative to the airplane. Determine (a) the power actually used to propel the airplane, (b) the total power developed by the engine, (c) the mechanical efficiency of the airplane.

SOLUTION dm 240 = = 7.4534 slugs/s dt 32.2 u = 2200 ft/s v = 570 mi/h = 836 ft/s dm (u − v) F= dt = (7.4534)(2200 − 836) = 10,166 lb

Data:

(a)

Power used to propel airplane: P1 = Fv = (10,166)(836) = 8.499 × 106 ft ⋅ lb/s Propulsion power = 15, 450 hp 

Power of kinetic energy of exhaust: 1 (Δm)(u − v) 2 2 1 dm P2 = (u − v) 2 2 dt 1 = (7.4534)(2200 − 836) 2 2 = 6.934 × 106 ft ⋅ lb/s

P2 ( Δt ) =

(b)

Total power:

P = P1 + P2 = 15.433 × 106 ft ⋅ lb/s Total power = 28,060 hp 

(c)

Mechanical efficiency:

P1 8.499 × 106 = P 15.433 × 106 = 0.551 Mechanical efficiency = 0.551 

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PROBLEM 14.81 In a Pelton-wheel turbine, a stream of water is deflected by a series of blades so that the rate at which water is deflected by the blades is equal to the rate at which water issues from the nozzle (Δm/Δt = Aρ v A ). Using the same notation as in Sample Problem 14.7, (a) determine the velocity V of the blades for which maximum power is developed, (b) derive an expression for the maximum power, (c) derive an expression for the mechanical efficiency.

SOLUTION Let u be the velocity of the stream relative to the velocity of the blade. u = (v − V ) Δm = ρ Av A Δt

Mass flow rate:

Principle of impulse and momentum: (Δm)u − Ft ( Δt ) = (Δm)u cos θ Δm Ft = u (1 − cos θ ) Δt = ρ Av A (v A − V )(1 − cos θ )

where Ft is the tangential force on the fluid. The force Ft on the fluid is directed to the left as shown. By Newton’s law of action and reaction, the tangential force on the blade is Ft to the right. Output power:

Pout = FV t = ρ Av A (v A − V )V (1 − cos θ )

(a)

V for maximum power output: dPout = ρ A (v A − 2V )(1 − cos θ ) = 0 dV

(b)

vA =

1 V  2

Maximum power: 1  1   ( Pout )max = ρ Av A  v A − v A  v A  (1 − cos θ ) 2  2   ( Pout )max =

1 ρ Av3A (1 − cos θ )  4

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PROBLEM 14.81 (Continued)

Input power = rate of supply of kinetic energy of the stream 1 1  (Δm)v A2  Δt  2  1 Δm 2 vA = 2 Δt 1 = ρ Av3A 2

Pin =

(c)

Efficiency:

η=

Pout Pin

η=

ρ Av A (v A − V )V (1 − cos θ ) 1 ρ Av3A 2 

η = 2 1 − 

V vA

V  (1 − cos θ )   vA

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PROBLEM 14.82 A circular reentrant orifice (also called Borda’s mouthpiece) of diameter D is placed at a depth h below the surface of a tank. Knowing that the speed of the issuing stream is v = 2 gh and assuming that the speed of approach v1 is zero, show that the diameter of the stream is d = D/ 2. (Hint: Consider the section of water indicated, and note that P is equal to the pressure at a depth h multiplied by the area of the orifice).

SOLUTION From hydrostatics, the pressure at section 1 is p1 = γ h = ρ gh. The pressure at section 2 is p2 = 0. Calculate the mass flow rate using section 2. mass = density × volume = density × area × length Δm = ρ A2 ( Δl ) = ρ A2 v(Δt ) dm Δm = = ρ A2 v dt Δt

Apply the impulse-momentum principle to fluid between sections 1 and 2.

(Δm)v1 + p1 A1 (Δt ) = (Δm)v dm dm v1 + p1 A1 = v dt dt dm p1 A1 = (v − v1 ) dt = ρ A2 v(v − v1 )

p1 = ρ gh and v = 2 gh

But v1 is negligible,

ρ ghA1 = ρ A2 (2 gh) or A1 = 2 A2 π

π  D2 = 2  d 2  4 4 

d=

D 2



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PROBLEM 14.83 Gravel falls with practically zero velocity onto a conveyor belt at the constant rate q = dm/dt. (a) Determine the magnitude of the force P required to maintain a constant belt speed v. (b) Show that the kinetic energy acquired by the gravel in a given time interval is equal to half the work done in that interval by the force P. Explain what happens to the other half of the work done by P.

SOLUTION (a)

We apply the impulse-momentum principle to the gravel on the belt and to the mass Δm of gravel hitting and leaving belt in interval Δt.

x comp.: mv + PΔt = mv + (Δm)v P=

(b)

Δm v = qv Δt

P = qv 

Kinetic energy acquired for unit time: 1 (Δm)v 2 2 ΔT 1 Δm 2 1 2 = v = qv Δt 2 Δt 2 ΔT =

(1)

Work done per unit time: ΔU PΔx = = Pv Δt Δt

Recalling the result of part a: ΔU = P(Δx) ΔU = ( qv)v = qv 2 Δt

(2)

Comparing Eqs. (1) and (2), we conclude that ΔT 1 ΔU = Δt 2 Δt

Q.E.D.

The other half of the work of P is dissipated into heat by friction as the gravel slips on the belt before reaching the speed v.





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PROBLEM 14.84* The depth of water flowing in a rectangular channel of width b at a speed v1 and a depth d1 increases to a depth d2 at a hydraulic jump. Express the rate of flow Q in terms of b, d1, and d2.

SOLUTION mass = density × volume

Mass flow rate:

= density × area × length Δm = ρ bd (Δl ) = ρ bdv(Δt ) dm Δm = = ρ bdv dt Δt 1 dm Q= = bdv ρ dt Q1 = Q2 = Q

Continuity of flow:

v1 =

Q bd1

v2

Q bd 2

Resultant pressure forces: p1 = γ d1

p2 = γ d 2

1 1 p1bd1 = γ bd12 2 2 1 1 F2 = p2bd 2 = γ bd 22 2 2 F1 =

Apply impulse-momentum principle to water between sections 1 and 2.

(Δm)v1 + F1 (Δt ) − F2 (Δt ) = (Δm)v2 Δm (v1 − v2 ) = F2 − F1 Δt

ρ Q 2 ( d 2 − d1 ) bd1d 2

 Q Q  1 2 2 −  = γ b d 2 − d1 bd bd 2 2   1

ρQ ⋅ 

(

)

1 = γ b(d1 + d 2 )( d 2 − d1 ) 2

Noting that γ = ρ g ,

Q=b

1 gd1d 2 (d1 + d 2 )  2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 959

PROBLEM 14.85* Determine the rate of flow in the channel of Problem 14.84, knowing that b = 12 ft, d1 = 4 ft, and d 2 = 5 ft. PROBLEM 14.84 The depth of water flowing in a rectangular channel of width b at a speed v1 and a depth d1 increases to a depth d2 at a hydraulic jump. Express the rate of flow Q in terms of b, d1, and d2.

SOLUTION mass = density × volume

Mass flow rate:

= density × area × length Δm = ρ bd (Δl ) = ρ bdv(Δt ) dm Δm = = ρ bdv dt Δt 1 dm Q= = bdv ρ dt Q1 = Q2 = Q

Continuity of flow:

v1 =

Resultant pressure forces:

Q bd1

p1 = γ d1

v2

Q bd 2

p2 = γ d 2

1 1 p1bd1 = γ bd12 2 2 1 1 F2 = p2bd 2 = γ bd 22 2 2 F1 =

Apply impulse-momentum principle to water between sections 1 and 2.

(Δm)v1 + F1 (Δt ) − F2 (Δt ) = (Δm)v2 Δm (v1 − v2 ) = F2 − F1 Δt

ρ Q 2 ( d 2 − d1 ) bd1d 2

 Q Q  1 2 2 −  = γ b d 2 − d1 bd bd 2 2   1

ρQ ⋅ 

(

)

1 = γ b(d1 + d 2 )( d 2 − d1 ) 2

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PROBLEM 14.85* (Continued)

1 gd1d 2 (d1 + d 2 ) 2

Noting that γ = ρ g ,

Q=b

Data:

g = 32.2 ft/s,2 Q = 12

b = 12 ft, d1 = 4 ft, d 2 = 5 ft

1 (32.2)(4)(5)(9) 2

Q = 646 ft 3 /s 

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PROBLEM 14.86 A chain of length l and mass m lies in a pile on the floor. If its end A is raised vertically at a constant speed v, express in terms of the length y of chain which is off the floor at any given instant (a) the magnitude of the force P applied at A, (b) the reaction of the floor.

SOLUTION Let ρ be the mass per unit length of chain. Apply the impulse-momentum to the entire chain. Assume that the reaction from the floor is equal to the weight of chain still in contact with the floor. Calculate the floor reaction. R = ρ g (l − y ) y  R = mg 1 −  l 

Apply the impulse-momentum principle.

ρ yv + P( Δt ) + R(Δt ) − ρ gL(Δt ) = ρ ( y + Δy )v PΔt = ρ (Δy )v + ρ gL( Δt ) − R(Δt ) P=ρ

(a)

Δy v + ρ gL − ρ ( L − y ) g Δt

= ρ v 2 + ρ gy

Let (b)

P=

m 2 (v + gy )  l

Δy dy = =v Δt dt y  R = mg 1 −   l 

From above,

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PROBLEM 14.87 Solve Problem 14.86, assuming that the chain is being lowered to the floor at a constant speed v. PROBLEM 14.86 A chain of length l and mass m lies in a pile on the floor. If its end A is raised vertically at a constant speed v, express in terms of the length y of chain which is off the floor at any given instant (a) the magnitude of the force P applied at A, (b) the reaction of the floor.

SOLUTION (a)

Let ρ be the mass per unit length of chain. The force P supports the weight of chain still off the floor. P = ρ gy

(b)

P=

mgy  l

Apply the impulse-momentum principle to the entire chain.

− ρ yv + P(Δt ) + R(Δt ) − ρ gL( Δt ) = − ρ g ( y + Δy )v R( Δt ) = ρ gL(Δt ) − P (Δt ) − ρ g ( Δy )v Δy R = ρ gL − ρ gy − ρ v Δt

Let Δt → 0. Then

Δy dy = = −v Δt dt R = ρ g (L − y) + ρ v 2

R=

m [ g ( L − y ) + v 2 ]  l

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PROBLEM 14.88 The ends of a chain lie in piles at A and C. When released from rest at time t = 0, the chain moves over the pulley at B, which has a negligible mass. Denoting by L the length of chain connecting the two piles and neglecting friction, determine the speed v of the chain at time t.

SOLUTION Let m be the mass of the portion of the chain between the two piles. This is the portion of the chain that is moving with speed v. The remainder of the chain lies in either of the two piles. Consider the time period between t and t + Δt and apply the principle impulse and momentum. Let Δm be the amount of chain that is picket up at A and deposited at C during the time period Δt. At time t, Δm is still in pile A while Δm has a downward at speed v just above pile C. The remaining mass (m − Δm) is moving with speed v. At time t + Δt , Δm is moving with speed v + Δv just above pile A and Δm is at rest in pile C. Over the time period an unbalanced weight of chain acts on the system. The weight is mgh L Apply the impulse-momentum principle to the system. W=

Consider moments about the pulley axle. r[(Δm)v + (m − Δm)v] + rW (Δt ) = r[( Δm)(v + Δv) + (m − Δm)(v + Δv)]

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PROBLEM 14.88 (Continued)

Dividing by r and canceling the terms (Δm)v and (m − Δm)v 0+

mgh (Δt ) = ( Δm)(v + Δv) + (m − Δm)( Δv) L = v( Δm) + m(Δv) m v ( Δt ) L

But

Δm =

Hence,

mgh mv 2 ( Δt ) = (Δt ) + m( Δv) L L

Solving for Δt ,

Δt =

L ( Δv ) gh − v 2

Letting c 2 = gh, and considering the limit as Δt and Δv become infinitesimal, gives dt =

L dv c − v2 2

Integrate, noting that v = 0 when t = 0 t=L tanh



v

0

v

dv L v = tanh −1 2 2 c c0 c −v

ct v = L c  gh  v = gh tanh  t   L   

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PROBLEM 14.89 A toy car is propelled by water that squirts from an internal tank at a constant 6 ft/s relative to the car. The weight of the empty car is 0.4 lb and it holds 2 lb of water. Neglecting other tangential forces, determine the top speed of the car.

SOLUTION Consider a time interval Δt. Let m be the mass of the car plus the water in the tank at the beginning of the interval and (m − Δm) the corresponding mass at the end of the interval. m0 is the initial value of m. Let v be the velocity of the car. Apply the impulse and momentum principle over the time interval.

Horizontal components

: mv + 0 = (Δm)( v − u cos 20°) + ( m − Δm)( v + Δv) Δm Δv = u cos 20° m − Δm

Let Δv be replaced by differential dv and Δm be replaced by the small differential − dm, the minus sign meaning that dm is the infinitesimal increase in m. dv = −u cos 20°

dm m

Integrating,

v = v0 − u cos 20° ln

Since v0 = 0,

v = u cos 20° ln

m m0

m0 m

The velocity is maximum when m = m f , the value of m when all of the water is expelled. vmax = u cos 20° ln

m0 mf

vmax = (6 ft/s) cos 20° ln

0.4 + 2 0.4

vmax = 10.10 ft/s 

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PROBLEM 14.90 A toy car is propelled by water that squirts from an internal tank. The weight of the empty car is 0.4 lb and it holds 2 lb of water. Knowing the top speed of the car is 8 ft/s, determine the relative velocity of the water that is being ejected.

SOLUTION Consider a time interval Δt. Let m be the mass of the car plus the water in the tank at the beginning of the interval and (m − Δm) the corresponding mass at the end of the interval. m0 is the initial value of m. Let v be the velocity of the car. Apply the impulse and momentum principle over the time interval.

Horizontal components

: mv + 0 = (Δm)( v − u cos 20°) + ( m − Δm)( v + Δv) Δm Δv = u cos 20° m − Δm

Let Δv be replaced by differential dv and Δm be replaced by the small differential − dm, the minus sign meaning that dm is the infinitesimal increase in m. dv = −u cos 20°

dm m

Integrating,

v = v0 − u cos 20° ln

Since v0 = 0,

v = u cos 20° ln

m m0

m0 m

The velocity is maximum when m = m f , the value of m when all of the water is expelled. vmax = u cos 20° ln 8 ft/s = u cos 20° ln

m0 mf 0.4 + 2 0.4

u = 4.75 ft/s 

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PROBLEM 14.91 The main propulsion system of a space shuttle consists of three identical rocket engines which provide a total thrust of 6 MN. Determine the rate at which the hydrogen-oxygen propellant is burned by each of the three engines, knowing that it is ejected with a relative velocity of 3750 m/s.

SOLUTION Thrust of each engine: Eq. (14.44):

1 P = (6 MN) = 2 × 106 N 3 dm u dt dm 2 × 106 N = (3750 m/s) dt P=

dm 2 × 106 N = dt 3750 m/s

dm = 533 kg/s  dt

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PROBLEM 14.92 The main propulsion system of a space shuttle consists of three identical rocket engines, each of which burns the hydrogen-oxygen propellant at the rate of 750 lb/s and ejects it with a relative velocity of 12000 ft/s. Determine the total thrust provided by the three engines.

SOLUTION From Eq. (14.44) for each engine: dm u dt (750 lb/s) = (12000 ft/s) 32.2 ft/s 2 = 279.50 × 103 lb

P=

For the 3 engines: Total thrust = 3(279.50 × 103 lb)

Total thrust = 839,000 lb 

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PROBLEM 14.93 A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine its acceleration (a) as it is fired, (b) as the last particle of fuel is being consumed.

SOLUTION From Eq. (14.44) of the textbook, the thrust is dm u dt (25 lb/s) = (13000 ft/s) 32.2 ft/s 2 = 10.093 × 103 lb

P=

Σ F = ma P − mg = ma

(a)

a=

(1)

W = W0 = 2600 lb

At the start of firing,

g = 32.2 ft/s 2 a=

From Eq. (1), (b)

P −g m

m=

2600 = 80.745 slug 32.2

10.093 × 103 lb − 32.2 = 92.80 ft/s 2 80.745 slug

a = 92.8 ft/s 2 

As the last particle of fuel is consumed, W = 2600 − 2200 = 400 lb g = 32.2 ft/s 2 (assumed)

From Eq. (1),

a=

m=

400 = 12.422 slug 32.2

10.093 × 103 lb − 32.2 = 780.30 ft/s 2 12.422

a = 780 ft/s 2 

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PROBLEM 14.94 A space vehicle describing a circular orbit at a speed of 24 × 103 km/h releases its front end, a capsule which has a gross mass of 600 kg, including 400 kg of fuel. If the fuel is consumed at the rate of 18 kg/s and ejected with a relative velocity of 3000 m/s, determine (a) the tangential acceleration of the capsule as the engine is fired, (b) the maximum speed attained by the capsule.

SOLUTION Thrust:

dm u dt = (18 kg/s)(3000 m/s)

P=

= 54 × 103 N (at )0 =

(a )

(b)

P 54 × 103 = = 90 m/s 2 m0 600

(at )0 = 90.0 m/s 2 

Maximum speed is attained when all the fuel is used up: v1 = v0 + = v0 +

 

t1 0 t1 0

at dt = v0 + u(

dm dt

m



P dt m

t1 0

) dt = v

0

+u



m1  m0

dm  −   m 

 m m  v1 = v0 + u  − ln 1  = v0 + u ln 0 m0  m1 

Data:

v0 = 24 × 103 km/h = 6.6667 × 103 m/s u = 3000 m/s m0 = 600 kg m1 = 600 − 400 = 200 kg v1 = 6.6667 × 103 + 3000 ln = 9.9625 × 103 m/s

600 200 v1 = 35.9 × 103 km/h 

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PROBLEM 14.95 A 540-kg spacecraft is mounted on top of a rocket with a mass of 19 Mg, including 17.8 Mg of fuel. Knowing that the fuel is consumed at a rate of 225 kg/s and ejected with a relative velocity of 3600 m/s, determine the maximum speed imparted to the spacecraft if the rocket is fired vertically from the ground.

SOLUTION See sample Problem 14.8 for derivation of v = u ln

Data:

m0 − gt m0 − qt

(1)

u = 3600 m/s q = 225 kg/s, mfuel = 17,800 kg m0 = 19, 000 kg + 540 kg = 19,540 kg

We have

mfuel = qt , 17,800 kg = (225 kg/s)t t=

17,800 kg = 79.111 s 225 kg/s

Maximum velocity is reached when all fuel has been consumed, that is, when qt = mfuel. Eq. (1) yields vm = u ln

m0 − gt m0 − mfuel

19,540 − (9.81 m/s 2 )(79.111 s) 19,540 − 17,800 = (3600 m/s) ln 11.230 − 776.1 m/s = 7930.8 m/s vm = 7930 m/s  = (3600 m/s) ln

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PROBLEM 14.96 The rocket used to launch the 540-kg spacecraft of Problem 14.95 is redesigned to include two stages A and B, each of mass 9.5 Mg, including 8.9 Mg of fuel. The fuel is again consumed at a rate of 225 kg/s and ejected with a relative velocity of 3600 m/s. Knowing that when stage A expels its last particle of fuel its casing is released and jettisoned, determine (a) the speed of the rocket at that instant, (b) the maximum speed imparted to the spacecraft.

SOLUTION dm = uq dt

Thrust force:

P=u

Mass of rocket + unspent fuel:

m = m0 − qt

Corresponding weight force:

W = mg a=

Acceleration:

F P −W P uq = = −g= −g m m m m0 − qt

Integrating with respect to time to obtain the velocity, v = v0 +



t 0

adt = v0 + u

= v0 − u ln

For each stage,



qdt − gt 0 m − qt 0 t

m0 − qt − gt m0

mfuel = 8900 kg q = 225 kg/s

For the first stage,

v0 = 0

(a)

v1 = 0 − 3600 ln

(1)

u = 3600 m/s mfuel 8900 = = 39.556 s q 225

t=

m0 = 540 + (2)(9500) = 19,540 kg 19,540 − 8900 − (9.81)(39.556) = 1800.1 m/s 19,540 v1 = 1800 m/s 

For the second stage, (b)

v0 = 1800.1 m/s,

m0 = 540 + 9500 = 10, 040 kg

v2 = 1800.1 − 3600 ln

10, 040 − 8900 − (9.81)(39.556) = 9244 m/s 10,040 v2 = 9240 m/s 

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PROBLEM 14.97 A communication satellite weighing 10,000 lb, including fuel, has been ejected from a space shuttle describing a low circular orbit around the earth. After the satellite has slowly drifted to a safe distance from the shuttle, its engine is fired to increase its velocity by 8000 ft/s as a first step to its transfer to a geosynchronous orbit. Knowing that the fuel is ejected with a relative velocity of 13,750 ft/s, determine the weight of fuel consumed in this maneuver.

SOLUTION Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time Δt.

: mv = (m − Δm)(v + Δv) + (Δm)(v + Δv − v) = mv + m( Δv) − ( Δm)v − (Δm)(Δv) + (Δm)v + (Δm)(Δv) − (Δm)v m(Δv) − u ( Δm) = 0 dm ( Δt ) dt u dm Δv dv = =− Δt dt m dt

Δm = −



v1 v0

dv = −



t1 0

u dm dt = − m dt

v1 − v0 = − u ln



m1 m0

u

dm m

m m1 = u ln 0 m0 m1

m0 v −v  = exp  1 0  m1  u 

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PROBLEM 14.97 (Continued)

Data:

v1 − v0 = 8000 ft/s u = 13, 750 ft/s m0 = 10,000 lb 10, 000 8000 = exp m1 13, 750 = 1.7893 m1 = 5589 kg mfuel = m0 − m1 = 10, 000 − 5589

mfuel = 4410 lb 

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PROBLEM 14.98 Determine the increase in velocity of the communication satellite of Problem 14.97 after 2500 lb of fuel has been consumed.

SOLUTION Data from Problem 14.95:

m0 = 10, 000 lb

u = 13, 750 ft/s

m1 = m0 − mfuel = 10,000 − 2500 = 7500 lb.

Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time Δt.

mv = (m − Δm)(v + Δv) + (Δm)(v + Δv − v) = mv + m(Δv) − ( Δm)v − (Δm)(Δv)

:

+ (Δm)v + (Δm)( Δv) − (Δm)v m(Δv) − u ( Δm) = 0 dm (Δt ) dt u dm Δv dv = =− m dt Δt dt

Δm = −



v1 v0

dv = −



t1 0

u dm dt = − m dt

v1 − v0 = −u ln



m1 m0

u

dm m

m1 m = u ln 0 m0 m1

Δv = v1 − v0 = 13, 750 ln

10,000 7500

Δv = 3960 ft/s 

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PROBLEM 14.99 Determine the distance separating the communication satellite of Problem 14.97 from the space shuttle 60 s after its engine has been fired, knowing that the fuel is consumed at a rate of 37.5 lb/s. PROBLEM 14.97 A communication satellite weighing 10,000 lb, including fuel, has been ejected from a space shuttle describing a low circular orbit around the earth. After the satellite has slowly drifted to a safe distance from the shuttle, its engine is fired to increase its velocity by 8000 ft/s as a first step to its transfer to a geosynchronous orbit. Knowing that the fuel is ejected with a relative velocity of 13,750 ft/s, determine the weight of fuel consumed in this maneuver.

SOLUTION Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time Δt.

:

mv = (m − Δm)(v + Δv) + (Δm)(v + Δv − v) = mv + m(Δv) − ( Δm)v − (Δm)(Δv) + (Δm)v + ( Δm)(Δv) − (Δm)v

m(Δv) − u (Δm) = 0 dm (Δt ) dt Δv dv u dm uq uq = =− =− =− Δt dt m dt m m0 − qt t uq t v = v0 + dt = v0 − u ln ( m0 − qt ) 0 0 m − qt 0

Δm = −



= v0 + u ln (m0 − qt ) + u ln m0  m − qt  = v0 − u ln  0   m0 

Set

dx dt

(1)

= v in Eq. (1) and integrate with respect to time. x = x0 + v0 t + u



 m − qt  ln  0  dt 0  m0  t

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PROBLEM 14.99 (Continued)

Call the last term x′ and let z=

m0 − qt m0

dz = −

x′ =

m0 u q



z z0

q dt m0

ln z dz =

or

dt = −

m0 dz q

m0u [( z ln z + z )]zz0 q

 m  m  m0 u  m0 − qt  m0 − qt − 1 − 0  ln 0 − 1    ln q  m0  m0  m0  m0     m u  qt  m0 − qt = 0  1 − − 1 + 1  ln q  m0  m0  

=

=

  m − qt  m0 u  m0 − qt − 1 + 1 − ut ln 0 − 1 ln q  m0 m0   

m u  m − qt = ut +  0 − ut  ln 0 m0  q   m  m0  = u t −  0 − t  ln   m0 − qt    q  m  m0  x = x0 + v0t + u t −  0 − t  ln   m0 − qt    q

Data:

x0 = 0

v0 = 0

(2)

q = 37.5 lb/s.

m0 = 10, 000 lb, t = 60 sec u = 13, 750 ft/s   10,000  10,000  x = 0 + 0 + (13, 750) 60 −  − 60  ln   37.5  10, 000 − (37.5)(60)  

= 100, 681 ft

x = 19.07 mi 

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PROBLEM 14.100 For the rocket of Problem 14.93, determine (a) the altitude at which all the fuel has been consumed, (b) the velocity of the rocket at that time. PROBLEM 14.93 A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine its acceleration (a) as it is fired, (b) as the last particle of fuel is being consumed.

SOLUTION See Sample Problem 14.8 for derivation of v = u ln

m0 m − qt − gt = −u ln 0 − gt m0 − qt m0

(1)

Note that g is assumed to be constant. Set

dy dt

= v in Eq. (1) and integrate with respect to time.

 m0 − gt  dt  u ln m0 − qt   t m − qt 1 = −u ln 0 dt − gt 2 0 m0 2

h=



h

0

dy =



t

0

vdt =



t

0



Let

z=

m0 − qt m0

h=

m0 u q



z z0

dz = − ln z dz −

q dt m0

or

dt = −

m0 dz q

1 2 m0 u 1 gt = [( z ln z + z )]zz0 − gt 2 q 2 2

=

 m  m  1 m0 u  m0 − qt  m0 − qt − 1 − 0  ln 0 − 1  − gt 2   ln q  m0  m0  m0  m0   2

=

  1 m0 u  qt  m0 − qt − 1 + 1 − gt 2  1 −  ln q  m0  m0   2

=

  m − qt  1 2 m0 u  m0 − qt − 1 + 1 − ut ln 0 − 1 − gt ln q  m0 m0    2

m u  m − qt 1 2 = ut +  0 − ut  ln 0 − gt m0 2  q   m  m0  1 2 h = u t −  0 − t  ln  − gt  m0 − qt  2   q

(2)

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PROBLEM 14.100 (Continued)

W0 = 2600 lb

Data:

qt = Wfuel = 2200 lb

q = 25 lb/s

u = 13000 ft/s g = 32.2 ft/s 2 W 2200 t = fuel = = 88 s q 25

(a)

From Eq. (2),

  1 2600  2600  2 h = (13000) 88 −  − 88  ln  − (32.2)(88) 25 2600 2200 2 −     = (13000)(88 − 16 ln 6.5) − 124680

= 630, 000 ft h = 119.3 mi 

(b)

From Eq. (1),

2600 − 2200 − (32.2)(88) 2600 = 13000 ln 6.5 − 2834 = 21500 ft/s

v = −13000 ln

v = 14,660 mi/h 

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PROBLEM 14.101 Determine the altitude reached by the spacecraft of Problem 14.95 when all the fuel of its launching rocket has been consumed.

SOLUTION See Sample Problem 14.8 for derivation of v = u ln

m0 m − qt − gt = −u ln 0 − gt m0 − qt m0

(1)

Note that g is assumed to be constant. Set

dy dt

= v in Eq. (1) and integrate with respect to time.  m0 − gt  dt  u ln m0 − qt   t m − qt 1 dt − gt 2 = −u ln 0 0 m0 2

h=



h

0

dy =



t

0

vdt =



t

0



Let

z=

m0 − qt m0

q dt m0

or

dt = −

m0 dz q

1 2 m0u 1 z gt = ( z ln z + z ) ]z − gt 2 [ 0 q 2 2   m  m  1 m u m − qt  m0 − qt = 0  0 − 1 − 0  ln 0 − 1  − gt 2  ln q  m0  m0  m0  m0   2

h=

m0 u q

dz = −



z

z0

ln z dz −

=

  1 m0 u  qt  m0 − qt − 1 + 1 − gt 2  1 −  ln q  m0  m0   2

=

  m − qt  1 2 m0 u  m0 − qt − 1 + 1 − ut ln 0 − 1 − gt ln q  m0 m0    2

m u  m − qt 1 2 = ut +  0 − ut  ln 0 − gt m0 2  q   m  m0  1 2 h = u t −  0 − t  ln  − gt  m0 − qt  2   q

(2)

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PROBLEM 14.101 (Continued)

Data:

u = 3600 m/s q = 225 kg/s

m0 = 19,000 + 540 = 19,540 kg mfuel = 17,800 kg

mfuel 17,800 = = 79.111 s q 225 m0 − qt = 1740 kg t=

From Eq. (2),

g = 9.81 m/s 2

  19,540  19,540  1 2 − 79.111 ln h = (3600) 79.111 −   − (9.81)(79.111) 1740  2  225   = 186, 766 m

h = 186.8 km 

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PROBLEM 14.102 For the spacecraft and the two-stage launching rocket of Problem 14.96, determine the altitude at which (a) stage A of the rocket is released, (b) the fuel of both stages has been consumed.

SOLUTION dm = uq dt

Thrust force:

P=u

Mass of rocket + unspent fuel:

m = m0 − qt

Corresponding weight force:

W = mg a=

Acceleration:

F P −W P uq = = −g= −g m m m m0 − qt

Integrating with respect to time to obtain velocity, v = v0 +



t

0

adt = v0 + u



qdt − gt 0 m − qt 0 t

m0 − qt − gt m0

= v0 − u ln

(1)

Integrating again to obtain the displacement, s = s0 + v0 t − u m0 − qt m0

Let

z=

Then

s = s0 + v0 t +



t

0

ln

m0 − qt 1 dt − gt 2 2 m0

dz = −

q dt m0

dt = −

m0 dz q

m0 u z 1 ln z dz − gt 2 2 q z0 mu 1 z = s0 + v0t + 0 ( z ln z + z ) z − gt 2 0 2 q



m0 u  m0 − qt m0 − qt m0 − qt m0 m m  1 ln ln 0 + 0  − gt 2 + −  q  m0 m0 m0 m0 m0 m0  2  m  m − qt  1 2 = s0 + v0t + u t +  0 − t  ln 0  − gt m0  2    q

= s0 + v0t +

(2)

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PROBLEM 14.102 (Continued)

For each stage,

mfuel = 8900 kg

u = 3600 m/s

q = 225 kg/s

For the first stage,

v0 = 0

t=

mfuel 8900 = = 39.556 s q 225

s0 = 0

m0 = 540 + (2)(9500) = 19,540 kg

From Eq. (1),

v1 = 0 − 3600 ln

19,540 − 8900 − (9.81)(39.556) 19,540

= 1800.1 m/s

From Eq. (2), (a)

  19,540  19,540 − 8900  1 2 − 39.556  ln s1 = 0 + 0 + 3600 39.556 +   − (9.81)(39.556) 225 19,540 2    

= 31, 249 m

For the second stage,

h1 = 31.2 km  v0 = 1800.1 m/s s0 = 31, 249 m m0 = 540 + 9500 = 10, 040 kg

From Eq. (2), (b)

  10,040  10,040 − 8900  − 39.556  ln s2 = 31, 249 + (1800.1)(39.556) + 3600 39.556 +   10, 040  225    1 − (9.81)(39.556) 2 2

= 197,502 m

h2 = 197.5 km 

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PROBLEM 14.103 In a jet airplane, the kinetic energy imparted to the exhaust gases is wasted as far as propelling the airplane is concerned. The useful power is equal to the product of the force available to propel the airplane and the speed of the airplane. If v is the speed of the airplane and u is the relative speed of the expelled gases, show that the mechanical efficiency of the airplane is η = 2v/(u + v). Explain why η = 1 when u = v.

SOLUTION Let F be the thrust force, and

dm dt

be the mass flow rate.

Absolute velocity of exhaust:

ve = u − v

Thrust force:

F=

Power of thrust force:

P1 = Fv =

Power associated with exhaust:

Total power supplied by engine:

dm (u − v) dt dm (u − v)v dt

1 1 (Δm)ve2 = (Δm)(u − v)2 2 2 1 dm (u − v) 2 P2 = 2 dt

P2 ( Δt ) =

P = P1 + P2 dm  1  (u − v)v − (u − v)2   dt  2  1 dm 2 = (u − v 2 ) 2 dt

P=

Mechanical efficiency:

useful power P1 = total power P 2(u − v)v η= 2 2 u −v

η=

η=

2v  (u + v)

η = 1 when u = v. The exhaust, having zero velocity, carries no power away.

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PROBLEM 14.104 In a rocket, the kinetic energy imparted to the consumed and ejected fuel is wasted as far as propelling the rocket is concerned. The useful power is equal to the product of the force available to propel the rocket and the speed of the rocket. If v is the speed of the rocket and u is the relative speed of the expelled fuel, show that the mechanical efficiency of the rocket is η = 2uv/(u 2 + v 2 ). Explain why η = 1 when u = v.

SOLUTION Let F be the thrust force and

dm dt

be the mass flow rate.

Absolute velocity of exhaust:

ve = u − v

Thrust force:

F=

Power of thrust force:

P1 = Fv =

Power associated with exhaust:

Total power supplied by engine:

Mechanical efficiency:

dm u dt dm uv dt

1 1 (Δm)ve2 = (Δm)(u − v)2 2 2 1 dm P2 = (u − v) 2 2 dt

P2 ( Δt ) =

P = P1 + P2 P=

1 dm   1 dm 2 (u − v 2 ) uv − (u − v) 2  =  2 dt   2 dt

η=

useful power P1 = P total power

η=

2uv  (u + v 2 ) 2

η = 1 when u = v. The exhaust, having zero velocity, carries no power away.

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PROBLEM 14.105 Three identical cars are being unloaded from an automobile carrier. Cars B and C have just been unloaded and are at rest with their brakes off when car A leaves the unloading ramp with a velocity of 5.76 ft/s and hits car B, which hits car C. Car A then again hits car B. Knowing that the velocity of car B is 5.04 ft/s after the first collision, 0.630 ft/s after the second collision, and 0.709 ft/s after the third collision, determine (a) the final velocities of cars A and C, (b) the coefficient of restitution for each of the collisions.

SOLUTION There are no horizontal forces acting. Horizontal momentum is conserved. (a)

Velocities: Event 1

2: Car A hits car B.

m(5.76) + 0 = m(v A ) 2 + m(5.04)

Event 2

3: Car B hits car C.

m (5.04) + 0 = m (0.630) + m(vC )3

Event 3

( vC )3 = 4.41 ft/s



( v A )4 = 0.641 ft/s



4: Car A hits car B again.

m(0.720) + m(0.630) = m(v A ) 4 + m(0.709)

(b)

( v A ) 2 = 0.720 ft/s

Coefficients of restitution: Event 1

2:

e1→ 2 =

(v A ) 2 − (vB ) 2 5.04 − 0.720 = (v A )1 − (vB )1 5.76 − 0

e1→ 2 = 0.750 

Event 2

3:

e2 →3 =

(vB )3 − (vC )3 4.41 − 0.630 = (vB )2 − (vC ) 2 5.04 − 0

e2 →3 = 0.750 

Event 3

4:

e3 → 4 =

(v A )4 − (vB ) 4 0.709 − 0.641 = (v A )3 − (vB )3 0.720 − 0.630

e3 → 4 = 0.756 

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PROBLEM 14.106 A 30-g bullet is fired with a velocity of 480 m/s into block A, which has a mass of 5 kg. The coefficient of kinetic friction between block A and cart BC is 0.50. Knowing that the cart has a mass of 4 kg and can roll freely, determine (a) the final velocity of the cart and block, (b) the final position of the block on the cart.

SOLUTION (a)

Conservation of linear momentum:

m0 v0 = (m0 + m A )v′ = ( m0 + mA + mC )vf (0.030 kg)(480 m/s) = (5.030 kg)v′ = (9.030 kg)vf 0.030 (480 m/s) = 2.863 m/s 5.030 0.030 vf = (480 m/s) = 1.5947 m/s 9.030 v′ =

(b)

v f = 1.595 m/s 

Work-energy principle: Just after impact:

Final kinetic energy:

Work of friction force:

1 (m0 + m A )v′2 2 1 = (5.030 kg)(2.863 m/s) 2 2 = 20.615 J 1 1 Tf = ( m0 + m A + mC ) v 2f 2 2 1 = (9.030 kg)(1.5947 m/s) 2 2 = 11.482 J T′ =

F = μk N = μk (m0 + m A ) g = 0.50(5.030)(9.81) = 24.672 N

Work = U = − Fx = −24.672 x

T ′ + U = T f : 20.615 − 24.672 x = 11.482

x = 0.370 m 

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PROBLEM 14.107 An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B which can slide along the floor of the car ( μk = 0.25). Knowing that the car was at rest with its brakes released and that it automatically coupled with the engine upon impact, determine the velocity of the car (a) immediately after impact, (b) after the load has slid to a stop relative to the car.

SOLUTION The masses are the engine (m A = 80 × 103 kg), the load (mB = 30 × 103 kg), and the flat car (mC = 20 × 103 kg). (v A )0 = 6.5 km/h

Initial velocities:

= 1.80556 m/s

(vB )0 = (vC )0 = 0.

No horizontal external forces act on the system during the impact and while the load is sliding relative to the flat car. Momentum is conserved. m A (v A )0 + mB (0) + mC (0) = m A (v A )0

Initial momentum:

(a)

(1)

Let v′ be the common velocity of the engine and flat car immediately after impact. Assume that the impact takes place before the load has time to acquire velocity. Momentum immediately after impact: m A v′ + mB (0) + mC v′ = (m A + mC )v′

(2)

Equating (1) and (2) and solving for v′, v′ =

m A (v A ) 0 mA + mC

(80 × 103 )(1.80556) (100 × 103 ) = 1.44444 m/s =

v′ = 5.20 km/h

(b)



Let vf be the common velocity of all three masses after the load has slid to a stop relative to the car. Corresponding momentum: m A v f + mB v f + mC v f = ( mA + mB + mC )vf

(3)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 989

PROBLEM 14.107 (Continued)

Equating (1) and (3) and solving for vf , vf =

m A (v A ) 0 m A + mB + mC

(80 × 103 )(1.80556) (130 × 103 ) = 1.11111 m/s =

vf = 4.00 km/h



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 990

PROBLEM 14.108 In a game of pool, ball A is moving with a velocity v0 when it strikes balls B and C which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and that v0 = 12 ft/s and vC = 6.29 ft/s, determine the magnitude of the velocity of (a) ball A, (b) ball B.

SOLUTION Conservation of linear momentum. In x direction: m(12 ft/s) cos 30° = mv A sin 7.4° + mvB sin 49.3° + m(6.29) cos 45° 0.12880vA + 0.75813vB = 5.9446

(1)

In y direction: 

m(12 ft/s) sin 30° = mvA cos 7.4° − mvB cos 49.3° + m(6.29)sin 45° 0.99167vA − 0.65210vB = 1.5523

(a)

Multiply (1) by 0.65210, (2) by 0.75813, and add: 0.83581 vA = 5.0533

(b) 

(2)

vA = 6.05 ft/s 

Multiply (1) by 0.99167, (2) by –0.12880, and add: 0.83581 vB = 5.6951 

vB = 6.81 ft/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 991

PROBLEM 14.109 Mass C, which has a mass of 4 kg, is suspended from a cord attached to cart A, which has a mass of 5 kg and can roll freely on a frictionless horizontal track. A 60-g bullet is fired with a speed v0 = 500 m/s and gets lodged in block C. Determine (a) the velocity of C as it reaches its maximum elevation, (b) the maximum vertical distance h through which C will rise.

SOLUTION Consider the impact as bullet B hits mass C. Apply the principle of impulse-momentum to the two particle system.

Σmv1 + ΣImp1→2 = Σmv 2

Using both B and C and taking horizontal components gives mB v0 cos θ + O = (mB + mC )v′ = mBC v′ v′ = =

mB v0 cos θ mBC (0.060 kg)(500 m/s) cos 20° = 6.9435 m/s (4.06 kg)

Now consider the system of mA and mBC after the impact, and apply to impulse momentum principle.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 992

PROBLEM 14.109 (Continued)

Σmv 2 + ΣImp 2→3 = Σmv3

Horizontal components: + mBC v′ + 0 = mA v A + mBC vcx vA = =

mBC (v′ − vcx ) mA 4.06 (6.9435 − vcx ) 5

v A = 5.6381 − 0.812vcx

(a)

(1)

in m/s

At maximum elevation. Both particles have the same velocity, thus vcx = v A v A = 5.6381 − 0.812 v A v A = 3.1115 m/s

(b)

Conservation of energy:

v A = 3.11 m/s 

T2 + V2 = T3 + V3 1 1 m A (0) + mBC (v′) 2 2 2 1 = (4.06)(6.9435) 2 = 97.871 J 2 (datum) V2 = 0 T2 =

1 1 2 2 2 m A v 2A + mBC (vBx + vBy ) 2 2 1 1 = (5)(3.1115) 2 + (4.06)[(3.1115) 2 + 0] = 43.857 J 2 2 V3 = mBC gh = (4.06)(9.81) h = 39.829 h T3 =

97.871 + 0 = 43.857 + 39.829 h h = 1.356 m 

Another method: We observe that no external horizontal forces are exerted on the system consisting of A, B, and C. Thus the horizontal component of the velocity of the mass center remains constant. m = mA + mB + mC = 5 + 0.06 + 4 = 9.06 kg vx =

(a)

mB v0 cos θ (0.060 kg)(500 m/s) cos 20° = = 3.1115 m/s m A + mB + mC 9.06 kg

At maximum elevation, vA and vBC are equal. v A = 3.1115 m/s

v A = 3.11 m/s



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PROBLEM 14.109 (Continued)

Immediately after the impact of B on C, the velocity vA is zero. (mB + mC )v′ = (m A + mB + mC )vx v′ =

(b)

Principle of work and energy:

m A + mB + mC 9.06 (3.1115 m/s) = 6.9435 m/s vx = 4.06 mB + mC

T2 + V2 = T3 + V3

T2 , V2 , and V3 are calculated as before.

For T3 we note that the velocities v′A and v′BC relative to the mass center are zero. Thus, T3 is given by T3 =

1 1 mv 2 = (9.06)(3.1115)2 = 43.857 J 2 2 h = 1.356 m 

As before, h is found to be

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 994

PROBLEM 14.110 A 15-lb block B is at rest and a spring of constant k = 72 lb/in. is held compressed 3 in. by a cord. After 5-lb block A is placed against the end of the spring, the cord is cut causing A and B to move. Neglecting friction, determine the velocities of blocks A and B immediately after A leaves B.

SOLUTION 5 = 0.15528 lb ⋅ s 2 /ft 32.2 15 = 0.46584 lb ⋅ s 2 /ft mB = 32.2 k = 72 lb/in = 864 lb/ft e = 3 in. = 0.25 ft h = 6 in. = 0.5 ft mA =

Conservation of linear momentum:

Horizontal components

0 + 0 = m Av A − mB vB

:

vB =

mA 1 vA = vA mB 3

Conservation of energy:

State 1:

1 2 1 ke = (864)(0.25) 2 = 27 ft ⋅ lb 2 2 V1g = 0 V1e =

T1 = 0

State 2:

V2e = 0 V2 g = WA h = (5)(0.5) = 2.5 ft ⋅ lb T2 =

1 1 m Av 2A + mB vB2 2 2 2

1 1 v  = (0.15528)v A2 + (0.46584)  A  = 0.10352v A2 2 2  3  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 995

PROBLEM 14.110 (Continued)

T1 + V1 = T2 + V2 :

0 + 27 = 0.10352v 2A + 2.5 v A2 = 236.67 ft 2

vA = 15.38 ft/s



vB = 5.13 ft/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 996

PROBLEM 14.111 Car A was at rest 9.28 m south of the Point O when it was struck in the rear by car B, which was traveling north at a speed vB . Car C, which was traveling west at a speed vC , was 40 m east of Point O at the time of the collision. Cars A and B stuck together and, because the pavement was covered with ice, they slid into the intersection and were struck by car C which had not changed its speed. Measurements based on a photograph taken from a traffic helicopter shortly after the second collision indicated that the positions of the cars, expressed in meters, were rA = −10.1i + 16.9j, rB = −10.1i + 20.4j, and rC = −19.8i − 15.2j. Knowing that the masses of cars A, B, and C are, respectively, 1400 kg, 1800 kg, and 1600 kg, and that the time elapsed between the first collision and the time the photograph was taken was 3.4 s, determine the initial speeds of cars B and C.

SOLUTION Mass center at time of first collision. (m A + mB + mC ) r1 = mA (rA )1 + mB (rB )1 + mC (rC )1 4800 r1 = (1400)(−9.28j) + (1800)(−12.8 j) + (1600)(40i) r1 = (13.3333 m)i − (7.5067 m) j

Mass center at time of photo. (m A + mB + mC ) r2 = m A (rA ) 2 + mB (rB ) 2 + mC (rC ) 2 4800 r2 = (1400)(−10.1i + 16.9 j) + (1800)(−10.1i + 20.4 j) + (1600)(−19.8i − 15.2 j) r2 = −(13.3333 m)i + (7.5125 m) j

Since no external horizontal forces act, momentum is conserved and the mass center moves at constant velocity. (m A + mB + mC ) v = mA ( v A )1 + mB ( v B )1 + mC ( vC )1 r2 − r1 = vt

(1) (2)

Combining (1) and (2), (m A + mB + mC )( r2 − r1) = [m A ( v A )1 + mB ( v B )1 + mC ( vC )1] t (4800) (−26.6667i + 15.0192 j) = [0 + (1800)(vB )1 j − (1600)(vC )1i](3.4)

Components.

j: 72092 = 6120(vB )1, i:

(vB )1 = 11.78 m/s,

− 128000 = −5440 (vC )1,

(vC )1 = 23.53 m/s,

vB = 42.4 km/h  vC = 84.7 km/h 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 997

PROBLEM 14.112 The nozzle shown discharges water at the rate of 200 gal/min. Knowing that at both B and C the stream of water moves with a velocity of magnitude 100 ft/s, and neglecting the weight of the vane, determine the force-couple system which must be applied at A to hold the vane in place (1 ft 3 = 7.48 gal).

SOLUTION Q=

200 gal/min (7.48 gal/ft 3 )(60 s/min)

= 0.44563 ft 3 /s dm γ Q = dt g (62.4 lb/ft 3 )(0.44563 ft 3 /s) 32.2 ft/s 2 = 0.8636 lb ⋅ s/ft vB = (100 ft/s) j vC = (100 ft/s)(sin 40° i + cos 40° j) =

Apply the impulse-momentum principle.

x components:

0 + Ax (Δt ) = (Δm)(100 sin 40°)

Δm (100sin 40°) Δt = (0.8636)(100 sin 40°)

Ax =

y components:

Ax = 55.5 lb

(Δm)(100) + Ay ( Δt ) = (Δm)(100 cos 40°) Δm (100)(cos 40° − 1) Δt = (0.8636)(100)(cos 40° − 1) = −20.2 lb

Ay =

Ay = 20.2 lb

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 998

PROBLEM 14.112 (Continued)

Moments about A:

 3  9    (Δm)(100) + M A (Δt ) =   (Δm)(100 cos 40°)  12   12   15  −   (Δm)(100 sin 40°)  12   Δm  MA =   (75 cos 40° − 125 sin 40° − 25)  Δt  = (0.8636)(−47.895) = −41.36 lb ⋅ ft MA = 41.4 lb ⋅ ft A = 59.1 lb





20.0° 

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PROBLEM 14.113 Prior to takeoff the pilot of a 6000-lb twin-engine airplane tests the reversible-pitch propellers with the brakes at Point B locked. Knowing that the velocity of the air in the two 6.6-ft-diameter slipstreams is 60 ft/s and that Point G is the center of gravity of the airplane, determine the reactions at Points A and B. Assume γ = 0.075 lb/ft 3 and neglect the approach velocity of the air.

SOLUTION Let F be the force exerted on the slipstream of one engine. F =

Calculation of

dm . dt

dm (vB − v A ) dt

mass = density × volume = density × area × length Δm = ρ AB (Δl ) = ρ ABvB (Δt ) = Δm γ AB vB = Δt g

γ AB vB (Δt ) g

dm γ  π 2  =  D  vB dt g 4 

or

Assume that v A , the velocity far upstream, is negligible. F =

γ π

g  4

  0.075  π  2 2 D 2  vB ( vB − 0 ) =   4  (6.6) (60) = 286.87 lb 32.2    

The force exerted by two slipstreams on the airplane is −2F .

−2F = 573.74 lb

Statics. Σ M B = 0: 0.9 W − 4.8 ( −2F ) − 9.3 A = 0

A=

1 [(0.9)(6000) − (4.8)(573.74)] 9.3

= 284.5 lb

A = 285 lb 

Fx = 0: − 2F − Bx = 0 Bx = −2 F = 573.74 lb

Σ Fy = 0:

A + B y − W = 284.5 + B y − (6000) = 0 B y = 5715.5 lb

B = [573.74 lb

] + [5715.5 lb ]

B = 5740 lb

84.3° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1000

PROBLEM 14.114 A railroad car of length L and mass m0 when empty is moving freely on a horizontal track while being loaded with sand from a stationary chute at a rate dm/dt = q. Knowing that the car was approaching the chute at a speed v0 , determine (a) the mass of the car and its load after the car has cleared the chute, (b) the speed of the car at that time.

SOLUTION Consider the conservation of the horizontal component of momentum of the railroad car of mass m0 and the sand mass qt.

m0 v0 = (m0 + qt )v

v=

m0 v0 m0 + qt

(1)

m0 v0 dx =v= dt m0 + qt x0 = 0

Integrating, using L= =



tL 0

vdt =



tL 0

and

x = L when t = t L ,

m0 v0 mv dt = 0 0 [ln ( m0 + qt L ) − ln m0 ] m0 + qt q

m0 v0 m0 + qt L ln q m0 ln

m0 + qt L qL = m0 m0 v0

(a)

Final mass of railroad car and sand

(b)

Using Eq. (1),

m0 + qt L = e qL/m0v0 m0 m0 + qt L = m0 eqL/m0v0 

vL =

m0 v0 mv = 0 0 e − qL/m0v0 m0 + qt L m0 vL = v0 e− qL/m0v0 

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PROBLEM 14.115 A garden sprinkler has four rotating arms, each of which consists of two horizontal straight sections of pipe forming an angle of 120° with each other. Each arm discharges water at a rate of 20 L/min with a velocity of 18 m/s relative to the arm. Knowing that the friction between the moving and stationary parts of the sprinkler is equivalent to a couple of magnitude M = 0.375 N ⋅ m, determine the constant rate at which the sprinkler rotates.

SOLUTION The flow through each arm is 20 L/min. Q=

20 L/min 1min × = 333.33 × 10−6 m3 /s 3 60 s 1000 L/m

dm = ρ Q = (1000 kg/m3 )(333.33 × 10−6 ) dt = 0.33333 kg/s

Consider the moment about O exerted on the fluid stream of one arm. Apply the impulse-momentum principle. Compute moments about O. First, consider the geometry of triangle OAB. Using first the law of cosines, (OA) 2 = 1502 + 1002 − (2)(150)(100) cos120° OA = 217.95 mm = 0.21795 m Law of sines:

sin β sin120° = 100 217.95

β = 23.413°, α = 60° − β = 36.587° Moments about O: (Δm)(vO )(0) + M O (Δt ) = (OA)( Δm)vs sin α − (OA)( Δm)(OA)ω

Δm [(OA)vs sin α − (OA) 2 ω ] Δt = (0.33333)[(0.21795)(18)sin 36.587° − (0.21795)2 ω ] = 0.77945 − 0.015834ω

MO =

Moment that the stream exerts on the arm is −M O . Balance of the friction couple and the four streams MF − 4 M O = 0 0.375 − 4(0.77945 − 0.015834ω ) = 0 ω = 43.305 rad/s

ω = 414 rpm 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1002

PROBLEM 14.116 A chain of length l and mass m falls through a small hole in a plate. Initially, when y is very small, the chain is at rest. In each case shown, determine (a) the acceleration of the first link A as a function of y, (b) the velocity of the chain as the last link passes through the hole. In case 1, assume that the individual links are at rest until they fall through the hole; in case 2, assume that at any instant all links have the same speed. Ignore the effect of friction.

SOLUTION Let ρ be the mass per unit length of chain. Assume that the weight of any chain above the hole is supported by the floor. It and the corresponding upward reaction of the floor are not shown in the diagrams. Case 1: Apply the impulse-momentum principle to the entire chain.

ρ yv + ρ gy Δt = ρ ( y + Δy )(v + Δv) = ρ yv + ρ ( Δy )v + ρ y (Δv) + ρ (Δy )( Δv) ρ gy = ρ

dy dv v+ ρy dt dt d = ρ ( yv) dt

ρ gy = ρ

Let Δt → 0.

Multiply both sides by yv. Let v =

dy on left hand side. dt

Integrate with respect to time.

Δy Δv (Δy )( Δv) v+ ρy +ρ Δt Δt Δt

ρ gy 2 v = ρ yv

d ( yv) dt

ρ gy 2

dy d = ρ yv ( yv) dt dt





ρ g y 2 dy = ρ ( yv) d ( yv) 1 1 ρ gy 3 = ρ ( yv) 2 3 2

Differentiate with respect to time.

2v

or

v2 =

2 gy 3

(1)

dv 2 dy 2 = g = gv dt 3 dt 3

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PROBLEM 14.116 (Continued)

a=

(a) (b)

Set y = l in Eq. (1).

v2 =

dv 1 = g dt 3

a = 0.333g 

2 gl 3

v = 0.817 gl



Case 2: Apply conservation of energy using the floor as the level from which the potential energy is measured. T1 = 0

V1 = 0

1 2 y mv V2 = − ρ gy 2 2 T1 + V1 = T2 + V2 T2 =

0=

Differentiating with respect to y, (a)

Acceleration:

(b)

Setting y = l in Eq. (2),

2v

1 2 1 mv − ρ gy 2 2 2

v2 =

ρ gy 2 m

=

gy 2 l

(2)

dv 2 gy = dy l a=v

dv gy = dy l

v 2 = gl

gy l



v = gl



a=

Note: The impulse-momentum principle may be used to obtain the force that the edge of the hole exerts on the chain.

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CHAPTER 15

PROBLEM 15.CQ1 A rectangular plate swings from arms of equal length as shown below. What is the magnitude of the angular velocity of the plate? (a) 0 rad/s (b) 1 rad/s (c) 2 rad/s (d ) 3 rad/s (e) Need to know the location of the center of gravity

SOLUTION Answer: (a) 

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PROBLEM 15.CQ2 Knowing that wheel A rotates with a constant angular velocity and that no slipping occurs between ring C and wheel A and wheel B, which of the following statements concerning the angular speeds are true? (a) ωa = ωb (b) ωa > ωb (c) ωa < ωb (d ) ωa = ωc (e) the contact points between A and C have the same acceleration

SOLUTION Answer: (b) 

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PROBLEM 15.1 The brake drum is attached to a larger flywheel that is not shown. The motion of the brake drum is defined by the relation θ = 36t − 1.6t 2 , where θ is expressed in radians and t in seconds. Determine (a) the angular velocity at t = 2 s, (b) the number of revolutions executed by the brake drum before coming to rest.

SOLUTION Given:

θ = 36t − 1.6t 2

radians

Differentiate to obtain the angular velocity.

ω=

dθ = 36 − 3.2t dt

(a)

At t = 2 s,

ω = 36 − (3.2)(2)

(b)

When the rotor stops,

ω = 0. 0 = 36 − 3.2t

rad/s

ω = 29.6 rad/s  t = 11.25 s

θ = (36)(11.25) − (1.6)(11.25) 2 = 202.5 radians In revolutions,

θ=

202.5 2π

θ = 32.2 rev 

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PROBLEM 15.2 The motion of an oscillating crank is defined by the relation θ = θ0 sin (π t /T ) − (0.5θ0 )sin (2 π t /T ), where θ is expressed in radians and t in seconds. Knowing that θ 0 = 6 rad and T = 4 s, determine the angular coordinate, the angular velocity, and the angular acceleration of the crank when (a) t = 0, (b) t = 2 s.

SOLUTION

ω=

dθ π 2π  πt   2π t  cos  = θ0 cos   − 0.5θ 0  dt T T T    T 

α=

dω π   πt   2π   2π t  sin  = −θ0   sin   + 0.5θ0    dt T  T   T   T 

2

(a)

2

θ =0 

t = 0:

ω=6

π 4

− 0.5(6)

2π 4

ω=0  α =0 

(b)

t = 2 s:  2π  4

θ = 6 sin 

  4π  − 0.5(6) sin  4  

 =6−0 

θ = 6.00 rad 

2π π   2π   4π  ω = 6   cos   − 0.5(6) cos   4 4 4      4  π 2π =6 =

4

(0) − 0.5(6)

4

(−1)

6π 4

ω = 4.71 rad/s  2

2

π   2π   2π   4π   sin   + 0.5(6)   sin   4 4 4        4 

α = −6 

2

2

π   2π  = −6   (1) + 3   (0) 4  4  3 = − π2 8



α = −3.70 rad/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1010

PROBLEM 15.3 The motion of a disk rotating in an oil bath is defined by the relation θ = θ0 (1 − e −t/ 4 ), where θ is expressed in radians and t in seconds. Knowing that θ 0 = 0.40 rad, determine the angular coordinate, velocity, and acceleration of the disk when (a) t = 0, (b) t = 3 s, (c) t = ∞.

SOLUTION

θ = 0.40(1 − e −t/4 ) dθ 1 ω= = (0.40)e− t/4 = 0.10e −t/ 4 dt 4 dω 1 α= = − (0.10)e −t/ 4 = −0.025e −t/ 4 dt 4 (a)

t = 0:

θ = 0.40(1 − e0 ) ω = 0.10e0

ω = 0.1000 rad/s 

α = −0.025e0 (b)

θ =0 

α = −0.0250 rad/s 2 

t = 3 s:

θ = 0.40(1 − e−3/ 4 ) = 0.40(1 − 0.4724)

ω = 0.10e−3/ 4 



= 0.10(0.4724) 



ω = 0.0472 rad/s 

α = −0.025e−3/4 



= −0.025(0.4724) 

 (c)

θ = 0.211 rad 

α = −0.01181 rad/s 2 

t = ∞:

θ = 0.40(1 − e −∞ ) = 0.40(1 − 0)

θ = 0.400 rad 

ω = 0.10e−∞

ω=0 

α = −0.025e−∞

α =0 

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PROBLEM 15.4 The rotor of a gas turbine is rotating at a speed of 6900 rpm when the turbine is shut down. It is observed that 4 min is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine (a) the angular acceleration, (b) the number of revolutions that the rotor executes before coming to rest.

SOLUTION

ω 0 = 6900 rpm = 722.57 rad/s t = 4 min = 240 s (a)

ω = ω 0 + α t; 0 = 722.57 + α (240) α = −3.0107 rad/s

(b)

α = −3.01 rad/s 2 

1 1 2 2 θ = 173, 416 − 86,708 = 86, 708 rad

θ = ω 0 t + α t 2 = (722.57)(240) + (−3.0107)(240) 2

 1 rev    2π rad 

θ = 86, 708 rad 

θ = 13,80 rev 

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PROBLEM 15.5 A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest.

SOLUTION For uniformly accelerated motion,

(a)

(1)

1 θ = θ 0 + ω0 t + α t 2 2

(2)

Data for start up:

θ 0 = 0, ω0 = 0,

At t = 5 s,

ω = 3600 rpm =

From Eq. (1),

(b)

ω = ω0 + α t

120π = α (5)

2π (3600) = 120π rad/s 60

α = 24π rad/s 2

From Eq. (2),

1 θ = 0 + 0 + (24π )(5)2 = 300π radians 2

In revolutions,

θ=

300π 2π

θ = 150 rev 

Data for coasting to rest:

θ 0 = 0, ω0 = 120π rad/s At t = 70 s,

ω=0

From Eq. (1),

0 = 120π − α (70)

α=

From Eq. (2),

θ = 0 + (120π )(70) −

In revolutions,

θ=

120π rad/s 70

(120π )(70) 2 = 4200π radians 2(70)

4200π 2π

θ = 2100 rev 

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PROBLEM 15.6 A connecting rod is supported by a knife-edge at Point A. For small oscillations the angular acceleration of the connecting rod is governed by the relation α = −6θ where α is expressed in rad/s2 and θ in radians. Knowing that the connecting rod is released from rest when θ = 20°, determine (a) the maximum angular velocity, (b) the angular position when t = 2 s.

SOLUTION Angular motion relations: dω ω dω = = −6θ dt dθ

α=

(1)

Separation of variables ω and θ gives

ω d ω = −6θ dθ Integrating, using ω = 0 when θ = θ 0 ,



ω 0

ω dω = −6

θ

θ θ dθ 0

1 2 ω = −3(θ 2 − θ02 ) = 3(θ02 − θ 2 ) 2

ω 2 = 6(θ02 − θ 2 ) (a)

ω = 6(θ02 − θ 2 )

ω is maximum when θ = 0. θ 0 = 20° = 0.34907 radians

Data:

2 ωmax = 6(0.34907 2 − 0) = 0.73108 rad 2 /s

(b)

From ω =

dθ we get dt

dt =



ω

=

1

ωmax = 0.855 rad/s 



6 θ02 − θ 2

Integrating, using t = 0 when θ = θ 0 ,



t

0

dt =

t=−



θ

1

6 θ 1 6

0

θ02

cos

2 −1

−θ 2

θ θ0

θ

=− θ0

1  1 θ −1 θ  cos −1 0 − cos =− θ0  θ0 6 6

θ = θ 0 cos( 6t ) = 0.34907 cos[( 6)2] = (0.34907)(0.18551) = 0.064756 radians θ = 3.71°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1014

PROBLEM 15.7 When studying whiplash resulting from rear end collisions, the rotation of the head is of primary interest. An impact test was performed, and it was found that the angular acceleration of the head is defined by the relation α = 700cosθ + 70sin θ where α is expressed in rad/s2 and θ in radians. Knowing that the head is initially at rest, determine the angular velocity of the head when θ = 30°.

SOLUTION Angular motion relations:

α=

dω ω dω = = 700 cos θ + 70sin θ dt dθ

Separating variables ω and θ gives

ω d ω = (700 cos θ + 70sin θ )dθ Integrating, using ω = 0 when θ = 0,



ω 0

ωdω =



θ 0

(700 cos θ + 70sin θ ) dθ θ

1 2 ω = (700sin θ − 70cos θ ) 2 0 = 700sin θ + 70(1 − cos θ )

ω = 1400sin θ + 140(1 − cos θ ) Data:

θ = 30° =

π 6

rad

With calculator set to “degrees” for trigonometric functions,

ω = 1400sin 30° + 140(1 − cos 30°) = 26.8 rad/s ω = 26.8 rad/s  With calculator set to “radians” for trigonometric functions,

ω = 1400sin(π /6) + 140(1 − cos(π /6)) = 26.8 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1015

PROBLEM 15.8 The angular acceleration of an oscillating disk is defined by the relation α = −kθ . Determine (a) the value of k for which ω = 8 rad/s when θ = 0 and θ = 4 rad when ω = 0, (b) the angular velocity of the disk when θ = 3 rad.

SOLUTION

α = −kθ dω ω= = −kθ dθ ω d ω = −kθ dθ (a)



0 8 rad/s

ω dω = −



0

4 rad 0

kθ dθ ;

4

1 2 1 ω = − kθ 2 2 2 8 0

1 1 (0 − 82 ) = − k (42 − 0) 2 2

(b)



ω

ω dω = −

8 rad/s



3 rad 0

kθ dθ ;

k = 4.00 s −2  ω

1 2 1 ω = − (4 s −1 )θ 2 2 2 0

1 2 1 (ω − 82 ) = − (4)(32 − 0) 2 2 2 ω − 64 = −36; ω 2 = 64 − 36 = 28

3

0

ω = 5.29 rad/s 

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PROBLEM 15.9 The angular acceleration of a shaft is defined by the relation α = −0.25ω , where α is expressed in rad/s2 and ω in rad/s. Knowing that at t = 0 the angular velocity of the shaft is 20 rad/s, determine (a) the number of revolutions the shaft will execute before coming to rest, (b) the time required for the shaft to come to rest, (c) the time required for the angular velocity of the shaft to be reduced to 1 percent of its initial value.

SOLUTION

α = −0.25ω dω ω = −0.25ω dθ d ω = −0.25dθ



(a)

0 20 rad/s

d ω = −0.25



θ 0

θ = (80 rad)

rev 2π rad



ω 20 rad/s



ω

= −0.25

t=−

For ω = 0 For Use Eq. (1):



t

θ = 12.73 rev 

dω = −0.25ω ; dt

α = −0.25ω ;

(b)

(c)

dθ ; (0 − 20) = −0.25θ ; θ = 80 rad

dt

0



ω

= −0.25dt

|ln ω |ω20 = −0.25t

1 (ln ω − ln 20) = 4(ln 20 − ln ω ) 0.25

t = 4 ln

20

t = 4 ln

20 = 4 ln ∞ 0

(1)

ω

t =∞ 

ω = 0.01ω 0 = 0.01(20) = 0.2 rad  20  t = 4 ln   = 4 ln100 = 4(4.605)  0.2 

t = 18.42 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1017

PROBLEM 15.10 The bent rod ABCDE rotates about a line joining Points A and E with a constant angular velocity of 9 rad/s. Knowing that the rotation is clockwise as viewed from E, determine the velocity and acceleration of corner C.

SOLUTION EA2 = 0.42 + 0.42 + 0.22 EA = 0.6 m rC/E = −(0.4 m)i + (0.15 m) j  EA = −(0.4 m)i + (0.4 m) j + (0.2 m)k  1 1 EA λ EA = = (−0.4i + 0.4 j + 0.2k ) = (−2i + 2 j + k ) 3 EA 0.6 1 ω = ω AE λ EA = (9 rad/s) (−2i + 2 j + k ) 3 ω = −(6 rad/s)i + (6 rad/s) j + (3 rad/s)k i j k 6 3 = −0.45i − 1.2 j + (−0.9 + 2.4)k vC = ω × rC/E = −6 −0.4 0.15 0 vC = −(0.45 m/s)i − (1.2 m/s) j + (1.5 m/s)k  aC = α AC × rC/E + ω × (ω × rC/E ) = α AC × rC/E + ω × vC i

j

k

aC = 0 +

−6 6 3 −0.45 −1.2 1.5 = (9 + 3.6)i + ( −1.35 + 9) j + (7.2 + 2.7)k aC = (12.60 m/s 2 )i + (7.65 m/s 2 ) j + (9.90 m/s 2 )k 

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PROBLEM 15.11 In Problem 15.10, determine the velocity and acceleration of corner B, assuming that the angular velocity is 9 rad/s and increases at the rate of 45 rad/s 2 . PROBLEM 15.10 The bent rod ABCDE rotates about a line joining Points A and E with a constant angular velocity of 9 rad/s. Knowing that the rotation is clockwise as viewed from E, determine the velocity and acceleration of corner C.

SOLUTION EA2 = 0.42 + 0.42 + 0.22 EA = 0.6 m rB/A = −(0.25 m) j  EA = −(0.4 m)i + (0.4 m) j + (0.2 m)k  EA  −0.4i + 0.4 j + 0.2k  1 λ EA = =  = ( −2i + 2 j + k ) EA  0.6  3 1 ω = ω AE λ EA = (9 rad/s) (−2 i + 2 j + k ) 3 ω = −(6 rad/s)i + (6 rad/s) j + (3 rad/s)k v B = ω × rB/A = (−6i + 6 j + 3k ) × (−0.25) j = 1.5k + 0.75i v B = (0.75 m/s)i + (1.5 m/s)k 

1 α = α AE λ EA = (45 rad/s 2 ) (−2i + 2 j + k ) 3 2 α = −(30 rad/s )i + (30 rad/s 2 ) j + (15 rad/s 2 )k a B = α × rB/A + ω × (ω × rB/A ) = α × rB/A + ω × v B i j k i j k 30 15 + −6 6 3 a B = −30 0 −0.25 0 0.75 0 1.5 = 3.75i + 7.5k + 9i + (2.25 + 9) j − 4.5k a B = (12.75 m/s 2 )i + (11.25 m/s 2 ) j + (3 m/s 2 )k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1019

PROBLEM 15.12 The assembly shown consists of the straight rod ABC which passes through and is welded to the rectangular plate DEFH. The assembly rotates about the axis AC with a constant angular velocity of 9 rad/s. Knowing that the motion when viewed from C is counterclockwise, determine the velocity and acceleration of corner F.

SOLUTION  AC = (14 in.)i − (8 in.) j + (8 in.)k AC = 18 in.  AC 14i − 8 j + 8k λ AC = = = (0.77778i − 0.44444 j + 0.44444k ) AC 18 ω = ωλ AC = (9 rad/s)(0.77778i − 0.44444 j + 0.44444k )

ω = (7 rad/s)i − (4 rad/s) j + (4 rad/s)k

Corner F:

α=0

rF /B = ( −7 in.)i + (4 in.)k = −(0.58333 ft)i + (0.33333 ft)k v F = ω × rF /B i j k = 7 −4 4 −0.58333 0 0.33333 = −1.3333i − 4.6667 j − 2.3333k v F = −(1.333 ft/s)i − (4.67 ft/s)j − (2.33 ft/s)k 

α =0 a F = α × rF /B + ω × (ω × rF /B ) = 0 + ω × vF aF = ω × v F =

i j k 7 −4 4 −1.3333 −4.6667 −2.3333

= (28.0)i + (11.0) j + (−38.0)k a F = (28.0 ft/s 2 )i + (11.00 ft/s 2 ) j − (38.0 ft/s 2 )k 

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PROBLEM 15.13 In Problem 15.12, determine the acceleration of corner H, assuming that the angular velocity is 9 rad/s and decreases at a rate of 18 rad/s 2 . PROBLEM 15.12 The assembly shown consists of the straight rod ABC which passes through and is welded to the rectangular plate DEFH. The assembly rotates about the axis AC with a constant angular velocity of 9 rad/s. Knowing that the motion when viewed from C is counterclockwise, determine the velocity and acceleration of corner F.

SOLUTION  AC = (14 in.)i − (8 in.) j + (8 in.)k AC = 18 in.  AC 14i − 8 j + 8k = = (0.77778i − 0.44444 j + 0.44444k ) λ AC = AC 18 ω = ωλ AC = (9 rad/s)(0.77778i − 0.44444 j + 0.44444k ) ω = (7 rad/s)i − (4 rad/s) j + (4 rad/s)k

α = −18 rad/s 2 ;

α = αλ AC = ( −18 rad/s 2 )(0.77778i − 0.44444 j + 0.44444k )

α = −(14 rad/s 2 )i + (8 rad/s 2 ) j − (8 rad/s 2 )k

Corner H:

rH /B = (7 in.)i + (4 in.)k = (0.58333 ft)i + (0.33333 ft)k i j k = 7 −4 4 0.58333 0 0.33333 = −1.3333i + 2.3333k

v H = −(1.333 ft/s)i + (2.33 ft/s)k 

a H = α × rH /B + ω × (ω × rH /B ) = α × rH /B + ω × v H i j k i j k aH = −14 8 −8 + 7 −4 +4 0.58333 0 0.33333 −1.3333 0 2.3333 = 2.6667i + 0 j − 4.46667k − 9.3333i − 21.667 j − 5.3333k a H = −(6.67 ft/s 2 )i − (21.7 ft/s 2 ) j − (10.00 ft/s 2 )k 

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PROBLEM 15.14 A circular plate of 120 mm radius is supported by two bearings A and B as shown. The plate rotates about the rod joining A and B with a constant angular velocity of 26 rad/s. Knowing that, at the instant considered, the velocity of Point C is directed to the right, determine the velocity and acceleration of Point E.

SOLUTION  BA = −(100 mm) j + (240 mm)k BA = 260 mm  BA −(100) j + (240)k = , λBA = α =0 BA 260  1  ω = ωλBA = 26   (−(100) j + (240)k )  260 

Point E:

rE /A = (120 mm)i − (80 mm) j − (120 mm)k v E = ω × rE /A i j k = 0 −10 24 120 −80 −120 = (3120 mm/s)i + (2880 mm/s) j + (1200 mm/s)k v E = (3.12 m/s)i + (2.88 m/s) j + (1.200 m/s)k  a E = α × rE/A + ω × (ω × rE/A ) = 0 + ω × v B aE = ω × vB i j k = 0 −10 24 3120 2880 1200 = −(81120 mm/s 2 )i + (74880 mm/s 2 ) j + (31200 mm/s 2 ) a E = −(81.1 m/s 2 )i + (74.9 m/s 2 ) j + (31.2 m/s 2 )k 

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PROBLEM 15.15 In Problem 15.14, determine the velocity and acceleration of Point E, assuming that the angular velocity is 26 rad/s and increases at the rate of 65 rad/s2.

SOLUTION See Problem 15.14 for λ BA and ω −(100) j + (240)k 260 ω = (10 rad/s)j = (24 rad/s)k

λ BA =

α = +65 rad/s 2 ;

 1  α = αλBA = (65 rad/s 2 )   − (100) j + (240)k  260  α = −(25 rad/s 2 ) j + (60 rad/s 2 )k

Point E:

rE /A = (120 mm)i − (80) j − (120)k v E = ω × rE /A i j k = 0 −10 24 120 −80 −120 = (3120 mm/s)i + (2880 mm/s) j + (1200 mm/s)k v E = (3.12 m/s)i + (2.88 m/s) j + (1.200 m/s)k 

a D = α × rE/A + ω × (ω × rE/A ) = α × rE/A + ω × v E i j k i j k a D = 0 −25 60 + 0 −10 24 120 −80 −120 3120 2880 1200 = −(7800 mm/s 2 )i + (7200 mm/s 2 ) j + (3000 mm/s 2 )k − (81120 mm/s 2 )i + (74880 mm/s 2 ) j + (31200 mm/s 2 )k a B = −(73320 mm/s 2 )i + (82080 mm/s 2 ) j + (34200 mm/s 2 )k = −(73.3 m/s 2 )i + (82.1 m/s 2 ) j + (34.2 m/s 2 )k



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PROBLEM 15.16 The earth makes one complete revolution around the sun in 365.24 days. Assuming that the orbit of the earth is circular and has a radius of 93,000,000 mi, determine the velocity and acceleration of the earth.

SOLUTION

ω=

2π rad (365.24 days)

( )( 24 h day

3600 s h

)

= 199.11 × 10−9 rad/s v = rω  5280 ft  (199.11 × 10−9 rad/s) = (93 × 106 mi)    mi  v = 97, 770 ft/s

v = 66, 700 mi/h 

a = rω 2 = (93 × 106 )(5280)(199.11 × 10−9 rad/s)2

a = 19.47 × 10−3 ft/s 2 

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PROBLEM 15.17 The earth makes one complete revolution on its axis in 23 h 56 min. Knowing that the mean radius of the earth is 3960 mi, determine the linear velocity and acceleration of a point on the surface of the earth (a) at the equator, (b) at Philadelphia, latitude 40° north, (c) at the North Pole.

SOLUTION 23 h 56 m = 23.933 h 2π rad ω= s (23.933 h) ( 3600 h ) = 72.925 × 10−6 rad/s  5280 ft  R = (3960 mi)    mi  = 20.91 × 106 ft r = radius of path = R cos φ

(a)

Equator: Latitude = φ = 0 v = rω = R(cos 0)ω

= (20.91 × 106 ft)(1)(72.925 × 10−6 rad/s)

v = 1525 ft/s 

a = rω 2

= R (cos 0)ω 2 = (20.91 × 106 ft)(1)(72.925 × 10−6 rad/s) 2

(b)

a = 0.1112 ft/s 2 

Philadelphia: Latitude = φ = 40° v = rω = R(cos 40°)ω

= (20.91 × 106 ft)( cos 40°)(72.925 × 10−6 rad/s)

v = 1168 ft/s 

a = rω 2

= R (cos 40°)ω 2 = (20.91 × 106 ft)(cos 40°)(72.925 × 10−6 rad/s) 2

(c)

a = 0.0852 ft/s 2 

North Pole: Latitude = φ = 0 r = R cos 0 = 0

v = a = 0 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1025

PROBLEM 15.18 A series of small machine components being moved by a conveyor belt pass over a 120 mm radius idler pulley. At the instant shown, the velocity of Point A is 300 mm/s to the left and its acceleration is 180 mm/s2 to the right. Determine (a) the angular velocity and angular acceleration of the idler pulley, (b) the total acceleration of the machine component at B.

SOLUTION vB = v A = 300 mm/s

rB = 120 mm

(aB )t = a A = 180 mm/s

(a)

(b)

vB = ω rB ,

ω=

vB 300 = = 2.5 rad/s rB 120

(aB )t = α rB ,

α=

(aB )t 180 = = 1.5 rad/s rB 120

ω = 2.50 rad/s



α = 1.500 rad/s 2



(aB ) n = rBω 2 = (120)(2.5) 2 = 750 mm/s 2 aB = (aB )t2 + (aB ) n2 = (180) 2 + (750) 2 = 771 mm/s 2

tan β =

750 , 180

β = 76.5° a B = 771 mm/s 2

76.5° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1026

PROBLEM 15.19 A series of small machine components being moved by a conveyor belt pass over a 120-mm-radius idler pulley. At the instant shown, the angular velocity of the idler pulley is 4 rad/s clockwise. Determine the angular acceleration of the pulley for which the magnitude of the total acceleration of the machine component at B is 2400 mm/s 2.

SOLUTION

ω B = 4 rad/s ,

rB = 120 mm

(aB ) n = rBωB2 = (120)(4) 2 = 1920 mm/s 2 aB = 2400 mm/s 2

(a B )t =

aB2 − (aB )2n =

(aB )t = rBα ,

α =

24002 − 19202 = ±1440 mm/s 2

(a B )t ±1440 = = ±12 rad/s 2 120 rB 12.00 rad/s 2

or



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1027

PROBLEM 15.20 The belt sander shown is initially at rest. If the driving drum B has a constant angular acceleration of 120 rad/s2 counterclockwise, determine the magnitude of the acceleration of the belt at Point C when (a) t = 0.5 s, (b) t = 2 s.

SOLUTION at = rα = (0.025 m)(120 rad/s 2 )

at = 3 m/s 2

(a)

t = 0.5 s:

ω = α t = (120 rad/s 2 )(0.5 s) = 60 rad/s an = rω 2 = (0.025 m)(60 rad/s)2

a n = 90 m/s 2 aB2 = at2 + an2 = 32 + 902

(b)

t = 2 s:

aB = 90.05 m/s 2 

ω = α t = (120 rad/s 2 )(2 s) = 240 rad/s an = rω 2 = (0.025 m)(240 rad/s) 2 an = 1440 m/s 2 aB2 = at2 + an2 = 32 + 14402

aB = 1440 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1028

PROBLEM 15.21 The rated speed of drum B of the belt sander shown is 2400 rpm. When the power is turned off, it is observed that the sander coasts from its rated speed to rest in 10 s. Assuming uniformly decelerated motion, determine the velocity and acceleration of Point C of the belt, (a) immediately before the power is turned off, (b) 9 s later.

SOLUTION

ω0 = 2400 rpm = 251.3 rad/s r = 0.025 m vC = rω = (0.025 m)(251.3 rad/s)

(a)

aC = rω 2 = (0.025 m)(251.3 rad/s) 2

(b)

When t = 10 s:

vC = 6.28 m/s  aC = 1579 m/s 2 

ω = 0. ω = ω0 + α t 0 = 251.3 rad/s + α (10 s)

α = −25.13 rad/s 2 When t = 9 s:

ω = ω0 + α t ω9 = 251.3 rad/s − (25.13 rad/s 2 )(9 s) = 25.13 rad/s vC = rω9

= (0.025 m)(25.13 rad/s)

v9 = 0.628 m/s 

(aC )t = rα n

= (0.025 m)( −25.13 rad/s 2 ) (aC )t = 0.628 m/s 2 (aC ) n = rω92

= (0.025 m)(25.13 rad/s)2 (aC ) n = 15.79 m/s 2

aC2 = ( aC )t2 + ( aC ) n2

= (0.628 m/s 2 ) 2 + (15.79 m/s 2 )2

aC = 15.80 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1029

PROBLEM 15.22 The two pulleys shown may be operated with the V belt in any of three positions. If the angular acceleration of shaft A is 6 rad/s2 and if the system is initially at rest, determine the time required for shaft B to reach a speed of 400 rpm with the belt in each of the three positions.

SOLUTION Angular velocity of shaft A: Belt speed: Angular speed of shaft B: Solving for t, Data: α A = 6 rad/s,

ω A = α At v = rAω A = rBωB

ωB = t=

v rAα At = rB rB rBωB rAα A

ωB = 400 rpm = 41.889 rad/s t=

rB 41.889 r d ⋅ = 6.9813 B = 6.9813 B rA rA dA 6

Belt at left:

d B 2 in. = d A 4 in.

t = 3.49 s 

Belt in middle:

d B 3 in. = d A 3 in.

t = 6.98 s 

Belt at right:

d B 4 in. = d A 2 in.

t = 13.96 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1030

PROBLEM 15.23 Three belts move over two pulleys without slipping in the speed reduction system shown. At the instant shown the velocity of Point A on the input belt is 2 ft/s to the right, decreasing at the rate of 6 ft/s2. Determine, at this instant, (a) the velocity and acceleration of Point C on the output belt, (b) the acceleration of Point B on the output pulley.

SOLUTION Left pulley. Inner radius

r1 = 2 in.

Outer radius

r2 = 4 in.

v A = 2 ft/s (a A )t = −6 ft/s 2 = 6 ft/s 2

ω 1=

vA 2 = = 6 rad/s r2 124

α1 =

( a A )t 6 = 4 = 18 rad/s 2 r2 12

Intermediate belt.  2 v1 = r1ω1 =   (6) = 1 ft/s  12   2 (a1)t = r1α1 =   (18) = 3 ft/s 2  12 

Right pulley. Inner radius

r3 = 2 in.

Outer radius

r4 = 4 in.

ω2 =

v1 1 = = 3 rad/s r4 ( 124 )

α2 =

( a1 )t 3 = 4 = 9 rad/s 2 r4 ( 12 )

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1031

PROBLEM 15.23 (Continued)

(a)

(b)

Velocity and acceleration of Point C.  2 vC = r3ω2 =   (3) = 0.5 ft/s  12 

vC = 0.5 ft/s



 2 (aC )t = r3α 2 =   (9) = 1.5 ft/s 2  12 

aC = 1.5 ft/s 2



Acceleration of Point B.  4 (aB ) n = r4ω22 =   (3) 2 = 3 ft/s 2  12 

(a B ) n = 3 ft/s 2

 4 (aB )t = r4α 2 =   (9) = 3 ft/s 2  12 

(a B )t = 3 ft/s 2

a B = 4.24 ft/s 2

45° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1032



PROBLEM 15.24 A gear reduction system consists of three gears A, B, and C. Knowing that gear A rotates clockwise with a constant angular velocity ω A = 600 rpm, determine (a) the angular velocities of gears B and C, (b) the accelerations of the points on gears B and C which are in contact.

SOLUTION

ω A = 600 rpm =

(a)

(600)(2π ) = 20π rad/s. 60

Let Points A, B, and C lie at the axles of gears A, B, and C, respectively. Let D be the contact point between gears A and B. vD = rD/ Aω A = (2)(20π ) = 40π in./s

ωB =

vD 40π 60 = = 10π rad/s = 10π ⋅ = 300 rpm 4 2π rD/B

ω B = 300 rpm



Let E be the contact point between gears B and C. vE = rE/BωB = (2)(10π ) = 20 π in./s

ωC =

vE 20π 60 = = 3.333π rad/s = (3.333π ) = 100 rpm 6 2π rE/C

ωC = 100 rpm



(b) Accelerations at Point E. On gear B :

On gear C :

aB =

aC =

vE2 (20π )2 = = 1973.9 in./s 2 2 rE/B

a B = 1974 in./s 2



aC = 658 in./s 2



vE2 (20π )2 = = 658 in./s 2 6 rE/C

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1033

PROBLEM 15.25 A belt is pulled to the right between cylinders A and B. Knowing that the speed of the belt is a constant 5 ft/s and no slippage occurs, determine (a) the angular velocities of A and B, (b) the accelerations of the points which are in contact with the belt.

SOLUTION (a)

(b)



Angular velocities. Disk A:

ωA =

vP 5 ft/s = rA (4/12) ft

ω A = 15.00 rad/s

Disk B:

ωB =

vP 5 ft/s = rB (8/12) ft

ω B = 7.50 rad/s

 

Accelerations of contact points. Disk A:

a A = ω A2 rA = (15.00 rad/s)2 ((4/12) ft)

a A = 75.00 rad/s 2 

Disk B:

aB = ωB2 rB = (7.50 rad/s) 2 ((8/12) ft)

a B = 37.5 rad/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1034

PROBLEM 15.26 Ring C has an inside radius of 55 mm and an outside radius of 60 mm and is positioned between two wheels A and B, each of 24-mm outside radius. Knowing that wheel A rotates with a constant angular velocity of 300 rpm and that no slipping occurs, determine (a) the angular velocity of the ring C and of wheel B, (b) the acceleration of the Points of A and B which are in contact with C.

SOLUTION  2π    60  = 31.416 rad/s rA = 24 mm

ω A = 300 rpm 

rB = 24 mm r1 = 60 mm r2 = 55 mm

[We assume senses of rotation shown for our computations.] (a)

Velocities: Point 1 (Point of contact of A and C) v1 = rAω A = r1ωC

ωC =

rA ωA r1

24 mm (300 rpm) 60 mm = 120 rpm =

ωC = 120 rpm 

Point 2 (Point of contact of B and C) v2 = rBωB = r2ωC

ωB =

r2 ωC rB

r2  rA   ωA rB  r1  55 mm  24 mm  =   300 rpm 24 mm  60 mm  =

ωB = 275 mm

ωB = 275 rpm 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1035

PROBLEM 15.26 (Continued)

(b)

Accelerations: Point on rim of A:

rA = 24 mm = 0.024 m a A = rAω A2 = (0.024 m)(31.416 rad/s) 2 = 23.687 m/s 2

Point on rim of B:

a A = 23.7 m/s 2 

 2π    60  = 28.798 rad/s

ωB = 275 rpm  aB = rBωB2

= (0.024 m)(28.798 rad/s) 2 = 19.904 m/s 2

a B = 19.90 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1036

PROBLEM 15.27 Ring B has an inside radius r2 and hangs from the horizontal shaft A as shown. Shaft A rotates with a constant angular velocity of 25 rad/s and no slipping occurs. Knowing that r1 = 12 mm, r2 = 30 mm, and r3 = 40 mm, determine (a) the angular velocity of ring B, (b) the accelerations of the points of shaft A and ring B which are in contact, (c) the magnitude of the acceleration of a point on the outside surface of ring B.

SOLUTION Let Point C be the point of contact between the shaft and the ring. vC = r1ω A

ωB = =

vC r2 r1ω A r2

ωB =

On shaft A:

a A = r1ω A2

On ring B:

 rω  aB = r2ωB2 = r2  1 A   r2 

r1ω A r2

a A = r1ω A2 2

aB =

r12ω A2 r2



Acceleration of Point D on outside of ring. r  aD = r3ωB2 = r3  1 ω A   r2 

2

2

r  aD = r3  1  ω A2  r2 

Data:

ω A = 25 rad/s r1 = 12 mm r2 = 30 mm r3 = 40 mm

(a)

r1 ωA r2 12 mm = (25 rad/s) 30 mm

ωB =

ω B = 10 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1037

PROBLEM 15.27 (Continued)

(b)

a A = r1ω A2 = (12 mm)(25 rad/s)2 = 7.5 × 103 mm/s 2

aB = =

r12 r2

a A = 7.50 m/s 2



a B = 3.00 m/s 2



a D = 4.00 m/s 2



ω A2

(12 mm) 2 (25 rad/s)2 (30 mm)

= 3 × 103 mm/s 2 2

(c)

r  aD = r3  1  ω A2  r2  2

 12 mm  2 = (40 mm)   (25 rad/s)  30 mm  aD = 4 × 103 mm/s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1038

PROBLEM 15.28 A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2 ft/s to v1 = 4 ft/s. Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum B, (b) the number of revolutions executed by drum B during the 4-s interval.

SOLUTION Belt motion:

v = v0 + a t 4 ft/s = 2 ft/s + a(4 s) a=

4 ft/s − 2 ft/s = 0.5 ft/s 2 = 6 in./s 2 4s

Since the belt does not slip relative to the periphery of the drum, the tangential accelation at the periphery of the drum is at = 6 in./s 2

(a)

Angular acceleration of drum B.

αB = (b)

at 6 in./s 2 = rB 15 in.

α B = 0.400 rad/s 2



Angular displacement of drum B. At t = 0,

ω0 =

v0 24 in./s = = 1.6 rad/s 15 in. rB

At t = 4 s,

ω1 =

v1 48 in./s = = 3.2 rad/s rB 15 in.

ω12 = ω02 + 2α Bθ B

In revolutions,

θB =

ω12 − ω02 (3.2)2 − (1.6)2 = = 9.6 radians 2α B (2)(0.400)

θB =

9.6 2π

θ B = 1.528 rev 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1039

PROBLEM 15.29 A pulley and two loads are connected by inextensible cords as shown. Load A has a constant acceleration of 300 mm/s2 and an initial velocity of 240 mm/s, both directed upward. Determine (a) the number of revolutions executed by the pulley in 3 s, (b) the velocity and position of load B after 3 s, (c) the acceleration of Point D on the rim of the pulley at t = 0.

SOLUTION

(a)

Motion of pulley:

( v E )0 = (v A )0 = 240 mm/s (a E )t = a A = 300 mm/s 2 (vE )0 = rω0 : 240 mm/s = (120 mm)ω0

ω0 = 2 rad/s

(aE )t = rα : 300 mm 2 /s = (120 mm)α

α = 2.5 rad/s 2

ω = ω0 + α t

For t = 3 s:

= 2 rad/s + (2.5 rad/s 2 )(3 s) = 9.5 rad/s 1 θ = ω0 t + α t 2 2 1 = (2 rad/s)(3 s) + (2.5 rad/s 2 )(3 s) 2 2  1  θ = 17.25 rad θ = 17.25    2π 

(b)

Load B:

θ = 2.75 rev 

r = 180 mm t =3s vB = rω = (0.180 m)(9.5 rad/s) = 1.710 m/s ΔyB = rθ = (0.180 m)(17.25 rad) = 3.105 m

v B = 1.710 m/s



ΔyB = 3.11 m



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PROBLEM 15.29 (Continued)

(c)

Point D:

r = 180 mm

t =0

(a D )t = rα = (180 mm)(2.5 rad/s 2 ) = 450 mm/s 2 (aD )n = rω02 = (180 mm)(2 rad/s) 2 = 720 mm/s 2 (a D )n = 720 mm/s 2

a D = 849 mm/s 2

32.0° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1041

PROBLEM 15.30 A pulley and two loads are connected by inextensible cords as shown. The pulley starts from rest at t = 0 and is accelerated at the uniform rate of 2.4 rad/s2 clockwise. At t = 4 s, determine the velocity and position (a) of load A, (b) of load B.

SOLUTION Uniformly accelerated motion.

ω0 = 0 α = 2.4 rad/s 2 ω = ω0 + α t = 0 + (2.4 rad/s 2 )(4 s)

At t = 4 s:

ω = 9.6 rad/s 2 1 2

θ = θ0 + ω0 t + α t 2 =0+0+

1 (2.4 rad/s 2 )(4 s) 2 2

θ = 19.20 rad (a)

Load A.

At t = 4 s:

rA = 120 mm

vA = rA ω = (120 mm)(9.6 rad/s) = 1152 mm/s

vA = 1.152 m/s



yA = 2.30 m



vB = 1.728 m/s



yA = rA θ = (120 mm)(19.2 rad) = 2304 mm

(b)

Load B.

At t = 4 s:

rB = 180 mm vB = rBω = (180 mm)(9.6 rad/s) = 1728 mm/s yB = rBθ = (180 mm)(19.2 rad) = 3456 mm

yB = 3.46 m



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PROBLEM 15.31 A load is to be raised 20 ft by the hoisting system shown. Assuming gear A is initially at rest, accelerates uniformly to a speed of 120 rpm in 5 s, and then maintains a constant speed of 120 rpm, determine (a) the number of revolutions executed by gear A in raising the load, (b) the time required to raise the load.

SOLUTION The load is raised a distance h = 20 ft = 240 in. For gear-pulley B, radius to rope groove is r1 = 15 in. Required angle change for B:

θB =

h 240 = = 16 radians r1 15

Circumferential travel of gears A and B: s = r2θ B = rAθ A

where r2 = 18 in. and rA = 3 in.

s = (18 in.)(16 radians) = 288 in.

(a)

(b)

Angle change of gear A:

θA =

s 288 = = 96 radians 3 rA

In revolutions,

θA =

96 2π

Motion of gear A.

ω0 = 0, ω f = 120 rpm = 4π rad/s

θ A = 15.28 rev 

Gear A is uniformly accelerated over the first 5 seconds.

α=

ω f − ω0 t

1 2

=

4π rad/s = 2.5133 rad/s 2 5s

1 2

θ A = α t 2 = (2.5133)(5)2 = 31.416 radians The angle change over the constant speed phase is Δθ = θ A − θ = 96 − 31.416 = 64.584 radians

For uniform motion,

Δθ = ω f (Δt ) Δt =

Total time elapsed:

Δθ

ωf

=

64.584 = 5.139 s 4π

t f = 5 s + Δt

t f = 10.14 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1043

PROBLEM 15.32 Disk B is at rest when it is brought into contact with disk A which is rotating freely at 450 rpm clockwise. After 6 s of slippage, during which each disk has a constant angular acceleration, disk A reaches a final angular velocity of 140 rpm clockwise. Determine the angular acceleration of each disk during the period of slippage.

SOLUTION (ω A )0 = 450 rpm = 47.124 rad/s

Disk A: When t = 6 s:

ω A = 140 rpm = 14.661 rad/s

ω A = (ω A )0 + α At 14.661 rad/s = 47.124 rad/s + α A (6 s)

α A = −5.41 rad/s

α A = 5.41 rad/s 2



ω0 = 0

Disk B: When t = 6 s: (end of slippage)

rAω A = rBωB : (3 in.)(14.661 rad/s) = (5 in.)(ωB )

ωB = 8.796 rad/s ωB = (ωB )0 + α B t 8.796 rad/s = 0 + α B (6 s) α B = 1.466 rad/s 2

α B = 1.466 rad/s 2



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PROBLEM 15.33 A simple friction drive consists of two disks A and B. Initially, disk A has a clockwise angular velocity of 500 rpm and disk B is at rest. It is known that disk A will coast to rest in 60 s. However, rather than waiting until both disks are at rest to bring them together, disk B is given a constant angular acceleration of 2.5 rad/s2 counterclockwise. Determine (a) at what time the disks can be brought together if they are not to slip, (b) the angular velocity of each disk as contact is made.

SOLUTION (ω A )0 = 500 rpm = 52.36 rad/s

Disk A: Disk A will coast to rest in 60 s.

ω A = (ω A )0 + α At; 0 = 52.36 rad/s + α A (60 s) α A = −0.87266 rad/s 2 At time t:

ω A = (ω A )0 + α At ω A = 52.36 − 0.87266 t Disk B:

α B = 2.5 rad/s 2 (ω B )0 = 0

At time t:

ωB = (ωB )0 + α B t; ωB = 2.5t

(a)

(1)

(2)

rAω A = rBω B

Bring disks together when:

(3 in.)(52.36 − 0.87266t ) = (5 in.)(2.5t ) 157.08 − 2.618t = 12.5t 157.08 = 15.118t

(b)

t = 10.39 s 

When contact is made (t = 10.39 s) Eq. (1):

ω A = 52.36 − 0.87266 (10.39) ω A = 43.29 rad/s

Eq. (2):

ω A = 413 rpm



ω B = 248 rpm



ωB = 2.5(10.39) ωB = 25.975 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1045

PROBLEM 15.34 A simple friction drive consists of two disks A and B. Initially, disk A has a clockwise angular velocity of 500 rpm and disk B is at rest. It is known that disk A will coast to rest in 60 s. However, rather than waiting until both disks are at rest to bring them together, disk B is given a constant angular acceleration of 2.5 rad/s2 counterclockwise. Determine (a) at what time the disks can be brought together if they are not to slip, (b) the angular velocity of each disk as contact is made.

SOLUTION (ω A )0 = 500 rpm = 52.36 rad/s

Disk A: Disk a will coast to rest in 60 s.

ω A = (ω A )0 + α At; 0 = 52.36 + α A (60 s) α A = −0.87266 rad/s 2 At time t:

ω A = (ω A )0 + α At ; ω A = 52.36 − 0.87266t Disk B:

α B = 2.5 rad/s 2 (ω B )0 = 0

At time t:

ωB = (ωB )0 + α B t ; ω B = 2.5t

(a)

Bring disks together when:

(1)

(2)

rAω A = rBω B

(80 mm)(52.36 − 0.87266t ) = (60 mm)(2.5t ) 4188.8 − 69.813t = 150t 4188.8 = 219.813t

t = 19.056 s

(b)

t = 19.06 s 

Contact is made: Eq. (1):

ω A = 52.36 − 0.87266(19.056) ω A = 35.73 rad/s

Eq. (2):

ω A = 341 rpm



ω B = 455 rpm



ωB = 2.5(19.056) ωB = 47.64 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1046

PROBLEM 15.35 Two friction disks A and B are both rotating freely at 240 rpm counterclockwise when they are brought into contact. After 8 s of slippage, during which each disk has a constant angular acceleration, disk A reaches a final angular velocity of 60 rpm counterclockwise. Determine (a) the angular acceleration of each disk during the period of slippage, (b) the time at which the angular velocity of disk B is equal to zero.

SOLUTION (a)

(ω A )0 = 240 rpm = 25.133 rad/s

Disk A: When t = 8 s,

ω A = 60 rpm = 6.283 rad/s

ω A = (ω A )0 + α At; 6.283 rad/s = 25.133 rad/s + α A (8 s) α A = −2.356 rad/s 2 α A = 2.36 rad/s 2



(ωB )0 = 240 rpm = 25.123 rad/s

Disk B:

When t = 8 s: (slippage stops) rAω A = rBω B (80 mm)(6.283 rad/s) = (60 mm)ωB

ωB = 8.378 rad/s For

:

ω B = 8.38 rad/s

ωB = (ω B )0 + α B t 8.375 rad/s = −25.133 rad/s + α B (8 s)

α B = 4.188 rad/s 2 (b)

α B = 4.19 rad/s 2



Time when ωB = 0 For

:

ωB = (ωB )0 + α B t 0 = −25.133 rad/s + (4.188 rad/s 2 )t

t = 6.00 s

t = 6.00 s 

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PROBLEM 15.36* Steel tape is being wound onto a spool which rotates with a constant angular velocity ω 0 . Denoting by r the radius of the spool and tape at any given time and by b the thickness of the tape, derive an expression for the acceleration of the tape as it approaches the spool.

SOLUTION Let one layer of tape be wound and let v be the tape speed. vΔt = 2π r and Δr = b bv bω Δr = = Δt 2π r 2π

For the spool:

d ω d  v  1 dv d 1 =  = +v   dt dt  r  r dt dt  r  a v dr a v bω − = − r r 2 dt r r 2 2π 1 bω 2  = a − =0 2π  r =

a=

bω02 2π



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PROBLEM 15.37* In a continuous printing process, paper is drawn into the presses at a constant speed v. Denoting by r the radius of the paper roll at any given time and by b the thickness of the paper, derive an expression for the angular acceleration of the paper roll.

SOLUTION Let one layer of paper be unrolled. vΔt = 2π r and Δr = −b Δr −bv dr = = Δt 2π r dt dω α= dt d v =   dt  r  1 dv d 1 = +v   r dt dt  r  v dr =0− 2 r dt  v  −bv  =  − 2    r  2π r  =

bv 2 2π r 3

α=

bv 2 2π r 3



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PROBLEM 15.CQ3 The ball rolls without slipping on the fixed surface as shown. What is the direction of the velocity of Point A? (a)

(b)

(c)

(d )

(e)

SOLUTION Answer: (b) 

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PROBLEM 15.CQ4 Three uniform rods, ABC, DCE and FGH are connected as shown. Which of the following statements are true? (a) ωABC = ωDCE = ωFGH (b) ωDCE > ωABC > ωFGH (c)

ωDCE < ωABC < ωFGH

(d ) ωABC > ωDCE > ωFGH (e)

ωFGH = ωDCE < ωABC

SOLUTION Answer: (a) 

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PROBLEM 15.38 An automobile travels to the right at a constant speed of 48 mi/h. If the diameter of a wheel is 22 in., determine the velocities of Points B, C, D, and E on the rim of the wheel.

SOLUTION vA = 48 mi/h = 70.4 ft/s

d = 22 in. r =

ω=

vC = 0 

d = 11 in. = 0.91667 ft 2

vA 70.4 = = 76.8 rad/s 0.91667 r

vB/A = vD/A = vE/A = rω

= (0.91667)(76.8) = 70.4 ft/s vB = vA + vB/A = [70.4 ft/s

] + [70.4 ft/s

] vB = 140.8 ft/s

vD = vA + vD/A = [70.4 ft/s

vE = vA + vE/A = [70.4 ft/s



] + [70.4 ft/s



30°] vD = 136.0 ft/s

15.0° 

vE = 99.6 ft/s

45.0° 

] + [70.4 ft/s ]

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PROBLEM 15.39 The motion of rod AB is guided by pins attached at A and B which slide in the slots shown. At the instant shown, θ = 40° and the pin at B moves upward to the left with a constant velocity of 6 in./s. Determine (a) the angular velocity of the rod, (b) the velocity of the pin at end A.

SOLUTION

vA = vB + vA/B

[ vA↑] = [6 in./s

15°] + [v A/B

40°]

Law of sines. v vA 6 in./s = A/B = sin 55° sin 75° sin 50° vA = 6.42 in./s

(b) vA/B = 7.566 in./s



40°

vA/B = ( AB)ω AB

AB = 20 in.

7.566 in./s = (20 in.)ω AB

(a)

ω AB = 0.3783 rad/s

ω AB = 0.378 rad/s



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PROBLEM 15.40 Collar B moves upward with a constant velocity of 1.5 m/s. At the instant when θ = 50°, determine (a) the angular velocity of rod AB, (b) the velocity of end A of the rod.

SOLUTION Draw a diagram showing the motion of rod AB.

Plane motion vA = v A

=

Translation

25°

vB = 1.5 m/s

+

Rotation vB/A = vB/A

50°

vB = v A + v B/A

[1.5 m/s↑] = [v A

25°] + [vB/A

50°]

Draw the velocity vector diagram. Interior angles of the triangle. 90° − 25° = 65° 90° − 50° = 40° 25° + 50° = 75°

vB = 1.5 m/s

Law of sines. vB/A vB vA = = sin 75° sin 40° sin 65°

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PROBLEM 15.40 (Continued)

(a)

Angular velocity of AB. sin 65° (1.5 m/s) = 1.4074 m/s sin 75° v 1.4074 m/s = B /A = l AB 1.2 m

vB /A = ω AB

(b)

ω AB = 1.173 rad/s



Velocity of end A. vA =

sin 40° (1.5 m/s) sin 75°

vA = 0.998 m/s

25° 

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PROBLEM 15.41 Collar B moves downward to the left with a constant velocity of 1.6 m/s. At the instant shown when θ = 40°, determine (a) the angular velocity of rod AB, (b) the velocity of collar A.

SOLUTION

vA = vB + vA/B [v A ] = [1.6 m/s

30°] + [v A/B

40°]

Law of sines. v vA 1.6 m/s = A/B = sin 70° sin 60° sin 50° v A = 1.963 m/s

(b) vA/B = 1.809 m/s 2



40°

AB = 0.5 m v A/B = ( AB)ω AB 1.809 m/s = (0.5 m)ω AB

(a)

ω AB = 3.618 rad/s

ω AB = 3.62 rad/s



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PROBLEM 15.42 Collar A moves upward with a constant velocity of 1.2 m/s. At the instant shown when θ = 25°, determine (a) the angular velocity of rod AB, (b) the velocity of collar B.

SOLUTION

vB = vA + vB/A [v B

30°] = [1.2 m/s

] + [vB/A

25°]

Law of sines. v vB 1.2 m/s = B/A = sin 65° sin 60° sin 55° v B = 1.328 m/s

(b) v B/A = 1.269 m/s

30° 

65°

vB/A = ( AB)ω AB 1.269 m/s = (0.5 m)ω AB ω AB = 2.538 rad/s ω AB = 2.54 rad/s

(a)



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PROBLEM 15.43 Rod AB moves over a small wheel at C while end A moves to the right with a constant velocity of 25 in./s. At the instant shown, determine (a) the angular velocity of the rod, (b) the velocity of end B of the rod.

SOLUTION Slope angle of rod.

tan θ = AC =

7 = 0.7, 10

θ = 35°

10 = 12.2066 in. cos θ

CB = 20 − AC = 7.7934 in.

Velocity analysis. v A = 25 in./s

,

vC /A = ACω AB

θ

θ

vC = vC

vC = v A + vC/A

Draw corresponding vector diagram. vC/A = v A sin θ = 25sin 35° = 14.34 in./s

(a)

ω AB =

vC/A AC

=

14.34 = 1.175 rad/s 12.2066 ω AB = 1.175 rad/s



vC = v A cos θ = 25cos θ = 20.479 in./s v B/ C = CBω AB = (7.7934)(1.175) = 9.1551 in./s

vB/C has same direction as vC/A. v B = vC + v B/C

Draw corresponding vector diagram. tan φ =

(b)

vB =

vB/C vC

=

9.1551 , φ = 24.09° 20.479

vC 20.479 = = 22.4 in./s = 1.869 ft/s cos φ cos 24.09°

φ + θ = 59.1° v B = 1.869 ft/s

59.1° 

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PROBLEM 15.44 The plate shown moves in the xy plane. Knowing that (v A ) x = 120 mm/s, (vB ) y = 300 mm/s, and (vC ) y = −60 mm/s, determine (a) the angular velocity of the plate, (b) the velocity of Point A.

SOLUTION rC/B = (180 mm)i + (360 mm) j ω =ωk vB = (vB ) x i + (300 mm/s) j vC = (vC ) x i − (60 mm/s) j vC = vB + vC/B

(a)

(vC ) x i − (60 mm/s) j = (vB ) x i + (300 mm/s) j + ω × rC/B (vC ) x i − 60 j = (vB ) x i + 300 j + ω k × (180i + 360 j) (vC ) x i − 60 j = (vB ) x i + 300 j + 180ω j − 360ω i

Coefficients of j:

−60 = 300 + 180ω

ω = −2 rad/s (b)

Velocity of A:

ω = 2 rad/s



rA/B = −(180 mm)i + (180 mm) j vA = vB + vA/B = vB + ω × rA/B 120i + (v A ) y j = (vB ) x i + 300 j + (−2k ) × (−180i + 180 j) 120i + (v A ) y j = (vB ) x i + 300 j + 360 j + 360i

Coefficients of j:

(v A ) y = 300 + 360 = 660 mm/s vA = (120 mm/s)i + (660 mm/s) j 

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PROBLEM 15.45 In Problem 15.44, determine (a) the velocity of Point B, (b) the point of the plate with zero velocity. PROBLEM 15.44 The plate shown moves in the xy plane. Knowing that (v A ) x = 120 mm/s, (vB ) y = 300 mm/s, and (vC ) y = −60 mm/s, determine (a) the angular velocity of the plate, (b) the velocity of Point A.

SOLUTION rB/A = (180 mm)i − (180 mm) j

From the answer of Problem 15.44, we have ω = −(2 rad/s)k vA = (120 mm/s)i + (660 mm/s) j

(a)

Velocity of B: vB = vA + vB/A = vA + ω × rB/A = 120i + 660 j − 2k × (180i − 180 j) = 120i + 660 j − 360 j − 360i vB = −(240 mm/s)i + (300 mm/s) j 

(b)

Point with v = 0: Let P = xi + yj be an arbitrary point. Thus

rP/A = (180 + x)i + yj

vP = vA + v P/A = vA + ω × rP/A vP = 120i + 660 j + (−2k ) × [(180 + x)i + yj] vP = 120i + 660 j − (360 + 2 x) j + 2 yi vP = (120 + 2 y )i + (300 − 2 x) j

For v P = 0:

120 + 2 y = 0 and 300 − 2 x = 0

v = 0 at:

y = −60 mm, x = 150 mm 

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PROBLEM 15.46 The plate shown moves in the xy plane. Knowing that (v A ) x = 250 mm/s, (vB ) x = −450 mm/s, and (vC ) x = −500 mm/s, determine (a) the angular velocity of the plate, (b) the velocity of Point A.

SOLUTION ω =ωk

Angular velocity:

rB/A = (150 mm)i

Relative position vectors:

rC/A = (200 mm)i + (150 mm) j v A = (250 mm/s)i + (v A ) y j

Velocity vectors:

v B = (vB ) x i − (450 mm/s) j vC = −(500 mm/s)i + (vC ) y j

Unknowns are ω, (v A ) y , (vB ) x , and (vC ) y . v B = v A + v B/A = v A + ω k × rB/A (vB ) x i − 450 j = 250i + (v A ) y j + ω k × 150i = 250i + (v A ) y j + 150ω j

i:

(vB ) x = 250

(1)

j:

−450 = (v A ) y + 150ω

(2)

vC = v A + v C/A = v A + ω k × rC/A −500i + (vC ) y j = 250i + (v A ) y j + ω k × (200i − 150 j) = 250i + (v A ) y j + 200ω j + 150ω i

(a)

i:

−500 = 250 + 150ω

(3)

j:

(vC ) y = (v A ) y + 150ω

(4)

Angular velocity of the plate. From Eq. (3),

ω=−

750 = −5 150

ω = −(5.00 rad/s)k = 5.00 rad/s

(b)



Velocity of Point A. From Eq. (2), (v A ) y = −450 − 150ω = −450 − (150)(−5) = 300 mm/s v A = (250 mm/s)i + (300 mm/s) j 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1061

PROBLEM 15.47 The plate shown moves in the xy plane. Knowing that (vA)x = 12 in./s, (vB)x = −4 in./s, and (vC)y = −24 in./s, determine (a) the angular velocity of the plate, (b) the velocity of Point B.

SOLUTION Angular velocity: Relative position vectors:

ω =ωk rA/B = −(2 in.)i + (4 in.) j rC/B = (6 in.)i − (2 in.) j

Velocity vectors:

v A = (12 in./s)i + (v A ) y j v B = −(4 in./s)i + (vB ) y j vC = (vC ) x i − (24 in./s) j

Unknowns are ω , (v A ) y , (vB ) y , and (vC ) x . v A = v B + v A /B = v B + ω k × rA/B 12i + (v A ) y = −4i + (vB ) y j + ω k × ( −2i + 4 j) = −4i + (vB ) y j − 2ω j − 4ω i

i:

12 = −4 − 4ω

(1)

j:

(v A ) y = (vB ) y − 2ω

(2)

v C = v B + vC /B = v B + ω k × rC/B (vC ) x i − 24 j = −4i + (vB ) y j + ω k × (6i − 2 j) = −4i + (vB ) y j + 6ω j + 2ω i

i: j:

(vC ) x = −4 + 2ω

(3)

−24 = (vB ) y + 6ω

(4)

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PROBLEM 15.47 (Continued)

(a)

Angular velocity of the plate. From Eq. (1),

ω=−

16 = −4 rad/s 4

ω = −(4.00 rad/s)k = 4.00 rad/s

(b)



Velocity of Point B. From Eq. (4),

(vB ) y = −24 − (6)( −4) = 0 v B = −(4.00 in./s)i 

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PROBLEM 15.48 In the planetary gear system shown, the radius of gears A, B, C, and D is a and the radius of the outer gear E is 3a. Knowing that the angular velocity of gear A is ωA clockwise and that the outer gear E is stationary, determine (a) the angular velocity of each planetary gear, (b) the angular velocity of the spider connecting the planetary gears.

SOLUTION Gear E is stationary.

vE = 0

Let A be the center of gear A and the spider. Since the motions of gears B, C, and D are similar, only gear B is considered. Let H be the effective contact point between gears A and B.

Gear A:

v H = aω A

(a)

v H = v E + v H /E

Planetary gears B, C, and D:

1 2

: aω A = 0 + (2a )ω B

ωB = ω A 1 ω B = ωC = ω D = ω A 2



v B = v E + v B/E 1  : vB = 0 + a  ω A  2 

(b)

Spider.

vB =

1 aω A 2

(1)

v B = (2a)ωs

(2)

Equating expressions (1) and (2) for vB, 1 aω A = (2a)ωs 2

1 4

ωs = ω A

1 ωs = ω A 4



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1064

PROBLEM 15.49 In the planetary gear system shown, the radius of gears A, B, C, and D is 30 mm and the radius of the outer gear E is 90 mm. Knowing that gear E has an angular velocity of 180 rpm clockwise and that the central gear A has an angular velocity of 240 rpm clockwise, determine (a) the angular velocity of each planetary gear, (b) the angular velocity of the spider connecting the planetary gears.

SOLUTION Since the motions of the planetary gears B, C, and D are similar, only gear B is considered. Let Point H be the effect contact point between gears A and B and let Point E be the effective contact point between gears B and E. Given angular velocities:

ω E = 180 rpm = 6π rad/s ω A = 240 rpm = 8π rad/s

Outer gear E:

radius = rE = 90 mm vE = rE ωE = (90 mm)(6π rad/s) = 540π mm/s v E = 540π mm/s

Gear A:

radius = rA = 30 mm vH = rAω A = (30 mm)(8π rad/s) = 240π mm/s v H = 240π mm/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1065

PROBLEM 15.49 (Continued)

Planetary gear B:

radius = rB = 30 mm,

ω B = ωB

v H = v E + v H /E

[(30 mm)ω A

] = [540π mm/s

] + [(60 mm)ωB

]

30ω A = 540π − 60ωB

ωB = (a)

540π + 30ω A 1 1 = 9π + ω A = 9π − (8π ) = 5π rad/s 60 2 2

Angular velocity of planetary grears: ω B = ωC = ω D = 5π rad/s

v B = v H + vB/H = [(30 mm)ω A

] + [30 mm ω B

= 150 rpm



= 195 rpm



]

vB = (30 mm)(8π rad/s) + (30 mm)(5π rad/s) = 390π mm/s

(b)

Spider:

arm = rs = 60 mm, ω s = ωs vB = rsωs

ωs =

vB 390π mm/s = = 6.5π rad/s 60 mm rs ω s = 6.5π rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1066

PROBLEM 15.50 Arm AB rotates with an angular velocity of 20 rad/s counterclockwise. Knowing that the outer gear C is stationary, determine (a) the angular velocity of gear B, (b) the velocity of the gear tooth located at Point D.

SOLUTION Arm AB:

Gear B:

(a)

BE = 0.05 m:

v B = vD + vB/E = 0 + ( BE )ωB

2.4 m/s = 0 + (0.05 m)ω B

ωB = 48 rad/s (b)

DE = (0.05 2):

ω B = 48 rad/s



vD = vE + vD/E = 0 + ( DE )ωB

vD = 0 + (0.05 2)(48) vD = 3.39 m/s

vD = 3.39 m/s

45° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1067

PROBLEM 15.51 In the simplified sketch of a ball bearing shown, the diameter of the inner race A is 60 mm and the diameter of each ball is 12 mm. The outer race B is stationary while the inner race has an angular velocity of 3600 rpm. Determine (a) the speed of the center of each ball, (b) the angular velocity of each ball, (c) the number of times per minute each ball describes a complete circle.

SOLUTION Data: ω A = 3600 rpm = 376.99 rad/s, rA =

ωB = 0

1 d A = 30 mm 2

d = diameter of ball = 12 mm

Velocity of point on inner race in contact with a ball. v A = rAω A = (30)(376.99) = 11310 mm/s

Consider a ball with its center at Point C. v A = vB + v A /B v A = 0 + ωC d

ωC =

v A 11310 = d 12

= 942.48 rad/s vC = vB + vC/B =0+

1 d ω = (6)(942.48) = 5654.9 mm/s 2 vC = 5.65 m/s 

(a) (b)

Angular velocity of ball.

ωC = 942.48 rad/s (c)

ωC = 9000 rpm 

Distance traveled by center of ball in 1 minute. lC = vC t = 5654.9(60) = 339290 mm

Circumference of circle:

2π r = 2π (30 + 6) = 226.19 mm

Number of circles completed in 1 minute: n=

l 2π r

=

339290 226.19

n = 1500 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1068

PROBLEM 15.52 A simplified gear system for a mechanical watch is shown. Knowing that gear A has a constant angular velocity of 1 rev/h and gear C has a constant angular velocity of 1 rpm, determine (a) the radius r, (b) the magnitudes of the accelerations of the points on gear B that are in contact with gears A and C.

SOLUTION Point where A contacts B: v1 = rAω A = rωB

ωB =

rAω A r

(1)

v2 = rBω B = rωC

ωC =

rB ωB r

(2)

Point where B contacts C:

From Eqs. (1) and (2),

ωC =

rA rB r2

r 2 = rA rB

r2 =

ω A 1 rev/h 1 = = ωC 1 rev/m 60

(0.6 in.)(0.36 in.) = 0.0036 in 2 60 r = 0.0600 in. 

Radius r: Angular velocity of B.

(b)

ωA ωC

rA = 0.6 in., rB = 0.36 in.

Data:

(a)

ωA

Point where B contacts A.

2π rad/s 60 0.060 2π r ωB = ωC = = 0.017453 rad/s rB 0.36 60

ωC = 1 rpm =

an = rωB2 = (0.0600 in.)(0.017453 rad/s) 2 an = 18.28 × 10−6 in./s 2 



Point where B contacts C.

an = rBωB2 = (0.36 in.)(0.017453 rad/s) 2 an = 109.7 × 10−6 in./s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1069

PROBLEM 15.53 Arm ACB rotates about Point C with an angular velocity of 40 rad/s counterclockwise. Two friction disks A and B are pinned at their centers to arm ACB as shown. Knowing that the disks roll without slipping at surfaces of contact, determine the angular velocity of (a) disk A, (b) disk B.

SOLUTION 

Arm ACB : Fixed axis rotation.



rA/C = 24 mm,

vA = rA/C ω AB = (24)(40) = 960 mm/s

rB/C = 18 mm,

vB = rB/C ω AB = (18)(40) = 720 mm/s

Disk B: Plane motion = Translation with B + Rotation about B. rB = 30 mm, v D = v B − v D/B



0 = 720 + 30 ωB mm/s

ωB =

720 = 24 rad/s 30

vE = vB + vE/B



= 720 + (30)(24) = 1440 mm/s

Disk A: Plane motion = Translation with A + Rotation about A. rA = 12 mm,

vE = vA − vE/ A

1440 = 960 + 12ω A

ωA =

1440 + 960 = 200 rad/s 12

(a)

ω A = 200 rad/s



(b)

ω B = 24.0 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1070

PROBLEM 15.54 Arm ACB rotates about Point C with an angular velocity of 40 rad/s counterclockwise. Two friction disks A and B are pinned at their centers to arm ACB as shown. Knowing that the disks roll without slipping at surfaces of contact, determine the angular velocity of (a) disk A, (b) disk B.

SOLUTION Arm ACB : Fixed axis rotation.



rA/C = 0.3 in.,

vA = rA/C ω AB = (0.3)(40) = 12 in./s

rB/C = 1.8 in.,

vB = rB/C ω AB = (1.8)(40) = 72 in./s

Disk B: Plane motion = Translation with B + Rotation about B.



rB = 0.6 in.,

vD = vB − vB/A

0 = 72 + 0.6ωB

ωB =

72 = 120 rad/s 0.6

vE = vB + vE/B

= 72 + (0.6)(120) = 144 in./s



Disk A: Plane motion = Translation with A + Rotation about A. rA = 1.5 in.,

vE = vA − vE/A

144 = 12 1.5ω A

ωA =

144 + 12 = 104 rad/s 1.5

(a)

ω A = 104.0 rad/s



(b)

ω B = 120.0 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1071

PROBLEM 15.55 Knowing that at the instant shown the velocity of collar A is 900 mm/s to the left, determine (a) the angular velocity of rod ADB, (b) the velocity of Point B.

SOLUTION Consider rod ADB.

v D = vD j, v A = −(900 mm/s)i rD /A = −(80 mm)i − (150 mm) j v D /A = ω AD × rD /A = ω AD k × (−80i − 150 j) = 150ω AD i − 80ω AD j v 0 = v A + v D /A vD j = −900i + 150ω AD i − 80ω AD j

Equate components. i: 0 = −900 + 150ω AD

(a)

ω AD = 6 rad/s ω AD = (6.00 rad/s)k = 6.00 rad/s

Angular velocity of ADB. By proportions,



150 + 60 rD/A = 1.4 rD/A 150 = −(112 mm)i − (210 mm) j

rB /A =

v B = v A + ω AD k × rB/A = −900i + 6k × (−112i − 210 j) = −900i − 672 j + 1260i

(b)

Velocity of B.

v B = (360 mm/s)i − (672 mm/s) j = 762 mm/s

61.8° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1072

PROBLEM 15.56 Knowing that at the instant shown the angular velocity of rod DE is 2.4 rad/s clockwise, determine (a) the velocity A, (b) the velocity of Point B.

SOLUTION ω DE = 2.4 rad

Rod DE: Point E is fixed.

vD = ωDB rDE = (2.4 rad/s)(120 mm) = 288 mm/s

v D = 288 mm/s = (288 mm/s) j rA/D = (80 mm)i + (150 mm) j, ω AD = ω AD k , v A = v A i

Rod ADB:

v A = v D + v D/A = v D + ω AD k × rA/D v A i = (288 mm/s) j + ω AD k × [(80 mm)i + (150 mm) j] v A i = 288 j + 80ω AD j − 150ω AD i

Equate components. i: j: From Eq. (2), From Eq. (1), (a)

Velocity of collar A.

(b)

Velocity of Point B. By proportions

v A = −150ω AD

(1)

0 = 288 + 80ω AD

ω AD = −

288 80

(2) ω AD = (−3.6 rad/s)k

v A = −(150)( −3.6) = 540 mm/s

v A = 540 mm/s

rB/D = −



60 rA/D = −(32 mm)i − 60 mm j 150

v B = v D + v B/D = v D + ω AD × rB/D = (288 mm/s) j + [ −(3.6 rad/s)k ] × [−(32 mm)i − (60 mm) j] = (288 mm/s) j + (115.2 mm/s) j − (216 mm/s)i v B = −(216 mm/s)i + (403.2 mm/s) j

v B = 457 mm/s

61.8° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1073

PROBLEM 15.57 A straight rack rests on a gear of radius r and is attached to a block B as shown. Denoting by ωD the clockwise angular velocity of gear D and by θ the angle formed by the rack and the horizontal, derive expressions for the velocity of block B and the angular velocity of the rack in terms of r, θ , and ωD .

SOLUTION

Gear D:

Rotation about D. Tooth E is in contact with rack AB. vE = rωD lEB =

Rack AB: Plane motion

=

θ

r tan θ

+

Translation with E

vB = vE + vB/E

[ vB

Rotation about E.

] = [ vE

θ ] + [vB/E θ ]

Draw velocity vector diagram. vB =

vE rωD = cos θ cos θ

vB =

rωD cos θ



vB/E = vE tan θ = rωD tan θ

ω AB = =

vB/E lEB rω D tan θ r tan θ

= ω D tan 2 θ

ω AB = ωD tan 2 θ



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1074

PROBLEM 15.58 A straight rack rests on a gear of radius r = 2.5 in. and is attached to a block B as shown. Knowing that at the instant shown the velocity of block B is 8 in./s to the right and θ = 25°, determine (a) the angular velocity of gear D, (b) the angular velocity of the rack.

SOLUTION Gear D:

Rotation about D. Tooth E is in contact with rack AB. v E = rωD lEB =

Rack AB: Plane motion

=

θ

r tan θ +

Translation with E

vB = vE + vB/E

Rotation about E.

[ vB

] = [ vE

θ ] + [vB/E θ ]

Draw velocity vector diagram. vB =

vE rωD = cos θ cos θ

vB =

rωD cos θ

vB/E = vE tan θ = rωD tan θ

ω AB =

vB/E lEB

=

rω D tan θ r tan θ

= ωD tan 2 θ

ω AB = ω D tan 2 θ Data: (a) (b)

r = 2.5 in. θ = 25° v B = 8 in./s

ωD =

vB cos θ 8 cos 25° = r 2.5

ω AB = 2.90 tan 2 25°

ω D = 2.90 rad/s ω AB = 0.631 rad/s

 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1075

PROBLEM 15.59 Knowing that at the instant shown the angular velocity of crank AB is 2.7 rad/s clockwise, determine (a) the angular velocity of link BD, (b) velocity of collar D, (c) the velocity of the midpoint of link BD.

SOLUTION ω AB = 2.7 rad/s

Crank AB: Point A is fixed.

vB = ω AB rAB = (2.7 rad/s)(5 in.) = 13.5 in./s

v B = 13.5 in./s

= −(13.5 in./s)i

rD /B = (12 in.)i − (9 in.) j, ω BD = ωBD

Link BD:

= ω AB k ,

v D = vD = vD j v D = v B + v B /D = v B + ωBD k × rB /D vD j = −(13.5 in./s)i + ω BD k × [(12 in.)i − (9 in.) j] = −13.5i + 12ωBD j + 9ωBD i

Equate components. i: j: (a)

vD = 12ωBD

(2)

ωBD =

13.5 9

ωBD = 1.500 rad/s



Velocity of collar D. From Eq. (2),

(c)

(1)

Angular velocity of link BD. From Eq. (1),

(b)

0 = −13.5 + 9ωBD

vD = (12)(1.5)

v D = 18.00 in./s 

Velocity of midpoint M of link BD. 1 rD /B = (6 in.)i − (4.5 in.) j 2 = v B + v M /B = v B + ω BD k × rM /D

rM /B = vM

= −13.5i + (1.500k ) × (6i − 4.5 j) = −13.5i + 9 j + 6.75i v M = −(6.75 in./s)i + (9.00 in./s) j = 11.25 in./s

53.1° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1076

PROBLEM 15.60 In the eccentric shown, a disk of 2-in.-radius revolves about shaft O that is located 0.5 in. from the center A of the disk. The distance between the center A of the disk and the pin at B is 8 in. Knowing that the angular velocity of the disk is 900 rpm clockwise, determine the velocity of the block when θ = 30°.

SOLUTION (OA) sin θ = ( AB )sin β

Geometry.

(OA) sin θ AB 0.5 sin 30° = , 8

sin β =

β = 1.79°

Shaft and eccentric disk. (Rotation about O)

ωOA = 900 rpm = 30π rad/s vA = (OA) ωOA = (0.5)(30π) = 15π in/s

Rod AB.

(Plane motion = Translation with A + Rotation about A.)

vB = vA + vB/ A Draw velocity vector diagram.

[v B

] = [v A

60°] + [v A/B

β]

90° − β = 88.21°

ϕ = 180° − 60° − 88.21° = 31.79°

Law of sines.

vB vA = sin φ sin(90° − β ) v A sin ϕ vB = sin (90° − β ) =



(15π )sin 31.79° sin 88.21°

= 24.837 in./s 

vB = 24.8 in./s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1077

PROBLEM 15.61 In the engine system shown, l = 160 mm and b = 60 mm. Knowing that the crank AB rotates with a constant angular velocity of 1000 rpm clockwise, determine the velocity of the piston P and the angular velocity of the connecting rod when (a) θ = 0, (b) θ = 90°.

SOLUTION

ω AB = 1000 rpm (a)

θ = 0°. Crank AB. (Rotation about A)

=

(1000)(2π ) = 104.72 rad/s 60

rB/A = 0.06 m

vB = vB/Aω AB = (0.06)(104.72) = 6.2832 m/s Rod BD.

(Plane motion = Translation with B + Rotation about B) vD = vB + vD/B vD

= [6.2832

] + [vD/B

]

vD = 0 vD/B = 6.2832 m/s vP = vD

ωBD = (b)

θ = 90°. Crank AB. (Rotation about A)

vP = 0 

vB 6.2832 = 0.16 l

ω BD = 39.3 rad/s



rB/ A = 0.06 m

vB = rB/ Aω AB = (0.06)(104.72) = 6.2832 m/s

Rod BD. (Plane motion = Translation with B + Rotation about B.) vD = vB + vD/B [vD ] ] = [6.2832] + [ vD /B

β

]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1078

PROBLEM 15.61 (Continued)

vD/B = 0,

ωBD =

vD = 6.2832 m/s

vD/B

ωBD = 0 

l

v P = v D = 6.2832 m/s

vP = 6.28 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1079

PROBLEM 15.62 In the engine system shown l = 160 mm and b = 60 mm. Knowing that crank AB rotates with a constant angular velocity of 1000 rpm clockwise, determine the velocity of the piston P and the angular velocity of the connecting rod when θ = 60°.

SOLUTION

ω AB = 1000 rpm = θ = 60°. Crank AB. (Rotation about A)

(1000)(2π ) = 104.72 rad/s 60

rB/A = 3 in.

30°

vB = rB /Aω AB = (0.06)(104.72) = 6.2832 m/s

60°

Rod BD. (Plane motion = Translation with B + Rotation about B.) Geometry.

l sin β = r sin θ r 0.06 sin β = sin θ = sin 60° 0.16 l β = 18.95° vD = vB + vD/B

[vD ] = [314.16 60°] + [vD /B β] Draw velocity vector diagram. ϕ = 180° − 30° − (90° − β ) = 78.95° Law of sines. vD/B vD vB = = sin ϕ sin 30° sin (90° − β )

vB sin ϕ cos β 6.2832 sin 78.95° = cos18.95° = 6.52 m/s

vD =

vP = vD

vP = 6.52 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1080

PROBLEM 15.62 (Continued)

vB sin 30° cos β 6.2832sin 30° = cos18.95° = 3.3216 m/s

vD/B =

ωBD =

vD/B l

=

3.3216 0.16

ωBD = 20.8 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1081

PROBLEM 15.63 Knowing that at the instant shown the angular velocity of rod AB is 15 rad/s clockwise, determine (a) the angular velocity of rod BD, (b) the velocity of the midpoint of rod BD.

SOLUTION ω AB = 15 rad/s

Rod AB:

vB = ( AB)ω AB = (0.200)(15) = 3 m/s

vB = 3 m/s

vB = −(3 m/s)i, v D = vD j, ω BD = ωBD k

Rod BD:

rB /D = (0.6 m)i + (0.25 m) j v B = v D + v B /D = v D + ω BD × rB /D −3i = vD j + ωBD k × (0.6i + 0.25 j) = vD j + 0.6ω BD j − 0.25ω BD i

Equate components. i: −3 = −0.25ωBD 0 = vD + 0.6ωBD

j: (a)

(2)

Angular velocity of rod BD. From Eq. (1), From Eq. (2),

(b)

(1)

3 0.25

ω BD = 12.00 rad/s

vD = −0.6ω BD

vD = −7.2 m/s

ωBD =



Velocity of midpoint M of rod BD. 1 rB /D = (0.3 m)i + (0.125 m) j 2 = v D + v M /D = vD j + ωBD k × rM /D

rM /D = vM

= −7.2 j + 12.00k × (0.3i + 0.125 j) = −(1.500 m/s)i − (3.60 m/s) j

vM = 3.90 m/s

67.4° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1082

PROBLEM 15.64 In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.

SOLUTION ω AB = 4 rad/s

Bar AB: (Rotation about A)

= −(4 rad/s)k

rB /A = −(175 mm)i

vB = ω AB × rB /A = (−4k ) × (−175i )

vB = (700 mm/s)j

Bar BD: (Plane motion = Translation with B + Rotation about B.) ω BD = ωBD k rD /B = −(200 mm)j

vD = vB + ω BD × rD /B = 700 j + (ωBD k ) × (−200 j) vD = 700 j + 200ωBD i ω DE = ω DE k

Bar DE: (Rotation about E)

rD /E = −(275 mm)i + (75 mm)j vD = ω DE × rD /E = (ωDE k ) × (−275i + 75 j) vD = −275ωDE j − 75ωDE i

Equating components of the two expressions for vD , 700 = −275ωDE

ωDE = −2.5455 rad/s

i : 200ωBD = −75ωDE

ω BD = − ωBD

j:

ω DE = 2.55 rad/s



3 8

3

ωBD = −   (−2.5455) = 0.95455 rad/s 8

ω BD = 0.955 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1083

PROBLEM 15.65 In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.

SOLUTION

β = tan −1

Bar AB:

0.4 = 26.56° 0.8

0.8 = 0.8944 m cos β vB = ( AB)ω AB = (0.8944 m)(4 m/s)

AB =

vB = 3.578 m/s

γ = tan −1

Bar DE:

26.56°

0.4 = 38.66° 0.5

0.5 = 0.6403 m cos γ vD = ( DE )ωDE

DE =

vD = (0.6403 m)ωDE

38.66°

Bar BD:

vD = vB + vD/B [ vD

γ ] = [ vB

β ] + [ v D/B ]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1084

PROBLEM 15.65 (Continued)

Law of sines. vD/B vD 3.578 m/s = = sin 63.44° sin 65.22° sin 51.34° vD = 4.099 m/s (0.6403 m)ωDE = 4.099 m/s

ω DE = 6.4 rad/s



ω BD = 5.2 rad/s



vD/B = 4.160 m/s (0.8 m)vBD = 4.16 m/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1085

PROBLEM 15.66 Robert’s linkage is named after Richard Robert (1789–1864) and can be used to draw a close approximation to a straight line by locating a pen at Point F. The distance AB is the same as BF, DF and DE. Knowing that the angular velocity of bar AB is 5 rad/s clockwise in the position shown, determine (a) the angular velocity of bar DE, (b) the velocity of Point F.

SOLUTION Bar AB:

ω AB = 5 rad/s

= − (5 rad/s)k

In inches,

rB /A = 3i + 122 − 32 j = 3i + 135 j v B = ω AB × rB /A = −5k × (3i + 135 j) = 5 135i − 15 j rD /B = (6 in.)i, rF /B = 3i − 135 j (in.), ω BD = ωBD k

Object BDF:

v D = v B + v B /D = v B + ωBD k × rD /B = 5 135i − 15 j + ωBD k × 6i = 5 135i − 15 j + 6ω BD j

(1)

ω DE = ωDE k , rD /E = −(3 in.)i + ( 135 in.) j,

Bar DE: Point E is fixed so v E = 0

v D = ω DE × rD /E = ω DE k × (−3i + 135 j) = − 135ωDE i − 3ωDE j

(2)

Equating like components of vD from Eqs. (1) and (2), i: j: (a)

5 135 = − 135ωDE

(3)

−15 + 6ωBD = −3ω DE

(4)

Angular velocity of bar DE. From Eq. (3),

ωDE = −5 rad/s

From Eq. (4),

ωBD = (15 − 3ωDE ) = (15 + 15)

1 6

ω DE = 5.00 rad/s 1 6



ω BD = 5.00 rad/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1086

PROBLEM 15.66 (Continued)

(b)

Velocity of Point F.

vF /B = 3i − 135 j vF = v B = v F /B = v B + ωBD k × rF /B = 5 135i − 15 j + 5k × (3i − 135 j) = 5 135i − 15 j + 15 j + 5 135i = 10 135i vF = 116.2 in./s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1087

PROBLEM 15.67 Robert’s linkage is named after Richard Robert (1789–1864) and can be used to draw a close approximation to a straight line by locating a pen at Point F. The distance AB is the same as BF, DF and DE. Knowing that the angular velocity of plate BDF is 2 rad/s counterclockwise when θ = 90°, determine (a) the angular velocities of bars AB and DE, (b) the velocity of Point F. When θ = 90°, determine (a) the angular velocity of bar DE (b) the velocity of Point F.

SOLUTION When θ = 90°, the configuration of the linkage is close to that shown at the right. Bar AB:

ω AB = ω AB k

In inches,

rB/A = 6i + 6 3 j

v B = ω AB × rB/A = ω AB k × (6i + 6 3 j) = −6 3ω AB i + 6ω AB j

Object BDF:

rD/B = 6i + 6(2 − 3) j rF /B = 6i − 6 3 j

ω BD = (2 rad/s)k

v D = v B + v D/B = v B = ωBD k × rD/B = −6 3ω AB i + 6ω AB j + 2k × [6i + 6(2 − 3) j] = −6 3ω AB i + 6ω AB j + 24i − 12 3i + 12 j

Bar DE:

ω DE = ωDE k ,

(1)

rD/E = 12 j

v D = ω DE × rD/E = ωDE k × 12 j = −12ω DE i

(2)

Equating like components of vD from Eqs. (1) and (2), i: j:

−6 3ω AB + 12 − 6 3 = −12ωDE 6ω AB + 12 = 0

ω AB = −2 rad/s From Eq. (3),

(3)

ω AB = 2.00 rad/s



−12ω DE = −(6 3)( −2) + 24 − 12 3

ωDE = 2 − 2 3 = −1.4641

ω DE = 1.464 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1088

PROBLEM 15.67 (Continued)

v D = −(12)( −1.4641)i = 17.569i v F = v D + ωBD k × rF /D = 17.569i + 2k × (−12 j) = (41.569 in./s)i v F = 41.6 in./s



Note: The exact configuration of the linkage when θ = 90° may be calculated from trigonometry using the figure given below. Applying the law of cosines to triangle ADB gives

γ = 13.5° so that angle EAB is 45° + 13.5° = 58.5°.

We used 60° in the approximate analysis. Point F then lies about 0.53 in. to the right of Point E.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1089

PROBLEM 15.68 In the position shown, bar DE has a constant angular velocity of 10 rad/s clockwise. Knowing that h = 500 mm, determine (a) the angular velocity of bar FBD, (b) the velocity of Point F.

SOLUTION Bar DE:

(Rotation about E)

ω DE = 10 rad/s

= −(10 rad/s)k

rD/E = −(0.1 m)i + (0.2 m)j vD = ω DE × rD/E = ( −10k ) × (− 0.1i + 0.2 j) = (1 m/s)j + (2 m/s)i

Bar FBD: (Plane motion = Translation with D + Rotation about D.) ω BD = ωBD k rB/D = −(0.3 m)i + (0.1 m)j

vB = vD + ω BD × rB/D = j + 2i + (ωBD k ) × (−0.3i + 0.1j) = j + 2i − 0.3ωBD j − 0.1ωBD i

Bar AB:

(Rotation about A) ω AB = ω AB k rB/A = (0.42 m)j

vB = ω AB × rB/A = (ω AB k ) × (0.42 j) = −0.42ω AB i

Equating components of the two expressions for vB , (a)

j:

1 − 0.3ωBD = 0

ωBD = 3.3333 rad/s

i:

2 − 0.1ω BD = −0.42ω AB

rF/D = CrB/D



2 − (0.1)(3.3333) = −0.42ω AB

ω AB = −3.9683 rad/s Bar FBD:

ω BD = 3.33 rad/s

ω AB = 3.97 rad/s

where C =

h + 0.3 0.3

vF = vB + ω BD × rF/D = j + 2i + C ( −0.3ωBD j − 0.1ω BD i )

With

(b)

= j + 2i + C ( − j − 0.33333i ) 0.8 h = 500 mm = 0.5 m, C = = 2.6667 0.3 vF = j + 2i − 2.6667 j − 0.88889i

vF = (1.11111 m/s)i − (1.66667 m/s)j

vF = 2.00 m/s

56.3° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1090

PROBLEM 15.69 In the position shown, bar DE has a constant angular velocity of 10 rad/s clockwise. Determine (a) the distance h for which the velocity of Point F is vertical, (b) the corresponding velocity of Point F.

SOLUTION Bar DE:

(Rotation about E)

ω DE = 10 rad/s

= −(10 rad/s)k

rD/E = −(0.1 m)i + (0.2 m)j vD = ω DE × rD /E = (−10k ) × (− 0.1i + 0.2 j) = (1 m/s)j + (2 m/s)i

Bar FBD: (Plane motion = Translation with D + Rotation about D.) ω BD = ωBD k rB/D = −(0.3 m)i + (0.1 m)j

vB = vD + ω BD × rB/D = j + 2i + (ωBD k ) × (−0.3i + 0.1j) = j + 2i − 0.3ωBD j − 0.1ωBD i

Bar AB:

(Rotation about A) ω AB = ω AB k rB/A = (0.42 m)j

vB = ω AB × rB/A = (ω AB k ) × (0.42 j) = −0.42ω AB i

Equating components of the two expressions for vB , (a)

j:

1 − 0.3ωBD = 0

ωBD = 3.3333 rad/s

i:

2 − 0.1ω BD = −0.42ω AB

2 − (0.1)(3.3333) = −0.42ω AB

ω AB = −3.9683 rad/s Bar FBD:

rF/D = CrB/D

ω AB = 3.97 rad/s

where C =

h + 0.3 0.3

vF = vB + ω BD × rF/D = j + 2i + C ( −0.3ωBD j − 0.1ω BD i ) = j + 2i + C ( − j − 0.33333i )

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1091

PROBLEM 15.69 (Continued)

But vF = vF j. Equating components of the two expressions for vF , i:

(a) (b)

0 = 2 − 0.33333C

C =6

h = 0.3C − 0.3 = (0.3)(6) − 0.3

j: vF = j − Cj = (1 − 6) j 

h = 1.500 m 

vF = 5.00 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1092

PROBLEM 15.70 Both 6-in.-radius wheels roll without slipping on the horizontal surface. Knowing that the distance AD is 5 in., the distance BE is 4 in. and D has a velocity of 6 in./s to the right, determine the velocity of Point E.

SOLUTION Disk D:

Velocity at the contact Point P with the ground is zero. v 0 = 6 in./s

ωD =

vD 6 in./s = = 1 rad/s rD/P 6 in.

ω D = 1 rad/s

v A = rA/PωD = (6 in. + 5 in.)(1 rad/s) = 11 in./s

At Point A,

v A = 11 in./s

Disk E:

Velocity at the contact Point Q with the ground is zero. ω E = ω E

= ωE k.

rB /Q = −(4 in.)i + (6 in.) j v B = v B /Q = ω E × rB /Q = ωE k × (−4i + 6 j) v B = −6ω E i − 4ωE j

Connecting rod AB:

(1)

rB /A = ( 142 − 52 )i − 5 j in inches.

v B /A = 171i − 5 j

ω AB = ω AB k

v B = v A + v B /A = v A + ω AB k × ( 171i − 5 j) = 11i + 5ω AB i + 171ω AB j

(2)

Equating expressions (1) and (2) for vB gives −6ωE i − 4ωE j = 11i + 5ω AB i + 171ω AB j

Equating like components and transposing terms, i: j:

5ω AB + 6ωE = −11

(3)

171ω AB + 4ωE = 0

(4)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1093

PROBLEM 15.70 (Continued)

Solving the simultaneous equaitons (3) and (4),

ω AB = 0.75265 rad/s, Velocity of Point E.

ωE = −2.4605 rad/s v E = ωE k × rE /Q = −2.4605k × 6 j v E = 14.76 in./s i = 14.76 in./s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1094

PROBLEM 15.71 The 80-mm-radius wheel shown rolls to the left with a velocity of 900 mm/s. Knowing that the distance AD is 50 mm, determine the velocity of the collar and the angular velocity of rod AB when (a) β = 0, (b) β = 90°.

SOLUTION (a) β = 0.

Wheel AD.

vC = 0,

ω AD =

vD = 45 in./s

vD 900 = = 11.25 rad/s CD 80

CA = (CD ) − ( DA) = 80 − 50 = 30 mm v A = (CA)ω AD = (30)(11.25) = 337.5 mm/s

vB = vA + vB/ A

Rod AB.

] = [337.5

[vB

ϕ]

] + [vB/ A

Wheel AD.

vC = 0, tan γ = CA =



ω AB = 0 

vB/ A = 0

(b) β = 90°.

v B = 338 mm/s

ω AD = 11.25 rad/s

DA 50 = , DC 80

γ = 32.005°

DC = 94.34 mm cos γ

v A = (CA)ω AD = (94.34)(11.25) = 1061.3 mm/s

v A = [1061.3 mm/s

Rod AB.

32.005°]

v B = vB sin ϕ =

80 , ϕ = 18.663° 250

Plane motion = Translation with A + Rotation about A.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1095

PROBLEM 15.71 (Continued)

vB = vA + vB/ A

] = [v A

[vB

γ ] + [vB /A

ϕ]

Draw velocity vector diagram.

δ = 180° − γ − (90° + ϕ ) = 90° − 32.005° − 18.663° = 39.332°

Law of sines. vB/ A vB vA = = sin δ sin γ sin (90° + ϕ ) vB =

v A sin δ (1061.3)sin 39.332° = sin (90° + ϕ ) sin 108.663°

= 710 mm/s vB/ A =

vB = 710 mm/s



ω AB = 2.37 rad/s



v A sin γ (1061.3) sin 32.005° = sin (90° + ϕ ) sin108.663°

= 593.8 mm/s

ω AB =

vB/ A AB

=

593.8 = 2.37 rad/s 250

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1096

PROBLEM 15.72* For the gearing shown, derive an expression for the angular velocity ωC of gear C and show that ωC is independent of the radius of gear B. Assume that Point A is fixed and denote the angular velocities of rod ABC and gear A by ω ABC and ω A respectively.

SOLUTION

Label the contact point between gears A and B as 1 and that between gears B and C as 2. Rod ABC:

ω ABC = ω ABC

Assume

for sketch.

vA = 0 vB = ( rA + rB )ω ABC vC = (rA + 2rB + rC )ω ABC

Gear A: Gear B:

ω A = 0,

v A = 0,

v1 = 0

v1 = vB − rB ω B = 0

(rA + rB )ω ABC − rB ωB = 0  rA + rB  rB

ωB = 

  ω ABC 

v2 = vB + rB ω B = 2(rA + rB )ω ABC

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1097

PROBLEM 15.72* (Continued)

v2 = vC − rC ωC

Gear C:

2(rA + rB )ω ABC = ( rA + 2rB + rC )ω ABC − rC ωC

ωC = ( rA − rC )ω ABC = − rC ωC



ωC = 1 − 

rA   ω ABC  rC 

Note that the result is independent of rB .

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1098

PROBLEM 15.CQ5 The disk rolls without sliding on the fixed horizontal surface. At the instant shown, the instantaneous center of zero velocity for rod AB would be located in which region? (a) region 1 (b) region 2 (c) region 3 (d ) region 4 (e) region 5 ( f ) region 6

SOLUTION Answer: (a) 

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PROBLEM 15.CQ6 Bar BDE is pinned to two links, AB and CD. At the instant shown the angular velocities of link AB, link CD and bar BDE are ωAB, ωCD, and ωBDE, respectively. Which of the following statements concerning the angular speeds of the three objects is true at this instant? (a) ωAB = ωCD = ωBDE (b) ωBDE > ωAB > ωCD (c) ωAB = ωCD > ωBDE (d ) ωAB > ωCD > ωBDE (e) ωCD > ωAB > ωBDE

SOLUTION Answer: (e) 

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PROBLEM 15.73 A juggling club is thrown vertically into the air. The center of gravity G of the 20 in. club is located 12 in. from the knob. Knowing that at the instant shown G has a velocity of 4 ft/s upwards and the club has an angular velocity of 30 rad/s counterclockwise, determine (a) the speeds of Point A and B, (b) the location of the instantaneous center of rotation.

SOLUTION Unit vectors: Relative positions:

i =1

,

j =1 , k =1

 8  rA/G = −(1 ft)i, rB/A =  ft  i  12 

Angular velocity:

ω = 30 rad/s

= (30 rad/s)k

Velocity at A:

vA = v G + v A/G = vG + ω × rA/G = (4 ft/s) j + (30 rad/s)k × (−1 ft)i = (4 ft/s) j − (30 ft/s) j = −(26 ft/s) j = 26 ft/s

vA = 26.0 ft/s 

Velocity at B:

vB = v G + v B/G = vG + ω × rB/G  8  = (4 ft/s) j + (30 rad/s)k ×  ft  i  12  = (4 ft/s) j + (20 ft/s) j = (24 ft/s) j = 24 ft/s vB = 24.0 ft/s 

Let rC/ G = x i bet the position of the instantaneous center C relative to G. vC = v G + vC/G = v G + ω × ( x i ) = (4 ft/s) j + (30 rad/s)k × ( x i ) = (4 ft/s) j + (30 ft/s) x j = 0 x=−

4 ft/s 4 = − ft = −1.6 in. 30 rad/s 30

Point C lies 1.6 in. to the left of G. 

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PROBLEM 15.74 A 10-ft beam AE is being lowered by means of two overhead cranes. At the instant shown, it is known that the velocity of Point D is 24 in./s downward and the velocity of Point E is 36 in./s downward. Determine (a) the instantaneous center of rotation of the beam, (b) the velocity of Point A.

SOLUTION

ω= lCE =

(a) (b)

vE − vD 3 − 2 1 = = rad/s lED 3 3 vD

ω

=

2 1 3

= 6 ft

l AC = 3 + 4 − 6 = 1 ft

1 vA = l AC ω = (1)   = 0.3333 ft/s  3

C lies 1 ft to the right of A.  vA = 4.00 in./s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1102

PROBLEM 15.75 A helicopter moves horizontally in the x direction at a speed of 120 mi/h. Knowing that the main blades rotate clockwise with an angular velocity of 180 rpm, determine the instantaneous axis of rotation of the main blades.

SOLUTION v 0 = 120 mi/h = 176 ft/s

ω = 180 rpm =

(180)(2π ) = 18.85 rad/s 60

Top view v0 = zω z=

v0

ω

=

176 = 9.34 ft 18.85

Instantaneous axis is parallel to the y axis and passes through the point

x=0  z = 9.34 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1103

PROBLEM 15.76 A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that end E of the cord is pulled to the left with a velocity of 120 mm/s, determine (a) the angular velocity of the drums, (b) the velocity of the center of the drums, (c) the length of cord wound or unwound per second.

SOLUTION Since the drum rolls without sliding, its instantaneous center lies at D. v E = v B = 120 mm/s v A = v A /Dω ,

(a)

ω=

vB = rB/Dω

vB 120 = = 3 rad/s vB/D 100 − 60

ω = 3.00 rad/s 

(b)

v A = (100)(3) = 300 mm/s

v A = 300 mm/s



Since v A is greater than vB , cord is being wound. v A − vB = 300 − 120 = 180 mm/s

Cord wound per second = 180.0 mm 

(c)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1104

PROBLEM 15.77 A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that end E of the cord is pulled to the left with a velocity of 120 mm/s, determine (a) the angular velocity of the drums, (b) the velocity of the center of the drums, (c) the length of cord wound or unwound per second.

SOLUTION Since the drum rolls without sliding, its instantaneous center lies at B. v E = v D = 120 mm/s v A = rA/Bω , vD = rD /Bω

ω=

(a)

vD 120 = = 3 rad/s rD /B 100 − 60

ω = 3.00 rad/s



v A = (60)(3.00) = 180 mm/s

(b)

v A = 180 mm/s



Since v A is to the right and v D is to the left, cord is being unwound. v A − vE = 180 + 120 = 300 mm/s

Cord unwound per second = 300 mm 

(c)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1105

PROBLEM 15.78 The spool of tape shown and its frame assembly are pulled upward at a speed vA = 750 mm/s. Knowing that the 80-mm-radius spool has an angular velocity of 15 rad/s clockwise and that at the instant shown the total thickness of the tape on the spool is 20 mm, determine (a) the instantaneous center of rotation of the spool, (b) the velocities of Points B and D.

SOLUTION v A = 750 mm/s

ω = 15 rad/s x=

(a)

vA

ω

=

750 = 50 mm 15

The instantaneous center lies 50 mm to the right of the axle.  CB = 80 + 20 − 50 = 50 mm

(b)

vB = (CB )ω = (50)(15) = 750 mm/s

vB = 750 mm/s  CD = 80 + 50 = 130 mm vD = (CD )ω = (130)(15) = 1950 mm/s

vD = 1.950 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1106

PROBLEM 15.79 The spool of tape shown and its frame assembly are pulled upward at a speed v A = 100 mm/s. Knowing that end B of the tape is pulled downward with a velocity of 300 mm/s and that at the instant shown the total thickness of the tape on the spool is 20 mm, determine (a) the instantaneous center of rotation of the spool, (b) the velocity of Point D of the spool.

SOLUTION vD = vA = 100 mm/s

(a)

Since v 0 and vB are parallel, instantaneous center C is located at intersection of BC and line joining end points of vD and vB . Similar triangles. OC BC OC + BC = = v0 vB v0 + vB OC =

v0 (OC + BC ) v0 + vB

100 mm/s (100 mm) (100 + 300) mm/s = 25 mm

OC =

(b)

v vD = 0 ; ( DO ) + (OC ) (OC )

vD 100 mm/s = (80 + 25) mm 25 mm

 vD = 420 mm/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1107

PROBLEM 15.80 The arm ABC rotates with an angular velocity of 4 rad/s counterclockwise. Knowing that the angular velocity of the intermediate gear B is 8 rad/s counterclockwise, determine (a) the instantaneous centers of rotation of gears A and C, (b) the angular velocities of gears A and C.

SOLUTION Contact points: 1

between gears A and B.

2

between gears B and C.

ω ABC = 4 rad/s

Arm ABC:

v A = (0.300)(4) = 1.2 m/s vC = (0.300)(4) = 1.2 m/s

ωB = 8 rad/s

Gear B:

v1 = (0.200)(8) = 1.6 m/s v2 = (0.100)(8) = 0.8 m/s

Gear A:

ωA =

v1 − v A 1.6 − 1.2 = 0.100 0.100

ω A = 4 rad/s A =

vA

ωA

=

1.2 = 0.3 m = 300 mm 4

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1108

PROBLEM 15.80 (Continued) Gear C:

ωC =

vC − v2 1.2 − 0.8 = 0.200 0.2

ωC = 2 rad/s C =

vC

ωC

=

1.2 = 0.6 m 2

(a) Instantaneous centers. Gear A: 300 mm left of A  Gear C: 600 mm left of C  (b) Angular velocities. ω A = 4.00 rad/s



ωC = 2.00 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1109

PROBLEM 15.81 The double gear rolls on the stationary left rack R. Knowing that the rack on the right has a constant velocity of 2 ft/s, determine (a) the angular velocity of the gear, (b) the velocities of Points A and D.

SOLUTION Since the rack R is stationary, Point C is the instantaneous center of the double gear. Given:

v B = 2 ft/s = 24 in./s

Make a diagram showing the locations of Points A, B, C, and D on the double gear. vB = ω lCB

ω=

vB 24 in./s = = 2.40 rad/s 10 in. lCB

ω = 2.40 rad/s

(a)

Angular velocity of the gear.

(b)

Velocity of Point A.

v A = l AC ω = (4 in.)(2.40 rad/s)

Geometry:

lCD = (4 in.) 2 + (6 in.) 2 = 52 in. tan β =

Velocity of Point D.

4 in. 6 in.



vA = 9.60 in./s = 0.800 ft/s 

β = 33.7°

v D = lCDω = 52(2.40) = 17.31 in./s

vD = 1.442 ft/s

33.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1110

PROBLEM 15.82 An overhead door is guided by wheels at A and B that roll in horizontal and vertical tracks. Knowing that when θ = 40° the velocity of wheel B is 1.5 ft/s upward, determine (a) the angular velocity of the door, (b) the velocity of end D of the door.

SOLUTION Locate instantaneous center at intersection of lines drawn perpendicular to vA and vB. (a)

Angular velocity. vB = ( BC )ω 1.5 ft/s = (3.214 ft)ω ω = 0.4667 rad/s

(b)

ω = 0.467 rad/s



Velocity of D: In ΔCDE:

6.427 = 59.2° 3.83 6.427 = 7.482 ft CD = sin β vD = (CD )ω

β = tan −1

= (7.482 ft)(0.4667 rad/s) = 3.49 ft/s vD = 3.49 ft/s

59.2° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1111

PROBLEM 15.83 Rod ABD is guided by wheels at A and B that roll in horizontal and vertical tracks. Knowing that at the instant shown β = 60° and the velocity of wheel B is 40 in./s downward, determine (a) the angular velocity of the rod, (b) the velocity of Point D.

SOLUTION Rod ABD:

We locate the instantaneous center by drawing lines perpendicular to vA and vD. (a)

Angular velocity. vB = ( BC )ω 40 in./s = (12.99 in.)ω ω = 3.079 rad/s

(b)

ω = 3.08 rad/s



Velocity of D: In ΔCDE:

γ = tan −1

7.5 25.98 = 16.1°; CD = = 27.04 in. 25.98 cos γ

vD = (CD )ω = (27.04 in.)(3.079 rad/s) = 83.3 in./s vD = 83.3 in./s

vD = 83.3 in./s

16.1°

73.9° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1112

PROBLEM 15.84 Rod BDE is partially guided by a roller at D which moves in a vertical track. Knowing that at the instant shown the angular velocity of crank AB is 5 rad/s clockwise and that β = 25°, determine (a) the angular velocity of the rod, (b) the velocity of Point E.

SOLUTION Crank AB:

ω AB = 5 rad/s

rB /A = 120 mm

vB = ω AB rB /A = (5)(0.120)

v B = 0.6 m/s

Rod BDE: Draw a diagram of the geometry of the rod and note that v B = 0.6 m/s

and v D = v D .

Locate Point C, the instantaneous center, by noting that BC is perpendicular to v B and DC is perpendicular to vD. Calculate lengths of BC and CD.

lBC = 500sin 25° = 211.31 mm lCD = 500cos 25° = 453.15.

(a)

Angular velocity of the rod.

ωBCD =

vB 0.6 m/s = = 2.8394 rad/s lBC 0.21131 m ω BCD = 2.84 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1113

PROBLEM 15.84 (Continued)

(b)

Velocity of Point E. Locate Point F on the diagram. CF = 700cos 25° mm

FE = 200sin 25°

FE 200sin 25° 2 = = tan 25° = 0.13323 CF 700cos 25° 7 γ = 7.6° β = 90° − γ = 82.4°

tan γ =

lCE = (CF ) 2 + ( FE )2 = 640.02 mm = 0.64002 m vE = lCE ω = (0.64002)(2.8394) v E = 1.817 m/s

82.4° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1114

PROBLEM 15.85 Rod BDE is partially guided by a roller at D which moves in a vertical track. Knowing that at the instant shown β = 30°, Point E has a velocity of 2 m/s down and to the right, determine the angular velocities of rod BDE and crank AB.

SOLUTION Crank AB: When AB is vertical, the velocity v B at Point B is horizontal. Rod BDE: Draw a diagram of the geometry of the rod and note that v B is horizontal and vD is vertical.

Locate Point C, the instantaneous center C, by noting that CB is vertical and CD is horizontal. From the diagram, with Point F added, CF = 700cos 30° mm

FE = 200sin 30° mm

CE = (CF ) 2 + ( FE )2 = 614.41 mm = 0.61441 m

Angular velocity of rod BDE

ωBDE =

vE 2 m/s = = 3.2552 rad/s (CE ) 0.61441 m ω BDE = 3.26 rad/s

Velocity of B.



CB = 500sin 30° mm = 250 mm = 0.250 m vB = (CB)ω BDE = (0.250)(3.2552) v B = 0.81379 m/s

Angular velocity of crank AB:

AB = 120 mm = 0.120 m

ω AB =

vB 0.81379 m/s = ( AB) 0.120 m

ω AB = 6.78 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1115

PROBLEM 15.86 Knowing that at the instant shown the velocity of collar D is 1.6 m/s upward, determine (a) the angular velocity of rod AD, (b) the velocity of Point B, (c) the velocity of Point A.

SOLUTION

We draw perpendiculars to vB and vD to locate instantaneous center C. (a)

Angular velocity: vD = (CD)ω

1.6 m/s = (0.31177 m)ω

ω = 5.132 rad/s

ω = 5.13 rad/s



vB = ( BC )ω = (180 mm)(5.132 rad/s)

(b)

vD = 923.76 mm/s

vD = 0.924 m/s



vA = ( AC )ω

(c) In triangle ACE:

tan β =

207.85 mm 300 mm

β = 34.72°

AC = (207.85) 2 + (300) 2

AC = 364.97 mm

vA = (364.97 mm)(5.132 rad/s) = 1873.0 mm/s vA = 1.870 m/s

34.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1116

PROBLEM 15.87 Knowing that at the instant shown the angular velocity of rod BE is 4 rad/s counterclockwise, determine (a) the angular velocity of rod AD, (b) the velocity of collar D, (c) the velocity of Point A.

SOLUTION Rod AD. v B = rB/E ω BE = (0.192)(4) = 0.768 m/s

(a)

Instantaneous center C is located by noting that CD is perpendicular to vD and CB is perpendicular to vB. rB/C = 0.360 sin 30° = 0.180 m



ω AD =

 vB 0.768 = = 4.2667 rB/C 0.180 ω AD = 4.27 rad/s

(b)

Velocity of D.



rD/C = 0.360 cos 30° = 0.31177 m vD = rD/C ω = (0.31177)(4.2667) vD = 1.330 m/s 

(c)

Velocity of A. l AE = 0.240 cos 30° = 0.20785 m lCE = 0.600sin 30° = 0.300 m tan β =

0.20785 0.300

β = 34.7°

lCA = (0.20785) 2 + (0.300) 2 = 0.36497 m vA = lCAω AD = (0.36497)(4.2667) = 1.557 m/s vA = 1.557 m/s

34.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1117

PROBLEM 15.88 Rod AB can slide freely along the floor and the inclined plane. Denoting by vA the velocity of Point A, derive an expression for (a) the angular velocity of the rod, (b) the velocity of end B.

SOLUTION Locate the instantaneous center at intersection of lines drawn perpendicular to vA and vB . Law of sines. AC BC = sin[90° − ( β − θ )] sin(90° − θ ) l = sin β AC BC = cos( β − θ ) cos θ l = sin β cos( β − θ ) AC = l sin β cos θ BC = l sin β

(a)

Angular velocity:

vA = ( AC )ω = l

cos( β − θ ) ω sin β

(b)

Velocity of B:

vB = ( BC )ω = l

cos θ  vθ sin β  ⋅ ⋅  sin β  l cos( β − θ ) 

ω=

vA sin β ⋅  l cos( β − θ )

vB = vA

cos θ  cos( β − θ )

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1118

PROBLEM 15.89 Rod AB can slide freely along the floor and the inclined plane. Knowing that θ = 20°, β = 50°, l = 2 ft, and vA = 8 ft/s, determine (a) the angular velocity of the rod, (b) the velocity of end B.

SOLUTION Locate the instantaneous center at intersection of lines draw perpendicular to vA and vB. Law of sines. AC BC = sin[90° − ( β − θ )] sin(90° − θ ) l = sin β AC BC = cos( β − θ ) cos θ l = sin β cos( β − θ ) AC = l sin β cos θ BC = l sin β

Angular velocity:

vA = ( AC )ω = l

ω= Velocity of B:

cos( β − θ ) ω sin β

vA sin β ⋅ l cos( β − θ )

vB = ( BC )ω = l vB = vA

cos θ sin β

v sin β  ⋅ A ⋅   l cos( β − θ ) 

cos θ cos( β − θ )

Data:

θ = 20°, β = 50°, l = 2 ft, v A = 8 ft/s

(a)

ω=

vA sin β 8 ft/s sin 50° = ⋅ l cos( β − θ ) 2 ft cos(50° − 20°)

ω = 3.5382 rad/s

ω = 3.54 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1119

PROBLEM 15.89 (Continued)

(b)

cos θ cos( β − θ ) cos 20° = (8 ft/s) cos(50° − 20°)

vB = vA

vB = 8.6805 ft/s

vB = 8.68 ft/s

50° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1120

PROBLEM 15.90 Two slots have been cut in plate FG and the plate has been placed so that the slots fit two fixed pins A and B. Knowing that at the instant shown the angular velocity of crank DE is 6 rad/s clockwise, determine (a) the velocity of Point F, (b) the velocity of Point G.

SOLUTION vE = ( DE )ωDE = (72 mm)(6 rad/s)

Crank DE:

vE = 432 mm/s

(vF ) y = vE = 432 mm/s

Rod EF: Plate FG:

vA and vB are velocities of points on the plate next to the pins A and B. We draw lines perpendicular to vA and vB to locate the instantaneous center C.

(a)

Velocity of Point F: vF = (CF )ω (vF ) y = [(CF )ω ]cos β = [(CF ) cos β ]ω

But

(vF ) y = 432 mm/s and (CF ) cos β = 480 mm: 432 mm/s = (480 mm)ω ω = 0.9 rad/s

ω = 0.9 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1121

PROBLEM 15.90 (Continued)

CF = 487.97 mm β = 10.37° vF = (CF )ω = (487.97 mm)(0.9 rad/s) = 439.18 mm/s vF = 439 mm/s

(b)

10.4°

v F = 439 mm/s

79.6° 

vG = 411 mm/s

20.5° 

Velocity of Point G: CG = 456.78 mm γ = 20.50° vG = (CG )ω = (456.78 in.)(0.9 rad/s) vG = 411.11 mm/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1122

PROBLEM 15.91 The disk is released from rest and rolls down the incline. Knowing that the speed of A is 1.2 m/s when θ = 0°, determine at that instant (a) the angular velocity of the rod, (b) the velocity of B. Only portions of the two tracks are shown.

SOLUTION Draw the slider, rod, and disk at θ = 0°. Let Point P be the contact point between the disk and the incline. It is the instantaneous center of the disk. vA is parallel to the incline. So that vA = vA

30°

Constraint of slider: vB = vB

To locate the instantaneous center C of the rod AB, extend the line AP to meet the vertical line through P at Point C. l AC = l AB / sin 30° lBC = l AB / tan 30°

(a)

Angular velocity of rod AB.

ω AB =

v A v A sin 30° (1.2 m/s) sin 30° = = l AC l AB 0.6 m ω AB = 1.000 rad/s

(b)

Velocity of Point B.



vB = lBC ω AB vB =

l AB v A sin 30° = v A cos30° = 1.2 cos 30° tan 30° l AB v B = 1.039 m/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1123

PROBLEM 15.92 Arm ABD is connected by pins to a collar at B and to crank DE. Knowing that the velocity of collar B is 400 mm/s upward, determine (a) the angular velocity of arm ABD, (b) the velocity of Point A.

SOLUTION

vB = 16 in./s

tan γ =

EF 125 = DF 300

v D = vD

γ

Locate the instantaneous center (Point C) of bar ABD by noting that velocity directions at Points B and D are known. Draw BC perpendicular to vB and DC perpendicular to vD.  125  CJ = ( DJ ) tan γ = (160)   = 66.667 mm  300  CB = JB − CJ = 320 − 66.667 = 253.33 mm

(a)

ω ABD =

vB 400 = = 1.57895 rad/s CB 253.33

ω ABD = 1.579 rad/s



CK = CB + BK = 253.33 + 180 = 433.33 mm KA 90 tan β = = , β = 11.733°, 90° − β = 78.3° CK 433.33 CK 433.33 AC = = = 442.58 mm cos β cos 11.733°

(b)

vA = ( AC )ω ABD = (442.58)(1.57895) = 699 mm/s

vA = 699 mm/s

78.3° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1124

PROBLEM 15.93 Arm ABD is connected by pins to a collar at B and to crank DE. Knowing that the angular velocity of crank DE is 1.2 rad/s counterclockwise, determine (a) the angular velocity of arm ABD, (b) the velocity of Point A.

SOLUTION

tan γ =

300 EF 125 FD = = = 325 mm , γ = 22.620°, ED = cos γ cos γ DF 300

vD = ( ED)ωDE = (325)(1.2) = 390 mm/s v D = 390 mm/s

γ

v B = vB

Locate the instantaneous center (Point C) of bar ABD by noting that velocity directions at Points B and D are known. Draw BC perpendicular to vB and DC perpendicular to vD. 160 DJ  125  = 66.667 mm, CD = = = 173.33 mm CJ = ( DJ ) tan γ = (160)   cos γ cos γ  300 

(a)

ω ABD =

vD 390 = = 2.25 rad/s CD 173.33

ω ABD = 2.25 rad/s



CK = KJ − CJ = 500 − 66.667 = 433.33 mm AK 90 tan β = = 90° − β = 78.3° β = 11.733° CK 433.33 CK 433.33 AC = = = 442.58 mm cos β cos β

(b)

vA = ( AC )ω ABD = (442.58)(2.25) = 996 mm/s

vA = 996 mm/s

78.3° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1125

PROBLEM 15.94 Two links AB and BD, each 25 in. long, are connected at B and guided by hydraulic cylinders attached at A and D. Knowing that D is stationary and that the velocity of A is 30 in./s to the right, determine at the instant shown (a) the angular velocity of each link, (b) the velocity of B.

SOLUTION Link DB: Point D is stationary. Assume ω BD = ωBD .

vB = ( DB)ωBD tan β =

7 in. 24 in.

v B is perpendicular to DB.

β = 16.3°

90° − β = 73.7°

Link AB: Draw the configuration. Locate the instantaneous center C of link AB by noting that the line BC is perpendicular to v B , i.e., along DB, and that AC is perependicular to v A = v A . (v A = 30 in./s).

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1126

PROBLEM 15.94 (Continued)

AC = AE + EC = 13 in. + (7 in.) BC =

ω AB =

9 in. = 15.625 in. 24 in.

15 in. (25 in.) = 15.625 in. 24 in. vA 30 in./s = = 1.92 rad/s ( AC ) 15.625 in.

vB = ( BC )ω AB = (15.625)(1.92) = 30 in./s

Returning to link DB, (a)

(b)

ωBD =

vB 30 in./s = = 1.20 rad/s ( DB) 25 in.

Angular velocities:

ω AB = 1.920 rad/s



ω BD = 1.200 rad/s



vB = 30.0 in./s

Velocity of Point B:

73.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1127

PROBLEM 15.95 Two 25-in. rods are pin-connected at D as shown. Knowing that B moves to the left with a constant velocity of 24 in./s, determine at the instant shown (a) the angular velocity of each rod, (b) the velocity of E.

SOLUTION

Rod AB:

Draw lines perpendicular to vA and vB to locate instantaneous center C AB . vB = ( BC AB )ω AB 24 in./s = (20 in.)ω AB





Velocity of D:

ω AB = 1.200 rad/s





DC AB = 12.5 in. vD = ( DC AB )ω AB = (12.5 in.)(1.2 rad/s) vD = 15 in./s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1128

PROBLEM 15.95 (Continued)

Rod DE:

Draw lines perpendicular to vD and vE to locate instantaneous center CDE . DCDE = (20)2 + (26.667)2 = 33.333 in. vD = ( DC DE )ω DE

(a)

15 in./s = (33.333 in.)ωDE ; ωDE = 0.45 rad/s

ω DE = 0.450 rad/s



vE = ( ECDE )ωDE = (11.667 in.)(0.45 rad/s)

(b)

vE = 5.25 in./s

vE = 5.25 in./s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1129

PROBLEM 15.96 Two rods ABD and DE are connected to three collars as shown. Knowing that the angular velocity of ABD is 5 rad/s clockwise, determine at the instant shown (a) the angular velocity of DE, (b) the velocity of collar E.

SOLUTION

ω ABC = 5 rad/s v B = vB

v A = vA v E = vE

Locate Point I, the instantaneous center of rod ABD by drawing IA perependicular to vA and IB perpendicular to vB. 200 φ = 26.565° 400 400 I ID = = 447.21 mm cos φ vD = ω ABD lID = (5)(447.21 mm)

tan φ =

vD = 2236.1 mm/s

φ

Locate Point J, the instantaneous center of rod DE by drawing JD perpendicular to vD and JE perpendicular to vE. l JD =

ωDE =

400 = 447.21 mm cos φ vD 2236.1 mm/s = = 5 rad/s l JD 447.21 ωDE = 5.00 rad/s 

(a) l JE = 200 + l JD cos φ = 600 mm vE = l JE ωDE = (600)(5) = 3000 mm/s

v E = 3.00 m/s 

(b)

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PROBLEM 15.97 Two collars C and D move along the vertical rod shown. Knowing that the velocity of collar C is 660 mm/s downward, determine (a) the velocity of collar D, (b) the angular velocity of member AB.

SOLUTION AB = 400 mm

Instantaneous centers:

at I for BC. at J for BD.

Geometry.  240  IC =   (220) = 165 mm  320   240  JD =   (420) = 315 mm  320   220  AI =   (400) = 275 mm  320  BI = AB − AI = 400 − 275 = 125 mm BJ = BI = 125 mm

Member BC.

vC = 660 mm/s

vC 660 = = 4 rad/s IC 165 vB = ( BI )ωBC = (125 mm)(4 rad/s) = 500 mm/s

ωBC =

Member BD. (a) (b)

ωBD =

vB 500 mm/s = = 4 rad/s 125 mm BJ

v D = ( JD)ωBD = (315 mm)(4 rad/s)

ω AB =

vB 500 mm/s = AB 400 mm

v D = 1260 mm/s 

ω AB = 1.250 rad/s



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PROBLEM 15.98 Two rods AB and DE are connected as shown. Knowing that Point D moves to the left with a velocity of 40 in./s, determine (a) the angular velocity of each rod, (b) the velocity of Point A.

SOLUTION

We locate two instantaneous centers at intersections of lines drawn as follows: C1:

For rod DE, draw lines perpendicular to vD and vE.

C2:

For rod AB, draw lines perpendicular to vA and vB.

Geometry:

BC1 = (8 in.) 2 = 8 2 in. DC1 = 16 in. BC2 = (9 in. + 8 in.) 2 = 17 2 in. AC2 = 25 in.

(a)

Rod DE:

vD = ( DC1 )ω DE

40 in./s = (16 in.)ω DE

ωDE = 2.5 rad/s

ω DE = 2.5 rad/s



vB = ( BC )ωDE

= (8 2 in.)(2.5 rad/s) vB = 20 2 in./s

45°

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1132

PROBLEM 15.98 (Continued)

Rod AB:

vB = ( BC2 )ω AB

20 2 in./s = (17 2 in.)ω AB

ω AB =

20 rad/s = 1.1765 rad/s 17 ω AB = 1.177 rad/s

(b)



vA = ( AC2 )ω AB

= (25 in.)(1.1765 rad/s) vA = 29.41 in./s

vA = 29.4 in./s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1133

PROBLEM 15.99 Describe the space centrode and the body centrode of rod ABD of Problem 15.83. (Hint: The body centrode need not lie on a physical portion of the rod.) PROBLEM 15.83 Rod ABD is guided by wheels at A and B that roll in horizontal and vertical tracks. Knowing that at the instant shown β = 60° and the velocity of wheel B is 40 in./s downward, determine (a) the angular velocity of the rod, (b) the velocity of Point D.

SOLUTION Draw x and y axes as shown with origin at the intersection of the two slots. These axes are fixed in space. vA = vA

vB = vB

,

Locate the space centrode (Point C) by noting that velocity directions at Points A and B are known. Draw AC perpendicular to vA and BC perpendicular to vB.  The coordinates of Point C are xC = −l sin β and yC = l cos β 

xC2 + yC2 = l 2 = (15 in.) 2



The space centrode is a quarter circle of 15 in. radius centered at O. 



Redraw the figure, but use axes x and y that move with the body. Place origin at A.

 

xC = ( AC ) cos β



= l cos 2 β =



l (1 + cos 2β ) 2

yC = ( AC ) sin β

= l cos β sin β = 2



l sin 2β 2

2

l  l 2 2 2 2  xC − 2  + yC =  2  = ( xC − 7.5) + yC = 7.5    

The body centrode is a semicircle of 7.5 in. radius centered midway between A and B.



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PROBLEM 15.100 Describe the space centrode and the body centrode of the gear of Sample Problem 15.2 as the gear rolls on the stationary horizontal rack.

SOLUTION

Let Points, A, B, and C move to A′, B′, and C′ as shown. Since the instantaneous center always lies on the fixed lower rack, the space centrode is the lower rack. space centrode: lower rack  Since the point of contact of the gear with the lower rack is always a point on the circumference of the gear, the body centrode is the circumference of the gear. body centrode: circumference of gear 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1135

PROBLEM 15.101 Using the method of Section 15.7, solve Problem 15.60. PROBLEM 15.60 In the eccentric shown, a disk of 2-in.-radius revolves about shaft O that is located 0.5 in. from the center A of the disk. The distance between the center A of the disk and the pin at B is 8 in. Knowing that the angular velocity of the disk is 900 rpm clockwise, determine the velocity of the block when θ = 30°.

SOLUTION

(OA) = 0.5 in. ωOA = 900 rpm = 30π rad/s vA = (OA)ωOA = (0.5)(30π ) = 15π in./s vA = vA

60°,

vB = vA

Locate the instantaneous center (Point C) of bar BD by noting that velocity directions at Point B and A are known. Draw BC perpendicular to vB and AC perpendicular to vA. sin β =

(OA) sin 30° 0.5sin 30° = , β = 1.79° AB 8

OB = (OA) cos 30° + ( AB) cos β = 0.5cos 30° + 8cos β = 8.4291 in. AC =

OB 8.4291 − OA = − 0.5 = 9.2331 in. cos 30° cos 30°

BC = (OB) tan 30° = 4.8665 in.

ω AB =

vA v = B AC BC

(4.8665)(15π )  BC  = 24.84 in./s vB =  vA =  9.2331  AC 

vB = 24.8 in./s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1136

PROBLEM 15.102 Using the method of Section 15.7, solve Problem 15.64. PROBLEM 15.64 In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.

SOLUTION ω AB = 4 rad/s

Bar AB: (Rotation about A)

vB = ω AB ( AB) = (4)(175)

AB = 175 mm v B = 700 mm/s

ω DE = ωDE

Bar DE: (Rotation about E)

DE = (275) 2 + (75)2 = 285.04 mm

β

v D = 285.04ωDE tan β =

Bar BD:

75 mm = 0.27273 275 mm

v B = 700 mm/s ,

v D = 285.04ωDE

β

Locate the instantaneous center of bar BD by drawing line BC perpendicular to vB and line DC perpendicular to vD. BD = 200 mm CB =

BD (200)(275) = = 733.3 mm tan β 75

CD =

BD (200)(285.04) = = 760.11 mm sin β 75

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1137

PROBLEM 15.102 (Continued)

ωBD =

vB 700 mm/s = = 0.95455 rad/s CB 733.33 mm

vD = ωBD (CD ) = (0.95455 rad/s)(760.11 mm) = 725.56 mm/s

ωDE =

vD 725.56 = = 2.5455 rad/s 285.04 285.04

Angular velocities:

ω BD = 0.955 rad/s



ω DE = 2.55 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1138

PROBLEM 15.103 Using the method of Section 15.7, solve Problem 15.65. PROBLEM 15.65 In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.

SOLUTION

β = tan −1

Bar AB:

0.4 m = 26.56° 0.8 m

0.8 m = 0.8944 m cosβ vB = ( AB )ω AB = (0.8944 m)(4 rad/s)

AB =

vB = 3.578 m/s

 

26.56°

0.4 m = 38.66° 0.5 m 0.5 m DE = = 0.6403 m cos γ vD = ( DE )ωDE

γ = tan −1

Bar DE:

vD = (0.6403 m)ωDE

 

38.66°

(1)

Bar BD: Locate instantaneous center at intersection of lines drawn perpendicular to vB and vD. Law of sines.

BC CD 0.8 m = = sin 51.34° sin 63.44° sin 65.22° BC = 0.688 m CD = 0.7881 m vB = ( BC )ωBD 3.578 m/s = (0.688 m)ω BD ;

ω BD = 5.2 rad/s



ω DE = 6.4 rad/s



vD = (CD )ω BD = (0.7881 m)(5.2 m/s)



= 4.098 m/s Eq. (1):

4.098 m/s = (0.6403 m)ωDE ;

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1139

PROBLEM 15.104 Using the method of section 15.7, solve Problem 15.38. PROBLEM 15.38 An automobile travels to the right at a constant speed of 48 mi/h. If the diameter of a wheel is 22 in., determine the velocities of Points B, C, D, and E on the rim of the wheel.

SOLUTION vA = 48 mi/h = 70.4 ft/s vC = 0 

d = 22 in., r =

1 d = 11 in. = 0.91667 ft 2

Point C is the instantaneous center. vA 70.4 = = 76.8 rad/s r 9.1667 CB = 2r = 1.8333 ft vB = (CB)ω = (1.8333)(76.8) = 140.8 ft/s

ω=

vB = 140.8 ft/s



1 2

γ = (30°) = 15° CD = 2r cos15° = (2)(0.91667) cos15° = 1.7709 ft vD = (CD )ω = (1.7709)(76.8) = 136.0 ft/s vD = 136.0 ft/s

15.0° 

CE = r 2 = 0.91667 2 = 1.2964 ft vE = (CE )ω = (1.2964)(76.8) = 99.56 ft/s vE = 99.6 ft/s

45.0° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1140

PROBLEM 15.CQ7 A rear wheel drive car starts from rest and accelerates to the left so that the tires do not slip on the road. What is the direction of the acceleration of the point on the tire in contact with the road, that is, Point A? (a)

(b)

(c)

(d)

(e)

SOLUTION The tangential acceleration will be zero since the tires do not slip, but there will be an acceleration component perpendicular to the ground. Answer: (c) 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1141

PROBLEM 15.105 A 3.5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow down the unwinding motion. At the instant considered, the deceleration of the cable attached at A is 4 m/s2, while that of the cable at B is 1.5 m/s2. Determine (a) the angular acceleration of the beam, (b) the acceleration of Point C.

SOLUTION a A = 4 m/s a B = 1.5 m/s

Assume ω = 0. (a)

Angular acceleration.

α = αk

a B = a A + a B /A 1.5 j = 4 j + α k × (3i) = 4 j + 3α j 1.5 − 4 = −0.83333 α= 3 α = −0.833k

(b)

α = 0.833 rad/s 2

Acceleration of Point C. Because the cables are unwinding at the same speed, ω = 0 aC = a A + aC/A = a A + α × rC/A = 4 j + (−0.83333k × 3.5i ) = 4 j − 2.9167 j = 1.0833j aC = 1.083 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1142

PROBLEM 15.106 The acceleration of Point C is 0.3 m/s 2 downward and the angular acceleration of the beam is 0.8 rad/s 2 clockwise. Knowing that the angular velocity of the beam is zero at the instant considered, determine the acceleration of each cable.

SOLUTION

ω=0 α = (−0.8 rad/s)k aC = 0.3 m/s

Acceleration of cable A. a A = aC + α × rA /C = −0.3j + [−0.8k × ( −3.5i )] = −0.3j + 2.8 j = (2.5 m/s 2 ) j

a A = 2.50 m/s 2 

Acceleration of cable B. a B = aC + α × rB /C = −0.3j + [ −0.8k × (−0.5i )] = −0.3j + 0.4 j = 0.1j = (0.1 m/s 2 ) j

a B = 0.100 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1143

PROBLEM 15.107 A 900-mm rod rests on a horizontal table. A force P applied as shown produces the following accelerations: aA = 3.6 m/s 2 to the right, α = 6 rad/s 2 counterclockwise as viewed from above. Determine the acceleration (a) of Point G, (b) of Point B.

SOLUTION 



]

aG = [3.6 m/s 2

] + [(0.45 m)(6 rad/s 2 )

aG = [3.6 m/s 2

] + [2.7 m/s 2

] + [(0.9 m)(6 rad/s 2 )

a B = [3.6 m/s 2

] + [5.4 m/s 2



aG = 0.9 m/s 2



a B = 1.8 m/s 2



]

a B = [3.6 m/s 2



]

]

] + [( AB)α

a B = a A + a B/ A = [a A

(b)



] + [( AG )α

aG = a A + aG/ A = [ a A

(a)

]

]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1144

PROBLEM 15.108 In Problem 15.107, determine the point of the rod that (a) has no acceleration, (b) has an acceleration of 2.4 m/s 2 to the right.

SOLUTION (a)

For aQ = 0: + ( AQ )α

aQ = a A + aQ/A = a A 0 = 3.6 m/s 2

y=

+ ( y )(6 rad/s 2 )

3.6 m/s 2 = 0.6 m 6 rad/s a = 0 at 0.6 m from A 

(b)

For aQ = 2.4 m/s 2

: ] + [( AQ )α

]

] + [( y )(6 rad/s 2 )

]

aQ = a A + aQ /A = [aA 2.4 m/s 2

= [3.6 m/s 2

1.2 m/s 2

= ( y )(6 rad/s 2 ) y = 0.2 m

a = 2.4 m/s 2

at 0.2 m from A 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1145

PROBLEM 15.109 Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of Point A, (b) of Point D.

SOLUTION Geometry.

Let β be angle BAC. sin β =

Velocity analysis.

4 in. 8 in.

β = 30°

ω BC = 45 rpm

= 4.7124 rad/s

vA = v A

vB = ( BC )ωBC = (4)(4.7124) = 18.8496 in./s

vB = 18.8496 in./s

vA and vB are parallel; hence, the instantaneous center of rotation of rod AD lies at infinity. ω AD = 0

α BC = 0

Acceleration analysis. Crank BC:

vA = vB = 18.8496 in./s

(aB )t = ( BC )α = 0 2 (aB ) n = ( BC )ωBC = (4)(4.7124) 2

a B = 88.827 in./s 2 α AD = α AD

Rod ABD:

a A = aA

a A = a B + (a A/B )t + (a A/B ) A [a A ] = [88.827

] + [8α AD

2 30°] + [8ω AD

60°]

Resolve into components. : (a)

0 = 88.827 + 8α AD cos 30° + 0

α AD = −12.821 rad/s 2

: a A = 8α AD sin 30° = (8)(−12.821)sin 30° = −51.284 in./s 2 a A = 51.3 in./s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1146

PROBLEM 15.109 (Continued)

(b)

a D = a B + (a D/B )t + (a D/B )t = [88.827

] + [8α BD

2 30°] + [8ωBD [

= [88.827

] + [(8)(−12.821)

= [88.827

] + [102.568

60°]

30°] + 0

30°] = [177.653

+ 51.284 ]

a D = 184.9 in./s 2

16.1° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1147

PROBLEM 15.110 End A of rod AB moves to the right with a constant velocity of 6 ft/s. For the position shown, determine (a) the angular acceleration of rod AB, (b) the acceleration of the midpoint G of rod AB.

SOLUTION Use units of ft and seconds. i =1

Geometry and unit vectors:

,

j =1 ,

k =1

rB /A = −(10cos 30°)i + (10sin 30°) j

rB/D = −4 j

Velocity analysis. v A = 6 ft/s

Rod AB:

,

v B = vB

Since vA and vB are parallel, the instantaneous center lies at infinity, so ω AB = 0 and vB = v A . Acceleration analysis. α AB = α AB

Rod AB: a A = 0 since vA is constant.

= α AB k

2 a B = a A + a B /A = a A + a AB × rB /A − ω AB rB /A

= 0 + α AB k × (−10 cos 30°i + 5 j) − 0 a B = −(10 cos 30°)α AB j − 5α AB j

Rod BD:

(1)

a D = 0, α BD = α BD k 2 a B = a D + a B /D = 0 + α BD × rB /D − ωBD rB /D

= α BD k × (−4 j) − (1.5) 2 (−4 j) = 4α BD i + 9 j

(2)

Equating the coefficients of j in the expressions (1) and (2) for a B . −(10 cos 30°)α AB = 9

α AB = −1.0392

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1148

PROBLEM 15.110 (Continued)

(a)

Angular acceleration of rod AB:

(b)

Acceleration of midpoint G of rod AB.

α AB = 1.039 rad/s 2



2 aG = a A + aG /A = a A + α AB k × rG /A − ω AB rG /A

= 0 − 1.0392k × (−5cos 30°i + 5sin 30° j) aG = (2.60 ft/s 2 )i + (4.50 ft/s 2 ) j = 5.20 ft/s 2

60°

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1149

PROBLEM 15.111 An automobile travels to the left at a constant speed of 72 km/h. Knowing that the diameter of the wheel is 560 mm, determine the acceleration (a) of Point B, (b) of Point C, (c) of Point D.

SOLUTION vA = 72 km/h ⋅

h 1000 m ⋅ = 20 m/s 3600 s km

Rolling with no sliding, instantaneous center is at C. vA = ( AC )ω ; 20 m/s = (0.28 m)ω

ω = 71.429 rad/s

Acceleration.

Plane motion = Trans. with A + Rotation about A aB /A = aC /A = aD /A = rω 2 = (0.280 m)(71.429 rad/s)2 = 1428.6 m/s2

(a)

a B = a A + a B /A = 0 + 1428.6 m/s 2

a B = 1430 m/s 2 

(b)

aC = a A + aC /A = 0 + 1428.6 m/s 2

aC = 1430 m/s 2 

(c)

a D = a A + a D /A = 0 + 1428.6 m/s 2

60° a D = 1430 m/s 2



60° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1150

PROBLEM 15.112 The 18-in.-radius flywheel is rigidly attached to a 1.5-in. -radius shaft that can roll along parallel rails. Knowing that at the instant shown the center of the shaft has a velocity of 1.2 in./s and an acceleration of 0.5 in./s 2 , both directed down to the left, determine the acceleration (a) of Point A, (b) of Point B.

SOLUTION Velocity analysis. Let Point G be the center of the shaft and Point C be the point of contact with the rails. Point C is the instantaneous center of the wheel and shaft since that point does not slip on the rails. vG = rω , ω =

vG 1.2 = = 0.8 rad/s r 1.5

Acceleration analysis. Since the shaft does not slip on the rails, aC = aC

20°

aG = [0.5 in./s 2

Also,

20°]

aC = aG + (aC/G )t + (aC/G )n

[aC

Components (a)

20°] = [0.5 in./s 2

20°:

0.5 = −1.5α

20°] + [1.5α

20°] + [1.5ω 2

20°]

α = 0.33333 rad/s 2

Acceleration of Point A. aA = aG + (aA/G )t + (aA/G ) n

= [0.5

(b)

20°] + [18α

] + [18ω 2 ]

= [0.4698

] + [0.1710 ] + [6

= [6.4698

] + [11.670 ]

] + [11.52 ] a A = 13.35 in./s 2

61.0° 

a B = 12.62 in./s 2

64.0° 

Acceleration of Point B. a B = aG + (a B/G )t + (a B/G ) n

= [0.5

20°] + [18α

] + [18ω 2 ]

= [0.4698

] + [0.1710 ] + [6

= [5.5302

] + [11.349 ]

] + [11.52 ]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1151

PROBLEM 15.113 A 3-in.-radius drum is rigidly attached to a 5-in.-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that at the instant shown end D of the cord has a velocity of 8 in./s and an acceleration of 30 in./s 2, both directed to the left, determine the accelerations of Points A, B, and C of the drums.

SOLUTION Velocity analysis.

v D = vA = 8 in./s

Instantaneous center is at Point B.

v A = ( AB)ω , 8 = (5 − 3)ω

ω = 4 rad/s Acceleration analysis.

a B = [aB ] for no slipping.

α =α a A = [30 in./s 2 aG = [aG

] + [( a A )n ]

]

a B = a A + (a B/A )t + (a B/A ) n

[aB ] = [30

Components

:

] + [( a A )n ] + [(5 − 3)α

0 = −30 + 2α

] + [5 − 3)ω 2 ]

α = 15 rad/s 2

a B = aG + (a B /G )t + (a B / G ) n

[aB ] = [aG

Components

: :

] + [5α

0 = − aG + 5α

] + [5ω 2 ] aG = 5α = 75 in./s 2

aB = (5)(4) 2 = 80 in./s 2

a B = 80.0 in./s 2 

a A = aG + (a A/G )t + (a A/G ) n = [75

] + [3α

] + [3ω 2 ]

= [75

] + [45

] + [48 ]

= [30 in./s 2

] + [48 in./s 2 ] a A = 56.6 in./s 2

58.0° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1152

PROBLEM 15.113 (Continued)

aC = aG + (aC/G )t + (aC/G )n = [75

] + [5α ] + [5ω 2

= [75

] + [75 ] + [80

= [155 in./s 2

] ]

] + [75 in./s 2 ] aC = 172.2 in./s 2



25.8° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1153

PROBLEM 15.114 A 3-in.-radius drum is rigidly attached to a 5-in.-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that at the instant shown end D of the cord has a velocity of 8 in./s and an acceleration of 30 in./s 2 , both directed to the left, determine the accelerations of Points A, B, and C of the drums.

SOLUTION Velocity analysis.

vD = vB = 8 in./s

Instantaneous center is at Point A.

vB = ( AB)ω ,

8 = (5 − 3)ω

ω = 4 rad/s Acceleration analysis.

a A = [a A ] for no slipping. α = α a B = [30 in./s 2 aG = [aG

] + [( aB )n ]

]

a A = a B + (a A/B )t + (a A/B ) n

[a A ] = [30

Components

:

] + [( aB )n ] + [(5 − 3)α

0 = −30 + 2α

] + [(5 − 3)]ω 2 ]

α = 15 rad/s 2

a A = aG + (a A/G )t + (a A/G ) n

] + [3α

a A = [aG

Components

: :

0 = aG − 3α

] + [3ω 2 ]

aG = 3α = 45 in./s 2

a A = 3ω 2 = (3)(4)2 = 48 in./s 2

a A = 48.0 in./s 2 

a B = aG + (a B/G )t + (a B/G ) n = [45

] + [5α

] + [5ω 2 ]

= [45

] + [75

] + [80 ]

= [30 in./s 2

] + [80 in./s 2 ] a B = 85.4 in./s 2

69.4° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1154

PROBLEM 15.114 (Continued) aC = aG + (aC/G )t + (aC/G )n = [45

] + [5α ] + [5ω 2

= [45

] + [75 ] + [80

= [35 in./s 2

] ]

] + [75 in./s 2 ]

aC = 82.8 in./s 2

65.0° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1155

PROBLEM 15.115 A carriage C is supported by a caster A and a cylinder B, each of 50-mm diameter. Knowing that at the instant shown the carriage has an acceleration of 2.4 m/s 2 and a velocity of 1.5 m/s, both directed to the left, determine (a) the angular accelerations of the caster and of the cylinder, (b) the accelerations of the centers of the caster and of the cylinder.

SOLUTION Rolling occurs at all surfaces of contact. Instantaneous centers are at points of contact with floor. r = 0.025 m

Caster:

a A = aC = 2.4 m/s 2 (aD ) x = 0 (rolling with no sliding)

a A = a D + a A/D ] = [(aD ) x

[a A

] + [(aD ) y ] + [rd A

] + [rω A2 ]

a A = 0 + rα A = (0.025 m)α A

2.4 m/s 2

α A = 96 rad/s 2

r = 0.025 m

Cylinder:

(a E ) x = aC = 2.4 m/s 2 ( aE ) x = 0

a E = a D + a E/D [( aE ) x

] + [(aE ) y ] = [(aD ) x

] + [(aD ) y ] + [2rαB

] + 2rωB2 ]

: (aE ) x = (aD ) y + 2rα B [2.4 m/s 2

] = 0 + 2(0.025 m)α B

α B = 48 rad/s 2 [ aB

:

] = [(aD ) x

] + [(aD ) y ] + [rα

] + [rω 2 ]

aB = 0 + rα B aB = (0.025 m)(48 rad/s 2 );

a B = 1.2 m/s 2

Answers: α A = 96.0 rad/s 2 ,

(a)

α B = 48.0 rad/s 2 ,

(b)

a A = 2.40 m/s 2



a B = 1.200 m/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1156

PROBLEM 15.116 A wheel rolls without slipping on a fixed cylinder. Knowing that at the instant shown the angular velocity of the wheel is 10 rad/s clockwise and its angular acceleration is 30 rad/ s 2 counterclockwise, determine the acceleration of (a) Point A, (b) Point B, (c) Point C.

SOLUTION r = 0.04 m

Velocity analysis.

ω = 10 rad/s

Point C is the instantaneous center of the wheel. v A = [(rω )

] = [(0.04)(10)

] = 0.4 m/s

]

α = 30 rad/s 2

Acceleration analysis.

ρ = R + r = 0.16 + 0.04 = 0.2 m.

Point A moves on a circle of radius

aC = aC

Since the wheel does not slip,

aC = a A + (aC/ A )t + (aC/ A )n [aC ] = [(a A )t

 v2 ]+  A ρ

  + [rα 

  + [(0.04)(30) 

= [(a A )t

 (0.4) 2 ]+   0.2

= [(a A )t

] + [0.8 ] + [1.2

Components.

] + [rω 2 ]

] + [(0.04)(10) 2 ]

] + [4 ]

: − (a A )t + 1.2 = 0

(a A )t = 1.2 m/s 2

: aC = −0.8 + 4.0

aC = 3.2 m/s 2

(a) Acceleration of Point A. a A = [1.2 m/s 2

] + [0.8 m/s 2 ]

a A = 1.442 m/s 2

33.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1157

PROBLEM 15.116 (Continued)

a B = a A + (a B/ A )t + (a B/ A )n

(b) Acceleration of Point B. a B = [1.2 = [1.2

] + [0.8 ] + [rα

] + [rω 2

]

] + [0.8 ] + [(0.04)(30) ] + [(0.04)(10) 2

= [2.8 m/s 2

] + [2 m/s 2 ]

(c) Acceleration of Point C.

] a B = 3.44 m/s 2

aC = aC

35.5° 

aC = 3.20 m/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1158

PROBLEM 15.117 The 100 mm radius drum rolls without slipping on a portion of a belt which moves downward to the left with a constant velocity of 120 mm/s. Knowing that at a given instant the velocity and acceleration of the center A of the drum are as shown, determine the acceleration of Point D.

SOLUTION Velocity analysis. v A = v B + v A/B

[180 mm/s

Components

] + [(100 mm)ω

] = [120 mm/s

]

: 180 = −120 = 100ω

ω = 3 rad/s

Acceleration analysis.

Point A moves on a path parallel to the belt. The path is assumed to be straight. a A = 720 mm/s 2

30°

Since the drum rolls without slipping on the belt, the component of acceleration of Point B on the drum parallel to the belt is the same as the belt acceleration. Since the belt moves at constant velocity, this component of acceleration is zero. Thus a B = aB

60°

Let the angular acceleration of the drum be α . a B = a A + (a B/ A )t + (a B/ A ) n

[a B Components

:

] = [720

] + [rα

] + [rω 2

]

0 = 720 − 100α

α = 7.2 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1159

PROBLEM 15.117 (Continued)

Acceleration of Point D. a D = a A + (a D/ A )t + (a D/ A ) n

30°] + [rα

= [a A = [720

Components:

30° :

Components:

60°:

60°] + [rω 2

30°] + [(100)(7.2)

30°]

60°] + [(100) (3)2

30°]

− 720 + 900 = 180 mm/s 2

720 mm/s 2

aD = 1802 + 7202 = 742.16 mm/s 2 tan β =

720 180

β = 76.0°

β − 30° = 46.0°

a D = 742 mm/s 2

46.0° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1160

PROBLEM 15.118 In the planetary gear system shown the radius of gears A, B, C, and D is 3 in. and the radius of the outer gear E is 9 in. Knowing that gear A has a constant angular velocity of 150 rpm clockwise and that the outer gear E is stationary, determine the magnitude of the acceleration of the tooth of gear D that is in contact with (a) gear A, (b) gear E.

SOLUTION Velocity. Gears:

T = Tooth of gear D in contact with gear A vT = rω A = (3 in.)ω A

Since vE = 0, E is instantaneous center of gear D. vT = 2rωD (3 in.)ω A = 2(3 in.)ωD 1 2

ωD = ω A 1 vD = rω D = (3 in.) ω A = (1.5 in.)ωA 2

Spider: vD = (6 in.)ωS (1.5 in.)ωA = (6 in.)ωS 1 4

ωS = ω A ω A = 150 rpm = 15.708 rad/s

1 ω D = ω A = 7.854 rad/s 2 1 ω S = ω A = 3.927 rad/s 4

Acceleration. Spider:

ωS = 3.927 rad/s aD = ( AD)ωS2 = (6 in.)(3.927 rad/s) 2 a D = 92.53 in./s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1161

PROBLEM 15.118 (Continued)

Gear D:

Plane motion (a)

=

Trans. with D

+

Rotation about D

Tooth T in contact with gear A. aT = a D + aT/D = aD + ( DT )ωD2 = 92.53 in./s 2

+ (3 in.)(7.854 rad/s) 2

= 92.53 in./s 2

+ 185.06 in./s 2

aT = 92.53 in./s 2

aT = 92.5 in./s 2 

(b)

Tooth E in contact with gear E. a E = a D + a E/D = a D + ( ED )ωD2



= 92.53 in./s 2

+ (3 in.)(7.854 rad/s) 2

= 92.53 in./s 2

+ 185.06 in./s 2

a E = 277.6 in./s 2

aE = 278 in./s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1162

PROBLEM 15.119 The 200-mm-radius disk rolls without sliding on the surface shown. Knowing that the distance BG is 160 mm and that at the instant shown the disk has an angular velocity of 8 rad/s counterclockwise and an angular acceleration of 2 rad/s2 clockwise, determine the acceleration of A.

SOLUTION m, m/s, m/s 2

Units:

Unit vectors: i = 1

, j =1 , k =1

Geometric analysis.

Let P be the point where the disk contacts the flat surface. rG /A = 0.200 j

rB / G = −0.16i

rA/B = − 0.600i − 0.200 j

Velocity analysis.

ωG = (8 rad/s)k , v P = 0, v A = v A i

v B = v P + vG /P = v P + ωG × rG /P = 0 + 8k × ( −0.160i + 0.200 j) = −1.6i − 1.28 j v A = v B + v A/B = v B + ω AB × rA/B v A i = −1.6i − 1.28 j + ω AB k × (− 0.600i − 0.200 j) = −1.6i − 1.28 j − 0.77460ω AB j + 0.2ω AB i

Resolve into components and transpose terms. j: Acceleration analysis:

0.77460ω AB = −1.28

ω AB = −1.6525 rad/s 

a A = a A j, aG = −2 rad/s 2k a P = (ωG2 r ) j = (8) 2 (0.2) j = (12.8 m/s 2 ) j a B = a P + a B /P = a P + aG × rB /P − ωG2 rB /P = 12.8 j + (−2k ) × (−0.160i + 0.200 j) − (8) 2 (−0.160i + 0.200 j) = 12.8 j + 0.32 j + 0.4i + 10.24i − 12.8 j = 10.64i + 0.32 j 2 a A = a B + a A /B = a B + α AB k × rA /B − ω AB rA /B

= 10.64i + 0.32 j + α AB k × ( − 0.600i − 0.200 j) − (1.6525) 2 (− 0.600i − 0.200 j) = 10.64i + 0.32 j − 0.77460α AB j + 0.2α AB i + 2.115i + 0.54615 j a A i = 12.755i + 0.86615 j + 0.2α AB i − 0.77460α AB j

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PROBLEM 15.119 (Continued)

Resolve into components and transpose terms. j: i:

0 = 0.86615 − 0.77460α AB

α AB = 1.1182

a A = 12.755 + 0.2α AB = 12.755 + (0.2)(1.1182) = 12.98 a A = (12.98 m/s 2 )i = 12.98 m/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1164

PROBLEM 15.120 Knowing that crank AB rotates about Point A with a constant angular velocity of 900 rpm clockwise, determine the acceleration of the piston P when θ = 60°.

SOLUTION sin β sin 60° = 0.05 0.15

Law of sines.

β = 16.779°

ω AB = 900 rpm = 30π rad/s

Velocity analysis.

v B = 0.05ω AB = 1.5π m/s

60°

ωBD = ωBD

v D = vD v D/B = 0.15 ω BD

β

v D = v B + v D/B [vD ] = [1.5π Components

:

β]

0 = 1.5π cos 60° − 0.15ωBD cos β

ωBD = Acceleration analysis.

60°] + [0.15ωBD

1.5π cos 60° = 16.4065 rad/s 0.15 cos β

α AB = 0 2 a B = 0.05ω AB = (0.05)(30π )2 = 444.13 m/s 2

30°

α BD = α BD

a D = aD a D/B = [0.15α AB = [0.15α BD a D = a B + a D/B

2 β ] + [0.15ω BD

β ] + [40.376

β] β]

Resolve into components.

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PROBLEM 15.120 (Continued)

:

0 = −444.13 cos 30° + 0.15α BD cos β + 40.376 sin β

α BD = 2597.0 rad/s 2 :

aD = 444.13 sin 30° − (0.15)(2597.0)sin β + 40.376 cos β = 148.27 m/s 2

aP = aD

a P = 148.3 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1166

PROBLEM 15.121 Knowing that crank AB rotates about Point A with a constant angular velocity of 900 rpm clockwise, determine the acceleration of the piston P when θ = 120°.

SOLUTION sin β sin120° = , 0.05 0.15

Law of sines.

β = 16.779°

ω AB = 900 rpm = 30π rad/s

Velocity analysis.

v B = 0.05ω AB = 1.5π m/s v D = vD

60°

ωBD = ωBD

v D/B = 0.15ωBD

β

v D = v B + v D/B [vD ] = [1.5π Components

:

β]

0 = −1.5π cos 60° − 0.15ω BD cos β

ωBD = − Acceleration analysis.

60°] + [0.15ωBD

1.5π cos 60° = 16.4065 rad/s 0.15 cos β

α AB = 0 2 a B = 0.05ω AB = (0.05)(30π )2 = 444.13 m/s 2

a D = aD

30°

α BD = α BD

a D/B = [0.15α AB = [6α BD

2 β ] + [0.15ω BD

β ] + [40.376

a D = a B + a D/B

β] β]

Resolve into components.

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PROBLEM 15.121 (Continued)

:

0 = −444.13 cos 30° + 0.15α BD cos β + 40.376 sin β

α BD = 2597.0 rad/s 2 :

aD = −444.13 sin 30° − (0.15)(2597.0)sin β + 40.376 cos β = −296 m/s 2

aP = aD

a P = 296 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1168

PROBLEM 15.122 In the two-cylinder air compressor shown the connecting rods BD and BE are each 190 mm long and crank AB rotates about the fixed Point A with a constant angular velocity of 1500 rpm clockwise. Determine the acceleration of each piston when θ = 0.

SOLUTION

Crank AB. vA = 0, a A = 0, ω AB = 1500 rpm = 157.08 rad/s , α AB = 0 vB = vA + vB/ A = 0 + [0.05 ωAB

45°] = [7.854 m/s

45°]

a B = aA + (a B/ A )t + (a B/ A )n = 0 + [0.05α AB

2 45°] + [0.05ω AB

= [(0.05)(157.08) 2 v D = vD

Rod BD.

45°

45°]

45°] = 1233.7 m/s

2

45°

ω BD = ωBD

v D = v B + v B/D vD

Components

45°] + [0.19ωBD

45° = [7.854

0 = 7.854 − 0.19ωBD

45°:

a D = aD

45°]

ωBD = 41.337 rad/s

45°

a D = a B + (a D/B )t + (a D/B )n [ aD

45°] = [1233.7

45°] + [0.19 α BD

2 45°] + [0.19 ωBD

45°]

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PROBLEM 15.122 (Continued)

Components

45°:

aD = 1233.7 + (0.19)(41.337) 2 = 1558.4 m/s 2 a D = 1558 m/s 2 sin γ =

Rod BE.

45° 

0.05 , γ = 15.258°, β = 45° − γ = 29.742° 0.19

v E = vE

45°

ωBE = 0.

Since v E is parallel to v B ,

a E = aE

2 45° (aB/E ) n = 0.19 ωBE =0

(a E /B )t = (aE /B )t

β

a E = a B + (a B /E )t

Draw vector addition diagram.

γ = 45° − β = 15.258° aE = aB tan γ = 1233.7 tan γ = 336.52 m/s 2

a E = 337 m/s 2

45° 

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PROBLEM 15.123 The disk shown has a constant angular velocity of 500 rpm counter-clockwise. Knowing that rod BD is 10 in. long, determine the acceleration of collar D when (a) θ = 90°, (b) θ = 180°.

SOLUTION ω A = 500 rpm

Disk A.

= 52.36 rad/s

α A = 0, ( AB) = 2 in. vB = ( AB)ω A = (2)(52.36) = 104.72 in./s aB = ( AB)ω A2 = (2)(52.36)2 = 5483.1 in./s 2

(a)

θ = 90°.

v B = 104.72 m/s , v D = vD

sin β =

2 in. = 0.4 5 in.

β = 23.58°

vD and vB are parallel.

ωBD = 0 a B = 5483.1 in./s 2 a D/B = [( BD )α BD

= [10 α BD a D = a B + a D/B

: :

,

a D = aD ,

2 β ] + [( BD )ωBD

α BD = α BD

β]

β]+ 0 Resolve into components.

0 = 5483.1 + (10 cos β )α BD

α BD = −598.26 rad/s 2

aD = 0 − (10 sinβ )( −598.26) + 0 = 2393.0 in./s 2

a D = 199.4 ft/s 2 

  

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PROBLEM 15.123 (Continued)

(b)

θ = 180°.

vB = 104.72 in./s

sin β =

6 in. = 0.6 10 in.

vB = 104.72 in./s

vD = vD

,

β = 36.87 vD = vD

,

Instantaneous center of bar BD lies at Point C.

ωBD =

vB 104.72 = = 13.09 rad/s ( BD ) 10 cos β

a B = 5483.1 in./s 2 , a D = aD , α BD = α BD a D/B = [( BD )α BD

= [10α BD a D = a B + a D/B

:

2 β ] + [( BD )ωBD

β ] + [1713.5

β]

β]

Resolve in components.

0 = 0 + (10 cos β )α BD + 1713.5 sinβ

α BD = −128.51 rad/s 2 :

aD = 5483.1 − (10 sin β )( −128.51) + 1713.5cos β = 7625.0 in./s 2

a D = 635 ft/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1172

PROBLEM 15.124 Arm AB has a constant angular velocity of 16 rad/s counterclockwise. At the instant when θ = 90°, determine the acceleration (a) of collar D, (b) of the midpoint G of bar BD.

SOLUTION 2 aB = ( AB)ω AB

Rod AB:

= (3 in.)(16 rad/s)2 a B = 768 in./s 2

Rod BD: instantaneous center is at OD; ω BD = 0 sin β = (3 in.)/(10 in.) = 0.3; β = 17.46°

Acceleration.

Plane motion

=

Trans. with B

+

Rotation about B

a D = a B + a D/B = aB + (a D/B )t + (a D/B )n

(a)

= [aB ] + [( BD )α

2 β ] + [( BD)ωBD

= [768 in./s 2 ↓] + [(10 in.) α

a D ↔ = [768 in./s ↓] + [(10 in.)α

β]

β ] + [(10 in.)(0) 2

β]

β]

Vector diagram: a D = (768 in./s 2 ) tan17.46°

= 241.62 in./s 2 a D = 242 in./s 2



(10 in.)α = (768 in./s 2 )/cos17.46° (10 in.)α = 805.08 in./s 2 α = 80.5 rad/s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1173

PROBLEM 15.124 (Continued)

aG = a B + aG /B = a B + (aG /B )t + (aG /B ) n

(b)

= [a B

] + [( BG )α

= [768 in./s 2 aG = [768 in./s 2

components:

2 β ] + [( BG )ωBD

] + [(5 in.)(80.5 rad/s 2 ) ] + [402.5 in./s 2

β] β ] + [( BG )(0) 2 ]

17.46°]

(aG ) x = (402.5 in./s 2 ) sin 17.46° (aG ) x = 120.77 in./s 2

components:

(aG ) y = 768 in./s 2 − (402.5 in./s 2 ) cos 17.46°

= 768 in./s 2 − 384 in./s 2 (aG ) y = 384 in./s 2

aG = 403 in./s 2

72.5° 

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PROBLEM 15.125 Arm AB has a constant angular velocity of 16 rad/s counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.

SOLUTION 3.403 in. 10 in. β = 19.89°

β = sin −1

Velocity.

v B = ( AB)ω AB = (3 in.)(16 rad/s) = 48 in./s

30°

Rod BD:

Plane motion

=

Trans. with B

+

Rotation about B

v D = v B + v D/B = v B + [( BD)ωBD v D ↔ = [48 in./s

components:

β]

30°] + [(10 in.)ωBD

19.89°]

(48 in./s)sin 30° − (10 in.)ω BD cos19.89°

ωBD =

(48 in./s)sin 30° = 2.552 rad/s (10 in.) cos19.890°

Acceleration. Rod AB:

2 a B = [( AB)ω AB

60°] = [(3 in.)(16 rad/s) 2

a B = 768 in./s 2

60°

60°]

Rod BD:

Plane motion

=

Trans. with B

+

Rotation about B

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1175

PROBLEM 15.125 (Continued)

a D = a B + a B/D = a B + (a D/B )t + (a D/B ) n aD ↔ = [ aB

60°] + [( BD )α BD

= [768 in./s 2

aD ↔ = [768 in./s 2

2 β ] + [( BD)ωDB

60°] + [(10 in.)α BD ] + [(10 in.)α BD

β]

β ] + [(10 in.)(2.552 rad/s 2 )

19.89°] + [65.14 in./s 2

β]

19.89°]

Vector diagram.

y components: 768sin 60° + 65.14sin19.89° − 10α BD cos19.89° = 0

:

α BD = 73.09 rad/s 2 x components: :

aD = 768cos 60° + 65.14 cos19.89° + (10)(73.09) sin19.89° aD = 693.9 in./s 2

a D = 694 in./s 2



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PROBLEM 15.126 A straight rack rests on a gear of radius r = 3 in. and is attached to a block B as shown. Knowing that at the instant shown θ = 20°, the angular velocity of gear D is 3 rad/s clockwise, and it is speeding up at a rate of 2 rad/s2, determine (a) the angular acceleration of AB, (b) the acceleration of block B.

SOLUTION Let Point P on the gear and Point Q on the rack be located at the contact point between them. Units: inches, in./s, in./s2 i =1

Unit vectors:

, j =1 , k =1

rP /D = 3(sin θ i + cos θ j)

Geometry:

3 (cos θ i − sin θ j) tan θ θ = 20°

rB /Q =

ωgear = 3 rad/s

Gear D:

α gear = 2 rad/s 2

vP = ωgear r = (3)(3) = 9 in./s

v P = 9 in./s

2 (aP ) n = ωgear r = (3) 2 (3) = 27 in./s 2

(aP )t = α gear r = (2)(3) = 6 in./s 2

θ

(a P ) n = 27 in./s 2

θ

(a P )t = 6 in./s 2

θ

Velocity analysis. Gear to rack contact: Rack AQB:

vQ = v P = 9 in./s

ω AB = ω AB , v B = vB

,

θ α AB = α AB αB = αB

v B = vQ + v B /Q = v Q + ω AB k × rB /Q vB i = 9(cos θ i − sin θ j) + ω AB k × (7.74535i − 2.81908 j)

= (9cos θ + 2.81907ω AB )i + (−9sin θ + 2.81907ω AB ) j

Equating like components, j:

0 = −9sin θ + 7.74535ω AB

ω AB = 0.39742

ω AB = 0.39742 rad/s

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PROBLEM 15.126 (Continued)

Acceleration analysis. Gear to rack contact:

θ

(aQ )t = (a P )t = 6 in./s 2 2 (aQ ) n = (a P ) n + rωrd

θ

ω rd = ω AB − ω D = 3.39742 rad/s

where

(aQ ) n = 27 in./s 2

20° + (3)(3.39742) 2 20°

= 7.6274 in./s 2 20° aQ = 6(cosθ i − sin θ j) + 7.6274(sin θ i + cos θ j)

Then,

= (8.2469 in./s 2 )i + (5.1153 in./s 2 ) j 2 a B = aQ + a B /Q = aQ + α AB k × rB /Q − ω AB rB /Q

aB i = 8.2469i + 5.1153j + α AB k × (7.74535i − 2.81908 j) − (0.39742) 2 (7.74535i − 2.81908 j) = (8.2469 + 2.81908α AB − 1.22332)i + (5.1153 + 7.74535α AB + 0.44526) j

Equating like components of a B , j:

0 = 5.1153 + 7.74535α AB + 0.44526

α AB = −0.71792 rad/s 2 i:

aB = 8.2469 + (2.81908)(−0.71792) − 1.22332 aB = 5.00 in./s 2

(a)

Angular acceleration of AB:

α AB = 0.718 rad/s 2



(b)

Acceleration of block B:

a B = 5.00 in./s 2



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PROBLEM 15.127 Knowing that at the instant shown rod AB has a constant angular velocity of 6 rad/s clockwise, determine the acceleration of Point D.

SOLUTION

ω AB = 6 rad/s

Velocity analysis.

v B = ( AB)ω AB = (90)(6) = 540 mm/s v B = vB

, v D = vD

The instantaneous center of bar BDE lies at ∞.

ωBD = 0 and vD = vB = 540 mm/s

Then

ωCD =

vD 540 = = 3 rad/s CD 180

α AB = 0

Acceleration analysis.

2 a B = ( AB)ω AB = [(90)(6)2 ] = 3240 mm/s 2

a D = [(CD)α CD

2 ] + [(CD )ωCD ] = [180 α CD

= [180α CD

] + [1620 mm/s 2 ]

a D/B = [90α BD

2 ] + [225α BD ] + [225ωBD

= [90α BD





Resolve into components.

−1620 = − 3240 + 225α BD ,

:

α BD = 7.2 rad/s 2

180α CD = 0 + (90)(7.2),

a D = [3240 ] + [(90)(7.2) = [648

2 ] + [90 ωBD ]

] + [225α BD ]

a D = a B + a D/B

:

] + [(180)(3) 2 ]

α CD = 3.6 rad/s 2 ] + [(225)(7.2) ]

] + [1620 mm/s 2 ]

= 1745 mm/s 2

68.2°

a D = 1.745 m/s 2

68.2° 

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PROBLEM 15.128 Knowing that at the instant shown rod AB has a constant angular velocity of 6 rad/s clockwise, determine (a) the angular acceleration of member BDE, (b) the acceleration of Point E.

SOLUTION

ω AB = 6 rad/s

Velocity analysis.

v B = ( AB)ω AB = (90)(6) = 540 mm/s v B = vB

, v D = vD

The instantaneous center of bar BDE lies at ∞.

ωBD = 0 and vD = vB = 540 mm/s

Then

ωCD =

vD 27 = = 3 rad/s CD 9

α AB = 0

Acceleration analysis.

2 a B = ( AB)ω AB = [(90)(6)2 ] = 3240 mm/s 2

a D = [(CD)α CD

2 ] + [(CD )ωCD ] = [180 α CD

= [180α CD

] + [1620 mm/s 2 ]

a D/B = [90α BD

2 ] + [225α BD ] + [225ωBD

= [90α BD

:

Resolve into components.

−1620 = − 3240 + 225α BD , a E/B = [180α BD

α BD = 7.20 rad/s 2

2 ] + [450 α BD ] + [450ωBD

= [(180)(7.2) = [1296 mm/s 2

(b)

2 ] + [90 ωBD ]

] + [225α BD ]

a D = a B + a D/B

(a)

] + [(180)(3) 2 ]

] + [(450)(7.2) ] + [0

] + [0 ]

] + [3240 mm/s 2 ]

a E = a B + a B/E = [3240 mm/s 2 ] + [1296 mm/s 2 = 1296 mm/s 2

2 ] + [180ω BD ]

] + [3240 mm/s 2 ]

a E = 1.296 m/s 2



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PROBLEM 15.129 Knowing that at the instant shown bar AB has a constant angular velocity of 19 rad/s clockwise, determine (a) the angular acceleration of bar BGD, (b) the angular acceleration of bar DE.

SOLUTION Velocity analysis.

ω AB = 19 rad/s vB = ( AB)ω AB = (8)(19) = 152 in./s v B = vB

, v D = vD

Instantaneous center of bar BD lies at C.

ω BD =

vB 152 = = 19 rad/s BC 8

vD = (CD)ωBD = (19.2)(19) = 364.8 in./s

ω DE = Acceleration analysis.

vD 364.8 = = 24 rad/s 2 DE 15.2

α AB = 0. 2 a B = [( AB)ω AB ] = [(8)(19) 2 ] = 2888 in./s 2 a D = [( DE )α DE ] + [( DE )ωDE

= [15.2α DE ] + [8755.2 in./s 2 (a D/B )t = [19.2α BD ] + [8α DB 2 (a D/B ) n = [19.2ωBD

= [6931.2 in./s 2

(b)

]

] + [2888 in./s 2 ] Resolve into components.

8755.2 = 0 + 8α BD + 6931.2

(a ) :

]

2 ] + [8ωBD ]

a D = a B + (a D/B )t + (aD/B ) n :

]

α BD = 228 rad/s 2



α DE = 92.0 rad/s 2



15.2α DE = −2888 + (19.2)(228) − 2888

α DE = −92 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1181

PROBLEM 15.130 Knowing that at the instant shown bar DE has a constant angular velocity of 18 rad/s clockwise, determine (a) the acceleration of Point B, (b) the acceleration of Point G.

SOLUTION Velocity analysis.

ω DE = 18 rad/s v D = ( DE )ωDE = (15.2)(18) = 273.6 in./s v D = vD , v B = vB

Point C is the instantaneous center of bar BD.

ω BD =

vD 273.6 = = 14.25 rad/s CD 19.2

vB = (CB)ωBD = (8)(14.25) = 114 in./s

ω AB = Acceleration analysis.

vB 114 = = 14.25 rad/s AB 8

α DE = 0 2 a D = [( DE )ω DE

] = [(15.2)(18) 2

a B = [( AB)α AB

2 ] + [( AB)ω AB ]

= [8α AB

]

] + [1624.5 in./s 2 ]

(a D/B )t = [19.2α BD ] + [8α BD 2 (a D/B ) n = [19.2ωBD

] = [4924.8 in./s 2

]

2 ] + [8ωBD ]

a D = a B + (a D/B )t + (a D/B ) n

Resolve into components.

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PROBLEM 15.130 (Continued) : 0 = 1624.5 − 19.2α BD + 1624.5,

α BD = 169.21875 rad/s 2 : 4924.8 = 8α AB + (8)(169.21875) + 3898.8

α AB = −40.96875 rad/s 2 (a )

a B = [(8)(−40.96875) = [327.75 in./s 2

(b)

aG = a B + aG/B = a B + = aB +

] + [1624.5 in./s 2 ] ] + [1624.5 in./s 2 ],

a B = 138.1 ft/s 2

78.6° 

aG = 203 ft/s 2

19.5° 

1 a D/B 2

1 1 (a D − a B ) = (a B + a D ) 2 2

 −327.75 + 4924.8 =  2 

= [2298.5 in./s 2 ]

 1624.5 + 2  

  

+ [812.25 in./s 2 ]

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PROBLEM 15.131 Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE.

SOLUTION rB /A = −(20 in.)i − (40 in.) j

Relative position vectors.

rD /B = (40 in.)i rD /E = (20 in.)i − (25 in.) j

Velocity analysis. Bar AB (Rotation about A): ω AB = 4 rad/s

= −(4 rad/s)k

rB /A = −(20 in.)i − (40 in.) j

v B = ω AB × rB /A = (−4k ) × (−20i − 40 j)

v B = −(160 in./s)i + (80 in./s) j

Bar BD (Plane motion = Translation with B + Rotation about B): ω BD = ω BD k

rD /B = (40 in.)i

v D = v B + ω BD × rD /B = v B + (ωBD k ) × (40i ) v D = −(160 in/s)i + (40ωBD + 80 in./s) j

Bar DE (Rotation about E): ω DE = ωDE k rD /E = (20 in.)i − (25 in.) j v D = ω DE × rD /E = (ωDE k ) × (20i − 25 j) v D = 20ω DE j + 25ω DE i

Equating components of the two expression for vD, i:

−160 = 25ωDE

ω DE = −6.4 rad/s

j:

40ωBE + 80 = 20ωDE

40ωBD + 80 = 20(−6.4)

ωBD = −5.2 rad/s

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PROBLEM 15.131 (Continued)

Summary of angular velocities:

ω AB = 4 rad/s

Acceleration analysis.

α AB = 0, α BD = α BD k , α DE = α DE k

Bar AB (Rotation about A):

ω DE = 6.4 rad/s

ω BD = 5.2 rad/s

2 a B = α AB × rB /A − ω AB rB /A

= 0 − (4) 2 (−20i − 40 j) = (320 in./s 2 )i + (640 in./s 2 ) j

Bar BD (Translation with B + Rotation about B): 2 a D = a B + α BD × rD /B − ωBD rD /B

= 320i + 640 j + α BD k × (40i ) − (5.2)2 (40)i = 320i + 640 j + 40α BD j − 1081.6i = −761.60i + (640 + 40α BD ) j

(1)

Bar DE (Rotation about E): 2 a D = α DE × rD /E − ωDE rD /E

= α DE k × (20i − 25 j) − (6.4)2 (20i − 25 j) = −20α DE j + 25α DE i − 819.20i + 1024 j = (25α DE − 819.20)i + (20α DE + 1024) j

(2)

Equate like components of aD expressed by Eqs. (1) and (2). i:

−761.60 = 25α DE − 819.20

α DE = 2.3040 rad/s 2

j:

640 + 40α BD = (20)(2.304) + 1024

α BD = 10.752 rad/s 2

(a)

Angular acceleration of bar BD.

α BD = 10.75 rad/s 2



(b)

Angular acceleration of bar DE.

α DE = 2.30 rad/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1185

PROBLEM 15.132 Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE.

SOLUTION Velocity analysis. Bar AB (Rotation about A):

ω AB = 4 rad/s

= −(4 rad/s)k

rB /A = −(175 mm)i

vB = ω AB × rB /A = (−4k ) × (−175i )

vB = (700 mm/s)j

Bar BD (Plane motion = Translation with B + Rotation about B): ω BD = ω BD k

rD /B = −(200 mm)j

vD = vB + ω BD × rD /B = 700 j + (ωBD k ) × (−200 j) vD = 700 j + 200ω BD i

Bar DE (Rotation about E):

ω DE = ω DE k rD /E = −(275 mm)i + (75 mm)j vD = ω DE × rD /E = (ωDE k ) × (−275i + 75 j) vD = −275ωDE j − 75ωDE i

Equating components of the two expressions for vD , j:

700 = −275ωDE

i : 200ωBD = −75ωBD

ωDE = −2.5455 rad/s 3 8

ωDE = − ω BD 3

ωBD = −   (−2.5455) = 0.95455 rad/s 8 Acceleration analysis. Bar AB: Bar BD:

ω DE = 2.55 rad/s

ω BD = 0.955 rad/s

α AB = 0 2 a B = −ω AB rB /A = −(4) 2 ( −175i ) = (2800 mm/s 2 )i

α BD = α BD k 2 a D = a B + α BD × rD /B − ω BD rD /B

= 2800i + α BD k × (−200 j) − (0.95455) 2 (−200 j) = (2800 + 200 α BD )i + 182.23j

(1)

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PROBLEM 15.132 (Continued)

α DE = α DE k

Bar DE:

2 a D = α DE × rD /E − ωDE rD /E

= α DE k × (−275i + 75 j) − (2.5455)2 ( −275i + 75 j) = −275α DE j − 75α DE i + 1781.8i − 485.95 j = (−75α DE + 1781.8)i − (275α DE + 485.95) j

(2)

Equate like components of aD expressed by Eqs. (1) and (2). j: i:

182.23 = −(275α DE + 485.95) (2800 + 200α BD ) = [−(75)( −2.4298) + 1781.8]

α DE = −2.4298 rad/s 2 α BD = −4.1795 rad/s 2

(a)

Angular acceleration of bar BD.

α BD = 4.18 rad/s 2



(b)

Angular acceleration of bar DE.

α DE = 2.43 rad/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1187

PROBLEM 15.133 Knowing that at the instant shown bar AB has an angular velocity of 4 rad/s and an angular acceleration of 2 rad/s2, both clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE by using the vector approach as is done in Sample Problem 15.8.

SOLUTION Relative position vectors.

rB /A = −(20 in.)i − (40 in.) j rD /B = (40 in.)i rD /E = (20 in.)i − (25 in.) j

Velocity analysis. Bar AB (Rotation about A): ω AB = 4 rad/s

= −(4 rad/s)k

rB /A = −(20 in.)i − (40 in.) j

v B = ω AB × rB /A = (−4k ) × (−20i − 40 j)

v B = −(160 in./s)i + (80 in./s) j

Bar BD (Plane motion = Translation with B + Rotation about B): ω BD = ω BD k

rD /B = (40 in.)i

v D = v B + ω BD × rD /B = v B + (ωBD k ) × (40i ) v D = −(160 in/s)i + (40ωBD + 80 in./s) j

Bar DE (Rotation about E): ω DE = ωDE k rD /E = (20 in.)i − (25in.) j v D = ω DE × rD /E = (ωDE k ) × (20i − 25 j) v D = 20ω DE j + 25ω DE i

Equating components of the two expression for vD, i:

−160 = 25ωDE

ω DE = −6.4 rad/s

j:

40ωBE + 80 = 20ωDE

40ωBD + 80 = 20(−6.4)

ωBD = −5.2 rad/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1188

PROBLEM 15.133 (Continued)

Summary of angular velocities:

ω AB = 4 rad/s

Acceleration analysis.

α AB = −(2 rad/s 2 )k , α BD = α BD k , α DE = α DE k

Bar AB (Rotation about A)

ω DE = 6.4 rad/s

ω BD = 5.2 rad/s

2 a B = α AB × rB /A − ω AB rB /A

= ( −2k ) × (−20i − 40 j) − (4) 2 (−20i − 40 j) = −(80 in./s 2 )i + (40 in./s 2 ) j + (320 in./s 2 )i + (640 in./s 2 ) j = (240 in./s 2 )i + (680 in./s 2 ) j

Bar BD (Translation with B + Rotation about B): 2 a D = a B + α BD × rD /B − ωBD rD /B

= 240i + 680 j + α BD k × (40i) − (5.2) 2 (40)i = 240i + 680 j + 40α BD j − 1081.6i = −841.60i + (680 + 40α BD ) j

Bar DE

(1)

(Rotation about E): 2 a D = α DE × rD /E − ωDE rD /E

= α DE k × (20i − 25 j) − (6.4)2 (20i − 25 j) = 20α DE j + 25α DE i − 819.20i + 1024 j = (25α DE − 819.20)i + (20α DE + 1024) j

(2)

Equate like components of aD expressed by Eqs. (1) and (2). i:

−841.60 = 25α DE − 819.20

j:

680 + 40α BD = (20)(−0.896) + 1024

α DE = −0.896 rad/s 2 α BD = 8.152 rad/s 2

(a)

Angular acceleration of bar BD.

α BD = 8.15 rad/s 2



(b)

Angular acceleration of bar DE.

α DE = 0.896 rad/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1189

PROBLEM 15.134 Knowing that at the instant shown bar AB has an angular velocity of 4 rad/s and an angular acceleration of 2 rad/s2, both clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE by using the vector approach as is done in Sample Problem 15.8.

SOLUTION Velocity analysis. ω AB = 4 rad/s

Bar AB (Rotation about A):

= −(4 rad/s)k

rB /A = −(175 mm)i vB = (700 mm/s)j

vB = ω AB × rB /A = (−4k ) × (−175i )

Bar BD (Plane motion = Translation with B + Rotation about B): ω BD = ω BD k rD /B = −(200 mm)j vD = vB + ω BD × rD /B = 700 j + (ωBD k ) × (−200 j) vD = 700 j + 200ω BD i ω DE rD /E vD vD

Bar DE (Rotation about E):

= ω DE k = −(275 mm)i + (75 mm)j = ω DE × rD /E = (ωDE k ) × (−275i + 75 j) = −275ωDE j − 75ωDE i

Equating components of the two expressions for vD , j:

700 = −275ωDE

i : 200ωBD = −75ωDE

ωDE = −2.5455 rad/s ωBD

3

ωBD = −   (−2.5455) = 0.95455 rad/s 8 Acceleration analysis. Bar AB: Bar BD:

ω DE = 2.55 rad/s

3 = − ω DE 8

ω BD = 0.955 rad/s

α AB = 2 rad/s 2 = −(2 rad/s 2 )k 2 a B = α AB × rB /A − ω AB rB /A

α BD

= ( −2k ) × (−175i) − (4)2 (−175i) = 2800 mm/s 2 i + 350 mm/s 2 j = α BD k

2 a D = a B + α BD × rD /B − ω BD rD /B

= 2800i + 350 j + α BD k × ( −200 j) − (0.95455) 2 (−200 j) = (2800 + 200 α BD )i + 532.23j

(1)

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PROBLEM 15.134 (Continued)

α DE = α DE k

Bar DE:

2 a D = α DE × rD /E − ωDE rD /E

= α DE k × (−275i + 75 j) − (2.5455)2 ( −275i + 75 j) = −275α DE j − 75α DE i + 1781.8i − 485.95 j = (−75α DE + 1781.8)i − (275α DE + 485.95) j

(2)

Equate like components of a D expressed by Eqs. (1) and (2). j: i:

532.23 = −(275α DE + 485.95) (2800 + 200α BD ) = [−(75)( −3.7025) + 1781.8]

α DE = −3.7025 rad/s 2 α BD = −3.7025 rad/s 2

(a)

Angular acceleration of bar BD.

α BD = 3.70 rad/s 2



(b)

Angular acceleration of bar DE.

α DE = 3.70 rad/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1191

PROBLEM 15.135 Robert’s linkage is named after Richard Robert (1789–1864) and can be used to draw a close approximation to a straight line by locating a pen at Point F. The distance AB is the same as BF, DF and DE. Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine (a) the angular acceleration of bar DE, (b) the acceleration of Point F.

SOLUTION Units:

inches, in./s, in./s2

Unit vectors:

i =1

, j =1 , k =1 .

Geometry:

rB /A = 3i + 122 − 32 j = 3i + 135 j rD /B = 6i

rF /B = 3i − 135 j

rD /E = −3i + 135 j ω AB = 4 rad/s 2

Velocity analysis:

= −4 k

Bar AB:

vB = ω AB × rB /A = −4k × (3i + 135 j) = 4 135i − 12 j

Object BDF:

vD = v B + v D /B = v B + ω BD × rD /B = 4 135i − 12 j + ωBD k × 6i = 4 135i − 12 j + 6ωBD j

(1)

v D = ωDE × rD /E = ωDE k × (−3i + 135 j)

Bar DE:

= − 135ω DE i − 3ω DE j

(2)

Equating like components of v D from Eqs. (1) and (2), i: j:

4 135 = − 135ωDE

(3)

−12 + 6ωBD = −3ωDE

(4)

From Eq. (3),

ωDE = −4

ωDE = 5 rad/s

From Eq. (4),

ωBD = (12 − 3ωDE ) = 5

1 6

ωBD = 4 rad/s

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PROBLEM 15.135 (Continued)

α AB = 0

Acceleration Analysis:

2 a B = α AB × rB /A − ω AB rB /A

Bar AB:

= 0 − (4)2 (3i + 135 j) = −48i − 16 135 j 2 a D = a B + a D /B = a B + α BD × rD /B − ω BD rD /B

Object BDF:

a D = −48i − 16 135 j + α BD k × (6i ) − (4) 2 (6i) = −144i − 16 135 j + 6α BD j

(5)

2 a D = α DE × rD /E − ωDE rD /E

Bar DE:

a D = α DE k × (−3i + 135 j) − (4) 2 (−3i + 135 j) = − 135α DE i − 3α DE j + 48i − 16 135 j

(6)

Equating like components of aD from Eqs. (5) and (6), i : − 144 = − 135α DE + 48

(7)

j: − 16 135 + 6α BD = −3α DE − 16 135

(8)

192

From Eq. (7),

α DE =

From Eq. (8),

α BD = − α BD = −

135 1 2

(a)

Angular acceleration of bar DE:

(b)

Acceleration of Point F:

96 135 α DE = 16.53 rad/s 2



2 a F = a B + a F /B = a B + α BD × rF /B − ωBD rF /B

 96  k  × (3i − 135 j) − (4) 2 (3i − 135 j) = −48i − 16 135 j +  −  135  288 j − 96i − 48i + 16 135 j = −48i − 16 135 j − 135 288 j = −192i − 135 a F = −(192.0 in./s 2 )i − (24.8 in./s 2 ) j = 193.6 in./s 2

7.36° 

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PROBLEM 15.136 For the oil pump rig shown, link AB causes the beam BCE to oscillate as the crank OA revolves. Knowing that OA has a radius of 0.6 m and a constant clockwise angular velocity of 20 rpm, determine the velocity and acceleration of Point D at the instant shown.

SOLUTION Units: meters, m/s, m/s2 Unit vectors:

i =1

Crank OA:

rOA = 0.6 m, ωOA = 20 rpm

= 2.0944 rad/s

v A = ωOA rOA = (2.0944)(0.6)

v A = 1.25664 m/s

,

j =1 ,

α OA = 0

k =1 .

( a A )t = 0

2 (a A ) n = ωOA rOA = (2.0944) 2 (0.6) = 2.6319 m/s 2

a A = 2.6319 m/s 2 v B = vA ,

Rod AB:

Since v B and v A are parallel, v A = v B and ω AB = 0. v B = 1.25664 m/s 2 a B = a A + a B /A = a A + α AB k × rB /A − ω AB rB /A

= 2.6319i + α AB k × (0.6i + 2 j) − 0 = (2.6319 − 2α AB )i + 0.6α AB j

(1)

Beam BCE: Point C is a pivot. vB = ωBCE rBC

ωBCE =

vB 1.25664 = = 0.41888 3 rBC

vE = ωBCE rCE = (0.41888)(3.3) = 1.38230 ω BCE = 0.41888 rad/s

v E = 1.38230 m/s

2 a B = α BCE × rB /C − ωBCE rB /C

= α BCE k × (−3i ) − (0.41888) 2 (−3i ) = 0.52638i − 3α BCE j

(2)

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PROBLEM 15.136 (Continued)

Equating like components of α B expressed by Eqs. (1) and (2), i:

2.6319 − 2α AB = 0.52638

α AB = 1.05276 rad/s

j:

0.6α AB = −3α BCE

α BCE = −0.21055 rad/s 2

α AB = 1.055276 rad/s

α BCE = 0.21055 rad/s 2

a E = (a E /C )t + (a E /C )n (a E /C )t = α BCE × rE /C = (−0.21055k ) × (3.3i)

= −(0.69482 m/s 2 ) j

String ED:

vD = vE

v D = 1.382 m/s 

a D = (a E /C )t = −(0.69482 m/s 2 ) j

a D = 0.695 m/s 2 

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PROBLEM 15.137 Denoting by rA the position vector of Point A of a rigid slab that is in plane motion, show that (a) the position vector rC of the instantaneous center of rotation is rC = rA +

ω × vA

ω2

where ω is the angular velocity of the slab and vA is the velocity of Point A, (b) the acceleration of the instantaneous center of rotation is zero if, and only if, aA =

α v + ω × vA ω A

where α = α k is the angular acceleration of the slab.

SOLUTION (a)

At the instantaneous center C,

vC = 0 vA = vC + ω × rA /C = ω × rA/C

ω × vA = ω × (ω × rA /C ) = −ω 2rA /C rA /C = − rC − rA = −

ω × vA

ω

2

= −rC/ A

or

rC/ A =

ω × vA

ω

ω × vA

ω2 rC = rA +

2

ω × vA

ω2



aA = aC + α × rA /C + ω × vA/ C

(b)

= aC − α k ×

ω × vA

ω2

+ ω × ( vA − vC )

αω k × (k × vA ) + ω × vA ω2 α = aC + vA + ω × vA ω = aC −

Set aC = 0.

aA =

α v + ω × vA  ω A

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PROBLEM 15.138* The drive disk of the scotch crosshead mechanism shown has an angular velocity ω and an angular acceleration α, both directed counterclockwise. Using the method of Section 15.9, derive expressions for the velocity and acceleration of Point B.

SOLUTION Origin at A. yB = l + yP = l + b sin θ v = y = b cos θθ = b cos θ ω B

B

vB = bω cos θ 

aB =  yB d vB dt d = (b cos θθ) dt aB = −b sin θθ 2 + b cos θθ

=

aB = bα cos θ − bω 2 sin θ 

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PROBLEM 15.139* The wheels attached to the ends of rod AB roll along the surfaces shown. Using the method of Section 15.9, derive an expression for the angular velocity of the rod in terms of vB , θ , l , and β .

SOLUTION Law of sines.

dB l = sin θ sin β l dB = sin θ sin β

d (d B ) dt l dθ = cos θ dt sin β l = cos θ ω sin β

vB =

ω=

vB sin β  l cos θ

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PROBLEM 15.140* The wheels attached to the ends of rod AB roll along the surfaces shown. Using the method of Section 15.9 and knowing that the acceleration of wheel B is zero, derive an expression for the angular acceleration of the rod in terms of vB , θ , l , and β.

SOLUTION Law of sines.

dB l = sin θ sin β l sin θ dB = sin β d (d B ) dt l dθ = cos θ sin β dt l = cos θ ω sin β v sin β ω= B l cos θ

vB =

Note that

dvB = 0. dt d ω vB sin β sin θ dθ α= = ⋅ ⋅ dt l cos 2θ dt

aB =

α=

vB sin β sin θ vB sin β ⋅ l cos θ l cos 2θ

2

 vB sin β  sin θ  cos3θ  l  

α =

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PROBLEM 15.141* A disk of radius r rolls to the right with a constant velocity v. Denoting by P the point of the rim in contact with the ground at t = 0, derive expressions for the horizontal and vertical components of the velocity of P at any time t.

SOLUTION x A = rθ ,

yA = r

xP = x A − r sin θ

= rθ − r sin θ yP = y A − r cos θ

= r − r cos θ x θ= A r v y A = 0, θ = r vt x A = vt , θ = r x A = v,

vt  v  x P = vx = rθ − r cos θθ = r 1 − cos  r r  vt  v  y P = v y = r sin θθ = r  sin  r r 

vt   vx = v 1 − cos   r   v y = v sin

vt  r

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PROBLEM 15.142* Rod AB moves over a small wheel at C while end A moves to the right with a constant velocity vA. Using the method of Section 15.9, derive expressions for the angular velocity and angular acceleration of the rod.

SOLUTION tan θ =

b xA

cot θ =

xA =u b

θ = cot −1 u u 1 + u2 (2uu )u

θ = − θ = But

(1 + u )

u 1 + u2

ω = θ and α = θ u=

Then



2 2

ω=

xA x v v , u = A = − A , u = − A = 0 b b b b vA b

1+

( ) xA b

( )( ) α=   1 + ( )  2

xA b

vA b

xA b

bvA , b + x A2

=

2

2

ω=

2

bvA b + x A2 2



2 2

−0=

(

2bxA v 2A b2 + x 2A

)

2

,

α=

(

2bxAv A2 b 2 + x A2

)

2



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PROBLEM 15.143* Rod AB moves over a small wheel at C while end A moves to the right with a constant velocity vA. Using the method of Section 15.9, derive expressions for the horizontal and vertical components of the velocity of Point B.

SOLUTION sin θ =

b

(b

)

1/ 2

+ x A2

2

cosθ =

,

(b

xA 2

+ x A2

)

1/2

xB = l cos θ − x A =

(b

lx A

yB = l sin θ = x B =

(

)

1/ 2

+ x A2

2

lb

(b

lx A b2 +

2

)

1/ 2 x A2

lb x A 2

=

(

b 2 + x A2

y B = −

But

(b

− xA

)

3/2

+ x A2 −

(

)

1/2

lx A x A x A b 2 + x A2

)

3/2

− x A

− x A

lbx A x A 2

x A = −v A ,

+ x A2

)

3/ 2

x B = (vB ) x ,

y B = (vB ) y

( vB ) x = v A −

(

lb 2v A b 2 + x A2

( vB ) y =

(

)



3/2

lbx Av A b 2 + x A2

)

3/2



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PROBLEM 15.144 Crank AB rotates with a constant clockwise angular velocity ω. Using the method of Section 15.9, derive expressions for the angular velocity of rod BD and the velocity of the point on the rod coinciding with Point E in terms of θ , ω , b, and l.

SOLUTION Law of cosines for triangle ABE. u 2 = l 2 + b 2 − 2bl cos(180° − θ ) = l 2 + b 2 + 2bl cos θ l + b cos θ cos ϕ = u b sin θ tan ϕ = l + b cos θ d (l + b cos θ )(b cos θ )θ + (b sin θ )(b cos θ )θ (tan ϕ ) = sec 2 ϕϕ = dt (l + b cos θ ) 2

ϕ = =

But,

θ = ω ,

(cos 2 ϕ )[bl cos θ + b 2 (cos 2 θ + sin 2 θ )]θ (l + b cos θ )2 bl cos θ + b 2  b(b + l cos θ )  θ= 2 θ 2 u l + b 2 + 2bl cos θ

ϕ = ωBD ,

and

vE = −u

ωBD =

Hence,

b(b + l cos θ ) ω l + b 2 + 2bl cos θ 2



Differentiate the expression for u 2 . 2uu = −2bl sin θθ bl sin θ ω vE = −u = 2 l + b 2 + 2bl cos θ vE =

bl sin θ ω l + b 2 + 2bl cos θ 2

 b sin θ  tan −1     l + b cos θ 

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PROBLEM 15.145 Crank AB rotates with a constant clockwise angular velocity ω. Using the method of Section 15.9, derive an expression for the angular acceleration of rod BD in terms of θ , ω , b, and l.

SOLUTION Law of cosines for triangle ABE. u 2 = l 2 + b 2 − 2bl cos(180° − θ )

= l 2 + b 2 + 2bl cos θ l + b cos θ cos ϕ = u b sin θ tan ϕ = l + b cos θ d (l + b cos θ )(b cos θ )θ + (b sin θ )(b cos θ )θ (tan ϕ ) = sec 2 ϕϕ = dt (l + b cos θ ) 2

ϕ = =

ϕ =

(cos 2 ϕ )[bl cos θ + b 2 (cos 2 θ + sin 2 θ )]θ (l + b cos θ ) 2 bl cos θ + b 2  b(b + l cos θ )  θ= 2 θ 2 u l + b 2 + 2bl cos θ b(b + l cos θ ) 

θ

l 2 + b 2 + 2bl cos θ (l 2 + b 2 + 2bl cos θ )(−bl sin θ ) − b(b + l cos θ )(−2bl sin θ )  2 + θ (l 2 + b 2 + 2bl cos θ ) 2

=

But,

θ = ω ,

b(b + l cos θ ) l 2 + b 2 + 2bl cos θ

θ = ω = 0,

θ −

bl (l 2 − b 2 )sin θ θ 2 (l 2 + b2 + 2bl cos θ ) 2

ϕ = α BD α BD =

bl (l 2 − b 2 )sin θ ω2 l 2 + b 2 + 2bl cos θ



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PROBLEM 15.146* Pin C is attached to rod CD and slides in a slot cut in arm AB. Knowing that rod CD moves vertically upward with a constant velocity v0, derive an expression for (a) the angular velocity of arm AB, (b) the components of the velocity of Point A; and (c) an expression for the angular acceleration of arm AB.

SOLUTION

b cos θ sin θ dy d  cos θ  dθ vC = v0 = C = b dt dθ  sin θ  dt

yC =

(a)

=b

θ = −

− sin 2 θ − cos 2 θ  b θ = − 2 θ 2 sin θ sin θ v0 sin 2 θ b

(1)

But ω = θ (b)

ω=

v0 sin 2 θ b



x A = l sin θ x A = l cos θθ = −

v0 l (sin 2 θ cos θ ) b

y A = l cos θ y A = −l sin θθ =

v0 l 3 sin θ b

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PROBLEM 15.146* (Continued)

vA =

Components: (c)

v0 l 2 sin θ cos θ b

+

v0l 3 sin θ  b

Differentiating Eq. (1),

θ =

d  v0 sin 2 θ − dt  b

 2v0 sin θ cos θ dθ  = − b dt 

 2v sin θ cos θ = − 0 b 

2 2   v0 sin θ  2v0 3  = 2 sin θ cos θ   − b   b

α = θ

α=

2v02 b

2

sin 3 θ cos θ



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PROBLEM 15.147* The position of rod AB is controlled by a disk of radius r which is attached to yoke CD. Knowing that the yoke moves vertically upward with a constant velocity v0, derive expression for the angular velocity and angular acceleration of rod AB.

SOLUTION From geometry,

But,

r sin θ dy r cos θ dθ =− dt sin 2θ dt y=

dy = −v0 dt

and

dθ =ω dt

r cos θ ω sin 2 θ v sin 2 θ ω= 0 r cos θ

v0 =

From geometry,

But,

ω=

v0 sin 2 θ r cos θ



r sin θ dy r cos θ dθ =− dt sin 2 θ dt y=

dy = − v0 dt

and

dθ =ω dt

r cos θ ω sin 2 θ v sin 2 θ ω= 0 r cos θ

v0 =

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PROBLEM 15.147* (Continued)

Angular acceleration.

α= =

d ω d ω dθ d ω ω = = dt dθ dt dθ v0 (2 cos 2 θ sin θ + sin 3 θ )  v0 sin 2 θ    r cos 2 θ  r cos θ  2

 v  (1 + cos 2 θ ) sin 3 θ = 0  cos3 θ  r  2

v  α =  0  (1 + cos 2 θ ) tan 3 θ  r 



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PROBLEM 15.148* A wheel of radius r rolls without slipping along the inside of a fixed cylinder of radius R with a constant angular velocity ω. Denoting by P the point of the wheel in contact with the cylinder at t = 0, derive expressions for the horizontal and vertical components of the velocity of P at any time t. (The curve described by Point P is a hypocycloid.)

SOLUTION Define angles θ and ϕ as shown.

θ = ω ,

θ = ωt

Since the wheel rolls without slipping, the arc OC is equal to arc PC. r (ϕ + θ ) = Rϕ rθ ϕ= R−r rθ rω = ϕ = R−r R−r rω t ϕ= R−r xP = ( R − r )sin ϕ − r sin θ (vP ) x = xP = ( R − r ) cos ϕϕ − r cos θθ rω t  rω   = ( R − r )  cos   − r (cos ωt )(ω ) − r  R − r  R  rω t   (vP ) x = rω  cos − cos ω t   R r −   yP = R − ( R − r ) cos ϕ − r cos θ (vP ) y = y P = ( R − r )sin ϕϕ + r sin θθ rω t  rω   = ( R − r )  sin   + r (sin ω t)(ω ) R − r  R − r   rω t   (vP ) y = rω  sin + sin ω t   R−r  

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PROBLEM 15.149* In Problem 15.148, show that the path of P is a vertical straight line when r = R/2. Derive expressions for the corresponding velocity and acceleration of P at any time t.

SOLUTION Define angles θ and ϕ as shown.

θ = ω ,

θ = ω t , θ = 0

Since the wheel rolls without slipping, the arc OC is equal to arc PC. r (ϕ + θ ) = Rθ = 2rθ ϕ =0 ϕ = θ = ω

ϕ = θ = 0 xP = ( R − r ) sin ϕ − r sin θ = r sin θ − r sin θ =0

The path is the y axis. 

yP = R − ( R − r ) cos ϕ − r cos θ = R − r cos θ − r cos θ = R(1 − cos θ ) v = y = R sin θθ P

v = ( Rω sin ω t ) j 

a = v

= ( R cos θθ 2 − sin θθ) = Rω 2 cos θ

a = ( Rω 2 cos ω t ) j 

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PROBLEM 15.CQ8 A person walks radially inward on a platform that is rotating counterclockwise about its center. Knowing that the platform has a constant angular velocity ω and the person walks with a constant speed u relative to the platform, what is the direction of the acceleration of the person at the instant shown? (a) Negative x (b) Negative y (c) Negative x and positive y (d ) Positive x and positive y (e) Negative x and negative y

SOLUTION The ω 2r term will be in the negative x-direction and the Coriolis acceleration will be in the negative y-direction. Answer: (e) 

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PROBLEM 15.150 Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in rod BD and by the collar that slides on rod AE. Knowing that at the instant considered the rods rotate clockwise with constant angular velocities, determine for the given data the velocity of pin P.

ω AE = 8 rad/s, ωBD = 3 rad/s

SOLUTION AB = 500 mm = 0.5 m, AP = 0.5 tan 30°, BP = ω AE = 8 rad/s

0.5 cos 30°

ω BD = 3 rad/s

,

Let P′ be the coinciding point on AE and u1 be the outward velocity of the collar along the rod AE. v P = v P′ + v P/ AE = [( AP)ω AE ] + [u1

]

Let P′′ be the coinciding point on BD and u2 be the outward speed along the slot in rod BD. v P = v P′′ + v P/BD = [( BP)ωBD

30°] + [u2

60°]

Equate the two expressions for v P and resolve into components. :

u1 = 1.5 + 0.5u2

or :

u2 =

From (1),

 0.5  u1 =   (3)(cos30°) + u2 cos 60°  cos30°  (1)

 0.5  −(0.5 tan 30°)(8) = −   (3)sin 30° + u2 sin 60°  cos30° 

1 [1.5 tan 30° − 4 tan 30°] = −1.66667 m/s sin 60°

u1 = 1.5 + (0.5)(−1.66667) = 0.66667 m/s v P = [(0.5 tan 30°)(8) ] + [0.66667

] = [2.3094 m/s ] + [0.66667 m/s

]

vP = − 2.30942 + 0.66667 2 = 2.4037 m/s tanβ =

2.3094 0.66667

β = 73.9° v P = 2.40 m/s

73.9° 

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PROBLEM 15.151 Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in rod BD and by the collar that slides on rod AE. Knowing that at the instant considered the rods rotate clockwise with constant angular velocities, determine for the given data the velocity of pin P.

ω AE = 7 rad/s, ωBD = 4.8 rad/s

SOLUTION AB = 500 mm = 0.5 m, AP = 0.5 tan 30°, BP = ω AE = 7 rad/s

0.5 cos 30°

ω BD = 4.8 rad/s

,

Let P′ be the coinciding point on AE and u1 be the outward velocity of the collar along the rod AE. v P = v P′ + v P/ AE = [( AP)ω AE ] + [u1

]

Let P′′ be the coinciding point on BD and u2 be the outward speed along the slot in rod BD. v P = v P′′ + v P/BD = [( BP)ωBD

30°] + [u2

60°]

Equate the two expressions for v P and resolve into components.  0.5  u1 =   (4.8)(cos30°) + u2 cos60°  cos30° 

:

u1 = 2.4 + 0.5u2

or

 0.5  −(0.5 tan 30°)(7) = −   (4.8)sin 30° + u2 sin 60°  cos30° 

: u2 =

From (1),

(1)

1 [2.4 tan 30° − 3.5 tan 30°] = −0.73333 m/s sin 60°

u1 = 2.4 + (0.5)(−0.73333) = 2.0333 m/s v P = [(0.5 tan 30°)(7)

] + [2.0333

] = [2.0207 m/s ] + [2.0333 m/s

]

vP = (2.0333)2 + (2.0207)2 = 2.87 m/s tanβ = −

2.0207 , 2.0333

β = −44.8° v P = 2.87 m/s

44.8°  

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PROBLEM 15.152 Two rotating rods are connected by slider block P. The rod attached at A rotates with a constant angular velocity ωA . For the given data, determine for the position shown (a) the angular velocity of the rod attached at B, (b) the relative velocity of slider block P with respect to the rod on which it slides. b = 8 in.,

ω A = 6 rad/s.

SOLUTION Dimensions: Law of sines. AP BP 8 in. = = sin 20° sin120° sin 40° AP = 4.2567 in. BP = 10.7784 in.

ω AP = 6 rad/s

Velocities. Note: P′ = Point of BE coinciding with P. vP = ( AP)ω AP = (4.2567 in.)(6 rad/s)

= 25.540 in./s

30°

vP = vP′ + vP/BE

[25.540

(a)

30°] = [vP′

30°]

vP′ = (25.54) cos 40°

ωBE

(b)

70°] + [vP/BE

= 19.565 in./s v = P′ BP 19.565 in./s = 10.7784 in. = 1.8152 rad/s

ω BE = 1.815 rad/s



vP/BE = (25.54)sin 40°

= 16.417 in./s

v P/BE = 16.42 in./s

20° 

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PROBLEM 15.153 Two rotating rods are connected by slider block P. The rod attached at A rotates with a constant angular velocity ωA . For the given data, determine for the position shown (a) the angular velocity of the rod attached at B, (b) the relative velocity of slider block P with respect to the rod on which it slides.

ω A = 10 rad/s.

b = 300 mm,

SOLUTION Dimensions: Law of sines. AP BP 300 mm = = sin 20° sin120° sin 40° AP = 159.63 mm BP = 404.19 mm

ω AD = 10 rad/s

Velocities. Note: P′ = Point of AD coinciding with P. vP′ = ( AP′)ω AD

= (159.63 mm)(10 rad/s) = 1596.3 mm/s

30°

vP = vP′ + vP/AD

[vP

(a)

70°] = [1596.3

60°]

vP = (1596.3)/cos 40°

ωBP

(b)

30°] + [vP/AD

= 2083.8 mm/s v = P BP 2083.8 mm/s = 404.19 mm = 5.155 rad/s

ω BD = 5.16 rad/s

vP /AD = (1596.3)tan 40° = 1339.5 mm/s

v P/AD = 1.339 m/s



60° 

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PROBLEM 15.154 Pin P is attached to the wheel shown and slides in a slot cut in bar BD. The wheel rolls to the right without slipping with a constant angular velocity of 20 rad/s. Knowing that x = 480 mm when θ = 0, determine the angular velocity of the bar and the relative velocity of pin P with respect to the rod for the given data. (a) θ = 0, (b) θ = 90°.

SOLUTION

Coordinates. x A = ( x A )0 + rθ , xB = 0, xC = x A ,

yA = r

yB = r yC = 0

xP = x A + e sin θ yP = r + e cos θ

Data:

( x A )0 = 480 mm = 0.48 m r = 200 mm = 0.20 m e = 140 mm = 0.14 m

Velocity analysis.

ω AC = ω AC

,

ω BD = ωBD

v P = v A + v P/ A = [rω AC

] + [eω AC

v P′ = [ xPωBD ] + [(e cos θ )]ωBD v P/F = [u cos β

,

θ]

]

] + [u sin β ]

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PROBLEM 15.154 (Continued)

Use v P = v P′ + v P/F and resolve into components.

(a)

θ = 0.

:

(r + e cos θ )ω AC = (e cos θ )ωBD + (cos β )u

(1)

:

(e sin θ )ω AC = xPω BD − (sin β )u

(2)

x A = 0.48 m, tanβ =

xP = 0.48 m,

e cos θ 0.14 , = xP 0.48

ω AC = 20 rad/s

β = 16.26°

Substituting into Eqs. (1) and (2), (0.20 + 0.14)(20) = 0.14ωBD + (cos16.26°)u

(1)

0 = 0.48ωBD − (sin16.26°)u

(2)

Solving simultaneously,

ω BD = 3.81 rad/s,

ω BD = 3.81 rad/s v P/F = 6.53 m/s

u = 6.53 m/s,

(b) θ = 90°.



16.26° 

π  xP = 0.48 + (0.20)   + 0.14 = 0.93416 m 2

β =0 Substituting into Eqs. (1) and (2), (0.20)(20) = u

(1)

u = 4 m/s (0.14)(20) = 0.93416ωBD

ωBD = 2.9973 rad/s,

(2) ω BD = 3.00 rad/s v P/F = 4.00 m/s

 

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PROBLEM 15.155 Bar AB rotates clockwise with a constant angular velocity of 8 rad/s and rod EF rotates clockwise with a constant angular velocity of 6 rad/s. Determine at the instant shown (a) the angular velocity of bar BD, (b) the relative velocity of collar D with respect to rod EF.

SOLUTION Bar AB.

(Rotation about A)

ω AB = 8 rad/s v B = (12)(8) = 96 in./s

Rod EF.

(Rotation about E )

ωEF = 6 rad/s. v D′ = (12)(6) = 72 in./s

Bar BD.

Assume angular velocity is ω BD . Plane motion = Translation with B + Rotation about B. v D = v B + v D /B = [96

Collar D.

] + [24ωBD ] + [12ωBD

]

(1)

Sliding on rotating rod EF with relative velocity u . v D = v D′ + v D /EF = [72

] + [u ]

(2)

Matching the expressions (1) and (2) for vD, Components

:

96 + 12ω BD = − 72

ω BD = −14 ω BD = 14.00 rad/s

(a) Components :

24ω BD = u



u = (24)(−14) = −336 in./s v D / EF = 28.0 ft/s 

(b)

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PROBLEM 15.156 Bar AB rotates clockwise with a constant angular velocity of 4 rad/s. Knowing that the magnitude of the velocity of collar D is 20 ft/s and that the angular velocity of bar BD is counterclockwise at the instant shown, determine (a) the angular velocity of bar EF, (b) the relative velocity of collar D with respect to rod EF.

SOLUTION Bar AB.

ω AB = 4 rad/s

(Rotation about A)

v B = (1 ft)(4 rad/s) = 4 ft/s

Bar BD.

Angular velocity is ω BD . Plane motion = Translation with B + Rotation about B. v D = v B + v D /B = [4

] + [2ωBD ] + [1ωBD

]

Magnitude of v D : vD = 20 ft/s vD2 = (4 + ωBD )2 + (2ωBD )2 = (20) 2 2 5ωBD + 8ω BD − 384 = 0

ωBD =

−8 ± 88 10

v D = [4

Rod EF.

Positive root ωBD = 8 rad/s

] + [(2)(8) ] + [(1)(8)

] = [12

] + [16 ]

Angular velocity = ω EF

(Rotation about E)

v D' = [(1)ωEF

Collar D.

(1)

]

Slides on rotating rod EF with relative velocity u . v D = v D' + v D /EF = [1ωEF

] + [u ]

(2)

Matching the expressions (1) and (2) for v D , (a) Component (b) Component :

:

12 = 1ω EF

ω EF = 12.00 rad/s

16 = u



v D /EF = 16 ft/s 

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PROBLEM 15.157 The motion of pin P is guided by slots cut in rods AD and BE. Knowing that bar AD has a constant angular velocity of 4 rad/s clockwise and bar BE has an angular velocity of 5 rad/s counterclockwise and is slowing down at a rate of 2 rad/s2, determine the velocity of P for the position shown.

SOLUTION Units: meters, m/s, m/s2 i =1

Unit vectors:

,

j =1 ,

k =1 .

Geometry: Slope angle θ of rod BE. tan θ =

0.15 = 0.5 0.3

θ = 26.565°



rP /A = 0.1i + 0.15 j

rP /B = −0.3i + 0.15 j

Angular velocities:

ω AD = −(4 rad/s)k

ω BE = (5 rad/s)k

Angular accelerations:

α AD = 0

α BE = −(2 rad/s 2 )k

Velocity of Point P′ on rod AD coinciding with the pin: v P′ = ω AD × rP /A = (−4k ) × (0.1i + 0.15 j) = 0.6i − 0.4 j

Velocity of the pin relative to rod AD: v P /AD = u1 = u1 j

v P = v P′ + v P /AD = 0.6i − 0.4 j + u1 j

Velocity of P:

Velocity of Point P″ on rod BE coinciding with the pin: v P′′ = ω BE × rP /B = 5k × (−0.3i + 0.15 j) = −0.75i − 1.5 j

Velocity of the pin relative to rod BE: v P /BE = u2

Velocity of P:

θ = −u2 cos θ i + u2 sin θ j

v P = v P′′ + v P /BE

= −0.75i − 1.5 j − u2 cos θ i + u2 sin θ j

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1220

PROBLEM 15.157 (Continued)

Equating the two expressions for vP and resolving into components, i:

0.6 = −0.75 − u2 cos θ u2 = −

j:

1.35 = −1.50965 cos 26.565°

−0.4 + u1 = −1.5 + u2 sin θ u1 = −1.1 + (−1.50935)sin 26.535° = −1.77500

Velocity of P: v P = 0.6i − 0.4 j − 1.775 j = 0.6i − 2.175 j v P = 2.26 m/s

74.6° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1221

PROBLEM 15.158 Four pins slide in four separate slots cut in a circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. If each pin maintains the same velocity relative to the plate when the plate rotates about O with a constant counterclockwise angular velocity ω , determine the acceleration of each pin.

SOLUTION For each pin:

a P = a P′ + a P/F + aC

Acceleration of the coinciding Point P′ of the plate. For each pin a P′ = rω 2 towards the center O.

Acceleration of the pin relative to the plate. For pins P1 , P2 and P4 ,

a P/F = 0

For pin P3 ,

a P/F =

u2 ← r

Coriolis acceleration aC . For each pin aC = 2ωu with aC in a direction obtained by rotating u through 90° in the sense of ω, i.e., . Then

a1 = [rω 2 →] + [2ωu ↓]

a1 = rω 2 i − 2ω uj 

a2 = [rω 2 ↓] + [2ωu →]

a 2 = 2ω ui − rω 2 j 

 2  a3 = [ rω 2 ←] +  u ←  + [2ωu ←]  r 



a4 = [rω 2 ↑] + [2ωu ↑]

  u2 + 2ω u  i  a3 = −  rω 2 + r   a4 = (rω 2 + 2ω u ) j 

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PROBLEM 15.159 Solve Problem 15.158, assuming that the plate rotates about O with a constant clockwise angular velocity ω . PROBLEM 15.158 Four pins slide in four separate slots cut in a circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. If each pin maintains the same velocity relative to the plate when the plate rotates about O with a constant counterclockwise angular velocity ω , determine the acceleration of each pin.

SOLUTION For each pin:

a P = a P′ + a P/F + aC

Acceleration of the coinciding Point P′ of the plate. For each pin a P′ = rω 2 towards the center O.

Acceleration of the pin relative to the plate. For pins P1 , P2 and P4 ,

a P/F = 0

For pin P3 ,

a P/F =

u2 ← r

Coriolis acceleration aC . For each pin aC = 2ωu with aC in a direction obtained by rotating u through 90° in the sense of ω. Then

a1 = [rω 2 →] + [2ω u ↑]

a1 = rω 2 i + 2ω uj 

a 2 = [rω 2 ↓] + [2ω u ←]

a2 = −2ω ui − rω 2 j 

 2  a3 = [ rω 2 ←] +  u ←  + [2ω u →]  r 



a 4 = [rω 2 ↑] + [2ω u ↓]

 u2  a3 =  2ω u − rω 2 −  i  r   a4 = (rω 2 − 2ω u ) j 

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PROBLEM 15.160 Pin P slides in the circular slot cut in the plate shown at a constant relative speed u = 500 mm/s. Assuming that at the instant shown the angular velocity of the plate is 6 rad/s and is increasing at the rate of 20 rad/s2, determine the acceleration of pin P when θ = 90°.

SOLUTION

θ = 90°

Units: meters, m/s, m/s2 i =1

Unit vectors:

,

j =1 , k =1

rP /A = (0.15i + 0.1j)

rP /C = 0.1j

Motion of Point P′ on the plate coinciding with P. ω = (6 rad/s)k

α = (20 rad/s2 )k

v P′ = ω × rP /A = 6k × (−0.15i + 0.1j) = −0.6i − 0.9 j a P′ = α × rP /A − ω 2 rB /A

= 20k × (−0.15i + 0.1j) − (6) 2 ( −0.15i + 0.1j) = −2i − 3 j + 5.4i − 3.6 j = 3.4i − 6.6 j

Motion of P relative to the plate AC. u = 500 mm/s = 0.5 m/s v P /AC = −ui = −0.5i a P /AC = −ui −

Coriolis acceleration:

u = 0

u (0.5) 2 j=0− j = −2.5 j R 0.1

2ω × v P /AC = (2)(6k ) × (−0.5i) = −6 j

Acceleration of P. a P = a P′ + a P /AC + 2ω × v P /AC = 3.4i − 6.6 j − 2.5 j − 6 j = (3.4 m/s 2 )i − (15.1 m/s) j

a P = 15.47 m/s 2

77.3° 

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PROBLEM 15.161 The cage of a mine elevator moves downward at a constant speed of 40 ft/s. Determine the magnitude and direction of the Coriolis acceleration of the cage if the elevator is located (a) at the equator, (b) at latitude 40° north, (c) at latitude 40° south.

SOLUTION Earth makes one revolution (2π radians) in 23.933 h (86,160 s). Ω=

2π j 86,160

= (72.926 × 10 −6 rad/s)j

Velocity relative to the Earth at latitude angle ϕ . v P/earth = 40(− cos ϕ i − sin ϕ j)

Coriolis acceleration aC . aC = 2Ω × v P/earth = (2)(72.926 × 10−6 j) × [40(− cos ϕ i − sin ϕ j)] = (5.8341 × 10−3 cos ϕ )k

(a)

ϕ = 0°,

cos ϕ = 1.000

aC = 5.83 × 10 −3 ft/s 2 west 

(b)

ϕ = 40°,

cos ϕ = 0.76604

aC = 4.47 × 10−3 ft/s2 west 

(c)

ϕ = −40°,

cos ϕ = 0.76604

aC = 4.47 × 10−3 ft/s2 west 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1225

PROBLEM 15.162 A rocket sled is tested on a straight track that is built along a meridian. Knowing that the track is located at latitude 40° north, determine the Coriolis acceleration of the sled when it is moving north at a speed of 900 km/h.

SOLUTION Earth makes one revolution (2π radians) in 23.933 h = 86,160 s. Ω=

2π 86,160

= (72.926 × 10 −6 rad/s)j

Speed of sled.

u = 900 km/h = 250 m/s

Velocity of sled relative to the Earth. v P/earth = 250(−sin ϕ i + cos ϕ j)

Coriolis acceleration.

aC = 2Ω × v P/earth aC = (2)(72.926 × 10−6 j) × [250(−sin ϕ i + cos ϕ j)]

At latitude ϕ = 40°,

= 0.036463sin ϕ k

aC = 0.036463 sin 40°k = (0.0234 m/s 2 )k



aC = 0.0234 m/s 2 west 

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PROBLEM 15.163 The motion of blade D is controlled by the robot arm ABC. At the instant shown, the arm is rotating clockwise at the constant rate ω = 1.8 rad/s and the length of portion BC of the arm is being decreased at the constant rate of 250 mm/s. Determine (a) the velocity of D, (b) the acceleration of D.

SOLUTION i =1

Unit vectors: Units: meters, m/s, m/s

, j =1 ,

k=

2

rD /A = (0.32 m)i − (0.24 m) j

Motion of Point D′ of extended frame AB. ω = −(1.8 rad/s)k α=0 v D′ = ω × rD /A = ( −1.8k ) × (0.32i − 0.24 j)

= −0.432i − 0.576 j a D′ = α × rD /A − ω 2 (rD /A ) = 0 − (1.8)2 (0.32i − 0.24 j) = −1.0368i + 0.7776 j

Motion of Point D relative to frame AB. v D /AB = 250 mm/s 25° = −(0.25cos 25°)i + (0.25sin 25°) j

= −0.22658i + 0.10565 j a D /AB = 0

Coriolis acceleration

2ω × v D /AB = (2)(−1.8k ) × (−0.22658i + 0.10565 j) = 0.38034i + 0.81569 j

(a)

Velocity of Point D. v D = v D′ + v D /AB = −0.432i − 0.576 j − 0.22658i + 0.10565 j = −0.65858i − 0.47035 j v D = (0.659 m/s)i − (0.470 m/s) j = 0.809 m/s

35.5° 

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PROBLEM 15.163 (Continued)

(b)

Acceleration of Point D. a D = a D′ + a D /AB + 2ω × v D /AB = −1.0368i + 0.7776 j + 0 + 0.38034i + 0.81569 j = −0.6565i + 1.5933 j

a D = −(0.657 m/s 2 )i + (1.593 m/s 2 ) j = 1.723 m/s 2

67.6° 

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PROBLEM 15.164 At the instant shown the length of the boom AB is being decreased at the constant rate of 0.2 m/s and the boom is being lowered at the constant rate of 0.08 rad/s. Determine (a) the velocity of Point B, (b) the acceleration of Point B.

SOLUTION Velocity of coinciding Point B′ on boom. v B′ = rω = (6)(0.08) = 0.48 m/s

60°

Velocity of Point B relative to the boom. v B/boom = 0.2 m/s

(a)

30°

Velocity of Point B. v B = v B′ + v B/boom : (vB ) x = 0.48cos 60° − 0.2cos30° = 0.06680 m/s : (vB ) y = −0.48sin 60° − 0.2sin 30° = −0.51569 m/s vB = 0.066802 + 0.515692

= 0.520 m/s 0.51569 tan β = , β = 82.6° 0.06680

v B = 0.520 m/s

82.6° 

Acceleration of coinciding Point B ′ on boom. a B′ = rω 2 = (6)(0.08) 2 = 0.0384 m/s 2

30°

Acceleration of B relative to the boom. a B/boom = 0

Coriolis acceleration. 2ωu = (2)(0.08)(0.2) = 0.032 m/s 2

60°

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PROBLEM 15.164 (Continued)

(b)

Acceleration of Point B. a B = a B + a B/boom + 2ω u : (aB ) x = −0.0384 cos 30° + 0 − 0.032cos 60° = −0.04926 m/s 2 : (aB ) y = −0.0384sin 30° + 0 + 0.032sin 60° = 0.008513 m/s2 aB = (0.04926)2 + (0.008513)2 = 0.0500 m/s 2 0.008513 tan β = , β = 9.8° 0.04926

a B = 50.0 mm/s 2

9.8° 

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PROBLEM 15.165 At the instant shown the length of the boom AB is being increased at the constant rate of 0.2 m/s and the boom is being lowered at the constant rate of 0.08 rad/s. Determine (a) the velocity of Point B, (b) the acceleration of Point B.

SOLUTION Velocity of coinciding Point B ′ on boom. v B′ = rω = (6)(0.08) = 0.48 m/s

60°

Velocity of Point B relative to the boom. v B/boom = 0.2 m/s

(a)

30°

Velocity of Point B. v B = v B′ + v B/boom : (vB ) x = 0.48 cos 60° + 0.2 cos 30° = 0.4132 m/s : (vB ) y = −0.48 sin 60° + 0.2 sin 30° = −0.3157 m/s

vB = (0.4132)2 + (0.3157) 2 = 0.520 m/s tan β = −

0.3157 , 0.4132

β = −37.4°

v B = 0.520 m/s

37.4° 

Acceleration of coinciding Point B′ on boom. a B′ = rω 2 = (6)(0.08) 2 = 0.0384 m/s 2

30°

Acceleration of B relative to the boom. a B/boom = 0

Coriolis acceleration. 2ωu = (2)(0.08)(2) = 0.032 m/s 2

60°

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PROBLEM 15.165 (Continued)

(b)

Acceleration of Point B. a B = a B′ + a B/boom + 2ω u : (aB ) x = −0.0384 cos 30° + 0.032 cos 60° = −0.017255 m/s 2 : (aB ) y = −0.0384 sin 30° − 0.032 sin 60° = −0.04691 m/s 2 aB = (0.017255)2 + (0.04691)2 = 0.0500 m/s 2 0.04691 tan β = , β = 69.8° 0.017255

a B = 50.0 mm/s 2

69.8° 

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PROBLEM 15.166 The sleeve BC is welded to an arm that rotates about A with a constant angular velocity ω. In the position shown rod DF is being moved to the left at a constant speed u = 400 mm/s relative to the sleeve. For the given angular velocity ω , determine the acceleration (a) of Point D, (b) of the point of rod DF that coincides with Point E.

ω = (3 rad/s)i.

SOLUTION (a)

Point D.

v D /F = v D /BC = (0.4 m/s)k ; aD /F = 0  AD = −(0.12 m)j + (0.3 m)k  a D′ = ω × (ω × AD ) = −ω 2 ( AD ) = −(3 rad/s) 2 AD = +(1.08 m/s 2 )j − (2.70 m/s 2 )k a c = 2ω × v D /F = 2[(3 rad/s)i ] × (0.40 m/s)k = −(2.4 m/s 2 ) j a D = a D′ + a D /F + ac = [+ (1.08 m/s 2 )j − (2.70 m/s 2 )k ] + 0 + [ −(2.4 m/s)j]

a D = −(1.32 m/s) j − (2.70 m/s)k 

(b)

Point P of DF that coincides with E. v P /F = v P /BC = (0.40 m/s)k ; a P /F = 0  AE = −(0.120 m)j  a P′ = ω × (ω × AE ) = −ω 2 AE = −(3 rad/s)2 AE = (1.08 m/s 2 ) j a c = 2ω × v P /F

= 2[(3 rad/s)i ] × (0.40 m/s)k = −(2.40 m/s 2 ) j a P = a P′ + a P /F + ac

= [(1.08 m/s 2 ) j] + 0 + [−(2.4 m/s 2 ) j] a P = −(1.32 m/s 2 ) j 

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PROBLEM 15.167 The sleeve BC is welded to an arm that rotates about A with a constant angular velocity ω . In the position shown rod DF is being moved to the left at a constant speed u = 400 mm/s relative to the sleeve. For the given angular velocity ω , determine the acceleration (a) of Point D, (b) of the point of rod DF that coincides with Point E.

ω = (3 rad/s)j.

SOLUTION (a)

Point D.

v D /F = v D /BC = (0.4 m/s)k ; aD /F = 0  AD = −(0.12 m)j + (0.3 m)k  a D′ = ω × (ω × AD) = 3j × (3 j × (−0.12 j + 0.3)k ) = −(2.70 m/s 2 )k a c = 2ω × v D /F = 2[(3 rad/s)j] × (0.40 m/s)k = (2.4 m/s 2 )i a D = a D′ + a D /F + a c = [−(2.70 m/s 2 )k ] + 0 + [(2.4 m/s)i ]

a D = (2.4 m/s)i − (2.70 m/s)k 

(b)

Point P of DF that coincides with E. v P /F = v P /BC = (0.40 m/s)k ; a P /F = 0  AE = −(0.120 m)j  a P′ = ω × (ω × AE ) = 3 j × (3j × (−0.12 j) = 0 a c = 2ω × v P /F = 2[(3 rad/s) j] × (0.40 m/s)k = (2.40 m/s2 )i a P = a P′ + a P /F + ac

= 0 + 0 + (2.4 m/s 2 )i

a P = (2.4 m/s 2 )i 

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PROBLEM 15.168 A chain is looped around two gears of radius 40 mm that can rotate freely with respect to the 320-mm arm AB. The chain moves about arm AB in a clockwise direction at the constant rate of 80 mm/s relative to the arm. Knowing that in the position shown arm AB rotates clockwise about A at the constant rate ω = 0.75 rad/s, determine the acceleration of each of the chain links indicated. Links 1 and 2.

SOLUTION Let the arm AB be a rotating frame of reference.

Link 1:

Ω = 0.75 rad/s

= −(0.75 rad/s)k :

r1 = −(40 mm)i, v1/AB = u ↑ = (80 mm/s) j a1′ = −Ω 2 r1 = −(0.75)2 ( −40) = (22.5 mm/s)i 802 160 mm/s → = (160 mm/s 2 )i 40 ρ = (2)( −0.75k ) × (80 j)

a1/AB = 2Ω × v P /AB

u2

=

= (120 mm/s)i a1 = a1′ + a1/AB + 2Ω × v1/AB = (302.5 mm/s 2 )i

Link 2:

a1 = 303 mm/s 2



r2 = (160 mm)i + (40 mm) j v 2/AB = u → = (80 mm/s)i a′2 = −Ω 2 r2 = −(0.75)2 (160i + 40 j) a 2/AB

= −(90 mm/s 2 )i − (22.5 mm/s 2 ) j =0

2Ω × v 2/AB = (2)(−0.75k ) × (80i) = −(120 mm/s 2 ) j a 2 = a′2 + a 2/AB + 2Ω × v 2/AB = −90i − 22.5 j − 120 j = −(90 mm/s 2 )i − (142.5 mm/s 2 ) j a 2 = (90) 2 + (142.5) 2 = 168.5 mm/s 2 tan β =

142.5 , β = 57.7° 90

a 2 = 168.5 mm/s 2

57.7° 

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PROBLEM 15.169 A chain is looped around two gears of radius 40 mm that can rotate freely with respect to the 320-mm arm AB. The chain moves about arm AB in a clockwise direction at the constant rate 80 mm/s relative to the arm. Knowing that in the position shown arm AB rotates clockwise about A at the constant rate ω = 0.75 rad/s, determine the acceleration of each of the chain links indicated. Links 3 and 4.

SOLUTION Ω = 0.75 rad/s

Let arm AB be a rotating frame of reference.

Link 3:

= −(0.75 rad/s)k

r3 = (360 mm)i v3/AB = u ↓ = −(80 mm/s) j a3′ = −Ω2 r3 = −(0.75)2 (360) = −(202.5 mm/s 2 )i a3/AB =

u2

ρ

=

(80) 2 = 160 mm/s 2 i ← = −(160 mm/s 2 )i 40

2Ω × v 3/AB = (2)(−0.75k ) × (−80 j) = −(120 mm/s 2 )i a3 = a3′ + a3/AB + 2Ω × v 3/AB = −(482.5 mm/s 2 )i

Link 4:

a3 = 483 mm/s 2



r4 = (160 mm)i − (40 mm) j v 4/AB = u ← = −(80 mm/s 2 )i a 4′ = −Ω 2 r4 = −(0.75) 2 (160i − 40 j) a 4/AB

= −(90 mm/s 2 )i + (22.5 mm/s 2 ) j =0

2Ω × v 4/AB = (2)( −0.75k ) × (−80i ) = (120 mm/s 2 ) j

a 4 = a 4′ + a 4/AB + 2Ω × v 4/AB

= −(90 mm/s 2 )i + (142.5 mm/s 2 ) j a 4 = (90)2 + (142.5) 2

= 168.5 mm/s 2 tan β =

142.5 , β = 57.7° 90

a 4 = 168.5 mm/s 2

57.7° 

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PROBLEM 15.170 A basketball player shoots a free throw in such a way that his shoulder can be considered a pin joint at the moment of release as shown. Knowing that at the instant shown the upper arm SE has a constant angular velocity of 2 rad/s counterclockwise and the forearm EW has a constant clockwise angular velocity of 4 rad/s with respect to SE, determine the velocity and acceleration of the wrist W.

SOLUTION Units: meters, m/s, m/s2 Unit vectors:

i =1

,

j =1 ,

k =1

Relative positions: rE/S = (0.35cos 30°)i + (0.35sin 30°) j = 0.30311i + 0.175 j rW /E = −(0.3cos80°)i + (0.3sin 80°) j = −0.05209i + 0.29544 j rW /S = rE/S + rW /E = 0.25101i + 0.47044 j

Use a frame of reference rotating with the upper arm SE with angular velocity  = 0) Ω = (2 rad/s)k (Ω

The motion of the wrist W relative to this frame is a rotation about the elbow E with angular velocity  = 0) ω = −(4 rad/s)k (ω

Motion of Point W ′ in the frame coinciding with W. vW ′ = Ω × rW /S = (2k ) × (0.25101i + 0.47044 j) = −0.94088i + 0.50204 j aW ′ = −Ω 2 rW /S = −(2) 2 (0.25101i + 0.47044 j) = −1.00408i − 1.88176 j

Motion of W relative to the frame. vW /SE = ω × rW /E = (−4k ) × (−0.05210i + 0.29544 j) = 1.18176i + 0.2084 j aW /SE = −ω 2rW /E = −(4) 2 (−0.05210i + 0.29544 j) = 0.8336i − 4.72708 j

Velocity of W:

vW = vW ′ + vW /SE = 0.24088i + 0.71044 j vW = 0.750 m/s

71.3° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1237

PROBLEM 15.170 (Continued)

Coriolis acceleration: 2Ω × vW /SE = (2)(2k ) × (1.18176i + 0.2084 j) = −0.8336i + 4.72704 j

Acceleration of W: aW = aW ′ + aW / SE + 2Ω × vW /SE = −1.00408i − 1.88176 j aW = 2.13 m/s 2

61.9° 

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PROBLEM 15.171 The human leg can be crudely approximated as two rigid bars (the femur and the tibia) connected with a pin joint. At the instant shown the veolcity and acceleration of the ankle is zero. During a jump, the velocity of the ankle A is zero, the tibia AK has an angular velocity of 1.5 rad/s counterclockwise and an angular acceleration of 1 rad/s2 counterclockwise. Determine the relative angular velocity and angular acceleration of the femur KH with respect to AK so that the velocity and acceleration of H are both straight up at the instant shown.

SOLUTION Units: inches, in./s, in./s2 Unit vectors: Relative positions:

i =1

j =1 ,

,

k =1

rK /A = (12 in.)i + (12 in.) j rH /K = −(14 in.)i + (14 in.) j rH /A = rK /A + rH /K = −(2 in.)i + (26 in.) j

Use a frame of reference moving with the lower leg AK with angular velocity

and angular acceleration

Ω = 1.5 rad/s

= (1.5 rad/s)k

 = 1.0 rad/s Ω

= (1.0 rad/s 2 )k

The motion of the hip H relative to this frame is a rotation about the knee K with angular velocity

ω = ωk and angular acceleration

α = αk

Both ω and α are measured relative to the lower leg AK. Motion of Point H ′ in the frame coinciding with H. v H ′ = Ω × rH /A = 1.5k × (−2i + 26 j) aH ′

= −(39 in./s)i − (3 in./s) j  × r − Ω2r =Ω H /A

H /A

= (1.0k ) × (−2i + 26 j) − (1.5) 2 (−2i + 26 j) = −26i − 2 j + 4.5i − 58.5 j = −(21.5 in./s 2 )i − (60.5 in./s 2 ) j

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PROBLEM 15.171 (Continued)

Motion of H relative to the frame. v H/AK = ω k × rH /K = ω k × (−14i + 14 j) = −14ω i − 14ω j a H/AK = α k × rH/K − ω 2 rH /K = α k × (−14i + 14 j) − ω 2 (−14i + 14 j) = −14α i − 14α j + 14ω 2 i − 14ω 2 j vH = vH

Velocity of H.

= vH j

v H = v H ′ + v H /AK

vH j = −39i − 3j − 14ω i − 14ω j

Resolve into components. 0 = −39 − 14ω

i:

ω=−

vH = −3 − 14ω

j:

vH = −36 in./s

ω = −(2.79 rad/s)k = 2.79 rad/s

Relative angular velocity: Coriolis acceleration:

39 = −2.7857 rad/s 14



2Ω × v H /AK = (2)(1.5k ) × (−14ω i − 14ω j) = 42ω i − 42ω j = −(117 in./s 2 )i + (117 in./s 2 ) j

Acceleration of H.

a H = a H = aH j a H = a H ′ + a H /AK + 2Ω × v H /AK

aH j = −21.5i − 60.5 j − 14α i − 14α j + 14ω 2 i − 14ω 2 j − 117i + 117 j

Resolve into components. 0 = −21.5 − 14α + (14)(−2.7857) 2 − 117

i:

α = −2.1327 rad/s 2 j:

aH = −60.5 − (14)( −2.1327) − (14)( −2.7857) 2 + 117 = −22.284 in./s 2

α = −(2.13 rad/s 2 )k = 2.13 rad/s 2

Relative angular acceleration:



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PROBLEM 15.172 The collar P slides outward at a constant relative speed u along rod AB, which rotates counterclockwise with a constant angular velocity of 20 rpm. Knowing that r = 250 mm when θ = 0 and that the collar reaches B when θ = 90°, determine the magnitude of the acceleration of the collar P just as it reaches B.

SOLUTION

ω = 20 rpm =

(20)(2π ) 2π = rad/s 60 3

α =0 θ = 90° =

radians

2

θ = θ0 + ω t

Uniform rotational motion.

t=

Uniform motion along rod.

π

θ − θ0 = ω

π

2 2π 3

= 0.75 s

r = r0 + ut r − r0 0.5 − 0.25 1 = = m/s, t 0.75 3 1 v P/ AB = m/s 3

u=

Acceleration of coinciding Point P′ on the rod.

(r = 0.5 m)

a P ′ = rω 2 2

 2π  = (0.5)    3  2π 2 = m/s 2 9 = 2.1932 m/s 2

Acceleration of collar P relative to the rod. Coriolis acceleration. Acceleration of collar P.

a P/ AB = 0

 2π  1  2 2ω × v P/ AB = 2ωu = (2)    = 1.3963 m/s  3  3  a P = a P′ + a P/ AB + 2ω × v P/ AB a P = [2.1932 m/s 2 ] + [1.3963 m/s 2 a P = 2.60 m/s

2

57.5°

] aP = 2.60 m/s 2 

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PROBLEM 15.173 Pin P slides in a circular slot cut in the plate shown at a constant relative speed u = 90 mm/s. Knowing that at the instant shown the plate rotates clockwise about A at the constant rate ω = 3 rad/s, determine the acceleration of the pin if it is located at (a) Point A, (b) Point B, (c) Point C.

SOLUTION

ω = 3 rad/s , α = 0, u = 90 mm/s = 0.09 m/s, u = 0 ρ = 100 mm u2

ρ

=

(90) 2 = 81 mm/s 2 = 0.081 m/s 2 100

ω 2 = 36 rad 2 /s 2 2ω u = (2)(3)(90) = 540 mm/s 2 = 0.54 m/s 2

(a)

Point A.

Coriolis acceleration.

rA = 0,

v A/F = 0.09 m/s ←

a A′ = 0,

a A/F =

u2

ρ

↑ = 0.081 m/s 2 ↑

2ω u ↑ = 0.54 m/s 2 ↑

a A = a A′ + a A/F + [2ω u ↑] = 0.621 m/s 2 ↑

(b)

Point B.

rB = 0.1 2 m

45°,

a B′ = −ω 2 rB = −(9)(0.1 2) a B/F =

a A = 0.621 m/s 2 ↑ 

v B/F = 0.09 m/s ↑

45° = 0.9 2 m/s 2

45°

u2

ρ

= 0.081 m/s 2 →

Coriolis acceleration.

2ω u = 0.54 m/s 2 →

a B = a B′ + a B/F + [2ω u →] = [1.521 m/s 2 →] + [0.9 m/s 2 ↓]

a B = 1.767 m/s 2

30.6° 

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PROBLEM 15.173 (Continued)

(c)

Point C.

rC = 0.2 m ↑ vC/F = 0.09 m/s → aC ′ = −ω 2 rC = −(9)(0.2 ↑) = 1.8 m/s 2 ↓ a C/ F =

Coriolis acceleration.

u2

ρ

= 0.081 m/s 2 ↓

2ω u = 0.54 m/s 2 ↓ aC = aC ′ + aC/F + [2ω u ↓] = 2.421 m/s 2 ↓

aC = 2.42 m/s 2 ↓ 

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PROBLEM 15.174 Pin P slides in a circular slot cut in the plate shown at a constant relative speed u = 90 mm/s. Knowing that at the instant shown the angular velocity ω of the plate is 3 rad/s clockwise and is decreasing at the rate of 5 rad/s 2 , determine the acceleration of the pin if it is located at (a) Point A, (b) Point B, (c) Point C.

SOLUTION

ω = 3 rad/s , α = 5 rad/s , u = 90 mm/s = 0.09 m/s, u = 0 ρ = 100 mm u2

ρ

=

(90) 2 = 81 mm/s2 = 0.081 m/s 2 100

ω 2 = 36 rad 2 /s 2 2ω u = (2)(3)(90) = 540 mm/s 2 = 0.54 m/s 2

(a)

rA = 0

Point A.

v A/F = 0.09 m/s ← a A′ = 0 a A /F =

Coriolis acceleration.

u2

ρ

↑ = 0.081 m/s 2 ↑

2ω u ↑ = 0.54 m/s 2 ↑ a A = a A′ + a A/F + [2ω u ↑]

a A = 0.621 m/s 2 ↑ 

= 0.621 m/s 2 ↑

(b)

Point B.

rB = 0.1 2 m

45°,

v B/F = 0.09 m/s ↑

a B′ = α k × rB − ω 2 rB = [(0.1 2)(5) = [0.5 2 m/s 2

a B/F =

Coriolis acceleration.

u2

ρ

45°] − [(9)(0.1 2)

45°] + [0.9 2 m/s 2

45°]

45°]

= 0.081 m/s 2 →

2ω u = 0.54 m/s 2 →

a B = a B′ + a B/F + [2ω u →] = [1.021 m/s2 →] + [1.4 m/s2 ↓]

a B = 1.733 m/s 2

53.9° 

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PROBLEM 15.174 (Continued)

(c)

Point C.

rC = 0.2 m ↑ vC/F = 0.09 m/s → aC ′ = α k × rC − ω 2rC = [(0.2)(5) ←] − [(9)(0.2 ↑)] = [1 m/s 2 ←] + [1.8 m/s 2 ↓] aC/ F =

u2

ρ

= 0.081 m/s 2 ↓

Coriolis acceleration.

2ω u = 0.54 m/s 2 ↓ aC = aC ′ + aC/F + 2ω u ↓ = [1 m/s 2 ←] + [2.421 m/s 2 ↓]

aC = 2.62 m/s2

67.6° 

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PROBLEM 15.175 Pin P is attached to the wheel shown and slides in a slot cut in bar BD. The wheel rolls to the right without slipping with a constant angular velocity of 20 rad/s. Knowing that x = 480 mm when θ = 0, determine (a) the angular acceleration of the bar and (b) the relative acceleration of pin P with respect to the bar when θ = 0.

SOLUTION Coordinates. x A = ( x A )0 + rθ ,

yA = r

xB = 0,

yB = r

xC = x A ,

yC = 0

xP = x A + e sin θ ,

yP = r + e cos θ

( xA )0 = 480 mm = 0.48 m

Data:

r = 200 mm = 0.20 m e = 140 mm = 0.14 m θ =0 xP = 480 mm = 0.48 m

Velocity analysis. ω AC = 20 rad/s ,

ω BD = ωBD

v P = (r + e)ω AC = (0.20 + 0.14)(20)

= 6.8 m/s v P′ = [ xPω BD ] + [eωBD v P/F = [u cos β tan β =

]

] + [u sin β ]

e 0.14 = xP 0.48

β = 16.260° Use v P = v P′ + v P/F and resolve into components. : 6.8 = 0.14ω BD + u cos β

(1)

0 = 0.48ω BD − u sin β

(2)

: Solving (1) and (2),

ωBD = 3.8080 rad/s,

u = 6.528 m/s

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PROBLEM 15.175 (Continued)

Acceleration analysis.

α AC = 0,

aA = 0

α BD = α BD 2 a P/ A = rω AB = (0.14)(20) 2 = 56 m/s 2

a P = a A + a P/ A = 56 m/s 2 a P′ = [ xPα BD ] + [eα B

2 ] + [ xPωBD

= [0.48α BD ] + [0.14α BD

2 ] + [eωBD ]

] + [(0.48)(3.8080)2

]

+ [(0.14)(3.8080) 2 ] = [0.48α BD ] + [0.14α BD

] + [6.9604 m/s 2

]

+ [2.0301 m/s 2 ]

a P/F = [u cos β

] + [u sin β ]

Coriolis acceleration. 2ωBD u = (2)(3.8080)(6.528) = [49.717 m/s 2

Use a P = a P′ + a P/F + [2ω BD u

β]

β ] and resolve into components. : 0 = 0.14α BD − 6.9604 + u cos β + 49.717 sin β

or

0.14α BD + u cos β = −6.9602

(3)

: 56 = 0.48α BD + 2.0301 + u sin β + 49.717 cos β

or

0.48α BD − u sin β = 6.2415

Solving (3) and (4),

α BD = 8.09 rad/s,

(4) u = −8.43 m/s

2

α BD = 8.09 rad/s 2

(a)

a P/F = 8.43 m/s 2

(b)



16.26° 

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PROBLEM 15.176 Knowing that at the instant shown the rod attached at A has an angular velocity of 5 rad/s counterclockwise and an angular acceleration of 2 rad/s2 clockwise, determine the angular velocity and the angular acceleration of the rod attached at B.

SOLUTION Geometry: Apply the law of sines to the triangle ABP to determine the lengths AP and BP. Angle PBA = 180° − 70° = 110°

Angle APB = 180° − 25° − 110° = 45° 0.2 AP BP = = sin 45° sin110° sin 25° BP = 0.119534

AB = 200 mm = 0.2 m AP = 0.265785 m

Unit vectors:

i =1

j =1 ,

,

k =1

Relative position vectors: rP /A = 0.265785(sin 25°i + cos 25° j) = 0.11233i − 0.24088 j rP /B = 0.119534(sin 70°i + cos 70° j) = 0.11233i + 0.04088 j

Angular velocities:

ω AP = 5 rad/s

= (5 rad/s)k

ω BP = ωBP k

Angular accelerations:

α AP = 2 rad/s 2

= − (2 rad/s 2 )k

α BP = α BP k

Velocity of P:

v P = ω AP × rP /A = 5k × (0.11233i + 0.24088 j) = −(1.2044 m/s)i + (0.56165 m/s) j

Acceleration of P:

2 a P = α AP × rP /A − ω AP rP /A

= (−2k ) × (0.11233i + 0.24088 j) − (5)2 (0.11233i + 0.24088 j) = −(2.3265 m/s 2 )i − (6.2467 m/s 2 ) j

Consider the slider P as a particle sliding along the rotating rod BP with a relative velocity v rel = u

20° = u (cos 20°i + sin 20° j)

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PROBLEM 15.176 (Continued)

and a relative acceleration a rel = u

20° = u (cos 20°i + sin 20° j)

Consider the rod BP as a rotating frame of reference. Motion of Point P′ on the rod currently at P. v P′ = ω BP × rP /B = ωBP k × (0.11233i + 0.04088 j) = −0.04088ωBP i + 0.11233ωBP j 2 rP /B a P′ = α BP × rP /B − ωBP 2 = α BP k × (0.11233i + 0.04088 j) − ωBP (0.11233i + 0.04088 j) 2 2 i − 0.04088ωBP j = −0.04088α BP i + 0.11233α BP j − 0.11233ωBP

v P = v P′ + v rel

Velocity of P:

Resolve into components.

i:

−1.2044 = −0.04088ωBP + u cos 20°

j:

0.56165 = 0.11233ωBP + u sin 20°

Solving the simultaneous equations for ωBP and u,

ωBP = 7.8612 rad/s

u = −0.93971 m/s

ω BP = 7.86 rad/s

Angular velocity of BP:



v rel = −0.93971(cos 20°i + sin 20° j)

Relative velocity: Coriolis acceleration:

2ω BP × v rel = (2)(7.8612k ) × (−0.93971cos 20°i − 0.93971sin 20° j) = (5.0532 m/s 2 )i − (13.8835 m/s 2 ) j

Acceleration of P:

a P′ = a P′ + arel + 2ω BP × v rel

Resolve into components.

i:

2 −2.3265 = −0.04088α BP − 0.11233ωBP + u cos 20° + 5.0532

−0.04088α BP + u cos 20° = −0.43788

j:

(1)

2 −6.2467 = 0.11233α BP − 0.04088ωBP + u sin 20° − 13.8835

0.11233α BP + u sin 20° = 10.1631

(2)

Solving Eqs. (1) and (2) simultaneously,

α BP = 81.146 rad/s2 u = 3.0641 m/s2 α BP = 81.1 rad/s 2

Angular acceleration of BP:



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PROBLEM 15.177 The Geneva mechanism shown is used to provide an intermittent rotary motion of disk S. Disk D rotates with a constant counterclockwise angular velocity ωD of 8 rad/s. A pin P is attached to disk D and can slide in one of the six equally spaced slots cut in disk S. It is desirable that the angular velocity of disk S be zero as the pin enters and leaves each of the six slots; this will occur if the distance between the centers of the disks and the radii of the disks are related as shown. Determine the angular velocity and angular acceleration of disk S at the instant when φ = 150°.

SOLUTION Geometry: r 2 = 1.252 + 2.502 − (2)(1.25)(2.50) cos30° r = 1.54914 in.

Law of cosines.

Law of sines.

sin β sin 30° = 1.25 r β = 23.794°

Let disk S be a rotating frame of reference. Ω = ωS

 =α Ω S

,

Motion of coinciding Point P′ on the disk. vP′ = rωS = 1.54914ωS

β

aP′ = −α S k × rP/O − ωS2 rP/O = [1.54914α S

β ] + [1.54914ωS2

β]

Motion relative to the frame. vP/S = u

β a P/S = u 2ωS u

Coriolis acceleration.

β

β

vP = vP′ + vP/S = [1.54914ωS

β ] + [u

β]

a P = a P + a P/S + 2ωS u = [1.54914α S

β ] + [1.54914ωS2

β ] + [u

β ] + [2ωS u

β]

Motion of disk D. (Rotation about B)



vP = ( BP)ωD = (1.25)(8) = 10 in./s

30°

aP = [( BP)α D

30°] = 0 + [(1.25)(8) 2

= 80 in./s 2

60°] + [( BP)ωS2

30°]

30°

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PROBLEM 15.177 (Continued)

Equate the two expressions for vP and resolve into components.

β : 1.54914ωS = 10cos(30° + β ) 10 cos 53.794° 1.54914 = 3.8130 rad/s

ωS =

ω S = 3.81 rad/s



β : u = 10sin(30° + β ) = 10sin 53.794° = 8.0690 in./s Equate the two expressions for a P and resolve into components.

β : 1.54914α S − 2ωS u = 80sin (30° + β ) 80sin 53.794° + (2)(3.8130)(8.0690) 1.54914 = 81.4 rad/s 2 

αS =

α S = 81.4 rad/s 2



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PROBLEM 15.178 In Problem 15.177, determine the angular velocity and angular acceleration of disk S at the instant when φ = 135°.

PROBLEM 15.177 The Geneva mechanism shown is used to provide an intermittent rotary motion of disk S. Disk D rotates with a constant counterclockwise angular velocity ωD of 8 rad/s. A pin P is attached to disk D and can slide in one of the six equally spaced slots cut in disk S. It is desirable that the angular velocity of disk S be zero as the pin enters and leaves each of the six slots; this will occur if the distance between the centers of the disks and the radii of the disks are related as shown. Determine the angular velocity and angular acceleration of disk S at the instant when φ = 150°.

SOLUTION Geometry: Law of cosines.

Law of sines.

r 2 = 1.252 + 2.502 − (2)(1.25)(2.50)cos 45° r = 1.84203 in. sin β sin 45° = 1.25 r β = 28.675°

Let disk S be a rotating frame of reference. Ω = ωS

 =α Ω S

,

Motion of coinciding Point P′ on the disk. vP′ = rωs = 1.84203ωs

β

aP′ = −α S k × rP/O − ωS2 rP/O = [1.84203α S

β ] + [1.84203ωS2

β]

Motion relative to the frame. vP/S = u

Coriolis acceleration.

β 2ωS u

aP/S = u

β

β

vP = vP′ + vP/S = [1.84203ωS

β ] + [u

β]

a P = a P + a P/S + 2ωS u = [1.84203α S

β ] + [1.84203ωS2

β ] + [u

β ] + [2ωS u

β]

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PROBLEM 15.178 (Continued)

Motion of disk D. (Rotation about B) vP = ( BP)ωD = (1.25)(8) = 10 in./s

30°

aP = [( BP)α D

45°] = 0 + [(1.25)(8)2

= 80 in./s 2

45°] + [( BP)ωS2

45°]

45°

Equate the two expressions for vP and resolve into components.

β : 1.84203ωS = 10cos(45° + β ) 10 cos 73.675° 1.84203 = 1.52595 rad/s

ωS =

ω S = 1.526 rad/s



α S = 57.6 rad/s 2



β : u = 10sin(45° + β ) = 10sin 73.675° = 9.5968 in./s Equate the two expressions for aP and resolve into components.

β : 1.84203α S − 2ωS u = 80sin (45° + β ) 80sin 73.675° + (2)(1.52595)(9.5968) 1.84203 2 = 57.6 rad/s

αS =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1253

PROBLEM 15.179 At the instant shown, bar BC has an angular velocity of 3 rad/s and an angular acceleration of 2 rad/s 2 , both counterclockwise. Determine the angular acceleration of the plate.

SOLUTION rB/D = (4 in.)i + (3 in.)j

Relative position vectors.

rB/C = −(6 in.)i + (3 in.)j

Velocity analysis.

ω BC = 3 rad/s

Bar BC (Rotation about C):

ω BC = (3 rad/s)k

vB = ω BC × rB/C = 3k × (−6i + 3j) = −(9 in./s)i − (18 in./s)j

ωP = ωP k

Plate (Rotation about D):

Let Point B ′ be the point in the plate coinciding with B. vB′ = ω P × rB/D = ωP k × (4i + 3 j) = −3ωP i + 4ωP j vB/ F = vrel j

Let plate be a rotating frame.

vB = vB′ + vB/F = −3ω P i + (4ω P + vrel ) j

Equate like components of vB .

i:

−9 = −3ωP

ω P = (3 rad/s)k

j : −18 = (4)(3) + vrel

Acceleration analysis.

α BC = 2 rad/s 2

Bar BC:

α BC = (2 rad/s 2 )k

v rel = −(30 in./s)j

2 a B = α BC × rB/C − ω BC rB/C

= 2k × (−6i + 3j) − (3)2 (−6i + 3j) = −12 j − 6i + 54i − 27 j = (48 in./s 2 )i − (39 in./s 2 ) j

Plate:

α P = α Pk a B′ = α P × rB/D − ωP2 rB/D = α P k × (4i + 3j) − (3)2 (4i + 3j) = −3α P i + 4α P j − 36i − 27j

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PROBLEM 15.179 (Continued)

Relative to the frame (plate), the acceleration of pin B is a B/F = (arel )t j −

2 vrel

ρ

i = (arel )t j −

30 2 i 4

= −(225 in./s 2 )i + (arel )t j

Coriolis acceleration.

2ωP × v P/F ac = 2(3k ) × (−30 j) = (180 in./s 2 )i

Then

a B = a B′ + a B/F + ac a B = −(3α P + 36)i + (4α P − 27) j − 225i + ( arel )t j + 180i a B = −(3α P + 81)i + [4α P + (arel )t − 27]j

Equate like components of a B . i: 48 = −(3α P + 81)

α P = −43 rad/s 2 

α P = 43.0 rad/s 2



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PROBLEM 15.180 At the instant shown bar BC has an angular velocity of 3 rad/s and an angular acceleration of 2 rad/s 2 , both clockwise. Determine the angular acceleration of the plate.

SOLUTION rB/D = (4 in.)i + (3 in.)j

Relative position vectors.

rB/C = −(6 in.)i + (3 in.)j

Velocity analysis.

ω BC = 3 rad/s

Bar BC (Rotation about C):

ω BC = −(3 rad/s)k

vB = ω BC × rB/C = (−3k) × (−6i + 3j) = (9 in./s)i + (18 in./s)j

ωP = ωP k

Plate (Rotation about D):

Let Point B ′ be the point in the plate coinciding with B: vB′ = ω P × rB/D = ωP k × (4i + 3j) = −3ω P i + 4ω P j vB/F = vrel j

Let the plate be a rotating frame.

vB = vB′ + vB/F = −3ω P i + (4ω P + vrel ) j

Equate like components of vB.

i:

9 = −3ωP

ω P = −(3 rad/s)k

j: 18 = (4)(3) + vrel

Acceleration analysis.

α BC = 2 rad/s 2

Bar BC:

α BC = −(2 rad/s 2 )k

vrel = (30 in./s)j

2 a B = α BC × rB/C − ωBC rB/C

= (−2k) × (−6i + 3j) − (3)2 (−6i + 3j) = 12 j + 6i + 54i − 27 j = (60 in./s 2 )i − (15 in./s 2 ) j

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PROBLEM 15.180 (Continued)

Plate:

α P = α Pk a B′ = α P × rB/D − ωP2 rB/D = α P k × (4i + 3j) − (3)2 (4i + 3j) = −3α P i + 4α P j − 36i − 27j

Relative to the frame (plate), the acceleration of pin B is a B/F = (arel )t j −

2 vrel

ρ

i

30 2 i 4 = −(225 in./s 2 )i + (arel )t j = (arel )t j −

Coriolis acceleration.

2ωP × v P/F ac = 2(−3k ) × (30 j) = (180 in./s 2 )i

Then

a B = a B′ + a B/F + ac a B = −(3α P + 36)i + (4α P − 27) j − 225i + ( arel )t j + 180i a B = −(3α P + 81)i + [4α P + (arel )t − 27]j

Equate like components of a B . i: 60 = −(3α P + 81)

α P = −47 rad/s 2 

α P = 47.0 rad/s 2



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PROBLEM 15.181* Rod AB passes through a collar which is welded to link DE. Knowing that at the instant shown block A moves to the right at a constant speed of 75 in./s, determine (a) the angular velocity of rod AB, (b) the velocity relative to the collar of the point of the rod in contact with the collar, (c) the acceleration of the point of the rod in contact with the collar. (Hint: Rod AB and link DE have the same ω and the same α.)

SOLUTION Let ω = ω and α = α be the angular velocity and angular acceleration of the link DE and collar rigid body. Let F be a frame of reference moving with this body. The rod AB slides in the collar relative to the 30° and relative acceleration u = u 30°. Note that this frame of reference with relative velocity u = u relative motion is a translation that applies to all points along the rod. Let Point A be moving with the end of the rod and A′ be moving with the frame. Point E is a fixed point. rA′/E =

Geometry.

6 in. = 12 in. sin 30°

vA = 75 in./s

Velocity analysis.

vA′ = 12ω vA = vA′ + u

Resolve into components.

: 75 = 0 + u cos30° u =

:

75 = 86.603 in./s cos 30°

0 = −12ω + u sin 30° ω =

u sin 30° = 3.6085 rad/s 12

ω = 3.61 rad/s

(a)

Angular velocity.

(b)

Velocity of rod AB relative to the collar.

(c)

Acceleration analysis.

u = 86.6 in./s



30° 

aA = 0 a A′ = [12α ] + [12ω 2

] = [12α ] + [156.25

]

Coriolis acceleration. ac = 2ω u

60° = 625.01 in./s 2

a A = a A′ + u + ac

:

60°

Resolve into components.

0 = 156.25 + u cos 30° − 625.01cos 60° u = 180.43 in./s 2

:

0 = −12α + u sin 30° + 625.01sin 60°

α = 52.624 rad/s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1258

PROBLEM 15.181* (Continued)

For rod AB,

ω AB = 3.6085 rad/s α AB = 52.624 rad/s 2

Let P be the point on AB coinciding with collar D. 30° = 10.392 in.

rP/A = 12 cos 30°

30°.

a P = a A + (a P/A )t + (a P/A )n

= 0 + [(10.392)(52.624) = [546.87

60°] + [(10.392)(3.6085)2

60°] + [135.32

30°] = [390.63

30°]

] + [405.94 ]

a P = 563 in./s 2



46.1° 

a P may also be determined from a P = a P′ + u + ac using the rotating frame. The already calculated vectors u and ac also apply at Points P ′ and P. a P′ = a D = 6α

30° + 6ω

= [315.74 in./s 2

Then

a P = [315.74

= (135.31

60°

30°] + [78.13 in./s 2

30°] + [78.13 30°] + [546.88

60°]

60°] + (180.43

30°] + [625.01

60°]

60°]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1259

PROBLEM 15.182* Solve Problem 15.181, assuming that block A moves to the left at a constant speed of 75 in./s. PROBLEM 15.181 Rod AB passes through a collar which is welded to link DE. Knowing that at the instant shown block A moves to the right at a constant speed of 75 in./s, determine (a) the angular velocity of rod AB, (b) the velocity relative to the collar of the point of the rod in contact with the collar, (c) the acceleration of the point of the rod in contact with the collar. (Hint: Rod AB and link DE have the same ω and the same α.)

SOLUTION Let ω = ω and α = α be the angular velocity and angular acceleration of the link DE and collar rigid body. Let F be a frame of reference moving with this body. The rod AB slides in the collar relative to the 30° and relative acceleration u = u 30°. Note that this frame of reference with relative velocity u = u relative motion is a translation that applies to all points along the rod. Let Point A be moving with the end of the rod and A′ be moving with the frame. Point E is a fixed point. rA′/E =

Geometry.

6 in. = 12 in. sin 30°

vA = 75 in./s

Velocity analysis.

vA′ = 12ω vA = vA′ + u

:

Resolve into components.

75 = −86.603 in./s cos30° u sin 30° 0 = −12ω + u sin 30° ω = = −3.6085 rad/s 12

− 75 = 0 + u cos 30° u =

:

ω = 3.61 rad/s

(a)

Angular velocity.

(b)

Velocity of rod AB relative to the collar.

u = 86.6 in./s



30° 

aA = 0

Acceleration analysis.

a A′ = [12α ] + [12ω 2 ac = 2ω u

Coriolis acceleration.

60° = 625.01 in./s 2

a A = a A′ + u + ac

:

] = [12α ] + [156.25

]

60°

Resolve into components.

0 = 156.25 + u cos 30° − 625.01cos 60° u = 180.43 in./s 2

:

0 = −12α + u sin 30° + 625.01sin 60°

α = 52.624 rad/s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1260

PROBLEM 15.182* (Continued)

For rod AB,

ω AB = 3.6085 rad/s α AB = 52.624 rad/s 2

Let P be the point on AB coinciding with collar D. rP/A = 12 cos 30°

30° = 10.392 in.

30°.

a P = a A + (a P/A )t + (a P/A )n

= 0 + [(10.392)(52.624) = [546.87

60°] + [(10.392)(3.6085)2

60°] + [135.32

30°] = [390.63

30°]

] + [405.94 ]

a P = 563 in./s 2



46.1° 

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PROBLEM 15.183* In Problem 15.157, determine the acceleration of pin P. PROBLEM 15.157 The motion of pin P is guided by slots cut in rods AD and BE. Knowing that bar AD has a constant angular velocity of 4 rad/s clockwise and bar BE has an angular velocity of 5 rad/s counterclockwise and is slowing down at a rate of 2 rad/s2, determine the velocity of P for the position shown.

SOLUTION Units: meters, m/s, m/s2 , j =1 , k =1

i =1

Unit vectors: From the solution of Problem 15.157,

θ = 26.565°

R = 0.100 m 

rP /A = 0.1i + 0.15 j

rP /B = −0.3i + 0.15 j

ω AD = −(4 rad/s)k

ω BE = (5 rad/s2 )k

α AD = 0

α BE = −(2 rad/s 2 )k

v P /AD = u1 j = −(1.775 m/s) j v P /BE = −u2 cos θ i + u2 sin θ j

= ( −1.50935)(− cos 26.565°i + sin 26.565°) j = (1.35 m/s)i + (0.675 m/s) j

Acceleration of Point P′ on rod AD coinciding with the pin: 2 a P′ = α AD × rP /A − ω AD rP /A

= 0 − (4)2 (0.1i + 0.15 j) = −1.6i − 2.4 j Acceleration of the pin relative to rod AD: u12 (1.775)2 i = u1 j − i R 0.15 = u1 j − 21.004i

a P /AD = u1 j −

Coriolis acceleration:

a1 = 2ω AD × v P /AD a1 = 2( −4k ) × ( −1.775 j) = −14.2i

Acceleration of P:

a P = a P′ + a P /AD + a1 a P = −36.804i − 2.4 j + u1 j

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PROBLEM 15.183* (Continued)

Acceleration of Point P″ on rod BE coinciding with the pin 2 a P′′ = α BE × rP /B − ω BE rP /E

= (−2k ) × (−0.3i + 0.15 j) − (5) 2 (−0.3i + 0.15 j) = 0.6 j + 0.3i + 7.5i − 3.75 j = 7.8i − 3.15 j

Acceleration of the pin relative to the rod BE: a P /BE = u2 (− cos θ i + sin θ j) a 2 = 2ω BE × v P /BE

Coriolis acceleration:

= (2)(5k ) × (1.35i − 0.675 j) = −13.5 j + 6.75i a P = a P′′ + a P /BE + a 2

Acceleration of P:

a P = 7.8i − 3.15 j + u2 (− cos θ i + sin θ j) + 13.5 j + 6.75i = 14.55i + 10.35 j − u2 cos θ i + u2 sin θ j Equating the two expressions for aP and resolving into components, i:

−36.804 = 14.55 − u2 cos θ u2 =

j:

36.804 + 14.55 = 57.415 m/s 2 cos 26.565°

−2.4 + u1 = 10.35 + u2 sin θ u1 = 12.75 + 57.415sin 26.565° = 38.426 m/s 2

Acceleration of the pin. a P = −36.804i − 2.4 j + 38.427 j

= −36.804i + 36.027 j a P − (36.8 m/s 2 )i + (36.0 m/s 2 ) j = 51.5 m/s 2

44.4° 

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PROBLEM 15.184 At the instant considered, the radar antenna shown rotates about the origin of coordinates with an angular velocity ω = ωx i + ω y j + ω z k. Knowing that (vA ) y = 300 mm/s, (vB ) y = 180 mm/s, and (vB )z = 360 mm/s, determine (a) the angular velocity of the antenna, (b) the velocity of Point A.

SOLUTION rA = (0.3 m)i − (0.25 m)k v A = (v A ) x i + (0.3 m/s) j + (v A ) z k i

j

k

v A = ω × rA : (v A ) x i + 0.3j + (v A ) z k = ω x

ωy

ωz

0.3

0

−0.25

(v A ) x i + 0.3j + (v A ) z k = −0.25ω y i + (0.3ωz + 0.25ω x ) j − 0.3ω y k i: (v A ) x = −0.25ω y

j:

(1)

0.3 = 0.3ωz + 0.25ωx

(2)

k: (v A ) z = −0.3ω y

(3)

rB = (0.3 m)i − (0.25 m)j v B = (vB ) x i + (0.18 m/s)j + (0.36 m/s)k i v B = ω × rB : (vB ) x i + 0.18 j + 0.36k = ω x

j

k

ωy

ωz

0.3 −0.25

0

(vB ) x i + 0.18 j + 0.36k = 0.25ωz i + 0.3ωz j − (0.25ω x + 0.3ω y )k

i: (vB )x = 0.3ωz j: k: From Eq. (5),

(4)

0.18 = 0.3ωz

(5)

0.36 = −0.25ωx − 0.3ω y

(6)

ωz = 0.6 rad/s

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PROBLEM 15.184 (Continued)

1 (0.3 − 0.3ω z ) 0.25 = 0.48 rad/s

From Eq. (2),

ωx =

From Eq. (6),

ωy = −

(a)

1 (0.36 + 0.25ω x ) 0.3 = −1.6 rad/s

ω = (0.480 rad/s)i − (1.600 rad/s)j + (0.600 rad/s)k 

Angular velocity.

From Eq. (1),

(vA ) x = −0.25ω y = 0.400 m/s

From Eq. (3),

(v A ) z = −0.3ω y = 0.480 m/s

(b)

Velocity of Point A.

v A = (0.400 m/s)i + (0.300 m/s)j + (0.480 m/s)k vA = (400 mm/s)i + (300 mm/s)j + (480 mm/s)k 

or

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PROBLEM 15.185 At the instant considered the radar antenna shown rotates about the origin of coordinates with an angular velocity ω = ωx i + ω y j + ω z k. Knowing that (vA )x = 100 mm/s, (vA ) y = −90 mm/s, and (vB ) z = 120 mm/s, determine (a) the angular velocity of the antenna, (b) the velocity of Point A.

SOLUTION rA = (0.3 m)i − 0.25 m)k v A = (0.1 m/s)i − (0.09 m/s)j + (v A ) z k i

j

k

v A = ω × rA : 0.1i − 0.09 j + (v A ) z k = ω x

ωy

ωz

0.3

0

−0.25

0.1i − 0.09 j + (v A ) z k = −0.25ω y i + (0.3ωz + 0.25ω x ) j − 0.3ω y k 0.1 = −0.25ω y

(1)

j: −0.09 = 0.3ωz + 0.25ωx

(2)

i:

k:

(v A ) z = −0.3ω y

(3)

rB = (0.3 m)i − (0.25 m)j v B = (vB ) x i + (vB ) y j + (0.12 m/s)k i v B = ω × rB : (vB ) x i + (vB ) y j + 0.12k = ω x

j

k

ωy

ωz

0.3 −0.25

0

(vB ) x i + (vB ) y j + 0.12k = 0.25ωz i + 0.3ω z j − (0.25ωx + 0.3)ω y k i:

(vB )x = 0.25ωz

(4)

j:

(vB ) y = 0.3ωz

(5)

k:

From Eq. (1),

0.12 = −0.25ωx − 0.3ω y

ωy = −

(6)

0.1 = −0.4 rad/s 0.25

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PROBLEM 15.185 (Continued)

From Eq. (6),

ωx = −

1 (0.12 + 0.3ω y ) 0.25

=0

From Eq. (2),

From Eq. (3),

1 (0.09 + 0.25ω x ) 0.25 = −0.36 rad/s

ωz = −

(v A ) z = −(0.3)(−0.4) = 0.12 m/s

(a)

Angular velocity.

(b)

Velocity of Point A.

ω = −(0.400 rad/s)j − (0.360 rad/s)k 

vA = (0.1 m/s)i − (0.09 m/s)j + (0.12 m/s)k vA = (100 mm/s)i − (90 mm/s)j + (120 mm/s)k 

or

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PROBLEM 15.186 Plate ABD and rod OB are rigidly connected and rotate about the ball-and-socket joint O with an angular velocity ω = ωx i + ωx j + ωz k. Knowing that vA = (80 mm/s)i + (360 mm/s)j + (vA)z k and ωx = 1.5 rad/s, determine (a) the angular velocity of the assembly, (b) the velocity of Point D.

SOLUTION ω x = 1.5 rad/s

ω = (1.5 rad/s)i + ω y j + ω z k

rA = −(160 mm)i + (120 mm)j + (80 mm)k rD = +(160 mm)i + (120 mm)j − (80 mm)k vA = ω × rA

(a)

i = 1.5

j

k

ωy

ωz

−160 +120 +80 vA = (80ω y − 120ω z )i + (−160ω z − 120) j + (180 + 160ω y )k

But we are given:

vA = (80 mm/s)i + (360 mm/s)j + (vA )z k (v A ) x : 80ω y − 120ωz = 80 (vA ) y : − 160ω x − 120 = 360

(1)

ωz = −3 rad/s

(v A ) z : 180 + 160ω y = (v A ) z

(2) (3)

Substitute ωz = −3.0 rad/s into Eq. (1): 80ω y − 120(−3) = 80

ω y = −3.5 rad/s Substitute ω y = −3.5 rad/s into Eq. (3): 180 + 160(−3.5) = (v A ) z (v A ) z = −380 in./s ω = (1.5 rad/s)i − (3.5 rad/s)j − (3.0 rad/s)k 

We have:  

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PROBLEM 15.186 (Continued)

(b)

Velocity of D. vD = ω × rD i

j

k

= 1.5 −3.5 −3.0 +160 +120 −80 = (360 + 280)i + ( −480 + 120) j + (180 + 560)k

vD = (640 mm/s)i − (360 mm/s)j + (740 mm/s)k 



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PROBLEM 15.187 The bowling ball shown rolls without slipping on the horizontal xz plane with an angular velocity ω = ωx i + ω y j + ωz k. Knowing that v A = (14.4 ft/s)i − (14.4 ft/s) j + (10.8 ft/s)k and v D = (28.8 ft/s)i + (21.6 ft/s)k, determine (a) the angular velocity of the bowling ball, (b) the velocity of its center C.

SOLUTION Radius of ball: 4.3 in. = 0.35833 ft At the given instant, the origin is not moving. v A = ω × rA : 14.4i − 14.4 j + 10.8k =

i

j

k

ωx

ωy

ωz

0.35833 0.35833 0

14.4i − 14.4 j + 10.8k = −0.35833ω z i + 0.35833ωz j + 0.35833(ω x − ω y )k − 0.35833ω z = 14.4

i:

ω z = − 40.186 rad/s

0.35833ωz = −14.4

j:

k: 0.35833(ωx − ω y ) = 10.8

ωz = − 40.186 rad/s ωx − ω y = 30.140 rad/s i

v D = ω × rD : 28.8i + 21.6k = ω x 0

j

k

ωy

ωz

0.71667

0

28.8i + 21.6k = − 0.71667ωzi + 0.71667ωxk i : − 0.71667ω z = 28.8 k:

0.71667ω x = 21.6

ω z = − 40.186 rad/s ω x = 30.140 rad/s

ω y = ω x − 30.140 = 0

ω = (30.1 rad/s)i − (40.2 rad/s)k 

(a) Angular velocity. (b) Velocity of Point C.

vC = ω × rC = (30.140i − 40.186k ) × 0.35833j = 14.4i + 10.8k

vC = (14.4 ft/s)i + (10.8 ft/s)k 

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PROBLEM 15.188 The rotor of an electric motor rotates at the constant rate ω1 = 1800 rpm. Determine the angular acceleration of the rotor as the motor is rotated about the y axis with a constant angular velocity ω 2 of 6 rpm counterclockwise when viewed from the positive y axis.

SOLUTION ω 1 = 1800 rpm = 60π rad/s

ω 2 = 6 rpm = 0.2π rad/s

Total angular velocity.

ω = ω 2 j + ω1k ω = (0.2π rad/s)j + (60π rad/s)k

Angular acceleration. Frame Oxyz is rotating with angular velocity Ω = ω2 j.  α=ω  Οxyz + Ω × ω =ω = 0 + ω2 j × (ω2 j + ω1k ) = ω 2ω 1i α = (0.2π )(60π )i = (12π 2 rad/s 2 )i

α = (118.4 rad/s2 )i 

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PROBLEM 15.189 The disk of a portable sander rotates at the constant rate ω1 = 4400 rpm as shown. Determine the angular acceleration of the disk as a worker rotates the sander about the z axis with an angular velocity of 0.5 rad/s and an angular acceleration of 2.5 rad/s2, both clockwise when viewed from the positive z axis.

SOLUTION Spin rate:

ω1 = 4400 rpm = 460.77 rad/s

Angular velocity of disk relative to the housing: ω1 = (460.77 rad/s) j

Angular motion of the housing:

ω2 = −(0.5 rad/s)k

 2 = −(2.5 rad/s2 )k ω

Consider a frame of reference rotating with angular velocity Ω = ω 2 k = − (0.5 rad/s)k

Angular velocity of the disk:

ω = ω1 + ω 2 = (460.77 rad/s)j − (0.5 rad/s)k

Angular acceleration of the disk: 1 + ω  2 + Ω × (ω1 + ω2 ) α=ω

= 0 − 2.5k + (−0.5k ) × (460.77 j − 0.5k ) = (230.38 rad/s 2 )i − (2.5 rad/s 2 )k

α = (230 rad/s2 )i − (2.5 rad/s2 )k 

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PROBLEM 15.190 Knowing that the turbine rotor shown rotates at a constant rate ω1 = 9000 rpm, determine the angular acceleration of the rotor if the turbine housing has a constant angular velocity of 2.4 rad/s clockwise as viewed from (a) the positive y axis, (b) the positive z axis.

SOLUTION Spin rate:

ω1 = 9000 rpm = 942.48 rad/s

Angular velocity of the rotor relative to the axle:

ω1 = −(942.48 rad/s) i (a)

Axle rotates with angular velocity ω 2 = −(2.4 rad/s) j Consider a frame of reference rotating with angular velocity Ω = ω2 j

Angular acceleration:  1i + ω  2 j + Ω × (ω1 + ω 2 ) α=ω = 0 + 0 + Ω × ω1 = (−2.4 j) × (−942.48i) α = − (2260 rad/s 2 )k 

(b)

Axle rotates with angular velocity ω 2 = −(2.4 rad/s)k. Ω = −(2.4 rad/s)k α = Ω × ω1 = (−2.4k ) × ( −942.48i)

α = (2260 rad/s2 ) j 

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PROBLEM 15.191 In the system shown, disk A is free to rotate about the horizontal rod OA. Assuming that disk B is stationary (ω 2 = 0), and that shaft OC rotates with a constant angular velocity ω 1 , determine (a) the angular velocity of disk A, (b) the angular acceleration of disk A.

SOLUTION Disk A (In rotation about O):

ω y = ω 1,

Since

ω A = ωx i + ω1 j + ωz k

Point D is point of contact of wheel and disk. rD /O = −rj − Rk v D = ω A × rD /O i

j

k

= ωx ω1

ωz

0

−r − R

v D = (− Rω 1 + rω z )i + Rω 1 j − rωx k

Since ω 2 = 0, vD = 0. Each component of v D is zero. (vD ) z = rωx = 0; ω x = 0 R (vD ) x = − Rω 1 + rω z = 0; ω z =   ω 1 r

R ω A = ω 1 j +   ω 1k  r

(a)

Angular velocity.

(b)

Angular acceleration. Disk A rotates about y axis at rate ω 1 . αA =

dωA R   = ω y × ω A = ω 1 j ×  ω 1 j + ω 1k  dt r  

αA =

R 2 ω 1i  r

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PROBLEM 15.192 In the system shown, disk A is free to rotate about the horizontal rod OA. Assuming that shaft OC and disk B rotate with constant angular velocities ω1 and ω 2 , respectively, both counterclockwise, determine (a) the angular velocity of disk A, (b) the angular acceleration of disk A.

SOLUTION Disk A (in rotation about O):

ω A = ωx i + ω1 j + ωz k

Since ω y = ω1 ,

Point D is point of contact of wheel and disk. rD/O = −rj − Rk i v D = ω A × rD/O = ω x 0

j

k

ω1 ω z −r − R

v D = (−Rω1 + rωz )i + Rωx j − rωxk

(1)

ωB = ω 2 j

Disk B:

v D = ωB × rD/O = ω 2 j × (−rj − Rk ) = −Rω 2i

(2)

v D = v D : (−Rω1 − rωz )i + Rωx j − rωxk = −Rω 2i

From Eqs. 1 and 2:

−rωx = 0;

Coefficients of k:

ωx = 0

(− Rω 1 + rω z ) = − Rω2 ; ωz =

Coefficients of i:

R (ω 1 − ω 2 ) r ω A = ω1 j +

(a)

Angular velocity.

(b)

Angular acceleration. Disk A rotates about y axis at rate ω1. αA =

dωA R   = ω y × ω A = ω1 j × ω1 j + (ω 1 − ω 2 )k  dt r  

αA =

R (ω 1 − ω 2 )k  r

R ω1 (ω 1 − ω 2 )i  r

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PROBLEM 15.193 The L-shaped arm BCD rotates about the z axis with a constant angular velocity ω1 of 5 rad/s. Knowing that the 150-mm-radius disk rotates about BC with a constant angular velocity ω 2 of 4 rad/s, determine (a) the velocity of Point A, (b) the acceleration of Point A.

SOLUTION ω = ω 2 j + ω1k

Total angular velocity.

ω = (4 rad/s) j + (5 rad/s)k Angular acceleration.

Frame Oxyz is rotating with angular velocity Ω = ω1k.

α = ω  Oxyz + Ω × ω =ω = 0 + ω1k × (ω 2 j + ω 1k ) = −ω 1ω 2 i

α = −(5)(4)i = −20i α = −(20.0 rad/s 2 )i

(a)

Velocity of Point A.

rA = (0.15 m)i + (0.12 m) j v A = ω × rA =

i j k 0 4 5 0.15 0.12 0

= −0.6i + 0.75 j − 0.6k

v A = −(0.600 m/s)i + (0.750 m/s) j − (0.600 m/s)k  (b)

Acceleration of Point A.

a A = α × rA + ω × v A i j k i j k = −20 0 0 + 0 4 5 −0.6 0.75 −0.6 0.15 0.6 0 = −2.4k − 6.15i − 3 j + 2.4k

a A = −(6.15 m/s 2 )i − (3.00 m/s2 ) j 

= −6.15i − 3 j

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PROBLEM 15.194 The cab of the backhoe shown rotates with the constant angular velocity ω1 = (0.4 rad/s) j about the y axis. The arm OA is fixed with respect tot he cab, while the arm AB rotates about the horizontal axle A at the constant rate ω 2 = d β /dt = 0.6 rad/s. Knowing that β = 30°, determine (a) the angular velocity and angular acceleration of AB, (b) the velocity and acceleration of Point B.

SOLUTION rA = 20i + 8 j

(ft)

rB /A = 7i − 12.12 j

(ft)

rB = 27i − 4.12 j

(ft)

OXYZ is fixed; Oxyz rotates with Ω = 0.40 j

Angular velocity of AB

ω2 = + (0.60 rad/s)k

With respect to rotating frame:

ω = ω1 + ω2 = (0.40 rad/s) j + (0.60 rad/s)k 

With respect to fixed frame: Angular acceleration of AB

 )Oxyz = (ω  )Oxyz + Ω × ω α = (ω

α = 0 + (0.40 j) × (0.40 j + 0.60k )

α = (0.24 rad/s 2 )i 

Motion of B relative to rotating frame Oxyz. Since A does not move relative to Oxyz, ( v B /F ) = (rB )Oxyz = (rA )Oxyz + (rB /A )Oxyz = 0 + ω′ × rB /A = (0.60k ) × (7i − 12.12 j) v B /F = (7.27 ft/s)i + (4.2 ft/s) j

(1)

(a B /F ) = (rA )Oxyz + (rB /A )Oxyz = 0 + ω′ × (ω′ × rB /A ) = (0.60k ) × (7.27i − 4.2 j) a B /F = (2.52 ft/s 2 )i + (4.36 ft/s 2 ) j

(2)

Motion of B′ of frame Oxyz which coincides with B. v B′ = Ω × rB = (0.40 j) × (27i − 4.12 j) v B′ = −(10.8 ft/s)k

(3)

a B′ = Ω × (Ω × rB ) = Ω × v B′ = (0.4 j) × ( −10.8k ) a B′ = −(4.32 ft/s 2 )i

(4)

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PROBLEM 15.194 (Continued)

Velocity of B using Equations (1) and (3): v B′ = v B′ + v B /F = −10.8k + 7.27i + 4.2 j v B = (7.27 ft/s)i + (4.2 ft/s) j − (10.8 ft/s)k 

Acceleration of B

aB = aB′ + aB /F + aC We first compute the Coriolis acceleration. aC

aC = 2Ω × vB /F = 2(0.40j) × (7.27i + 4.2j) Recalling Equations (2) and (4), we now write

aB = −4.32i − 2.52i + 4.36j − 5.82k a B = −(6.84 ft/s2 )i + (4.36 ft/s2 ) j − (5.82 ft/s 2 )k 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1278

PROBLEM 15.195 A 3-in.-radius disk spins at the constant rate ω 2 = 4 rad/s about an axis held by a housing attached to a horizontal rod that rotates at the constant rate ω 1 = 5 rad/s. For the position shown, determine (a) the angular acceleration of the disk, (b) the acceleration of Point P on the rim of the disk if θ = 0, (c) the acceleration of Point P on the rim of the disk if θ = 90°.

SOLUTION ω = ω1i + ω2k

Angular velocity.

ω = (5 rad/s)i + (4 rad/s)k

(a)

Angular acceleration.

Frame Oxyz is rotating with angular velocity Ω = ω1i.

α = ω  Oxyz + Ω × ω =ω = 0 + ω 1i × (ω1i + ω2 k ) = −ω 1ω2 j = −(4)(5) j = −20 j

(b)

α = −(20.0 rad/s2 ) j 

θ = 0. Acceleration at Point P. rP = (3 in.)i = (0.25 ft)i v P = ω × rP = (5i + 4k ) × 0.25i = (1 ft/s) j a P = α × rP + ω × v P = −20 j × 0.25i + (5i + 4k ) × (1 ft/s) j = 5k + 5k − 4i = −4i + 10k

a P − (4.00 ft/s2 )i + (10.00 ft/s2 )k 

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PROBLEM 15.195 (Continued)

(c)

θ = 90°. Acceleration at Point P. rP = (0.25 ft) j v P = ω × rP = (5i + 4k ) × 0.25 j = −(1.25 ft/s)i + (1 ft/s) j a P = α × rP + ω × v P = −20 j × 0.25 j + (5i + 4k ) × (−1.25i + j) = 0 + 0 − 6.25 j − 4 j + 0 = −10.25 j

a P = −(10.25 ft/s2 ) j 

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PROBLEM 15.196 A 3-in.-radius disk spins at the constant rate ω 2 = 4 rad/s about an axis held by a housing attached to a horizontal rod that rotates at the constant rate ω1 = 5 rad/s. Knowing that θ = 30°, determine the acceleration of Point P on the rim of the disk.

SOLUTION ω = ω1i + ω2k

Angular velocity.

ω = (5 rad/s)i + (4 rad/s)k

Angular acceleration. Frame Oxyz is rotating with angular velocity Ω = ω1i.  =ω  Oxyz + Ω × ω α=ω = 0 + ω1i × (ω1i + ω2k ) = −ω1ω2 j = −(4)(5) j = −20 j α = −(20.0 rad/s 2 ) j

Geometry.

θ = 30°,

rP = (3 in.)(cos30°i + sin 30° j) = (0.25 ft)(cos30°i + sin 30° j)

Velocity of Point P.

v P = ω × rP =

i 5

j 0

k 4

0.25cos 30° 0.25sin 30° 0 = −(0.5 ft/s)i + (0.86603 ft/s) j + (0.625 ft/s)k

Acceleration of Point P.

a P = α × rP + ω × v P i =

j

0 −20 0.25cos30° 0.25sin 30°

k

i

j

k

0 + 5 0 4 0 −0.5 0.86603 0.625

= 4.3301k − 3.4641i − 5.125 j + 4.3301k

a P = −(3.46 ft/s2 )i − (5.13 ft/s2 ) j + (8.66 ft/s2 )k 

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PROBLEM 15.197 A 30 mm-radius wheel is mounted on an axle OB of length 100 mm. The wheel rolls without sliding on the horizontal floor, and the axle is perpendicular to the plane of the wheel. Knowing that the system rotates about the y axis at a constant rate ω1 = 2.4 rad/s, determine (a) the angular velocity of the wheel, (b) the angular acceleration of the wheel, (c) the acceleration of Point C located at the highest point on the rim of the wheel.

SOLUTION l = 100 mm = 0.1 m b = 30 mm = 0.03 m b tan β = = 0.3 l β = 16.699° rA = −l sec β i

Geometry.

rB = −l cos β i + b cos β j

(a)

Angular velocities.

For the system,

Ω = ω1 j = (2.4 rad/s) j

For the wheel,

ω = ωx i + ω y j + ωz k v A = ω × rA = (ω x i + ω y j + ω z k ) × (−l sec β i ) = 0

−(lω z sec β ) j − (lω y sec β )k = 0

ω y = 0, ω z = 0

ω = ωx i

v B = ω × rB = ω x i × (−l cos β i + b cos β j) = (ω x b cos β )k

For the system,

v B = Ω × rB = ω 1 j × (−l cos β i + b cos β j) = (ω 1l cos β )k

Matching the two expressions for v B ,

ωxb cos β = ω 1l cos β or

ωx =

ω 1l b

=

(2.4)(30) = 8 rad/s 100

ω = (8.00 rad/s)i 

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PROBLEM 15.197 (Continued)

(b)

Angular acceleration.  α=ω  Oxyz + Ω × ω =ω = (0 + 2.4 j) × 8i

α = −(19.20 rad/s 2 )k 

= −(19.2 rad/s 2 )k

(c)

Conditions at Point C. rC = −(l cos β − b sin β )i + 2b cos β j = (−87.162 mm)i + (57.47 mm) j vC = ω × rC = 8i × ( −87.162i + 57.47 j) = (459.76 mm/s)k aC = α × rC + ω × v C = −19.2k × (−87.162i + 57.47 j) + 8i × 459.76k = (1103.4 mm/s 2 )i − (2004.6 mm/s 2 ) j

aC = (1.103 m/s2 )i − (2.005 m/s2 ) j 

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PROBLEM 15.198 At the instant shown, the robotic arm ABC is being rotated simultaneously at the constant rate ω1 = 0.15 rad/s about the y axis, and at the constant rate ω 2 = 0.25 rad/s about the z axis. Knowing that the length of arm ABC is 1 m, determine (a) the angular acceleration of the arm, (b) the velocity of Point C, (c) the acceleration of Point C.

SOLUTION Angular velocity:

ω = ω1 j + ω 2k = (0.15 rad/s) j + (0.25 rad/s)k

Consider a frame of reference rotating with angular velocity Ω = ω 1 j = (0.15 rad/s) j

(a) Angular acceleration of the arm. 1 +ω  2 + Ω × (ω1 + ω 2 ) α=ω

= 0 + 0 + (0.15 j) × (0.15 j + 0.25k )

α = (0.0375 rad/s2 )i 

Arm ABC rotates about the fixed Point A. rC /A = (1 m)(cos35°i + sin 35°) j

= (0.81915 m)i + (0.57358 m) j (b)

Velocity of Point C.

vC = ω × rC /A i j k vC = 0 0.15 0.25 0.81915 0.57358 0 = −0.14340i + 0.20479 j − 0.12287k vC = −(0.1434 m/s)i + (0.204 m/s) j − (0.1229 m/s)k 

(c)

Acceleration of C:

aC = α × rC /A + ω × vC i j k i j k aC = 0.0375 0 0 + 0 0.15 0.25 0.81915 0.57358 0 −0.14340 0.20479 −0.12287 = 0.021509k + 0.06962i + 0.03585 j + 0.02151k

aC = −(0.0696 m/s 2 )i + (0.0359 m/s2 ) j + (0.0430 m/s 2 )k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1284

PROBLEM 15.199 In the planetary gear system shown, gears A and B are rigidly connected to each other and rotate as a unit about the inclined shaft. Gears C and D rotate with constant angular velocities of 30 rad/s and 20 rad/s, respectively (both counterclockwise when viewed from the right). Choosing the x axis to the right, the y axis upward, and the z axis pointing out of the plane of the figure, determine (a) the common angular velocity of gears A and B, (b) the angular velocity of shaft FH, which is rigidly attached to the inclined shaft.

SOLUTION Place origin at F. Point 1:

r1 = −(80 mm)i + (260 mm) j

Point 2:

r2 = + (80 mm)i + (50 mm) j ω E = + (30 rad/s)i ωG = + (20 rad/s)i v1 = ω E × r1 = (30i ) × (−80i + 260 j)

v1 = (7800 mm/s)k

(1)

v 2 = ωG × r2 = (20i ) × (80i + 50 j) v 2 = (1000 mm/s)k

(2)

ω = ωx i + ω y j + ωz k

Motion of gear unit AB:

i

j

k

v1 = ω × r1 = ω x

ωy

ωz

−80 +260 0 = −260ω z i − 80ω z j + (260ω x + 80ω y )k

Recall from Eq. (1) that v = 7800k. 7800k = −260ω z i − 80ω z j + (260ωx + 80ω y )k Equate coefficients of unit vectors.

ωz = 0 7800 = 260ωx + 80ω y

(3)

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PROBLEM 15.199 (Continued)

v 2 = ω × r2 i

j

= ωx

ωy

k 0

80 50 0 = (50ω x − 80ω y )k

Recall from Eq. (2) that v2 = 1000k, and write 1000 = 50ωx − 80ω y

(4)

Add Eqs. (3) and (4):

8800 = 310ωx

ωx = 28.387 rad/s

Eq. (4):

1000 = 50(28.387) − 80ω y

ω y = 5.242 rad/s

(a)

Common angular velocity of unit AB. ω = (28.387 rad/s)i + (5.242 rad/s) j ω = (28.4 rad/s)i + (5.24 rad/s) j 

(b)

Angular velocity of shaft FH. (See figure in text.) Point N is at nut, which is a part of unit AB and also is a part of shaft GH. 1 yN 2 rN = xN i + y N j

xN =

1 yN i + yN j 2 ω = (28.387 rad/s)i + (5.242 rad/s)j

rN =

Nut N as a part of unit AB:

v N = ω × rN

1  = (28.387i + 5.242 j) ×  yN i + yN j  2  v N = (28.387 yN − 2.621y N )k = + (25.766 y N )k

(5)

Nut N as a part of shaft FH. ωFH = ωFH i v N = ω FH × rN 1  = (ωFH i ) ×  y N i + y N j  2   = ωFH y N k

(6)

Equating expressions for v N from Eqs. (5) and (6), +(25.766 yN )k = ωFH y N k ω FH = (25.766 rad/s)i

ωFH = (25.8 rad/s)i 

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PROBLEM 15.200 In Problem 15.199, determine (a) the common angular acceleration of gears A and B, (b) the acceleration of the tooth of gear A which is in contact with gear C at Point I.

SOLUTION See the solution to part (a) of Problem 15.199 for the calculation of the common angular velocity of unit AB.

ω = (28.387 rad/s)i + (5.242 rad/s) j The angular velocity vector ω rotates about the x-axis with angular velocity ωFH . See part (b) of Problem 15.199 for the calculation of ωFH .

ωFH = (25.776 rad/s)i (a)

Common angular acceleration of unit AB. α = ω FH × ω = (25.776i ) × (28.387i + 5.242 j) = 135.12k

α = 135.1 rad/s2k 

The position and velocity vectors of a tooth at the contact Point 1 of gears A and C are r1 = −(80 mm)i + (260 mm) j v1 = (7800 mm/s)k as determined in part (a) of Problem 15.199. (b)

Acceleration of the tooth at Point 1 of gear A. a1 = α × r1 + ω × v1 = (135.12k ) × (−80i + 260 j) + (28.387i + 5.242 j) × 7800k = −10810 j − 35131i − 221419 j + 40888i = (5757 mm/s 2 )i − (232229 mm/s 2 ) j

a1 = (5.8 m/s 2 )i − (232 m/s2 ) j 

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PROBLEM 15.201 Several rods are brazed together to form the robotic guide arm shown which is attached to a ball-and-socket joint at O. Rod OA slides in a straight inclined slot while rod OB slides in a slot parallel to the z axis. Knowing that at the instant shown v B = (9 in./s)k , determine (a) the angular velocity of the guide arm, (b) the velocity of Point A, (c) the velocity of Point C.

SOLUTION

Since rod at D slides in slot which is of slope 1:2, (vD ) x = −2(vD ) y (v A ) x = −2(v A ) y

and (a)

Angular velocity.

ω = ωx i + ω y j + ω z k v B = ω × rB : (9 in./s)k = (ωx i + ω y j + ωz k ) × (12 in.)j 9k = 12ω x k − 12ωz i

Coefficients of k: Coefficients of i:

9 = 12ωx ωx = 0.75 rad/s 0 = −12ω z

ωz = 0

v A = ω × rA v A = (0.75i + ω y j) × (10k )

(v A ) x i + (v A ) y j + (v A ) z k = −7.5 j + 10ω y i

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PROBLEM 15.201 (Continued)

Coefficients of j:

(v A ) y = −7.5

Coefficients of i:

(v A ) x = 10ω y

Coefficients of k:

(v A ) z = 0

Recall the Equations

(vA ) x = −2(vA ) y

and

10ω y = −2(−7.5)

So,

ω y = 1.5 rad/s and (v A ) x = 15 in./s ω = (0.75 rad/s)i + (1.5 rad/s) j 

(b)

Velocity of A:

(c)

Velocity of C:

vA = (15 in./s)i − (7.5 in./s) j  rC = 5i + 4 j + 2k v C = ω × rC i j k = 0.75 1.5 0 5 4 2 = 3i − 1.5 j + (3 − 7.5)k

vC = (3 in./s)i − (1.5 in./s) j − (4.5 in./s)k 

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PROBLEM 15.202 In Problem 15.201 the speed of Point B is known to be constant. For the position shown, determine (a) the angular acceleration of the guide arm, (b) the acceleration of Point C. PROBLEM 15.201 Several rods are brazed together to form the robotic guide arm shown, which is attached to a ball-and-socket joint at O. Rod OA slides in a straight inclined slot while rod OB slides in a slot parallel to the z-axis. Knowing that at the instant shown vB = (9 in./s)k , determine (a) the angular velocity of the guide arm, (b) the velocity of Point A, (c) the velocity of Point C.

SOLUTION

Since rod at D slides in slot which is of slope 1:2, (vD ) x = −2(vD ) y and Angular velocity.

(v A ) x = −2(v A ) y ω = ω x i + ω y j + ωz k v B = ω × rB : (9 in./s)k = (ωx i + ω y j + ωz k ) × (12 in.)j

9k = 12ωx k − 12ωz i Coefficients of k: Coefficients of i:

9 = 12ωx ωx = 0.75 rad/s 0 = −12ω z

ωz = 0

v A = ω × rA v A = (0.75i + ω y j) × (10k )

(v A ) x i + (v A ) y j + (v A ) z k = −7.5j + 10ω y i Coefficients of j:

(v A ) y = −7.5

Coefficients of i:

(v A ) x = 10ω y

Coefficients of k:

(v A ) z = 0

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PROBLEM 15.202 (Continued) Recall the Equations

(vA ) x = −2(v A ) y

and

10ω y = −2(−7.5)

So,

ω y = 1.5 rad/s and (v A ) x = 15 in./s ω = (0.75 rad/s)i + (1.5 rad/s) j

Velocity of A:

v A = (15 in./s)i − (7.5 in./s) j

Velocity of C:

rC = 5i + 4 j + 2k v C = ω × rC i j k = 0.75 1.5 0 5 4 2 = 3i − 1.5 j + (3 − 7.5)k v C = (3 in./s)i − (1.5 in./s) j − (4.5 in./s)k a B = α × rB + ω × (ω × rC ) = α × rC + ω × v B i aB = α x 0

j

k

i j k α y α z + 0.75 1.5 0 0 0 9 12 0

= −12α z i + 12α x k + 13.5i − 6.75 j

a B = (13.5 − 12α z )i − 6.75 j + 12α x k (aB ) x = 13.5 − 12α z = 0 (aB ) y = −6.75

(1)

α z = 1.125 rad/s 2

(aB ) y = −6.75 in./s 2

(aB ) z = 12α x = 0

αx = 0

a A = α × rA + ω × (ω × rA ) = α × rA + ω × v A i

j

k

i

j

k

a A = 0 α y 1.125 + 0.75 1.5 0 15 −7.5 0 0 0 10 = 10α y i + (−5.625 − 22.5)k a A = 10α y i − 28.125k

Thus,

(a A ) x = 10α y (a A ) y = 0 ( a A ) z = −28.125 in./s 2

But

(a A ) x = −2(a A ) y = 0

Therefore,

(a A ) x = 10α y = 0 α y = 0

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PROBLEM 15.202 (Continued) 

(a)

Angular acceleration:

(b)

Acceleration of C:

α = (1.125 rad/s2 )k  aC = α × rC + ω × (ω × rC )



= α × rC + ω × v C rC = (5 in.)i + (4 in.) j + (2 in.)k



i j k i j k 0 aC = 0 0 1.125 + 0.75 1.5 5 4 2 3 −1.5 −4.5 = −4.5i + 5.625 j − 6.75i + 3.375 j + ( −1.125 − 4.5)k

aC = −(11.3 in./s2 )i + (9 in./s 2 ) j − (5.63 in./s 2 )k 

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PROBLEM 15.203 Rod AB of length 25 in. is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves toward Point E at a constant speed of 20 in./s, determine the velocity of collar A as collar B passes through Point D.

SOLUTION 2 = x A2 /B + y A2 /B + z A2/B : 252 = (−12) 2 + y A2 /B + (−20) 2 l AB

Geometry.

y A/B = 9 in. rA/B = ( −12 in.)i + (9 in.) j − (20 in.)k rD/C = (12 in.)i − (9 in.) j ,

lCD = (12)2 + ( −9) 2 = 15 in.

Velocity of collar B.

Velocity of collar A.

v B = vB

rD/C

lCD (12i − 9 j) = (16 in./s)i − (12 in./s) j v B = (20) 15 vA = v A j vA = v B + vA/B

v A/B = ω AB × rA/B

where

Noting that vA/B is perpendicular to rA/B , we get rB/ A ⋅ v B/ A = 0. Forming rA/B ⋅ vA , we get

rA/B ⋅ vA = rA/B ⋅ ( vB + vA/B ) = rA/B ⋅ v B + rA/B ⋅ vA/B

or From Eq. (1),

rA/B ⋅ vA = rA/B ⋅ vB

(1)

(−12i + 9 j − 2 0k ) ⋅ (vA j) = (−12i + 9 j − 80k ) ⋅ (16i − 12 j) 9vA = (−12)(16) + (9)(−12)

or

vA = −33.333 in./s

vA = −(33.3 in./s) j 

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PROBLEM 15.204 Rod AB, of length 11 in., is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves downward at a constant speed of 54 in./s, determine the velocity of collar A when c = 2 in.

SOLUTION 2 l AB = x A2/B + y A2 /B + z A2/B : (11)2 = (6) 2 + (−2) 2 + ( z A /B ) 2

Geometry.

z A/B = 9 in. rA/B = (−6 in.)i − (2 in.) j + (9 in.)k

Velocity of collar B.

v B = −vB j = −(54 in./s) j

Velocity of collar A.

vA = v Ak vA = v B + vA/B ,

v A/B = ω AB × rA/B

where

Noting that vA/B is perpendicular to rA/B , we get rB/ A ⋅ v B/ A = 0. Forming rA/B ⋅ vA , we get

rA/B ⋅ vA = rA/B ⋅ ( vB + vA/B ) = rA/B ⋅ v B + rA/B ⋅ vA/B

or

rA/B ⋅ vA = rA/B ⋅ vB

From Eq. (1), or

(1)

(−6 i − 2 j + 9 k ) ⋅ vAk = (−6i − 2 j + 9k ) ⋅ (−54 j) 9 vA = 108

vA = (12.00 in./s)k 

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PROBLEM 15.205 Rods BC and BD are each 840 mm long and are connected by ball-and-socket joints to collars which may slide on the fixed rods shown. Knowing that collar B moves toward A at a constant speed of 390 mm/s, determine the velocity of collar C for the position shown.

SOLUTION Geometry.

2 lBC = xC2 /B + yC2 /B + zC2 /B

(840) 2 = c 2 + (640 − 480) 2 + (200)2 c = 800 mm rC/B = (800 mm)i + (160 mm) j − (200 mm)k

Velocity of B.

Velocity of C.

5   12 v B = vB  − j − k   13 13  5   12 = (390 mm/s)  − j − k   13 13  = −(360 mm/s) j − (150 mm/s)k vC = vC i vC = v B + vC/B

where

vC/B = ωBC × rC/B

Noting that vC /B is perpendicular to rC/B , we get rC/B ⋅ vC/B = 0 So that

rC/B ⋅ vC = rC/B ⋅ vB

(800i + 160 j − 200 k ) ⋅ (−360 j − 150k ) = (800i + 160 j − 200k ) ⋅ (vC i) (160)(−360) + (−200)(−150) = 800 vC vC = −34.5 mm/s

vC = −(34.5 mm/s)i 

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PROBLEM 15.206 Rod AB is connected by ball-and-socket joints to collar A and to the 16-in.-diameter disk C. Knowing that disk C rotates counterclockwise at the constant rate ω 0 = 3 rad/s in the zx plane, determine the velocity of collar A for the position shown.

SOLUTION rB/C = (− 8 in.)k

Geometry.

rA/B = − (25 in.)i + (20 in.) j − (8 in.)k v B = ω0 j × rB/C

Velocity at B.

= 3j × (−8k ) = −(24 in./s)i

v A = vA j

Velocity of collar A.

v A = v B + v A /B

v A /B = ω AB × rA /B

where

Noting that vA /B is perpendicular to rA/B , we get rB / A ⋅ v B / A = 0. Forming rA /B ⋅ vA , we get

rA /B ⋅ vA = rA /B ⋅ ( v B + vA /B ) = rA /B ⋅ v B + rA /B ⋅ vA /B

or From Eq. (1),

rA /B ⋅ v A = rA /B ⋅ v B

(1)

(− 25i + 20 j − 8k ) ⋅ (v A j) = (− 25i + 20 j − 8k ) ⋅ (24i) 20v A = − 600

or

vA = −30 in./s

vA = −(30.0 in./s) j 

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PROBLEM 15.207 Rod AB of length 29 in. is connected by ball-and-socket joints to the rotating crank BC and to the collar A. Crank BC is of length 8 in. and rotates in the horizontal xz plane at the constant rate ω 0 = 10 rad/s. At the instant shown, when crank BC is parallel to the z axis, determine the velocity of collar A.

SOLUTION rB/C = (−8 in.)k ,

Geometry.

rA/B = (−12 in.)i + (21 in.) j + (16 in.)k v B = ω0 j × rB/C

Velocity at B.

= 10 j × (−8k ) = (−80 in./s)i

Velocity of collar A.

vA = v A j vA = v B + vA /B

v A /B = ω AB × rA /B

where

Noting that vA/B is perpendicular to rA /B , we get rB / A ⋅ vB / A = 0. Forming rA/B ⋅ vA , we get

rA /B ⋅ vA = rA /B ⋅ ( v B + vA /B ) = rA /B ⋅ vB + rA /B ⋅ vA /B

or From Eq. (1),

rA /B ⋅ v A = rA/B ⋅ v B

(1)

(−12i + 21j + 16k ) ⋅ (vA j) = (−12i + 21j + 16k ) ⋅ (−80i ) 21vA = 960

or

vA = 45.714 in./s

vA = (45.7 in./s) j 

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PROBLEM 15.208 Rod AB of length 300 mm is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves toward Point D at a constant speed of 50 mm/s, determine the velocity of collar A when c = 80 mm.

SOLUTION Geometry.

rA = yj , rD = (90 mm)i rC = (180 mm)k

rD/C = rD − rC = (40 mm)i − (180 mm)k lCD = (90) 2 + (180) 2

rB /C

= 201.246 mm c(rD/C ) = 180 80(90i − 180k ) = 180 = (40 mm)i − (80 mm)k

rB = rC + rB/C

rA /B

= 180k + 40i − 80k = (40 mm)i + (100 mm)k = rA − rB = −(40 mm)i + ( y mm) j − (100 mm)k

2 l AB = x A2/B + y 2 + z 2A/B : 3002 = (−40) 2 + y 2 + (−100) 2

y = 280 mm, rA/B = (−40 mm)i + (280 mm) j − (100 mm)k

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PROBLEM 15.208 (Continued)

vB = vB

Velocity of collar B.

rD/C

lCD

(50)(90i − 180k ) 201.246 = (22.3607 mm/s)i − (44.7214 mm/s)k

vB =

vA = vA j

Velocity of collar A.

vA = vB + vA /B

vA /B = ωAB × rA/B

where

Noting that vA/B is perpendicular to rA/B , we get rB/ A ⋅ vB/ A = 0. Forming rA/B ⋅ vA , we get

rA /B ⋅ vA = rA /B ⋅ ( vB + vA /B ) = rA /B ⋅ vB + rA /B ⋅ vA /B

or From Eq. (1),

rA /B ⋅ vA = rA /B ⋅ vB

(1)

(− 40i + 280 j − 100k ) ⋅ (vA j) = (−40i + 280 j − 100k ) ⋅ (22.3607i − 44.7214k ) 280vA = (−40)(22.3607) + (−100)(− 44.7214)

or

vA = 12.7775 mm/s

vA = (12.78 mm/s) j 

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PROBLEM 15.209 Rod AB of length 300 mm is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves toward Point D at a constant speed of 50 mm/s, determine the velocity of collar A when c = 120 mm.

SOLUTION Geometry.

rA = yj , rD = (90 mm)i rC = (180 mm)k rD/C = rD − rC = (90 mm)i − (180 mm)k

lCD = (90) 2 + (−180) 2

rB/C

= 201.246 mm c(rD/C ) = 180 120(90i − 180k ) = 180 = (60 mm)i − (120 mm)k

rB = rC + rB/C

rA/B

= 180k + 60i − 120k = (60 mm)i + (60 mm)k = rA − rB = − 60i + yj − 60k

2 l AB = x A2/B + y 2 + z 2A/B : 3002 = 602 + y 2 + 602

y = 287.75 mm, rA/B = (−60 mm)i + (287.75 mm) j − (60 mm)k

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PROBLEM 15.209 (Continued)

Velocity of collar B.

vB = vB

rD/C

lCD

(50)(90i − 180k ) 201.246 = (22.3607 mm/s)i − (44.7214 mm/s)k

vB =

Velocity of collar A.

vA = vA j vA = vB + vA /B

where

vA /B = ω AB × rA /B

Noting that vA/B is perpendicular to rA/B , we get rB / A ⋅ vB / A = 0. Forming rA/B ⋅ vA , we get

rA /B ⋅ vA = rA /B ⋅ ( vB + vA /B ) = rA /B ⋅ vB + rA /B ⋅ vA /B

or

rA /B ⋅ vA = rA /B ⋅ vB

(1)

From Eq. (1), (−60i + 287.75 j − 60k ) ⋅ (vA j) = (−60i + 287.75 j − 60k ) ⋅ ( 22.3607i − 44.7214j)

287.75vA = (−60)(22.3607) + (−60)(−44.7214) or

vA = 4.6626 mm/s

vA = (4.66 mm/s) j 

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PROBLEM 15.210 Two shafts AC and EG, which lie in the vertical yz plane, are connected by a universal joint at D. Shaft AC rotates with a constant angular velocity ω1 as shown. At a time when the arm of the crosspiece attached to shaft AC is vertical, determine the angular velocity of shaft EG.

SOLUTION ω AC = ω1k

Angular velocity of shaft AC.

Let ω3 j be the angular velocity of body D relative to shaft AD.

ωD = ω 1k + ω 3 j

Angular velocity of body D.

ωEG = ω2 (cos 25°k − sin 25° j)

Angular velocity of shaft EG.

Let ω4 i be the angular velocity of body D relative to shaft EG.

ωD = ω2 (cos 25°k − sin 25° j) + ω4 i

Angular velocity of body D.

Equate the two expressions for ω D and resolve into components.

i:

From Eq. (3),

0 = ω4

(1)

j: ω3 = −ω2 sin 25°

(2)

k : ω1 = ω2 cos 25°

(3)

ω2 =

ω1 cos 25°

ω EG =

ω2 cos 25°

(− sin 25° j + cos 25°k ) 

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PROBLEM 15.211 Solve Problem 15.210, assuming that the arm of the crosspiece attached to the shaft AC is horizontal. PROBLEM 15.210 Two shafts AC and EG, which lie in the vertical yz plane, are connected by a universal joint at D. Shaft AC rotates with a constant angular velocity ω1 as shown. At a time when the arm of the crosspiece attached to shaft AC is vertical, determine the angular velocity of shaft EG.

SOLUTION ω AC = ω1k

Angular velocity of shaft AC.

Let ω 3i be the angular velocity of body D relative to shaft AD.

ω D = ω1k + ω3i

Angular velocity of body D.

ω EG = ω 2 (cos 25°k − sin 25° j)

Angular velocity of shaft EG.

Let ω4 λ be the angular velocity of body D relative to shaft EG, where λ is a unit vector along the clevis axle attached to shaft EG. λ = cos 25° j + sin 25°k

ω4 λ = ω4 cos 25° j + ω4 sin 25°k Angular velocity of body D.

ω D = ω EG + ω4 λ ω D = (ω4 cos 25° − ω2 sin 25°) j

+ (ω4 sin 25° + ω2 cos 25°)k

Equate the two expressions for ω D and resolve into components. i:

From Eqs. (2) and (3),

ω3 = 0

(1)

j:

0 = ω4 cos 25° − ω 2 sin 25°

(2)

k:

ω1 = ω4 sin 25° + ω 2 cos 25°

(3)

ω2 = ω1 cos 25°

ωEG = ω1 cos 25°(− sin 25° j + cos 25°k ) 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1303

PROBLEM 15.212 In Problem 15.206, the ball-and-socket joint between the rod and collar A is replaced by the clevis shown. Determine (a) the angular velocity of the rod, (b) the velocity of collar A.

SOLUTION Geometry.

rB/C = (8 in.)k rA/B = −(25 in.)i + (20 in.) j − (8 in.)k

Velocity of collar B.

vB = ω0 j × rB/C = −3j × 8k = (24 in./s)i

Velocity of collar A.

vA = vA j

ωA = ω 1j

Angular velocity of collar A.

The axle of the clevis at A is perpendicular to both the y axis and the rod AB. A vector p along this axle is p = j × rA /B = j × (−25i + 20 j − 8k ) = −8i + 25k p = 82 + 252 = 26.2488

Unit vector λ along axle:

λ=

p = −0.30478i + 0.95242k p

Let ω2 be the angular velocity of the rod AB relative to collar A.

ω2 = ω 2 λ = −0.30478ω 2 i + 0.95242ω 2k ω AB = ω A + ω2

Angular velocity of rod AB.

ω AB = −0.30478ω 2 i + ω1 j + 0.95242ω 2k

(1)

vA = vB + vA/B = vB + ω AB × rA/B i vA j = 24i + −0.30478ω2 −25

j

k

ω1 0.95242ω2 20

−8

Resolving into components,

i:

0 = 24 − 8ω1 − 19.0484ω 2

j : vA = −26.2487ω 2 k:

0 = 25ω1 − 6.0956ω 2

(2) (3) (4)

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PROBLEM 15.212 (Continued)

Solving Eqs. (2), (3) and (4) simultaneously, vA = −30 in./s,

ω 1 = 0.27867 rad/s, ω 2 = 1.1429 rad/s (a)

Angular velocity of rod AB. From Eq. (1)

ω AB = −(0.30478)(1.1429)i + 0.27867 j + (0.95242)(1.1429)k ω AB = −(0.348 rad/s)i + (0.279 rad/s) j + (1.089 rad/s)k 

(b)

vA = −(30.0 in./s) j 

Velocity of A.

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PROBLEM 15.213 In Problem 15.205, the ball-and-socket joint between the rod and collar C is replaced by the clevis connection shown. Determine (a) the angular velocity of the rod, (b) the velocity of collar C.

SOLUTION rC = xC i + (640 mm) j, rB = (480 mm) j + (200 mm)k

Geometry.

l AB =

4802 + 2002 = 520 mm

rC/B = xC i + (160 mm) j − (200 mm)k

Length of rod BC.

2 lBC = 8402 = xC2 + 1602 + 2002

xC = 800 mm

Solving for xC ,

rC/B = (800 mm)i + (160 mm) j − (200 mm)k vB =

Velocity.

390 (−480 j − 200k ) = (−360 mm) j − (150 mm/s)k 520

vC = vC i Angular velocity of collar C.

ωC = ωC i

The axle of the clevis at C is perpendicular to the x-axis and to the rod BC. A vector along this axle is

p = i × rC/B p = i × (800i + 160 j − 200k ) = (200 mm) j + (160 mm)k p=

Let λ be a unit vector along the axle.

200 2 + 1602 = 256.125 mm

λ=

p = 0.78087 j + 0.62470k p

Let ωs = ω sλ be the angular velocity of rod BC relative to collar C.

ωs = 0.78087ωs j + 0.62470ωsk Angular velocity of rod BC.

ωBC = ωC + ωs ωBC = ωC i + 0.78087ω s j + 0.62470ω sk vC = v B + ωBC × rC/B i

j k vC i = −360 j − 150k + ωC 0.78087ω s 0.62470ω s 800 160 −200

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PROBLEM 15.213 (Continued) Resolving into components,

i: vC = − 256.126ωs

(1)

j:

0 = −360 + 200ωC + 499.76ωs

(2)

k:

0 = −150 + 160ωC − 624.70ωs

(3)

Solving the simultaneous equations (1), (2), and (3),

ωC = 1.4634 rad/s, ω s = 0.13470 rad/s, vC = −34.50 mm/s (a)

Angular velocity of rod BC.

ωBC = 1.4634i + (0.78087)(0.13470) j + (0.62470)(0.13470)k ωBC = (1.463 rad/s)i + (0.1052 rad/s) j + (0.0841 rad/s)k  (b)

vC = −(34.5 mm/s)i 

Velocity of collar C.

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PROBLEM 15.214 In Problem 15.204, determine the acceleration of collar A when c = 2 in. PROBLEM 15.204 Rod AB of length 11 in., is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves downward at a constant speed of 54 in./s, determine the velocity of collar A when c = 2 in.

SOLUTION 2 l AB = x A2 /B + y 2A/B + z 2A /B : (11)2 = (6) 2 + (−2) 2 + ( z A /B ) 2

Geometry.

z A /B = 9 in. rA/B = (−6 in.)i − (2 in.) j + (9 in.)k

Velocity of collar B.

v B = −vB j = −(54 in./s) j (2.5)(4.5i − 9k ) vB = = (1.11803 in.)i − (2.23607 in./s)k 10.0623

Velocity of collar A.

vA = v Ak vA = v B + vA/B ,

v A/B = ω AB × rA/B

where

Noting that vA /B is perpendicular to rA/B , we get rB / A ⋅v B / A = 0 Forming rA /B ⋅ vA , we get

rA /B ⋅ vA = rA /B ⋅ ( vB + vA /B ) = rA/B ⋅ v B + rA /B ⋅ vA/B

or

rA /B ⋅ vA = rA/B ⋅ vB

From Eq. (1),

(1)

(−6 i − 2j + 9 k ) ⋅ vAk = (−6i − 2j + 9k ) ⋅ (−54j)

or

9 vA = 108

Relative velocity

vA/B = v A − v B

vA = (12.00 in./s)k 

v A /B = (54 in./s) j + (12.00 in./s)k (v A /B ) 2 = (54) 2 + (12.00) 2 = 3060 in 2 /s 2

Acceleration of collar B.

aB = 0

Acceleration of collar A.

a A = a Ak a A = a B + a A /B

where

a A/B = α AB × rA/ B + ω AB × v A/B

Noting that α AB × rA/B is perpendicular to rA/B , we get rA/ B ⋅ α AB × rA/B = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1308

PROBLEM 15.214 (Continued)

We note also that Then,

rA /B ⋅ ω AB × v A /B = v A /B ⋅ rA /B × ω A /B = − v A /B ⋅ v A /B = − (v A /B ) 2

rA/B ⋅ a A/B = 0 − (v A/B )2 = −(v A/B )2

Forming rA/ B ⋅ a A , we get

rA/B ⋅ a A = rA/B ⋅ (a A + a A/B ) = rA/ B ⋅ a B + rA/B ⋅ a A/B

or

rA/B ⋅ a A = rA /B ⋅ a B − (v A/B )2

From Eq. (2)

(−6i − 2i + 9k ) ⋅ a Ak = 0 − 3060 9a A = −3060

(2)

a A = −(340 in./s2 )k 

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PROBLEM 15.215 In Problem 15.205, determine the acceleration of collar C. PROBLEM 15.205 Rod BC and BD are each 840 mm long and are connected by ball-and-socket joints to collars which may slide on the fixed rods shown. Knowing that collar B moves toward A at a constant speed of 390 mm/s, determine the velocity of collar C for the position shown.

SOLUTION 2 lBC = xC2 /B + yC2 /B + zC2 /B

Geometry.

c = 800m (840) 2 = c 2 + (640 − 480)2 + (200) 2 rC/B = (800 mm)i + (160 mm) j − (200 mm)k 5   12 v B = vB  − j − k  13 13   5   12 = (390 mm/s)  − j − k   13 13  = −(360 mm/s) j − (150 mm/s)k

Velocity of B.

Velocity of C.

vC = vC i

where

vC = v B + vC/B

vC/B = ω BC × rC/B Noting that v C /B is perpendicular to rC /B , we get rC /B ⋅ vC = rC /B ⋅ v B

(800i + 160 j − 200k ) ⋅ (−360 j + 150k ) = (800i + 160 j − 200k ) ⋅ (vC i ) (160)(−360) + (−200)(−150) = 800vC

vC = −(34.5 mm/s) i 

vC = −34.5 mm/s

Relative velocity:

vC /B = vC − v B v C /B = −(34.5 mm/s)i + (360 mm/s) j + (150 mm/s)k vC2 /B = (34.5) 2 + (360) 2 + (150) 2 = 153290 mm 2 /s 2

Acceleration of collar B:

aB = 0

Acceleration of collar C:

aC = aC i a C = a B + a C/B

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PROBLEM 15.215 (Continued)

where

aC/B = αCB × rC/B + ωCB × vC/B

Noting that αCB × rC /B is perpendicular to rC/B , we get rC/B ⋅ αCB × rC / B = 0 We note also that

rB / C ⋅ ω CB × v C/B = v C/B ⋅ rC/B × ω C/B = − v C/ B ⋅ v C/B = −(vC/B ) 2

Then,

rC/B ⋅ aC = 0 − (vC/B )2 = −(vC/B )2

Forming rC /B ⋅ aC , we get

rC/ B ⋅ aC = rC/B ⋅ (a B + aC/ B ) = rB / C ⋅ a B + rB / C ⋅ aC/ B

so that

rC/B ⋅ aC = rC/B ⋅ a B − (vC/B )2

From Equation (2),

(800i + 160 j − 200k ) ⋅ aC i = 0 − 153290

(2)

800aC = −153290

aC = −191.6 mm/s 2

aC = −(191.6 mm/s 2 )i 

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PROBLEM 15.216 In Problem 15.206, determine the acceleration of collar A. PROBLEM 15.206 Rod AB is connected by ball-and-socket joints to collar A and to the 16-in.-diameter disk C. Knowing that disk C rotates counterclockwise at the constant rate ω0 = 3 rad/s in the zx plane, determine the velocity of collar A for the position shown.

SOLUTION rB/C = (−8 in.)k

Geometry.

rA/B = −(25 in.)i + (20 in.) j − (8 in.)k vB = ω0 j × rB/C

Velocity at B.

= 3j × (−8k ) = (24 in./s)i vA = vA j

Velocity of collar A.

vA = v B + vA/B vA/B = ω AB × rA/B

where

Noting that vA /B is perpendicular to rA/B , we get rB/A ⋅ vB/A = 0. Forming rA /B ⋅ vA , we get

rA/B ⋅ vA = rA/B ⋅ ( vB + vA/B ) = rA/B ⋅ v B + rA/B ⋅ vA/B

or From Eq. (1)

rA/B ⋅ vA = rA/B ⋅ vB

(1)

(−25i + 20 j − 8k ) ⋅ (v A j) = (−25i + 20 j − 8k ) ⋅ (24i) 20vA = −600

or

Relative velocity.

vA = −30 in./s vA/B = v A − v B vA/B = (−30 in./s ) j + (24 in./s)i (vA /B ) 2 = (−30) 2 + (24)2 = 1476 in 2 /s 2

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PROBLEM 15.216 (Continued)

Acceleration at B.

a B = ω0 j × v B = 3 j × 24i = −(72 in./s 2 )k

Acceleration of collar A.

a A = aA j a A = a B + a A/B

where

a A/B = α AB × rA/B + ω AB × v A/B

Noting that α AB × rA/B is perpendicular to rA/B , we get rA/B ⋅ α AB × rA/B = 0 We note also that

rA/B ⋅ ω AB × vA/B = vA/B ⋅ rA/B × ω A/B = − vA/B ⋅ vA/B = −(vA/B ) 2

Then

rA/B ⋅ a A/B = 0 − (vA/B ) 2 = −(vA/B ) 2

Forming rA/B ⋅ a A , we get

rA/B ⋅ a A = rA/B ⋅ (a A + a A/B ) = rA/B ⋅ a B + rA/B ⋅ a A/B

or From Eq. (2),

rA/B ⋅ a A = rA/B ⋅ a B − (vA/B )2

(2)

(−25i + 20 j − 8k ) ⋅ (a A j) = (−25i + 20 j − 8k ) ⋅ (−72k ) − 1476 20a A = 576 − 1476 = −45 in./s 2

a A = −(45.0 in./s2 ) j 

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PROBLEM 15.217 In Problem 15.207, determine the acceleration of collar A. PROBLEM 15.207 Rod AB of length 29 in. is connected by balland-socket joints to the rotating crank BC and to the collar A. Crank BC is of length 8 in. and rotates in the horizontal xz plane at the constant rate ω0 = 10 rad/s. At the instant shown, when crank BC is parallel to the z axis, determine the velocity of collar A.

SOLUTION rB/C = (−8 in.)k ,

Geometry.

rA/B = (−12 in.)i + (21 in.) j + (16 in.)k vB = ω0 j × rB/C

Velocity at B.

= 10 j × ( −8k ) = (−80 in./s)i vA = vA j

Velocity of collar A.

vA = vB + vA/B vA/B = ω AB × rA/B

where

Noting that vA/B is perpendicular to rA/B , we get rB/A ⋅ vB/A = 0. Forming rA/B ⋅ vA , we get

rA/B ⋅ vA = rA/B ⋅ ( vB + vA/B ) = rA/B ⋅ vB + rA/B ⋅ vA/B

or From Eq. (1)

rA/B ⋅ vA = rA/B ⋅ vB

(1)

(−12i + 21j + 16k ) ⋅ (v A j) = (−12i + 21j − 16k ) ⋅ (−80i) 21v A = 960

or

v A = 45.714 in./s

v A = (45.7 in./s) j  

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PROBLEM 15.217 (Continued)

vA/B = vA − v B

Relative velocity.

vA/B = (45.714 in./s ) j + (80 in./s)i (vA/B ) 2 = (45.714) 2 + (80) 2 = 8489.8 (in./s 2 )

Acceleration of Point B.

a B = ω0 j × vB = 10 j × ( −80)i = (800 in./s 2 )k

Acceleration of collar A.

a A = aA j a A = a B + a A/B a A/B = α AB × rA/B + ω AB × vA/B

where

Noting that α AB × rA/B is perpendicular to rA/B , we get rA/B ⋅ α AB × rA/B = 0. rA/B ⋅ ω AB × vA/B = vA/B ⋅ rA/B × ω A/B

We note also that

= − vA/B ⋅ vA/B = −(vA/B )2

rA/B ⋅ a A/B = 0 − (vA/B ) 2

Then

= −(vA/B ) 2

Forming rA/B ⋅ a A , we get

rA/B ⋅ a A = rA/B ⋅ (a A + a A/B ) = rA/B ⋅ a B + rA/B − a A/B

or From Eq. (2)

rA/B ⋅ a A = rA/B ⋅ a B − (vA/B )2

(2)

(−12i + 21j + 16k ) ⋅ (a A j) = ( −12i + 21j + 16k ) ⋅ (800k ) − 8489.8 21a A = 12,800 − 8489.8

a A = (205 in./s 2 ) j 

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PROBLEM 15.218 In Problem 15.208, determine the acceleration of collar A. PROBLEM 15.208 Rod AB of length 300 mm is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves toward Point D at a constant speed of 50 mm/s, determine the velocity of collar A when c = 80 mm.

SOLUTION Geometry.

rA = yj, rD = (90 mm)i rC = (180 mm)k rD /C = rD − rC = (90 mm)i − (180 mm)k lCD = (90)2 + (180) 2 = 201.246 mm c(rD / C ) 80(90i − 180k ) = = (40 mm)i − (80 mm)k 180 180 rB = rC + rB /C = 180k + 40i − 80k = (40 mm)i + (100 mm)k

rB /C =

rA /B = rA − rB = −(40 mm)i + ( y mm) j − (100 mm)k 2 l AB = x A2 / B + y 2 + z A2 / B : 300 2 = (−40) 2 + y 2 + (−100) 2

y = 280 mm, rA /B = ( −40 mm)i + (280 mm) j − (100 mm)k

Velocity of collar B.

vB = vB

rD/C lCD

(50)(90i − 180k ) 201.246 = (22.3607 mm/s)i − (44.7214 mm/s)k

vB =

Velocity of collar A.

vA = vA j vA = vB + vA/B

where

vA/B = ω AB × rA/B

Noting that vA/B is perpendicular to rA/B , we get rB/A ⋅ vB/A = 0.

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PROBLEM 15.218 (Continued)

Forming rA/B ⋅ vA , we get

rA/B ⋅ vA = rA/B ⋅ ( vB + vA/B ) = rA/B ⋅ v B + rA/B ⋅ vA/B rA/B ⋅ vA = rA/B ⋅ vB

or From (1)

(1)

(−40i + 280 j − 100k ) ⋅ (v A j) = (−40i + 280 j − 100k ) ⋅ (22.3607i − 44.7214k ) 280v A = (−40)(22.3607) + (−100)( −44.7214) v A = 12.7775 mm/s

or

vA = (12.7775 mm/s) j

vA/B = vA − vB

Relative velocity.

= (12.7775 mm/s)i − (22.3607 mm/s) j + (44.7214 mm/s)k vA/B ⋅ vA/B = (12.7775) 2 + (22.3607)2 + (44.7214)2 = 2663.3 mm 2 /s 2

Acceleration of collar B.

aB = 0

Acceleration of collar A.

a A = aA j a A = a B + aA/B a A/B = α AB × rA/B + ω AB × vA/B

where

Noting that α AB × rA/B is perpendicular to rA/B , we get rA/B ⋅ α AB × rA/B = 0 rA/B ⋅ ω AB × vA/B = vA/B ⋅ rA/B × ωA/B

We note also that

= − vA/B ⋅ vA/B = −(vA/B ) 2

rA/B ⋅ aA/B = 0 − (vA/B ) 2

Then

= −(vA/B ) 2

Forming rA/B ⋅ a A , we get

rA/B ⋅ a A = rA/B ⋅ (a A + a A/B ) = rA/B ⋅ a B + rA/B ⋅ a A/B

or From Eq. (2)

rA/B ⋅ a A = rA/B ⋅ a B − (vA/B )2

(2)

(−40i + 280 j − 100k ) ⋅ (a A j) = 0 − 2663.3 280a A = −2663.3

a A = −9.512 mm/s 2 a A = −(9.51 mm/s 2 ) j 

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PROBLEM 15.219 In Problem 15.209, determine the acceleration of collar A. PROBLEM 15.209 Rod AB of length 300 mm is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing that collar B moves toward Point D at a constant speed of 50 mm/s, determine the velocity of collar A when c = 120 mm.

SOLUTION Geometry.

rA = yj, rD = (90 mm)i rC = (180 mm)k rD /C = rD − rC = (40 mm)i − (180 mm)k lCD = (90)2 + ( −180)2 = 201.246 c(rD /C ) 120(90i − 180k ) = = (60 mm)i − (120 mm)k 180 180 rB = rC + rB /C = 180k + 60i − 120k = (60 mm)i + (60 mm)k

rB /C =

rA /B = rA − rB = −60i + y j − 60k 2 l AB = x A2 / B + y 2 + z 2A /B : 3002 = 602 + y 2 + 602

y = 287.75 mm, rA / B = ( −60 mm)i + (287.75 mm) j − (60 mm)k

Velocity of collar B.

vB = vB

rD/C lCD

(50)(90i − 180k ) 201.246 = (22.3607 mm/s)i − (44.7214 mm/s)k

vB =

Velocity of collar A.

vA = v A j vA = v B + vA/B

where

vA/B = ω AB × rA/B

Noting that vA/B is perpendicular to rA/B , we get rB/A ⋅ v B/A = 0.

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PROBLEM 15.219 (Continued)

Forming rA/B ⋅ vA , we get

rA/B ⋅ vA = rA/B ⋅ ( vB + vA/B ) = rA/B ⋅ vB + rA/B ⋅ vA/B rA/B ⋅ vA = rA/B ⋅ vB

or From Eq. (1)

(1)

(−60i + 287.75 j − 60k ) ⋅ (v A j) = (−60i + 287.75 j − 60k ) ⋅ (1.11803i − 2.23607 j) 287.75v A = (−60)(22.3607) + (−60)(−44.7214) v A = 4.6626 mm/s

or

vA = (4.6626 mm/s)j

vA/B = vA − v B

Relative velocity.

= −(22.3607 mm/s)i + (4.6626 mm/s) j + (44.7214 mm/s)k vA/B ⋅ vA/B = (22.3607) 2 + (4.6626)2 + (44.7214) 2 = 2521.7

Acceleration of collar B.

aB = 0

Acceleration of collar A.

a A = aA j a A = a B + a A/B a A/B = α AB × rA/B + ω AB × vA/B

where

Noting that α AB × rA/B is perpendicular to rA/B , we get rA/B ⋅ α AB × rA/B = 0. rA/B ⋅ ω AB × vA/B = vA/B ⋅ rA/B × ω A/B

We note also that

= − vA/B ⋅ vA/B = −(vA/B )2

rA/B ⋅ a A/B = 0 − (vA/B ) 2

Then

= −(v A/B )2

Forming rA/B ⋅ a A , we get

rA/B ⋅ a A = rA/B ⋅ (a A + a A/B ) = rA/B ⋅ a B + rA/B ⋅ a A/B

or From Eq. (2),

rA/B ⋅ a A = rA/B ⋅ a B − (vA/B )2

(2)

(−60i + 287.75 j − 60k ) ⋅ ( a A j) = 0 − 2521.7 287.75a A = −2521.7

a A = −8.764 mm/s 2 a A = −(8.76 mm/s 2 ) j 

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PROBLEM 15.220 A square plate of side 18 in. is hinged at A and B to a clevis. The plate rotates at the constant rate ω 2 = 4 rad/s with respect to the clevis, which itself rotates at the constant rate ω 1 = 3 rad/s about the Y axis. For the position shown, determine (a) the velocity of Point C, (b) the acceleration of Point C.

SOLUTION Geometry.

rC = (18 in.)(cos 20°i − sin 20° j)

 = 0. Then Let frame Oxyz rotate about the y axis with angular velocity Ω = ω 1 j and angular acceleration Ω the motion relative to the frame consists of rotation with angular velocity ω2 = ω2 k and angular acceleration α 2 = 0 about the z axis.

(a)

vC ′ = Ω × rC = 3j × (18cos 20°i − 18sin 20° j) = −54 cos 20°k vC/F = ω2 × rC = 4k × (18cos 20°i − 18sin 20° j) = 72sin 20°i + 72cos 20° j vC = vC ′ + vC/F = 72sin 20°i + 72cos 20° j − 54 cos 20°k

vC = (24.6 in./s)i + (67.7 in./s) j − (50.7 in./s)k 

(b)

aC ′ = Ω × vC ′ = 3 j × (−54cos 20°k ) = −162 cos 20°i aC/F = ω2 × vC/F = 4k × (72sin 20°i + 72 cos 20° j) = − 288cos 20°i + 288sin 20° j 2Ω × vC/F = (2)(3j) × (72sin 20°i + 72 cos 20° j) = − 432sin 20°k aC = aC ′ + aC/F + 2Ω × vC/F = −(162 + 288) cos 20°i + 288sin 20° j − 432sin 20°k aC = −(423 in./s 2 )i + (98.5 in./s 2 ) j − (147.8 in./s2 )k 

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PROBLEM 15.221 A square plate of side 18 in. is hinged at A and B to a clevis. The plate rotates at the constant rate ω 2 = 4 rad/s with respect to the clevis, which itself rotates at the constant rate ω 1 = 3 rad/s about the Y axis. For the position shown, determine (a) the velocity of corner D, (b) the acceleration of corner D.

SOLUTION Geometry.

rD = (18 in.)(cos 20°i − sin 20° j) + (9 in.)k

 = 0. Then Let frame Oxyz rotate about the y axis with angular velocity Ω = ω1 j and angular acceleration Ω the motion relative to the frame consists of rotation with angular velocity ω2 = ω2 k and angular acceleration α 2 = 0 about the z axis.

(a)

vD′ = Ω × rD = 3j × (18cos 20°i − 18sin 20° j + 9k ) = 27i − 54 cos 20°k vD/F = ω2 × rD = 4k × (18cos 20°i − 18sin 20° j + 9k ) = 72sin 20°i + 72cos 20° j vD = vD′ + vD/F = (27 + 72sin 20°)i + 72cos 20° j − 54cos 20°k vD = (51.6 in./s)i + (67.7 in./s) j − (50.7 in./s)k 

(b)

a D′ = Ω × vD′ = 3j × (27i − 54cos 20°k ) = −162 cos 20°i − 81k a D/F = ω2 × vD/F = 4k × (72sin 20°i + 72 cos 20° j) = − 288cos 20°i + 288sin 20° j 2Ω × vD/F = (2)(3j) × (72sin 20°i + 72cos 20° j) = − 432sin 20°k a D = a D′ + a D/F + 2Ω × vC/F = − (162 + 288) cos 20°i + 288sin 20° j − (81 + 432sin 20°)k a D = − (423 in./s 2 )i + (98.5 in./s2 ) j − (229 in./s2 )k 

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PROBLEM 15.222 The rectangular plate shown rotates at the constant rate ω 2 = 12 rad/s with respect to arm AE, which itself rotates at the constant rate w1 = 9 rad/s about the Z axis. For the position shown, determine the velocity and acceleration of the point of the plate indicated. Corner B.

SOLUTION Geometry. With the origin at A,

rB = (0.135 m) j

Let frame AXYZ rotate about the Y axis with constant angular velocity Ω = ω 1k = (9 rad/s)k. Then the motion relative to the frame consists of rotation about the X axis with constant angular velocity ω2 = ω2 i = (12 rad/s)i. Motion of coinciding Point B ′.

vB′ = Ω × rB = 9k × 0.135 j = −(1.215 m/s)i a B′ = α × rB + Ω × vB′ = 0 + 9k × (−1.215i ) = −(10.935 m/s 2 ) j

Motion relative to the frame.

vB/F = ω2 × rB = 12i × 0.135 j a B/F

= (1.62 m/s)k = α 2 × rB + ω2 × vB/F = 0 + 12i × 1.62k = −(19.44 m/s 2 ) j

Velocity of Point B. Coriolis acceleration.

vB = vB′ + vB/F

vB = − (1.215 m/s)i + (1.620 m/s)k 

2Ω × vB/F 2Ω × vB/F = (2)(9k ) × 1.62k = 0

Acceleration of Point B.

a B = a B′ + a B/F + 2Ω × vB/F a B = − (30.375 in./s 2 ) j

a B = − (30.4 m/s 2 ) j 

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PROBLEM 15.223 The rectangular plate shown rotates at the constant rate ω 2 = 12 rad/s with respect to arm AE, which itself rotates at the constant rate ω1 = 9 rad/s about the Z axis. For the position shown, determine the velocity and acceleration of the point of the plate indicated. Corner C.

SOLUTION Geometry. With the origin at A,

rC = (0.135 m) j + (0.09 m)k

Let frame AXYZ rotate about the Y axis with constant angular velocity Ω = ω 1k = (9 rad/s)k. Then the motion relative to the frame consists of rotation about the X axis with constant angular velocity ω2 = ω 2 i = (12 rad/s)i. Motion of coinciding Point C ′ in the frame. vC ′ = Ω × rC

aC ′

= 9k × (0.135 j + 0.09k ) = −(1.215 m/s)i = α × rC + Ω × vC ′ = 0 + 9k × (1.215i ) = −(10.935 m/s 2 ) j

Motion relative to the frame. vC/F = ω2 × rC

a C /F

= 12i × (0.135 j + 0.09k ) = − (1.08 m/s) j + (1.62 m/s)k = α 2 × rC + ω2 × vC/F = 0 + 12i × (−1.08 j + 1.62k ) = − (19.44 m/s 2 ) j − (12.96 m/s 2 )k

Velocity of Point C.

vC = vC ′ + vC/F vC = −(1.215 m/s)i − (1.080 m/s) j + (1.620 m/s)k 

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PROBLEM 15.223 (Continued)

Coriolis acceleration.

2Ω × vC/F 2Ω × vC/F = (2)(9k ) × (−1.08 j + 1.62k ) = (19.44 m/s 2 )i

Acceleration of Point C.

aC = aC ′ + aC/F + 2Ω × vC/F aC = (19.44 m/s 2 )i − (30.375 m/s 2 ) j − (12.96 m/s 2 )k aC = (19.44 m/s 2 )i − (30.4 m/s 2 ) j − (12.96 m/s 2 )k 

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PROBLEM 15.224 Rod AB is welded to the 0.3-m-radius plate, which rotates at the constant rate ω 1 = 6 rad/s. Knowing that collar D moves toward end B of the rod at a constant speed u = 1.3 m, determine, for the position shown, (a) the velocity of D, (b) the acceleration of D.

SOLUTION Geometry.

rB/ A = (0.6 m)i + (0.25 m)j

rC /A = (0.3 m)i

0.5 rB/ A 0.6 rD/C = rD/ A − rC/ A

rD/ A =

= (0.2 m)i + (0.20833 m) j l AB = 0.62 + 0.252 = 0.65 m

Unit vector along AB:

λ AB = =

rB/ A l AB 12 5 i+ j 13 12

Let Oxyz be a frame of reference currently coinciding with OXYZ, but rotating with angular velocity Ω = ω 1 j = (6 rad/s) j

(a)

Velocity of D.

v D = v D′ + v D/ AB v D′ = Ω × rD/C = 6 j × (0.2i + 0.20833 j) = −(1.2 m/s)k

v D/ AB = uλ AB 5   12 = 1.3  i + j   13 13  = (1.2 m/s)i + (0.5 m/s) j v D = (1.2 m/s)i + (0.5 m/s) j − (1.2 m/s)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1325

PROBLEM 15.224 (Continued)

(b)

Acceleration of D.

a D = a D′ + a D/ AB + 2Ω × v D/ AB a D′ = ω1 × (ω1 × rD/C ) = 6 j × (6 j × (0.2i + 0.20833j)) = −(7.2 m/s 2 )i a D/ AB = 0 2Ω × v D/F = (2)(6 j) × ((1.2)i + (0.5) j) = −(14.4 m/s 2 )k

a D = −(7.2 m/s 2 )i − (14.4 m/s 2 )k 

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PROBLEM 15.225 The bent rod ABC rotates at the constant rate ω 1 = 4 rad/s. Knowing that collar D moves downward along the rod at a constant relative speed u = 65 in./s, determine, for the position shown, (a) the velocity of D, (b) the acceleration of D.

SOLUTION Units: inches, in./s, in./s2 Geometry.

rE = 3k rB = −12 j + 8k rB/E = −12 j + 5k rD =

1 (rE + rB ) = −6 j + 5.5k 2

lEB = 122 + 52 = 13

Unit vector along EB:

λ=

rB/E lEB

=−

12 5 j+ k 13 13

Use a rotating frame of reference that rotates with angular velocity Ω = ω1 = (4 rad/s) j

Motion of Point D′ in the frame currently at D. v D′ = ω1 × rD = 4 j × (−6 j + 5.5k ) = (22 in./s)i  1 j × rD + ω1 × v D′ a D′ = ω = 0 + (4 j) × (22i ) = −(88 in./s 2 )k

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PROBLEM 15.225 (Continued)

Motion of collar D relative to the frame. 5   12 v D/F = u λ = (65 in./s)  − j + k  13 13   = −(60 in./s) j + (25 in./s)k a D/F = 0

(a)

Velocity of D.

(Constant speed on straight path)

v D = v D′ + v D/F v D = 22i − 60 j + 25k v D = (22 in./s)i − (60 in./s) j + (25 in./s)k 

Coriolis acceleration.

2Ω × v D/F = (2)(4 j) × (−60 j + 25k ) = (200 in./s 2 )i

(b)

Acceleration of Point D.

a D = a D′ + a D/F + 2Ω × v D a D = (200 in./s 2 )i − (88 in./s 2 )k 

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PROBLEM 15.226 The bent pipe shown rotates at the constant rate ω 1 = 10 rad/s. Knowing that a ball bearing D moves in portion BC of the pipe toward end C at a constant relative speed u = 2 ft/s, determine at the instant shown (a) the velocity of D, (b) the acceleration of D.

SOLUTION With the origin at Point A,

rD = (8 in.)i + (12 in.) j − (6 in.)k rC/B = (8 in.)i − (6 in.)k ,

lBC =

82 + 62 = 10 in.

Let the frame Axyz rotate with angular velocity (a)

Velocity of D.

Ω = ω 1i = (10 rad/s)i

v D′ = Ω × rD = 10i × (8i + 12 j − 6k ) = (60 in./s) j + (120 in./s)k

u = 2 ft/s = 24 in./s, 24 (8i − 6k ) 10 = (19.2 in./s)i − (14.4 in./s)k

v D/F =

v D = v D′ + v D/F = (19.2 in./s)i + (60 in./s) j + (105.6 in./s)k v D = (1.600 ft/s)i + (5.00 ft/s) j + (8.80 ft/s)k 

(b)

Acceleration of D. a D′ = Ω × v D′ = 10i × (60 j + 120k ) = −(1200 in./s 2 ) j + (600 in./s 2 )k a D/F = 0

2Ω × v D/F = (2)(10i) × (19.2i − 14.4k ) = (288 in./s 2 ) j a D = a D′ + a D/F + 2Ω × v D/F = −(912 in./s 2 ) j + (600 in./s 2 )k a D = −(76.0 ft/s 2 ) j + (50.0 ft/s 2 )k 

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PROBLEM 15.227 The circular plate shown rotates about its vertical diameter at the constant rate ω 1= 10 rad/s. Knowing that in the position shown the disk lies in the XY plane and Point D of strap CD moves upward at a constant relative speed u = 1.5 m/s, determine (a) the velocity of D, (b) the acceleration of D.

SOLUTION Geometry.

rD/C = (0.2 m)(cos 30°i − sin 30° j) = (0.1 3 m)i − (0.1 m) j

Let frame Cxyz, which at the instant shown coincides with CXYZ, rotate with angular velocity Ω = ω1 j = (10 rad/s) j.

Motion of coinciding Point D′ in the frame. vD′ = Ω × rD/C = 10 j × (0.1 3i + 0.1j) = − ( 3 m/s)k aD′ = − Ω 2 ( r cos 30°)i = −102 (0.1 3)i = − (10 3 m/s 2 )i

Motion of Point D relative to the frame.

u = 1.5 m/s

vD/F = u (sin 30°i + cos 30° j) = (0.75 m/s)i + (0.75 3 m/s) j aD/F =

u2

ρ

⋅ (− cos 30°i + sin 30° j)

1.52 (− cos 30°i + sin 30° j) 0.2 = − (5.625 3 m/s 2 )i + (5.625 m/s 2 ) j =

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PROBLEM 15.227 (Continued)

(a)

Velocity of Point D.

vD = vD′ + vD/F vD = (0.75 m/s)i + (0.75 3 m/s) j − ( 3 m/s)k vD = (0.750 m/s)i + (1.299 m/s) j − (1.732 m/s)k 

Coriolis acceleration.

2Ω × vD/F 2Ω × vD/F = (2)(10 j) × (0.75i + 0.75 3 j) = − (15 m/s 2 )k

(b)

Acceleration of Point D.

a D = a D′ + a D/F + 2Ω × vD/F a D = − (15.625 3 m/s 2 )i + (5.625 m/s 2 ) j − (15 m/s 2 )k a D = (27.1 m/s 2 )i + (5.63 m/s 2 ) j − (15.00 m/s 2 )k 

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PROBLEM 15.228 Manufactured items are spray-painted as they pass through the automated work station shown. Knowing that the bent pipe ACE rotates at the constant rate ω 1 = 0.4 rad/s and that at Point D the paint moves through the pipe at a constant relative speed u = 150 mm/s, determine, for the position shown, (a) the velocity of the paint at D, (b) the acceleration of the paint at D.

SOLUTION Use a frame of reference CE rotating about the x-axis with angular velocity Ω = ω1 = (0.4 rad/s)i rD/F = (250 mm)(− cos 60°i + sin 60° j)

Geometry:

= −(125 mm)i + (216.51 mm) j

Motion of Point D′ fixed in the frame CE but coinciding with Point D at the instant considered. vD′ = Ω × rD/F = (0.4i ) × (−125i + 216.51j) = (86.603 mm/s)k  i×r + Ω× v a D′ = Ω D/F D′ = 0 + (0.4i ) × (86.603k ) = − (34.641 mm/s 2 ) j

Motion of D relative to the frame CE. v D/CE = u (cos 30°i + sin 30° j) = (150 mm/s)(cos 30°i + sin 30° j) = (129.90 mm/s)i + (75 mm/s) j a D/CE = u (cos 30°i + sin 30° j) +

u2 (sin 30°i − cos 30° j) R

(150) 2 (sin 30°i − cos 30° j) 250 = (45 mm/s 2 )i − (77.94 mm/s 2 ) j =0+

(a)

Velocity of D.

v D = v D′ + v D/CE vD = (129.9 mm/s)i + (75.0 mm/s) j + (86.6 mm/s)k 

Coriolis acceleration

2Ω × v D/CE 2Ω × v D/CE = (2)(0.4i ) × (129.90i + 75 j) = (60 mm/s 2 )k

(b)

Acceleration of D.

a D = a D′ + a D/CE + 2Ω × v D/CE a D = (45.0 mm/s 2 )i − (112.6 mm/s 2 ) j + (60.0 mm/s 2 )k 

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PROBLEM 15.229 Solve Problem 15.227, assuming that at the instant shown the angular velocity ω1 of the plate is 10 rad/s and is decreasing at the rate of 25 rad/s2, while the relative speed u of Point D of strap CD is 1.5 m/s and is decreasing at the rate of 3 m/s 2. PROBLEM 15.227 The circular plate shown rotates about its vertical diameter at the constant rate ω 1= 10 rad/s. Knowing that in the position shown the disk lies in the XY plane and Point D of strap CD moves upward at a constant relative speed u = 1.5 m/s, determine (a) the velocity of D, (b) the acceleration of D.

SOLUTION Geometry.

rD/C = (0.2 m)(cos 30°i − sin 30° j) = (0.1 3 m)i − (0.1 m) j

Let frame Cxyz, which at the instant shown coincides with CXYZ, rotate with angular velocity and angular acceleration Ω = ω 1 j = (10 rad/s) j.  = −(25 rad/s 2 ) j. Ω

Motion of coinciding Point D′ in the frame vD′ = Ω × rD/C = 10 j × (0.1 3i + 0.1j) = − ( 3 m/s)k  ×r aD′ = − Ω 2 ( r cos 30°)i + Ω D/C = −102 (0.1 3)i − 25 j × (0.1 3i + 0.1j) = − (10 3 m/s 2 )i + (2.5 3 m/s 2 )k

Motion of Point D relative to the frame.

u = 1.5 m/s

u = −3 m/s 2

vD/F = u (sin 30°i + cos 30° j) = (0.75 m/s)i + (0.75 3 m/s) j

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PROBLEM 15.229 (Continued) aD/F =

u2

ρ

⋅ (− cos 30°i + sin 30° j) + u (sin 30°i + cos 30° j)

1.52 (− cos 30°i + sin 30° j) − 3(sin 30°i + cos 30° j) 0.2 = − (5.625 3 m/s 2 )i + (5.625 m/s 2 ) j − (1.5 m/s 2 )i − (1.5 3 m/s 2 ) j =

(a)

Velocity of Point D.

vD = vD′ + vD/F vD = (0.75 m/s)i + (0.75 3 m/s) j − ( 3 m/s)k vD = (0.750 m/s)i + (1.299 m/s) j − (1.732 m/s)k 

Coriolis acceleration.

2Ω × vD/F 2Ω × vD/F = (2)(10 j) × (0.75i + 0.75 3 j) = − (15 m/s 2 )k

(b)

Acceleration of Point D.

a D = a D′ + a D/F + 2Ω × vD/F a D = − (10 3 m/s 2 )i + (2.5 3 m/s 2 )k − (5.625 3 m/s 2 )i + (5.625 m/s 2 ) j − (1.5 m/s 2 )i − (1.5 3 m/s 2 ) j − (15 m/s 2 )k a D = − (28.6 m/s 2 )i + (3.03 m/s 2 ) j − (10.67 m/s 2 )k 

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PROBLEM 15.230 Solve Problem 15.226, assuming that at the instant shown the angular velocity ω 1 of the pipe is 10 rad/s and is decreasing at the rate of 15 rad/s 2 , while the relative speed u of the ball bearing is 2 ft/s and is increasing at the rate of 10 ft/s 2. PROBLEM 15.226 The bent pipe shown rotates at the constant rate ω 1 = 10 rad/s. Knowing that a ball bearing D moves in portion BC of the pipe toward end C at a constant relative speed u = 2 ft/s, determine at the instant shown (a) the velocity of D, (b) the acceleration of D.

SOLUTION With the origin at Point A,

rD = (8 in.)i + (12 in.) j − (6 in.)k rC/B = (8 in.)i − (6 in.)k , lBC =

82 + 62 = 10 in.

Let the frame Axyz rotate with angular velocity Ω = ω1i = (10 rad/s)i and angular acceleration  = ω i = − (15 rad/s 2 )i. Ω 1

(a)

Velocity of D.

v D′ = Ω × rD = 10i × (8i + 12 j − 6k ) = (60 in./s) j + (120 in./s)k u = 2 ft/s = 24 in./s,

24 (8i − 6k ) 10 = (19.2 in./s)i − (14.4 in./s)k

u=

v D = v D′ + u = (19.2 in./s)i + (60 in./s) j + (105.6 in./s)k v D = (1.600 ft/s)i + (5.00 ft/s) j + (8.80 ft/s)k 

(b)

Acceleration of D.

a D′ = Ω × rD + Ω × v D′ = −15i × (8i + 12 j − 6k ) + 10i × (60 j + 120k ) = −90 j − 180k − 1200 j + 600k = − (1290 in./s 2 ) j + (420 in./s 2 )k

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PROBLEM 15.230 (Continued)

arel = 10 ft/s 2 = 120 in./s 2 a rel =

120 (8i − 6k ) = (96 in./s 2 )i − (72 in./s 2 )k 10

2Ω × u = (2)(10i) × (19.2i − 14.4k ) = (288 in./s 2 ) j a D = a D′ + a rel + 2Ω × u = (96 in./s)i − (1002 in./s) j + (348 in./s)k a D = (8.00 ft/s 2 )i − (83.5 ft/s 2 ) j + (29.0 ft/s 2 )k 

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PROBLEM 15.231 Using the method of Section 15.14, solve Problem 15.192. PROBLEM 15.192 In the system shown, disk A is free to rotate about the horizontal rod OA. Assuming that shaft OC and disk B rotate with constant angular velocities ω1 and ω 2 , respectively, both counterclockwise, determine (a) the angular velocity of A, (b) the angular acceleration of disk A.

SOLUTION Moving frame Axyz rotates with angular velocity Ω = ω1 j ωdisk/F = ω x i + ω z k rD/A = − rj − Rk

(a)

Total angular velocity of disk A: ω = ω 1 j + ωdisk/F = ωx i + ω 1 j + ωz k

(1)

Denote by D point of contact of disks Consider disk B:

v D = ω2 j × ( − Rk ) = − Rω2 i

(2)

Consider system OC, OA and disk A. v D′ = Ω × rB /A = ω1 j × ( −rj − Rk ) = − Rω1i v B /F = ωdisk/F × rD /A = (ω 2 i + ω z k ) × (−rj − Rk ) = − rω x k + Rω x j + rω z i v D = v D′ + v D /F = − Rω1i − rω x k + Rω x j + rω2 i

(3)

Equate v D = v D from Eq. (2) and Eq. (3). − Rω 2 i = − Rω1 + rω2 + Rω x j − rω x k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1337

PROBLEM 15.231 (Continued)

Coefficient of j:

0 = Rω x → ω x = 0 − Rω2 = − Rω1 + rω z ;

Coefficient of i:

ωz =

R (ω1 − ω2 ) r ω = ω1 j +

Eq. (3): (b)

R (ω1 − ω2 )k  r

Disk A rotates about y axis at rate ω1 . α = ω1 × ω R   = ω1 j × ω j + (ω1 − ω2 )  k r  

α = ω1 (ω1 − ω2 )

R j  r

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PROBLEM 15.232 Using the method of Section 15.14, solve Problem 15.196. PROBLEM 15.196 A 3-in.-radius disk spins at the constant rate ω 2 = 4 rad/s about an axis held by a housing attached to a horizontal rod that rotates at the constant rate ω 1 = 5 rad/s. Knowing that θ = 30°, determine the acceleration of Point P on the rim of the disk.

SOLUTION Let frame Oxyz rotate with angular velocity Ω = ω1i = (5 rad/s)i

The motion relative to the frame is the spin

ω2 k = (4 rad/s)k θ = 30° rP = (3 in.)(cos 30°i + sin 30° j) v P′ = ω1i × rP = 5i × (3cos 30°i + 3sin 30° j) = (7.5 in./s)k v P/F = ω2 k × rP = 4k × (3cos 30°i + 3sin 30° j) = − (6 in./s)i + (10.392 in./s) j a P′ = ω1i × rP + ω1i × v P′ = 0 + 5i × 7.5k a P/F

= −(37.5 in./s 2 )j = ω 2 k × rP = ω2 k × v P/F = 0 + 4k × (−6i + 10.392 j)

= −(41.569 in./s 2 )i − (24 in./s 2 ) j ac = 2Ω × v P/F = (2)(5i ) × (−6i + 10.392 j) = (103.92 in./s 2 )k

Acceleration at Point P.

a P = a P′ + a P/F + ac a P = −(41.6 in./s 2 )i − (61.5 in./s 2 ) j + (103.9 in./s 2 )k 

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PROBLEM 15.233 Using the method of Section 15.14, solve Problem 15.198. PROBLEM 15.198 At the instant shown, the robotic arm ABC is being rotated simultaneously at the constant rate ω1 = 0.15 rad/s about the y axis, and at the constant rate ω 2 = 0.25 rad/s about the z axis. Knowing that the length of arm ABC is 1 m, determine (a) the angular acceleration of the arm, (b) the velocity of Point C, (c) the acceleration of Point C.

SOLUTION Geometry:

Dimensions in meters. rC/A = (1.0 cos 35°)i + (1.0sin 35°) j = 0.81915i + 0.57358 j ω1 = ω 1 j = (0.15 rad/s) j

Angular velocities:

ω2 = ω2 k = (0.15 rad/s)k

(ω 1 = 0) (ω 2 = 0)

Use a frame of reference rotating about the y-axis. Its angular velocity is (a)

Ω = ω 1 j = (0.15 rad/s) j

Angular acceleration: α = ω 1 j + ω 2 k + Ω × ω 2

= 0 + 0 + (0.15 j)(0.25k ) α = (0.0375 rad/s 2 )i 

Motion of coinciding Point C. vC ′ = Ω × rC/A = 0.15 j × (0.81915i + 0.57358 j) aC ′

= −(0.12287 m/s)k = Ω × (Ω × rC/A ) = (0.15 j) × ( −0.12287k ) = − (0.018431 m/s 2 )i

Motion of C relative to the frame. vC/F = ω 2 k × rC/A = 0.25k (0.81915i + 0.57358 j) aC/ F

= − (0.14339 m/s)i + (0.20479 m/s) j = ω1k × vC/F = 0.25k × ( −0.14339i + 0.20479 j) = − (0.051198 m/s 2 )i − (0.035848 m/s 2 ) j

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PROBLEM 15.233 (Continued)

(b)

Velocity of C.

v C = v C ′ + v C /F vC − (0.143 m/s)i + (0.205 m/s) j − (0.123 m/s)k 

Coriolis acceleration.

2Ω × v C / F 2Ω × vC/F = (2)(0.15 j) × (−0.14339i + 0.20479 j) = (0.043017 m/s 2 )k

(c)

Acceleration of C.

a C = a C ′ + a C / F + 2Ω × v C / F aC = −0.01843i − 0.051198i − 0.035848 j + 0.043017k aC = − (0.0696 m/s 2 )i − (0.0358 m/s 2 ) j + (0.0430 m/s 2 )k 

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PROBLEM 15.234 A disk of radius 120 mm rotates at the constant rate ω 2 = 5 rad/s with respect to the arm AB, which itself rotates at the constant rate ω1 = 3 rad/s. For the position shown, determine the velocity and acceleration of Point C.

SOLUTION Geometry.

rC/ A = (0.195 m)i − (0.1) j rC/B = (0.12 m)i

Let frame Axyz , which coincides with the fixed frame AXYZ at the instant shown, be rotating about the y axis with constant angular velocity Ω = ω1 j = (3 rad/s) j. Then the motion relative to the frame consists of rotation about the axle B with a constant angular velocity ω2 = ω 2 k = (5 rad/s)k. Motion of the coinciding Point C ′ in the frame. vC ′ = Ω × rC/ A

aC ′

= 3j × (0.195i − 0.14 j) = −(0.585 m/s)k = Ω × vC ′ = 3j × (−0.585k ) = −(1.755 m/s 2 )i

Motion relative to the frame.

vC/F = ω2 × rC/B

a C /F

= 5k × 0.12i = (0.6 m/s) j = ω 2 × v C /F = 5k × ( −0.6 j) = −(3 m/s 2 )i

Velocity of Point C. Coriolis acceleration. Acceleration of Point C.

v C = v C ′ + v C /F

vC = (0.600 m/s) j − (0.585 m/s)k 

2Ω × v C/F = (2)(3j) × (0.6 j) = 0 a C = a C ′ + a C / F + 2Ω × v C / F aC = −1.7551i − 3i + 0

aC = −(4.76 m/s 2 )i 

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PROBLEM 15.235 A disk of radius 120 mm rotates at the constant rate ω 2 = 5 rad/s with respect to the arm AB, which itself rotates at the constant rate ω1 = 3 rad/s. For the position shown, determine the velocity and acceleration of Point D.

SOLUTION Geometry.

rD/ A = (0.075 m)i − (0.26 m) j rD/B = −(0.12 m) j

Let frame AXY Z , which coincides with the fixed frame AXYZ at the instant shown, be rotating about the y axis with constant angular velocity Ω = ω1 j = (3 rad/s) j. Then the motion relative to the frame consists of rotation about the axle B with a constant angular velocity ω2 = ω2 k = (5 rad/s)k. Motion of the coinciding Point D′ in the frame. v D′ = Ω × rD/ A

a D′

= 3j × (0.075i − 0.26 j) = −(0.225 m/s)k = Ω × v D′ = 3j × (−0.225k ) = −(0.675 m/s 2 )i

Motion relative to the frame.

v D/F = ω2 × rD/B

a D/F

= 5k × (−0.12 j) = (0.6 m/s)i = ω 2 × v C /F = 5k × (0.6i ) = (3 m/s 2 ) j

Velocity of Point D. Coriolis acceleration. Acceleration of Point D.

v D = v D′ + v D/F

v D = (0.600 m/s)i − (0.225 m/s)k 

2Ω × v D/F = (2)(3j) × (0.6i) = −(3.6 m/s 2 )k a D = a D′ + a D/F + 2Ω × v D/F a D = −(0.675 m/s 2 )i + (3.00 m/s 2 ) j − (3.60 m/s 2 )k 

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PROBLEM 15.236 The arm AB of length 16 ft is used to provide an elevated platform for construction workers. In the position shown, arm AB is being raised at the constant rate dθ /dt = 0.25 rad/s; simultaneously, the unit is being rotated about the Y axis at the constant rate ω 1 = 0.15 rad/s. Knowing that θ = 20°, determine the velocity and acceleration of Point B.

SOLUTION Frame of reference. Let moving frame Axyz rotate about the Y axis with angular velocity Ω = ω1j = (0.15 rad/s) j.

Geometry. rB/ A = −16 cos 20°i + 16sin 20° j = −(15.035 ft)i + (5.4723 ft) j

Place Point O on Y axis at same level as Point A. rB/O = rB/ A + rA/O = rB/ A + (2.5 ft)i = −(12.535 ft)i + (5.4723 ft) j

Motion of corresponding Point B′ in the frame. v B′ = Ω × rB/O = (0.15 j) × −(12.535i + 5.4723 j) = (1.8803 ft/s)k a B′ = Ω × v B′ = (0.15 j) × (1.8803k ) = (0.28204 ft/s 2 )i

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PROBLEM 15.236 (Continued)

Motion of Point B relative to the frame. dθ k dt = −(0.25 rad/s)k

ω2 = −

v B/F = ω2 × rB/ A

a B/F

= (−0.25)k × ( −15.035i + 5.4723 j) = (1.36808 ft/s)i + (3.7588 ft/s) j = ω2 × v B/F = (−0.25k ) × (1.36808i + 3.7588 j) = (0.93969 ft/s 2 )i − (0.34202 ft/s 2 ) j

Velocity of Point B.

v B = v B′ + vB/F v B = −(1.37 ft/s)i + (3.76 ft/s) j + (1.88 ft/s)k 

Coriolis acceleration.

2Ω × v B/F = (2)(0.15 j) × (1.36808i + 3.7588 j) = −(0.41042 ft/s 2 )k

Acceleration of Point B.

a B = a B′ + a B/F + 2Ω × vB/F a B = (1.22 ft/s 2 )i − (0.342 ft/s 2 ) j − (0.410 ft/s 2 )k 

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PROBLEM 15.237 The remote manipulator system (RMS) shown is used to deploy payloads from the cargo bay of space shuttles. At the instant shown, the whole RMS is rotating at the constant rate ω 1 = 0.03 rad/s about the axis AB. At the same time, portion BCD rotates as a rigid body at the constant rate ω 2 = d β /dt = 0.04 rad/s about an axis through B parallel to the X axis. Knowing that β = 30°, determine (a) the angular acceleration of BCD, (b) the velocity of D, (c) the acceleration of D.

SOLUTION At the instant given, Points A, B, C, and D lie in a plane which is parallel to the YZ plane. The plane ABCD is rotating with angular velocity. Ω = ω1j = (0.03 rad/s) j

(ω 1 = 0)

Body BCD is rotating about an axis through B parallel to the x-axis at angular velocity. ω2 =

(a)

dβ i = (0.04 rad/s)i dt

(ω 2 = 0)

Angular acceleration of BCD. α BCD = ω 1 j + ω 2 i + Ω × ω 2

= 0 + 0 + (0.03j) × (0.04i ) α BCD = − (0.0012 rad/s 2 )k 

Let the plane of BCD be a rotating frame of reference rotating about AB with angular velocity Ω. Geometry:

β = 30° rD/B = (6.5 m)(sin β j − cos β k ) + (2.5 m) j = (5.75 m) j − (5.6292 m)k

Motion of Point D′ in the frame. v D′ = Ω × rD/B = 0.03 j × (3.25 j − 5.6292k ) a D′

= − (0.168875 m/s)i = Ω × v D′ = (0.03j) × (−0.168875i ) = (0.0050662 m/s 2 )k

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PROBLEM 15.237 (Continued)

Motion of D relative to the frame: This motion is a rotation about B with angular velocity. ω2 = (0.04 rad/s)i v D/frame = ω2 × rD/B

a D/frame

= (0.04i ) × (5.75 j − 5.6292k ) = (0.22517 m/s) j + (0.23 m/s)k = ω2 × rD/frame = (0.04i ) × (0.22517 j + 0.23k ) = − (0.0092 m/s 2 ) j + (0.009007 m/s 2 )k

(b)

Velocity of D.

v D = v D′ + v D/frame v D = − (0.169 m/s)i + (0.225 m/s) j + (0.230 m/s)k 

Coriolis acceleration :

2Ω × v D/frame 2Ω × v D/frame = (2)(0.03j) × (0.22517 j + 0.23k ) = (0.0138 m/s 2 )i

(c)

Acceleration of D.

a D = a D′ + a D/frame + 2Ω × v D/ frame a D = (0.0138 m/s 2 )i − (0.0092 m/s 2 ) j + (0.0141 m/s 2 )k 

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PROBLEM 15.238 The body AB and rod BC of the robotic component shown rotate at the constant rate ω 1 = 0.60 rad/s about the Y axis. Simultaneously a wire-and-pulley control causes arm CD to rotate about C at the constant rate ω 2 = dβ /dt = 0.45 rad/s. Knowing that β = 120°, determine (a) the angular acceleration of arm CD, (b) the velocity of D, (c) the acceleration of D.

SOLUTION

Ω = ω1 ω D/F

= (0.6 rad/s) j = ω2

= −(0.45 rad/s)k ω = ω1 + ω2 = (0.6 rad/s) j − (0.45 rad/s)k

(a)

Angular acceleration of CD. α = Ω×ω = (0.6 rad/s) j × [(0.6 rad/s) j − (0.45 rad/s)k ] α = −(0.27 rad/s 2 )i 

For β = 120° :

rD /C = (400 mm) sin 30°i + (400 mm) cos 30° j rD /B

= (200 mm)i + (346.41 mm) j = (500 mm)i + rD /C = (700 mm)i + (346.41 mm) j

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PROBLEM 15.238 (Continued)

(b)

v D = v D′ + v D/F

Velocity of D.

v D = Ω × rD /B = (0.6 rad/s) j × [(700 mm)i + (346.41 mm) j] = − (420 mm/s)k v D /F = ω D /F × rD /C = −(0.45 rad/s)k × [(200 mm)i + (346.41 mm) j] = −(90 mm/s) j + (155.88 mm/s)i v D = v D′ + v D /F :

(c)

Acceleration of D.

v D = (156 mm/s)i − (90 mm/s) j − (420 mm/s)k 

a D = a D′ + a D/F + ac

a D′ = Ω × (Ω × rD /B ) = Ω × v D′ = (0.6 rad/s) j × ( −420 mm/s)k = −(252 mm/s 2 )i a D /F = ω D /F × (ω D /F × rD /C ) = ω D /F × v D /F = −(0.45 rad/s)k × [−(90) j + (155.88)i] = −(40.5 mm/s 2 )i − (70.148 mm/s 2 ) j ac = 2Ω × v D /F = 2(0.6 rad/s) j × [−(90) j + (155.88)i ] = −(187.06 mm/s 2 )k a D = a D′ + a D /F + ac = −(252 mm/s 2 )i − (40.5 mm/s 2 )i − (70.148 mm/s 2 ) j − (187.06 mm/s 2 )k a D = −(293 mm/s 2 )i − (70.1 mm/s 2 ) j − (187 mm/s 2 )k 

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PROBLEM 15.239 The crane shown rotates at the constant rate ω1 = 0.25 rad/s; simultaneously, the telescoping boom is being lowered at the constant rate ω 2 = 0.40 rad/s. Knowing that at the instant shown the length of the boom is 20 ft and is increasing at the constant rate u = 1.5 ft/s, determine the velocity and acceleration of Point B.

SOLUTION rB/ A = rB

Geometry.

= (20 ft)(sin 30° j + cos 30°k ) = (10 ft) j + (10 3 ft)k

Method 1

Let the unextending portion of the boom AB be a rotating frame of reference. Ω = ω 2i + ω1j

Its angular velocity is

= (0.40 rad/s)i + (0.25 rad/s) j.

Its angular acceleration is

α = ω1j × ω 2i = −ω 1ω 2k = −(0.10 rad/s 2 )k.

Motion of the coinciding Point B′ in the frame. v B′ = Ω × rB = (0.40i + 0.25 j) × (10 j + 10 3k ) = (2.5 3 ft/s)i − (4 3 ft/s) j + (4 ft/s)k a B′ = α × rB + Ω × v B′ i = 0

j 0

k i −0.10 + 0.40

0 10 10 3

j 0.25

k 0

2.5 3 −4 3

4

i + i − 1.6 j − 3.8538k = (2 ft/s )i − (1.6 ft/s ) j − (3.8538 ft/s 2 )k 2

Motion relative to the frame.

2

v B/F = u (sin 30° j + cos 30°k ) a B/F

= (1.5 ft/s) sin 30° j + (1.5 ft/s) cos30°k =0

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1350

PROBLEM 15.239 (Continued)

Velocity of Point B.

v B = v B′ + v B/F v B = 2.5 3i − 4 3 j + 4k + 1.5sin 30° j + 1.5cos 30°k v B = (4.33 ft/s)i − (6.18 ft/s) j + (5.30 ft/s)k 

Coriolis acceleration.

2Ω × v B/F 2Ω × v B/F = (2)(0.40i + 0.25 j) × (1.5sin 30° j + 1.5cos 30°k ) = (0.64952 ft/s 2 )i − (1.03923 ft/s 2 ) j + (0.6 ft/s 2 )k

Acceleration of Point B.

a B = a B′ + a B/F + 2Ω × v B/F a B = (2 + 0.64952)i − (1.6 + 1.03923j) + ( −3.8538 + 0.6)k a B = (2.65 ft/s 2 )i − (2.64 ft/s 2 ) j − (3.25 ft/s 2 )k 

Method 2

Let frame Axyz , which at the instant shown coincides with AXYZ, rotate with an angular velocity Ω 1 = ω 1 j = (0.25 rad/s)j. Then the motion relative to this frame consists of turning the boom

relative to the cab and extending the boom. Motion of the coinciding Point B′ in the frame. v B′ = Ω × rB = 0.25 j × (10 j + 10 3k ) a B′

= (2.5 3 m/s)i = Ω × v B′ = 0.25 j × (2.5 3i ) = −(0.625 3 m/s 2 )k

Motion of Point B relative to the frame.

Let the unextending portion of the boom be a rotating frame with constant angular velocity Ω 2 = ω 2 i = (0.40 rad/s)i. The motion relative to this frame is the extensional motion with speed u. v B′′ = Ω 2 × rB = 0.40i × (10 j + 10 3k ) a B′′

= −(4 3 ft/s) j + (4 ft/s)k = Ω 2 × v B′′ = 0.40i × (−4 3 j + 4k ) = −(1.6 ft/s 2 ) j − (1.6 3 ft/s 2 )k

v B/boom = u (sin 30° j + cos 30°k ) a B/boom

= (1.5 ft/s)sin 30° j + (1.5 ft/s) cos 30°k =0

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PROBLEM 15.239 (Continued)

2Ω 2 × v B /boom = (2)(0.40i) × (1.5sin 30° j + 1.5cos 30°k ) v B /F

= −(1.03923 ft/s 2 ) j + (0.6 ft/s 2 )k = v B′′ + v B/boom = −4 3 j + 4k + 1.5sin 30° j + 1.5cos 30°k

a B/F

= −(6.1782 ft/s) j + (5.299 ft/s)k = a B′′ + a B/boom + 2Ω 2 × v B/boom = −1.6 j − 1.6 3k + 0 − 1.03923j + 0.6k = −(2.6392 ft/s 2 ) j − (2.1713 ft/s 2 )k

Velocity of Point B.

v B = v B′ + v B/F v B = 2.5 3i − 6.1782 j + 5.299k v B = (4.33 ft/s)i − (6.18 ft/s) j + (5.30 ft/s)k 

Coriolis acceleration.

2Ω 1 × v B/F 2Ω1 × v B /F = (2) (0.25 j) × (−6.1782 j + 5.299k ) = (2.6495 ft/s 2 )i

Acceleration of Point B.

a B = a B′ + a B/F + 2Ω 1 × vB/F a B = −0.625 3k − 2.6392 j − 2.1713k + 2.6495i a B = (2.65 ft/s 2 )i − (2.64 ft/s 2 ) j − (3.25 ft/s 2 )k 

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PROBLEM 15.240 The vertical plate shown is welded to arm EFG, and the entire unit rotates at the constant rate ω1 = 1.6 rad/s about the Y axis. At the same time, a continuous link belt moves around the perimeter of the plate at a constant speed u = 4.5 in./s. For the position shown, determine the acceleration of the link of the belt located (a) at Point A, (b) at Point B.

SOLUTION Let the moving frame of reference be the unit, less the pulleys and belt. It rotates about the Y axis with constant angular velocity Ω = ω 1 j = (1.6 rad/s)j. The relative motion is that of the pulleys and belt with speed u = 90 mm/s.

(a)

Acceleration at Point A. rA = −(5 in.)i + (19 in.)j vA′ = Ω × rA = 1.6 j × (−5i + 19 j) = (8 in./s)k a A′ = Ω × v A′ = 1.6 j × 8k vA/F

= (12.8 in./s 2 )i = uk = (4.5 in./s)k

 u2  a A/F = −   j  ρ   4.52  = −   j  3 

(

)

= − 6.75 in./s 2 j 2Ω × vA/F = (2)(1.6 j) × (4.5k ) = (14.4 in./s 2 )i a A = a A′ + a A/F + 2Ω × vA/F = 12.8i − 6.75 j + 14.4i

a A = (27.2 in./s 2 )i − (6.75 in./s 2 ) j 

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PROBLEM 15.240 (Continued)

(b)

Acceleration of Point B. rB = −(5 in.)i + (10 in.)j + (3 in.)k vB′ = Ω × rB = 1.6 j × (−5i + 10 j + 3k ) = (4.8 in./s)i + (8 in./s)k a B′ = Ω × vB′ = 1.6 j × (4.8i + 8k ) = (12.8 in./s 2 )i − (7.68 in./s 2 )k vB/F = −uj a B/F

= −(3 in./s)j =0

2Ω × vB/F = (2)(1.6 j) × (4.5 j) = 0 a B = a B′ + a B/F + 2Ω × vB/F = 12.8i − 7.68k + 0 + 0

a B = (12.80 in./s 2 )i − (7.68 in./s 2 )k 

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PROBLEM 15.241 The vertical plate shown is welded to arm EFG, and the entire unit rotates at the constant rate ω1 = 1.6 rad/s about the Y axis. At the same time, a continuous link belt moves around the perimeter of the plate at a constant speed u = 4.5 in./s. For the position shown, determine the acceleration of the link of the belt located (a) at Point C, (b) at Point D.

SOLUTION Let the moving frame of reference be the unit, less the pulleys and belt. It rotates about the Y axis with constant angular velocity Ω = ω 1 j = (1.6 rad/s)j. The relative motion is that of the pulleys and belt with speed u = 90 mm/s.

(a)

Acceleration at Point C. rC = −(5 in.)i + (4 in.)j vC ′ = Ω × rA = 1.6 j × ( −5i + 4 j) = (8 in./s)k aC ′ = Ω × vA′ = 1.6 j × 8k vC/F

= (12.8 in./s 2 )i = −uk = −(4.5 in./s)k

 u2  aC/F =   j  ρ   4.52  =   j  3  = (6.75 in./s 2 ) j 2Ω × vC/F = (2)(1.6 j) × (−4.5k )

= −(14.4 in./s 2 )i aC = aC ′ + aC/F + 2Ω × vC/F = 12.8i + 6.75 j − 14.4i

aC = −(1.600 in./s 2 )i + (6.75 in./s 2 ) j 

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PROBLEM 15.241 (Continued)

(b)

Acceleration at Point D. rD = −(5 in.)i + (10 in.)j − (3 in.)k vD′ = Ω × rB

a D′

= 1.6 j × (−5i + 10 j − 3k ) = −(4.8 in./s)i + (8 in./s)k = Ω × vB′ = 1.6 j × (−4.8i + 8k )

vD /F

= (12.8 in./s 2 )i + (7.68 in./s 2 )k = uj = (4.5 in./s)j

a D/F = 0 2Ω × vD/F = (2)(1.6) × (−4.5 j) = 0 a D = a D′ + a D/F + 2Ω × vD/F = 12.8i + 7.68k + 0 + 0

a D = (12.80 in./s 2 )i + (7.68 in./s 2 )k 

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PROBLEM 15.242 A disk of 180-mm radius rotates at the constant rate ω 2 = 12 rad/s with respect to arm CD, which itself rotates at the constant rate ω1 = 8 rad/s about the Y axis. Determine at the instant shown the velocity and acceleration of Point A on the rim of the disk.

SOLUTION Geometry.

rA/D = (0.15 m)i + (0.18 m) j − (0.36 m)k rA/C = (0.18 m) j

Let frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the y axis with constant angular velocity Ω = ω1 j = (8rad/s) j. Then the motion relative to the frame consists of a rotation of the disk AB about the bent axle CD with constant angular velocity ω2 = ω2 k = (12 rad/s)k. Motion of the coinciding Point A′ in the frame. v A′ = Ω × rA/D = 8 j × (0.15i + 0.18 j − 0.36k ) = −(2.88 m/s)i − (1.2 m/s)k a A′ = Ω × v A′ = 8 j × (−2.88i − 1.2k ) = −(9.6 m/s 2 )i + (23.04 m/s 2 )k

Motion of Point A relative to the frame. vA/F = ω 2 × rA/D = 12k × 0.18 j = −(2.16 m/s)i a A/F = ω 2 × v A/F = 12k × (−2.16i ) = −(25.92 m/s 2 ) j

Velocity of Point A.

vA = vA′ + vA/F vA = −2.88i − 1.2k − 2.16i vA = −(5.04 m/s)i − (1.200 m/s)k 

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PROBLEM 15.242 (Continued)

Coriolis acceleration.

2Ω × vA/F 2Ω × vA/F = (2)(8 j) × (−2.16i ) = (34.56 m/s 2 )k

Acceleration of Point A.

a A = a A′ + a A/F + 2Ω × vA/F a A = −9.6i + 23.04k − 25.92 j + 34.56k a A = −(9.60 m/s 2 )i − (25.9 m/s 2 ) j + (57.6 m/s 2 )k 

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PROBLEM 15.243 A disk of 180-mm radius rotates at the constant rate ω 2 = 12 rad/s with respect to arm CD, which itself rotates at the constant rate ω1 = 8 rad/s about the Y axis. Determine at the instant shown the velocity and acceleration of Point B on the rim of the disk.

SOLUTION Geometry.

rB/D = (0.15 m)i − (0.18 m) j − (0.36 m)k rB/C = −(0.18 m) j

Let frame Dxyz, which coincides with the fixed frame DXYZ at the instant shown, be rotating about the Y axis with constant angular velocity Ω = ω1 j = (8 rad/s) j . Then the motion relative to the frame consists of a rotation of the disk AB about the bent axle CD with constant angular velocity ω2 = ω2 k = (12 rad/s)k. Motion of the coinciding Point B′ in the frame. vB′ = Ω × rB/D = 8 j × (0.15i − 0.18 j − 0.36k ) = −(2.88 m/s)i − (1.2 m/s)k a B′ = Ω × v B′ = 8 j × (−2.88i − 1.2k ) = −(9.6 m/s 2 )i + (23.04 m/s 2 )k

Motion of Point B relative to the frame. vB/F = ω2 × rB/D = 12k × (−0.18 j) = (2.16 m/s)i a B/F = ω2 × vB/F = 12k × 2.16i = (25.92 m/s 2 ) j

Velocity of Point B.

vB = vB′ + vB/F vB = −2.88i − 1.2k + 2.16i vB = −(0.720 m/s)i − (1.200 m/s)k 

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PROBLEM 15.243 (Continued)

Coriolis acceleration.

2Ω × v B/F 2Ω × v B/F = (2)(8 j) × (2.16i ) = −(34.56)k

Acceleration of Point B.

a B = a B′ + a B/F + 2Ω × vB/F a B = −9.6i + 23.04k + 25.92 j − 34.56k a B = −(9.60 m/s 2 )i + (25.9 m/s 2 ) j − (11.52 m/s 2 )k 

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PROBLEM 15.244 A square plate of side 2r is welded to a vertical shaft which rotates with a constant angular velocity ω 1. At the same time, rod AB of length r rotates about the center of the plate with a constant angular velocity ω 2 with respect to the plate. For the position of the plate shown, determine the acceleration of end B of the rod if (a) θ = 0, (b) θ = 90°, (c) θ = 180°.

SOLUTION Use a frame of reference moving with the plate. Its angular velocity is Ω = ω 1 j

 = 0) (Ω

Geometry :

rA/O = r (sin 30° j − cos 30°k ) rB/O = rA/O + rB/A

Acceleration of coinciding Point B′ in the frame. a B′ = Ω × (Ω × rB/O )

Motion relative to the frame. (Rotation about A with angular velocity ω 2 ). ω 2 = ω2 (cos 30° j + sin 30°k )

(ω 2 = 0)

v B/F = ω 2 × rB/A a B/F = ω 2 × v B/F = −ω22rB/A

Coriolis acceleration:

2Ω × v B/F

Acceleration of B.

a B = a B′ + a B /F + 2Ω × v B /F

θ =0

rB /A = r (− sin 30° j + cos 30°k )

(a)

rB /O = 0 a B′ = 0 v B /F = rω 2 i a B /F = −ω 22 rB /A = rω22 (sin 30° j − cos 30°k ) 2Ω × v B /F = 2(ω 1 j) × ( rω 2 i) = −2rω 1ω 2k a B = a B′ + a B /F + aC = 0 + rω 22 (sin 30° j − cos 30°k ) − 2rω 1ω 2k a B = rω 22 sin 30° j − ( rω 22 cos 30° + 2rω 1ω 2 )k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1361

PROBLEM 15.244 (Continued)

(b)

θ = 90° rB/ A = r i rB/O = r i + r sin 30° j − r cos 30°k

aB′ = Ω × (Ω × rB/O ) = ω 1 j × [ω 1 j × (r i + r sin 30° j − r cos 30°k )] = − rω 12 i + rω 12 cos 30°k v B /F = rω 2 (sin 30° j − cos 30°k ) a B /F = − rω 22 i 2Ω × v B /F = 2(ω1 j) × ( rω 2 )(sin 30° j − cos 30°k ) = −2rω 1ω 2 cos 30°i a B = a B′ + a B/F + aC = [− rω 12 i + rω12 cos 30°k ] − rω 22 i − 2rω 1ω 2 cos 30°i a B = −r (ω 12 + ω 22 + 2ω1ω 2 cos 30°)i + rω 12 cos 30°k 

(c)

θ = 180° rB/ A = r (sin 30° j − cos 30°k ) rB/O = 2r (sin 30° j − cos 30°k ) a B′ = +(2r cos 30°)ω 12 k = +2rω 12 cos 30°k v B/F = −rω2 i a B/F = rω 22 (− sin 30° j + cos 30°k ) 2Ω × v B/F = 2(ω 1 j) × (− rω2 i ) = 2rω 1ω 2 k a B = a B′ + a B/F + 2Ω × v B/F = 2rω 12 cos 30°k + rω 22 (− sin 30° j + cos 30°k ) + 2rω 1ω 2 k a B = −rω22 sin 30° j + r (2ω12 cos 30° + ω22 cos 30° + 2ω1ω2 )k 

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PROBLEM 15.245 Two disks, each of 130-mm radius, are welded to the 500-mm rod CD. The rod-and-disks unit rotates at the constant rate ω 2 = 3 rad/s with respect to arm AB. Knowing that at the instant shown ω 1 = 4 rad/s, determine the velocity and acceleration of (a) Point E, (b) Point F.

SOLUTION Let the frame of reference BXYZ be rotating about the Y axis with angular velocity Ω = ω2 j = (4 rad/s) j . The motion relative to this frame is a rotation about the X axis with angular velocity ω x i = (3 rad/s)i. (a)

Point E.

rE/B = (0.25 m)i + (0.13 m) j rE/D = (0.13 m) j

Motion of Point E′ in the frame. vE ′ = Ω × rE/B = 4 j × (0.25i + 0.13j) = −(1 m/s)k a E ′ = Ω × vE ′ = 4 j × ( −k ) = −(4 m/s 2 )i

Motion of Point E relative to the frame. vE/F = ω x i × rE/D

a E/F

= 3i × 0.13j = (0.39 m/s)k = ω x i × vE/F = 3i × (0.39k ) = −(1.17 m/s 2 ) j

Coriolis acceleration.

ac = 2Ω × vE /F ac = (2)(4 j) × (0.39k ) = (3.12 m/s 2 )i

Velocity of Point E.

vE = vE ′ + vE/F vE = −k + 0.39k = 0.61k

Acceleration of Point E.

vE = (0.610 m/s)k 

a E = a E ′ + a E/F + ac a E = −4i − 1.17 j + 3.12i

a E = −(0.880 m/s 2 )i + (1.170 m/s 2 ) j 

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PROBLEM 15.245 (Continued)

(b)

Point F.

rF/B = (0.25 m)i + (0.13 m)k rF/D = (0.13 m)k

Motion of Point F′ in the frame. vF ′ = Ω × rF/B = 4 j × (0.25i + 0.13k ) = (0.52 m/s)i − (1 m/s)k aF ′ = Ω × vF ′ = (4 j) × (0.52i − k ) = −(4 m/s 2 )i − (2.08 m/s 2 )k

Motion of Point F relative to the frame. vF/F = ω x i × rF/D

aF/F

= 3i × (0.13k ) = −(0.39 m/s) j = ω x i × vF/F = 3i × (−0.39 j) = −(1.17 m/s 2 )k

Coriolis acceleration.

ac = 2Ω × vF/ F ac = (2)(4 j) × (−0.39 j) = 0

Velocity of Point F.

vF = vF ′ + vF/F vF = 0.52i − k − 0.39 j vF = (0.520 m/s)i − (0.390 m/s) j − (1.000 m/s)k 

Acceleration of Point F.

a F = a F ′ + a F/F + ac a F = −4i − 2.08k − 1.17k + 0 a F = −(4.00 m/s 2 )i − (3.25 m/s 2 )k 

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PROBLEM 15.246 In Problem 15.245, determine the velocity and acceleration of (a) Point G, (b) Point H. PROBLEM 15.245 Two disks, each of 130-mm radius, are welded to the 500-mm rod CD. The rod-and-disks unit rotates at the constant rate ω 2 = 3 rad/s with respect to arm AB. Knowing that at the instant shown ω 1= 4 rad/s, determine the velocity and acceleration of (a) Point E, (b) Point F.

SOLUTION Let the frame of reference BXYZ be rotating about the Y axis with angular velocity Ω = ω 2 j = (4 rad/s) j . The motion relative to this frame is a rotation about the X axis with angular velocity ω x i = (3 rad/s)i. (a)

Point G.

rG/B = −(0.25 m)i + (0.13 m) j rG/C = (0.13 m) j

Motion of Point G′ in the frame. vG′ = Ω × rG/B

aG ′

= 4 j × ( −0.25i + 0.13j) = (1 m/s)k = Ω × vG′ = 4j × k = (4 m/s 2 )i

Motion of Point G relative to the frame. vG/F = ω x i × rG/C

aG/F

= 3i × 0.13j = (0.39 m/s)k = ω x i × vG/F = 3i × (0.39k ) = −(1.17 m/s 2 ) j

Coriolis acceleration.

ac = 2Ω × vG/F ac = (2)(4 j) × (0.39k ) = (3.12 m/s 2 )i

Velocity of Point G.

vG = vG ′ + vG /F vG = k + 0.39k

vG = (1.390 m/s)k 

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PROBLEM 15.246 (Continued)

aG = aG ′ + aG/F + ac

Acceleration of Point G.

aG = 4i − 1.17 j + 3.12i

(b)

Point H.

aG = (7.12 m/s 2 )i − (1.170 m/s 2 ) j 

rH/B = −(0.25 m)i + (0.13 m)k rH/C = (0.13 m)k

Motion of Point H′ in the frame. vH ′ = Ω × rH/B

aH ′

= 4 j × ( −0.25i + 0.13k ) = (0.52 m/s)i + (1 m/s)k = Ω × vH ′ = 4 j × (0.52i + k ) = (4 m/s 2 )i − (2.08 m/s 2 )k

Motion of Point H relative to the frame. v H /F = ω x i × rH /C

a H /F

= 3i × (0.13k ) = −(0.39 m/s) j = ω x i × vH /F = 3i × ( −0.39 j) = −(1.17 m/s 2 )k

Coriolis acceleration.

ac = 2Ω × vH /F ac = (2)(4 j) × (0.39 j) = 0

Velocity of Point H.

vH = vH ′ + vH/F vH = 0.52i + k − 0.39 j vH = (0.520 m/s)i − (0.390 m/s) j + (1.000 m/s)k 

Acceleration of Point H.

a H = a H ′ + a H/F + ac a H = 4i − 2.08k − 1.17k + 0

a H = (4.00 m/s 2 )i − (3.25 m/s 2 )k 

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PROBLEM 15.247 The position of the stylus tip A is controlled by the robot shown. In the position shown the stylus moves at a constant speed u = 180 mm/s relative to the solenoid BC. At the same time, arm CD rotates at the constant rate ω 2 = 1.6 rad/s with respect to component DEG. Knowing that the entire robot rotates about the X axis at the constant rate ω 1 = 1.2 rad/s, determine (a) the velocity of A, (b) the acceleration of A.

SOLUTION Geometry:

rD /G = − (500 mm)i + (300 mm) j + (600 mm)k rA/D = − (250 mm)i + (600 mm)k rA/G = rA/D + rD /G = − (750 mm)i + (300 mm) j + (600 mm)k

Angular velocities:

Stylus motion:

ω1 = ω1i = (1.2 rad/s)i

(ω 1 = 0)

ω 2 = ω 2 j = (1.6 rad/s) j

(ω 2 = 0)

u = −ui = −(180 mm/s)i

(u = 0)

Method 1

Let the rigid body BCD be a rotating frame of reference. Its angular velocity is

ωCD = ω1 + ω 2 = (1.2 rad/s)i + (1.6 rad/s) j

Its angular acceleration is

α CD = ω1 × ωCD = (1.2 rad/s)i × [(1.2 rad/s)i + (1.6 rad/s) j] = (1.92 rad/s 2 )k

Motion of the coinciding Point A′ in the frame. vA′ = v D + v A′ /D vA′ = ω1 × rD /G + (ω1 + ω 2 ) × rA/D = (1.2 rad/s)i × [ − (500 mm)i + (300 mm) j] + [(1.2 rad/s)i + (1.6 rad/s) j] × [− (250 mm)i + (600 mm)k ] = (360 mm/s)k − (720 mm/s) j + (400 mm/s)k + (960 mm/s)i v A′ = (960 mm/s)i − (720 mm/s) j + (760 mm/s)k a D = ω 1 × (ω1 × rD /G ) = (1.2 rad/s)i × {(1.2 rad/s)i × [− (500 mm)i + (300 mm) j]} = − (432 mm/s 2 ) j a A′ /D = α CD × rA/D + ωCD × (ωCD × rA /D ) = (1.92 rad/s 2 )k × [− (250 mm)i + (600 mm)k ] + ωCD × {[(1.2 rad/s)i + (1.6 rad/s) j] × [− (250 mm)i + (600 mm)k ]} PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1367

PROBLEM 15.247 (Continued)

a A′ /D = − (480 mm/s 2 ) j + ωCD × [− (720 mm/s) j + (400 mm/s)k + (960 mm/s)i] i j k = −480 j + 1.2 1.6 0 960 −720 400 = −480 j + 640i − 480 j − 864k − 1536k a A′ /D = (640 mm/s 2 )i − (960 mm/s 2 ) j − (2400 mm/s 2 )k a A′ = a D + a A′ /D a A′ = (640 mm/s 2 )i − (1392 mm/s 2 ) j − (2400 mm/s 2 )k

Motion of Point A relative to the frame. v A/F = u = − (180 mm/s)i a A /F = 0

(a)

Velocity of A.

v A = v A′ + v A/F vA = (960 mm/s)i − (720 mm/s) j + (760 mm/s)k − (180 mm/s)i vA = (0.78 m/s)i − (0.72 m/s) j + (0.76 m/s)k 

Coriolis acceleration:

ac = 2ωCD × v A/F ac = 2[(1.2 rad/s)i + (1.6 rad/s) j] × ( −180 mm/s)i = +(576 mm/s 2 )k

(b)

Acceleration of A.

a A = a A′ + a A /F + ac a A = (640 mm/s 2 )i − (1392 mm/s 2 ) j − (2400 mm/s 2 )k + (576 mm/s2 )k a A = (0.64 m/s 2 )i − (1.392 m/s 2 ) j − (1.824 m/s 2 )k 

Method 2 Use a frame of reference rotating about the x axis with angular velocity. ω1 = ω 1i = (1.2 rad/s)i (ω 1 = 0) Motion of coinciding Point A′ in the frame. v A′ = ω1 × rA /G = (1.2 rad/s)i × [− (500 mm)i + (300 mm) j + (600 mm)k ] = (360 mm/s)k − (720 mm/s) j a A′ = ω 1 × (ω1 × rA/G ) = ω1 × v A′ = (1.2 rad/s)i × [(360 mm/s)k − (720 mm/s) j] = − (432 mm/s 2 ) j − (864 mm/s 2 )k

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PROBLEM 15.247 (Continued)

Motion of Point A relative to the frame. v A/F = ω 2 × rA/D + u = (1.6 rad/s) j × [− (250 mm)i + (600 mm)k ] − (180 mm/s)i = (400 mm/s)k + (960 mm/s)i − (180 mm/s)i a A /F :

(Since A moves on CD, which rotates at rate ω 2 , we have a Coriolis term here). a A /F = ω 2 × (ω 2 × rA/D ) + 2ω 2 × u = ω 2 × {(1.6 rad/s) j × [− (250 mm)i + (600 mm)k ]} + 2ω 2 × u = (1.6 rad/s) j × [(400 mm/s)k + (960 mm/s)i ] + 2(1.6 rad/s) j × (−180 mm)i = (640 mm/s 2 )i − (1536 mm/s 2 )k + (576 mm/s 2 )k = (640 mm/s 2 )i − (960 mm/s 2 )k

(a)

Velocity of A.

v A = v A′ + v A/F v A = 360k − 720 j + 400k + 960i − 180i v A = (0.78 m/s)i − (0.72 m/s) j + (0.76 m/s)k 

Coriolis acceleration:

ac = 2ω1 × v A/ F ac = 2(1.2 rad/s)i × [(400 mm)k + (780 mm/s)i ] = −(960 mm/s 2 ) j

(b)

Acceleration of A.

a A = a A′ + a A/F + ac a A = −432 j − 864k + 640i − 960k − 960 j a A = (0.64 m/s 2 )i − (1.392 m/s 2 ) j − (1.824 m/s 2 )k 

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PROBLEM 15.248 The angular acceleration of the 600-mm-radius circular plate shown is defined by the relation α = α 0 e −t . Knowing that the plate is at rest when t = 0 and that α 0 = 10 rad/s 2 , determine the magnitude of the total acceleration of Point B when (a) t = 0, (b) t = 0.5 s, (c) t = ∞.

SOLUTION ω t dω = α 0 e −t ; d ω = α 0 e−t dt 0 0 dt −t t ω = α 0 | − e |0 ω = α 0 (1 − e−t )



α=



at = rα = rα 0 e−t = (0.6 m)(10 rad/s 2 )e−t = 6e−t an = rω 2 = rα 02 (1 − e −t )2 = (0.6)(10) 2 (1 − e −t ) 2 = 60(1 − e−t ) 2

(a)

t = 0:

at = 6e0 = 6 m/s 2 an = 60(1 − e0 ) 2 = 0 aB2 = at2 + an2 = 62

(b)

t = 0.5 s:

aB = 6.00 m/s 2 

at = 6e −0.5 = 6(0.6065) = 3.639 m/s 2 an = 60(1 − e −0.5 ) 2 = 60(1 − 0.6065)2 = 9.289 m/s 2 aB2 = at2 + an2 = (3.639)2 + (9.289)2

(c)

t = ∞:

aB = 9.98 m/s 2 

at = 6e −∞ = 0 an = 60(1 − e −∞ ) 2 = 60 m/s 2 aB2 = at2 + an2 = 0 + 602

aB = 60.0 m/s 2 

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PROBLEM 15.249 Cylinder A is moving downward with a velocity of 9 ft/s when the brake is suddenly applied to the drum. Knowing that the cylinder moves 18 ft downward before coming to rest and assuming uniformly accelerated motion, determine (a) the angular acceleration of the drum, (b) the time required for the cylinder to come to rest.

SOLUTION Block A:

v 2 − v02 = 2as 0 − (9 ft/s) 2 = 2a (18 ft) a = −2.25 ft/s 2

a = 2.25 ft/s 2

v A = rω0

Drum:

9 ft/s = (0.75 ft)ω

ω0 = 12 rad/s (a)

a = rα −(2.25 ft/s 2 ) = (0.75 ft)α

α = −3 rad/s 2 (b)

Uniformly accelerated motion.

α = 3.00 rad/s 2



ω = 0 when t = t1 ω = ω0 + α t : 0 = (12 rad/s) − (3 rad/s 2 )t1

t1 = 4.00 s 

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PROBLEM 15.250 A baseball pitching machine is designed to deliver a baseball with a ball speed of 70 mph and a ball rotation of 300 rpm clockwise. Knowing that there is no slipping between the wheels and the baseball during the ball launch, determine the angular velocities of wheels A and B.

SOLUTION Let Point G be the center of the ball, A its contact point with wheel A, and B its contact point with wheel B. Given:

 m   1 h   5280 ft  vG =  70     = 102.667 ft/s = 1232 in./s  h   3600 s   mi   2π rad   1 min   = 10π rad/s   rev   60 s 

ω ball = (300 rpm) 

ω ball = 10π rad/s

vG = 1232 in./s

Unit vectors: Relative positions:

i=1

j = 1,

k=1

1 (−3 j) = −(1.5 in.) j 2 1 = (3j) = (1.5 in.) j 2

rA/G = rB /G

Velocities at A and B.

,

v A = vG + ω ball × rA/G = 1232i + ( −10π k ) × (1.5 j) = (1232 + 47.12)i = 1279.12 in./s v B = vG + ω ball × rB /G = 1232i + ( −10π k ) × (1.5π j) = (1232 − 47.12)i = 1184.88 in./s

Angular velocity of A.

ωA =

v A 1184.88 = = 169.27 rad/s rA 7 ω A = 1616 rpm



ω B = 1745 rpm



Angular velocity of B.

ωB =

vB 1279.12 = = 182.73 rad/s rB 7

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PROBLEM 15.251 Knowing that inner gear A is stationary and outer gear C starts from rest and has a constant angular acceleration of 4 rad/s2 clockwise, determine at t = 5 s (a) the angular velocity of arm AB, (b) the angular velocity of gear B, (c) the acceleration of the point on gear B that is in contact with gear A.

SOLUTION Angular velocity of gear C at t = 5 s.

ω c = α c t = (4 rad/s 2 )(5 s) = 20 rad/s

ω c = 20 rad/s

Let Point 1 be the contact point between gears A and B. Let Point 2 be the contact point between gears B and C. Points A, B, and C are the centers, respectively, of gears A, B, and C. Positions:

Take x axis along the straight line A1 B2. x A = xC = 0 x1 = rA = 80 mm xB = (80 mm + 40 mm) = 120 mm x2 = 120 mm + 80 mm = 200 mm

Velocity at 1.

Since gear A is stationary, v1 = 0.

Velocity at 2.

v2 = x2ωC = (200)(20) v 2 = 4000 mm/s

Point 1 is the instantaneous center of gear B. v2 = ( x2 − x1 )ω B = (120 mm)ωB v2 4000 = = 33.333 rad/s 120 120 vB = ( xB − x1 )ω B = (40 mm)(33.333 rad/s) = 1333.33 mm/s

ωB =

ω AB = (a)

Angular velocity of arm AB:

(b)

Angular velocity of gear B:

vB 1333.33 = = 11.111 rad/s xB 120 ω AB = 11.11 rad/s ω B = 33.3 rad/s

 

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PROBLEM 15.251 (Continued)

Calculate tangential accelerations: (a1 )t = 0 (a2 )t = (aC )t = rCα 2 = (200 mm)(4 rad/s 2 ) (a2 )t = 800 mm/s 2 (a2 )t = ( x2 − x1 )α B = (120 mm)α B ( a2 )t 800 = = 6.667 rad/s 2 120 120 (aB )t = ( xB − x1 )α B = (40 mm)(6.667 rad/s 2 )

αB =

= 266.67 mm/s 2

α AB = Angular accelerations:

(aB )t 266.67 = = 2.2222 rad/s 2 rB 120

α AB = 2.2222 rad/s 2

α B = 6.667 rad/s

Acceleration of Point B. 2 a B = [α AB xB ] + [ω AB xB

]

= [(2.2222)(120) ] + [(11.11) 2 (120)

]

= (266.67 mm/s ) + (14815 mm/s ) 2

(c)

2

Acceleration of Point 1 on gear B. a1 = a B + [α B ( xB − x1 ) ] + [ωB2 ( xB − x1 )

]

= [266.67 ] + [14815

] + [(6.667)(40) ] + [(33.333) 2 (40)

= [266.67 ] + [14815

] + [266.67 ] + [44444

= 29629 mm/s

]

]

a1 = 29.6 m/s 2



Note that the tangential component of acceleration is zero as expected.

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PROBLEM 15.252 Knowing that at the instant shown bar AB has an angular velocity of 10 rad/s clockwise and it is slowing down at a rate of 2 rad/s2, determine the angular accelerations of bar BD and bar DE.

SOLUTION Velocity Analysis

ω AB = 10 rad/s vB = ( AB )ω AB = (0.200)(10) = 2 m/s

vB = vB vD = vD

Locate the instantaneous center (Point C) of bar BD by noting that velocity directions at Points B and D are known. Draw BC perpendicular to vB and DC perpendicular to vD.

ωBD =

vB 2 = = 8 rad/s BC 0.25

vD = (CE )ωBD = (0.6)(8) = 4.8 m/s

ωDE = Acceleration Analysis: Unit vectors:

vD 4.8 = = 24 rad/s DE 0.2

α AB = 2 rad/s 2 , i =1

,

ω BD = 8.00 rad/s

v D = 4.8 m/s ω DE = 24 rad/s

ω AB = 10 rad/s

j =1 ,

k =1

2 a B = a A + α AB × rB /A − ω AB rB /A

= 0 + (2k ) × (−0.2) j − (10) 2 (−0.2 j) = (0.4 m/s 2 )i + (20 m/s 2 ) j

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PROBLEM 15.252 (Continued) 2 a D = a B + α BD × rD /B − ωBD rD /B

= 0.4i + 20 j + α BD k × (−0.6i − 0.25 j) − (8) 2 ( −0.6i − 0.25 j)

a D = (38.8 + 0.25α BD )i + (36 − 0.6α BD ) j

(1)

2 a D = a E + α DE × rD /E − ωDE rD /E

= 0 + α DE k × (−0.2i ) − (24) 2 (−0.2i) = 115.2i − 0.2α DE j

(2)

Equate like components of a D from Equations (1) and (2). i: j:

38.8 + 0.25α BD = 115.2

α BD = 305.6 rad/s 2

36 − 0.6α BD = −0.2α DE

α DE = 3α BD − 180 = 736.8 rad/s 2 Angular acceleration:

α BD = 306 rad/s 2



α DE = 737 rad/s 2



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PROBLEM 15.253 Knowing that at the instant shown rod AB has zero angular acceleration and an angular velocity of 15 rad/s counterclockwise, determine (a) the angular acceleration of arm DE, (b) the acceleration of Point D.

SOLUTION 3 , β = 36.87° 4 4 AB = = 5 in. cos β 4 DE = = 5 in. cos β vB = ( AB)ω AB = (5)(15)

tan β =

= 75 in./s v B = vB β , v D = v D

β

Point C is the instantaneous center of bar BD. CB =

v 5 75 = 6.25 in. ωBD = B = = 12 rad/s cos β CB 6.25

CD =

5 = 6.25 in. vD = (CD)ω BD = (6.25)(12) = 75 in./s cos β

ωDE = Acceleration analysis.

vD 75 = = 15 rad/s DE 5

α AB = 0 a B = [( AB)α AB

2 β ] + [( AB)ω AB

= 0 + [(5)(15)2

β ] = 1125 in./s 2

2 a D/B = [( BD)α BD ] + [( BD )ωBD

= [10α BD ] + [(10)(12) 2 = [10α BD ] + [1440 in./s 2 a D = [( DE )α DE

β ] + [(5)(15) 2

= [5α DE

β ] + [1125 in./s 2

β

] ] ]

2 β ] + [( DE )ω DE

= [5α DE

a D = a B + a D/B

β]

β]

β] β]

Resolve into components.

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PROBLEM 15.253 (Continued)

(a )

:

5α DE sin β + 1125cos β = −1125cos β − 1440

α DE = −1080 rad/s 2 (b)

α DE = 1080 rad/s 2

a D = [(5)(−1080)

β ] + [1125 in./s 2

= [5400 in./s2

β ] + [1125 in./s 2



β] β]

1125 5400 γ = 11.77°

tan γ =

aD = 54002 + 11252 = 5516 in./s 2 = 460 ft/s 2 90° − β + γ = 64.9°

a D = 460 ft/s 2

64.9° 

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PROBLEM 15.254 Rod AB is attached to a collar at A and is fitted with a wheel at B that has a radius r = 15 mm. Knowing that when θ = 60° the colar has velocity of 250 mm/s upward and the speed of the collar is decreasing at a rate of 150 mm/s2, determine (a) the angular acceleration of rod AB, (b) the angular acceleration of the wheel.

SOLUTION Geometry.

We note that Point B moves on a circle of radius R = 300 mm centered at C. It is useful to use the angle β, angle ACB of the triangle ABC, which indicates the motion of B along this curve. Applying the law of sines to the triangle ABC,

or

  AB BC = (θ = 60°) sin(π − θ ) sin β  AB 200 mm sin β =  sin θ = sin 60° 300 mm BC β = 35.264°

With meters as the length unit and the unit vectors defined as i =1

,

j =1 ,

and k = 1

The relative position vectors are rB /A = (0.2 m)(sin 60°i − cos 60° j) rB / C = (0.3 m)(sin β i − cos β j)

Let Point P be the contact point where the wheel rolls on the fixed surface. rP /B = (0.015 m)(sin β i − cos β j)

Velocity analysis.

v A = 0.250 m/s v B = vB

β

vP = 0 ω AB = ω AB k

ω BP = ωBP k

v B = v A + v B /A = v A + ω AB × rB /A

= 0.250 j + (ω AB )k × (0.2sin 60°i − 0.2 cos 60° j) vB (cos β i + sin β j) = 0.25 j + 0.1ω AB i + 0.173205ω AB j

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PROBLEM 15.254 (Continued)

Resolve into components: i:

vB cos β = +0.1ω AB

(1)

j:

vB sin β = 0.25 − 0.173205ω AB

(2)

Solving the simultaneous Equations (1) and (2),

ω AB = −2.43913

vB = −0.29873 m/s v B = 0.29873 m/s

β

ω AB = 2.43913 rad/s

Since the wheel rolls without slipping, Point P is its instantaneous center. 



Acceleration analysis

| vB | = rωBP

ω PB =

0.29873 0.15

ω BP = 1.992 rad/s

a A = 150 mm/s 2

(a P )t = 0 2 a B = a A + a B/A = a A + α AB k × rB /A − ω AB rB /A

= −0.150 j + α AB k × (0.2sin 60°i − 0.2 cos 60° j) − (2.43913)2 (0.2sin 60°i − 0.2cos 60° j) = −0.150 j + (0.2sin 60°)α AB j + (0.2cos 60°)α AB i − 1.03046i + 0.59494 j = (0.2 cos 60°)α AB i + (0.2sin 60°)α AB j − 1.03046i + 0.44494 j

Consider the motion of B along its circular path. a B = [( aB )t

 vB2

β]+ 

 R



β 

vB2 (− sin β i + cos β j) R (0.29873) 2 = (aB )t cos β i + (aB )t sin β j − (− sin β i + cos β j) 0.3 = (aB )t cos β i + (aB )t sin β j − 0.17174i + 0.24288 j = (aB )t (cos β i + sin β j) +

Equate the two expression for aB and resolve into components. i:

(aB )t cos β − 0.17174 = (0.2cos 60°)α AB − 1.03046

(3)

j:

(aB )t sin β + 0.24288 = 0.2sin 60°α AB + 0.44494

(4)

Solving Eqs. (3) and (4) simultaneously, (aB )t = −2.0187 m/s 2

α AB = −7.8956 rad/s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1380

PROBLEM 15.254 (Continued)

(a)

Angular acceleration of AB.

(b)

Angular acceleration of the wheel.

α AB = 7.90 rad/s 2



α BP = 134.6 rad/s 2



(aP )t = (aB )t + rα BP = 0

α BP =

( aB ) t ( −2.0187) =− = 134.6 rad/s 2 0.015 r

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PROBLEM 15.255 Water flows through a curved pipe AB that rotates with a constant clockwise angular velocity of 90 rpm. If the velocity of the water relative to the pipe is 8 m/s, determine the total acceleration of a particle of water at Point P.

SOLUTION or ω = 9.4248 rad/s .

Let the curved pipe be a rotating frame of reference. Its angular velocity is 90 rpm Motion of the frame of reference at Point P′. vP′ = ( AP)ω

45° = (0.5 2)(9.4248)

45° = 6.6643 m/s

45°

a P′ = ( AP)ω 2

45° = (0.5 2 )(9.4248) 2

45° = 62.81 m/s 2

45°

Motion of water relative to the frame at Point P. vP/F = 8 m/s ← a P/F = =

(vP/F )2

ρ



(8 m/s)2 ↓ 0.5 m

= 128 m/s 2 ↓

Coriolis acceleration.

ac = 2ω vP/F = (2)(9.4248)(8) = 150.797 m/s 2 ac = 150.797 m/s 2 ↑

Acceleration of water at Point P.

a P = a P′ + a P/F + ac a P = [62.81

45°] + [128 ↓] + [150.797 ↑]

= [44.413 m/s 2 →] + [21.616 m/s 2 ↓] a P = 49.4 m/s 2



26.0° 

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PROBLEM 15.256 A disk of 0.15-m radius rotates at the constant rate ω 2 with respect to plate BC, which itself rotates at the constant rate ω 1 about the y axis. Knowing that ω 1 = ω 2 = 3 rad/s, determine, for the position shown, the velocity and acceleration (a) of Point D, (b) of Point F.

SOLUTION Frame AXYZ is fixed. Moving frame, Exyz, rotates about y axis at Ω = ω1 j = (3 rad/s)j

(a)

Point D:

ω2 = ω2 j = (3 rad/s)j rD/A = 0 rD/E = −(0.15 m)i vD′ = Ω × rD/A = 0 vD/F = ω A × rD/E

= (3 rad/s)j × ( −0.15 m)i = (0.45 m/s)k vD = vD′ + v D/F

vD = (0.45 m/s)k 

a D′ = Ω × vD′ = 0 a D/F = ω1 × vD/F = (3 rad/s)j × (0.45 m/s)k = (1.35 m/s 2 )i ac = 2Ω × vD/F = 2(3 rad/s)j × (0.45 m/s)k = (2.70 m/s 2 )i a D = a D′ + a D/F + ac = 0 + (1.35 m/s 2 )i + (2.70 m/s 2 )i

a D = (4.05 m/s 2 )i 

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PROBLEM 15.256 (Continued)

(b)

Point F:

ω2 = ω2 j = (3 rad/s)j rF/A = (0.3 m)i; rF/E = (0.15 m)i vF ′ = Ω × rF/A = (3 rad/s)j × (0.3 m)i vF/F

= −(0.9 m/s)k = ω2 × rF/E

= (3 rad/s)j × (0.15 m)i = −(0.45 m/s)k vF = vF ′ + vF/F = −(0.9 m/s)k − (0.45 m/s)k

vF = −(1.35 m/s)k 

aF′ = Ω × vF′ = (3 rad/s)j × (−0.9 m/s)k a F/F

= −(2.7 m/s 2 )i = ω2 × vF/F = (3 rad/s)j × (−0.45 m/s)k

= −(1.35 m/s 2 )i ac = 2Ω × vF/F = 2(3 rad/s)j × (−0.45 m/s)k = −(2.7 m/s 2 )i a F = a F ′ + a F/F + ac = −(2.7 m/s 2 )i − (1.35 m/s 2 )i − (2.7 m/s 2 )i a F = −(6.75 m/s 2 )i 



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PROBLEM 15.257 Two rods AE and BD pass through holes drilled into a hexagonal block. (The holes are drilled in different planes so that the rods will not touch each other.) Knowing that rod AE has an angular velocity of 20 rad/s clockwise and an angular acceleration of 4 rad/s2 counterclockwise when θ = 90°, determine, (a) the relative velocity of the block with respect to each rod, (b) the relative acceleration of the block with respect to each rod.

SOLUTION Geometry:

When θ = 90°, Point H is located as shown in the sketch. Apply the law of sines to the triangle ABH. r1 r2 100 mm = = sin 60° sin 30° sin 90° r1 = 57.735 mm r2 = 115.470 mm

The angle at H remains at 60° so that rods AE and BD have a common angular velocity ω = ω common angular acceleration α = α , where

and a

ω = −20 rad/s and

α = 4 rad/s

Consider the double slider H as a particle sliding along the rotating rod AH with relative velocity u 1 relative acceleration u 1 .

and

Let H ′ be the point on rod AE that coincides with H. v H ′ = [r1ω ] = [(57.735 mm)(−20 rad/s) ] = [1154.7 mm/s ] a H ′ = [r1α ] + [r1ω 2

]

= [(57.735 mm)(4 rad/s 2 ) ] + [(57.735 mm)(20 rad/s) 2 = [230.9 mm/s ] + [23094 mm/s 2

2

]

]

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PROBLEM 15.257 (Continued)

The corresponding Coriolis acceleration is a1 = [2ω u 1 ] = [(2)(−20)u1 ] = 40u1 v H = v H ′ + u1

= [1154.7 mm ]+[u1

a H = aH ′ + [u1

] + [40u1 ] ] + [u 1

= [230.9 mm/s ] + [23094 mm/s 2

(1)

] ] + [40u 1 ]

(2)

Now consider the double slider H as a particle sliding along the rotating rod BD with relative velocity u 2 and relative acceleration u 2 60°.

60°

Let H ′′ be the point on rod BD that coincides with H. v H ′′ = r2ω

30° = (115.47 mm)(−20 rad/s)

a H ′′ = r2α

30° + r2ω 2

= 461.9 mm/s

30°

60°

= (115.47 mm)(4 rad/s 2 ) 2

30° = 2309.4 mm/s

30° + (115.47 mm)(20 rad/s) 2

30° + 46188 mm/s

2

60°

60°

The corresponding Coriolis acceleration is a 2 = 2ω u 2

30° = (2)(−20 rad/s)u 2

v H = v H ′′ + u2

60° = 2309.4 mm/s

a H = a H ′′ + u2

60° + 40u2

= [(461.9 mm/s ) 2

60° = 40u2 30° + u2

30° 60°

(3)

30° 60°] + [u2

30°] + [(46188 mm/s 2 )

60°] + [40u2

30° ]

(4)

Equate expression (1) and (3) for vH and resolve into components. 1154.7 mm/s = −(2309.4 mm/s) sin 30° + u2 sin 60°

:

u2 =

2309.4sin 30° − 1154.7 sin 60°

u1 = 2309.4cos 30° + 0

:

u2 = 0 u1 = 2000 mm/s

Substitute the values for u1 and u2 into the Coriolis acceleration terms, and equate expressions (2) and (4) for aH, and resolve into components. :

230.9 mm/s 2 − (40 rad/s)(2000 mm/s) = (461.9 mm/s 2 ) sin 30° + (46188 mm/s 2 ) sin 60° + u2 sin 60° − (40)(0)sin 30°

u2 =

230.9 − 80000 − 230.9 + 40000 sin 60°

u2 = −46188 mm/s 2

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PROBLEM 15.257 (Continued)

:

u1 − 23094 mm/s 2 = −(461.4 mm/s 2 ) cos 30° − (46188 mm/s 2 ) cos 60° −(46188 mm/s 2 ) cos 60° + 0

u1 = −23494 mm/s 2

(a)

Relative velocities:

AE: u1

= 2.00 m/s

BD: u2

(b)

60° = 0 

AE: u1 = 23.5 m/s 2

Relative accelerations: BD: u2

60° = 46.2 m/s 2



 60° 

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PROBLEM 15.258 Rod BC of length 24 in. is connected by ball-and-socket joints to a rotating arm AB and to a collar C that slides on the fixed rod DE. Knowing that the length of arm AB is 4 in. and that it rotates at the constant rate ω 1 = 10 rad/s, determine the velocity of collar C when θ = 0.

SOLUTION lBC = 24 in. rC/B = (8 in.)i − (16 in.)j + zC/B k

Geometry.

242 zC/B rC/B rB/A

= 82 + 162 + zC2 /B = 16 in. = (8 in.)i − (16 in.)j + (16 in.)k = −(4 in.)i

v B = ω1k × rB/A

Velocity at B.

= 10k × (−4i ) = −(40 in./s)j

Velocity of collar C.

vC = vC k v C = v B + v C/ B vC/B = ω BC × rB/C

where

Noting that vC/B is perpendicular to rC/B , we get rC/B ⋅ v C/B = 0 Forming rC/B ⋅ vC , we get

rC/B ⋅ vC = rC/B ⋅ ( vB + vC/B ) = rC/B ⋅ vB + rC/B ⋅ vC/B

or From Eq. (1)

rC/B ⋅ vC = rC/B vB

(1)

(8i − 16j + 16k ) ⋅ (vC k ) = (8i − 16 j + 16k ) ⋅ (−40 j) 16vC = (−16)(−40)

or

vC = 40 in./s

vC = (40.0 in./s)k 

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PROBLEM 15.259 In the position shown, the thin rod moves at a constant speed u = 3 in./s out of the tube BC. At the same time, tube BC rotates at the constant rate ω 2 = 1.5 rad/s with respect to arm CD. Knowing that the entire assembly rotates about the X axis at the constant rate ω1 = 1.2 rad/s, determine the velocity and acceleration of end A of the rod.

SOLUTION Geometry.

rA = (6 in.) j + (9 in.)k

Method 1 Let the rigid body DCB be a rotating frame of reference. Its angular velocity is

Ω = ω 1i + ω 2 k = (1.2 rad/s)i − (1.5 rad/s)k.

Its angular acceleration is

α = ω1i × ω2 k = −ω1ω2 j = (1.8 rad/s 2 ) j.

Motion of the coinciding Point A′ in the frame. vA′ = Ω × rA = (1.2i − 1.5k ) × (6 j + 9k ) = 7.2k − 10.8 j + 9i = (9 in./s)i − (10.8 in./s) j + (7.2 in./s)k aA′ = α × rA + Ω × vA′ i j k i j k = 0 1.8 0 + 1.2 −1.5 0 0 6 9 9 −10.8 7.2 = 16.2i − 16.2i − 22.14 j − 12.96k = −(22.14 in./s 2 ) j − (12.96 in./s 2 )k

Motion of Point A relative to the frame. vA/F = uj = (3 in./s) j, aA/F = 0

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PROBLEM 15.259 (Continued)

Velocity of Point A.

vA = vA′ + vA/F vA = 9i − 10.8 j + 7.2k + 3j vA = (9.00 in./s)i − (7.80 in./s) j + (7.20 in./s)k 

Coriolis acceleration.

2Ω × vA/F = (2)(1.2i − 1.5k ) × 3j = (9 in./s 2 )i + (7.2 in./s 2 )k

Acceleration of Point A.

aA = aA′ + aA /F + 2Ω × vA /F aA = − 22.14 j − 12.92k + 9i + 7.2k aA = (9.00 in./s 2 )i − (22.1 in./s 2 ) j − (5.76 in./s 2 )k 

Method 2 Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity Ω = ω 1i = 1.2i rad/s. Then the motion relative to the frame consists of the rotation of body DCB about the Z axis with angular velocity ω 2 k = −(1.5 rad/s)k plus the sliding motion u = ui = (3 in./s) j of the rod AB relative to the body DCB. Motion of the coinciding Point A′ in the frame. vA′ = Ω × rA = 1.2i × (6 j + 9k ) = −(10.8 in./s) j + (7.2 in./s)k aA′ = Ω × vA′ = 1.2i × (−10.8 j + 7.2k ) = −(8.64 in./s 2 ) j − (12.96 in./s 2 )k

Motion of Point A relative to the frame. vA/F = ω2 k × rA + uj

aA/F

= (−1.5k ) × (6 j + 9k ) + 3j = (9 in./s)i + (3 in./s) j = α 2 k × rA + ω 2k × (ω 2 k × rA ) + uj + 2ω 2 k × (uj) = 0 + ( −1.5k ) × (9i ) + 0 + (2) (−1.5k ) × (3j) = −13.5 j + 9i = (9 in./s 2 )i − (13.5 in./s 2 ) j

Velocity of Point A.

vA = vA′ + vA/F vA = −10.8 j + 7.2k + 9i + 3j vA = (9.00 in./s)i − (7.80 in./s) j + (7.20 in./s)k 

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PROBLEM 15.259 (Continued)

Coriolis acceleration.

2Ω × vA /F = (2)(1.2i ) × (−9i + 3j) = (7.2 in./s 2 )k

Acceleration of Point A.

aA = aA′ + aA /F + 2Ω × vA /F aA = −8.64 j − 12.96k + 9i − 13.5 j + 7.2k aA = (9.00 in./s 2 )i − (22.1 in./s 2 ) j − (5.76 in./s 2 )k 

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CHAPTER 16

PROBLEM 16.CQ1 Two pendulums, A and B, with the masses and lengths shown are released from rest. Which system has a larger mass moment of inertia about its pivot point? (a) A (b) B (c) They are the same.

SOLUTION Answer: (b)

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PROBLEM 16.CQ2 Two pendulums, A and B, with the masses and lengths shown are released from rest. Which system has a larger angular acceleration immediately after release? (a) A (b) B (c) They are the same.

SOLUTION Answer: (a)

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PROBLEM 16.CQ3 Two solid cylinders, A and B, have the same mass m and the radii 2r and r respectively. Each is accelerated from rest with a force applied as shown. In order to impart identical angular accelerations to both cylinders, what is the relationship between F1 and F2? (a) F1 = 0.5F2 (b) F1 = F2 (c) F1 = 2F2 (d ) F1 = 4F2 (e) F1 = 8F2

SOLUTION Answer: (c) Fr = I α

α=

1 2

F (2r ) Fr FR = 1 = 2 2 2 2 1 1 mR m(2r ) mr 2 2

F1 2 F2 = mr mr

F1 = 2F2 

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PROBLEM 16.F1 A 6-ft board is placed in a truck with one end resting against a block secured to the floor and the other leaning against a vertical partition. Draw the FBD and KD necessary to determine the maximum allowable acceleration of the truck if the board is to remain in the position shown.

SOLUTION Answer:

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PROBLEM 16.F2 A uniform circular plate of mass 3 kg is attached to two links AC and BD of the same length. Knowing that the plate is released from rest in the position shown, in which lines joining G to A and B are, respectively, horizontal and vertical, draw the FBD and KD for the plate.

SOLUTION Answer:

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PROBLEM 16.F3 Two uniform disks and two cylinders are assembled as indicated. Disk A weighs 20 lb and disk B weighs 12 lb. Knowing that the system is released from rest, draw the FBD and KD for the whole system.

SOLUTION Answer:

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PROBLEM 16.F4 The 400-lb crate shown is lowered by means of two overhead cranes. Knowing the tension in each cable, draw the FBD and KD that can be used to determine the angular acceleration of the crate and the acceleration of the center of gravity.

SOLUTION Answer:

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PROBLEM 16.1 A conveyor system is fitted with vertical panels, and a 15-in. rod AB weighing 5 lb is lodged between two panels as shown. If the rod is to remain in the position shown, determine the maximum allowable acceleration of the system.

SOLUTION Geometry:

Mass:

10 in. = 10.642 in. cos 20° 1 b = (15 in.)sin 20° = 2.5652 in. 2 1 c = (15 in.) cos 20° = 7.0477 in. 2 5 lb W m= = = 0.15528 slug g 32.2 ft/s 2

d=

Kinetics:

ΣM B = Σ( M B )eff : Cd − Wb = − mac

Maximum allowable acceleration. This occurs at loss of contact when C = 0. ma =

Wb (5 lb)(2.5652 in.) = = 1.8199 lb c 7.0477 in. a=

ma 1.8199 lb = m 0.15528 slug

a = 11.72 ft/s 

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PROBLEM 16.2 A conveyor system is fitted with vertical panels, and a 15-in. rod AB weighing 5 lb is lodged between two panels as shown. Knowing that the acceleration of the system is 3 ft/s 2 to the left, determine (a) the force exerted on the rod at C, (b) the reaction at B.

SOLUTION 10 in. = 10.642 in. cos 20° 1 b = (15 in.) sin 20° = 2.5652 in. 2 1 c = (15 in.) cos 20° = 7.0477 in. 2

CB = d =

Geometry:

W 5 lb = = 0.15528 slug g 32.2 ft/s 2

Mass:

m=

Kinetics:

m a = (0.15528 slug)(3 ft/s 2 ) = 0.46588 lb

(a)

Force at C. ΣM B = Σ( M B )eff : Cd − Wb = − mac C=

Wb mac (5 lb)(2.5652 in.) (0.46588 lb)(7.0477 in.) − = − d d 10.642 in. 10.642

C = 0.89669 lb

C = 0.897 lb

20° 

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PROBLEM 16.2 (Continued)

(b)

Reaction at B. + ΣFy = Σ( Fy )eff : B y − W + C sin 20° = 0

B y = 5 lb − (0.89669 lb)sin 20° = 4.6933 lb ΣFx = Σ( Fx )eff : Bx − C cos 20° = ma

Bx = (0.89669 lb) cos 20° + 0.46588 = 1.3085 lb B = 1.3085 lb

+4.6933 lb

B = 4.87 lb

74.4° 

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PROBLEM 16.3 Knowing that the coefficient of static friction between the tires and the road is 0.80 for the automobile shown, determine the maximum possible acceleration on a level road, assuming (a) four-wheel drive, (b) rear-wheel drive, (c) front-wheel drive.

SOLUTION (a)

Four-wheel drive:

+ ΣFy = 0: N A + N B − W = 0

Thus:

N A + N B = W = mg

FA + FB = μk NA + μ k NB = μ k ( NA + NB ) = μ kW = 0.80mg ΣFx = Σ( Fx )eff : FA + FB = ma

0.80mg = ma a = 0.80 g = 0.80(32.2 ft/s 2 )

(b)

a = 25.8 ft/s 2



a = 12.27 ft/s 2



Rear-wheel drive:

ΣM B = Σ( M B )eff : (40 in.)W − (100 in.) NA = −(20 in.)ma

NA = 0.4W + 0.2ma

Thus:

FA = μk N B = 0.80(0.4W + 0.2ma ) = 0.32mg + 0.16ma ΣFx = Σ( Fx )eff : FA = ma

0.32mg + 0.16ma = ma 0.32 g = 0.84a a=

0.32 (32.2 ft/s 2 ) 0.84

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PROBLEM 16.3 (Continued) (c)

Front-wheel drive:

ΣM A = Σ( M A )eff : (100 in.) N B − (60 in.)W = −(20 in.) ma

N B = 0.6W − 0.2ma

Thus:

FB = μk N B = 0.80(0.6W − 0.2ma ) = 0.48mg − 0.16ma ΣFx = Σ( Fx )eff : FB = ma

0.48mg − 0.16ma = ma 0.48 g = 1.16a a=

0.48 (32.2 ft/s 2 ) 1.16

a = 13.32 ft/s 2



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PROBLEM 16.4 The motion of the 2.5-kg rod AB is guided by two small wheels which roll freely in horizontal slots. If a force P of magnitude 8 N is applied at B, determine (a) the acceleration of the rod, (b) the reactions at A and B.

SOLUTION

ΣFx = Σ( Fx )eff : P = ma

(a)

a=

P 8N = = 3.20 m/s 2 m 2.5 kg a = 3.20 m/s 2

2r    2r  ΣM B = Σ( M B )eff : W  r −  − Ar = ma   π   π 

(b)

2 2  2  2 A = W 1 −  − ma   = mg 1 −  − P    π π   π π  2  2 = (2.5 kg)(9.81 m/s 2 ) 1 −  − (8 N)    π π  = 8.912 N − 5.093 N = 3.819 N







A = 3.82 N 

ΣFy = 0: A + B − W = 0 B = W − A = (2.5)(9.81) − 3.819,

B = 20.71 N 

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PROBLEM 16.5 A uniform rod BC of mass 4 kg is connected to a collar A by a 250-mm cord AB. Neglecting the mass of the collar and cord, determine (a) the smallest constant acceleration aA for which the cord and the rod lie in a straight line, (b) the corresponding tension in the cord.

SOLUTION Geometry and kinematics: Distance between collar and floor = AD = 250 mm + 350 mm = 600 mm When cord and rod lie in a straight line: AC = AB + BC = 250 mm + 400 mm = 650 mm cos θ =

AD 600 mm = AC 650 mm

θ = 22.62° Kinetics (a)

Acceleration at A. ΣM C = Σ( M C )eff :

W (CG )sin θ = ma (CG ) cos θ ma = mg tan θ a = g tan θ = (9.81 m/s 2 ) tan 22.62° aA = a = 4.09 m/s 2

(b)



Tension in the cord. ΣFx = Σ( Fx )eff :

T sin θ = ma = mg tan θ T=

mg (4 kg)(9.81) = cos θ cos 22.62°

T = 42.5 N 

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PROBLEM 16.6 A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel drive truck is 1 m/s2, determine (a) the reaction at each of the front wheels, (b) the force between the boulder and the pallet.

SOLUTION Kinematics:

Acceleration of truck: aT = 1 m/s 2

.

When the truck moves 1 m to the left, the boulder B and pallet A are raised 0.5 m. Then,

a A = 0.5 m/s 2

Kinetics:

Let T be the tension in the cable.

Pallet and boulder:

a B = 0.5 m/s 2

ΣFy = Σ( Fy )eff :

2T − (m A + mB ) g = (m A + mB )aB

2T − (450 kg)(9.81 m/s 2 ) = (450 kg)(0.5 m/s 2 ) T = 2320 N

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PROBLEM 16.6 (Continued)

M R = Σ( M R )eff :

Truck:

− N F (3.4 m) + mT g (2.0 m) − T (0.6 m) = mT aT (1.0 m)

(2.0 m)(2000 kg)(9.81 m/s 2 ) (0.6 m)(2320 N) (1.0 m)(2000 kg)(1.0 m/s) − + 3.4 m 3.4 m 3.4 m = 11541.2 N − 409.4 N − 588.2 N = 10544 N

NF =

ΣFy = Σ( Fy )eff :

N F + N R − mT g = 0

10544 N + N R − (2000 kg)(9.81 m/s 2 ) = 0 N R = 9076 N ΣFx = Σ( Fx )eff : FR − T = mT aT FR = 2320 N + (2000 kg) (1.0 m/s 2 ) = 4320 N

(a)

Reaction at each front wheel: 1 NF : 2

5270 N 

Reaction at each rear wheel: 1 FR 2

(b)

1 + NR 2

5030 N

64.5°

Force between boulder and pallet.

Boulder

ΣFy = Σ( Fy )eff :

N B + M B g − mB aB

N B = (400 kg)(9.81 m/s 2 ) + (400 kg)(0.5 m/s 2 ) = 4124 N

4120 N (compression) 

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PROBLEM 16.7 The support bracket shown is used to transport a cylindrical can from one elevation to another. Knowing that μ s = 0.25 between the can and the bracket, determine (a) the magnitude of the upward acceleration a for which the can will slide on the bracket, (b) the smallest ratio h/d for which the can will tip before it slides.

SOLUTION (a)

Sliding impends ΣFy = Σ( Fx )eff :

F = ma cos 30°

ΣFy = Σ( Fy )eff : N − mg = ma sin 30° N = m( g + a sin 30°)

μs =

F N

ma cos 30° m( g + a sin 30°) g + a sin 30° = 4a cos 30° 0.25 =

a 1 = g 4 cos 30° − sin 30°

(b)

a = 0.337g

30° 

Tipping impends h d ΣM G = Σ( M G )eff : F   − N   = 0 2   2 F d = N h

μ=

F ; N

0.25 =

d ; h

h = 4.00  d

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PROBLEM 16.8 Solve Problem 16.7, assuming that the acceleration a of the bracket is directed downward. PROBLEM 16.7 The support bracket shown is used to transport a cylindrical can from one elevation to another. Knowing that μ s = 0.25 between the can and the bracket, determine (a) the magnitude of the upward acceleration a for which the can will slide on the bracket, (b) the smallest ratio h/d for which the can will tip before it slides.

SOLUTION (a)

Sliding impends: ΣFx = Σ( Fx )eff :

F = ma cos 30°

ΣFy = Σ( Fy )eff : N − mg = − ma sin 30° N = m( g − a sin 30°)

μs =

F N

ma cos 30° m( g − a sin 30°) g − a sin 30° = 4a cos 30° 0.25 =

a 1 = = 0.25226 g 4 cos 30° + sin 30°

a = 0.252g

30°

 (b)

Tipping impends: h d ΣM G = Σ( M G )eff : F   = W   2 2

μ=

F ; N

F d = N h 0.25 =

d ; h

h = 4.00  d

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PROBLEM 16.9 A 20-kg cabinet is mounted on casters that allow it to move freely ( μ = 0) on the floor. If a 100-N force is applied as shown, determine (a) the acceleration of the cabinet, (b) the range of values of h for which the cabinet will not tip.

SOLUTION

(a)

Acceleration ΣFx = Σ( Fx )eff : 100 N = ma 100 N = (20 kg)a a = 5.00 m/s 2

(b)

For tipping to impend :



A=0

ΣM B = Σ( M B )eff : (100 N) h − mg (0.3 m) = ma (0.9 m) (100 N)h − (20 kg)(9.81 m/s 2 )(0.3 m) = (100 N)(0.9 m) h = 1.489 m

For tipping to impend

: B=0

ΣM A = Σ( M A ) eff : (100 N) h + mg (0.3 m) = ma (0.9 m) (100 N)h + (20 kg)(9.81 m/s 2 )(0.3 m) = (100 N)(0.9 m) h = 0.311 m 0.311 m ≤ h ≤ 1.489 m 

Cabinet will not tip:

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PROBLEM 16.10 Solve Problem 16.9, assuming that the casters are locked and slide on the rough floor ( μk = 0.25). PROBLEM 16.9 A 20-kg cabinet is mounted on casters that allow it to move freely ( μ = 0) on the floor. If a 100-N force is applied as shown, determine (a) the acceleration of the cabinet, (b) the range of values of h for which the cabinet will not tip.

SOLUTION (a)

Acceleration. + ΣFy = 0:

NA + N B − W = 0 N A + N B = mg

But, F = μ N , thus FA + FB = μ (mg ) ΣFx = Σ( Fx )eff : 100 N − ( FA + FB ) = ma 100 N − μ mg = ma 100 N − 0.25(20 kg)(9.81 m/s 2 ) = (20 kg)a a = 2.548 m/s 2

(b)

a = 2.55 m/s 2



Tipping of cabinet.

W = mg = (20 kg)(9.81 m/s 2 ) W = 196.2 N

For tipping to impend : N A = 0. ΣM B = Σ( M B )eff : (100 N)h − W (0.3 m) = ma (0.9 m) (100 N)h − (196.2 N)(0.3 m) = (20 kg)(2.548 m/s 2 )(0.9 m) h = 1.047 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1414

PROBLEM 16.10 (Continued)

For tipping to impend :

NB = 0

ΣM A = Σ( M A )eff : (100 N)h + W (0.3 m) = ma (0.9 m) (100 N)h + (196.2 N)(0.3 m) = (20 kg)(2.548 m/s 2 )(0.9 m) h = −0.130 m (impossible) h ≤ 1.047 m 

Cabinet will not tip:

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PROBLEM 16.11 A completely filled barrel and its contents have a combined mass of 90 kg. A cylinder C is connected to the barrel at a height h = 550 mm as shown. Knowing μ s = 0.40 and μk = 0.35, determine the maximum mass of C so the barrel will not tip.

SOLUTION Kinematics:

Assume that the barrel is sliding, but not tipping.

α =0 Since the cord is inextensible, Kinetics:

aG = a

aC = a

Draw the free body diagrams of the barrel and the cylinder. Let T be the tension in the cord.

The barrel is sliding.

FF = μk N = 0.35 N

Assume that tipping is impending, so that the line of action of the reaction on the bottom of the barrel passes through Point B. e = 250 mm

For the barrel.

ΣFy = 0: N − WB = 0

N = WB = mB g = 882.90 N

ΣM G = 0: Ne − 100T − (450)(0.35 N ) = 0 T=

250 − 157.5 e − (450)(0.35) (882.90) = 816.68 N N= 100 100

ΣFx = mB a : T − 0.35 N = mB a a=

816.68 882.90 − 0.35 = 5.6407 m/s 2 90 90

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PROBLEM 16.11 (Continued)

For the cylinder:

ΣF = mC a : WC − T = mC a mC g − T = mC a mC =

T 816.68 = = 195.88 kg g − a 9.81 − 5.6407 mC = 195.9 kg 

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PROBLEM 16.12 A 40-kg vase has a 200-mm-diameter base and is being moved using a 100-kg utility cart as shown. The cart moves freely ( μ = 0) on the ground. Knowing the coefficient of static friction between the vase and the cart is μs = 0.4, determine the maximum force F that can be applied if the vase is not to slide or tip.

SOLUTION Vase:

ΣFy = Σ( Fy )eff : N − mV g = 0 N = mV g = (40 kg)(9.81 m/s 2 ) = 392.4 N Ff = μS N

For impending sliding,

F f = (0.4)(392.4 N) = 156.96 N ΣFx = Σ( Fx )eff : F f = mV a a=

Ff mV

=

156.96 N = 3.924 m/s 2 40

This is the limiting value of a for sliding. M B = Σ( M B )eff:

mV ge = mV ah a=

100 mm e (9.81 m/s 2 ) = 1.635 m/s 2 g= 600 mm h

This is the limiting value of a for tipping. The smaller value governs.

a = 1.635 m/s 2

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PROBLEM 16.12 (Continued)

Cart and vase:

ΣFx = Σ( Fx )eff : F = mC a + mV a F = (100 kg)(1.635 m/s 2 ) + (40 kg)(1.635 m/s 2 ) F = 229 N 

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PROBLEM 16.13 The retractable shelf shown is supported by two identical linkage-and-spring systems; only one of the systems is shown. A 20-kg machine is placed on the shelf so that half of its weight is supported by the system shown. If the springs are removed and the system is released from rest, determine (a) the acceleration of the machine, (b) the tension in link AB. Neglect the weight of the shelf and links.

SOLUTION The links AB and CE keep the line AC on the shelf parallel to the fixed line BE. Thus the shelf moves in curvilinear translation.  α =0 and aG is perpendicular to AB Since link AB is a massless two-force member, the force at A is along link AB. Since link CDE is massless and the spring DF is removed, the force at C is directed along the link CDE. Consider the kinetics of the shelf.

Force perpendicular to link AB.

30°

mg cos 30° = ma a = g cos 30° = (9.81 m/s) cos 30° = 8.4957 m/s 2

(a)

a = 8.50 m/s 2

Acceleration of the machine:

60° 

ΣM C = Σ( M C )eff : ( FA cos 30°)(0.080) + ( FA sin 30°)(0.100) − mg (0.150) = (ma sin 30°)(0.160) + (ma cos 30°)(0.150) 0.11928FA − 0.150 mg = −0.04990 mg cos 30°

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PROBLEM 16.13 (Continued)

(b)

Tension in link AB.

FA = 89525 mg

Taking mg to be half the weight of the machine, mg =

1 (20 kg)(9.81 m/s 2 ) = 98.1 N 2 FA = (0.89522)(98.1 N)

F = 87.8 N 

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PROBLEM 16.14 A uniform rectangular plate has a mass of 5 kg and is held in position by three ropes as shown. Knowing that θ = 30°, determine, immediately after rope CF has been cut, (a) the acceleration of the plate, (b) the tension in ropes AD and BE.

SOLUTION

(a)

Acceleration

+

30°ΣF = ΣFeff : mg sin 30° = ma a = 0.5 g = 4.905 m/s 2

(b)

a = 4.91 m/s 2

30° 

Tension in ropes ΣM A = Σ( M A )eff : (TB cos 30°)(0.3 m) − mg (0.15 m) = − ma (cos 30°)(0.12 m) − ma (sin 30°)(0.15 m) 0.2598TB − (5 kg)(9.81 m/s 2 )(0.15 m) = −(5 kg)(4.905 m/s 2 )(0.1039 + 0.075) 0.2598TB − 7.3575 = −4.388 TB = +11.43 N

TBE = 11.43 N 

10°ΣF = ΣFeff : TA + 11.43 N − mg cos30° = 0 TA + 11.43 N − (5 kg)(9.81) cos 30° = 0 TA + 11.43 N − 42.48 N = 0 TA = 31.04 N

TAD = 31.0 N 

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PROBLEM 16.15 A uniform rectangular plate has a mass of 5 kg and is held in position by three ropes as shown. Determine the largest value of θ for which both ropes AD and BE remain taut immediately after rope CF has been cut.

SOLUTION

ΣF = ΣFeff : mg sin θ = ma a = g sin θ ΣM B = Σ( M B )eff : mg (0.15 m) = ma cos θ (0.12 m) + ma sin θ (0.15 m)

mg (0.15) = m ( g sin θ )(0.12 cos θ + 0.15sin θ ) 1 = 0.8sin θ cos θ − sin 2 θ 1 − sin 2 θ = 0.8sin θ cos θ cos 2 θ = 0.8sin θ cos θ sin θ 1 = 0.8 cos θ tan θ = 1.25;

θ = 51.3° 

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PROBLEM 16.16 Three bars, each of mass 3 kg, are welded together and are pin-connected to two links BE and CF. Neglecting the weight of the links, determine the force in each link immediately after the system is released from rest.

SOLUTION

Mass center of ABCD is at G y=

Σmi yi 3(0.225) + 3(0.225) + 3(0) = = 0.15 m 9 Σmi W = mg

40° ΣF = ΣFeff : mg cos 50° = ma (9.81 m/s 2 ) cos 50° = a a = 6.3057 m/s 2

40°

ΣM B = Σ( M B )eff : ( FCF sin 50°)(0.450 m) − (9 kg)(9.81 m/s 2 )(0.225 m) = − ma sin 40°(0.225 m) − ma cos 40°(0.150 m) 0.34472 FCF − 19.8653 = −ma (0.14463 + 0.11491) 0.34472 FCF − 19.8653 = −9(6.3057)(0.25953) FCF = +14.8998 N

FCF = +14.90 N compression 

50° ΣF = ΣFeff : FBE + 14.9 lb − (9 kg)(9.81)sin 50° = 0 FBF = +52.734 N

FBE = +52.7 N compression 

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PROBLEM 16.17 Members ACE and DCB are each 600 mm long and are connected by a pin at C. The mass center of the 10-kg member AB is located at G. Determine (a) the acceleration of AB immediately after the system has been released from rest in the position shown, (b) the corresponding force exerted by roller A on member AB. Neglect the weight of members ACE and DCB.

SOLUTION Analysis of linkage Since members ACE and DCB are of negligible mass, their effective forces may also be neglected and the methods of statics may be applied to their analysis. Free body: Entire linkage: ΣM D = 0:

( By − E ) (0.6 cos 30°) − Bx (0.6 sin 30°) = 0 ( By − E ) cos 30° − Bx sin 30° = 0

(1)

Free body: member ACE ΣM C = 0: A (0.3 cos30°) − E (0.3 cos 30°) = 0;

E=A

Carrying into Eq. (1): ( By − A) cos 30° − Bx sin 30° = 0

(2)

Equations of Motion for Member AB

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PROBLEM 16.17 (Continued)

(a)

+

60° ΣFt − Σ( Ft )eff :

( By − A) cos30° − Bx sin 30° + W cos 30° = ma

Recalling equation (2), we have, W cos 30° = ma

a =

W cos 30° = g cos 30° m

a = (9.81 m/s 2 ) cos 30°

(b)

a = 8.50 m/s 2

60° 

ΣM B = Σ(M B )eff : W (0.15 m) − A(0.6 m) cos 30° = (ma sin 60°) (0.15 m) − (ma cos 60°) (0.05 m)

But,

ma = W cos30°

0.15 W − 0.6 A cos 30° = W cos 30°(0.15 sin 60° − 0.05 cos 60°)   1 1 1 A=W cos 60°  = 0.11384 W = 0.11384 mg − sin 60° + 12  4 cos 30° 4 

Recalling that m = 10 kg so that mg = 98.1 N, A = 0.11384(98.1) = 11.168 N A = 11.17 N 

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PROBLEM 16.18 The 15-lb rod BC connects a disk centered at A to crank CD. Knowing that the disk is made to rotate at the constant speed of 180 rpm, determine for the position shown the vertical components of the forces exerted on rod BC by the pins at B and C.

SOLUTION We first determine the acceleration of Point B of disk.

ω = 180 rpm = 18.85 rad/s Since ω = constant aB = (aB ) n = rω 2  8  =  ft  (18.85 rad/s)2  12  a B = 236.9 ft/s 2

60°

Since rod BC is in translation. a = aB = 236.9 ft/s 2

60°

Vertical components of forces at B and C. ΣM B = Σ( M B )eff : C y (30 in.) − W (15 in.) = −ma sin 60°(15 in.)

Since

15 lb (236.9) = 110.36 lb 32.2 30C y − (15)(15) = −110.36sin 60°(15) = −95.57(15)

W = 15 lb and ma =

2C y = −95.57 + 15 = −80.57, C y = −40.285 lb ΣFy = Σ( Fy )eff :

C y = 40.3 lb 

By − W + C y = −ma sin 60° By = W − C y −

15 (236.9)sin 60° = 15 + 40.285 − 95.57 32.2

By = −40.285 lb

B y = 40.3 lb 

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PROBLEM 16.19 The triangular weldment ABC is guided by two pins that slide freely in parallel curved slots of radius 6 in. cut in a vertical plate. The weldment weighs 16 lb and its mass center is located at Point G. Knowing that at the instant shown the velocity of each pin is 30 in./s downward along the slots, determine (a) the acceleration of the weldment, (b) the reactions at A and B.

SOLUTION v = 30 in./s

Slot:

an =

v 2 (30 in./s)2 = = 150 in./s 2 r 6 in.

a n = 12.5 ft/s 2 at = at

Weldment is in translation

30°

60°

an = 12.5 ft/s 2

60° ΣF = ΣFeff : mg cos 30° = mat at = 27.886 ft/s 2

(a)

60°

Acceleration

β = tan −1

an 12.5 = tan −1 = 24.14° 27.886 at

a 2 = at2 + an2 = (27.886) 2 + (12.5) 2 a = 30.56 ft/s 2

84.1°

a = 30.6 ft/s 2

84.1° 

  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1428

PROBLEM 16.19 (Continued)

(b)

Reactions ma =

16 lb (30.56 ft/s 2 ) = 15.185 lb 2 32.2 ft/s

ΣM A = Σ( M A )eff : B cos 30°(9 in.) − (16 lb)(6 in.) = (15.185 lb)(cos84.14°)(3 in.) − (15.185 lb)(sin 84.14°)(6 in.) 7.794 B − 96 = + 4.651 − 90.634 B = +1.285 lb

B = 1.285 lb

30° 

ΣFx = Σ( Fx )eff : A cos 30° + B cos 30° = ma cos84.14° A cos 30° + (1.285 lb) cos 30° = (15.185 lb) cos84.14° A cos 30° + 1.113 lb = 1.550 lb A = +0.505 lb

A = 0.505 lb

30° 

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PROBLEM 16.20 The coefficients of friction between the 30-lb block and the 5-lb platform BD are μs = 0.50 and μk = 0.40. Determine the accelerations of the block and of the platform immediately after wire AB has been cut.

SOLUTION Assume that the block does not slide relative to the platform. Draw the free body diagram of the platform and block.

ΣFt = Σ( Ft )eff : W sin 30° = ma a=

W 1 sin 30° = g = 16.1 ft/s 2 m 2

a = 16.1 ft/s 2

30°

Check whether or not the block will slide relative to the platform. Draw the free body diagram of the block alone.

1  1 ma = m  g  = W = 15 lb 2  2 ΣFx = Σ( Fx )eff : F = (15 lb) cos 30° = 12.99 lb ΣFy = Σ( Fy )eff : 30 lb − N = (15 lb) sin 30° N = 30 − 7.5 = 22.5 lb

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PROBLEM 16.20 (Continued)

Thus,

Fm = μ s N = 0.50 (22.5 lb) = 11.25 lb

Since

F > Fm ,

the block will slide

Now assume that the block slides relative to the platform. Equations of motion for block: (assuming sliding)

ΣFy = Σ( Fy )eff : 30 − N =

30 ( ab ) y g

N  (ab ) y = g 1 −  30   ΣFx = Σ( Fx )eff : 0.40 N =

(1)

30 (ab ) x g

N (ab ) x = g    75 

(2)

Equations of motion for platform.

ΣFt = Σ( Ft )eff : ( N − 5)sin 30° − (0.40 N) cos 30° = a p = g (0.5 + 0.030718 N)

5 aP g

(3)

If contact is maintained between block and platform, we must have (ab ) y = (a p ) y = a p sin 30°

(4)

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PROBLEM 16.20 (Continued)

Substituting from (1) and (3) into (4): N  g 1 −  = g (0.5 + 0.030718 N) sin 30°  30  (0.015359 + 0.033333) N = 0.75 N = 15.403 lb

Substituting for N in (2) and (1): (ab ) x = (32.2 ft/s 2 )

15.403 , (ab ) x = 6.61 ft/s 2 75

 15.403  (ab ) y = (32.2 ft/s 2 ) 1 − , (ab ) y = 15.67 ft/s 2 30  

ab = 17.01 ft/s 2

67.1° 

Substituting for N in (3): aP = (32.2 ft/s 2 )(0.5 + 0.030718 × 15.403) = 31.335 ft/s 2

a P = 31.3 ft/s 2

30° 

Note: Since N > 0, we check that contact between block and platform is maintained.

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PROBLEM 16.21 Draw the shear and bending-moment diagrams for the vertical rod AB of Problem 16.16. PROBLEM 16.16 Three bars, each of mass 3 kg, are welded together and are pin-connected to two links BE and CF. Neglecting the weight of the links, determine the force in each link immediately after the system is released from rest.

SOLUTION From the solution of Problem 16.16, the acceleration of all points of vertical rod AB is a = 6.3057 m/s 2

40°

or

a = 4.8304 m/s 2

+4.0532 m/s 2

Mass of rod AB:

m = 3 kg

Mass per unit length:

m 3 kg = = 6.6667 kg/m l 0.450 m

Effective force per length:

m a l (6.6667 kg/m)(4.8304 m/s 2 ) 32.203 N/m

+ (6.6667 kg/m)(4.0532 m/s 2 )

+ 27.021 N/m

Only the horizontal component contributes to the shear and bending moment. Let x be a vertical coordinate (positive down) with its origin at A. Draw the free body diagram of the portion of the rod AB lying above the section defined by x.

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PROBLEM 16.21 (Continued)

ΣF = Σ( F j )eff : V = 32.203x ΣM J = Σ( M j )eff : M = (32.203x)

x 2

= 16.101x 2

Shear and bending moment diagrams.

Vmax = (32.203 N/m)(0.450 m)

Vmax = 14.49 N 

M max = (16.101 N/m)(0.450 m) 2

M max = 3.26 N ⋅ m 

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PROBLEM 16.22* Draw the shear and bending-moment diagrams for the connecting rod BC of Problem 16.18. PROBLEM 16.18 The 15-lb rod BC connects a disk centered at A to crank CD. Knowing that the disk is made to rotate at the constant speed of 180 rpm, determine for the position shown the vertical components of the forces exerted on rod BC by the pins at B and C.

SOLUTION We first determine the acceleration of Point B of disk:

ω = 180 rpm = 18.85 rad/s Since ω = constant aB = (aB ) n = rω 2

 8  =  ft  (18.85 rad/s)2  12  a B = 236.9 ft/s 2

60°

Since rod BC is in translation. a = a B = 236.9 ft/s 2

60°

ΣM B = Σ( M B )eff : C y (30 in.) − W (15 in.) = −ma sin 60°(15 in.) Since W = 15 lb

and

ma =

15 lb (236.9) = 110.36 lb: 32.2

30C y − (15)(15) = −110.36sin 60°(15) = −95.57(15) 2C y = −95.57 + 15 = −80.57, C y = −40.285 lb

C y = 40.3 lb ΣFy = Σ( Fy )eff : B y − W + C y = −ma sin 60° By = W − C y −

15 (236.9)sin 60° = 15 + 40.285 − 95.57 32.2

B y = −40.285 lb

B y = 40.3 lb

a y = 236.7 sin 60°

a y = 205.2 ft/s 2

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PROBLEM 16.22* (Continued)

Distributed weight per unit length = w =

Distributed mass per unit length = Vertical component of effective forces =

15 lb = 6 lb/ft  30    ft  12 

w 6 6 = = lb ⋅ s 2 /ft 2 g g 32.2 w 6 ay = (205.2) g 32.2

= 38.236 lb/ft

ΣFy = Σ( Fy )eff : By + C y + (2.5 ft)(6 lb/ft) = (2.5 ft)(38.236 lb/ft) By + C y = 80.59 lb

From symmetry.

By = C y = 40.285 lb

B y = C y = 40.3 lb

Maximum value of bending moment occurs at G, where V = 0: | M |max = Area under V-diagram from B to G =

1 (40.285 lb)(1.25 ft) 2

| M |max = 25.2 lb ⋅ ft  VB = −40.3 lb  \

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PROBLEM 16.23 For a rigid slab in translation, show that the system of the effective forces consists of vectors (Δmi ) a attached to the various particles of the slab, where a is the acceleration of the mass center G of the slab. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a single vector m a attached at G.

SOLUTION Since slab is in translation, each particle has same acceleration as G, namely a. The effective forces consist of (Δmi ) a.

The sum of these vectors is: or since

Σ( Δmi ) a = (ΣΔmi ) a ΣΔmi = m, Σ(Δmi ) a = m a

The sum of the moments about G is:

Σri′× (Δmi ) a = (ΣΔmi ri′) × a

(1)

But, ΣΔmi ri′ = mr ′ = 0, because G is the mass center. It follows that the right-hand member of Eq. (1) is zero. Thus, the moment about G of ma must also be zero, which means that its line of action passes through G and that it may be attached at G.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1437

PROBLEM 16.24 For a rigid slab in centroidal rotation, show that the system of the effective forces consists of vectors −(Δmi )ω 2 r'i and (Δmi )(α × r'i ) attached to the various particles Pi of the slab, where ω and α are the angular velocity and angular acceleration of the slab, and where r'i denotes the position vector of the particle Pi relative to the mass center G of the slab. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a couple Ι α .

SOLUTION For centroidal rotation: Effective forces are:

ai = (ai )t + (ai )n = α × ri′ − ω 2 ri′

(Δmi )ai = ( Δmi )(α × ri ) − (Δmi )ω 2 ri′(Δmi )(α × ri )

Σ( Δmi )ai = Σ(Δmi )(α × ri′) − Σ( Δmi )ω 2 ri′ = α × Σ(Δmi )ri′ − ω 2 Σ(Δmi )ri′

Since G is the mass center,

Σ(Δmi ) ri′ = 0

effective forces reduce to a couple, Summing moments about G, Σ(r1′ × Δmi ai ) = Σ[ri′ × (Δmi )(α × ri′ )] − Σri′ × ( Δmi )ω 2 ri′

But, and, Thus, Since

ri′ × (Δmi )ω 2ri′ = ω 2 (Δmi )(ri′ × ri′) = 0 ri′ × (Δmi )(α × ri′) = ( Δmi )r1′2 α

Σ(ri′ × Δmi ai ) = Σ(Δmi ) r12α = Σ( Δmi )r ′i 2 α Σ(Δm) r i′2 = I ,

the moment of the couple is I α.

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PROBLEM 16.25 The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 50-kg rotor, which has a centroidal radius of gyration of 180 mm, then coasts to rest. Knowing that kinetic friction results in a couple of magnitude 3.5 N ⋅ m exerted on the rotor, determine the number of revolutions that the rotor executes before coming to rest.

SOLUTION I = mk 2 = (50)(0.180) 2 = 1.62 kg ⋅ m 2 M = I α:

3.5 N ⋅ m = (1.62 kg ⋅ m 2 )α

α = 2.1605 rad/s 2 (deceleration)  2π    60  = 120π rad/s

ω0 = 3600 rpm  ω 2 = ω02 + 2αθ

0 = (120π rad/s) 2 + 2(−2.1605 rad/s 2 )θ

θ = 32.891 × 103 rad = 5235.26 rev

θ = 5230 rev 

or

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PROBLEM 16.26 It takes 10 min for a 6000-lb flywheel to coast to rest from an angular velocity of 300 rpm. Knowing that the radius of gyration of the flywheel is 36 in., determine the average magnitude of the couple due to kinetic friction in the bearings.

SOLUTION m=

W 6000 = = 186.335 lb ⋅ s 2 /ft g 32.2

k = 36 in. = 3 ft

I = mk 2 = (186.336)(3) 2 = 1677 lb ⋅ s 2 ⋅ ft  2π   = 10π rad/s  60 

ω0 = 300 rpm  ω = ω0 + α t

0 = 10π rad/s + α (600 s)

α = −0.05236 rad/s 2 M = I α = (1677 lb ⋅ s 2 ⋅ ft)(−0.05236 rad/s 2 ) = 87.81 lb ⋅ ft M = 87.8 lb ⋅ ft 

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PROBLEM 16.27 The 8-in.-radius brake drum is attached to a larger flywheel that is not shown. The total mass moment of inertia of the drum and the flywheel is 14 lb ⋅ ft ⋅ s 2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.35. Knowing that the angular velocity of the flywheel is 360 rpm counterclockwise when a force P of magnitude 75 lb is applied to the pedal C, determine the number of revolutions executed by the flywheel before it comes to rest.

SOLUTION Lever ABC: Static equilibrium (friction force ) F = μk N = 0.35 N ΣM A = 0: N (10 in.) − F (2 in.) − (75 lb)(9 in.) = 0 10 N − 2(0.35 N ) − 675 = 0 N = 72.58 lb

F = μk N = 0.35(72.58 lb) = 25.40 lb

Drum: 2 ft 3  2π  ω0 = 360 rpm    60  ω0 = 12π rad/s

r = 8 in. =

ΣM D = Σ( M D )eff : Fr = I α 2  (25.4 lb)  ft  = (14 lb ⋅ ft ⋅ s 2 )α 3 

α = 1.2097 rad/s 2 (deceleration) ω 2 = ω02 + 2αθ : 0 = (12π rad/s) 2 + 2(−1.2097 rad/s 2 )θ θ = 587.4 rad  1  2π

θ = 587.4 rad 

  = 93.49 rev 

θ = 93.5 rev 

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PROBLEM 16.28 Solve Problem 16.27, assuming that the initial angular velocity of the flywheel is 360 rpm clockwise. PROBLEM 16.27 The 8-in.-radius brake drum is attached to a larger flywheel that is not shown. The total mass moment of inertia of the drum and the flywheel is 14 lb ⋅ ft ⋅ s 2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.35. Knowing that the angular velocity of the flywheel is 360 rpm counterclockwise when a force P of magnitude 75 lb is applied to the pedal C, determine the number of revolutions executed by the flywheel before it comes to rest.

SOLUTION Lever ABC: Static equilibrium (friction force ) F = μ k N = 0.35 N ΣM A = 0: N (10 in.) + F (2 in.) − (75 lb)(9 in.) = 0

10 N + 2(0.35 N ) − 675 = 0 N = 63.08 lb F = μk N = 0.35(63.08 lb) = 22.08 lb

Drum: 2 ft 3  2π  ω0 = 360 rpm    60  ω0 = 12π rad/s r = 8 in. =

ΣM D = Σ( M D )eff = 0 : Fr = I α 2  (22.08 lb)  ft  = (14 lb ⋅ ft ⋅ s 2 )α 3 

α = 1.0515 rad/s 2 (deceleration) ω 2 = ω02 + 2αθ : 0 = (12π rad/s) 2 + 2( −1.0515 rad/s 2 )θ θ = 675.8 rad θ = 675.8 rad = 107.56 rev

θ = 107.6 rev 

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PROBLEM 16.29 The 100-mm-radius brake drum is attached to a flywheel which is not shown. The drum and flywheel together have a mass of 300 kg and a radius of gyration of 600 mm. The coefficient of kinetic friction between the brake band and the drum is 0.30. Knowing that a force P of magnitude 50 N is applied at A when the angular velocity is 180 rpm counterclockwise, determine the time required to stop the flywheel when a = 200 mm and b = 160 mm.

SOLUTION Equilibrium of lever AD (Dimensions in mm) ΣM C = 0; (50 N)(200) + T2 (40) − T1(160) = 0 4 T1 − T2 = 250 N

(1)

Equation of Motion for flywheel and drum I =mk

2

= (300 kg)(0.600 m) 2 = 108 kg ⋅ m 2 r = 0.100 m ΣM C = Σ(M C )eff : T2r − T1r = Iα (T2 − T1)(0.100 m) = 108 α

α = (925.93 × 10−6 )(T2 − T1)

(2)

Belt Friction Using μk instead of μs since the brake band is slipping: T2 = e μk β T1

Making μk = 0.30

or

T2 = T1 e μk β

(3)

T2 = 2.5663T1

(4)

and β = 180° = π rad in (3): T2 = T1 e0.30π

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PROBLEM 16.29 (Continued)

Substituting for T2 from (4) into (1):

From (1):

4T1 − 2.5663T1 = 250 N

T1 = 174.38 N

T2 = 4(174.38) − 250

T2 = 447.51 N

Substituting for T1 and T2 into (2):

α = (925.93 × 10−6 )(447.51 − 174.38), α = 0.2529 rad/s 2 Kinematics

ω 0 = + 18.850 rad/s

ω0 = 180 rpm

α = − 0.2529 rad/s 2 

α = 0.2529 rad/s 2

ω = ω0 + α t : 0 = 18.850 − 0.2529t t = 74.5 s 

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PROBLEM 16.30 The 180-mm radius disk is at rest when it is placed in contact with a belt moving at a constant speed. Neglecting the weight of the link AB and knowing that the coefficient of kinetic friction between the disk and the belt is 0.40, determine the angular acceleration of the disk while slipping occurs.

SOLUTION Belt:

F = μk N

Disk:

ΣFx = Σ( Fx )eff :

N − FAB cos θ = 0 FAB cos θ = N

(1)

ΣFy = Σ( Fy )eff : μ k N + FAB sin θ − mg = 0

FAB sin θ = mg − μ k N Eq. (2) : Eq. (1)

tan θ =

(2)

mg − μk N N

N tan θ = mg − μ k N mg N= tan θ + μk mg μk F = μk N = tan θ + μk

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PROBLEM 16.30 (Continued)

ΣM A = Σ( M A )eff : Fr = I α

r I

α= F mg μk r ⋅ 1 2 tan θ + μk mr 2 μk 2g = ⋅ r tan θ + μk =

Data:

r = 0.18 m

θ = 60° μk = 0.40 α=

2(9.81 m/s 2 ) 0.40 ⋅ 0.18 m tan 60° + 0.40

α = 20.4 rad/s 2



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PROBLEM 16.31 Solve Problem 16.30, assuming that the direction of motion of the belt is reversed. PROBLEM 16.30 The 180-mm disk is at rest when it is placed in contact with a belt moving at a constant speed. Neglecting the weight of the link AB and knowing that the coefficient of kinetic friction between the disk and the belt is 0.40, determine the angular acceleration of the disk while slipping occurs.

SOLUTION F = μk N

Belt: Disk:

ΣFx = Σ( Fx )eff : N − FAB cos θ ; FAB cos θ = N

(1)

ΣFy = Σ( Fy )eff : FAB sin θ − mg − μk N = 0 FAB sin θ = mg + μk N

(2)

mg + μk N Eq. (2) : tan θ = Eq. (1) N N tan θ = mg + μ k N ; N =

mg tan θ − μk

ΣM A = Σ( M A )eff : Fr = I α

α= Data:

μ r μk 2g Fr μk Nr mg = = k 2⋅ = ⋅ 1 r I r tan θ − μk mr tan θ − μk 2

r = 0.18 m, θ = 60°, μk = 0.40

α=

2(9.81 m/s 2 ) 0.40 ⋅ 0.18 m tan 60° − 0.40

α = 32.7 rad/s 2



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PROBLEM 16.32 In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wrapped around the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate bearing friction from the computation, a second block of mass 24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.

SOLUTION Kinematics:

Kinetics:

ΣM B = Σ( M B )eff : (m A y )r − M f = I α + (m A a)r

m A gr − M f = I

a + mA ar r

(1)

y =3m t = 4.6 s

Case 1:

1 2 at 2 1 3 m = a(4.6 s) 2 2 a = 0.2836 m/s 2 y=

m A = 12 kg

Substitute into Eq. (1)  0.2836 m/s 2 (12 kg)(9.81 m/s 2 )(0.6 m) − M f = I  0.6 m 

 2  + (12 kg)(0.2836 m/s )(0.6 m) 

70.632 − M f = I (0.4727) + 2.0419

(2)

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PROBLEM 16.32 (Continued)

y=3m

Case 2:

t = 3.1 s 1 y = at 2 2 1 3 m = a(3.1 s) 2 2 a = 0.6243 m/s 2 m A = 24 kg

Substitute into Eq. (1):  0.6243 m/s 2 (24 kg)(9.81 m/s 2 )(0.6 m) − M f = I  0.6 m 

 2  + (24 kg)(0.6243 m/s )(0.6 m) 

141.264 − M f = I (1.0406) + 8.9899

(3)

Subtract Eq. (1) from Eq. (2) to eliminate Mf 70.632 = I (1.0406 − 0.4727) + 6.948 63.684 = I (0.5679) I = 112.14 kg ⋅ m 2

I = 112.1 kg ⋅ m 2 

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PROBLEM 16.33 The flywheel shown has a radius of 20 in. a weight of 250 lbs, and a radius of gyration of 15 in. A 30-lb block A is attached to a wire that is wrapped around the flywheel, and the system is released from rest. Neglecting the effect of friction, determine (a) the acceleration of block A, (b) the speed of block A after it has moved 5 ft.

SOLUTION Kinematics:

Kinetics:

ΣM B = Σ( M B )eff : (mA g )r = I α + (m A a )r

a m A gr = mF k 2   + m A ar r WA a= 2 mA + mF ( kr )

(a)

Acceleration of A

a=

(

(30 lb)

30 lb 32.2 ft/s 2

)+(

250 lb 32.2 ft/s2

)( )

15 in. 2 20 in.

aA = 5.66 ft/s 2 

or (b)

= 5.6615 ft/s 2

Velocity of A

v A2 = (v A )02 + 2a A s

For s = 5 ft

v A2 = 0 + 2(5.6615 in./s 2 )(5 ft) = 56.6154 ft 2 /s 2 v A = 7.5243 ft/s v A = 7.52 ft/s 

or

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PROBLEM 16.34 Each of the double pulleys shown has a mass moment of inertia of 15 lb ⋅ ft ⋅ s 2 and is initially at rest. The outside radius is 18 in., and the inner radius is 9 in. Determine (a) the angular acceleration of each pulley, (b) the angular velocity of each pulley after Point A on the cord has moved 10 ft.

SOLUTION Case 1: (a)

(b)

 9  ΣM 0 = Σ( M 0 )eff : (160 lb)  ft  = (15 lb ⋅ ft ⋅ s 2 )α  12 

θ=

(a)



ω = 14.61 rad/s



10 ft = 13.333 rad ( 129 ft )

ω 2 = 2αθ = 2(8 rad/s 2 )(13.33 rad) Case 2:

α = 8 rad/s 2

 9   9 ΣM 0 = Σ( M 0 )eff : (160)   = 15α + ma    12   12 

160  9  9   α   32.2  12  12  120 = (15 + 2.795)α 120 = 15α +

α = 6.7435 rad/s 2

(b)

θ=

α = 6.74 rad/s 2



ω = 13.41 rad/s



10 ft = 13.333 rad ( 129 ft )

ω 2 = 2αθ = 2(6.7435 rad/s 2 )(13.333 rad)

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PROBLEM 16.34 (Continued)

Case 3:

ΣM 0 = Σ( M 0 )eff :

(a)

460  9  300  9   9   9 (460)   − (300)   = 15α + a a + 32.2  12  32.2  12   12   12  2

120 = 15α +

2

460  9  300  9  α+ α   32.2  12  32.2  12 

α = 4.2437 rad/s 2

(b)

θ=

(a)



ω = 10.64 rad/s 2



10 ft = 13.333 rad 9 12

ω 2 = 2αθ = 2(4.2437)(13.333)

Case 4:

α = 4.24 rad/s 2

80  18   18  ΣM 0 = Σ( M 0 )eff : (80)   = 15α + a 32.2  12   12  2

80  18  α 32.2  12  120 = (15 + 5.5901)α 120 = 15α +

α = 5.828 rad/s 2 α = 5.83 rad/s 2

(b)

θ=



10 ft = 6.6667 rad ( 1812 ft )

ω 2 = 2αθ = 2(5.828 rad/s 2 )(6.6667 rad)

ω = 8.82 rad/s



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PROBLEM 16.35 Each of the gears A and B has a mass of 9 kg and has a radius of gyration of 200 mm; gear C has a mass of 3 kg and has a radius of gyration of 75 mm. If a couple M of constant magnitude 5 N-m is applied to gear C, determine (a) the angular acceleration of gear A, (b) the tangential force which gear C exerts on gear A.

SOLUTION Kinematics: We express that the tangential components of the accelerations of the gear teeth are equal: at = 0.25α A

= 0.25α B = 0.1α C

αB = α A α C = 2.5α A

(1)

Kinetics: I A = m A k A2 = 9(0.20)2

Gear A:

= 0.36 kg ⋅ m 2 ΣM A = Σ( M A )eff : F (0.25) = 0.36α A F = 1.44α A

(2)

Because of symmetry, gear C exerts an equal force F on gear B. I C = mC kC2

Gear C:

= 3(0.075) 2 = 0.016875 kg ⋅ m 2 ΣM C = Σ( M C )eff : M − 2 FrC = I C α C 5 N ⋅ m − 2 F (0.1 m) = 0.016875α C

(a)

Angular acceleration of gear A. Substituting for α C from (1) and for F from (2): 5 − 2(1.44α A )(0.1) = 0.016875(2.5α A ) 5 − 0.288α A = 0.04219α A 5 = 0.3302α A

α A = 15.143 rad/s 2 (b)

α A = 15.14 rad/s 2



Tangential force F. Substituting for α A into (2):

F = 1.44(15.143)

FA = 21.8 N



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PROBLEM 16.36 Solve Problem 16.35, assuming that the couple M is applied to disk A. PROBLEM 16.35 Each of the gears A and B has a mass of 9 kg and has a radius of gyration of 200 mm; gear C has a mass of 3 kg and has a radius of gyration of 75 mm. If a couple M of constant magnitude 5 N-m is applied to gear C, determine (a) the angular acceleration of gear A, (b) the tangential force which gear C exerts on gear A.

SOLUTION αA = αA

Kinematics:

αA = αA

αB = α B

At the contact point between gears A and C, at = rAα A = rCα C

αC =

rA 0.25 αA = αA 0.1 rC

At the contact point between gears B and C, at = rBα B = rCα C

αB =

rC 0.1 0.25 αC = ⋅ αA = αA 0.25 0.1 rA

Kinetics: Gear B:

FBC rB = I Bα B = I Bα A

ΣM B = Σ( M B )eff :

FBC =

Gear C:

IB αA rB

ΣM C = Σ( M C )eff : FBC rC + FAB rC = I C α C rC 0.25 I Bα A + FAB rC = I Cα A rB 0.1 FAC =

Gear A:

M A = Σ( M A )eff : MA −

rA rC

1 rC

0.25   0.1  0.25 I B + 0.1 I C  α A  

(1)

M A − rA FAB = I Aα A

0.25   0.1  0.25 I B + 0.1 I C  α A = I Aα A   2    0.25  M A =  I A + IB +  IC  α A     0.1   

(2)

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PROBLEM 16.36 (Continued)

Data: From Eq. (2)

m A = mB = 9 kg k A = k B = 0.2 m I A = I B = m A k A2 = 9(0.2) 2 = 0.36 kg ⋅ m 2 mC = 3 kg kC = 0.075 m I C = 3(0.075)2 = 0.016875 kg ⋅ m 2 MA = 5 N⋅m 2    0.25  5 = 0.36 + 0.36 +  (0.016875)  α A   0.1   

(a)

Angular acceleration.

(b)

Tangential gear force. From Eq. (1)

α A = 6.0572 rad/s 2

FAC =

α A = 6.06 rad/s 2



 1  0.1   0.25    (0.36) +   (0.016875)  (6.0572) 0.1  0.25   0.1  

= 11.278

FAC = 11.28 N



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PROBLEM 16.37 Gear A weighs 1 lb and has a radius of gyration of 1.3 in; Gear B weighs 6 lb and has a radius of gyration of 3 in.; gear C weighs 9 lb and has a radius of gyration of 4.3 in. Knowing a couple M of constant magnitude of 40 lb ⋅ in is applied to gear A, determine (a) the angular acceleration of gear C, (b) the tangential force which gear B exerts on gear C.

SOLUTION Masses and moments of inertia. mA =

1 lb = 0.031056 lb ⋅ s 2 /ft 2 32.2 ft/s

mB =

6 lb = 0.18634 lb ⋅ s 2 /ft 32.2 ft/s 2

mC =

9 lb = 0.27950 lb ⋅ s 2 /ft 32.2 ft/s 2

 1.3 in.  I A = mA k A2 = (0.031056 lb ⋅ s 2 /ft)    12 in./ft 

2

= 0.36448 × 10−3 lb ⋅ s 2 ⋅ ft IB =

mB k B2

 3 in.  = (0.18634 lb ⋅ s /ft)    12 in./ft 

2

2

= 11.646 × 10−3 lb ⋅ s 2 ⋅ ft  4.3 in.  I C = mC kC2 = (0.27950 lb ⋅ s 2 /ft)    12 in./ft 

2

= 35.889 × 10−3 lb ⋅ s 2 ⋅ ft

Kinematics.

Gear A:

rA = 2 in.

Gear B:

r1 = 4 in., r2 = 2 in.

Gear C:

rC = 6 in.

Point of contact between A and B. at = rAα A = r1α B

αA =

r1 4 in. αB = αB rA 2 in.

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PROBLEM 16.37 (Continued)

Point of contact between B and C. at = r2α B = rCα C

αB = Summary.

rC 6 in. αC = αC r2 2 in.

α B = 3α C

(1)

α A = 2α B = 6α C

(2)

Kinetics. Applied couple: M = 40 lb ⋅ in. = 3.3333 lb ⋅ ft Gear A: ΣM A = Σ( M A )eff :

M − FAB rA = I Aα A FAB =

M I A (6α C ) − rA rA

3.3333 lb ⋅ ft (0.36448 × 10−3 )(6) αC − (2/12) ft (2/12) = 20 lb − 0.013121α C =

(3)

Gear B: ΣM B = Σ( M B )eff :

FAB r1 − FBC r2 = I Bα C

FBC = FAB =

r1 I Bα B − r2 r2

= 2 FAB −

3I B αC r2

= 2[20 lb − 0.013121α C ] −

(3)(11.646 × 10−3 ) αC (2/12)

= 40 lb − 0.23587α C

Gear C: ΣM C = Σ( M C )eff :

FBC rC = I Cα C

 6 (40 − 0.23587α C )   = (35.889 × 10−3 )α C  12  20 = 0.153824α C αC = 130.0 rad/s 2

(a)

Angular acceleration of gear C.

(b)

Tangential force which gear B exerts on gear C. FBC = 40 lb − (0.23587)(130.0) = 9.33 lb



9.33 lb 

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PROBLEM 16.38 Disks A and B are bolted together, and cylinders D and E are attached to separate cords wrapped on the disks. A single cord passes over disks B and C. Disk A weighs 20 lb and disks B and C each weigh 12 lb. Knowing that the system is released from rest and that no slipping occurs between the cords and the disks, determine the acceleration (a) of cylinder D, (b) of cylinder E.

SOLUTION Masses and moments of inertia. mD =

WD 15 lb = = 0.46584 lb ⋅ s 2 /ft 2 g 32.2 ft/s

mE =

WE 18 lb = 0.55901 lb ⋅ s 2 /ft g 32.2 ft/s 2

Assume disks have uniform thickness. 2

IA =

1 1 20 lb  8  2 m A rA2 =  ft  = 0.138026 lb ⋅ s ⋅ ft 2 2 32.2 ft/s 2  12  2

1 1 12 lb  6  ft  = 0.046584 lb ⋅ s 2 ⋅ ft I B = mB rB2 = 2  2 2 32.2 ft/s  12  I C = I B = 0.046584 lb ⋅ s 2 ⋅ ft I AB = I A + I B = 0.184610 lb ⋅ s 2 ⋅ ft

Kinematics:

a D = aD , a E = aE , α AB = α AB

, αC = αC

For inextensible cord between disk A and cylinder D,  8  aD = (at ) A = r1α AB =  ft  α AB = 0.66667α AB  12 

(1)

For inextensible cord between disks B and C, r2α AB = r3α C

αC =

 6 in.  r2 α AB =   α AB = α AB r3  6 in. 

For inextensible cord between disk C and cylinder E,  6  aE = (at ) D =  ft  α C = 0.5α AB  12 

(2)

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PROBLEM 16.38 (Continued)

Kinetics. Let T be the tension in the cord between disks B and C. ΣM B = Σ( M B )eff :

rW 1 D − r2T = I ABα AB + r1mD aD T=

r1 I α rm a WD − AB AB − 1 D D r2 r2 r2

I r1 m r2  WD −  AB + D 1  α AB r2 r2   r2  0.184610 (0.46584)(8/12) 2  8 in. (15 lb) −  = +  α AB 6 in. 6/12  6/12 

T=

T = 20 lb − 0.78330α AB ΣM C = Σ( M C )eff :

(3)

r3T − r3WE = I Cα C + r3 mE aE I  T = WE +  C + mE r3  α AB  r3 

 0.046584  + (0.55901)(6/12)  α AB T = 18 lb +  6/12   T = 18 lb + 0.37267α AB

(4)

Subtracting Eq. (4) from Eq. (3) to eliminate T, 0 = 2 lb − 1.15597α AB

(a)

Acceleration of cylinder D. From Eq. (1),

(b)

α AB = 1.7301 rad/s 2

aD = (0.66667)(7.7301)

a D = 1.153 ft/s 2 

Acceleration of cylinder E. From Eq. (2),

aE = (0.5)(1.7301)

a E = 0.865 ft/s 2 

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PROBLEM 16.39 A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are μs = 0.50 and μk = 0.40. For P = 3.6 lb, determine (a) whether slipping occurs between the belt and either cylinder, (b) the angular acceleration of each cylinder.

SOLUTION Assume that no slipping occurs. Then: Cylinder A.

1 2

abelt = (4 in.) α A = (8 in.) α B

αB = α A

(1)

 4  ΣM G = Σ( M G )eff : FA  ft  = I Aα A  12  2

 4  1 5 4  FA   =   αA  12  2 g  12  5 αA FA = 6 g

Cylinder B.

(2)

 8  ΣM G = Σ( M G )eff : FB  ft  = I Bα B  12  2

 8  1 20  8   1  FB   =    αA   12  2 g  12   2  20 α A FB = 6 g

Belt

ΣFA = 0: P − FA − FB = 0 3.60 −

(3) (4)

5 α A 20 α A − =0 6 g 6 g (3.60)6 g 25 = 0.864 g

αA =

α A = 27.82 rad/s 2

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PROBLEM 16.39 (Continued)

Check that belt does not slip. 5 (0.864) = 0.720 lb 6

From (2):

FA =

From (4):

Fe = P − FA = 3.60 − 0.720 = 2.88 lb

But

Fm = μ s N = 0.50(5 lb) = 2.50 lb

Since Fe > Fm , assumption is wrong. Slipping occurs between disk B and the belt.  1   αB ≠ 2 α A   

We redo analysis of B, assuming slipping.

FB = μk N = 0.40(5 lb) = 2 lb 2

 8  1 20  8  ΣM G = Σ( M G )eff : (2 lb)  ft  =   αB  12  2 g  12 

α B = 0.3g , Belt Eq. (4):

α B = 9.66 rad/s 2



P − FA − FB = 0 FA = P − FB = 3.60 − 2 = 1.60 lb

Since FA < Fm ,

There is no slipping between A and the belt. 

Our analysis of disk A, therefore is valid. Using Eq. (2), We have

1.60 lb =

5 αA 6 g

α A = 1.92 g

α A = 61.8 rad/s 2



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PROBLEM 16.40 Solve Problem 16.39 for P = 2.00 lb. PROBLEM 16.39 A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are μ s = 0.50 and μk = 0.40. For P = 3.60 lb, determine (a) whether slipping occurs between the belt and either of the cylinders, (b) the angular acceleration of each cylinder.

SOLUTION Assume that no slipping occurs. Then: Cylinder A

1 2

abelt = (4 in.) α A = (8 in.) α B

αB = α A

(1)

 4  ΣM G = Σ( M G )eff : FA  ft  = I Aα A  12  2

 4  1 5 4  FA   =   αA  12  2 g  12  5 αA FA = 6 g

Cylinder B

(2)

 4  ΣM G = Σ( M G )eff : FB  ft  = I Bα B  12  2

 8  1 20  8   1  FB   =    αA   12  2 g  12   2  20 α A FB = 6 g

Belt

(3)

ΣFA = 0: P − FA − FB = 0 2.00 −

5 α A 20 α A − =0 6 g 6 g (2.00)6 g 25 = 0.480 g

αA =

α A = 15.46 rad/s 2

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PROBLEM 16.40 (Continued)

From (2):

FA =

5 (0.480) = 0.400 lb 6

From (4):

FB =

20 (0.480) = 1.600 lb 6

But

FM = μ s N = 0.50(5 lb) = 2.50 lb

Thus, FA and FB are both < Fm . Our assumption was right: There is no slipping between cylinders and belt From (1):

α A = 15.46 rad/s 2



α B = 7.73 rad/s 2



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PROBLEM 16.41 Disk A has a mass of 6 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3 kg and is initially at rest. The disks are brought together by applying a horizontal force of magnitude 20 N to the axle of disk A. Knowing that μk = 0.15 between the disks and neglecting bearing friction, determine (a) the angular acceleration of each disk, (b) the final angular velocity of each disk.

SOLUTION While slipping occurs, a friction force F is applied to disk A, and F to disk B. Disk A: 1 mA rA2 2 1 = (6 kg)(0.08 m)2 2 = 0.0192 kg ⋅ m 2

IA =

ΣF : N = P = 20 N F = μ N = 0.15(20) = 3 N ΣM A = Σ( M A )eff : FrA = I Aα A (3 N)(0.08 m) = (0.0192 kg ⋅ m 2 )α A

α A = 10.227

α A = 12.50 rad/s 2



α B = 33.3 rad/s 2



Disk B: 1 mB rB2 2 1 = (3 kg)(0.06 m) 2 2 = 0.0054 kg ⋅ m 2

IB =

ΣM B = Σ( M B )eff : FrB = I Bα B (3 N)(0.06 m) = (0.0054 kg ⋅ m 2 )α B

α B = 33.333 rad/s 2 (ω A )0 = 360 rpm = 12π rad/s

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PROBLEM 16.41 (Continued)

Sliding stops when vC = vC ′ or ω A rA = ωB rB rA [(ω A )0 − α At ] = rBα B t (0.08 m)[12π rad/s − (12.5 rad/s 2 )t ] = (0.06 m)(33.333 rad/s 2 )t t = 1.00531 s

ω A = (ω A )0 − α At = 12π rad/s − (12.5 rad/s 2 )(1.00531 s) = 25.132 rad/s

ω A = (25.132 rad/s) = 240.00 rpm

or

ω A = 240 rpm



ω B = 320 rpm



ωB = α B t = 33.333 rad/s 2 (1.00531 s) = 33.510 rad/s

ωB = (33.510 rad/s) = 320.00 rpm

or

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PROBLEM 16.42 Solve Problem 16.41, assuming that initially disk A is at rest and disk B has an angular velocity of 360 rpm clockwise. PROBLEM 16.41 Disk A has a mass of 6 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3 kg and is initially at rest. The disks are brought together by applying a horizontal force of magnitude 20 N to the axle of disk A. Knowing that μk = 0.15 between the disks and neglecting bearing friction, determine (a) the angular acceleration of each disk, (b) the final angular velocity of each disk.

SOLUTION While slipping occurs, a friction force F is applied to disk A, and F to disk B. Disk A:

1 mA rA2 2 1 = (6 kg)(0.08 m)2 2 = 0.0192 kg ⋅ m 2

IA =

ΣF : N = P = 20 N F = μ N = 0.15(20) = 3 N ΣM A = Σ( M A )eff : FrA = I Aα A (3 N)(0.08 m) = (0.0192 kg ⋅ m 2 )α A

α A = 10.227 Disk B:

α A = 12.50 rad/s 2



α B = 33.3 rad/s 2



1 mB rB2 2 1 = (3 kg)(0.06 m) 2 2 = 0.0054 kg ⋅ m 2

IB =

ΣM A = Σ( M B )eff : FrB = I Bα B (3 N)(0.06 m) = (0.0054 kg ⋅ m 2 )α B

α B = 33.333 rad/s 2

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PROBLEM 16.42 (Continued)

Sliding starts when vC = vC ′ . That is when

ω A rA = ω B rB (α At )rA = [(ω B )0 − α B t ]rB [(12.5 rad/s 2 )t ](0.08 m) = [12π rad/s − (33.333 rad/s 2 )(t )](0.06 m) t = 4.02124 t = 0.75398 s

ω A = α At = (12.5 rad/s2 )(0.75398 s) = 9.4248 rad/s ω A = (9.4248 rad/s) = 90.00 rpm ω A = 90.0 rpm

or



ωB = (ωB )0 − α B t = 12π rad/s − (33.333 rad/s 2 )(0.75398 s) = 12.566 rad/s

ωB = (12.566 rad/s) = 120.00 rpm ω B = 120.0 rpm

or



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PROBLEM 16.43 Disk A has a mass mA = 3 kg, a radius rA = 300 mm, and an initial angular velocity ω0 = 300 rpm clockwise. Disk B has a mass mB = 1.6 kg, a radius rB = 180 mm, and is at rest when it is brought into contact with disk A. Knowing that μk = 0.35 between the disks and neglecting bearing friction, determine (a) the angular acceleration of each disk, (b) the reaction at the support C.

SOLUTION (a)

Disk B. ΣFy = Σ( Fy )eff : N − WB = 0 N = WB = mB g F = μ k N = μk mB g

Thus,

ΣM B = Σ( M B )eff : FrB = I Bα B 1 2

μk mB grB = mB rB2α B αB =

For the given data: α B =

2μk g rB

(1)

2(0.35)(9.81 m/s 2 ) = 38.15 rad/s 2 0.180 m

α B = 38.2 rad/s 2



WB = mB g = (1.6 kg)(9.81 m/s 2 ) = 15.696 N F = μ k mB g = (0.35)(15.696) = 5.4936 N ΣFx = Σ( Fx )eff : F − RB = 0

R B = 5.4936 N



Disk A: ΣM A = Σ( M A )eff : FrA = I Aα A 1 2

μk mB grA = mA rA2α A αA =

2μk g mB rA mA

(2)

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PROBLEM 16.43 (Continued)

For the given data:

αA =

2(0.35)(9.81 m/s 2 ) 1.6 kg = 9.156 rad/s 2 0.300 m 4 kg

ΣFx = Σ( Fx )eff : ( RA ) x − F = 0





α A = 9.16 rad/s 2

(R A ) x = 5.4936 N

 

ΣFy = Σ( Fy )eff : ( RA ) y − mA g − mB g = 0 ( RA ) y = (m A + mB ) g = (4 kg + 1.6 kg)(9.81 m/s 2 ) (R A ) y = 54.936 N

= 54.936 N

(b)

Reaction at C. ΣFx = 0: Cx = 0

ΣFy = 0: C y − 54.936 N = 0

C y = 54.936 N

ΣM C = 0: M C − (5.4936 N)(0.480 m) = 0

C = 54.9 N 

M C = 2.64 N ⋅ m



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PROBLEM 16.44 Disk B is at rest when it is brought into contact with disk A, which has an initial angular velocity ω 0. (a) Show that the final angular velocities of the disks are independent of the coefficient of friction μ k between the disks as long as μk ≠ 0. (b) Express the final angular velocity of disk A in terms of ω 0 and the ratio mA/mR of the masses of the two disks.

SOLUTION (a)

Disk B. ΣFy = Σ( Fy )eff : N − wB = 0 N = wB = mB g

Thus,

F = μ k N = μk mB g ΣM B = Σ( M B )eff : FrB = I Bα B 1 2

μk mB grB = mB rB2α B αB =

2μk g rB

(1)

Disk A. ΣM A = Σ( M A )eff : FrA = I Aα A 1 2

μk mB grA = mA rA2α A αA =

2μk g mB rA mA

Disk A.

ω A = ω 0 − α A t = ω0 −

Disk B.

ωB = α B t =

(2) 2μ k g mB t rA m A

2 μk g t rB

(3) (4)

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PROBLEM 16.44 (Continued)

When disks have stopped slipping: vP = ω A rA = ω B rB

ω0 rA − (2μk g ) t=

mB t = 2μk gt mA

ω0 rA 1 2 μk g 1 + mmB

A

t=

mA ω0 rA 2 μk g mA + mB

(5)

Substituting for t from (5) into (3) and (4):

ω A = ω0 −

m A + mB − mB m A + mB

ω A = ω0

mA m A + mB

2μ k g ω0 rA mA rA 2 μk g mA + mB

ω B = ω0

rA mA rB m A + mB

ω A = ω0 ωB = (a)

2μ k g mB ω0 rA ω0 mB mA = ω0 − rA m A 2μ k g m A + mB mA + mB



(7)

From Eqs. (6) and (7), it is apparent that ωA and ωB are independent of μ k. However, if μk = 0, we have from Eqs. (3) and (4) ω A = ω0

(b)

(6)

and ωB = 0.

ω A = ω0 /(1 + mB /m A )

We can write (6) in the form



which shows that ωA depends only upon ω 0 and mB/mA.

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PROBLEM 16.45 Cylinder A has an initial angular velocity of 720 rpm clockwise, and cylinders B and C are initially at rest. Disks A and B each weigh 5 lb and have a radius r = 4 in. Disk C weighs 20 lb and has a radius of 8 in. The disks are brought together when C is placed gently onto A and B. Knowing that μk = 0.25 between A and C and no slipping occurs between B and C, determine (a) the angular acceleration of each disk, (b) the final angular velocity of each disk.

SOLUTION Assume Point C, the center of cylinder C, does not move. This is true provided the cylinders remain in contact as shown. Slipping occurs initially between disks A and C and ceases when the tangential velocities at their contact point are equal. We first determine the angular accelerations of each disk while slipping occurs. Masses and moments of inertia: m A = mB = mC =

WA 5 lb = = 0.15528 lb ⋅ s 2 /ft g 32.2 ft/s 2

WC 20 lb = = 0.62112 lb ⋅ s 2 /ft g 32.2 ft/s 2 2

1 1  4 I A = I B = m A r 2 = (0.15528)   = 0.0086266 lb ⋅ s 2 ⋅ ft 2 2  12  2

IC =

1 1  8  mC (2r ) 2 = (0.62112)   = 0.138027 lb ⋅ s 2 ⋅ ft 2 2  12 

Kinematics: No slipping at contact BC. (at ) BC = (at ) BC

30°

(at ) BC = rα B = 2rα C

α B = 2α C

FAC = μ k N AC

Friction condition:

(1) (2)

Kinetics: Disk B:

ΣFB = Σ( FB )eff : FBC r = I Bα B IB 2I α B = B αC r r (2)(0.0086266) = αC 4/12 = 0.051760α C

FBC = FBC

(3)

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PROBLEM 16.45 (Continued)

Disk C: ΣM C = Σ( M C )eff : FAC (2r ) − FBC (2r ) = I Cα C FAC = FBC +

IC αC 2r

= 0.051760α C +

0.138027 αC 8/12

FAC = 0.25880α C

From Eq. (2),

N AC =

(4)

FAC = 1.03520α C 0.25

(5)

30°ΣF = ΣFeff : WB sin 30° + FBC − FAC cos 60° − N AC sin 60° = 0

+

(20)sin 30° + (0.051760 − 0.25880 cos 60° − 1.03520sin 60°)α C = 0 10 − 0.97415α C = 0

α C = 10.2654 rad/s 2 FBC = (0.051760)(10.2654) = 0.53134 lb. FAC = (0.25880)(10.2654) = 2.6567 lb. N AC = (1.03520)(10.2654) = 10.6267 lb.

Check that N BC > 0. +

60°ΣF = ΣFeff : N BC − WC cos 30° + N AC cos 60° − FAC sin 60° = 0 N BC = (20 lb) cos 30° − (10.6267 lb) cos 60° + (2.6567 lb) sin 60° = 0 N BC = 14.3079 lb.

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PROBLEM 16.45 (Continued)

Disk A: ΣM A = Σ( M A )eff : FAC r = I Aα A

αA = (a)

FAC r (2.6567)(4/12) = = 102.69 rad/s 2 0.0086266 IA

Angular accelerations of disks. From Eq. (1),

(b)

α A = 102.7 rad/s 2



α B = 20.5 rad/s 2



α C = 10.27 rad/s 2



Final angular velocities. Disk A:

ω0 = 720 rpm = 24π rad/s ω A = ω0 − α A t = 24π − 102.69t (vt ) AC = rω A =

4 (24π − 102.69t ) 12

( vt ) AC = (8π − 34.23) ft/s

Disk C:

ωC = α C t

30°

= 10.2654t

 8  ( vt )CA = 2rωC =   (10.2654)t  12  ( vt )CA = 6.8436t

30°

Time when tangential velocities are equal. 8π − 34.23t = 6.8436t

t = 0.6119 s

ω A = 24π − (102.69)(0.6119) = 12.562 rad/s ωC = (10.2654)(0.6119) = 6.2813 rad/s

ω A = 120.0 rpm



ωC = 60.0 rpm



ω B = 120.0 rpm



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PROBLEM 16.46 Show that the system of the effective forces for a rigid slab in plane motion reduces to a single vector, and express the distance from the mass center G of the slab to the line of action of this vector in terms of the centroidal radius of gyration k of the slab, the magnitude a of the acceleration of G, and the angular acceleration α .

SOLUTION We know that the system of effective forces can be reduced to the vector m a at G and the couple I α. We further know from Chapter 3 on statics that a force-couple system in a plane can be further reduced to a single force.

The perpendicular distance d from G to the line of action of the single vector m a is expressed by writing ΣM G = Σ( M G )eff : I α = (ma )d d=

I α mk 2α = ma ma

d=

k 2α  a

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PROBLEM 16.47 For a rigid slab in plane motion, show that the system of the effective forces consists of vectors (Δmi ) a , −(Δmi )ω 2ri′ , and (Δmi )(α × ri′ ) attached to the various particles Pi of the slab, where a is the acceleration of the mass center G of the slab, ω is the angular velocity of the slab, α is its angular acceleration, and ri′ denotes the position vector of the particle Pi , relative to G. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a vector m a attached at G and a couple I α.

SOLUTION Kinematics:

The acceleration of PL is ai = a + a Pi /G ai = a + α × ri′ + ω × (ω × ri′ )

= a + α × ri′ − ω 2ri′

Note: that α × ri′ is ⊥ to ri′ Thus, the effective forces are as shown in Figure P16.47 (also shown above). We write (Δmi )ai = ( Δmi ) a + ( Δmi )(α × ri′) − (Δmi )ω 2 ri′

The sum of the effective forces is Σ( Δmi )ai = Σ(Δmi ) a + Σ(Δmi )(α × ri′) − Σ(Δmi )ω 2 ri′ Σ( Δmi )ai = a Σ( Δmi ) + α × Σ( Δmi ) ri′ − ω 2 Σ( Δmi )ri′

We note that Σ( Δmi ) = m. And since G is the mass center, Σ( Δmi )ri′ = m ri ′ = 0 Σ( Δmi )ai = m a

Thus,

(1)

The sum of the moments about G of the effective forces is: Σ(ri′ × Δmi ai ) = Σri′ × Δmi a + Σri′ × (Δmi )(α × ri′) − Σri′ × (Δmi )ω 2 ri′ Σ(ri′ × Δmi ai ) = (Σri′Δmi ) a + Σ[ri′ × (α × ri′) Δmi ] − ω 2 Σ(ri′ × ri′)Δmi

Since G is the mass center, Also, for each particle, Thus,

Σri′Δmi = 0 ri′ × ri′ = 0

Σ(ri′ × Δmi ai ) = Σ [ri′ × (α i × ri′)Δmi ]

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PROBLEM 16.47 (Continued)

Since α ⊥ ri′, we have ri′ × (α × ri′) = ri2α and Σ(rii′ × Δmi ai ) = Σri′ 2 (Δmi )α = (Σri′ 2 Δmi )α

Since Σri′ 2 Δmi = I

Σ(ri′ × Δmi ai ) = I α

(2)

From Eqs. (1) and (2) we conclude that system of effective forces reduce to m a attached at G and a couple I α.

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PROBLEM 16.48 A uniform slender rod AB rests on a frictionless horizontal surface, and a force P of magnitude 0.25 lb is applied at A in a direction perpendicular to the rod. Knowing that the rod weighs 1.75 lb, determine (a) the acceleration of Point A, (b) the acceleration of Point B, (c) the location of the point on the bar that has zero acceleration.

SOLUTION W g 1 W 2 I= L 12 g

m=

ΣFx = Σ( Fx )eff : P = ma = a= ΣM G = Σ( M G )eff : P

W a g

P 0.25 lb 1 g= g= g W 1.75 lb 7

a=

1 g 7

L 1 W 2 Lα = Iα = 2 12 g

α =6

P 5 0.25 lb g 6 g =6 ⋅ = W L 1.75 lb L 7 L

α=

6g 7L

We calculate the accelerations immediately after the force is applied. After the rod acquires angular velocity, there will be additional normal accelerations. (a)

Acceleration of Point A. aA = a +

(b)

L 1 L 6 4 4 α = g + ⋅ g = g = (32.2 ft/s2 ) 2 7 2 7 7 7

a A = 18.40 ft/s 2



a B = 9.20 ft/s 2



Acceleration of Point B. aB = a −

L 1 L 6 2 2 α = g − ⋅ g = − g = − (32.2 ft/s2 ) 2 7 2 7 7 7

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PROBLEM 16.48 (Continued)

(c)

Point of zero acceleration. aP = 0 a − ( z P − zG )α = 0 z P − zG =

Since zG =

a

α

=

1 g 7 6 g ⋅ 7 L

=

1 L 6

1 L 2 1 1 2 L+ L= L 2 6 3 2 z P = (36 in.) 3 zP =

z P = 24.0 in. 

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PROBLEM 16.49 (a) In Problem 16.48, determine the point of the rod AB at which the force P should be applied if the acceleration of Point B is to be zero. (b) Knowing that P = 0.25 lb, determine the corresponding acceleration of Point A.

SOLUTION ΣFx = Σ( FX )eff : P = ma = a=

P g W

ΣM G = Σ( M G )eff : Ph = I α =

α=

(a)

1 W 2 Lα 12 g

12Ph g WL2

aB = a −

Position of force P.

W a g

L α 2

P L 12 Ph g− ⋅ g W 2 WL2 L 36 in. h= = = 6 in. 6 6 0=

Thus, P is located 12 in. from end A.  For (b)

h=

12 P ( L6 ) L P g : α= g=2 ⋅ 6 W L WL2

Acceleration of Point A. aA = a + aA = 2

L P L P g P α = g + ⋅2 =2 g 2 2 W L W W

0.25 lb (32.2 ft/s 2 ) 1.75 lb

a A = 9.20 ft/s 2



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PROBLEM 16.50 A force P of magnitude 3 N is applied to a tape wrapped around a thin hoop of mass 2.4 kg. Knowing that the body rests on a frictionless horizontal surface, determine the acceleration of (a) Point A, (b) Point B.

SOLUTION I = mr 2

Hoop:

ΣFx = Σ( Fx )eff : P = ma a=

P m

ΣM G = Σ( M G )eff : Pr = I α = mr 2α α=

(a)

P mr

Acceleration of Point A. a A = a + rα = aA = 2

(b)

P P  P  + r =2m m mr  

3N = 2.5 m/s 2 2.4 kg

a A = 2.50 m/s 2



Acceleration of Point B. aB = a − rα =

P  P  − r =0 m  mr 

aB = 0 

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PROBLEM 16.51 A force P is applied to a tape wrapped around a uniform disk that rests on a frictionless horizontal surface. Show that for each 360° rotation of the disk the center of the disk will move a distance π r.

SOLUTION I =

Disk:

1 2 mr 2

ΣFx = Σ( Fx )eff : P = ma a=

P m

ΣM G = Σ( M G )eff : Pr = I α Pr =

1 2 mr α 2

α=

2P mr

Let t1 be time required for 360° rotation. 1 θ = α t12 2 1  2P  2 2π rad =  t1 2  mr  2π mr t12 = P

Let x1 = distance G moves during 360° rotation. x1 =



1 2 1 P  2π mr  at1 = 2 2 m  P 

x1 = π r 

Q.E.D. 

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PROBLEM 16.52 A 250-lb satellite has a radius of gyration of 24 in. with respect to the y axis and is symmetrical with respect to the zx plane. Its orientation is changed by firing four small rockets A, B, C, and D, each of which produces a 4-lb thrust T directed as shown. Determine the angular acceleration of the satellite and the acceleration of its mass center G (a) when all four rockets are fired, (b) when all rockets except D are fired.

SOLUTION

W = 250 lb,

(a)

m=

 250 lb I = mk y2 =  2  32.2 ft/s

W g

  24 in. 2 2    = 31.056 slug ⋅ ft 12   

With all four rockets fired: ΣF = ΣFeff :

0 = ma

a=0 

ΣM G = Σ( M G )eff : 4Tr = I α  32 in.  −4(4 lb)   = 31.056α  12 

α = −1.3739 rad/s 2 α = −(1.374 rad/s 2 ) j 

(b)

With all rockets except D: ΣFx = Σ( Fx )eff : − 4 lb =

250 ax 32.2

ax = −0.51520 ft/s 2

ΣFz = Σ( Fz )eff :

0=

250 a 32.2

az = 0

ΣM G = Σ( M G )eff :

3Tr = I α

 32 in.  −3(4 lb)   = 31.056α   12 

a = −(0.515 ft/s 2 )i 

α = −(1.030 rad/s 2 ) j 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1483

PROBLEM 16.53 A rectangular plate of mass 5 kg is suspended from four vertical wires, and a force P of magnitude 6 N is applied to corner C as shown. Immediately after P is applied, determine the acceleration of (a) the midpoint of edge BC, (b) corner B.

SOLUTION

I =

1 1 m(b 2 + h 2 ) = (5 kg)[(0.4 m)2 + (0.3 m)2 ] = 0.10417 kg ⋅ m 2 12 12

ΣF = ΣFeff :

P = ma

6 N = (5 kg)a

a = + (1.2 m/s 2 )k

ΣM G = Σ( M G )eff : P(0.2 m) = I α (6 N)(0.2 m) = (0.10417 kg ⋅ m 2 )α

(a)

α = −(11.52 rad/s 2 ) j

a E = a + α × rE/G

= + (1.2 m/s 2 )k − (11.52 rad/s 2 ) j × (0.2 m)i = + (1.2 m/s 2 )k + (2.304 m/s 2 )k

a E = (3.50 m/s 2 )k 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1484

PROBLEM 16.53 (Continued)

(b)

a B = a + α × rB/G = +(1.2 m/s 2 )k − (11.52 rad/s)j × [(0.2 m)i +(0.15 m)k ] = +(1.2 m/s 2 )k + (2.304 m/s 2 )k + (1.728 m/s 2 )i a B = (1.728 m/s 2 )i + (3.5 m/s 2 )k 

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PROBLEM 16.54 A uniform slender L-shaped bar ABC is at rest on a horizontal surface when a force P of magnitude 4 N is applied at Point A. Neglecting friction between the bar and the surface and knowing that the mass of the bar is 2 kg, determine (a) the initial angular acceleration of the bar, (b) the initial acceleration of Point B.

SOLUTION (a)

Mass center at G

x=

(m /2) x1 + ( m /2) x2 m

x=

1(0.15) + 1(0) = 0.075 m 2

x=z

 1 m  m (0.075) 2 + (0.075) 2  = 0.0375 kg ⋅ m 2 I = 2    (0.3)2 + 2 12  2  

(

ΣFx = 0 = maGx , ΣFz = − 4 = maGz ,

)

aGx = 0 aGz = − 2 m/s 2

ΣM G = −4(0.075) = 0.0375α ,

(b)

α = − (8 rad/s 2 ) j 

a B = aG + a B/G = −2k − 8 j × (−0.075i − 0.075k )

a B = (0.6 m/s 2 )i − (2.6 m/s 2 )k 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1486

PROBLEM 16.55 By pulling on the string of a yo-yo, a person manages to make the yo-yo spin, while remaining at the same elevation above the floor. Denoting the mass of the yo-yo by m, the radius of the inner drum on which the string is wound by r, and the centroidal radius of gyration of the yo-yo by k , determine the angular acceleration of the yo-yo.

SOLUTION

ΣFy = Σ( Fy )eff : T − mg = 0; T = mg ΣM G = Σ( M G )eff :

Tr = Iα mgr = mk 2α

α=

rg k2

α=

rg k2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1487

PROBLEM 16.56 The 80-g yo-yo shown has a centroidal radius of gyration of 30 mm. The radius of the inner drum on which a string is wound is 6 mm. Knowing that at the instant shown the acceleration of the center of the yo-yo is 1 m/s2 upward, determine (a) the required tension T in the string, (b) the corresponding angular acceleration of the yo-yo.

SOLUTION

W = mg W = 0.080 kg (9.81 m/s 2 ) = 0.7848 N ΣFy = Σ( Fy )eff : T − W =

W a g

T − (0.08 kg)(9.81 m/s 2 ) = (0.08 kg)(1 m/s 2 ) T = 0.8648 N

(a)

T = 0.865 N 

Tension in the string. ΣM G = Σ( M G )eff : Tr = Iα (0.8648 N)(0.006 m) = mk 2α 5.1888 × 10−3 N ⋅ m = (0.08 kg)(0.03 m)2 α

(b)

Angular acceleration.

α = 72.067 rad/s 2

α = 72.1 rad/s 2



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PROBLEM 16.57 A 6-lb sprocket wheel has a centroidal radius of gyration of 2.75 in. and is suspended from a chain as shown. Determine the acceleration of Points A and B of the chain, knowing that TA = 3 lb and TB = 4 lb.

SOLUTION m=

W g

I = mk 2  6 lb =  2  32.2 ft/s

  2.75 in.       12 in./ft 

2

= 9.7858 × 10−3 slug ⋅ ft 2 r = 3.5 in. ΣFy = Σ( Fy )eff : TA + TB − W = ma  (6 lb) TA + TB − 6 lb =  2  32.2 ft/s

  a 

a = 5.3667(TA + TB − 6)

(1)

 3.5   3.5  ft  − TA  ft  = I α ΣM G = Σ( M G )eff : TB   12   12   3.5  (TB − TA )  ft  = (9.7858 × 10−3 slug ⋅ ft 2 )α  12 

α = 29.805 (TB − TA ) Given data:

(2)

TA = 3 lb, TB = 4 lb

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1489

PROBLEM 16.57 (Continued)

Eq. (1): Eq. (2):

a = 5.3667(3 + 4 − 6) = 5.3667 ft/s 2

α = 29.805(4 − 3) = 29.805 rad/s 2 a A = ( a A )t = a + rα  3.5  = 5.3667 −   (29.805)  12 

= −3.3264 ft/s 2

a A = 3.33 ft/s 2 

a A = ( a A )t = a + rα  3.5  = 5.3667 +   (29.805)  12 

= +14.06 ft/s 2

a B = 14.06 ft/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1490

PROBLEM 16.58 The steel roll shown has a mass of 1200 kg, a centriodal radius of gyration of 150 mm, and is lifted by two cables looped around its shaft. Knowing that for each cable TA = 3100 N and TB = 3300 N, determine (a) the angular acceleration of the roll, (b) the acceleration of its mass center.

SOLUTION m = 1200 kg

Data:

I = mk 2 = (1200)(0.150) 2 = 27 kg ⋅ m 2 1 1 r = d = (0.100) = 0.050 m 2 2 TA = 3100 N TB = 3300 N

(a)

Angular acceleration. ΣM G = Σ( M G )eff : 2TB r − 2TA r = I α 2(TB − TA )r I (2)(3300 − 3100)(0.050) = 27

α=

(b)

α = 0.741 rad/s 2



Acceleration of mass center. ΣFy = Σ( Fy )eff : 2TA + 2TB − mg = ma 2(TA + TB ) −g m 2(3100 + 3300) = − 9.81  1200

a=

a = 0.857 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1491

PROBLEM 16.59 The steel roll shown has a mass of 1200 kg, has a centriodal radius of gyration of 150 mm, and is lifted by two cables looped around its shaft. Knowing that at the instant shown the acceleration of the roll is 150 mm/s 2 downward and that for each cable TA = 3000 N, determine (a) the corresponding tension TB , (b) the angular acceleration of the roll.

SOLUTION m = 1200 kg

Data:

I = mk 2 = (1200)(0.150) 2 = 27 kg ⋅ m 2 1 1 r = d = (0.100) = 0.050 m 2 2 TA = 3000 N a = 0.150 m/s 2

(a)

Tension in cable B. ΣFy = Σ( Fy )eff : 2TA + 2TB − mg = −ma mg − ma − TA 2 m( g − a ) = − TA 2 (1200)(9.81 − 0.150) = − 3000 2 = 2796 N

TB =

(b)

TB = 2800 N 

Angular acceleration. ΣM G = Σ( M G )eff : 2TB r − 2TA r = I α 2(TB − TA )r I (2)(2796 − 3000) = 27 = −15.11 rad/s 2 

α=

α = 15.11 rad/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1492

PROBLEM 16.60 A 15-ft beam weighing 500 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration of cable A is 20 ft/s2 and the deceleration of cable B is 2 ft/s2, determine the tension in each cable.

SOLUTION Kinematics:

aB = a A + (15 ft)α

2 = 20 + 15α α = 1.2 rad/s 2 a=

1 1 ( a A + aB ) = (2 + 20) 2 2

a = 11 ft/s 2

Kinetics:

1 mL2 12 1 500 (15 ft) 2 = 12 32.2 ft/s 2 = 291.15 lb ⋅ ft ⋅ s 2

I =

ΣM B = Σ( M B )eff : TA (15 ft) − W (2.5 ft) = ma (7.5 ft) + I α 500 lb (11 ft/s 2 )(7.5 ft) 32.2 ft/s 2 + (291.15 lb ⋅ ft/s)(1.2 rad/s 2 ) 15TA − 3750 = 1281 + 349.3

TA (15 ft) − (500 lb)(7.5 ft) =

TA = 358.7 lb

TA = 359 lb 

ΣF = ΣFeff : TA + TB − W = ma 500 lb (11 ft/s 2 ) 2 32.2 ft/s TB = 312.2 lb

358.7 lb + TB − 500 lb =

TB = 312 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1493

PROBLEM 16.61 A 15-ft beam weighing 500 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration of cable A is 20 ft/s2 and the deceleration of cable B is 2 ft/s 2, determine the tension in each cable.

SOLUTION Kinematics:

aB = a A + 12α 2 = 20 + 12α α = 1.5 rad/s 2

a = a A + 7.5α = 20 + (7.5)(1.5)

a = 8.75 ft/s 2 I =

1 2 1 500 mL = (15) 2 = 291.15 lb ⋅ ft ⋅ s 2 2 12 32.2

Kinetics:

ΣM B = Σ( M B )eff : TA (12 ft) + W (4.5 ft) = ma (4.5 ft) + I α 500 lb (8.75 ft/s 2 )(4.5 ft) 32.2 ft/s 2 + (291.15)(1.5 rad/s 2 ) 12TA − 2250 = 611.4 + 436.7

TA (12 ft) − (500 lb)(4.5 ft) =

TA = 275 lb 

ΣF = ΣFeff : TA + TB − W = ma 275 lb + TB − 500 =

500 (8.75) 32.2

TB = 361 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1494

PROBLEM 16.62 Two uniform cylinders, each of weight W = 14 lb and radius r = 5 in., are connected by a belt as shown. If the system is released from rest, determine (a) the angular acceleration of each cylinder, (b) the tension in the portion of belt connecting the two cylinders, (c) the velocity of the center of the cylinder A after it has moved through 3 ft.

SOLUTION Kinematics Let a A = a A be the acceleration of the center of cylinder A, a AB = a AB be acceleration of the cord between the disks, α A = α A

be the angular acceleration of disk A, and α B = α B

be the angular acceleration of disk B.

a A = rα A

(1)

a AB = a A + rα A = 2rα A = rα B

(2)

m A = mB = m

(3)

Masses and moments of inertia

I A = IB =

Kinetics: Disk A:

1 2 mr 2

(4)

Let TAB be the tension in the portion of the cable between disks A and B. ΣM P = Σ( M P )eff : rWA − 2rTAB = rmA a A + I Aα A

(5)

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PROBLEM 16.62 (Continued)

ΣM B = Σ( M B )eff : rTAB = I Bα B

Disk B:

(6)

Add 2 × Eq. (6) to Eq. (5) to eliminate TAB. rWA = rm A a A + I Aα A + 2 I Bα B

(7)

Use Eqs. (1) and (2) to eliminate a A and α B rWA = rm A ( rα A ) + I Aα A + 2 I B ⋅ (2α A ) = (m A r 2 + I A + 4 I B )α A  1 1  =  mr 2 + mr 2 + 4  mr 2   α A 2 2   = 3.5 mr 2α A rWA 1g = 2 3.5r 3.5 mr 2g α B = 2α A = 3.5r

αA =

From Eq. (2),

TAB

I α = B B = r =

Data:

W = 14 lb,

(

1 2

)

mr 2 (2 g ) 3.5r 2

mg W = 3.5 3.5

g = 32.2 ft/s2, r = 5 in. =

5 ft 12

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1496

PROBLEM 16.62 (Continued)

(a)

Angular accelerations.

αA = 



(b)

Tension TAB.

α B = 2α A

TAB =

(c)

32.2 ft/s 2 = 22.08 rad/s (3.5) ( 125 ft )

W 14 lb = 3.5 3.5

α A = 22.1 rad/s 2



α B = 44.2 rad/s 2



TAB = 4.00 lb 

Velocity of the center of A. 5 5 α A = (22.08) = 9.20 ft/s 2 12 12 v A2 = [(v A )0 ]2 + 2a A d A

aA =

= 0 + (2)(9.20 ft/s)(3 ft) = 55.2 ft 2 /s 2 v A = 7.43 ft/s

v A = 7.43 ft/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1497

PROBLEM 16.63 A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the bar, (b) the acceleration of Point A, (c) the acceleration of Point B.

SOLUTION 1 1 T1 = T2 = W = mg 2 2

Statics: (a)

Angular acceleration:

L ΣM G = Σ( M G )eff : T   = I α 2 1 L 1 mg   = mL2α 2  2  12

α= ΣFy = Σ( Fy )eff :

(b)

α=

3g L



W − T1 = ma

mg − a=

3g L

1 mg = ma 2

1 g 2

a=

1 g 2

Acceleration of A: a A = aG + a A/G aA = =

1 L g− α 2 2 1 L  3g  g−  2 2  L 

aA = − g

(c)

aA = g 

Acceleration of B: a B = aG + a B/G aB = a +

L 1 L  3g  α = g +   = + 2g 2 2 2 L 

a B = 2 g 

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PROBLEM 16.64 A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the beam, (b) the acceleration of Point A, (c) the acceleration of Point B.

SOLUTION

ΣFy =

2mg = maG 3

1  mg  L mL2α ΣM G =  =   3  2 12

aG =

(a)

(b)

aA =

2g 2 g L + 3 L 2

=

2g 3

,

α =

2g L

g 3 , g 3



5g 3



aA =

(c)

aB =

2g 2 g L + 3 L 2

=



5g 3

aB =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1499

PROBLEM 16.65 A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the bar, (b) the acceleration of Point A, (c) the acceleration of Point B.

SOLUTION Before spring 1 breaks: ΣFy = 0: T1 sin 30° + T2 sin 30° − W = 0

Since T1 = T2 by symmetry, 2T1 sin 30° = W = mg

T1 = mg

30°

Immediately after spring 2 breaks, elongation of spring 1 is unchanged. Thus, we still have T1 = mg

(a)

30°

Angular acceleration: ΣM G = Σ( M G )eff : (T1 sin 30°) (mg sin 30°) ΣFx = Σ( Fx )eff :

L = Iα 6 L 1 = mL2α 6 12

g L



T1 cos 30° = max mg cos 30° = ma ,

ΣFy = Σ( Fy )eff :

α=

ax = 0.866g

W − T1 sin 30° = ma y mg − mg sin 30° = ma y

a y = 0.5g

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PROBLEM 16.65 (Continued)

Accelerations of A and B

Translation (b)

+

Acceleration of A: a A = aG + a A/G = [0.866g = [0.866g

(c)

Rotation about G  g  L  ] + [0.5g ] +     L  2 

]+ 0

a A = 0.866g



Acceleration of B: a B = aG + a B/G = [0.866g = [0.366g

 g  L  ] + [0.5g ] +     L  2 

] + [g ]

a B = 1.323g

49.1° 

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PROBLEM 16.66 A thin plate of the shape indicated and of mass m is suspended from two springs as shown. If spring 2 breaks, determine the acceleration at that instant (a) of Point A, (b) of Point B. A square plate of side b.

SOLUTION 1 m(b 2 + b 2 ) 12

I =

1 2 mb 6

1 1 T1 = T2 = W = mg 2 2

Statics: Kinetics:

I =

ΣM G = Σ( M G )eff :

T

b = Iα 2

1 b 1 mg   = mb 2α 2 2 6 α=

3g 2b

ΣFy = Σ( Fy )eff : W − T1 = ma mg −

1 mg = ma 2

a=

1 g 2

Kinematics:

Plane motion

=

Translation

+

Rotation

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PROBLEM 16.66 (Continued)

(a)

a A = aG + a A/G = a + aA =

(b)

b  3g  g g = +   2  2b  2 4

a B = aG + a B/G = a + aB =

b α 2

aA =

1 g  4

aB =

5 g  4

b α 2

b  3g  1 5 = g g +  2  2b  2 4

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PROBLEM 16.67 A thin plate of the shape indicated and of mass m is suspended from two springs as shown. If spring 2 breaks, determine the acceleration at that instant (a) of Point A, (b) of Point B. A circular plate of diameter b.

SOLUTION 2

I =

1 1 T1 = T2 = W = mg 2 2

Statics: Kinetics:

1 b 1 m = mb 2 2  2  8

ΣM G = Σ( M G )eff :

T1

b = Iα 2

1 b 1 mg   = mb 2α 2 2 8 α=2

g b

ΣFy = Σ( Fy )eff : W − T1 = ma mg −

1 mg = ma 2

a=

1 g 2

Kinematics:

Plane motion

=

Translation

+

(a)

a A = aG + a A/G = a +

b g  b 1 α = g + 2  2 b  2 2

(b)

a B = aG + a B/G = a +

b 1 α = g 2 2

+

b g  2 2  b 

Rotation aA =

1 g  2

aB =

3 g  2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1504

PROBLEM 16.68 A thin plate of the shape indicated and of mass m is suspended from two springs as shown. If spring 2 breaks, determine the acceleration at that instant (a) of Point A, (b) of Point B. A rectangular plate of height b and width a.

SOLUTION 1 m( a 2 + b 2 ) 12

Moment of inertia.

I=

Statics:

ax = 0,

a y = 0,

α =0

Draw the force triangle showing equilibrium. T1 = T2 = mg sin 45°

Kinetics:

T2 = 0

Since there is no time for displacements to occur, the tension in spring 1 remains equal to T1 = mg sin 45°

Then

1 T1 =  mg 2

  1  +  2 mg  

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1505

PROBLEM 16.68 (Continued)

1 +ΣF = m a : [ mg ] +  mg 2 1 a= g 2

  1  +  2 mg = m a  

 1   +  2 g  

ΣM G = Σ( M G )eff :

1 a 1 b mg   + mg   = I α 2 2 2 2 1 1 mg (a + b) = m(a 2 + b2 )α 4 12

In vector notation,

Kinematics.

α=

3 g ( a + b) a 2 + b2

1 g (i − j) 2 3 g ( a + b) α=− 2 k a + b2 a=

a P = a + α × rP /G − ω 2 rP/G

Since there is no time to acquire angular velocity, ω 2 = 0 (a)

Acceleration at A.

1 rA/G = − ai 2

aA =

(b)

Acceleration at B.

rB/G = aB =

1  3 g ( a + b)   1  g (i − j) +  − 2 k  ×  − ai  2 2  a +b   2  aA =

1 3 g ( a + b) a g (i − j) + j 2 a 2 + b2

aB =

1 3 g ( a + b) a g (i − j) − j 2 a 2 + b2

1 ai 2 1  3 g ( a + b)   1  g (i − j) +  − 2 k  ×  ai  2 2  a +b  2 

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PROBLEM 16.69 A sphere of radius r and mass m is projected along a rough horizontal surface with the initial velocities indicated. If the final velocity of the sphere is to be zero, express, in terms of v0, r, and μk , (a) the required magnitude of ω 0, (b) the time t1 required for the sphere to come to rest, (c) the distance the sphere will move before coming to rest.

SOLUTION I = mk 2

Kinetics: ΣFx = Σ( Fx )eff :

F = ma

μk mg = ma a = μk g ΣM G = Σ( M G )eff : Fr = I α

( μk mg )r = mk 2α

α=

Kinematics:

μk gr k2

v = v0 − at v = v0 − μk gt

For v = 0 when t = t1

0 = v0 − μk gt1 ;

t1 =

v0 μk g

(1)

ω = ω0 − α t μ gr ω = ω0 − k 2 t k

For ω = 0 when t = t1 Set Eq. (1) = Eq. (2) Distance traveled:

μk gr

0 = ω0 −

k2

v0 k2 = ω ; μ k g μ k gr 0 s1 = v0 t1 −

t1 ;

t1 =

ω0 =

k2 ω μk gr 0

(2)

r v0 k2

(3)

1 2 at1 2 2

 v  1  v  s1 = v0  0  − ( μk g )  0  ;  μk g  2  μk g 

s1 =

v02 2 μk g

(4)

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PROBLEM 16.69 (Continued)

For a solid sphere

k2 =

(a)

Eq. (3):

ω0 =

(b)

Eq. (1)

(c)

Eq. (4)

2 2 r 5 2 5

r 5 v0 v = 2 0 2 r r

ω0 =

5 v0 2 r

t1 = s1 =



v0  μk g v02

2 μk g



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PROBLEM 16.70 Solve Problem 16.69, assuming that the sphere is replaced by a uniform thin hoop of radius r and mass m. PROBLEM 16.69 A sphere of radius r and mass m is projected along a rough horizontal surface with the initial velocities indicated. If the final velocity of the sphere is to be zero, express, in terms of v0, r, and μk , (a) the required magnitude of ω 0, (b) the time t1 required for the sphere to come to rest, (c) the distance the sphere will move before coming to rest.

SOLUTION I = mk 2

Kinetics: ΣFx = Σ( Fx )eff :

F = ma

μk mg = ma a = μk g ΣM G = Σ( M G )eff : Fr = Iα ( μk mg )r = mk 2α

α=

Kinematics:

μk gr k2

v = v0 − at v = v0 − μk gt

For v = 0 when t = t1

0 = v0 − μk gt1 ;

t1 =

v0

(1)

μk g

ω = ω0 − α t μ gr ω = ω0 − k 2 t k

For ω = 0 when t = t1 Set Eq. (1) = Eq. (2) Distance traveled:

μk gr

0 = ω0 −

k2

v0 k2 = ω ; μk g μk gr 0 s1 = v0 t1 −

t1 ;

t1 =

ω0 =

k2 ω μk gr 0

(2)

r v0 k2

(3)

1 2 at1 2 2

 v  1  v  s1 = v0  0  − ( μ k g )  0  ;  μk g  2  μk g 

s1 =

v02 2 μk g

(4)

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PROBLEM 16.70 (Continued)

For a hoop, (a)

Eq. (3):

(b)

Eq. (1):

(c)

Eq. (4):

k =r

ω0 =

v r r = 0 2 0 r r

ω0 = t1 = s1 =

v0 r



v0  μk g v02

2 μk g



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PROBLEM 16.71 A bowler projects an 8-in.-diameter ball weighing 12 lb along an alley with a forward velocity v0 of 15 ft/s and a backspin ω0 of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the ball will have traveled at time t1.

SOLUTION Kinetics: ΣFx = Σ( Fx )eff : μk mg = ma

a = μk g ΣM G = Σ( M G )eff :

Fr = Iα

( μk mg )r =

α=

Kinematics:

2 2 mr α 5 5 μk g 2 r

When the ball rolls, the instant center of rotation is at C, and when t = t1 v = rω

(1)

v = v0 − at = v0 − μ k gt

(2)

ω = −ω0 + α t = −ω0 +

5 μk g t 2 r

When t = t1: Eq. (1) : v = rω :

5 μk g   v0 − μ k gt1 =  −ω0 + t1  r 2 r   v0 − μ k gt1 = −ω0 r + t1 =

5 μk gt1 2

2 (v0 + rω0 ) g μk g

(3)

1 v0 = 15 ft/s, ω0 = 9 rad/s, r = 4 in. = ft 3

(a)

1   15 + (9)   2 3  t1 =  = 1.5972 s 7 0.1(32.2)

t1 = 1.597 s 

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PROBLEM 16.71 (Continued)

(b)

Eq. (2):

v1 = v0 − μk gt1 = 15 − 0.1(32.2)(1.5972) v1 = 15 − 5.1429 v1 = 9.86 ft/s 

= 9.857 ft/s

(c)

a = μk g = 0.1(32.2 ft/s 2 ) = 3.22 ft/s 2

s1 = v0 t1 −

1 2 at1 2

1 = (15 ft/s)(1.597 s) − (3.22 ft/s 2 )(1.597 s) 2 2 = 23.96 − 4.11 = 19.85 ft

s1 = 19.85 ft 

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PROBLEM 16.72 Solve Problem 16.71, assuming that the bowler projects the ball with the same forward velocity but with a backspin of 18 rad/s. PROBLEM 16.71 A bowler projects an 8-in.-diameter ball weighing 12 lb along an alley with a forward velocity v0 of 15 ft/s and a backspin ω0 of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the ball will have traveled at time t1.

SOLUTION Kinetics: ΣFx = Σ( Fx )eff : μk mg = ma

a = μk g ΣM G = Σ( M G )eff : Fr = Iα ( μk mg )r = α=

2 2 mr α 5 5 μk g 2 r

Kinematics: When the ball rolls, the instant center of rotation is at C, and when t = t1 v = rω

(1)

v = v0 − at = v0 − μk g t

(2)

ω = −ω0 + α t = −ω0 +

5 μk g t 2 r

When t = t1: Eq. (1) v = rω :

5 μk g   v0 − μ k gt1 =  −ω0 + t1  r 2 r   v0 − μ k gt1 = −ω0 r + t1 =

5 μk gt1 2

2 (v0 + rω0 ) 7 μk g

v0 = 15 ft/s, ω0 = 18 rad/s, r =

(3) 1 ft 3

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PROBLEM 16.72 (Continued) 1 2 (15 + 3 (18) ) = 1.8634 s 7 0.1(32.2)

(a)

Eq. (3):

t1 =

(b)

Eq. (2):

v1 = v0 − μk g t

t1 = 1.863 s 

= 15 − 0.1(32.2)(1.8634) v1 = 15 − 6.000 v1 = 9 ft/s 

= 9 ft/s

(c)

a = μk g = 0.1(32.2 ft/s 2 ) = 3.22 ft/s 2 s1 = v0 t1 −

1 2 a t1 2

1 = (15 ft/s)(1.8634 s) − (3.22 ft/s 2 )(1.8634 s)2 2 = 27.95 − 5.59 = 22.36 ft

s1 = 22.4 ft 

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PROBLEM 16.73 A uniform sphere of radius r and mass m is placed with no initial velocity on a belt that moves to the right with a constant velocity v1. Denoting by μ k the coefficient of kinetic friction between the sphere and the belt, determine (a) the time t1 at which the sphere will start rolling without sliding, (b) the linear and angular velocities of the sphere at time t1.

SOLUTION Kinetics:

ΣFx = Σ( Fx )eff : F = ma

μk mg = ma a = μk g ΣM G = Σ( M G )eff : Fr = I α ( μk mg )r =

α=

Kinematics:

2 2 mr α 5 5 μk g 2 r

v = at = μ k gt

ω = αt =

(1)

5 μk g t 2 r

(2)

Point C is the point of contact with belt.  5 μk g  vC = v + ω r = μk gt +  t r 2 r  7 vC = μk gt 2

(a)

When sphere starts rolling (t = t1), we have vC = v1 v1 =

(b)

7 μk gt1 2

t1 =

2 v1  7 μk g

Velocities when t = t1 Eq. (1):

 2 v1  v = μg    7 μk g 

Eq. (2):

ω =

 5 μk g   2 v1     2 r   7 μk g 

v=

2 v1 7

ω=

5 v1 7 r





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PROBLEM 16.74 A sphere of radius r and m has a linear velocity v0 directed to the left and no angular velocity as it is placed on a belt moving to the right with a constant velocity v1. If after first sliding on the belt the sphere is to have no linear velocity relative to the ground as it starts rolling on the belt without sliding, determine in terms of v1 and the coefficient of kinetic friction μk between the sphere and the belt (a) the required value of v0, (b) time t1 at which the sphere will start rolling on the belt, (c) the distance the sphere will have moved relative to the ground at time t1.

SOLUTION Kinetics:

ΣFx = Σ( Fx )eff : F = ma

μk mg = ma a = μk g ΣM G = Σ( M G )eff : Fr = Iα ( μk mg )r =

α=

5 2 mr α 2 5 μk g 2 r

v = v0 − at = v0 − μk gt

Kinematics:

ω = αt =

(1)

5 μk g t 2 r

(2)

Point C is the point of contact with belt. vC = −v + rω vC = −v + r vC = −v +

5 μk g t 2 r

5μk g t 2

(3)

But, when t = t1, v = 0 and vc = v1 5μ k g t1 2

Eq. (3):

v1 =

Eq. (1):

v = v0 − μk gt

When t = t1 , v = 0,

 2v1  0 = v0 − μk g    5μ k g 

t1 =

2v1  5 μk g

v0 =

2 v1  5

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PROBLEM 16.74 (Continued)

Distance when t = t1:

s = v0 t1 −

1 2 at1 2

 2v1   2   2v1  1 s =  v1    − (μk g )    5   5μ k g  2  5μ k g 



s=

v12  4 2  − ;  μk g  25 25 

2

s=

2 v12 25 μk g



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PROBLEM 16.CQ4 A cord is attached to a spool when a force P is applied to the cord as shown. Assuming the spool rolls without slipping, what direction does the spool move for each case? Case 1:

(a) left

(b) right

(c) It would not move.

Case 2:

(a) left

(b) right

(c) It would not move.

Case 3:

(a) left

(b) right

(c) It would not move.

SOLUTION Answer: Case 1: (a) Case 2: (a) Case 3: (b)

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PROBLEM 16.CQ5 A cord is attached to a spool when a force P is applied to the cord as shown. Assuming the spool rolls without slipping, in what direction does the friction force act for each case? Case 1:

(a) left

(b) right

(c) The friction force would be zero.

Case 2:

(a) left

(b) right

(c) The friction force would be zero.

Case 3:

(a) left

(b) right

(c) The friction force would be zero.

SOLUTION Answer: Case 1: (b) Case 2: (b) Case 3: (b)

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PROBLEM 16.CQ6 A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the front tires? (a) left (b) right (c) The friction force is zero.

SOLUTION Answer: (b)

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PROBLEM 16.CQ7 A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tires? (a) left (b) right (c) The friction force is zero.

SOLUTION Answer: (a)

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PROBLEM 16.F5 A uniform 6 × 8-in. rectangular plate of mass m is pinned at A. Knowing the angular velocity of the plate at the instant shown is ω, draw the FBD and KD.

SOLUTION Answer:



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PROBLEM 16.F6 Two identical 4-lb slender rods AB and BC are connected by a pin at B and by the cord AC. The assembly rotates in a vertical plane under the combined effect of gravity and a couple M applied to rod AB. Knowing that in the position shown the angular velocity of the assembly is ω, draw the FBD and KD that can be used to determine the angular acceleration of the assembly and the tension in cord AC.

SOLUTION Answer:



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PROBLEM 16.F7 The 4-lb uniform rod AB is attached to collars of negligible mass which may slide without friction along the fixed rods shown. Rod AB is at rest in the position θ = 25° when an horizontal force P is applied to collar A causing it to start moving to the left. Draw the FBD and KD for the rod.

SOLUTION Answer:

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PROBLEM 16.F8 A uniform disk of mass m = 4 kg and radius r = 150 mm is supported by a belt ABCD that is bolted to the disk at B and C. If the belt suddenly breaks at a point located between A and B, draw the FBD and KD for the disk immediately after the break.

SOLUTION Answer:



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PROBLEM 16.75 Show that the couple Iα of Figure 16.15 can be eliminated by attaching the vectors m at and m an at a Point P called the center of percussion, located on line OG at a distance GP = k 2 / r from the mass center of the body.

SOLUTION

OG = r

at = rα

We first observe that the sum of the vectors is the same in both figures. To have the same sum of moments about G, we must have ΣM G = ΣM G : Iα = ( mat )(GP) mk 2α = mr α (GP)

GP =

k2 r

(Q.E.D.) 

Note: The center of rotation and the center of percussion are interchangeable. Indeed, since OG = r , we may write GP =

k2 GO

or GO =

k2 GP

Thus, if Point P is selected as center of rotation, then Point O is the center of percussion.

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PROBLEM 16.76 A uniform slender rod of length L = 900 mm and mass m = 4 kg is suspended from a hinge at C. A horizontal force P of magnitude 75 N is applied at end B. Knowing that r = 225 mm, determine (a) the angular acceleration of the rod, (b) the components of the reaction at C.

SOLUTION (a)

Angular acceleration. a = rα

I =

1 mL2 12

L  ΣM C = Σ( M C )eff : P  r +  = (ma )r + Iα 2  1 mL2α 12 L   2 1 2 P  r +  = m  r + L α 2 12    = (mrα )r +

Substitute data:

0.9 m  1    = (4 kg) (0.225 m) 2 + (0.9 m)2  α (75 N) 0.225 m +  2  12    50.625 = 0.4725α

α = 107.14 rad/s 2 (b)

α = 107.1 rad/s 2



Components of reaction at C. ΣFy = Σ( Fy )eff : C y − W = 0

C y = W = mg = (4 kg)(9.81 m/s 2 )

C y = 39.2 N 

ΣFx = Σ( Fx )eff : C x − P = −ma

Cx = P − ma = P − m(r α ) = 75 N − (4 kg)(0.225 m)(107.14 rad/s 2 )

Cx = 75 N − 96.4 N



= −21.4 N



C x = 21.4 N



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PROBLEM 16.77 In Problem 16.76, determine (a) the distance r for which the horizontal component of the reaction at C is zero, (b) the corresponding angular acceleration of the rod.

SOLUTION (a)

a = rα 1 I = mL2 12

Distance r .

ΣFx = Σ( Fx )eff : P = ma

P = m(rα ) P α= mr ΣM G = Σ( M G )eff : P

(1)

L = Iα 2

L 1 = mL2α 2 12 L 1  P  P = mL2   2 12  mr  P

L L2 = 2 12r 1 900 mm r = L r= 6 6

(b)

r = 150 mm 

Angular acceleration. Eq. (1):

α=

P P 6P = = L mr m ( 6 ) mL

α=

6(75 N) = 125 rad/s 2 (4 kg)(0.9 m)

α = 125 rad/s 2



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PROBLEM 16.78 A uniform slender rod of length L = 36 in. and weight W = 4 lb hangs freely from a hinge at A. If a force P of magnitude 1.5 lb is applied at B horizontally to the left (h = L), determine (a) the angular acceleration of the rod, (b) the components of the reaction at A.

SOLUTION 1 a= α 2 ΣM A = Σ( M A )eff : PL = (ma )

I =

1 mL2 12

L + Iα 2

 L L 1 =  m α  + mL2α  2  2 12 1 PL = mL2α 3

(a)

α=

Angular acceleration.

=

3P mL 3(1.5 lb)

(

4 lb 32.2 ft/s2

) (3 ft)

= 12.08 rad/s 2

(b)

α = 12.08 rad/s 2



Components of the reaction at A. ΣFy = Σ( Fy )eff : Ay − W = 0 Ay = W = 4 lb

A y = 4.00 lb 

ΣFx = Σ( Fx )eff : Ax − P = −ma L  3P  P L  Ax = P − m  α  = P − m  =− 2  mL  2 2  P 1.5 lb Ax = − = − = −0.75 lb  2 2

A x = 0.750 lb



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PROBLEM 16.79 In Problem 16.78, determine (a) the distance h for which the horizontal component of the reaction at A is zero, (b) the corresponding angular acceleration of the rod. PROBLEM 16.78 A uniform slender rod of length L = 36 in. and weight W = 4 lb hangs freely from a hinge at A. If a force P of magnitude 1.5 lb is applied at B horizontally to the left (h = L), determine (a) the angular acceleration of the rod, (b) the components of the reaction at A.

SOLUTION L α 2 1 I = mL2 12

a=

ΣFx = Σ( Fx )eff :

P = ma L  P = m α  2 

(b)

α=

Angular acceleration.

α=

ΣM G = Σ( M G )eff :

(a)

 Ph −   Ph −   h − 

2P mL

(

2(1.5 lb) 4 lb 32.2 ft/s 2

) (3 ft)

α = 8.05 rad/s 2



L  = Iα : P(h − L) 2 L 1  2 P  PL = mL2   = 2  12  mL  6 L L = ; 2 6

h=

L L 2 + = L 2 6 3

h = 24 in. 

Distance h.

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PROBLEM 16.80 The uniform slender rod AB is welded to the hub D, and the system rotates about the vertical axis DE with a constant angular velocity ω, (a) Denoting by ω the mass per unit length of the rod, express the tension in the rod at a distance z from end A in terms of w, l, z, and ω, (b) Determine the tension in the rod for w = 0.3 kg/m, l = 400 mm, z = 250 mm, and ω = 150 rpm.

SOLUTION Consider motion in the horizontal plane. Since ω is constant, the angular acceleration is zero. Only normal acceleration, i.e., along the rod occurs. For a section defined by the coordinate z, the acceleration of the mass center of the portion extending from z to the section is a = rω 2 = (l − z /2)ω 2

Kinetics.

The mass of the section is m = wz

ΣF = ΣFeff : T = ma

(a)

 = ( wz )  l − 

(b)

Data:

 z2  T = w  lz −  ω 2  2  

z 2 ω 2 

(150)(2π ) = 5π rad/s 60 z = 0.250 m, l = 0.400 m, w = 0.3 kg/m

ω = 150 rpm =

 (0.25 m) 2  2 T = (0.3 kg/m) (0.4 m)(0.25 m) −  (5π rad/s) = 5.09 N 2   T = 5.09 N 



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PROBLEM 16.81 The shutter shown was formed by removing one quarter of a disk of 0.75-in. radius and is used to interrupt a beam of light emanating from a lens at C. Knowing that the shutter weighs 0.125 lb and rotates at the constant rate of 24 cycles per second, determine the magnitude of the force exerted by the shutter on the shaft at A.

SOLUTION See inside front cover for centroid of a circular sector. r = r =

2r sin α 3α 2(0.75 in.)sin ( 34 π ) 3( 34 π )

r = 0.15005 in. an = rω 2

ω = 24 rad/s = 24(2π ) rad/s ω = 150.8 rad/s ΣF = ΣFeff : R = m an = mr ω 2 =

(0.125 lb)  0.15005  ft  (150.8 rad/s) 2  32.2 ft/s 2  12 

R = 1.1038 lb

Force on shaft is

R = 1.104 lb R = 1.104 lb 

Magnitude:

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PROBLEM 16.82 A 6-in.-diameter hole is cut as shown in a thin disk of 15-in.-diameter. The disk rotates in a horizontal plane about its geometric center A at the constant rate of 480 rpm. Knowing that the disk has a mass of 60 lb after the hole has been cut, determine the horizontal component of the force exerted by the shaft on the disk at A.

SOLUTION Determination of mass center of disk We determine the centroid of the composite area:

xA = x1 A1 − x2 A2 or x ( A1 − A2 ) = x1 A1 − x2 A2 x =

xA1 − x2 A2 0 − (8) π (3)2 (8)(3) 2 = = − = − 0.33333 in. A1 − A2 π (15) 2 − π (3) 2 (15)2 − (3) 2

Kinetics

Mass center G coincides with centroid C

ω = 480 rpm = 50.265 rad/s

 0.33333  2 2 a = rω2 =   (50.265 rad/s) = 70.183 ft/s  12   60 lb  2 ΣF = Σ( F )eff : A = m a =   (70.183 ft/s ) 32.2  

A = 130.8 N



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PROBLEM 16.83 A turbine disk of mass 26 kg rotates at a constant rate of 9600 rpm. Knowing that the mass center of the disk coincides with the center of rotation O, determine the reaction at O immediately after a single blade at A, of mass 45 g, becomes loose and is thrown off.

SOLUTION  2π    60  ω = 320π rad/s

ω = 9600 rpm 

Consider before it is thrown off. ΣF = ΣFeff : R = man = mF ω 2 = (45 × 10−3 kg)(0.3 m)(320 π ) 2 R = 13.64 kN

Before blade was thrown off, the disk was balanced ( R = 0). Removing vane at A also removes its reaction, so disk is unbalanced and reaction is R = 13.64 kN



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PROBLEM 16.84 A uniform rod of length L and mass m is supported as shown. If the cable attached at end B suddenly breaks, determine (a) the acceleration of end B, (b) the reaction at the pin support.

SOLUTION w=0 a =

ΣM A = Σ( M A )eff : W

(b)

L α 2

L L = I α + ma 2 2

mg

L 1 L L = mL2α + m  α  2 12 2 2

mg

L 1 2 3g = mL α α = 2 3 2L

Reaction at A. L ΣFy = Σ( Fy )eff : A − mg = −ma = −m α 2

 L  3 g  A − mg = −m     2  2 L  3 A − mg = − mg 4 A=

(a)

1 mg 4

A=

1 mg 4



3 g 2



Acceleration of B. a B = a n + a B/A = 0 + Lα 3 g 3 aB = L  = g 2 L 2

aB =

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PROBLEM 16.85 A uniform rod of length L and mass m is supported as shown. If the cable attached at end B suddenly breaks, determine (a) the acceleration of end B, (b) the reaction at the pin support.

SOLUTION

ω=0

a=

L α 4

L L = I α + ma 4 4 L 1 L L mg = mL2α + m  α  4 12 4 4 L 7 mg = mL2α 4 48

ΣM C = Σ( M C )eff : W

(b)

12g 7L

C=

4 mg  7

Reaction at C. L ΣFy = Σ( Fy )eff : C − mg = − ma = −m α 4  L  12g  C − mg = −m     4  7 L  3 C − mg = − mg 7 4 C = mg 7

(a)

α=

Acceleration of B. aB = aC + aC/B = 0 + aB =

3L  12 g  9 = g 4  7 L  7

3L α 4

aB =

9 g  7

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PROBLEM 16.86 A 12-lb uniform plate rotates about A in a vertical plane under the combined effect of gravity and of the vertical force P. Knowing that at the instant shown the plate has an angular velocity of 20 rad/s and an angular acceleration of 30 rad/s2 both counterclockwise, determine (a) the force P, (b) the components of the reaction at A.

SOLUTION  6  at = rα =  ft  (30 rad/s 2 ) = 15 ft/s 2  12   6  an = rω 2 =  ft  (20 rad/s) 2 = 200 ft/s 2  12 

Kinematics.

Mass and moment of inertia.

I=

m=

W 12 lb = = 0.37267 lb ⋅ s 2 /ft g 32.2 ft/s 2

2 2 m  10   20     +    = (0.37267 lb ⋅ s 2 /ft)(0.28935 ft 2 ) = 0.10783 lb ⋅ s 2 ⋅ ft 12  12   12  

Kinetics.

(a)

Force P.

 16   6   6  ΣM A = Σ( M A )eff : P  ft  − W  ft  = mat  ft  + I α  12   12   12  4 1 1 P = (12)   + (0.37267)(15)   + (0.10783)(30) 3 2 2 P = 9.0224 lb.

P = 9.02 lb 

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PROBLEM 16.86 (Continued)

(b)

Reaction at A. ΣFy = Σ( Fx )eff : Ax = − man = −

12 (200) 32.2

Ax = −74.53 lb

A x = 74.5 lb



ΣFy = Σ( Fy )eff : Ay + P − W = mat Ay = W + mat − P = 12 +

12 (15) − 9.02 = 8.57 lb 32.2

A y = 8.57 lb 

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PROBLEM 16.87 A 1.5-kg slender rod is welded to a 5-kg uniform disk as shown. The assembly swings freely about C in a vertical plane. Knowing that in the position shown the assembly has an angular velocity of 10 rad/s clockwise, determine (a) the angular acceleration of the assembly, (b) the components of the reaction at C.

SOLUTION Kinematics: an = (CG )ω 2 = (0.14 m)(10 rad/s 2 )

an = 14 m/s 2 at = (CG )α = (0.14 m)α

I AB

(a)

1 mdisk (CG ) 2 2 1 = (5 kg)(0.08 m) 2 2 = 16 × 10−3 kg ⋅ m 2 1 = m AB ( AB) 2 12 1 = (1.5 kg)(0.12 m) 2 12 = 1.8 × 10−3 kg ⋅ m 2

I disk =

Kinetics:

Angular acceleration. ΣM C = Σ( M C )eff : WAB (0.14 m) = I diskα + m AB at (0.14 m) + I ABα (1.5 kg)(9.81 m/s 2 )(0.14 m) = I diskα + (1.5 kg)(0.14 m) 2 α + I ABα 2.060 N ⋅ m = (16 × 10−3 + 29.4 × 10−3 + 1.8 × 10−3 )α 2.060 N ⋅ m = (47.2 × 10−3 kg ⋅ m 2 )α

α = 43.64 rad/s 2

α = 43.6 rad/s 2



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PROBLEM 16.87 (Continued)

(b)

Components of reaction of C. ΣFx = Σ( Fx )eff : Cx = −m AB an = −(1.5 kg)(14 m/s 2 ) Cx = −21.0 N ΣFy = Σ( Fy )eff :

C x = 21.0 N



at = (0.14 m)(α )

C y − mdisk g − m AB g = −m AB at C y − (5 kg)9.81 − (1.5 kg) 9.81 = −(1.5 kg)(0.14 m)(43.64 rad/s 2 ) C y − 49.05 N − 14.715 N = −9.164 N C y = +54.6 N

Cy = 54.6 N 

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PROBLEM 16.88 Two uniform rods, ABC of weight 6-lb and DCE of weight 8-lb, are connected by a pin at C and by two cords BD and BE. The T-shaped assembly rotates in a vertical plane under the combined effect of gravity and of a couple M which is applied to rod ABC. Knowing that at the instant shown the tension in cord BE is 2 lb and the tension in cord BD is 0.5 lb, determine (a) the angular acceleration of the assembly, (b) the couple M.

SOLUTION We first consider the entire system and express that the external forces are equivalent to the effective forces of both rods. (a AC )t = 0.75α (aDE )t = 1.5α I AC =

1  6 lb  (1.5 ft) 2   12  32.2 

= 34.938 × 10−3 slug ⋅ ft 2 I DE =

1  8 lb  (2 ft)2 12  32.2 

= 82.816 × 10 −3 slug ⋅ ft 2 6 8 (a AC )t (0.75) + I DEα + (aDE )t (1.5) 32.2 32.2 6 8 (0.75α )(0.75) + 82.816 × 10 −3 α + (1.5α )(1.5) M = 34.938 × 10−3 α + 32.2 32.2

ΣM A = Σ( M A )eff : M = I ACα +

M = 0.78157α

(1)

We now consider rod DE alone:

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PROBLEM 16.88 (Continued)

3  3  ΣM C = Σ( M C )eff :  TBE  (1 ft) −  TBD  (1 ft) = I DEα 5 5     0.6(TBE − TBD ) = 82.816 × 10−3 α TBE − TBD = 0.13803 α

(2)

TBD = 0.5 lb

Given data:

TBE = 2 lb

(a)

Angular acceleration. Substitute into (2):

(b)

2 − 0.5 = 0.13803α

α = 10.87 rad/s 2



Couple M. Carry value of α into (1):

M = 0.78157 (10.868) = 8.4938 ft ⋅ lb

M = 8.49 ft ⋅ lb



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PROBLEM 16.89 The object ABC consists of two slender rods welded together at Point B. Rod AB has a weight of 2 lb and bar BC has a weight of 4 lb. Knowing the magnitude of the angular velocity of ABC is 10 rad/s when θ = 0, determine the components of the reaction at Point C when θ = 0.

SOLUTION Masses and lengths:

WAB = 2 lb, LAB = 1 ft, WBC = 4 lb, LBC = 2 ft 1 1 2  2 −3 2 mAB L2AB =   (1) = 5.1760 × 10 slug ⋅ ft 12 12  32.2  1 1  4  2 2 −3 = mBC L2BC =   (2) = 41.408 × 10 slug ⋅ ft 12 12  32.2 

I AB =

Moments of inertia:

I BC

rCD = 22 + 0.52 = 2.0616 ft

Geometry:

1 LAB = 1 ft 2 0.5 tan β = 2 β = 14.036° rCE =

Kinematics:

Let α = α

be the angular acceleration of object ABC. ( a AB )t = rCDα

β

( a AB )n = rCDω 2

β

( aBC )t = rCEα ( aBC ) n = rCE ω 2

Kinetics:

ΣM C =

Σ( M C )eff :

WAB

LAB = I ABα + rCD m AB (a AB )t 2 + I BCα + rCE mBC (aBC )t

(

)

2 2 = I AB + mAB rCD + I BC + mBC rCE α

  2   4  2 (2)(0.5) =  (5.1760 × 10−3 ) +  (2.0616) 2 + (41.408 × 10−3 ) +    (1)  α  32.2   32.2   

α = 2.3 rad/s 2

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PROBLEM 16.89 (Continued)

ΣFx =

Σ( Fx )eff :

Cx = m AB ( a AB )t cos β + mAB (a AB )n sin β + mBC (aBC )t = m AB rCD (α cos β + ω 2 sin β ) + mBC rCEα

 2   4  2 =  (2.0616)(2.3cos14.035° + 10 sin14.035°) +   (1)(2.3)  32.2   32.2  = 3.6770 lb ΣFy =

Σ( Fy )eff :

C y − mAB g − mBC g = −m AB (a AB )t sin β + m AB (a AB )n cos β + mBC ( aBC ) n C y = (WAB + WBC ) + mAB rCD (ω 2 cos β − α sin β ) + mBC rCE ω

 2  2 Cy = 6 +   (2.0616)(10 cos14.035° − 2.3sin14.035°) 32.2   4   2 +  (1)(10)  32.2  C y = 30.773 lb

Reaction at C.

C = 3.67702 + 30.7732 = 30.992 lb 30.773 3.6770 φ = 83.186°

tan φ =

C = 31.0 lb

83.2° 

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PROBLEM 16.90 A 3.5-kg slender rod AB and a 2-kg slender rod BC are connected by a pin at B and by the cord AC. The assembly can rotate in a vertical plane under the combined effect of gravity and a couple M applied to rod BC. Knowing that in the position shown the angular velocity of the assembly is zero and the tension in cord AC is equal to 25 N, determine (a) the angular acceleration of the assembly, (b) the magnitude of the couple M.

SOLUTION (a)

Angular acceleration. Rod AB:

ΣM B = Σ( M B )eff :

1 mAB L2AB 12 1 = 3.5 kg(0.8 m) 2 12 = 0.18667 kg ⋅ m 2

I AB =

3 (25 N)(0.8 m) − (3.5 kg)(9.81 m/s 2 )(0.4 m) = I ABα 5 I ABα = −1.7340 N ⋅ m

(1)

(0.18667 kg ⋅ m 2 )α = −1.7340 N ⋅ m α = −9.2893 rad/s 2

α = 9.29 rad/s 2



Entire assembly: Since AC is taut, assembly rotates about C as a rigid body. Kinematics: CB = (0.3)2 + (0.4)2 = 0.5 m CGBC =

1 CB = 0.25 m 2

aBC = (0.25 m)α a AB = (0.3 m)α

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PROBLEM 16.90 (Continued)

1 mBC (CB ) 2 12 1 = (2 kg)(0.5 m) 2 12 = 0.041667 kg ⋅ m 2

I BC =

Kinetics:

(b)

Couple M. ΣM C = Σ( M C )eff : M − mBC g (0.2 m) = mBC aBC (0.25 m) + I BCα + m AB a AB (0.3 m) + I ABα M − (2 kg)(9.81 m/s 2 )(0.2 m) = 2 kg(0.25 m) 2 α + (0.041667 kg ⋅ m 2 )α + 3.5 kg(0.3 m) 2 α + I ABα

Substitute α = 9.29 rad/s 2

IABα = −1.7340 N ⋅ m M − 3.9240 = (0.125)(9.2893) + (0.041667)(9.2893) + (0.315)(9.2893) + (0.18667)(9.2893) M − 3.9240 = 1.1612 + 0.3871 + 2.9261 + 1.7340 M − 3.9240 = 6.2084 M = +10.132 N ⋅ m

M = 10.13 N ⋅ m



M − mBC g (0.2 m) = I Cα 2 1  Since C is fixed, we could also use: M − (3.5 kg)(9.81 m/s 2 )(0.2 m) =  mCB CB + I AB + 3.5 kg(0.3 m)2  α 3  

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PROBLEM 16.91 A 9-kg uniform disk is attached to the 5-kg slender rod AB by means of frictionless pins at B and C. The assembly rotates in a vertical plane under the combined effect of gravity and of a couple M which is applied to rod AB. Knowing that at the instant shown the assembly has an angular velocity of 6 rad/s and an angular acceleration of 25 rad/s2, both counterclockwise, determine (a) the couple M, (b) the force exerted by pin C on member AB.

SOLUTION We first consider the entire system and express that the external forces are equivalent to the effective forces of the disk and the rod. mR = 5 kg,

mD = 9 kg

WR = mD g = (5 kg)(9.81 m/s 2 ) = 49.05 N WD = mD g = (9 kg)(9.81 m/s 2 ) = 88.29 N 1 1 mR L2AB = (5 kg)(0.5 m)2 = 0.104167 kg ⋅ m 2 12 12 1 1 I D = mD rD2 = (9 kg)(0.2 m) 2 = 0.18 kg ⋅ m 2 2 2 aR = (0.25 m)α IR =

aD = (0.5 m)α

(a)

ΣM A = Σ( M A )eff : M − WR (0.125) − WD (0.25) = I Rα + mR (aR )t (0.25) + I Dα + mD ( aD )t (0.5) M − (49.05)(0.125) − (88.29)(0.25) = 0.104167α + (5)(0.25α )(0.25) + 0.18α + (9)(0.5α )(0.5) M − 6.1312 − 22.073 = (0.10417 + 0.3125 + 0.18 + 2.25)α M − 28.204 = (2.8467)(25) M = 99.370 N-m

M = 99.4 N-m



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PROBLEM 16.91 (Continued)

(b)

Consider now the disk alone:

ΣM B = Σ( M B )eff : C (0.15) = I Dα = (0.18)(25)

C = 30.0 N

30° 

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PROBLEM 16.92 Derive the equation ΣM C = I Cα for the rolling disk of Figure 16.17, where ΣM C represents the sum of the moments of the external forces about the instantaneous center C, and I C is the moment of inertia of the disk about C.

SOLUTION

ΣM C = Σ( M C )eff : ΣM C = (ma )r + I α = (mrα ) r + I α = (mrα )r + I α

But, we know that

I C = mr 2 + I ΣM C = I C α

Thus:

(Q.E.D.) 

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PROBLEM 16.93 Show that in the case of an unbalanced disk, the equation derived in Problem 16.92 is valid only when the mass center G, the geometric center O, and the instantaneous center C happen to lie in a straight line.

SOLUTION Kinematics: a = aC + aG/C = aC + α × rG/C + ω × (ω × rG/C )

or, since ω ⊥ rG/C a = aC + α × rG/C − ω2 rG/C

(1)

Kinetics:

ΣM C = Σ( M C )eff : ΣM C = Iα + rG/C × m a

Recall Eq. (1):

ΣM C = I α + rG/C × m(aC + α × rG/C − ω2 rG/C ) ΣM C = I α + rG/C × maC + mrG/C × (α × rG/C ) − mω2 rG/C × rG/C

But:

rG/C × rG/C = 0 and α ⊥ rG/C rG/C × m(α × rG/C ) = mrG2/Cα

Thus: Since

(

)

ΣM C = I + mrG2/C α + rG/C × maC I C = I + mrG2/C ΣM C = I Cα + rG/C × maC

(2)

Eq. (2) reduces to ΣM C = I Cα when rG/C × maC = 0; that is, when rG/C and aC are collinear. Referring to the first diagram, we note that this will occur only when Points G, O, and C lie in a straight line. (Q.E.D.) 

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PROBLEM 16.94 A wheel of radius r and centroidal radius of gyration k is released from rest on the incline and rolls without sliding. Derive an expression for the acceleration of the center of the wheel in terms of r, k , β , and g.

SOLUTION I α = mk 2α ma = mrα

ΣM C = Σ( M C )eff : (W sin β )r = (ma )r + I α (mg sin β ) r = (mrα )r + mk 2α

rg sin β = (r 2 + k 2 )α

α=

rg sin β r2 + k 2

a = rα = r

rg sin β r2 + k2

a=

r2 g sin β  r2 + k 2

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PROBLEM 16.95 A homogeneous sphere S, a uniform cylinder C, and a thin pipe P are in contact when they are released from rest on the incline shown. Knowing that all three objects roll without slipping, determine, after 4 s of motion, the clear distance between (a) the pipe and the cylinder, (b) the cylinder and the sphere.

SOLUTION I = mk 2

General case:

a = rα

ΣM C = Σ( M C )eff : ( w sin β )r = I α + mar mg sin β r = mk 2α + mr 2α

α=

rθ sin β r2 + k 2

a = rα = r

For pipe: For cylinder:

For sphere:

(a)

rg sin β r2 + k 2

a=

k =r

aP =

1 2

aC =

2 5

aS =

k2 =

k2 =

r2 g sin β r2 + k 2

(1)

r2 1 g sin β = g sin β 2 2 r +r 2

r2 r2 +

2

r 2

g sin β =

2 g sin β 3

r2 5 g sin β = g sin β 2 7 r2 + r2 5

Between pipe and cylinder. 1 2 1 aC/P = aC − aP =  −  g sin β = g sin β 6 3 2

xC/P =

1 11  aC /P t 2 =  g sin β  t 2 2 26 

SI units:

xC/P =

11  9.81 m/s 2  sin10°(4 s) 2 = 2.27 m  2  6 

US units:

xC/P =

11  32.2 ft/s 2  sin10°(4 s)2 = 7.46 ft  2  6 

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PROBLEM 16.95 (Continued)

(b)

Between sphere and cylinder. 1 5 2 aS/C = aS − aC =  −  g sin β = g sin β 21 7 3

xS/C =

1 1 1  aS/C t 2 =  g sin β  t 2 2 2  21 

xS/C =

SI units:

1 1  9.81 m/s 2  sin10°(4 s) 2 = 0.649 m  2  21 

xS/C =

US units:

1 1  32.2 ft/s 2  sin10°(4 s) 2 = 2.13 ft  2  21 

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PROBLEM 16.96 A 40-kg flywheel of radius R = 0.5 m is rigidly attached to a shaft of radius r = 0.05 m that can roll along parallel rails. A cord is attached as shown and pulled with a force P of magnitude 150 N. Knowing the centroidal radius of gyration is k = 0.4 m, determine (a) the angular acceleration of the flywheel, (b) the velocity of the center of gravity after 5 s.

SOLUTION Mass and moment of inertia:

m = 40 kg I = mk 2 = (40 kg)(0.4 m)2 = 6.4 kg ⋅ m 2

Kinematics: (no slipping)

a = rα

15°

Kinetics:

Let Point C be the contact point between the flywheel and the rails. ΣM C = Σ( M C )eff : P ( R + r ) − mgr sin15° = I α + (ma )r P ( R + r ) − mgr sin15° = ( I + mr 2 )α

(a)

Angular acceleration of the flywheel. P ( R + r ) − mgr sin15° I + mr 2 (150 N)(0.55 m) − (40 kg)(9.81 m/s 2 )(0.05 m) sin15° = (6.4 kg ⋅ m 2 ) + (40 kg)(0.05 m) 2

α=

= 11.911 rad/s 2

α = 11.91 rad/s 2



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PROBLEM 16.96 (Continued)

(b)

Velocity of center of gravity after 5 s. a = rα = (0.05 m)(11.911 rad/s 2 ) = 0.59555 m/s 2 a = 0.59555 m/s 2

15°

v = v 0 + at = 0 + (0.59555 m/s 2 )(5 s) v = 2.98 m/s

15° 

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PROBLEM 16.97 A 40-kg flywheel of radius R = 0.5 m is rigidly attached to a shaft of radius r = 0.05 m that can roll along parallel rails. A cord is attached as shown and pulled with a force P. Knowing the centroidal radius of gyration is k = 0.4 m and the coefficient of static friction is μ s = 0.4, determine the largest magnitude of force P for which no slipping will occur.

SOLUTION Mass and moment of inertia:

m = 40 kg I = mk 2 = (40 kg)(0.4 m)2 = 6.4 kg ⋅ m 2

Kinematics: (no slipping)

a = rα

15°

Kinetics:

+ ΣFn = Σ( Fn )eff : N − mg cos15° = 0 N = (40 kg)(9.81 m/s 2 ) cos15° = 379.03 N

For impending slipping,

F = μ s N = (0.4)(379.03) = 151.61 N

+

15° ΣF = Σ( F )eff : P − F − mg sin15° = ma P − 151.61 N = (40 kg)(9.81 m/s 2 )sin15° = (40 kg)(0.05 m)α P − 253.17 N = (2 kg ⋅ m)α

(1)

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PROBLEM 16.97 (Continued)

Let C be the contact point between the flywheel and the rails. ΣM C = Σ( M C )eff : P ( R + r ) − mgr sin15° = Ia + (ma )r P ( R + r ) − mgr sin15° = ( I + mr 2 )α (0.55 m) P − (40 kg)(9.81 m/s 2 )(0.05 m) sin15°

= [6.4 kg ⋅ m 2 + (40 kg)(0.05 m) 2 ]α

(2)

Solving Eqs. (1) and (2) simultaneously, P = 303 N

α = 24.8 rad/s2 P = 303 N 

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PROBLEM 16.98 A drum of 60-mm radius is attached to a disk of 120-mm radius. The disk and drum have a total mass of 6 kg and a combined radius of gyration of 90 mm. A cord is attached as shown and pulled with a force P of magnitude 20 N. Knowing that the disk rolls without sliding, determine (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion.

SOLUTION a = rα = (0.12 m)α I = mk 2 = (6 kg)(0.09 m) 2 I = 48.6 × 10−3 kg ⋅ m 2 ΣM C = Σ( M C )eff : (20 N)(0.12 m) = (ma ) r + I α 2.4 N ⋅ m = (6 kg)(0.12 m) 2 α + 48.6 × 10−3 kg ⋅ m 2 2.4 = 135.0 × 10−3 α

α = 17.778 rad/s 2

(a)

α = 17.78 rad/s 2



a = 2.13 m/s 2



a = rα = (0.12 m)(17.778 rad/s 2 ) = 2.133 m/s 2

(b)

ΣFy = Σ( Fy )eff :

N − mg = 0 N = (6 kg)(9.81 m/s 2 )

N = 58.86 N

ΣFx = Σ( Fx )eff : 20 N − F = ma 20 N − F = (6 kg)(2.133 m/s 2 ) F = 7.20 N ( μ s ) min =

F 7.20 N = N 58.86 N

( μ s ) min = 0.122 

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PROBLEM 16.99 A drum of 60-mm radius is attached to a disk of 120-mm radius. The disk and drum have a total mass of 6 kg and a combined radius of gyration of 90 mm. A cord is attached as shown and pulled with a force P of magnitude 20 N. Knowing that the disk rolls without sliding, determine (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion.

SOLUTION a = rα = (0.12 m) α I = mk 2 = (6 kg)(0.09 m)2 I = 48.6 × 10−3 kg ⋅ m 2 ΣM C = Σ( M C )eff : (20 N)(0.18 m) = (ma )r + Iα 3.6 N ⋅ m = (6 kg)(0.12 m) 2 α + 48.6 kg ⋅ m 2 3.6 = 135 × 10−3 α

(a)

α = 26.667 rad/s 2

α = 26.7 rad/s 2



a = 3.20 m/s 2



a = rα = (0.12 m)(26.667 rad/s 2 ) = 3.2 m/s 2

(b)

ΣFy = Σ( Fy )eff : N − mg = 0 N = (6 kg)(9.81 m/s 2 )

N = 58.86 N

ΣFx = Σ( Fx )eff : 20 N − F = ma 20 N − F = (6 kg)(3.2 m/s 2 ) ( μ s ) min =

F 0.8 N = N 58.86 N

F = 0.8 N

( μ s )min = 0.0136 

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PROBLEM 16.100 A drum of 60-mm radius is attached to a disk of 120-mm radius. The disk and drum have a total mass of 6 kg and a combined radius of gyration of 90 mm. A cord is attached as shown and pulled with a force P of magnitude 20 N. Knowing that the disk rolls without sliding, determine (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion.

SOLUTION a = rα = (0.12 m)α I = mk 2 = (6 kg)(0.09 m)2 I = 48.6 kg ⋅ m 2 ΣM C = Σ( M C )eff : (20 N)(0.06 m) = (ma )r + Iα 1.2 N ⋅ m = (6 kg)(0.12 m)2 α + 48.6 × 10−3 kg ⋅ m 2 1.2 = 135 × 10−3α

α = 8.889 rad/s 2

(a)

α = 8.89 rad/s 2



a = rα = (0.12 m)(8.889 rad/s 2 ) = 1.0667 m/s 2

(b)

ΣFy = Σ( Fy )eff :



N − mg = 0 N = (6 kg)(9.81 m/s 2 )

ΣFk = Σ( Fx )eff :

a = 1.067 m/s 2

N = 58.86 N

20 N − F = ma (20 N) − F = (6 kg)(1.0667 m/s 2 ) ( μ s ) min =

F 13.6 N = N 58.86 N

F = 13.6 N

( μ s ) min = 0.231 

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PROBLEM 16.101 A drum of 60-mm radius is attached to a disk of 120-mm radius. The disk and drum have a total mass of 6 kg and a combined radius of gyration of 90 mm. A cord is attached as shown and pulled with a force P of magnitude 20 N. Knowing that the disk rolls without sliding, determine (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion.

SOLUTION a = rα = (0.12 m)α I = mk 2 = (6 kg)(0.09 m)2 I = 48.6 kg ⋅ m 2 ΣM C = Σ( M C )eff : (20 N)(0.06 m) = (ma )r + I α 1.2 N ⋅ m = (6 kg)(0.12 m) 2 α + (48.6 × 10−3 kg ⋅ m 2 )α 1.2 = 135 × 10−3α

α = 8.889 rad/s 2

(a)

α = 8.89 rad/s 2



a = rα = (0.12 m)(8.889 rad/s 2 ) = 1.0667 m/s 2

(b)

a = 1.067 m/s 2



ΣFy = Σ( Fy )eff : N + 20 N − mg = 0 N + 20 N − (6 kg)(9.81 m/s 2 ) ΣFx = Σ( Fx )eff :

N = 38.86 N

F = ma F = (6 kg)(1.0667 m/s 2 )

( μ s ) min =

F 6.4 N = N 38.86 N

F = 6.4 N

( μ s ) min = 0.165 

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PROBLEM 16.102 A drum of 4-in. radius is attached to a disk of 8-in. radius. The disk and drum have a combined weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 lb. Knowing that the coefficients of static and kinetic friction are μ s = 0.25 and μk = 0.20, respectively, determine (a) whether or not the disk slides, (b) the angular acceleration of the disk and the acceleration of G.

SOLUTION Assume disk rolls:

 8  a = rα =  ft  α  12  I = mk 2 =

10 lb  6  ft 32.2  12 

2

I = 0.07764 lb ⋅ ft ⋅ s 2

 8  ΣM C = Σ( M C )eff : (5 lb)  ft  = (ma )r + I α  12  2

10 lb  8  3.333 lb ⋅ ft = ft α + 0.07764α 32.2  12  3.333 = 0.21566α

α = 15.456 rad/s 2

α = 15.46 rad/s 2

 8  a = rα =  ft  (15.456 rad/s 2 )  12 

a = 10.30 ft/s 2

ΣFx = Σ( Fx )eff : − F + 5 lb = ma 10 lb (10.30 ft/s 2 ) 32.2 F = 1.80 lb

− F + 5 lb =

ΣFy = Σ( Fy )eff :

N − 10 lb = 0

N = 10 lb

Fm = μ s N = 0.25(10 lb) = 2.5 lb

(a)

Since F < Fm ,

(b)

Angular acceleration of the disk.

α = 15.46 rad/s 2



Acceleration of G.

a = 10.30 ft/s 2



Disk rolls without sliding 

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PROBLEM 16.103 A drum of 4-in. radius is attached to a disk of 8-in. radius. The disk and drum have a combined weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 lb. Knowing that the coefficients of static and kinetic friction are μ s = 0.25 and μk = 0.20, respectively, determine (a) whether or not the disk slides, (b) the angular acceleration of the disk and the acceleration of G.

SOLUTION  8  a = rα =  ft  α  12 

Assume disk rolls:

I = mk 2 =

10 lb  6  ft 32.2  12 

2

I = 0.07764 lb ⋅ ft ⋅ s 2 ΣM C = Σ( M C )eff : (5 lb)(1 ft) = (ma )r + I α 2

5 lb ⋅ ft =

10 lb  8  ft α + 0.07764α 32.2  12 

5 = 0.21566α

α = 23.184 rad/s 2

α = 23.2 rad/s 2

 8  a = rα =  ft  (23.184 rad/s 2 )  12  ΣFx = Σ( Fx )eff :

−F + 5 lb = ma −F + 5 lb =

ΣFy = Σ( Fy )eff :

a = 15.46 ft/s 2

10 lb (15.46 ft/s 2 ); F = 0.20 lb 32.2

N − 10 lb = 0

N = 10 lb

Fm = μ s N = 0.25(10 lb) = 2.5 lb

(a)

Since F < Fm ,

(b)

Angular acceleration of the disk.

Disk rolls without sliding  α = 23.2 rad/s 2 a = 15.46 ft/s 2

Acceleration of G.

 

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PROBLEM 16.104 A drum of 4-in. radius is attached to a disk of 8-in. radius. The disk and drum have a combined weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 lb. Knowing that the coefficients of static and kinetic friction are μ s = 0.25 and μk = 0.20, respectively, determine (a) whether or not the disk slides, (b) the angular acceleration of the disk and the acceleration of G.

SOLUTION  8  a = rα =  ft  α  12 

Assume disk rolls:

I = mk 2 =

10 lb  6  ft 32.2  12 

2

I = 0.07764 lb ⋅ ft ⋅ s 2

 4  ΣM C = Σ( M C )eff : (5 lb)  ft  = (ma )r + I α  12  2

10 lb  8  1.6667 lb ⋅ ft = ft α + 0.07764α 32.2  12  1.6667 = 0.21566α

α = 7.728 rad/s 2  8  a = rα =  ft  7.728 rad/s 2  12  ΣFx = Σ( Fx )eff :

α = 7.73 rad/s 2 a = 5.153 ft/s 2

−F + 5 lb = ma − F + 5 lb =

10 lb (5.153 ft/s 2 ) 32.2

F = 3.40 lb ΣFy = Σ( Fy )eff :

N − 10 lb = 0

N = 10 lb

Fm = μ s N = 0.25(10 lb) = 2.5 lb

(a)

Since F < Fm ,

Disk slides 

Knowing that disk slides F = μ k N = 0.20(10 lb) = 2 lb

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PROBLEM 16.104 (Continued)

(b)

Angular acceleration.  8   4  ΣM G = Σ( M G )eff : F  ft  − (5 lb)  ft  = I α  12   12   8  (2 lb)  ft  − 1.6667 lb ⋅ ft = (0.07764 lb ⋅ ft ⋅ s 2 )α  12  −0.3333 = 0.07764α

α = −4.29 rad/s 2 (c)

α = 4.29 rad/s 2



a = 9.66 ft/s 2



Acceleration of G. ΣFx = Σ( Fx )eff : −F + 5 lb = ma −2 lb + 5 lb =

10 lb a 32.2

a = 9.66 ft/s 2

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PROBLEM 16.105 A drum of 4-in. radius is attached to a disk of 8-in. radius. The disk and drum have a combined weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 lb. Knowing that the coefficients of static and kinetic friction are μ s = 0.25 and μk = 0.20, respectively, determine (a) whether or not the disk slides, (b) the angular acceleration of the disk and the acceleration of G.

SOLUTION  8  a = rα =  ft  α  12 

Assume disk rolls:

I = mk 2 =

10 lb  6  ft 32.2  12 

2

I = 0.07764 lb ⋅ ft ⋅ s 2 ΣM C = Σ( M C )eff :

 4  (5 lb)  ft  = (ma )r + I α  12  2

10 lb  8  1.6667 lb ⋅ ft = ft α + 0.07764α 32.2  12  1.6667 = 0.21566α

α = 7.728 rad/s 2

α = 7.73 rad/s 2

 8  a = rα =  ft  (7.728 rad/s 2 )  12  ΣFx = Σ( Fx )eff :

F = ma F=

ΣFy = Σ( Fy )eff :

a = 5.153 ft/s 2

10 lb (5.153 ft/s 2 ); F = 1.60 lb 32.2

N − 10 lb + 5 lb = 0

N = 5 lb

Fm = μ s N = 0.25(5 lb) = 1.25 lb

(a)

Since F > Fm ,

Disk slides 

Knowing that disk slides F = μ k N = 0.2(5) F = 1.00 lb

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PROBLEM 16.105 (Continued)

(b)

Angular acceleration.  4   8  ΣM S = Σ( M S )eff : (5 lb)  ft  − F  ft  = I α  12   12   4   8  (5 lb)  ft  − (1.00 lb)  ft  = 0.07764α  12   12  1.000 = 0.07764α

α = 12.88 rad/s 2 (c)

α = 12.88 rad/s 2



Acceleration of G. ΣFx = Σ( Fx )eff : F = ma 1.00 lb =

10 lb a 32.2

a = 3.22 ft/s 2

a = 3.22 ft/s 2



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PROBLEM 16.106 A 12-in.-radius cylinder of weight 16 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 4 lb is applied. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine (a) the acceleration of the carriage, (b) the acceleration of Point A, (c) the distance the cylinder has rolled with respect to the carriage after 0.5 s.

SOLUTION Masses and moments of inertia. mA =

16 lb = 0.49689 lb ⋅ s 2 /ft 32.2 ft/s 2

IA =

1 1 mA r 2 = (0.49689 lb ⋅ s 2 /ft)(1 ft) 2 = 0.24895 lb ⋅ s 2 ⋅ ft 2 2

Kinematics: Let a A = a A

mB =

6 lb = 0.18634 lb ⋅ s 2 /ft 32.2 ft/s 2

, a B = aB

a A = a B + a B /A a A /B = (a A − aB )

α=

a A /B r

(1)

Kinetics: Carriage and cylinder

ΣFx = Σ( Fx )eff : P = mA a A + mB aB

4 lb = (0.49689 lb ⋅ s 2 /ft) a A + (0.18634 lb ⋅ s 2 /ft) aB

Cylinder alone.

(2)

Point C is contact point.

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PROBLEM 16.106 (Continued)

ΣM C = Σ( M C )eff : 0 = I Aα + mA a A r

Substituting from Eqs. (1) and (2), a A − aB + mA a Ar r I  I 0 =  A + m A r  a A − A aB r  r  0 = IA

Data:

I A 0.24895 lb ⋅ s 2 ⋅ ft = = 0.24895 lb ⋅ s 2 1 ft r m A r = (0.49689 lb ⋅ s 2 /ft)(1 ft) = 0.49689 lb ⋅ s 2 0 = 0.74584 a A − 0.24895 aB

(3)

Solving Eqs. (2) and (3) simultaneously, a A = 3.7909 ft/s 2

(a)

Acceleration of the carriage.

(b)

Acceleration of Point A.

(c)

Relative displacement after 0.5 s.

aB = 11.3574 ft/s 2 a B = 11.36 ft/s 2



a A = 3.79 ft/s 2



x B/A = 0.946 ft



a A /B = 11.3574 ft/s 2 − 3.7909 ft/s 2 = 7.5665 ft/s 2 x A /B = =

1 (a A/B )t 2 2 1 (7.5665 ft/s)2 (0.5 s) 2 2

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PROBLEM 16.107 A 12-in.-radius cylinder of weight 16 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 4 lb is applied. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine (a) the acceleration of the carriage, (b) the acceleration of Point A, (c) the distance the cylinder has rolled with respect to the carriage after 0.5 s.

SOLUTION Masses and moments of inertia. mA =

16 lb = 0.49689 lb ⋅ s 2 /ft 2 32.2 ft/s

IA =

1 1 mA r 2 = (0.49689 lb ⋅ s 2 /ft)(1 ft) 2 = 0.24895 lb ⋅ s 2 ⋅ ft 2 2

Kinematics: Let a A = a A

mB =

6 lb = 0.18634 lb ⋅ s 2 /ft 2 32.2 ft/s

, a B = aB a A = a B + a B /A

a A /B = (aB − aB )

α=

a A /B r

(1)

Kinetics: Carriage and cylinder

ΣFx = Σ( Fx )eff : P = mA a A + mB aB 4 lb = (0.49689 lb ⋅ s 2 /ft) a A + (0.18634 lb ⋅ s 2 /ft)aB

Cylinder alone.

(2)

Point C is contact point.

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PROBLEM 16.107 (Continued)

ΣM C = Σ( M B )eff : Pr = I Aα + mA a A r

Substituting from Eqs. (1) and (2), Pr = I A

a A − aB + mA a A r r

I  I Pr =  A + m A r  a A − A aB r r   I A 0.24895 lb ⋅ s 2 ⋅ ft = = 0.24895 lb ⋅ s 2 1 ft r

Data:

m A r = (0.49689 lb ⋅ s 2 /ft)(1 ft) = 0.49689 lb ⋅ s 2

(4 lb)(1 ft) = 0.74584 a A − 0.24895 aB

(3)

Solving Eqs. (2) and (3) simultaneously, a A = 6.6284 ft/s 2

aB = 3.7909 ft/s 2

(a)

Acceleration of the carriage.

a B = 3.79 ft/s 2



(b)

Acceleration of Point A.

a A = 6.63 ft/s 2



(c)

Relative displacement after 0.5 s.

x A/B = 0.355 ft



a A /B = 6.6284 ft/s 2 − 3.7909 ft/s 2 = 2.8375 ft/s 2 x A /B = =

1 (a A/B )t 2 2 1 (2.8375 ft/s 2 )(0.5 s) 2 2

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PROBLEM 16.108 Gear C has a mass of 5 kg and a centroidal radius of gyration of 75 mm. The uniform bar AB has a mass of 3 kg and gear D is stationary. If the system is released from rest in the position shown, determine (a) the angular acceleration of gear C, (b) the acceleration of Point B.

SOLUTION Kinematics:

Since gear D is fixed, we have for Point E of gear C: But

(a E )t = 0

a E = a B + a E/ B

+ (aE )t = (aB )t + (a E/B )t 0 = 0.2 α AB − 0.1 α C

α AB =

1 αC 2

(1)

Gear C ΣM B = Σ( M B )eff : Q (0.1m) = I Cα C = (5 kg) (0.075 m) 2 α C Q = 0.28125 α C

(2)

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PROBLEM 16.108 (Continued)

Bar AB and gear C

(a)

ΣM A = Σ (M A )eff : WAB (0.1) + WC (0.2) − Q (0.3) = (m AB a ) 0.1 + I ABα AB + (mC aB ) 0.2 − I Cα C

(3) g (0.1) + (5) g (0.2) − Q(0.3) = 3(0.1α AB )0.1 +

1 (3)(0.2)2 α AB 12

+ 5(0.2 α AB ) 0.2 − 5(0.075) 2 α C

(1.3) g − 0.3 Q = 0.24α AB − 0.028125α C

Substituting for α AB and Q from (2) and (1): 1  1.3 g − 0.3(0.28125α C ) = 0.24  α C  − 0.028125 α C 2  1.3g = 0.17625 α C

α C = 7.3759(9.81) α C = 72.36

(b)

α C = 72.4 rad/s 2



1  aB = 0.2α AB = 0.2  α C  = 0.1α C = 0.1(72.36) 2 

a B = 7.24 m/s 2 

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PROBLEM 16.109 Two uniform disks A and B, each of mass of 2 kg, are connected by a 1.5 kg rod CD as shown. A counterclockwise couple M of moment 2.5 N-m is applied to disk A. Knowing that the disks roll without sliding, determine (a) the acceleration of the center of each disk, (b) the horizontal component of the force exerted on disk B by pin D.

SOLUTION r = 150 mm = 0.15 m

Geometry:

b = AC = BD = 50 mm = 0.05 m r + b = 0.20 m

Masses:

m A = mB = m = 2 kg

Moment of inertia:

I A = IB = I =

Kinematics:

α CD = 0

mCD = 1.5 kg

1 2 mr 2

a A = aB = a = a

Angular accelerations of disks:

For rod CD,

α=

a r

aC = a A + bα

+bω 2

a D = a B + bα

+bω 2

a = (a + bα )

+bω 2

 b = 1 +  a r 

+bω 2

Kinetics: Disk A:

ΣM P = Σ( M P )eff : M − (r + b)C x = mar + I α  =  mr + 

Disk B:

I (a ) r 

(1)

ΣM Q = Σ( M Q )eff : − ( r + b) Dx = mar + I α  =  mr + 

I a r 

(2)

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PROBLEM 16.109 (Continued)

Rod CD:

 b ΣFx = Σ( Fx )eff : C x + Dx = mCD 1 +  a r 

Multiply by (r + b) 2

 b (r + b)(C x + Dx ) = mCD r 1 +  a r 

(3)

Add Eqs. (1), (2), and (3) to eliminate C x and Dx 2

I   b M = 2  mr +  a + mCD r 1 +  a r r  

Apply the numerical data.

mr +

I 1 (0.15 m) 2 = (2 kg)(0.15 m) + (2 kg) = 0.45 kg ⋅ m 2 0.15 m r −2

2

 b  0.05  mCD r 1 +  = (1.5 kg)(0.15 m) 1 +  = 0.40 kg ⋅ m r   0.15 

2.5 N ⋅ m = 2(0.45 kg ⋅ m) a + (0.40 kg ⋅ m) a a = 1.9231 m/s 2

(a)

Acceleration of the center of each disk: a A = a B = 1.923 m/s 2

(b)



Horizontal component of the force exerted on disk B by pin D. From Eq. (2),

1  I  mr +  a r +b r 1 =− (0.45 kg ⋅ m) (1.9231 m/s 2 ) = −4.33 N 0.20 m

Dx = −

D x = 4.33 N



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PROBLEM 16.110 A 10-lb cylinder of radius r = 4 in. is resting on a conveyor belt when the belt is suddenly turned on and it experiences an acceleration of magnitude a = 6ft/s2. The smooth vertical bar holds the cylinder in place when the belt is not moving. Knowing the cylinder rolls without slipping and the friction between the vertical bar and the cylinder is negligible, determine (a) the angular acceleration of the cylinder, (b) the components of the force the conveyor belt applies to the cylinder.

SOLUTION Mass and moment of inertia. m=

W 10 lb = = 0.31056 lb ⋅ s 2 /ft g 32.2 ft/s 2

I=

1 2 1  4  mr = (0.31056 lb ⋅ s 2 /ft)  ft  = 0.017253 lb ⋅ s 2 ⋅ ft 2 2  12 

2

Kinematics:

The cylinder rolls without slipping on the belt which is accelerating at 6 ft/s 2 aG = (6 ft/s 2 − rα )

where α = α

  4   5° = 6 ft/s 2 −  ft  α   12   

5°.



is the angular acceleration of the cylinder.

Kinetics: Let Point C be the contact point between the belt and the cylinder.

ΣM C = Σ( M C )eff : Wr sin β = −rmaG + I α

  4   4   4   (10 lb)  ft  sin 5° = −  ft  (0.31056 lb ⋅ s 2 /ft)  6 ft/s 2 −  ft  α  + (0.017253 lb ⋅ s 2 ft)α  12   12   12    0.29052 lb ⋅ ft = −0.62112 lb ⋅ ft + (0.051760 lb ⋅ s 2 ⋅ ft)α

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PROBLEM 16.110 (Continued) (a)

Angular acceleration.

α = 17.613 rad/s 2

α = 17.61 rad/s 2



 4  ft  (17.613 rad/s 2 ) = 0.129 ft/s 2 aG = 6 ft/s 2 −  12  

(b)

Components of contact force: 5° ΣF = Σ( Feff ) : F − W sin 5° = maG F = W sin 5° + maG = (10 lb)sin 5° + (0.31056 lb ⋅ s 2 /ft)(0.129 ft/s 2 ) F = 0.912 lb



5° 

85° ΣF = ΣF = ΣFeff : N − W cos 5° = 0 N = W cos 5° = (10 lb) cos 5° N = 9.96 lb

85° 

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PROBLEM 16.111 A hemisphere of weight W and radius r is released from rest in the position shown. Determine (a) the minimum value of μs for which the hemisphere starts to roll without sliding, (b) the corresponding acceleration of Point B. [Hint: Note that OG = 83 r and that, by the parallel-axis theorem, I = 52 mr 2 − m(OG) 2.]

SOLUTION

Kinematics:

ω=0

aO = a A + aO/ A = 0 + [rα

]

a = aG = aO + aG/O

= [rα

Thus,

ax = rα

] + [ xα ] ,

a y = xα

(1)

Kinetics:

ΣM A = Σ(M A )eff :

W x = (max )r + (ma y ) x + I α mg x = (mrα ) r + (mxα ) x + mk 2 α

α=

gx r + x2 + k 2 2

(2)

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PROBLEM 16.111 (Continued)

F = mrα

ΣFx = Σ ( Fx )eff :

F = max

ΣFy = Σ ( Fy )eff :

N − W = − ma y

μmin =

F mrα = N mg − x α

N = mg − mxα

μmin =

rα g − xα

(3)

For a hemisphere: x = OG =

3 r 8

I = I O − mx 2 =

I =

2 2 9 mr − mr 2 5 64

α=

( 38 r ) = 9 2 2 9 r2 + r + ( 52 − 64 r ) 64

k2 =

2 3  mr 2 − m  r  5 8 

2

I 9  2 2 = − r m  5 64 

Substituting into (2)

(a)

3 8 7 5

g r

α=

15 g 56 r

Substituting into (3)

μmin =

(b)

g

1−

15 56 3 8

( )( ) 15 56

=

0.26786 0.89955

 15g  30 g aB = (2r ) α = (2r )   = 56  56r 

Note:

μmin = 0.298  a B = 0.536 g



In this problem we cannot use the equation ΣM A = I A α , since Points A, O, and G are not aligned.

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PROBLEM 16.112 Solve Problem 16.111, considering a half cylinder instead of a hemisphere. [Hint. Note that OG = 4r /3π and that, by the parallel-axis theorem, I = 12 mr 2 − m(OG ) 2.]

SOLUTION

Kinematics:

ω=0

aO = a A + aO/ A = 0 + [rα

]

a = aG = aO + aG/O

= [rα

Thus,

a x = rα

] + [ xα ] ,

a y = xα

(1)

Kinetics:

ΣM A = Σ(M A )eff :

W x = (max )r + (ma y ) x + I α mg x = (mrα ) r + (mxα ) x + mk 2 α

α=

gx r2 + x 2 + k 2

(2)

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PROBLEM 16.112 (Continued)

F = mrα

ΣFx = Σ ( Fx )eff :

F = max

ΣFy = Σ ( Fy )eff :

N − W = − ma y

μmin =

F mrα = N mg − x α

N = mg − mxα

μmin =

rα g − xα

(3)

For a half cylinder: x = OG =

4r 3π

I = I O − mx 2 = k2 =

1  4r  mr 2 − m   2  3π 

I 1 2  4r  = r −  m 2  3π 

2

2

Substituting into (2):

α= (a)

r2 +

( 34πr )

2

+

1 2 r − 2

( 34πr )

2

=

4 3π 3 2

g r

α=

8 g 9π r

Substituting into (3):

μmin =

(b)

( 34πr )

g

1−

8 9π 4 3π

( )( ) 8 9π

=

0.28294 0.87992

 8 g  16 g aB = (2r ) α = (2r )  =  9π r  9π

μmin = 0.322  aB = 0.566 g 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1581

PROBLEM 16.113 The center of gravity G of a 1.5-kg unbalanced tracking wheel is located at a distance r = 18 mm from its geometric center B. The radius of the wheel is R = 60 mm and its centroidal radius of gyration is 44 mm. At the instant shown the center B of the wheel has a velocity of 0.35 m/s and an acceleration of 1.2 m/s2, both directed to the left. Knowing that the wheel rolls without sliding and neglecting the mass of the driving yoke AB, determine the horizontal force P applied to the yoke.

SOLUTION Kinematics: Choose positive vB and a B to left.

Trans. with B

+

Rotation about B

a = [ a B + rω 2 ]

=

Rolling motion

r  +  aB  R 

Kinetics:

ΣM C = Σ( M C )eff : PR − Wr = (ma y )r + ( max ) R + Iα a r  PR − mgr = m  aB  r + m(aB + rω 2 ) R + mk 2 B R R  2  r2 k2   vB  = maB  + R +  + mr   R R   r  R  r2 + k 2  r 2 r P = mg   + maB 1 +  + m 2 vB 2 R R R  

(1)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1582

PROBLEM 16.113 (Continued)

m = 1.5 kg r = 0.018 m

Substitute:

R = 0.06 m k = 0.044 mm and g = 9.81 m/s 2 in Eq. (1)

P = 1.5(9.81)

 0.0182 + 0.0442 0.018 + 1.5( aB ) 1 + 0.06 0.062 

 0.018 2 vB  + 1.5 0.062 

P = 4.4145 + 2.4417aB + 7.5vB2

Data:

vB = 0.35 m/s

; vB = +0.35 m/s

a B = 1.2 m/s 2

; aB = +1.2 m/s 2

Substitute in Eq. (2):

(2)

P = 4.4145 + 2.4417(+1.2) + 7.5( +0.35) 2 = 4.4145 + 2.9300 + 0.9188 = +8.263 N

P = 8.26 N



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1583

PROBLEM 16.114 A small clamp of mass mB is attached at B to a hoop of mass mh . The system is released from rest when θ = 90° and rolls without sliding. Knowing that mh = 3mB , determine (a) the angular acceleration of the hoop, (b) the horizontal and vertical components of the acceleration of B.

SOLUTION Kinematics:

a A = rα

, aB/A = rα

aB = a A

+ aB/A

a B = rα

+ rα

(a B ) x = rα

(aB ) y = rα

Kinetics:

mh = 3mB I = mh r 2 = 3mB r 2

(a)

Angular acceleration. ΣM C = Σ( M C )eff : WB r = I α + mh a A r + mB (aB ) x r + mB (aB ) y r mB gr = 3mB r 2α + (3mB )r 2α + mB r 2α + mB r 2α gr = 8r 2α

(b)

1g 8r



Components of acceleration of B. (a B ) x = rα =



α=

1 g 8

(a B ) y = rα =

1 g  8 (a B ) x =



1 g 8

(a B ) y =

1 g 8

 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1584

PROBLEM 16.115 A small clamp of mass mB is attached at B to a hoop of mass mh. Knowing that the system is released from rest and rolls without sliding, derive an expression for the angular acceleration of the hoop in terms of mB, mh, r, and θ .

SOLUTION Kinematics:

Kinetics:

a A = rα

aB/ A = rα

θ

a B = aA

+ aB/ A

θ

a B = rα

+ rα

θ

I = mh r 2

ΣM C = Σ( M C )eff : WB r sin θ = I α + mh a A r + mB rα ( r + r cos θ ) + mB rα sin θ (r sin θ ) + mB rα cos θ (r + r cos θ ) mB gr sin θ = mh r 2α + mh (rα )r + mB rα (1 + cos θ )(r + r cos θ ) + mB rα sin θ (r sin θ ) mB gr sin θ = 2mh r 2α + mB r 2α [(1 + cos θ ) 2 + sin 2 θ ]



= 2mh r 2α + mB r 2α [1 + 2 cos θ + cos 2 θ + sin 2 θ ] mB sin θ g mB gr sin θ = r 2α [2mh + mB (2 + 2 cos θ )]   α= 2r mh + mB (1 + cos θ )

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1585

PROBLEM 16.116 A 4-lb bar is attached to a 10-lb uniform cylinder by a square pin, P, as shown. Knowing that r = 16 in., h = 8 in., θ = 20°, L = 20 in. and ω = 2 rad/s at the instant shown, determine the reactions at P at this instant assuming that the cylinder rolls without sliding down the incline.

SOLUTION Masses and moments of inertia. Bar:

mB =

WB 4lb = = 0.12422 slug g 32.2 2

1 1  4 lb  20 in.  2 mB L2 =    = 0.028755 slug ⋅ ft 12 12  32.2  12  W 10 lb mD = D = = 0.31056 slug g 32.2 IB =

Disk:

2

1 1  10 lb  16 in  2 I D = mC r 2 =    = 0.27605 slug ⋅ ft 2 2  32.2  12  Kinematics. Let Point C be the point of contact between the cylinder and the incline and Point G be the mass center of the cylinder without the bar. Assume that the mass center of the bar lies at Point P. α =α

For rolling without slipping,

. aG = aG

20°

(aC )t = aG − rα = 0 aG = rα (aP )t = aG + hα (aP )t = (r + h)α (aP ) h = 0 + hω 2 = hω 2

Kinetics: Using the cylinder plus the bar as a free body, ΣM C = +Σ( M C )eff : mD gr sin θ + mB g (r + h) sin θ = I Dα + mD aG r + I Bα + mB (aP )t (r + h) = [ I D + mD r 2 + I B + mB (r + h) 2 ]α

α= =

[mD gr + mB g (r + h)]sin θ I D + mD r 2 + I B + mB (r + h) 2 24  (10) ( 16 + (4) ( 12 ) sin 20° 12 ) 

10 0.27605 + ( 32.2 )( 1216 ) + (0.028755) + ( 32.24 ) ( 1224 ) 2

2

= 5.3896 rad/s 2

24 (aP )t = ( 12 ) (5.3896) = 10.7791 ft/s2

(aP ) n = ( 128 ) (2) 2 = 2.6667 ft/s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1586

PROBLEM 16.116 (Continued)

Using the bar alone as a free body, ΣFx = Σ( Fx )eff :

Px = mB ( aP )t cos 20° − mB (aP ) n sin 20°

 4   4  =  (10.779) cos 20° −  32.2  (2.6667)sin 20° 32.2     Px = 1.1450 lb ± ΣFy = Σ( Fy )eff : Py − mB g = −mB (aP )t sin 20° − mB (aP ) n cos 20°  4   4  Py = (4) −   (10.779)sin 20° −   (2.6667) cos 20° 32.2    32.2  Py = 3.2307 lb P = 1.1452 + 3.2307 2 = 3.4276 lb

3.2307 1.145 β = 70.5°

tan β =

P = 3.43 lb

70.5° 

Recognizing that P is the CG of the bar. ± ΣM P = Σ( M P )eff :

M P = I Bα

= (0.028755)(5.3896) 

M P = 0.1550 ft ⋅ lb



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PROBLEM 16.117 The ends of the 20-lb uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods. If the rod is released from rest when θ = 25°, determine immediately after release (a) the angular acceleration of the rod, (b) the reaction at A, (c) the reaction at B.

SOLUTION Kinematics: Assume α

ω=0

] + [4α

a B = a A + a B/A = [ a A

25°]

aB = (4α ) cos 25° = 3.6252α a A = (4α )sin 25° = 1.6905α

] + [2α

aG = a A + aG/ A = [ a A aG = [1.6905α

] + [2α

ax = (aG ) x = [1.6905α

25°]

25°] ] + [0.84524α

]

ax = 0.84524α

a y = [2α cos 25° ] = 1.8126α

We have found for α

ax = 0.84524α a y = 1.8126α

Kinetics:

I =

1 1 mL2 = m(4 ft) 2 12 12

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PROBLEM 16.117 (Continued)

(a)

Angular acceleration. ΣM E = Σ( M E )eff :

mg (1.8126 ft) = I α + max (0.84524 ft) + ma y (1.8126 ft) mg (1.8126) =

1 m(4) 2 α + m(0.84524)2 α + m(1.8126) 2α 12

g (1.8126) = 5.3333α

(b)

α = 0.33988g



ΣFy = Σ( Fy )eff : A − mg = −ma y = m(1.8126α )  20  A − 20 = −   (1.8126)(10.944)  32.2  A = 20 − 12.321 = 7.6791 lb

(c)

α = 10.944 rad/s 2

ΣFx = Σ( Fx )eff :

A = 7.68 lb 

B = max = m(0.84524α ) 20 (0.84524)(10.944) 32.2 B = 5.7453 lb B=



B = 5.75 lb



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1589

PROBLEM 16.118 The ends of the 20-lb uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods. A vertical force P is applied to collar B when θ = 25°, causing the collar to start from rest with an upward acceleration of 40 ft/s 2 . Determine (a) the force P, (b) the reaction at A.

SOLUTION Kinematics: ω = 0

a B = a A + a B/ A ; [40 ft/s 2 ] = [a A

] + [4α

25°]

aB = aB/ A cos 25° 40 ft/s 2 = (4α ) cos 25°

α = 11.034 rad/s 2

a A = 40 tan 25° = 18.6523 ft/s 2 aG = a A + aG/ A : aG = [18.652 aG = [18.652

] + [2α

25°]

] + [2(11.034)

25°]

ax = (aG ) x = 9.3262 ft/s 2 a y = (aG ) y = 20.00 ft/s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1590

PROBLEM 16.118 (Continued)

Kinetics:

(a)

I =

1 1 mL2 = m(4) 2 12 12

ΣM E = Σ( M E )eff : P(3.6252) − W (1.8126) = I α + max (0.84524) + ma y (1.8126)

W = mg = 20 lb 1 I α = mL2α 12 1  20  2 =  (4) (11.034) 12  32.2  = 9.1377 ft ⋅ lb  20  max =   (9.3262) = 5.7926 lb  32.2   20  ma y =   (20) = 12.4224 lb  32.2  P (3.6252) − (20)(1.8126) = 9.1377 + (5.7926)(0.84524) + (12.422)(1.8126) P (3.6252) − 36.252 = 9.1377 + 4.8962 + 22.517 P(3.6252) = 72.8031 P = 20.082 lb

(b)

P = 20.1 lb 

ΣFy = Σ( Fy )eff : A − W + P = ma y A − 20 + 20.082 = 12.4224 lb

A = 12.34 lb 

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PROBLEM 16.119 The motion of the 3-kg uniform rod AB is guided by small wheels of negligible weight that roll along without friction in the slots shown. If the rod is released from rest in the position shown, determine immediately after release (a) the angular acceleration of the rod, (b) the reaction at B.

SOLUTION Kinematics:

aG /A = (0.4 m)α a B /A = (0.8 m)α 30°] + [0.8α

a B = a A + a B/A : [aB ] = [a A aA =

0.8α = 0.92376α cos 30°

]

30°

a B = (0.8α ) tan 30° = 0.46188α a = aG = a A + aG /A ; a = [0.92376α ax = [0.8α

] + [0.4α

30° ] + [0.4α +

]

] = 0.4α

a y = [0.46188α ] = 0.46188α

We have:

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PROBLEM 16.119 (Continued)

ax = 0.4α

; a y = 0.46188α

1 1 ⋅ mL2 = ⋅ (3 kg)(0.8 m) 2 12 12 I = 0.16 kg ⋅ m 2 I =

(a)

Angular acceleration. ΣM E = Σ( M E )eff : mg (0.46188 m) = I α + max (0.4 m) + ma y (0.46188 m) 3(9.81)(0.46188) = 0.16α + 3(0.4)2α + 3(0.46188) 2 α 13.593 = 1.28α

α = 10.620 rad/s 2 (b)

α = 10.62 rad/s 2



Reaction at B. ΣM A = Σ( M A )eff : B(0.8 m) = − I α + max (0.4 m) 0.8B = −(0.16)(10.620) + 3(0.4)(10.620)(0.4) 0.8B = −1.6991 + 5.0974 B = 4.2479 N

B = 4.25 N



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PROBLEM 16.120 A beam AB of length L and mass m is supported by two cables as shown. If cable BD breaks, determine at that instant the tension in the remaining cable as a function of its initial angular orientation θ.

SOLUTION Kinematics:

At the instant just after the cable break,

ω=0

α =α

aG = a A +

Kinetics:

I=

L α = αA 2

L 2

θ+ α

1 mL2 12

L 1 = I α = mL2α 2 12 6T sin θ α= mL mL θ ΣF = ΣFeff : T − mg sin θ = − α 2 mL 6T sin θ =− ⋅ mL 2 = − 3T sin θ ΣM G = Σ( M G )eff : (T sin θ )

T=

Solving for T,

mg sin θ  1 + 3sin θ

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PROBLEM 16.121 End A of a uniform 10-kg bar is attached to a horizontal rope and end B contacts a floor with negligible friction. Knowing that the bar is released from rest in the position shown, determine immediately after release (a) the angular acceleration of the bar, (b) the tension in the rope, (c) the reaction at B.

SOLUTION α =α

Kinematics: [a B

 Lα aG = a A +   2

] = [a A ] + [ Lα

a A = 0.7071 Lα

45°]

  Lα 45° = [0.7071 α ] +    2

 Lα 45° = 2 

45°

+ ΣM P = m g

L (0.7071) 2

=

1 L L m L2 α + m α   2 2 12

 3g   2L 

α = 0.7071 

g = 9.81 m/s 2 ,

L = 1 m,

(a) α = 10.405 rad/s 2 ΣFx = T =



mLα m L (0.5)  3g  3 (0.7071) =   = mg = 36.788 N 2 2  L  8

3 ΣFy = N − mg = − mg , 8

N =

(b)

T = 36.8 N 

(c)

N = 61.3 N 

5 mg = 61.313 N 8

Alternate solution: Kinematics (realizing that the rope constrains Point A to the j-direction at the instant shown): a A j = aB i + α k x( L cos 45°i + L sin 45° j) = aB i + ( L cos 45°)α j − ( L sin 45°)α i PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1595

PROBLEM 16.121 (Continued)

x-components 0 = aB − ( L sin 45°)α

aB = ( L sin 45°)α

y-components a A = ( L cos 45°)α

Acceleration of CG aG = a B + α × rG /B − ω 2rG /B = a B − α × rG /B

L L  aG = ( L sin 45°)α i + α k ×  cos(45°)i + sin(45°) j  2 2  L L  = ( L sin 45)α i +  cos(45°)α j − sin(45°)α i  2 2  L L aG = sin 45°α i + cos 45°α j 2 2

Kinetics:

Equations of motion: ΣFx = max

L  −T = m  sin 45°  α 2 

(1)

ΣFy = ma y L  N − mg = m  cos 45°  α 2  L  N = mg + m  cos 45  α 2 

(2)

ΣM G = T α 1 L L  T   sin 45° − N  cos 45°  = mL2α 2 2  12

(3)

Solving the three simultaneous equations gives the same results as above.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1596

PROBLEM 16.122 End A of the 8-kg uniform rod AB is attached to a collar that can slide without friction on a vertical rod. End B of the rod is attached to a vertical cable BC. If the rod is released from rest in the position shown, determine immediately after release (a) the angular acceleration of the rod, (b) the reaction at A.

SOLUTION

ω=0

Kinematics:

a B = a A + a B/A ; [aB

] = [ a A ] +[Lα

θ]

0 = a A − Lα cos θ a A = Lα cos θ a = a A + aG/A

L a = [ Lα cos θ ] +  α 2 ax =

L α sin θ 2

; ay =



θ 

L α cos θ 2

Kinetics:

ΣM E = Σ( M E )eff : mg

L L  L  cos θ = I α + max  sin θ  + ma y  cos θ  2 2 2     2

2

mg

L 1 L  L  cos θ = mL2α + m  sin θ  α + m  cos θ  α 2 12 2 2    

mg

L 1 cos θ = mL2α 2 3

α=

3g cos θ 2L

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PROBLEM 16.122 (Continued)

L ΣFx = Σ( Fx )eff : A = max = m α sin θ 2 A=m A=

L3 g  cos θ  sin θ  22 L 

3 mg sin θ cos θ 4

m = 8 kg, θ = 30°, L = 0.75 m

Data: (a)

Angular acceleration.

α=

3 9.81 m/s 2 cos 30° 2 0.75 m

(b)

Reaction at A.

A=

3 (8 kg)(9.81 m/s 2 ) sin 30° cos 30° 4

α = 16.99 rad/s 2 A = 25.5 N

 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1598

PROBLEM 16.123 A uniform thin plate ABCD has a mass of 8 kg and is held in position by three inextensible cords AE, BF, and CG. If cord AE is cut, determine at that instant (a) if the plate is undergoing translation or general plane motion, (b) the tension in cords BF and CG.

SOLUTION Immediately after cord AE breaks, ω = 0. (a)

Assume that the cords BF and CG constrain the plate to undergo curvilinear translation, making α = 0. aG = a

+

30°

30°ΣF = ΣFeff : mg sin 30° = maG aG = g sin 30°

ΣM B = Σ( M B )eff : 0.075mg + 0.200TCG cos 60° = mg sin 30°(0.100 + 0.075sin 30°) 0.100TCG = −0.00625mg TCG = −0.0625mg

Since TC is negative, the cord becomes slack so that the plate undergoes general plane motion with TCG = 0. general plane motion  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1599

PROBLEM 16.123 (Continued)

(b)

a B = aA

Kinematics:

30°

α =α

aG = a B + 0.100α

+ 0.075α

Kinetics:

m = 8 kg I=

1 (8 kg)[(0.150 m) 2 + (0.200 m) 2 ] = 0.041667 kg ⋅ m 2 12

ΣFx = Σ( Fx )eff : TBF cos 60° = (8 kg)(0.100α ) + (8 kg)aB cos 30°

(1)

ΣFy = Σ( Fy )eff : TBF sin 60° − (8 kg)(9.81 m/s 2 ) = −(8 kg)(0.075α ) − (8 kg)aB sin 30°

(2)

ΣM G = Σ( M G )eff : 0.075TBF sin 60° − 0.100TBF cos 60° = I α 0.014952 TBF = 0.041667α

(3)

Solving Eqs. (1), (2), and (3) simultaneously, TBF = 65.168 N, α = 23.385 rad/s 2 , aB = 2.0028 m/s 2 TBF = 65.2 N 

Tension in cords:

TCG = 0 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1600

PROBLEM 16.124 The 4-kg uniform rod ABD is attached to the crank BC and is fitted with a small wheel that can roll without friction along a vertical slot. Knowing that at the instant shown crank BC rotates with an angular velocity of 6 rad/s clockwise and an angular acceleration of 15 rad/s2 counterclockwise, determine the reaction at A.

SOLUTION

Crank BC: BC = 0.1 m

Rod ABD:

(aB )t = ( BC )α = (0.1 m)(15 rad/s 2 ) = 1.5 m/s 2

(a B )t = 1.5 rad/s 2

(aB ) n = ( BC )ω 2 = (0.1 m)(6 rad/s)2 = 3.6 m/s 2

(a B ) n = 3.6 m/s 2

θ = sin −1

BC 0.1 m = sin −1 = 30° AB 0.2 m

a A = a B + a A /G

[a A ] = [1.5 + 3.6

] + [0.2α

β]

0 = 3.6 − (0.2α ) cos β

α=

3.6 18 = = 20.78 rad/s 2 0.2 cos β cos 30°

α = 20.78 rad/s 2

Kinetics: ΣM G = Σ( M G )eff : A(0.1732 m) = I α = =

1 m L2α 12

1 (4 kg)(0.4 m) 2 (20.78 rad/s 2 ) 12

A = 6.399 N

A = 6.40 N



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PROBLEM 16.125 The 7-lb uniform rod AB is connected to crank BD and to a collar of negligible weight, which can slide freely along rod EF. Knowing that in the position shown crank BD rotates with an angular velocity of 15 rad/s and an angular acceleration of 60 rad/s2, both clockwise, determine the reaction at A.

SOLUTION Crank BD:

 4  ω BD = 15 rad/s, v B =  ft  (15 rad/s) = 5 ft/s  12  α BD = 60 rad/s 2  4  (a B ) x =  ft  (60 rad/s 2 ) = 20 ft/s 2  12   4  (a B ) y =  ft  (15 rad/s) 2 = 75 ft/s 2 12  

Rod AB: Velocity: Instantaneous center at C.  25  CB =  ft  / tan 30° = 3.6084 ft  12  ω AB =

vB 5 ft/s = = 1.3856 rad/s CB 3.6084 ft

Acceleration:

(a A/B )t = ( AB )α AB =

25 α AB 12

 25  2 (a A /B ) n = ( AB)ω AB =   (1.3856)2 = 4 ft/s 2  12  (aG/B )t = (GB)α AB =

12.5 α AB 12

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PROBLEM 16.125 (Continued)

(aG /B ) n

 12.5  2 2 2 = (GB)ω AB =  (1.3856) = 2 ft/s 12  

a A = a B + a B/A = a B + (a B/A )t + (aG/A ) n [a A

30°] = [20

a A cos 30° = 20 − 4; (18.475) sin 30° = 75 −

 25 ] + [75 ] +  α AB  12 a A = 18.475 ft/s 2

  + [4 

]

30°

25 α AB ; α AB = 31.566 rad/s 2 12

a = a B + aG/B = a B + (aG/B )t + (aG/B )n a = [20

] + [75 ] + [ 12.5 (31.566) ] + [2 12

ax = 20 − 2 = 18;

]

ax = 18 ft/s 2

a y = 75 − 32.881 = 42.119; a y = 42.119 ft/s 2 2

Kinetics:

I =

1 7 lb  25  m( AB )2 = ft = 0.078628 slug ⋅ ft 2 12 12(32.2)  12 

 25   12.5   12.5  ft  − mg  ft  = − I α AB + ma y  ft  ΣM B = Σ( M B )eff : ( A sin 60°)   12   12   12   12.5   7   12.5  1.8042 A − (7 lb)  ft  = −(0.078628 slug ⋅ ft 2 )(31.566 rad/s 2 ) +  slug  (42.119 ft/s 2 )  ft   12   32.2   12  1.8042 A − 7.2917 = −2.4820 + 9.5378 A = 7.9522 lb

A = 7.95 lb

60° 

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PROBLEM 16.126 In Problem 16.125, determine the reaction at A, knowing that in the position shown crank BD rotates with an angular velocity of 15 rad/s clockwise and an angular acceleration of 60 rad/s2 counterclockwise. PROBLEM 16.125 The 7-lb uniform rod AB is connected to crank BD and to a collar of negligible weight, which can slide freely along rod EF. Knowing that in the position shown crank BD rotates with an angular velocity of 15 rad/s and an angular acceleration of 60 rad/s2, both clockwise, determine the reaction at A.

SOLUTION Crank BD:

ω BD = 15 rad/s,  4  ft  (15 rad/s) = 5 ft/s vB =   12  α BD = 60 rad/s 2  4  (a B ) x =  ft  (60 rad/s 2 ) = 20 ft/s 2  12   4  (a B ) y =  ft  (15 rad/s) 2 = 75 ft/s 2 12  

Rod AB: Velocity: Instantaneous center at C.  25  CB =  ft  /tan 30° = 3.6084 ft  12  ω AB =

vB 5 ft/s = = 1.3856 rad/s CB 3.6084 ft

Acceleration:

(a A /B )t = ( AB) α AB =

25 α AB 12

 25  2 (a A /B ) n = ( AB )ω AB =   (1.3856)2 = 4 ft/s 2  12 

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PROBLEM 16.126 (Continued)

(aG /B )t = (GB)α AB =

12.5 α AB 12

 12.5  2 2 2 (aG /B ) n = (GB )ω AB =  (1.3856) = 2 ft/s 12   a A = a B + a B /A = a B + (a B /A )t + (a B /A ) n 30°] = [20

[a A

: a A cos 30° = 20 + 4;

 25 ] + [75 ] +  α AB  12

a A = 27.713 ft/s 2

: (27.713) sin 30° = −75 +

25 α AB ; 12

  + [4 

]

30°

α AB = 42.651 rad/s 2

a = a B + aG /B = a B + (aG /B )t + (aG /A ) n a = [20

12.5 (42.651) ] + [75 ] +   12

ax = 20 + 2 = 22;

  + [2 

]

a x = 22 ft/s 2

a y = 75 − 44.428 = 30.572;

a y = 30.6 ft/s 2

2

Kinetics:

I =

7 lb  25  ft = 0.078628 slug ⋅ ft 2 12 (32.2)  12 

 25   12.5   12.5  ft  − mg  ft  = − Ia AB + m a y  ft  ΣM B = Σ( M B )eff : ( A sin 60°)  12 12      12   12.5   7   12.5  1.8042 A − (7 lb)  ft  = −(0.078628 slug ⋅ ft 2 )(42.651 rad/s 2 ) +  slug  (30.572 ft/s 2 )  ft   12   32.2   12  1.8042 A − 7.2917 = −3.3536 + 6.9230 A = 6.0198 lb

A = 6.02 lb

60° 

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PROBLEM 16.127 The 250-mm uniform rod BD, of mass 5 kg, is connected as shown to disk A and to a collar of negligible mass, which may slide freely along a vertical rod. Knowing that disk A rotates counterclockwise at a constant rate of 500 rpm, determine the reactions at D when θ = 0.

SOLUTION Kinematics:

(α A = 0)

For disk A:

ω A = 500 rpm = 52.36 rad/s vB = rω A = (0.05 m)(52.36 rad/s) = 2.618 m/s aB = rω A2 = (0.05 m)(52.36 rad/s 2 ) = 137.08 m/s

For rod (velocities) BC = (0.25) 2 − (0.15) 2 = 0.20 m 0.15 3 = 0.25 5 0.20 4 = sin β = 0.25 5

cos β =

ω=

vB 2.618 m/s = BC 0.20 m

ω = 13.09 rad/s

Kinematics of rod (accelerations) a B + a D/B = a D [aB ] + [(aD / B )n

β ] + [( aD /B )t

[137.08 ] + [0.25(13.09) 2

β ] = aD

β ] + [0.25α

β ] = aD

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PROBLEM 16.127 (Continued)

x components: 3 4 (42.84) − (0.25α ) = 0 5 5 25.704 − 0.2α = 0

α = 128.52 rad/s 2

a = aG = a B + aG/B = [137.08 ] + [0.125(13.09)2 = [137.08 ] + [21.42

] + [0.125(128.52)

β ] + [16.065

]

β]

3 4 ax = (21.42) − (16.065) = 0 5 5 4 3 a y = 137.08 − (21.42) − (16.065) = 110.31 m/s 2 5 5

Kinetics I =

1 1 ml 2 = (5 kg)(0.25 m) 2 12 12 = 0.026042 kg ⋅ m 2

ΣM B = Σ( M B )eff : D(0.20 m) + W (0.075 m) = ma (0.075) − Iα 0.2 D + 5(9.81)(0.075) = 551.6(0.075) − 3.347 0.2 D = 41.370 − 3.347 − 3.679 = 34.344 D = 171.7 N

D = 171.7 N



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PROBLEM 16.128 Solve Problem 16.127 when θ = 90°. PROBLEM 16.127 The 250-mm uniform rod BD, of mass 5 kg, is connected as shown to disk A and to a collar of negligible mass, which may slide freely along a vertical rod. Knowing that disk A rotates counterclockwise at a constants rate of 500 rpm, determine the reactions at D when θ = 0.

SOLUTION Kinematics:

For disk A:

(α A = 0)

ω A = 500 rpm = 52.36 rad/s vB = rω A = (0.05 m)(52.36 rad/s) = 2.618 m/s aB = rω A2 = (0.05 m)(52.36 rad/s)2 = 137.08 m/s 2

For rod (velocities) Since vD is parallel to, vB we have ω = 0 0.10 0.25 φ = 66.42°

cos φ =

We also note that

For rod (accelerations) (aD/B ) n = 0

ω=0

Since

a B + a D /B = a D 137.08 → +0.25α

φ = aD

(0.25α ) sin 66.42° = 137.08

α = 598.3 rad/s 2 a = aG = a B + aG/B ax = 137.08 − 0.125(598.3) sin 66.42° = +68.54 m/s 2 a y = 0.125(598.3) cos 66.42° = +29.92 m/s 2

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PROBLEM 16.128 (Continued)

Summary of kinematics:

α = 598.3 rad/s 2

,

a y = 29.92 m/s 2

0.2291D = 39.256 − 7.480 − 15.581 − 2.453 = 13.742

D = 60.0 N

,

ax = 68.54 m/s 2

1 1 ml 2 = (5 kg)(0.25 m) 2 12 12 = 0.026042 kg ⋅ m 2

I =

Kinetics:

We recall that φ = 66.42°. Thus: h = (0.25 m)sin φ = 0.2291 m h ΣM B = Σ( M B )eff : Dh + W (0.05) = max   − ma y (0.05) − Iα 2 D (0.2291) + 5(9.81)(0.05) = (342.7)

0.2291 − (149.6)(0.05) − 15.581 2



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PROBLEM 16.129 The 4-kg uniform slender bar BD is attached to bar AB and a wheel of negligible mass which rolls on a circular surface. Knowing that at the instant shown bar AB has an angular velocity of 6 rad/s and no angular acceleration, determine the reaction at Point D.

SOLUTION Kinematics: Since ( v D ) n = 0, (vD ) x cos 60° = (vD ) y cos 30° (vD ) x = (vB ) x = ( AB) ωBA = (0.75 m)(6 rad/s) (vD ) y = (vD ) x

cos 60° cos 30°

= (0.75)(6)

( 12 )

( ) 3 2

=

0.75 (6) 3

But (vD ) y = ωBD ( BD)

ωBD =

(vD ) y BD

=

0.75 (6) 6 rad/s = 0.75 3 3

2 a B = AB ω BA = (0.75 m)(6 rad/s) 2 = 27 m/s 2

a D = a B + a D /B

30°] + [(1.5)(6/ 3) 2

[(a D )t

60°

:

60°] = [27 ] + [0.75 αAB ] + [(0.75)(6/ 3)2

]

0 + 18 = −27 sin 60° + (0.75sin 60°)α BD + 9sin 30° 18 + 27sin 60° − 9sin 30° = 56.7846 rad/s 0.75sin 60° aG = a B + aG /B

α BD =

= [27 ] + [0.375α BD ] + [(0.375)(6/ 3)2 = [5.7058 m/s

2

] + [4.5 m/s

2

]

]

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PROBLEM 16.129 (Continued)

Kinetics: m = 4 kg

IG =

1 1 mL2 = (4)(0.75)2 = 0.1875 kg ⋅ m 2 12 12

ΣM B = Σ( M B )eff :

  1  − (39.24 N)(0.375 m) + D cos 30° (0.75 m) = (0.1875)(56.7846) − (54)(0.375) 1 −  N ⋅ m 3    0.64952 D = (14.715 + 10.6471 − 8.5587) N ⋅ m

D = 25.87 N

60° D = 25.9 N

or

60° 

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PROBLEM 16.130 The motion of the uniform slender rod of length L = 0.5 m and mass m = 3 kg is guided by pins at A and B that slide freely in frictionless slots, circular and horizontal, cut into a vertical plate as shown. Knowing that at the instant shown the rod has an angular velocity of 3 rad/s counter-clockwise and θ = 30°, determine the reactions at Points A and B.

SOLUTION m = 3 kg

Mass and moment of inertia: I =

1 1 mL2 = (3 kg)(0.5 m) 2 = 0.0625 kg ⋅ m 2 12 12

ω = 3 rad/s

Kinematics:

v A = vA

v B = vB

Locate the instantaneous center C by drawing line AC perpendicular to v A and line BC perpendicular to v B . v A = ( L cos 30°)ω

= (0.5 m cos 30°)(3 rad/s) = 1.29904 m/s vB = ( L sin 30°)ω

= (0.5 m sin 30°)(3 rad/s) = 0.75 m/s

α =α

a A = aA aB =

vB2 R

+ ( aB ) y =

(0.75 m/s) 2 0.3 m

= 1.875 m/s 2

( aB ) y + ( aB ) y

a B = a A + a B /A = a A + (a B /A )t + (a B /A ) n

where

(a B /A )t = Lα

and

(a B /A ) n = Lω 2

30° = 0.5α

30°

60° = (0.5)(3) 2

60° = 4.5 m/s 2

60°

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PROBLEM 16.130 (Continued)

Equating the two expressions for a B gives + (aB ) y = a A + 0.5α

1.875 m/s 2

:

30° + 4.5 m/s 2

60°

−1.875 = a A + 0.5α cos 30° + 2.25 a A = −0.5α cos 30° − 4.125

aG = a A + (aG /A )t + (aG /A ) n = (0.5α cos 30° + 4.125)

L + α 2

= (0.5α cos 30° + 4.125)

+

= (0.21651α + 3.00)

Kinetics:

0.5 α 2

30° +

L 2 ω 2

30° +

0.5 2 (3) 2

60° 60°

+ (1.94856 − 0.125α )

maG = (3 kg) aG

= [0.64952α + 9.00]

+ [5.8457 − 0.375α ]

I α = 0.0625α

ΣM C = Σ( M C )eff : mg (3)(9.81)

L L L sin 30° = I α + ( maG ) x cos 30° − (maG ) y sin 30° 2 2 2

0.5 sin 30° = 0.0625α 2 0.5 cos 30° 2 0.5 − (5.8457 − 0.375α ) sin 30° 2 3.67875 = 0.25α + 1.21784 + (0.64952α + 9.00)

α = 9.8436 rad/s 2 maG = [(0.64952)(9.8436) + 9.00]

+ [5.8457 − (0.375)(9.8436)]

= [15.394 N]

[2.1544 N]

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PROBLEM 16.130 (Continued)

ΣF = ΣFeff = maG : A + B

+ mg = maG

: A − (3 kg)(9.81 m/s) = 2.1544 N : B = −15.894 N

A = 31.6 N  B = 15.89 N



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PROBLEM 16.131 At the instant shown, the 20 ft long, uniform 100-lb pole ABC has an angular velocity of 1 rad/s counterclockwise and Point C is sliding to the right. A 120-lb horizontal force P acts at B. Knowing the coefficient of kinetic friction between the pole and the ground is 0.3, determine at this instant (a) the acceleration of the center of gravity, (b) the normal force between the pole and the ground.

SOLUTION l = 20 ft, W = 100 lb, P = 120 lb, μk = 0.3

Data:

I =

1 1  100  2 2 ml 2 =   (20) = 103.52 slug ⋅ ft 12 12  32.2 

α =α

Kinematics:

aC = aC

aG = aC + (aG/C )t + (aC/G ) n = [aC

Kinetics: Sliding to the right:

ΣM G =

Σ( M G )eff : N

l ]+  α 2

 l 10° +  ω 2  2

80° 

(1)

F f = μk N

l l l  sin10° − F f cos10° + P  cos10° − h  = Iα 2 2 2  

l l l  sin10° + μk N cos10° = P  cos10° − h  2 2 2  103.519α − (10)(sin10° − 0.3cos10°) N = 120(10 cos10° − 6) Iα − N

(2)

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PROBLEM 16.131 (Continued)

ΣFy =

ml  ml  Σ( Fy )eff : N − mg = −  α  sin10° − ω 2 cos10° 2  2  ml 2  ml  ω cos10°  sin10°  α + N = mg − 2  2 

 100    100  (10)sin10° α + N = (100) −  (10)(1) 2 cos10°     32.2   32.2  

(3)

Solving Eqs. (2) and (3) simultaneously,

α = 3.8909 rad/s 2 ΣFx =

Σ( Fx )eff : P − F f = maC −

N = 48.433 lb

ml ml α cos10° + ω 2 sin10° 2 2

ml ml α cos10° − ω 2 sin10° 2 2 100 100    100  2 (10)(3.8909) cos10° −  aC = 120 − (0.3)(48.433) +    (10)(1) sin10° 32.2 32.2 32.2     maC = P − μk N +

aC = 70.542 ft/s 2

Using Eq. (1),

aG = 70.542 + [(10)(3.8909)

10°] + [(10)(1)2

80°]

(aG ) x = 70.542 − 38.909 cos10° + 10cos80° = 33.961 ft/s 2 (aG ) y = −38.909sin10° − 10sin 80° = −16.605 ft/s 2 aG = (33.961) 2 + (16.605) 2 = 37.803 ft/s 2 tan β =

(a)

Acceleration at Point G.

(b)

Normal force.

16.605 33.961

β = 26.055° aG = 37.8 ft/s 2

26.1° 

N = 48.4 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1616

PROBLEM 16.132 A driver starts his car with the door on the passenger’s side wide open (θ = 0). The 80-lb door has a centroidal radius of gyration k = 12.5 in., and its mass center is located at a distance r = 22 in. from its vertical axis of rotation. Knowing that the driver maintains a constant acceleration of 6 ft/s2, determine the angular velocity of the door as it slams shut (θ = 90°).

SOLUTION a = aA

Kinematics:

(aG/ A )t = r α

θ

Kinetics:

ΣM A = Σ( M A )eff : 0 = I α + (mr α )r − ma A ( r cos θ ) mk 2α + mr 2α = ma A r cos θ

α= Setting α = ω ddωθ , and using r =

22 12

aA r k +r2 2

cos θ

ft, k = 12.5 ft 12

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1617

PROBLEM 16.132 (Continued)

ω

( 1222 ft )a A dω cos θ = dθ  12.5 ft 2 + 22 ft 2  ( 12 ) ( 12 )  = 0.41234a A cos θ

ω d ω = 0.41234a A cos θ dθ



ωf 0

ωdω =

1 2 ω 2

ωf



π /2 0

(0.4124a A ) cos θ dθ

= 0.41234a A |sin θ | π0 /2

0

ω 2f = 0.82468a A Given data:

(1)

a A = 6 ft/s 2

ω 2f = 0.82468(6) = 4.948 rad 2 /s 2

ω f = 2.22 rad/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1618

PROBLEM 16.133 For the car of Problem 16.132, determine the smallest constant acceleration that the driver can maintain if the door is to close and latch, knowing that as the door hits the frame its angular velocity must be at least 2 rad/s for the latching mechanism to operate.

SOLUTION a = aA

Kinematics:

(aG/ A )t = r α

θ

Kinetics:

ΣMA = Σ( MA )eff :

0 = I α + ( mr α )r − ma A (r cos θ )

mk 2α + mr 2α = ma A r cos θ

α= Setting α = ω ddωθ , and using r =

22 12

aA r k +r2 2

cos θ

ft, k = 12.5 ft 12

ω

( 1222 ft ) a A dω = cos θ dθ  12.5 ft 2 + 22 ft 2  ( 12 ) ( 12 )  = 0.41234a A cos θ

ω d ω = 0.41234a A cos θ dθ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1619

PROBLEM 16.133 (Continued)



ωf 0

ωdω =

1 2 ω 2

ωf



π /2 0

(0.4124a A ) cos θ dθ

= 0.41234a A | sin θ | π0 /2

0

ω 2f = 0.82468aA Given data:

ω f = 2 rad/s

Eq. (1):

ω 2f = 0.82468aA

(1)

(2) 2 = 0.82468aA



a A = 4.85 ft/s 2 

a A = 4.85 ft/s 2



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1620

PROBLEM 16.134 Two 8-lb uniform bars are connected to form the linkage shown. Neglecting the effect of friction, determine the reaction at D immediately after the linkage is released from rest in the position shown.

SOLUTION Kinematics: Bar AC:

Rotation about C  15  a = ( BC )α =  ft  α  12  a = 1.25α

sin θ =

15 in. θ = 30° 30 in.

Bar BC:

a D/B = Lα

Must be zero since aD

α BD = 0 and aBD = a

Kinetics: Bar BD

ΣFy = Σ( Fy )eff : By − W = − ma 8 lb (1.25α ) 32.2 By = 8 − 0.3105α

By − 8 lb = −

(1)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1621

PROBLEM 16.134 (Continued)

ΣM B = Σ( M B )eff : D (2.165 ft) − W (0.625 ft) = − ma (0.625 ft) 8 lb (1.25α )(0.625 ft) 32.2 D = 2.309 − 0.08965α

D (2.165 ft) − (8 lb)(0.625 ft) = −

(2)

1 m( AC ) 2 12 1 8 lb = (2.5 ft) 2 12 32.2 = 0.1294 lb ⋅ ft ⋅ s 2

I =

Bar AC:

ΣM C = Σ( M C )eff : W (1.25 ft) + By (1.25 ft) = I α + m(1.25α )(1.25)

Substitute from Eq. (1) for By 8 (1.25) 2 α 32.2 10 + 10 − 0.3881α = 0.1294α + 0.3882α 20 = 0.9057α

8(1.25) + (8 − 0.3105α )(1.25) = (0.1294)α +

α = 22.08 rad/s 2 Eq. (2),

D = 2.309 − 0.08965α = 2.309 − 0.08965(22.08) = 2.309 − 1.979



D = 0.330 lb 

D = 0.330 lb



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1622

PROBLEM 16.135 The 6-kg rod BC connects a 10-kg disk centered at A to a 5-kg rod CD. The motion of the system is controlled by the couple M applied to disk A. Knowing that at the instant shown disk A has an angular velocity of 36 rad/s clockwise and no angular acceleration, determine (a) the couple M, (b) the components of the force exerted at C on rod BC.

SOLUTION Kinematics: Velocity analysis. Disk AB:

v B = AB ω AB

Rod BC:

vC = vC

ω AB = 36 rad/s

30° = (0.200)(36)

30° = 7.2 m/s

30°

30°

Since vC is parallel to v B , bar BC is in translation. vC = 7.2 m/s

Rod CD:

ωCD =

ω BC = 0

30°

vC 7.2 m/s = = 28.8 rad/s lCD 0.25 m

ωCD = 28.8 rad/s

Acceleration analysis: Disk AB:

α AB = 0 2 a B = α AB × rB /A − ω AB rB /A

= 0 − (36) 2 (0.2)

Rod BC:

60° = 259.2 m/s 2

60°

α BC = α BC 2 aC = a B + (aC /B )t − ω BC rC /B

aC = 259.2

Rod CD:

60° + 0.4α AB + 0

(1)

α CD = α CD 2 α C = (aC /D )t − ωCD rC /D = [0.25α CD

aC = [0.25α CD

30°] − [207.36

30°] − [(28.8) 2 (0.25) 60°]

60°]

(2)

Equate components of two expressions (1) and (2) for aC . :

−259.2 cos 60° = −0.25α CD cos 30° − 207.36 cos 60°

α CD = 119.719 rad/s 2 :

α CD = 119.719 rad/s 2

−259.2sin 60° + 0.4α BC = 0.25α CD sin 30° − 207.36sin 60°

α BC = 149.649 rad/s 2

α BC = 149.649 rad/s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1623

PROBLEM 16.135 (Continued)

Accelerations of the mass centers. a AB = a A = 0

Disk AB: Rod BC:

rP /B = (0.2 m)

Mass center at Point P.

2 a P = a B + α BC × rP /B − ωBC rP /B

= [259.2

Rod CD:

60°] + [0.2α BC ] + 0 = [259.2

60°] + [29.9298 ]

rQ /D = 0.125 m

Mass center at Point Q.

60°

2 aQ = α CD × rQ /D − ωCD rQ /D

= [0.125 α CD

30°] − [(28.8)2 (0.125)

= [14.964875

30°] − [103.60

m AB = 10 kg, mBC = 6 kg,

Masses:

60°]

60°]

mCD = 5 kg

Effective forces at mass centers. Disk AB:

m AB a A = 0

Rod BC:

mBC a P = 6a P = [1555.2 N

60°] + [179.58 N ]

Rod CD:

mCD a a = 5aQ = [74.82 N

30°] + [518 N

60°]

Moments of inertia: Disk AB:

I AB =

1 1 2 m AB rAB = (10)(0.2) 2 = 0.2 kg ⋅ m 2 2 2

Rod BC:

I BC =

1 1 2 mBC lBC = (6)(0.4) 2 = 0.08 kg ⋅ m 2 12 12

Rod CD:

I CD =

1 1 2 mCD lCD = (5)(0.25)2 = 0.0260417 kg ⋅ m 2 12 12

Effective couples at mass centers. Disk AB:

I AB α AB = 0

Rod BC:

I BC α BC = (0.08)(149.649)

Rod CD:

I CD α CD = (0.0260417)(119.719)

= 11.97192 N ⋅ m = 3.11768 N ⋅ m

Summary of effective forces and couples

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1624

PROBLEM 16.135 (Continued)

Kinetics Rod BC:

1 ΣM B = Σ( M B )eff : lBC C y − lBC mg = Σ( M B )eff 2 (0.4 m) C y − (0.2 m)(6 kg)(9.81 m/s 2 )

= 11.9719 N ⋅ m + (0.2 m)(179.58 N) −(0.2 m)(1555.2 N) sin 60° C y = −524.27 N

Rod CD:

ΣM D = Σ( M D )eff : Cx lCD cos 30° + (524.27 N)(0.125 m) − mCD g (0.0625 m) = Σ( M D )eff : (0.25 m)Cx cos 30° + 65.534 N ⋅ m

− (5 kg)(9.81 m/s 2 )(0.0625 m) = 3.11768 N ⋅ m + (0.125 m)(74.82 N) Cx = −230.93 N

Disk AB and rod BC:

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1625

PROBLEM 16.135 (Continued)

ΣM A = Σ( M A )eff : −M − (524.27)(0.4 + 0.2sin 30°) + (230.93)(0.2 cos 30°) − (6 kg)(9.81 m/s 2 )(0.2 + 0.2sin 30°) = 11.9719 + (179.58)(0.2 + 0.2sin 30°) − (1555.2 cos 30°)(0.2) − M − 262.135 + 40.0 − 17.658 = 11.9719 + 53.874 − 269.369 − M = 36.27 N ⋅ m

(a)

Couple applied to disk A.

(b)

Components of force exerted at C on rod BC.

M = 36.3 N ⋅ m C = 231 N



+ 524 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1626

PROBLEM 16.136 The 6-kg rod BC connects a 10-kg disk centered at A to a 5-kg rod CD. The motion of the system is controlled by the couple M applied to disk A. Knowing that at the instant shown disk A has an angular velocity of 36 rad/s clockwise and an angular acceleration of 150 rad/s2 counterclockwise, determine (a) the couple M, (b) the components of the force exerted at C on rod BC.

SOLUTION Kinematics:

ω AB = 36 rad/s

Velocity analysis.

Disk AB:

v B = ABω AB

Rod BC:

vC = vC

30° = (0.200)(36)

30° = 7.2 m/s

30°

30°

Since vC is parallel to v B , bar BC is in translation vC = 7.2 m/s

Rod CD:

ωCD =

30°

ω BC = 0

vC 7.2 m/s = = 28.8 rad/s lCD 0.25 m

ωCD = 28.8 rad/s

Acceleration analysis. Disk AB:

α AB = 0 2 a B = α AB × rB /A − ω AB rB /A

= [(150)(0.2) = [30 m/s 2

Rod BC:

30°] − [(36)2 (0.2) 30°] + [259.2 m/s 2

60°] 60°]

α BC = 150 rad/s 2 2 aC = a B + (aC /B )t − ω BC rC /B

aC = [30

Rod CD:

] + [259.2

60°] + [0.4α AB ] + 0

(1)

α CD = α CD 2 aC = (aC /D )t − ωCD rC /D = [0.25α CD

aC = [0.25α CD

30°] − [207.36

30°] − [(28.8) 2 (0.25) 60°]

60°]

(2)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1627

PROBLEM 16.136 (Continued)

Equate components of two expressions (1) and (2) for aC . : −30cos 30° − 259.2 cos 60° = −0.25α CD cos 30° − 207.36 cos 60°

αCD = 239.719 rad/s 2

α CD = 239.719 rad/s 2

: 30cos 30° − 259.2sin 60° + 0.4α BC = 0.25α CD sin 30° − 207.36sin 60°

α BC = 149.649 rad/s 2

α BC = 149.649 rad/s 2

Accelerations of the mass centers. a AB = a A = 0

Disk AB: Rod BC:

rP /B = (0.2 m)

Mass center at Point P.

2 a P = a B + α BC × rP /B − ωBC rP /B

Rod CD:

= [30

30°] + [259.2

60°] + [0.2α BC ] + 0

= [30

30°] + [259.2

60°] + [29.9298 ]

rQ / D = 0.125 m

Mass center at Point Q.

60°

2 aQ = α CD × rQ /D − ωCD rQ /D

Masses:

= [0.125 α CD

30°] − [(28.8)2 (0.125)

= [29.964875

30°] − [103.60

m AB = 10 kg, mBC = 6 kg,

60°]

60°]

mCD = 5 kg

Effective forces at mass centers. Disk AB:

m AB a A = 0

Rod BC:

mBC a P = 6a P = [180 N

Rod CD:

mCD a a = 5aQ = [149.82 N

30°] + [1555.2 N 30°] + [518 N

60°] + [179.58 N ] 60°]

Moments of inertia: Disk AB:

I AB =

1 1 2 m AB rAB = (10)(0.2) 2 = 0.2 kg ⋅ m 2 2 2

Rod BC:

I BC =

1 1 2 mBC lBC = (6)(0.4) 2 = 0.08 kg ⋅ m 2 12 12

Rod CD:

I CD =

1 1 2 mCD lCD = (5)(0.25)2 = 0.0260417 kg ⋅ m 2 12 12

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1628

PROBLEM 16.136 (Continued)

Effective couples at mass centers. Disk AB:

I AB α AB = (0.2)(150) = 30 N ⋅ m

Rod BC:

I BC α BC = (0.08)(149.649) = 11.97192 N ⋅ m

Rod CD:

I CD αCD = (0.0260417)(239.719) = 6.24268 N ⋅ m

Summary of effective forces and couples.

Kinetics Rod BC:

1 ΣM B = Σ( M B )eff : lBC C y − lBC mg = Σ( M B )eff 2 (0.4 m) C y − (0.2 m)(6 kg)(9.81 m/s 2 ) = (0.2 m)(180 N sin 30°) + 11.9719 N ⋅ m + (0.2 m)(179.58 N) −(0.2 m)(1555.2 N) sin 60° C y = −479.27 N

Rod CD:

ΣM D = Σ( M D )eff : C x lCD cos 30° + (479.27 N)(0.125 m) − mCD g (0.0625 m) = Σ( M D )eff : (0.25 m)C x cos 30° + 59.909 N ⋅ m − (5 kg)(9.81 m/s 2 )(0.0625 m) = 6.24268 N ⋅ m + (0.125 m)(149.82 N) C x = −147.22 N

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1629

PROBLEM 16.136 (Continued)

Disk AB and rod BC:

ΣM A = Σ( M A )eff : −M − (479.27)(0.4 + 0.2sin 30°) + (147.22)(0.2 cos 30°) − (6 kg)(9.81 m/s2 )(0.2 + 0.2sin 30°) = (180)(0.2 + 0.2sin 30°) + 11.9719 + (179.58)(0.2 + 0.2sin 30°) − (1555.2 cos 30°)(0.2) − M − 239.635 + 25.50 − 17.658 = 54 + 11.9719 + 53.874 − 269.369 − M = 82.27 N ⋅ m

(a)

Couple applied to disk A.

(b)

Components of force exerted at C on rod BC.

M = 82.3 N ⋅ m C = 147.2 N



+ 479 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1630

PROBLEM 16.137 In the engine system shown l = 250 mm and b = 100 mm. The connecting rod BD is assumed to be a 1.2-kg uniform slender rod and is attached to the 1.8-kg piston P. During a test of the system, crank AB is made to rotate with a constant angular velocity of 600 rpm clockwise with no force applied to the face of the piston. Determine the forces exerted on the connecting rod at B and D when θ = 180°. (Neglect the effect of the weight of the rod.)

SOLUTION Kinematics: Crank AB:  2π   = 62.832 rad/s  60 

ω AB = 600 rpm 

2 aB = ( AB)ω AB = (0.1 m)(62.832 rad/s 2 )

a B = 394.78 m/s 2 vB = ( AB)ω AB = (0.1 m)(62.832 rad/s) = 6.2832 m/s

Also: Connecting rod BD: Velocity

Instantaneous center at D.

ωBD =

vB 6.2832 m/s = = 25.133 rad/s 0.25 m BD

Acceleration:

a D = a B + a D/B = [aB

2 ] + [( BD )ωBD

]

a D = [394.78 m/s 2

] + [(0.25 m)(25.133 rad/s) 2

a D = [397.78 m/s 2

] + [157.92 m/s 2

aBD =

1 1 (a B + a D ) = (394.78 2 2

]

] = 236.86 m/s 2

+ 236.86

) = 315.82 m/s 2

Kinetics of piston

D = 426.35 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1631

PROBLEM 16.137 (Continued)

Force exerted on connecting rod at D is: D = 426.35

Kinetics of connecting rod:

(neglect weight)

ΣFx = Σ( Fx )eff : B − D = mBD aBD

B − 426.34 N = (1.2 kg)(315.82 m/s 2 ) B = 426.35 N + 378.48 N = 805.33 N

Forces exerted on connecting rod.

B = 805 N





D = 426 N



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PROBLEM 16.138 Solve Problem 16.137 when θ = 90°. PROBLEM 16.137 In the engine system shown l = 250 mm and b = 100 mm. The connecting rod BD is assumed to be a 1.2-kg uniform slender rod and is attached to the 1.8-kg piston P. During a test of the system, crank AB is made to rotate with a constant angular velocity of 600 rpm clockwise with no force applied to the face of the piston. Determine the forces exerted on the connecting rod at B and D when θ = 180°. (Neglect the effect of the weight of the rod.)

SOLUTION Geometry:

l = 0.25 m, b = 0.1 m, x = l 2 − b 2 = 0.2291 m rB/A = (0.1 m)j, rD/B = −(0.2291 m)i − (0.1 m)j

Kinematics:

ω AB = 600 rev/min = 62.832 rad/s

Velocity:

ω AB = −(62.832 rad/s)k vB = ω AB × rB /A = −(62.832k ) × (0.1i) = (6.2832 m/s)i vD = vB + ω BD × rD /B vD i = 6.2832i + ωBD k × (−0.2291i − 0.1j)

= 6.2832i + 0.1ωBD j − 0.2291ωBD i

Equating components gives vD = 6.2832 m/s, ωBD = 0.

Acceleration:

α AB = 0,

α BD = α BD k

a D = aD i

2 a B = α AB × rB /A − ω AB rB /A = 0 − (62.832) 2 (0.1j) = −(394.78 m/s 2 ) j 2 a D = a B + α BD × rB /D − ωBD rD /B

aD i = −394.78 j + α AB k × (−0.2291i − 0.1j) − 0 = −394.78 j − 0.2291α AB j + 0.1α AB i

Equate like components. j: i:

0 = −394.78 − 0.2291α AB α AB = −1723 rad/s 2 aD = (0.1)(−1723) = −172.3 m/s 2 α AB = −(1723 rad/s 2 )k

a D = −(172.3 m/s 2 )i 

 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1633

PROBLEM 16.138 (Continued)

Acceleration of mass center G of bar BD 1 (0.2291i + 0.1j) 2 = (0.11455 m)i + (0.05 m)j

rG /D =

2 aG = a D + α BD × rG /D − ω BD rG /D

= −172.3i + (−1723k ) × (0.11455i + 0.05 j) − 0 = −172.3i − 197.34 j + 86.15i = −(86.15 m/s 2 )i − (197.34 m/s 2 ) j

Force on bar BD at P in B. Use piston + bar BD as a free body

mD a D = (1.8)( −172.3i ) = −(310.14 N)i mBD aG = (1.2)(−86.15i − 197.34 j) I BDα BD

= −(103.38 N)i − (236.88 N) j 1 = (1.2)(0.25) 2 (−1723) = −10.769 N ⋅ m 12

ΣFx = Σ( Fx )eff : Bx = mD aD + ( mBD aG ) x Bx = −310.14 − 103.38 Bx = −413.52 N +ΣM B = Σ( M B )eff : − xN k = I BDα BD k + rD /B × (mD a D ) + rG /B × (mBD aG )

−0.2291 N k = −10.769k + (0.1)(−310.14)k + (−0.11455i − 0.05 j) × (−103.38i − 236.38 j) = −10.769k − 31.014k + 27.038k − 5.169k N = 86.923 N +ΣFy = Σ( Fy )eff : N + By = (mBD aG ) y 86.923 + By = −236.88 By = −323.80 N B = 413.522 + 323.802 = 525.2 N 323.80 tan β = β = 38.1° 413.52

B = 525 N

38.1° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1634

PROBLEM 16.138 (Continued)

Force exerted by bar BD as piston D. Use piston D as a free body

ΣF = ΣFeff



D + Nj = mD aD i D = mD aD i − Nj = −(310.14 N)i + (86.923 N)j

Force exerted by the piston on bar BD. By Newton’s third law, D′ = −D = (310.14 N)i − (86.923 N)j 

D′ = 322 N

15.7° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1635

PROBLEM 16.139 The 4-lb rod AB and the 6-lb rod BC are connected as shown to a disk that is made to rotate in a vertical plane at a constant angular velocity of 6 rad/s clockwise. For the position shown, determine the forces exerted at A and B on rod AB.

SOLUTION Kinematics: Velocity

ω AB = 0 vB = vA

= (0.25 ft)(6 rad/s) = 1.5 ft/s 2

ωBC =

vB 1.5 m/s 2 = 0.75 ft 0.75 ft

ω BC = 2 rad/s

Acceleration

a A = (0.25 ft)(6 rad/s) 2

a A = 9 ft/s 2 a B = (0.75 ft)(2 rad/s) 2 = 3 ft/s 2 aBC = (0.375 ft)(2 rad/s)2 aBC = 1.5 ft/s 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1636

PROBLEM 16.139 (Continued)

a AB =

1 1 (a A + aB ) = (9 + 3) 2 2

a AB = 6 ft/s 2 a A = aB + (0.5 ft)α AB 9 ft/s 2 = 3 ft/s 2 + (0.5 ft)α AB α AB = 12 rad/s 2 1 1 4 lb m AB ( AB)2 = (0.5 ft) 2 12 12 32.2 = 2.588 × 103 lb ⋅ ft ⋅ s 2

I AB =

Kinetics:

I AB

Rod BC: Since α BC = 0, a = 0 ΣM C = 0 yields Bx = 0

Rod AB:

ΣFx = Σ( Fx )eff : Ax = 0 ΣM A = Σ( M A )eff : By (0.5 ft) − WAB (0.25 ft) = I ABα AB − mAB a AB (0.25 ft)

0.5By − (4 lb)(0.25 ft) = (2.588 × 10−3 lb ⋅ ft ⋅ s 2 )(12 rad/s) 4 lb (6 ft/s2 )(0.25 ft) 32.2 0.5By − 1 = 0.03106 − 0.1863 −

0.5 By = 0.8447 By = 1.689 lb

B = 1.689 lb 

ΣFy = Σ( Fy )eff : Ay − WAB + By = − mAB a AB 4 lb (6 ft/s 2 ) 32.2 Ay = 1.565 lb

Ay − 4 lb + 1.689 lb = −

A = 1.565 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1637

PROBLEM 16.140 The 4-lb rod AB and the 6-lb rod BC are connected as shown to a disk that is made to rotate in a vertical plane. Knowing that at the instant shown the disk has an angular acceleration of 18 rad/s 2 clockwise and no angular velocity, determine the components of the forces exerted at A and B on rod AB.

SOLUTION Kinematics:

Velocity of all elements = C

Acceleration: a B = a A = (0.25 ft)(18 rad/s 2 ) = 4.5 ft/s 2

α BC =

aB 4.5 ft/s 2 = 0.75 ft 0.75 ft

α BC = 6 rad/s 2

aBC = (0.375 ft)(6 rad/s 2 ) = 2.25 ft/s 2 a AB = a A = a B = 4.5 ft/s 2 1 1 6 lb (0.75 ft) 2 mBC ( BC ) 2 = 12 12 32.2

Kinetics:

I BC =

Rod BC:

I BC = 8.734 × 10 −3 lb ⋅ ft ⋅ s 2

ΣM A = Σ( M A )eff : Bx (0.75 ft) = I BC α BC + mBC aBC (0.375 ft)

0.75 By = (8.734 × 10 −3 lb ⋅ ft ⋅ s 2 )(6 rad/s 2 )  6 lb  2 +  (2.25 ft/s )(0.375 ft) 32.2   0.75Bx = 0.0524 + 0.1572 Bx = 0.2795 lb (on AB) B x = 0.280 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1638

PROBLEM 16.140 (Continued)

Rod AB:

ΣFx = Σ( Fx )eff : Ax − Bx = m AB a AB  4 lb  2 Ax − 0.2795 lb =   (4.5 ft/s )  32.2  Ax − 0.2795 lb = 0.5590 lb Ax = 0.8385 lb

A x = 0.839 lb

ΣM A : B y = 2 lb ΣMB : A y = 2 lb



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1639

PROBLEM 16.141 Two rotating rods in the vertical plane are connected by a slider block P of negligible mass. The rod attached at A has a weight of 1.6 lb and a length of 8 in. Rod BP weighs 2 lb and is 10 in. long and the friction between block P and AE is negligible. The motion of the system is controlled by a couple M applied to rod BP. Knowing that rod BP has a constant angular velocity of 20 rad/s clockwise, determine (a) the couple M, (b) the components of the force exerted on AE by block P.

SOLUTION Unit vectors: Geometry:

Kinematics: Velocity analysis. Rod BP:

i =1

, j =1 , k =1 .

 10  rP /A = −  sin 30° ft  j 12    10   10  rP /B = −  cos 30° ft  i −  sin 30° ft  j 12 12     8   ft  j rP /A = −   12  ω BP = −(20 rad/s) k ,

α BP = 0

v P = ω BP × rP /B 10  10  = −20k ×  − cos 30°i − sin 30° j  12  12  = −(8.3333 ft/s)i + (14.4338 ft/s)j

Rod AE: Use a frame of reference rotating with angular velocity ω AE = ω AE k. The collar P slides on the rod with relative velocity v P /A = uj v P = v P′ + v P /AE = ω AE × rP /A + u j  10  = ω AE k ×  − sin 30° j  + u j = 0.41667 ω AE i + u j  12 

Equate the two expressions for vP and resolve into components. i: − 8.3333 = 0.41667 ω AE

ω AE = −20 rad/s

j: 14.4388 = u

u = 14.4338 ft/s

Acceleration analysis. Rod BP:

2 a P = α BP × rP /B − ω BP rP /B

10  10  = 0 − (20)2  − cos 30°i − sin 30° j  12  12  = (288.68 ft/s 2 ) i + (166.67 ft/s 2 ) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1640

PROBLEM 16.141 (Continued)

α AE = α AE k ,

Rod AE:

a P /AE = u j

a P = a P′ + a P /AE + 2ω × v P /AE 2 a P′ = α AB × rP /A − ω AE rP /A

where

 10   10  = α AE k ×  − sin 30° j  − (20)2  − sin 30° j  12 12    

= 0.41667α AE i + (166.67 ft/s 2 ) j

and

2ω AE × v P /AE = (2)(−20k ) × (14.4338 j) = 577.35 ft/s 2 i

Equate the two expressions for a P and resolve into components. i: 288.68 = 0.41667α AE + 577.35

α AE = −692.8 rad/s 2 ω BP = −(20 rad/s)k , ω AE = −(20 rad/s)k

Summary:

α BP = 0,

α AE = −(692.8 rad/s 2 )k

Masses and moments of inertia. WAE 1.6 lb = = 0.049689 lb ⋅ s 2 /ft g 32.2 ft/s 2 W 2 lb = BP = = 0.062112 lb ⋅ s 2 /ft g 32.2 ft/s 2

m AE = mBP

2

I AE =

1 1  8  2 ft  = 1.8403 × 10−3 lb ⋅ s 2 ⋅ ft = (0.049689 lb ⋅ s 2 /ft)  m AE l AE 12 12 12  

I BP =

1 1  10  2 ft  = 3.5944 × 10−3 lb ⋅ s 2 ⋅ ft = (0.062112 lb ⋅ s 2 /ft)  mBP lBP 12 12  12 

2

Mass centers: Let Point G be the mass center of rod AE and Point H be that of rod BP.  4  rG /A = −  ft  j  12   10   10  rH /B = −  cos 30° ft  i −  sin 30° ft  j  12   12 

Acceleration of mass centers. 2 aG = α AE × rG /A − ω AE rG /A

= (−692.8 k ) × (−0.33333j) − (20)2 (−0.33333 j) = −(230.93 ft/s 2 ) i + (133.33 ft/s 2 ) j 2 a H = α BP × rH /B − ωBP rH /B

= 0 − (20) 2 (−0.72169i − 0.41667 j) = (288.68 ft/s 2 )i + (166.67 ft/s 2 ) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1641

PROBLEM 16.141 (Continued)

Effective forces at mass centers. Rod AE:

m AE aG = (0.049689)(−230.93i + 133.33j) = −(11.475 lb)i + (6.625 lb) j

Rod BP:

mBP a H = (0.062112)(288.68i + 166.67 j) = (17.930 lb)i + (10.352 lb) j

Effective couples at mass centers. Rod AE:

I AE α AE = (1.8403 × 10−3 )(−692.8 k ) = −(1.2750 lb ⋅ ft) k

Rod BP:

I BP α BP = 0

Kinetics. Rod AE: ΣM A = Σ(M A )eff : rP /A × ( − Pi ) = rG /A × (m AE aG ) + I AE α AE  5   4   − 12 j  × (− Pi ) =  − 12 j  × (−11.475i + 16.625 j) − 1.2750k     5 − Pk = −3.8249k − 1.2750k P = 12.240 lb 12

Rod BP: ΣM B = Σ(M B )eff: rP/B × Pi + rH /B × (−WBP j) + M k = rH /B × (mBP a H ) + I BP α BP 5  5cos 30°  Pk +  ft  (2 lb)k + M k 12  12  = (−0.36084i − 0.20833j) × (17.930i + 10.352 j) + 0

5.1k + 0.72169k + M k = −3.7354k + 3.7354k M = −5.82 lb ⋅ ft

(a)

Couple M.

M = 5.82 lb ⋅ ft



(b)

Force exerted on AE by block P.

P = 12.24 lb



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1642

PROBLEM 16.142 Two rotating rods in the vertical plane are connected by a slider block P of negligible mass. The rod attached at A has a mass of 0.8 kg and a length of 160 mm. Rod BP has a mass of 1 kg and is 200 mm long and the friction between block P and AE is negligible. The motion of the system is controlled by a couple M applied to bar AE. Knowing that at the instant shown rod BP has an angular velocity of 20 rad/s clockwise and an angular acceleration of 80 rad/s2 clockwise, determine (a) the couple M, (b) the components of the force exerted on AE by block P.

SOLUTION rP /A = (0.200 m) sin 30° = 0.100 m

Geometry:

rP /B = 0.200 m

30°

rE /A = 0.160 m

Kinematics:

ω BP = 20 rad/s α BP = 80 rad/s 2

Velocity analysis. v P = ω BP × rP /B = (20 rad/s) ⋅ (0.200 m)

Rod BP: Rod AE:

60° = 4 m/s

60°

Use a frame of reference rotating with angular velocity ω AE = ω AE . The collar slides on the rod with relative velocity v P /AE = u . v P = v P′ + v P /AE = ω AE × rP /A + u = 0.100 ω AE

+u

Equate the two expressions for v P using a triangle construction for vector addition. −0.100ω AE = 4 cos 60°

ω AE = −20 rad/s ω AE = 20 rad/s u = 4sin 60° = 3.461 m/s

Acceleration analysis. Rod BP:

v P /AE = 3.4641 m/s

α BP = α BP 2 a P = α BP × rP /B − ωBP rP /B

= [(80 rad/s 2 )(0.200 m) = [16 m/s 2

60°] + [(20 rad/s)2 (0.200 m)

60°] + [80 m/s 2

30°]

30°]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1643

PROBLEM 16.142 (Continued)

α AE = α AE ,

Rod AE:

a P /AE = u

α P = a P′ + a P /AE + 2ω AE × v P /AE 2 a P′ = α AE × rP /A − ω AE rP /A

where

= [(0.100 m)α AE = [0.100α AE

] + [(20 rad/s 2 )(0.100 m) ] ] + [40 m/s 2 ]

2ω AE × v P /AE = [(2)(20 rad/s)(3.4641 m/s)

and

] = 138.564 m/s 2

Equate the two expressions for a P and resolve into components. : −(16 m/s 2 ) cos 60° + (80 m/s 2 ) cos 30° = 0.100α AE + 138.564 m/s 2

α AE = −772.82 rad/s 2

α AE = 772.82 rad/s 2

Masses, weights, and moments of inertia. m AE = 0.8 kg WAE = mAE g = (0.8 kg)(9.81 m/s 2 ) = 7.848 N mBP = 1.0 kg WBP = mBP g = (1.0 kg)(9.81 m/s2 ) = 9.81 N 1 1 2 m AE l AE = (0.8 kg)(0.16 m)2 = 1.70667 × 10−3 kg ⋅ m 2 12 12 1 1 2 = mBP lBP = (1.0 kg)(0.20 m)2 = 3.3333 × 10−3 kg ⋅ m 2 12 12

I AE = I BP

Mass centers: Let Point G be the mass center of rod AE and Point H be that of rod BP. rG /A = 0.08 m

rH /B = 0.10 m

30°

Accelerations of mass centers. 2 aG = α AE × rG /A − ω AE rG /A

= (772.82 rad/s 2 )(0.08 m) = [61.826 m/s 2

+ (20 rad/s 2 )(0.08 m)

] + [32 m/s 2 ]

2 a H = α BP × rH /A − ω BP rH /B

= [(80 rad/s 2 )(0.10 m) = [8 m/s 2

60°] + [(20 rad/s) 2 (0.10 m)

60°] + [40 m/s 2

30°]

30°]

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1644

PROBLEM 16.142 (Continued)

Effective forces at mass centers. Rod AE:

m AE aG = [49.460 N

Rod BP:

mBP a H = [8 N

] + [25.6 N ]

60°] + [40 N

30°]

Effective couples at mass centers. Rod AE:

I AE α AE = 1.3189 N ⋅ m

Rod BP:

I BP α BP = 0.2667 N ⋅ m

Kinetics: Rod AE:

ΣM A = Σ( M A )eff : − 0.10 P = −(0.08)(49.460) − 1.3189

Rod BP:

ΣM B = Σ( M B )eff : (52.757 N)(0.1 m) + (9.81 N)(0.086603 m) + M

P = 52.757 N = −(8 N)(0.1 m) − 0.2667 N ⋅ m M = 7.1919 N ⋅ m

(a)

Couple M.

(b)

Force exerted on AE by force by block P.

M = 7.19 N ⋅ m



P = 52.8 N



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1645

PROBLEM 16.143* Two disks, each of mass m and radius r are connected as shown by a continuous chain belt of negligible mass. If a pin at Point C of the chain belt is suddenly removed, determine (a) the angular acceleration of each disk, (b) the tension in the left-hand portion of the belt, (c) the acceleration of the center of disk B.

SOLUTION Kinematics:

Assume α A

and α B

ω A = ωB = 0 a D = rα A a E = aD = rα A aB = aE + aB/E = (rα A + rα B ) aB = r (α A + α B )

Kinetics: Disk A:

Disk B:

ΣM A = Σ( M A )eff : Tr = I α A Tr =

1 2 mr α A 2

αA =

2T mr

(1)

ΣM B = Σ( M B )eff : Tr = I α B Tr =

1 2 mr α B 2

αB =

2T mr

(2)

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PROBLEM 16.143* (Continued)

From Eqs. (1) and (2) we note that α A = α B ΣM E = Σ( M E )eff : Wr = I α B + (maB )r Wr =

1 2 mr α B + mr (α A + α B )r 2 5 2

α A = α B : Wr = mr 2α A

Substitute for α A into (1):

2 g 2T = 5 r mr

αA =

2g 5 r



αB =

2g 5 r



1 T = mg  5

aB = r (α A + α B ) = r (2α A ) 2 g = 2r   5 r 

aB =

4 g 5



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PROBLEM 16.144* A uniform slender bar AB of mass m is suspended as shown from a uniform disk of the same mass m. Neglecting the effect of friction, determine the accelerations of Points A and B immediately after a horizontal force P has been applied at B.

SOLUTION

ω=0

Kinematics: Cylinder:

Rolling without sliding

(aC ) x = 0

a A = (aC ) x + a A/C = 0 + rα A a A = rα A

Rod AB:

αA =

aA r

aA =

L α AB − a AB 2

ΣM C = Σ( M C )eff : Ax r = ma A r + I α A

Kinetics:

Ax r = ma A r +

Cylinder:

1 2  aA  mr   2  r 

3 ma A 2 3 L  Ax = m  α AB − a AB  2 2  Ax =

(1)

Rod AB: ΣM A = Σ( M A )eff :



ΣFx = Σ( Fx )eff :

PL = ma AB

L + I α AB 2

PL = ma AB

L m 2 + L α AB 2 12

(2)

P + Ax = ma AB 

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PROBLEM 16.144* (Continued)

Substitute from (1): P+

Multiply by (4) + (2):

3 L  m  α AB − a AB  = ma AB 2 2  5 3 P = ma AB − mLα AB 2 4 1 5L 1 PL = ma AB − mL2α AB 9 18 12

L : 9

(4)

10 7 1 5  PL =  +  mLa AB = mLa AB 9 9  2 18  a AB =

(5)

(3)

10 P 7 m

(5)

5  10 P  3 m  − mLα AB 2  7 m 4 25 3 P= P − mLα AB 7 4 P=

(3)



:

18 3 P = mLα AB 7 4

24 P 7 mL

L α AB − a AB 2 L  24 P  10 P =  − 2  7 mL  7 m

aA =

 12 10  P aA =  −  7 m  7

:

α AB =

aA =

2P 7m



L α AB + a AB 2 L  24 P  10 P =  + 2  7 mL  7 m

aB =

 12 10  P aB =  +  7 m  7

aB =

22 P 7 m



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PROBLEM 16.145 A uniform rod AB, of mass 15 kg and length 1 m, is attached to the 20-kg cart C. Neglecting friction, determine immediately after the system has been released from rest, (a) the acceleration of the cart, (b) the angular acceleration of the rod.

SOLUTION Kinematics: We resolve the acceleration of G into the acceleration of the cart and the acceleration of G relative to A: aR = aG = a A + aG/A aR = aC + aG/A aG/A =

where

1 Lα 2

Kinetics: Cart and rod

mR = 15 kg mC = 20 kg L =1 m 1 I R = mR L2 12 1 aG /A = (1)α = 0.5α 2 ΣFx = Σ( Fx )eff : ( mC + mR ) g sin 25° = ( mC + mR )aC − mR aG /A cos 25°

 mg g sin 25° = aC −   mC + mR

 L    α  cos 25°  2 

 15  aC = (9.81)sin 25° +   (0.5cos 25°)α  20 + 15  aC = (9.81)sin 25° + 0.19421α

(1)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1650

PROBLEM 16.145 (Continued)

Rod

ΣM A = Σ( M A )eff : 0 = I α + (mR aG/R )

L L − (mR aC cos 25°) 2 2

1 15(1) 2α + 15(0.5α )(0.5) − (15aC cos 25°)(0.5) = 0 12 1.25α + 3.75α − (7.5cos 25°)aC = 0

α = (1.50 cos 25°) aC (a)

(2)

Acceleration of the cart. Substitute for α from (2) into (1): aC = (9.81) sin 25° + 0.19421(1.5cos 25°)aC 9.81sin 25° aC = 1 − 0.19421(1.5cos 25°) = 5.6331 m/s 2

(b)

aC = 5.63 m/s 2

25° 

Angular acceleration. From (2):

α = (1.50 cos 25°)(5.6331) = 7.6580 rad/s 2

α = 7.66 rad/s 2



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PROBLEM 16.146* The 5-kg slender rod AB is pin-connected to an 8-kg uniform disk as shown. Immediately after the system is released from rest, determine the acceleration of (a) Point A, (b) Point B.

SOLUTION

ω=0

Kinematics:

a B = rα C a = a B + aG/B L   a =  rα C + α AB  2  

Kinetics: Disk

ΣM C = Σ( M C )eff : Br = I α C

1 mC r 2α C 2 1 B = mC rα C 2

Br =

Rod AB:

ΣM G = Σ( M G )eff : B

L = I α AB 2

1 L 1 2  mC rα C  = m AB L α AB 2  2 12 1 m AB L ⋅ α AB αC = 3 mC r

(1)

ΣFy = Σ( Fy )eff : mAB g − B = m AB a m AB g −

1 L   mC rα C = m AB  rα C + α AB  2 2   g=

 1 mC  L + 1 rα C α AB +  2  2 m AB 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1652

PROBLEM 16.146* (Continued)

 1 mC  r  1 m AB L  g 1 = α AB +  + 1 ⋅  ⋅  α AB L 2  2 m AB  L  3 mC r  m AB  g  1 1 1 m AB  1 = + +  α AB =  2 +  α AB 3 L  2 6 3 mC  mC  3g 1 α AB = L 2 + mAB

(

mC

(2)

)

m AB = 5 kg, mC = 8 kg, r = 0.1 m, L = 0.25 m 3(9.81)m/s 2 1 ⋅ = 44.846 rad/s 2 kg 0.25 m 2 + 58 kg

Eq. (1):

α AB =

Eq. (2):

αC =

(b)

aB = rα C

Acceleration of B.

1 5 kg 0.25 m ⋅ ⋅ (44.846 rad/s 2 ) = 23.357 rad/s 2 3 8 kg 0.1 m

= (0.1 m)(23.357 rad/s 2 ) = 2.336 m/s 2

(a)

Acceleration of A.

a B = 2.34 m/s 2 

a A = a B + a B/A a A = a B + [ Lα AB ]

a A = 2.336 m/s 2 + (0.25 m)(44.846 rad/s 2 ) a A = 2.336 + 11.212

a A = 13.55 m/s 2 

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PROBLEM 16.147* The 6-lb cylinder B and the 4-lb wedge A are held at rest in the position shown by cord C. Assuming that the cylinder rolls without sliding on the wedge and neglecting friction between the wedge and the ground, determine, immediately after cord C has been cut, (a) the acceleration of the wedge, (b) the angular acceleration of the cylinder.

SOLUTION Kinematics: We resolve a B into a A and a component parallel to the incline a B = a A + a B/A

Where aB/A = rα , since the cylinder rolls on wedge A. aB/A = (0.25 ft)α

Kinetics: Cylinder and wedge

ΣFx = Σ( Fx )eff : 0 = mA a A + mB a A − mB aB/A cos 20° (4 + 6)lb 6 lb  3  aA − ft α cos 20° g g  12  an = (0.15cos 20°)α 0=

(1)

Cylinder

1 W 2 1 6 lb (0.25 ft) 2 r = 2 g 2 g 3 I = 16 g I =

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PROBLEM 16.147* (Continued)

(b)

Angular acceleration of the cylinder. ΣM E = Σ( M E )eff : (6 lb)sin 20°(0.25 ft) = I α + mB aB/A (0.25 ft) − mB a A cos 20°(0.25 ft) 3 6 lb (0.25α )(0.25) α+ 16(32.2) 32.2 6 lb a A cos 20°(0.25) − 32.2 0.51303 = 0.00582α + 0.01165α − 0.04378a A

1.5sin 20° =

Substitute from (1):

0.51303 = 0.01747α − 0.04378(0.15 cos 20°)α 0.51303 = (0.01747 − 0.00617)α

α = 45.41 rad/s 2 (a)

α = 45.4 rad/s 2



Acceleration of the wedge. Eq. (1):

a A = (0.15cos 20°)α = (0.15cos 20°)(45.41) a A = 6.401 ft/s 2

a A = 6.40 ft/s 2



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PROBLEM 16.148* The 6-lb cylinder B and the 4-lb wedge A are held at rest in the position shown by cord C. Assuming that the cylinder rolls without sliding on the wedge and neglecting friction between the wedge and the ground, determine, immediately after cord C has been cut, (a) the acceleration of the wedge, (b) the angular acceleration of the cylinder.

SOLUTION Kinematics: We resolve a B into a A and a horizontal component a B/A a B = a A + a B/A

Where aB/A = rα , since the cylinder B rolls on wedge A. aB/A = (0.25 ft)α

Kinetics: Cylinder and wedge:

ΣFx = Σ( Fx )eff : (WA + WB )sin 20° = (m A + mB )a A − mB aB/A cos 20°

 10  6 (10 lb)sin 20° =   a A −   (0.25α ) cos 20°  g  g 6 a A = g sin 20° + (0.25) cos 20°α 10 a A = g sin 20° + 0.15 cos 20°α

Cylinder:

(1)

ΣM E = Σ( M E )eff : 0 = I α + (mB aB/A )(0.25 ft) − (mB a A cos 20°)(0.25 ft)

1 6 lb 6 lb (0.25 ft) 2 α + (0.25α )(0.25) 2 g g 6 lb − a A cos 20°(0.25) g 1 0 = [0.1875α + 0.325α − 1.4095a A ] g 0 = 0.5625α − 1.4095a A α = 2.506a A 0=

(2)

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PROBLEM 16.148* (Continued)

(a)

Acceleration of the wedge. Substitute for α from (2) into (1): a A = g sin 20° + 0.15cos 20°(2.506a A ) a A = 11.013 + 0.3532a A (1 − 0.3532)a A = 11.013 a A = 17.027 ft/s 2

(b)

a A = 17.03 ft/s 2

20° 

Angular acceleration of the cylinder. Eq. (2):

α = 2.506a A = 2.506(17.027)

α = 42.7 rad/s 2

α = 42.7 rad/s 2



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PROBLEM 16.149* Each of the 3-kg bars AB and BC is of length L = 500 mm. A horizontal force P of magnitude 20 N is applied to bar BC as shown. Knowing that b = L (P is applied at C), determine the angular acceleration each bar.

SOLUTION Kinematics: Assume α AB

α BC

and ω AB = ωBC = 0

Kinetics: Bar BC

ΣM B = Σ( M B )eff : PL = I α BC + (maBC )

L 2

m 2 L  L L α BC + m  Lα AB + α BC  12 2  2 1 1 P = mLα AB + mLα BC 2 3 =

(1)

ΣFx = Σ( Fx )eff : P − Bx = maBC 1   P − Bx = m  Lα AB + Lα BC  2  

(2)

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PROBLEM 16.149* (Continued)

Bar AB:

L 2 L  L + m  α AB  2 2

ΣM A = Σ( M A )eff : Bx L = I α AB + (ma AB ) m 2 L α AB 12 1 Bx = mLα AB 3 =

(3)

Add (2) and (3):

P=

4 1 mLα AB + mLα BC 3 2

Subtract (1) from (4)

0=

5 1 mLα AB + mLα BC 6 6

(4)

α BC = −5α AB Substitute for α BC in (1):

P=

1 1 7 mLα AB + mL(−5α AB ) = − mLα AB 2 3 6

α AB = − Eq. (5) Data:

(5)

6 P 7 mL  6 P    7 mL 

α BC = −5  −

(6)

α BC =

30 P 7 mL

(7)

L = 500 mm = 0.5 m, m = 3 kg, P = 20 N

α AB = −

6 P 6 20 N =− 7 mL 7 (3 kg)(0.5 m)

= −11.249 rad/s 2

α BC =

α AB = 11.43 rad/s 2



α BC = 57.1 rad/s 2



30 P 30 20 N = 7 mL 7 (3 kg)(0.5 m)

= 57.143 rad/s 2

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PROBLEM 16.150* Each of the 3-kg bars AB and BC is of length L = 500 mm. A horizontal force P of magnitude 20 N is applied to bar BC. For the position shown, determine (a) the distance b for which the bars move as if they formed a single rigid body, (b) the corresponding angular acceleration of the bars.

SOLUTION Kinematics: We choose α = α AB = α BC

Kinetics: Bars AB and BC (acting as rigid body)

m ABC = 2m 1 (2m)(2 L)2 12 2 I = mL2 3 I =

ΣM A = Σ( M A )eff : P( L + b) = I ABC α + mABC aB L 2 2 mL α + (2m)( Lα ) L 3 8 P( L + b) = mL2α 3 P ( L + b) =

(1)

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PROBLEM 16.150* (Continued)

Bar BC:

ΣM B = Σ( M B )eff : Pb = I BC α + ( maBC )

L 2

m 2 3 L L α + m  Lα  12 2  2 5 Pb = mL2α 6 6 Pb α= 5 mL2 =

Substitute for α into (1):

(2)

8  6 Pb  P( L + b) = mL2  2  3  5 mL  16 PL + Pb = Pb 5 11  16  L =  − 1 b = b 5  5  5 b= L 11 6 P 5  L 5 mL2  11 

Eq. (2)

α=

Data:

L = 500 mm = 0.5 m

α=

6 P 11 mL

m = 3 kg P = 20 N

(a)

b=

5 5 L = (0.5) = 0.22727 m 11 11

(b)

α=

6 P 6 (20 N) = = 7.2727 rad/s 2  11 mL 11 (3 kg)(0.5 m)

b = 227 mm  α = 7.27 rad/s 2



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PROBLEM 16.151* (a) Determine the magnitude and the location of the maximum bending moment in the rod of Problem 16.78. (b) Show that the answer to Part a is independent of the weight of the rod.

SOLUTION a=

Rod AB:

L α 2

ΣM A = Σ( M A )eff : PL = (ma )

L + Iα 2

 1 L 1 =  m α  + m L2α  2  2 12 α=

3P mL

(1)

ΣFx = Σ( Fx )eff : Ax − P = −ma L L  3P  P =− Ax = P − m α = P − m   2 2  mL  2

Ax =

1 P 2

Portion AJ of Rod: External forces: Ax , WAJ , axial force FJ, shear VJ , and bending moment MJ. Effective forces: Since acceleration at any point is proportional to distance from A, effective forces are linearly distributed. Since mass per unit length is m/L, at Point J we find m m  L  aJ = L ( xα )  

Using (1):

m m  3P  aJ =   L L  mL  m 3Px aJ = 2 L L 1  3Px  2 x  ΣM J = Σ( M J )eff : M J − Ax x = −  2   2  L  3  MJ =

For M max :

1 1 P 3 Px − x 2 2 L2

(2)

dM J P 3 P 2 = − x =0 2 2 L2 dx x=

L

(3)

3

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PROBLEM 16.151* (Continued)

Substituting into (2) 3

( M J ) max ( M J ) max

1 PL 1 P  L  1 PL  2  = −  =   2  2 3 2 L  3 2 3 3 PL = 3 3

(4)

Note: Eqs. (3) and (4) are independent of W. Data:

L = 36 in., P = 1.5 lb

Eq. (3):

x=

Eq. (4):

( M J ) max =

L 3

=

36 in. 3

= 20.78 in.

(1.5 lb)(36 in.)

3 3 = 10.392 lb ⋅ in. M max = 10.39 lb ⋅ in. located 20.8 in. below A 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1663

PROBLEM 16.152* Draw the shear and bending-moment diagrams for the beam of Problem 16.84 immediately after the cable at B breaks.

SOLUTION From answers to Problem 16.84: aB =

3 g 2

α=

aB L

A=

1 mg 4

We now find =

a J = xα =

3g 2L

3g x 2L

Portion AJ of rod: External forces: Reaction A, distributed load per unit length mg/L, shear VJ , bending moment M J . Effective forces: Since a is proportional to x, the effective forces are linearly distributed. The effective force per unit length at J is: m m 3g 3mg aJ = ⋅ x= 2 x L L 2L 2L

ΣFy = Σ( Fy )eff :

mg mg 1  3mg  x− + VJ =  2 x  x L 4 2  2L  VJ =

 mg ΣM J = Σ( M J )eff :   L

mg mg 3 mg 2 − x+ x L 4 4 L2

1  3mg  x mg x − x + MJ =  2 2 4 2  2L  MJ =

Find Vmin :

dVJ mg 3 mg =− + x = 0; 2 L2 dx L

x=

 x x x   3

mg 1 mg 2 1 mg 3 x− x + x 4 2 L 4 L2 2 L 3 2

Vmin =

mg mg  2  3 mg  2  − L + L ; 4 L  3  4 L2  3 

Vmin = −

mg 12

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PROBLEM 16.152* (Continued)

Find M max where

VJ = 0: VJ =

mg mg 3 mg 2 − x+ x =0 4 L 4 L

3x 2 − 4 Lx + L2 = 0 (3x − L)( x − L) = 0

x=

L and x = L 3 2

3

M max =

mg  L  1 mg  L  1  mg  L  mgL − +  =       4 3 2 L 3 4  L  3  27

M min =

mg 1 mg 2 1 mg 3 L− L + L =0 4 2 L 4 L

M max =

mgL L at from A  27 3

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1665

PROBLEM 16.153 A cyclist is riding a bicycle at a speed of 20 mph on a horizontal road. The distance between the axles is 42 in., and the mass center of the cyclist and the bicycle is located 26 in. behind the front axle and 40 in. above the ground. If the cyclist applies the brakes only on the front wheel, determine the shortest distance in which he can stop without being thrown over the front wheel.

SOLUTION v0 = 20 mph

When cyclist is about to be thrown over the front wheel, N A = 0

ΣM B = Σ( M B )eff : mg (26 in.) = ma (40in.) a=

26 26 g= (32.2 ft/s 2 ) = 20.93 ft/s 2 40 40

Uniformly accelerated motion: v0 = 20 mph = 29.333 ft/s v − v02 = 2as : 0 − (29.333 ft/s)2 = 2( −20.93 ft/s 2 ) s 2

s = 20.555 ft

s = 20.6 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1666

PROBLEM 16.154 The forklift truck shown weighs 2250 lb and is used to lift a crate of weight W = 2500 lb. The truck is moving to the left at a speed of 10 ft/s when the brakes are applied on all four wheels. Knowing that the coefficient of static friction between the crate and the fork lift is 0.30, determine the smallest distance in which the truck can be brought to a stop if the crate is not to slide and if the truck is not to tip forward.

SOLUTION

Assume crate does not slide and that tipping impends about A. ( B = 0) ΣM A = Σ( M A )eff :

  a a (2500 lb)(3 ft) − (2250 lb)(4 ft) = −  2500  (4 ft) −  2250  (3 ft) g g   a 7500 − 9000 = −(10, 000 + 6750) g a a = 2.884 ft/s 2 = 0.09; a = 0.09(32.2 ft/s 2 ) g

Uniformly accelerated motion v 2 = v02 + 2ax; 0 = (10 ft/s) 2 − 2(2.884 ft/s 2 ) x

Check whether crate slides

x = 17.34 ft 

N =W F = ma =

μreq =

W a g

F a 2.884 ft/s 2 = = N g 32.2 ft/s 2

μreq = 0.09 < 0.30. The crate does not slide.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1667

PROBLEM 16.155 A 5-kg uniform disk is attached to the 3-kg uniform rod BC by means of a frictionless pin AB. An elastic cord is wound around the edge of the disk and is attached to a ring at E. Both ring E and rod BC can rotate freely about the vertical shaft. Knowing that the system is released from rest when the tension in the elastic cord is 15 N, determine (a) the angular acceleration of the disk, (b) the acceleration of the center of the disk.

SOLUTION (a)

Angular acceleration of the disk. Disk:

I disk =

1 1 mdisk r 2 = (5 kg)(0.075 m) 2 2 2

I disk = 14.06 × 10−3 kg ⋅ m 2 ΣM A = Σ( M A )eff : (15 N)(0.075 N) = I disk α disk 1.125 N ⋅ m = (14.06 × 103 kg ⋅ m 2 )α disk

α disk = 80.0 rad/s 2 (b)

α disk = 80.0 rad/s 2



Acceleration of center of disk. Entire assembly I BC =

1 1 mBC ( BC ) 2 = (3 kg)(0.15 m) 2 = 5.625 × 10−3 kg ⋅ m 2 12 12

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1668

PROBLEM 16.155 (Continued)

Assume

α BC is

a A = (0.15 m)α BC a = (0.075 m)α BC

ΣM C = Σ( M C )eff : 0 = I diskα disk − mdisk a A (0.15 m) − mBC a (0.075 m) − I BC α BC 0 = (14.06 × 10−3 kg ⋅ m 2 )(80 rad/s 2 ) − (5 kg)(0.15 m) 2 α BC − (3 kg)(0.075 m)2 α BC − (5.625 × 103 kg ⋅ m 2 )α BC 0 = 1.125 − 0.1125α BC − 16.875 × 10−3 α BC − 5.625 × 10−3 α BC 0 = 1.125 − 0.135α BC

α BC = +8.333 rad/s 2

α BC = 8.33 rad/s 2

a A = ( AC )α BC = (0.15 m)(8.333 rad/s 2 ) a A = +1.25 m/s 2

a A = 1.250 m/s 2 

Note: Answers can also be written: α disk = (80 rad/s 2 ) j

aA = −(1.25 m/s 2 )i

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1669

PROBLEM 16.156 Identical cylinders of mass m and radius r are pushed by a series of moving arms. Assuming the coefficient of friction between all surfaces to be μ < 1 and denoting by a the magnitude of the acceleration of the arms, derive an expression for (a) the maximum allowable value of a if each cylinder is to roll without sliding, (b) the minimum allowable value of a if each cylinder is to move to the right without rotating.

SOLUTION (a)

Cylinder rolls without sliding a = rα or α =

a r

P is horizontal component of force that the arm exerts on cylinder. ΣM A = Σ( M A )eff : Pr − ( μk P ) r = I α + (m a ) r 1 2a mr  2 r 3 ma P= 2 (1 − μ )

P (1 − μ )r =

ΣFy = 0: ΣFx = Σ( Fx )eff :

  + (ma ) r 

(1)

N − μ P − mg = 0

(2)

P − μ N = ma

(3)

Solve (2) for N and substitute for N into (3). P − μ 2 P − μ mg = ma

Substitute P from (1):

(1 − μ 2 )

3 ma −μ mg = ma 2 (1 − μ )

3(1 + μ )a − 2μ g = 2a a (1 + 3μ ) − 2μ g = 0

a=

2μ g  1 + 3μ

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1670

PROBLEM 16.156 (Continued)

(b)

Cylinder translates:

α =0

Sliding occurs at A:

Ax = μ N

Assume sliding impends at B:

By = μ P

ΣM A = Σ( M A )eff :

Pr − μ Pr = (ma )r P (1 − μ )r = m a r P=

ΣFx = Σ( Fx )eff : ΣFy = Σ( Fy )eff :

ma 1− μ

P − μ N = ma N − μ P − mg = 0

(4) (5) (6)

Solve (5) for N and substitute for N into (6). P − P μ 2 − μ mg = ma

Substitute for P from (4): ma (1 − μ 2 ) − μ mg = ma 1− μ a (1 + μ ) − μ g = a aμ − μg = 0

Summary:

a<

a = g 

2μ g: Rolling 1 + 3μ

3μ g < a < g : Rotating and sliding 1 + 3μ a > g: Translation

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PROBLEM 16.157 The uniform rod AB of weight W is released from rest when β = 70°. Assuming that the friction force between end A and the surface is large enough to prevent sliding, determine immediately after release (a) the angular acceleration of the rod, (b) the normal reaction at A, (c) the friction force at A.

SOLUTION We note that rod rotates about A. ω = 0 1 mL2 12 L a= α 2 I =

ΣM A = Σ( M A )eff :

L L  mg  cos β  = I α + (ma ) 2 2  1 1  L L mgL cos β = mL2α +  m α  2 12  2 2 1 = mL2α 3

α= ΣFx = Σ( Fx )eff :

3 g cos β 2 L

(1)

FA = ma sin β L L  3 g cos β  FA = m α sin β = m  sin β 2 2  2 L  FA =

ΣFy = Σ( Fy )eff :

3 mg sin β cos β 4

(2)

L  NA − mg = −ma cos β = − m  α  cos β 2 

L  3 g cos β  cos β 2  2 L   3  NA = mg 1 − cos 2 β   4 

NA − mg = −m

(3)

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PROBLEM 16.157 (Continued)

For β = 70°:

α=

3 g cos 70° 2 L

(a)

Eq. (1):

(b)

Eq. (3):

 3  NA = mg 1 − cos 2 70°   4 

(c)

Eq. (2)

FA =

3 mg sin 70° cos 70° 4

α = 0.513

g L



N A = 0.912mg  FA = 0.241mg



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PROBLEM 16.158 The uniform rod AB of weight W is released from rest when β = 70°. Assuming that the friction force is zero between end A and the surface, determine immediately after release (a) the angular acceleration of the rod, (b) the acceleration of the mass center of the rod, (c) the reaction at A.

SOLUTION ΣFx = Σ( Fx )eff :

a = max a A = ay +

0 = ay − ay =

ax = 0

L α 2

β

L α cos β 2

L α cos β 2

L  mg − A = ma y = m  α cos β  2 

ΣFy = Σ( Fy )eff :

(1)

1 L  ΣM G = Σ( M G )eff : A  cos β  = I α = mL2α 2 12   A=

mL α 6 cos β

(2)

Substitute (2) into (1): mg −

mL α L = m α cos β 6 cos β 2 L L  g =  cos β + α 6cos β  2 1  L g =  3cos β + α 6 cos β 

g=

L  3cos 2 β + 1   α 6  cos β 

α=

6 g  cos β    L  1 + 3cos 2 β 

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PROBLEM 16.158 (Continued)

a=

A=

 cos 2 β L L  6g cos β  α cos β =  cos β 3 g ⋅ =    1 + 3cos 2 β 2 2  L 1 + 3cos 2 β  

mL α mL  6 g cos β ⋅ = ⋅ ⋅ 6 cos β 6  L 1 + 3cos 2 β

  

 1 1 = mg  1 + 3cos 2 β  cos β

For β = 70°: 6g cos 70° L 1 + 3cos 2 70°

(a)

α=

(b)

a = 3g

(c)

A = mg

cos 2 70 1 + 3cos 2 70° 1 1 + 3cos 2 70°

α = 1.519

g L



a = 0.260g  A = 0.740 mg



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PROBLEM 16.159 A bar of mass m = 5 kg is held as shown between four disks, each of mass m′ = 2 kg and radius r = 75 mm. Knowing that the normal forces on the disks are sufficient to prevent any slipping, for each of the cases shown determine the acceleration of the bar immediately after it has been released from rest. (a)

(b)

(c)

SOLUTION (a) Configuration (a) Kinematics: a r Kinetics of one disk

a = rα

α=

Kinetics of bar

ΣM C = Σ( M C )eff : Fr = I α 1 a m′r 2   2 r 1 F = m′a 2

Fr =

ΣFy = Σ( Fy )eff : W − 4 F = ma

(1)

(2)

Substitute for F from (2) into (1). 1  mg − 4  m′a  = ma 2 

mg = (m + 2m′) a

a=

mg m + 2 m′

m = 5 kg, m′ = 2 kg: a =

5 g 5 + 2(2)

a=

5 g  9

(b) Configuration (b) Kinematics: Disk is rolling on vertical wall a′ = aC = rα a = ar = 2rα

Therefore:

α=

a 2r

a′ = rα =

1 a 2

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PROBLEM 16.159 (Continued) Kinetics of one disk

Kinetics of bar

I =

1 m′r 2 2

ΣM D = Σ( MD )eff :

ΣFy = Σ( Fy )eff :

F (2r ) + m′gr = I α + m′a′r

W − 4 F = ma mg − 2(2 F ) = ma

(1)

2 Fr = m′gr =

a 1  a  m′r 2   + m′ r 2 2  2r 

1  1 2 F = m′  a + a  − m′g 2  4

(2)

Substitute for 2F from (2) into (1): 3  mg − 2m′  a  + 2m′g = ma 4  3    m + m′  a = (m + 2m′) g 2   m = 5 kg, m′ = 2 kg : a =

a=

m + 2 m′ g 3 m + m′ 2

5+4 g 5+3

a=

9 g  8

(c) Configuration (c) Kinematics: Disk is rolling on vertical wall Kinetics of bar

a = a′ = rα

Kinetics of one disk Iα =

1 a m′r 2   2 r

ΣFy = Σ( Fy )eff : W − 4 F = ma mg − 4 F = ma

ΣM D = Σ( M D )eff :

(1)

( F + w′)r = I α + m′a′r 1 a m′r ′   + m′at 2 r 3 F = m′a − m′g 2

( F + m′g )r =

(2)

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PROBLEM 16.159 (Continued)

Substitute for F from (2) into (1): mg − 6m′a + 4m′g = ma (m + 6m′)a = (m + 4m′) g m = 5 kg, m′ = 2 kg : a =

a=

5+8 5 + 12

m + 4 m′ g m + 6 m′

a=

13 g  17

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PROBLEM 16.160 A uniform plate of mass m is suspended in each of the ways shown. For each case determine immediately after the connection at B has been released (a) the angular acceleration of the plate, (b) the acceleration of its mass center.

SOLUTION (1)

Plate attached to pins Kinematics: Assume

α

(ω = 0)

a = rα

x comp: y comp:

Kinetics:

(a)

ax = rα sin θ = (r sin θ )α a y = rα cos θ = (r cos θ )α

ax =

Thus:

θ

I =

1 cα 4

;

ay =

1 cα 2

(1)

2 1  2 c  5 m c +    = mc 2 12   2   48

c c c ΣM A = Σ( M A )eff : W   = I α + (max )   + (ma y )   2 4 2 1 5  1  c   1  c  mgc = mc 2α + m  cα   + m  cα   2 48  4  4   2  2  1 20 2 mgc = mc α 2 48

α = 1.2

g c

α = 1.2

g c



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PROBLEM 16.160 (Continued)

(b)

(2)

From (1):

ax =

1 1 cα = (1.2 g ) 4 4

ax = 0.3g

ay =

1 1 cα = (1.2 g ) 2 2

a y = 0.6 g

a = 0.671g

63.4° 

Plate suspended from wires. Kinematics: Assume

α

(ω = 0)

a = aC = a A + aG/ A

θ

= a A ↔ + rα

y comp. a y = 0 − rα cos θ = − ( r cos θ )α 1 c 2

r cos θ =

1 a y = − cα 2

Thus:

I =

Kinetics:

ay =

1 cα 2

(2)

2 1  2 c  I m c +    = mc 2 12   2   48

ΣFx = Σ( Fx )eff :

0 = ma2 ax = 0

ΣM A = Σ( M A )eff :

Recalling (1):

1  1  W  c  = I α + ( ma y )  c  2  2 

1 5 1  1  mgc = mc 2α +  mcα  c  2 48 2   2  1 17 2 24 g mgc = mc α α = 2 48 17 c ay =

1 1  24 g  12 cα = c  = g 2 2  17 c  17

α=

24 g 17 c

a=

12 g 17

 

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PROBLEM 16.160 (Continued)

(3)

Plate suspended from springs. Immediately after spring B is released, the tension in spring A is still 1 mg since its elongation is unchanged. 2

(a)

Angular acceleration. 1  1  ΣM G = Σ( M G )eff :  mg  c  = I α 2  2  1 5 mgc = mc 2α 4 48

(b)

α = 2.4

g c



Acceleration at mass center. ΣFx = Σ( Fx )eff : 0 = max ΣFy = Σ( Fy )eff :

mg −

ax = 0 1 mg = ma y 2 1 ay = g 2

a = 0.5g



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PROBLEM 16.161 A cylinder with a circular hole is rolling without slipping on a fixed curved surface as shown. The cylinder would have a weight of 16 lb without the hole, but with the hole it has a weight of 15 lb. Knowing that at the instant shown the disk has an angular velocity of 5 rad/s clockwise, determine (a) the angular acceleration of the disk, (b) the components of the reaction force between the cylinder and the ground at this instant.

SOLUTION Geometry: Let the mass center G of the cylinder lie a distance b below the geometric center for the position shown. Let C, the contact point between the cylinder and the fixed curved surface, be the origin of a coordinate system, as shown. The position vector of a point is r = x i + yj Let r be the radius of the cylinder and R that of the fixed curved surface Kinematics: The acceleration a P at a point is given by a P = aC + α × rP/C − ω 2 rP/ C

Let

α =α

Then, using the coordinate system a P = [(aC ) x +[ yα +[ yω

] + [(aC ) y ]

] + [xα ] 2

] + [xω 2

]

Since the cylinder rolls without slipping on a fixed surface, (aC ) x = 0 For Points G and A, aG = [( aC ) y ] + [( r − b)α a A = [(aC ) y ] + [rα

] + [(r − b)ω 2 ]

] + [rω ] 2

(1) (2)

Subtract Eq. (2) from Eq. (1) to eliminate (aC ) y aG = a A = [bα

] + [bω 2 ]

aG = a A + [bα

] + [bω 2 ] ] + [(a A ) y ] + [bα

= [(a A ) x = [(r − b)α

] + [bω 2 ]

] + [(a A ) y ] + [bω 2 ]

(3)

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PROBLEM 16.161 (Continued)

Point C is the instantaneous center, so that v A = rω

Point A is constrained to move on a circle of radius

ρ =R+r so its vertical component of acceleration is (a A ) y =

v 2A

ρ

=

r 2ω 2

ρ

aG = [( r − b)α

Using Eq. 3,

 r2  ] +  − b  ρ 

The effective force at the mass center is maG = [m(r − b)α

 r2  ] +  − b  ω 2 ρ 

Kinetics: ΣM C = Σ( M C )eff : 0 + I α + (r − b)m(r − b)α = [ I + m(r − b) 2 ]α

(a)

Angular acceleration.

(b)

Force components at C.

α=0 

ΣFx = Σ( Fx )eff : C x = m(r − b)α = 0

Cx = 0 

 r2  ΣFy = Σ( Fy )eff : C y − W = − m  − b  ω 2 ρ  2  W r C y = W −  − b  ω 2 gρ  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1683

PROBLEM 16.161 (Continued)

It remains to determine the distance b from the mass distribution of the cylinder. The mass center G coincides with the centroid of a circular cylinder of area A1 = π r 2 with a circular cut out of area A2 = 161 A, with its center located

(1) −

(2) Σ

Data:

2 3

r above the center of A1.

A1

y1

A1 y1

π r2

0

0

1 π r2 16

2 r 3

15 2 πr 16



1 π r3 24



1 π r3 24

Y Σ A = ΣAy1 15 2 1 π r Y = − π r3 16 24 2 Y =− r 45 2 b= r 45

r = 12 in. = 1 ft, R = 36 in., ρ = 48 in. b=

2 (12) = 0.53333 in. 45

 r2  144 − 0.53333 = 2.4667 in. = 0.20556 ft  − b  = ρ  48 15 lb (0.20556 ft)(5 rad/s) 2 C y = 15 lb − 32.2 ft/s 2

C = 12.61 lb 

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PROBLEM 16.162 The motion of a square plate of side 150 mm and mass 2.5 kg is guided by pins at corners A and B that slide in slots cut in a vertical wall. Immediately after the plate is released from rest in the position shown, determine (a) the angular acceleration of the plate, (b) the reaction at corner A.

SOLUTION AG =

Kinematics:

Plane motion

a = [ Lα sin 30°

L L 2= 2 2

=

aG/ A = ( AB)α =

+

Translation

aB = a A

+ aB/ A

60°

aB = a A

+ Lα

60°

a = aA

+ aG/ A

15°

 Lα ]+   2

 15° = [0.5Lα 

Lα 2

Rotation

] + [0.707Lα

15°]

Law of cosines a 2 = a A2 ⋅ aG2 / A − 2a A aG/ A cos15° a 2 = (0.5Lα ) 2 + (0.707 Lα ) 2 −2(0.5Lα )(0.707 Lα ) cos15° a 2 = L2α 2 (0.25 + 0.5 − 0.68301) a 2 = L2α 2 (0.06699) a = 0.25882Lα

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PROBLEM 16.162 (Continued)

Law of sines. aG/ A aG/ A 0.707 Lα a = ; sin β = sin15° = sin15° sin15° sin β 0.25950 Lα a sin β = 0.707; β = 135°

a = 0.2583Lα

Kinetics:

45°

(ω = 0)

We find the location of Point E where lines of action of A and B intersect. MG =

L

2 EAG = 15°

( EG ) y = 0.6829 L − 0.5 L = 0.183L ( EG ) = (0.183L) 2 = 0.2588L I =

1 2 mL 6

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PROBLEM 16.162 (Continued)

(a)

Angular acceleration. ΣM A = Σ( M A )eff : mg (0.183L) = I α + ( ma )(0.2588 L) 1 2 mL α + m(0.2588Lα )(0.2588 L) 6 1  0.183gL = L2α  + 0.06698  6  g g 0.183 = 0.2336α ; α = 0.7834 L L

0.183mgL =

α = 0.7834 (b)

9.81 m/s 2 0.15 m

α = 51.2 rad/s 2



Reaction at corner A. ΣFy = Σ( Fy )eff : A − mg = − ma sin 45° = − m (0.2588Lα )sin 45° g  = − m (0.2588L)  0.7834  sin 45° L   A − mg = 0.1434mg A = 0.8566mg

= 0.8566(2.5 kg)(9.81 m/s 2 ) = 21.01 N

A = 21.0 N 

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PROBLEM 16.163 Solve Problem 16.162, assuming that the plate is fitted with a single pin at corner A. Problem 16.162 The motion of a square plate of side 150 mm and mass 2.5 kg is guided by pins at corners A and B that slide in slots cut in a vertical wall. Immediately after the plate is released from rest in the position shown, determine (a) the angular acceleration of the plate, (b) the reaction of corner A.

SOLUTION Since both A and mg are vertical, ax = 0 and a is Kinematics: AG =

L 2

a = aA

15° aG/ A = ( AG )α

+ aG/ A

a = 0.183Lα

Kinetics:

I =

15°

15°

15°

1 2 mL 6

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PROBLEM 16.163 (Continued)

(a)

Angular acceleration. ΣM A = Σ( M A )eff : mg ( AG ) sin15° = I α + ma ( AG )sin15°  L   L  1 2 mg   sin15° = mL α + m(0.183Lα )   sin15° 6  2  2 g 1  0.183 =  + 0.033494  α L 6  g 0.183 = 0.2002α L g α = 0.9143 L 9.81 m/s 2 = 0.9143 0.15 m

(b)

α = 59.8 rad/s 2



Reaction at corner A. ΣFy = Σ( Fy )eff : A − mg = − ma A − mg = − m(0.183Lα ) g  = − m(0.183L)  0.9143  L  A − mg = − 0.1673mg A = 0.8326mg A = 0.8326(2.5 kg)(9.81 m/s 2 )

A = 20.4 N 

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PROBLEM 16.164 The Geneva mechanism shown is used to provide an intermittent rotary motion of disk S. Disk D weighs 2 lb and has a radius of gyration of 0.9 in. and disk S weighs 6 lb and has a radius of gyration of 1.5 in. The motion of the system is controlled by a couple M applied to disk D. A pin P is attached to disk D and can slide in one of the six equally spaced slots cut in disk S. It is desirable that the angular velocity of disk S be zero as the pin enters and leaves each of the six slots; this will occur if the distance between the centers of the disks and the radii of the disks are related as shown. Knowing disk D rotates with a constant counterclockwise angular velocity of 8 rad/s and the friction between the slot and pin P is negligible, determine when φ = 150° (a) the couple M, (b) the magnitude of the force pin P applies to disk S.

SOLUTION Geometry:

r 2 = 1.252 + 2.502 − (2)(1.25)(2.50) cos30°

Law of cosines.

r = 1.54914 in. Law of sines.

sin β sin 30° = 1.25 r β = 23.794°

Let disk S be a rotating frame of reference.

Ω = ωS

,

 =α Ω S

Motion of coinciding Point P′ on the disk.

vP′ = rωS = 1.54914ωS

β

aP′ = −α S k × rP/O − ωS2 rP/O = [1.54914α S

β ] + [1.54914ωS2

β]

Motion relative to the frame. vP/S = u Coriolis acceleration.

β

a P/S = u

2ωS u

β

β

vP = vP′ + vP/S = [1.54914ωS

β ] + [u

β]

a P = a P + a P/S + 2ωS u

= [1.54914α S

β ] + [1.54914ωS2

β ] + [u

β ] + [2ωS u

β]

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PROBLEM 16.164 (Continued)

Motion of disk D. (rotation about B)

vP = ( BP)ωD = (1.25)(8) = 10 in./s aP = [( BP)α D 

= 80 in./s2

60°] + [( BP)ωS2

30° 30°] = 0 + [(1.25)(8)2

30°]

30°

Equate the two expressions for vP and resolve into components.

β : 1.54914ωS = 10cos(30° + β ) 10cos 53.794° 1.54914 = 3.8130 rad/s

ωS =

β : u = 10sin(30° + β ) = 10sin 53.794° = 8.0690 in./s Equate the two expressions for aP and resolve into components.

β : 1.54914α S − 2ωS u = 80sin (30° + β ) 80sin 53.794° + (2)(3.8130)(8.0690) 1.54914 2 = 81.391 rad/s

αS =

Kinetics:

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PROBLEM 16.164 (Continued)

2

 6 lb  1.5  ft  (81.391 rad/s 2 ) = 0.23697 lb ⋅ ft I sα s =  2   32.2 ft/s  12  since α D = 0 I αD = 0

Disk S:

ΣM 0 = ΣM eff : Pr = I S α S  1.54914  ft  = 0.23697 lb ⋅ ft P  12 

Disk D:

P = 1.8356 lb

r = 90° − 30° − β = 36.206°

 1.25  ΣM B = Σ( M B )eff : M − ( P sin r )   ft  12  (1.25)(1.8356sin 36.206°) = 0.11294 lb ⋅ ft M= 12

(a)

Couple M.

(b)

Magnitude of contact force.

M = 1.355 lb ⋅ in



P = 1.836 lb 

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CHAPTER 17

PROBLEM 17.CQ1 A round object of mass m and radius r is released from rest at the top of a curved surface and rolls without slipping until it leaves the surface with a horizontal velocity as shown. Will a solid sphere, a solid cylinder or a hoop travel the greatest distance c? (a) (b) (c) (d )

A solid sphere A solid cylinder A hoop They will all travel the same distance.

SOLUTION Answer: (a) It has the smallest mass moment of inertia, so it will have the greatest speed at the bottom of the surface.

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PROBLEM 17.CQ2 A solid steel sphere A of radius r and mass m is released from rest and rolls without slipping down an incline as shown. After traveling a distance d the sphere has a speed v. If a solid steel sphere of radius 2r is released from rest on the same incline, what will its speed be after rolling a distance d? (a)

0.25 v

(b)

0.5 v

(c)

v

(d)

2v

(e)

4v

SOLUTION Answer: (c) Using conservation of energy you can show that the speed after traveling a distance d will be independent of the mass and the radius.

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PROBLEM 17.CQ3 Slender bar A is rigidly connected to a massless rod BC in Case 1 and two massless cords in Case 2 as shown. The vertical thickness of bar A is negligible compared to L. In both cases A is released from rest at an angle θ = θ0. When θ = 0° which system will have the larger kinetic energy? (a)

Case 1

(b)

Case 2

(c)

The kinetic energy will be the same.

SOLUTION Answer: (c)

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PROBLEM 17.CQ4 In Problem 17.CQ3, how will the speeds of the centers of gravity compare for the two cases when θ = 0°? (a)

Case 1 will be larger.

(b)

Case 2 will be larger.

(c)

The speeds will be the same.

SOLUTION Answer: (b) Case 1 will also have rotational kinetic energy, so the speed will be smaller.

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PROBLEM 17.CQ5 Slender bar A is rigidly connected to a massless rod BC in Case 1 and two massless cords in Case 2 as shown. The vertical thickness of bar A is not negligible compared to L. In both cases A is released from rest at an angle θ = θ 0 . When θ = θ ° which system will have the largest kinetic energy? (a)

Case 1

(b)

Case 2

(c)

The kinetic energy will be the same.

SOLUTION Answer: (a) Case 1 will have a greater change in gravitational potential energy, so the kinetic energy will be larger.

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PROBLEM 17.1 The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 50-kg rotor then coasts to rest after 5000 revolutions. Knowing that the kinetic friction of the rotor produces a couple of magnitude 4 N · m, determine the centroidal radius of gyration of the rotor.

SOLUTION

ω1 = 3600

Angular velocities:

rev 1min 2π rad ⋅ ⋅ = 120π rad/s min 60 s rev

ω2 = 0 Angular displacement:

5000 rev = 10000 π rad

Principle of work and energy: T1 + U1→2 = T2 : 1 1 I ω12 = I (120π ) 2 = 71.061 × 103 I 2 2 1 T2 = I ω22 = 0 2 U1→2 = − M θ = −(4 N ⋅ m)(10000π rad) = −40000π N ⋅ m T1 =

71.061 × 103 I − 40000π = 0 I = 1.76839 kg ⋅ m 2 I = mk 2

Centroidal radius of gyration.

k =

I 1.76839 kg ⋅ m 2 = = 0.1881 m m 50 kg

k = 188.1 mm 

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PROBLEM 17.2 It is known that 1500 revolutions are required for the 6000-lb flywheel to coast to rest from an angular velocity of 300 rpm. Knowing that the radius of gyration of the flywheel is 36 in., determine the average magnitude of the couple due to kinetic friction in the bearings.

SOLUTION Angular velocity:

Moment of inertia:

ω0 = 300 rpm = 10 π rad/s ω2 = 0 I = mk 2 =

6000 lb (3 ft) 2 32.2 ft/s 2

= 1677 lb ⋅ ft ⋅ s 2

Kinetic energy:

Work:

1 I ω02 2 1 = (1677)(10 π ) 2 2 = 827, 600 ft ⋅ lb T2 = 0 T1 =

U1→2 = − M θ

= − M (1500 rev)(2π rad/rev) = −9424.7 M

Principle of work and energy:

T1 + U1→2 = T2 827,600 − 9424.7 M = 0

Average friction couple:

M = 87.81 lb ⋅ ft

M = 87.8 lb ⋅ ft 

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PROBLEM 17.3 Two disks of the same material are attached to a shaft as shown. Disk A has a weight of 30 lb and a radius r = 5 in. Disk B is three times as thick as disk A. Knowing that a couple M of magnitude 15 lb ⋅ ft is to be applied to disk A when the system is at rest, determine the radius nr of disk B if the angular velocity of the system is to be 600 rpm after 4 revolutions.

SOLUTION For any disk:

m = ρ (π r 2 t ) 1 2 mr 2 1 = πρ tr 4 2

I =

Moment of inertia. Disk A: Disk B:

1 I A = πρ br 4 2

1 I B = πρ (3b)(nr ) 4 2 1  = 3n4  πρ br 4  2  = 3n4 I A I total = I A + I B = (1 + 3n 4 ) I A

Angular velocity:

ω1 = 0 ω2 = 600 rpm = 20π rad/s

Rotation:

θ = 4 rev = 8 π rad

Kinetic energy:

T1 = 0 T2 =

Work:

(1)

1 I total ω22 2

U1→2 = M θ = (15 lb ⋅ ft)(8π rad) = 376.99 lb ⋅ ft

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PROBLEM 17.3 (Continued)

Principle of work and energy:

T1 + U1→2 = T2 1 I total (20π ) 2 2 = 0.19099 slug ⋅ ft 2

0 + 376.991 = I total

But,

IA = =

1 mA rA2 2 1  30 lb  5  ft  2  32.2   12 

2

= 0.080875 slug ⋅ ft 2

From (1)

0.19099 = (1 + 3n 4 )(0.080875) n 4 = 0.45383 n = 0.82078

Radius of disk B:

rB = nrA = (0.82078)(5 in.) = 4.1039 in.

rB = 4.10 in. 

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PROBLEM 17.4 Two disks of the same material are attached to a shaft as shown. Disk A is of radius r and has a thickness b, while disk B is of radius nr and thickness 3b. A couple M of constant magnitude is applied when the system is at rest and is removed after the system has executed 2 revolutions. Determine the value of n which results in the largest final speed for a point on the rim of disk B.

SOLUTION m = ρ (π r 2 t )

For any disk:

1 2 mr 2 1 = πρ tr 4 2

I =

Moment of inertia. 1 I A = π ρ b r4 2

Disk A:

1 I B = π ρ (3b)(nr ) 4 2 1  = 3n 4  πρ br 4  2  

Disk B:

= 3n 4 I A I total = I A + I B = (1 + 3n 4 ) I A T1 = 0

Work-energy.

T2 = T1 + U1→ 2 = T2 : 0 + M (4π ) =

ω22 =

U1→2 = M θ = M (4π rad)

1 I total ω22 2 1 (1 + 3n4 ) I Aω22 2 8π M (1 + 3n 4 ) I A

For Point D on rim of disk B vD = (nr )ω2 or vD2 = n 2 r 2ω22 =

8π Mr 2 n2 ⋅ IA 1 + 3n 4

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PROBLEM 17.4 (Continued)

Value of n for maximum final speed. For maximum vD :

d  n2  dn  1 + 3n 4

  = 0 

1 [n 2 (12n3 ) − (1 + 3n 4 )(2n)] = 0 (1 + 3n 4 )2 12n5 − 2n − 6n5 = 0 2n(3n 4 − 1) = 0 1 n = 0 and n =    3

0.25

n = 0.760 

= 0.7598

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PROBLEM 17.5 The flywheel of a small punch rotates at 300 rpm. It is known that 1800 ft ⋅ lb of work must be done each time a hole is punched. It is desired that the speed of the flywheel after one punching be not less that 90 percent of the original speed of 300 rpm. (a) Determine the required moment of inertia of the flywheel. (b) If a constant 25-lb ⋅ ft couple is applied to the shaft of the flywheel, determine the number of revolutions which must occur between each punching, knowing that the initial velocity is to be 300 rpm at the start of each punching.

SOLUTION

ω1 = 300 rpm = 10π rad/s ω2 = 0.90ω1 = 9π rad/s

Angular velocities:

T1 + U1→ 2 = T2

Principle of work and energy:

1 1 I ω12 = I (10π ) 2 2 2 1 1 T2 = I ω22 = I (9π ) 2 2 U1→ 2 = −1800 ft ⋅ lb T1 =

1 1 I (10π ) 2 − 1800 = I (9π ) 2 2 2

(a)

Required moment of inertia. I =

(b)

2(1800) = 19.198 lb ⋅ ft ⋅ s 2 π (100 − 81) 2

I = 19.20 lb ⋅ ft ⋅ s 2 

Number of revolution between each punching. Definition of work:

U 2→1 = M θ : 1800 ft ⋅ lb = (25 lb ⋅ ft)θ

θ = 72 rad = 11.459 rev

θ = 11.46 rev 

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PROBLEM 17.6 The flywheel of a punching machine has a mass of 300 kg and a radius of gyration of 600 mm. Each punching operation requires 2500 J of work. (a) Knowing that the speed of the flywheel is 300 rpm just before a punching, determine the speed immediately after the punching. (b) If a constant 25-N ⋅ m couple is applied to the shaft of the flywheel, determine the number of revolutions executed before the speed is again 300 rpm.

SOLUTION I = mk 2

Moment of inertia.

= (300 kg)(0.6 m) 2 = 108 kg ⋅ m 2

ω1 = 300 rpm = 10π rad/s

Kinetic energy. Position 1.

1 2 I ω1 2 1 = (108)(10π ) 2 2 = 53.296 × 103 J

T1 =

T2 =

Position 2. Work.

1 2 I ω2 = 54ω22 2

U1→2 = −2500 J

Principle of work and energy for punching. T1 + U1→ 2 = T2 : 53.296 × 103 − 2500 = 54ω22

ω22 = 940.66

(a)

ω2 = 30.67 rad/s

ω2 = 293 rpm 

Principle of work and energy for speed recovery. T2 + U 2→1 = T1 U 2→1 = 2500 J M = 25 N ⋅ m U 2→1 = M θ 2500 = 25θ

θ = 100 rad θ = 15.92 rev 

(b)

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PROBLEM 17.7 Disk A, of weight 10 lb and radius r = 6 in., is at rest when it is placed in contact with belt BC, which moves to the right with a constant speed v = 40 ft/s. Knowing that μk = 0.20 between the disk and the belt, determine the number of revolutions executed by the disk before it attains a constant angular velocity.

SOLUTION Work of external friction force on disk A. Only force doing work is F. Since its moment about A is M = rF , we have U1→2 = M θ = rFθ = r ( μ k mg )θ

Kinetic energy of disk A. Angular velocity becomes constant when

v r T1 = 0

ω2 =

T2 =

1 I ω22 2

=

1  1 2  v   mr   22  r 

=

mv 2 4

T1 + U1− 2 = T2 : 0 + r μk mgθ =

mv 2 4

2

Principle of work and energy for disk A.

v2 rad 4 r μk g

Angle change

θ=

Data:

r = 0.5 ft

θ=

v2 rev 8 π r μk g

μk = 0.20 v = 40 ft/s

θ=

(40 ft/s) 2 8π (0.5 ft)(0.20)(32.2 ft/s 2 )

θ = 19.77 rev 

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PROBLEM 17.8 Disk A is of constant thickness and is at rest when it is placed in contact with belt BC, which moves with a constant velocity v. Denoting by μk the coefficient of kinetic friction between the disk and the belt, derive an expression for the number of revolutions executed by the disk before it attains a constant angular velocity.

SOLUTION Work of external friction force on disk A. Only force doing work is F. Since its moment about A is M = rF , we have U1→2 = M θ = rFθ = r ( μk mg )θ

Kinetic energy of disk A. Angular velocity becomes constant when

v r T1 = 0

ω2 =

T2 =

1 I ω22 2

=

1  1 2  v   mr   22  r 

=

mv 2 4

T1 + U1→2 = T2 : 0 + r μk mg θ =

mv 2 4

2

Principle of work and energy for disk A.

Angle change.

θ=

v2 rad 4r μ k g

θ=

v2 rev  8π r μk g

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PROBLEM 17.9 The 10-in.-radius brake drum is attached to a larger flywheel which is not shown. The total mass moment of inertia of the flywheel and drum is 16 lb ⋅ ft ⋅ s2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.40. Knowing that the initial angular velocity is 240 rpm clockwise, determine the force which must be exerted by the hydraulic cylinder if the system is to stop in 75 revolutions.

SOLUTION Kinetic energies.

ω1 = 240 rpm = 8π rad/s I = 16 lb ⋅ ft ⋅ s 2 1 1 T1 = I ω12 = (16)(8π ) 2 = 5053 ft ⋅ lb 2 2 ω2 = 0 T2 = 0

θ = 75 rev = 75(2π ) = 150π rad

Angular displacement. Work.

  10   U1− 2 = −M θ = −  F  ft   (150π rad) = −392.7 F   12  

Principle of work and energy.

T1 + U1− 2 = T2 :

5053 − 392.7 F = 0 F = μk N : 12.868 = (0.40) N

F = 12.868 lb N = 32.17 lb

Free body brake arm: ΣM A = 0: B(6 in.) + F (6 in.) − N (18 in.) = 0 B(6 in.) + (12.868 lb)(6 in.) − (32.17 lb)(18 in.) = 0 B = 83.64 lb B = 83.6 lb 

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PROBLEM 17.10 Solve Problem 17.9, assuming that the initial angular velocity of the flywheel is 240 rpm counterclockwise. PROBLEM 17.9 The 10-in.-radius brake drum is attached to a larger flywheel which is not shown. The total mass moment of inertia of the flywheel and drum is 16 lb ⋅ ft ⋅ s2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.40. Knowing that the initial angular velocity is 240 rpm clockwise, determine the force which must be exerted by the hydraulic cylinder if the system is to stop in 75 revolutions.

SOLUTION Kinetic energies.

ω1 = 240 rpm = 8π rad/s I = 16 lb ⋅ ft ⋅ s 2 1 1 T1 = I ω12 = (16)(8π ) 2 = 5053 ft ⋅ lb 2 2 ω2 = 0 T2 = 0

Angular displacement. Work.

θ = 75 rev = 75(2π ) = 150π rad

  10   U1− 2 = −M θ = −  F  ft   (150π rad) = −392.7 F   12  

Principle of work and energy.

T1 + U1− 2 = T2 :

5053 − 392.7 F = 0 F = μk N : 12.868 = (0.40) N

F = 12.868 lb N = 32.17 lb

Free body brake arm: ΣM A = 0: B(6 in.) − F (6 in.) − N (18 in.) = 0 B(6 in.) − (12.868 lb)(6 in.) − (32.17 lb)(18 in.) = 0 B = 109.37 lb B = 109.4 lb 

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PROBLEM 17.11 Each of the gears A and B has a mass of 2.4 kg and a radius of gyration of 60 mm, while gear C has a mass of 12 kg and a radius of gyration of 150 mm. A couple M of constant magnitude 10 N ⋅ m is applied to gear C. Determine (a) the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm, (b) the corresponding tangential force acting on gear A.

SOLUTION Moments of inertia. Gears A and B:

I A = I B = mk 2 = (2.4)(0.06) 2 = 8.64 × 10−3 kg ⋅ m 2

Gear C:

I C = (12)(0.15)2 = 270 × 10−3 kg ⋅ m 2

Kinematics.

rAω A = rBωB = rC ωC 200 ωC = 2.5ωC 80 θ A = θ B = 2.5θC

ω A = ωB =

Kinetic energy. Position 1.

T=

1 2 Iω : 2

10 π rad/s 3 25 ω A = ωB = 250 rpm = π rad/s 3

ωC = 100 rpm =

2

Gear A:

(T1 ) A =

1  25π  (8.64 × 10−3 )   = 2.9609 J 2  3 

Gear B:

(T1 ) B =

1  25π  (8.64 × 10−3 )   = 2.9609 J 2  3 

Gear C:

(T1 )C =

1  10π  (270 × 10−3 )   = 14.8044 J 2  3 

2

2

System: Position 2.

Gear A:

T1 = (T1 ) A + (T1 ) B + (T1 )C = 20.726 J

ωC = 450 rpm = 15π rad/s ω A = ωB = 37.5π rad/s (T2 ) A =

1 (8.64 × 10−3 )(37.5π ) 2 = 59.957 J 2

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PROBLEM 17.11 (Continued)

Gear B:

(T2 ) B =

1 (8.64 × 10−3 )(37.5π ) 2 = 59.957 J 2

Gear C:

(T2 )C =

1 (270 × 10−3 )(15π ) 2 = 299.789 J 2

T2 = (T2 ) A + (T2 ) B + (T2 )C = 419.7 J

System:

U1→2 = M θC = 10θC

Work of couple.

Principle of work and energy for system. T1 + U1→ 2 = T2 : 20.726 + 10θC = 419.7

θC = 39.898 radians (a)

θC = 6.35 rev 

Rotation of gear C. Rotation of gear A.

θ A = (2.5)(39.898) = 99.744 radians

Principle of work and energy for gear A. (T1 ) A + M Aθ A = (T2 ) A : 2.9609 + M A (99.744) = 59.957 M A = 0.57142 N ⋅ m

(b)

Ft =

Tangential force on gear A.

M A 0.57142 = 0.08 rA

Ft = 7.14 N 

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PROBLEM 17.12 Solve Problem 17.11, assuming that the 10-N ⋅ m couple is applied to gear B. PROBLEM 17.11 Each of the gears A and B has a mass of 2.4 kg and a radius of gyration of 60 mm, while gear C has a mass of 12 kg and a radius of gyration of 150 mm. A couple M of constant magnitude 10 N ⋅ m is applied to gear C. Determine (a) the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm, (b) the corresponding tangential force acting on gear A.

SOLUTION Moments of inertia. Gears A and B:

I A = I B = mk 2 = (2.4)(0.06) 2 = 8.64 × 10−3 kg ⋅ m 2

Gear C:

I C = (12)(0.15)2 = 270 × 10−3 kg ⋅ m 2

Kinematics.

rAω A = rBωB = rC ωC 200 ωC = 2.5ωC 80 θ A = θ B = 2.5θC

ω A = ωB =

Kinetic energy. Position 1.

T=

1 2 Iω : 2

10 π rad/s 3 25 ω A = ωB = 250 rpm = π rad/s 3

ωC = 100 rpm =

2

Gear A:

(T1 ) A =

1  25π  (8.64 × 10−3 )   = 2.9609 J 2  3 

Gear B:

(T1 ) B =

1  25π  (8.64 × 10−3 )   = 2.9609 J 2  3 

Gear C:

(T1 )C =

1  10π  (270 × 10−3 )   = 14.8044 J 2  3 

2

2

System:

T1 = (T1 ) A + (T1 ) B + (T1 )C = 20.726 J

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PROBLEM 17.12 (Continued)

ωC = 450 rpm = 15 π rad/s ω A = ω B = 37.5 π rad/s

Position 2.

Gear A:

(T2 ) A =

1 (8.64 × 10−3 )(37.5π ) 2 = 59.957 J 2

Gear B:

(T2 ) B =

1 (8.64 × 10−3 )(37.5π ) 2 = 59.957 J 2

Gear C:

(T2 )C =

1 (270 × 10−3 )(15π ) 2 = 299.789 J 2

T2 = (T2 ) A + (T2 ) B + (T2 )C = 419.7 J

System:

U1→2 = M θ B = 10θ B

Work of couple.

Principle of work and energy for system. T1 + U1→ 2 = T2 : 20.726 + 10θ B = 419.7

θ B = 39.898 radians (a)

39.898 = 15.959 radians 2.5

Rotation of gear C.

θC =

Rotation of gear A.

θ A = θ B = 39.898 radians

θC = 2.54 rev 

Principle of work and energy for gear A. (T1 ) A + M Aθ A = (T2 ) A : 2.9609 + M A (39.898) = 59.957 M A = 1.4285 N ⋅ m

(b)

Tangential force on gear A.

Ft =

M A 1.4285 = 0.08 rA

Ft = 17.86 N 

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PROBLEM 17.13 The gear train shown consists of four gears of the same thickness and of the same material; two gears are of radius r, and the other two are of radius nr. The system is at rest when the couple M0 is applied to shaft C. Denoting by I0 the moment of inertia of a gear of radius r, determine the angular velocity of shaft A if the couple M0 is applied for one revolution of shaft C.

SOLUTION Mass and moment of inertia: For a disk of radius r and thickness t:

m = ρ (π r 2 )t = ρπ tr 2 1 1 1 I 0 = mr 2 = ( ρπ tr 2 )r 2 = ρπ tr 4 2 2 2

For a disk of radius nr and thickness t,

I =

Kinematics:

1 ρπ t (nr ) 4 2

I = n4 I0

If for shaft A we have ω A Then, for shaft B we have ωB = ω A /n And, for shaft C we have ωC = ω A /n 2

Principle of work-energy: Couple M 0 applied to shaft C for one revolution. θ = 2π radians,

T1 = 0,

U1− 2 = M 0θ = M 0 (2π radians) = 2π M 0 T2 =

1 1 1 ( I shaft A ) wA2 + ( I shaft B )ωB2 + ( I shaft C )ωC2 2 2 2 2

1 1 1 ω  ω  = I 0 wA2 + ( I 0 + n 4 I 0 )  A  + ( n 4 I 0 )  2A  2 2 2  n  n  1 1   = I 0ω A2  n 2 + 2 + 2  2 n   =

1 1  I 0ω A2  n +  2 n 

T1 + U1− 2 = T2 : 0 + 2π M 0 =

1 1  I 0ω A2  n +  2 n 

ω A2 =

4π M 0 1 I 0 ( n + 1 )2 n

Angular velocity.

2

2

2

ωA =

π M0 2n  I0 n +1 2

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PROBLEM 17.14 The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm. Cylinder A and block B are attached to cords that are wrapped on the pulleys as shown. The coefficient of kinetic friction between block B and the surface is 0.2. Knowing that the system is at rest in the position shown when a constant force P = 200 N is applied to cylinder A, determine (a) the velocity of cylinder A as it strikes the ground, (b) the total distance that block B moves before coming to rest.

SOLUTION Kinematics. Let rA be the radius of the outer pulley and rB that of the inner pulley. v A = rAωC

vB = rBωC =

s A = rAθC

sB =

rB vA rA

rB sA rA

Use the principle of work and energy with position 1 being the initial rest position and position 2 being when cylinder A strikes the ground. T1 + U1→2 = T2 :

where

T1 = 0

and

T2 =

with

m A = 5 kg, mB = 15 kg, I C = mC kC2 = (15 kg)(0.160 m) 2 = 0.384 kg ⋅ m 2 T2 = =

1 1 1 m A v 2A + mB vB2 + I C ωC2 2 2 2

mB rB2 I C  2 1 mA + 2 + 2  vA 2  rA rA  1 (15 kg)(0.150 m) 2 0.384 kg ⋅ m 2  2 + 5 kg +  vA 2 (0.250 m) 2 (0.250 m) 2 

= (8.272 kg)v A2

Principle of work and energy applied to the system consisting of blocks A and B and the double pulley C. Work. where

U1→2 = Ps A + mA g s A − FF sB − mB g sB sin 30° sA = 1 m

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PROBLEM 17.14 (Continued)

sB =

and

rB 0.150 m sA = (1 m) = 0.6 m rA 0.250 m

To find Ff use the free body diagram of block B. 60° ΣF = 0: N B − mB g cos 30° = 0 N B = mB g cos 30° = (15 kg)(9.81 m/s) cos 30° = 127.44 N F f = μk N B = (0.2)(127.44 N) = 25.487 N U1→2 = (200 N)(1 m) + (5 kg)(9.81 m/s 2 )(1 m) − (25.487 N)(0.6 m) − (15 kg)(9.81 m/s 2 )(0.6 m) sin 30° = 189.613 J

Work-energy: (a)

0 + 189.613 J = (8.272 kg)v 2A

Velocity of A.

v A = 4.7877 m/s

v A = 4.79 m/s 

when the cylinder strikes the ground, vB =

rB 0.150 m (4.7877 m/s) = 2.8726 m/s vA = rA 0.250 m

ωC =

v A 4.7877 m/s = = 19.1508 rad/s rA 0.250 m

After the cylinder strikes the ground use the principle of work and energy applied to a system consisting of block B and double pulley C. Let T3 be its kinetic energy when A strikes the ground. 1 1 mB vB2 + I C ωC2 2 2 1 1 = (15 kg)(2.8726 m/s) 2 + (0.384 kg ⋅ m 2 )(19.1508 rad/s) 2 2 2 = 132.305 J

T3 =

When the system comes to rest,

T4 = 0

U 3→4 = −(25.487 N) sB′ − (15 kg)(9.81 m/s 2 )( sB′ sin 30°) = −(99.062 N) sB′

where sB′ is the additional travel of block B. T3 + U 3→4 = T4 : 132.305 J − (99.062 N) sB′ = 0 sB′ = 1.3356 m

(b)

sB + sB′ = 1.936 m 

Total distance:

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PROBLEM 17.15 Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a radius of gyration of 75 mm; gear C has a mass of 9 kg and a radius of gyration of 100 mm. The system is at rest when a couple M0 of constant magnitude 4 N · m is applied to gear C. Assuming that no slipping occurs between the gears, determine the number of revolutions required for disk A to reach an angular velocity of 300 rpm.

SOLUTION I = mk 2

Moments of inertia: Gear A:

I A = (1 kg)(0.030 m)2 = 0.9 × 10−3 kg ⋅ m 2

Gear B:

I B = (4 kg)(0.075 m) 2 = 22.5 × 10−3 kg ⋅ m 2

Gear C:

I C = (9 kg)(0.100 m) 2 = 90 × 10−3 kg ⋅ m 2

Let rA be the radius of gear A, r1 the outer radius of gear B, r2 the inner radius of gear B, and rC the radius of gear C. rA = 50 mm, r1 = 100 mm, r2 = 50 mm, rC = 150 mm

At the contact point between gears A and B, r1ωB = rAω A : ωB =

rA ω A = 0.5ω A r1

At the contact point between gear B and C. rC ωC = r2ωB : ωC =

r2 ω B = 0.33333ωB rC

ωC = 0.16667ω A Kinetic energy:

1 1 1 I Aω A2 + I BωB2 + IC ωC2 2 2 2 1 T = [0.9 × 10−3 ω A2 + (22.5 × 10−3 )(0.5ω A ) 2 + (90 × 10−3 )(0.16667ω A ) 2 ] 2 = (4.5125 × 10−3 kg ⋅ m 2 )ω A2 T=

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PROBLEM 17.15 (Continued)

Use the principle of work and energy applied to the system of all three gears with position 1 being the initial rest position and position 2 being when ω A = 300 rpm. 2π rad 1min rev ⋅ ⋅ 300 = 31.416 rad/s rev 60 s min T1 = 0

ωA =

T2 = (4.5125 × 10−3 kg ⋅ m 2 )(31.416 rad/s) 2 = 4.4565 J U1→2 = M θC = (4 N ⋅ m)θC

Principle of work and energy. T1 + U1→ 2 = T2 : 0 + 4.4565 J = 4(N ⋅ m)θC

θC = 1.11413 rad θC θA = = 6θC = 6.6848 rad 0.16667 6.6848 rad θA = 2π rad/rev

θ A = 1.063 rev 

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PROBLEM 17.16 A slender rod of length l and weight W is pivoted at one end as shown. It is released from rest in a horizontal position and swings freely. (a) Determine the angular velocity of the rod as it passes through a vertical position and determine the corresponding reaction at the pivot, (b) Solve part a for W = 1.8 lb and l = 3 ft.

SOLUTION v1 = 0

Position 1:

ω1 = 0 T1 = 0 l v2 = ω2 2

Position 2:

T2 =

1 1 mr22 + I ω22 2 2 2

1 l 1 1   m ω2 +  ml 2  ω22 2  2  2  12  1 2 2 T2 = ml ω2 6 =

U1→2 = mg

Work: Principle of work and energy:

T1 + U1→2 = T2

0 + mg

(a)

l 2

l 1 2 2 = ml ω2 2 6

Expressions for angular velocity and reactions. 3g l l 2 l 3g 3 a = ω2 = ⋅ = g 2 2 l 2

ω22 =

ω2 =

3g l



ΣF = Σ( F )eff : A − W = ma

3 A − mg = m g 2 5 A = mg 2

5 A= W  2

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PROBLEM 17.16 (Continued)

(b)

Application of data: W = 1.8 lb, l = 3 ft 3g 3g ω22 = = = 32.2 rad 2 /s 2 l 3 5 5 A = W = (1.8 lb) 2 2

ω2 = 5.67 rad/s



A = 4.5 lb 

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PROBLEM 17.17 A slender rod of length l is pivoted about a Point C located at a distance b from its center G. It is released from rest in a horizontal position and swings freely. Determine (a) the distance b for which the angular velocity of the rod as it passes through a vertical position is maximum, (b) the corresponding values of its angular velocity and of the reaction at C.

SOLUTION Position 1.

v = 0,

ω =0

Elevation:

h=0

V1 = mgh = 0

Position 2.

v2 = bω2

T1 = 0

1 ml 2 12 1 1 T2 = mv22 + I ω22 2 2 1  1  = m  b 2 + l 2  ω22 2  12  I=

Elevation:

h = −b

V2 = − mgb

Principle of conservation of energy. 1  2 1 2 2 m b + l  ω2 − mgb 2  12 

T1 + V1 = T2 + V2 : 0 + 0 =

2gb b + 121 l 2

ω22 = (a)

2

Value of b for maximum ω2 .

(

)

2 2  b + 121 l − b ( 2b ) d  b =0  2 1 2 = 2 db  b + 12 l  b 2 + 121 l 2

(b)

Angular velocity.

ω22 =

(

2g l2 12

)

b2 =

1 2 l 12

b=

l 12



l 12 2

l + 12

= 12

ω2 = 121/4

g l g l

ω2 = 1.861

g  l

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PROBLEM 17.17 (Continued)

an = bω22

Reaction at C.

=

l 12

12

g l

=g ΣFy = man : C y − mg = mg C y = 2mg ΣM C = mbat + I α :

0 = (mb2 + I )α

α = 0, at = 0 ΣFx = mat :

Cx = − mat = 0

C = 2mg 

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PROBLEM 17.18 A slender 9 lb rod can rotate in a vertical plane about a pivot at B. A spring of constant k = 30 lb/ft and of unstretched length 6 in. is attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated through 90°.

SOLUTION

Position 1: Unstretched Length

Spring:

 x1 = CD − ( 6 in. ) = 14.866 − 6 = 8.8661 in. = 0.73884 ft

Ve =

Gravity:

Kinetic energy:

1 2 1 kx1 = (30 lb/ft)(0.73884) 2 = 8.1882 lb ⋅ ft 2 2

 7  Vg = Wh = (9 lb)  ft  = 5.25 lb ⋅ ft  12  V1 = Ve + Vg = 8.1882 lb ⋅ ft + 5.25 lb ⋅ ft = 13.438 lb ⋅ ft

T1 = 0

Position 2: Spring:

x2 = 9 in. − 6 in. = 3 in. = 0.25 ft Ve =

1 2 1 kx2 = (30 lb/ft)(0.25 ft) 2 = 0.9375 lb ⋅ ft 2 2

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PROBLEM 17.18 (Continued)

Gravity:

Vg = Wh = 0 V2 = Ve + Vg = 0.9375 lb ⋅ ft

Kinetic energy:

 7  v2 = rω2 =  ft  ω2  12  1 1  9 lb  I = mL2 =  (2 ft)2 = 0.093168 slug ⋅ ft 2 12 12  32.2  1 1 T2 = mv22 + I ω22 2 2 2

=

1  9 lb    7   1 2     ft  ω2  + (0.093168)ω2 2  32.2    12   2

T2 = 0.094138ω22

Conservation of energy: T1 + V1 = T2 + V2 0 + 13.438 = 0.094138 ω22 + 0.9375

ω22 = 132.79 ω2 = 11.524 rad/s

ω2 = 11.52 rad/s



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PROBLEM 17.19 A slender 9 lb rod can rotate in a vertical plane about a pivot at B. A spring of constant k = 30 lb/ft and of unstretched length 6 in. is attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated through 90°.

SOLUTION

Position 1:

CD = 142 + 52 = 14.866 in. Unstretched Length

Spring:

 x1 = CD − ( 6 in. ) = 14.866 − 6 = 8.8661 in. = 0.73884 ft

Ve =

Gravity:

Kinetic energy:

1 2 1 kx1 = (30 lb/ft)(0.73884)2 = 8.1882 lb ⋅ ft 2 2

 7  Vg = Wh = 9 lb =  − ft  = −5.25 lb ⋅ ft  12  V1 = Ve + Vg = 8.1882 lb ⋅ ft − 5.25 lb ⋅ ft = 2.9382 lb ⋅ ft

T1 = 0

Position 2: Spring:

x2 = 9 in. − 6 in. = 3 in. = 0.25 ft Ve =

1 2 1 kx2 = (30 lb/ft)(0.25 ft) 2 = 0.9375 lb ⋅ ft 2 2

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PROBLEM 17.19 (Continued)

Gravity:

Vg = Wh = 0 V2 = Ve + Vg = 0.9375 lb ⋅ ft

Kinetic energy:

 7  v2 = rω2 =  ft  ω2  12  1 1  9 lb  I = mL2 =  (2 ft)2 = 0.093168 slug ⋅ ft 2 12 12  32.2  1 1 T2 = mv22 + I ω22 2 2 2

=

1  9 lb    7   1 2     ft  ω2  + (0.093168)ω2 2  32.2    12   2

T2 = 0.094138ω22

Conservation of energy: T1 + V1 = T2 + V2 0 + 2.9382 = 0.094138ω22 + 0.9375

ω22 = 21.253 ω2 = 4.6101 rad/s

ω2 = 4.61 rad/s



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PROBLEM 17.20 A 160-lb gymnast is executing a series of full-circle swings on the horizontal bar. In the position shown he has a small and negligible clockwise angular velocity and will maintain his body straight and rigid as he swings downward. Assuming that during the swing the centroidal radius of gyration of his body is 1.5 ft, determine his angular velocity and the force exerted on his hands after he has rotated through (a) 90°, (b) 180°.

SOLUTION Position 1. (Directly above the bar). Elevation:

h1 = 3.5 ft

Potential energy:

V1 = Wh1 = (160 lb)(3.5 ft) = 560 ft ⋅ lb

Speeds:

ω1 = 0, v1 = 0

Kinetic energy:

T1 = 0

(a)

Position 2. (Body at level of bar after rotating 90°). Elevation:

h2 = 0.

Potential energy:

V2 = 0

Speeds:

v2 = 3.5ω2 .

Kinetic energy:

T2 =

1 1 mv22 + mk 2ω22 2 2 1  160  1  160  T2 =  (3.5ω2 )2 +  (1.5)2ω22   2  32.2  2  32.2  = 36.025ω22

Principle of conservation of energy. T1 + V1 = T2 + V2 : 0 + 560 = 36.025ω22

ω22 = 15.545

ω2 = 3.94 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1729

PROBLEM 17.20 (Continued)

at = 3.5α

Kinematics:

an = 3.5ω22 = (3.5)(15.545) = 54.407 ft/s 2  160   160  ΣM 0 = Σ ( M 0 )eff : (3.5)(160) =  (3.5)(3.5α ) +  (1.5) 2 α    32.2   32.2 

α = 7.7724 rad/s 2 ΣFx = man : ΣFy = −mat :

at = 27.203 ft/s 2

 160  Rx =   (54.407) = 270.35 lb  32.2   160  Ry − 160 = −   (27.203)  32.2 

Ry = 24.83 lb

(b)

R = 271 lb

5.25° 

Position 3. (Directly below bar after rotating 180°). Elevation:

h3 = −3.5 ft.

Potential energy:

V3 = Wh3 = (160)(−3.5) = −560 ft ⋅ lb

Speeds:

v3 = 3.5ω3 .

Kinetic energy:

T3 = 36.025ω32

Principle of conservation of energy. T1 + V1 = T3 + V3 : 0 + 560 = 36.025ω32 − 560

ω32 = 31.09 Kinematics: From

ω3 = 5.58 rad/s



an = (3.5)(31.09) = 108.81 ft/s 2 ΣM 0 = Σ( M 0 )eff

α = 0,

and

at = 0

ΣFx = 0, Rx = 0

 160  ΣFy = man : Ry − 160 =   (108.81)  32.2 

Ry = 700.62 lb

R = 701 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1730

PROBLEM 17.21 A collar with a mass of 1 kg is rigidly attached at a distance d = 300 mm from the end of a uniform slender rod AB. The rod has a mass of 3 kg and is of length L = 600 mm. Knowing that the rod is released from rest in the position shown, determine the angular velocity of the rod after it has rotated through 90°.

SOLUTION Kinematics. L ω 2

Rod

vR =

Collar

vC = d ω

Position 1.

ω=0 T1 = 0 V1 = 0

Position 2.

T2 =

1 1 1 mR vR2 + I Rω 2 + mC vC2 2 2 2 2

1 1 1 1 L   = mR  ω  +  mR L2  ω 2 + mC d 2ω 2 2 2  12 2 2   1 1 = mR L2ω 2 + mC d 2ω 2 6 2 L V2 = −WC d − WR 2 1 L 1  T1 + V1 = T2 + V2 : 0 + 0 =  mR L2 + mC d 2  ω 2 − WC d − WR 2 2 6 

ω2 =

3(2WC d + WC L) 3mC d + mR L 2

2

=

3g (2mC d + mR L)

(1)

3mC d 2 + mR L2

Data:

mC = 1 kg, d = 0.3 m, mR = 3 kg, L = 0.6 m

From Eq. (1),

ω 2 = 3(9.81) 

 (2)(1)(0.3) + 3(0.6)  2 2  3(1)(0.3) + 3(0.6) 

= 52.32

ω = 7.23 rad/s



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PROBLEM 17.22 A collar with a mass of 1 kg is rigidly attached to a slender rod AB of mass 3 kg and length L = 600 mm. The rod is released from rest in the position shown. Determine the distance d for which the angular velocity of the rod is maximum after it has rotated 90°.

SOLUTION Kinematics. L ω 2

Rod

vR =

Collar

vC = d ω

Position 1.

ω=0 T1 = 0 V1 = 0

Position 2.

T2 =

1 1 1 mR vR2 + I Rω 2 + mC vC2 2 2 2 2

1 1 1 1 L   = mR  ω  +  mR L2  ω 2 + mC d 2ω 2 2 2  12 2 2   1 1 = mR L2ω 2 + mC d 2ω 2 6 2 L V2 = −WC d − WR 2 1 L 1  T1 + V1 = T2 + V2 : 0 + 0 =  mR L2 + mC d 2  ω 2 − WC d − WR 2 2 6 

ω2 = Let

x=

3(2WC d + WC L) 3mC d + mR L 2

2

=

3g (2mC d + mR L) 3mC d 2 + mR L2

(1)

d . L

mR mC 3g ω2 = ⋅ m L 3x 2 + R mC 2x +

Data:

mC = 1 kg, mR = 3 kg Lω 2 2x + 3 = 2 3g 3x + 3

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PROBLEM 17.22 (Continued)

Lω 2 /3g is maximum. Set its derivative with respect to x equal to zero. d  Lω 2  dx  3 g

 (3x 2 + 3)(2) − (2 x + 3)(6 x) =0  = (3x 2 + 3) 2 

−6 x 2 − 18 x + 6 = 0

Solving the quadratic equation x = −3.30 and

x = 0.30278

d = 0.30278 L = (0.30278)(0.6)  = 0.1817 m

d = 181.7 mm 

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PROBLEM 17.23 Two identical slender rods AB and BC are welded together to form an L-shaped assembly. The assembly is pressed against a spring at D and released from the position shown. Knowing that the maximum angle of rotation of the assembly in its subsequent motion is 90° counterclockwise, determine the magnitude of the angular velocity of the assembly as it passes through the position where rod AB forms an angle of 30° with the horizontal.

SOLUTION Moment of inertia about B.

Position 2.

1 1 I B = m AB l 2 + mBC l 2 3 3

θ = 30° V2 = WAB (hAB ) 2 + WBC (hBC ) 2 l  l  sin 30° + WBC  − cos 30°  2  2  1 1 T2 = I Bω22 = (m AB + mBC )l 2ω22 2 6 = WAB

Position 3.

θ = 90° V3 = WAB

l T3 = 0 2

Conservation of energy. T2 + V2 = T3 + V3 : 1 l l l ( mAB + mBC )l 2ω22 + WAB sin 30° − WBC cos 30° = 0 + WAB 6 2 2 2 3 W (1 − sin 30°) + WBC cos 30° ω22 = ⋅ AB l mAB + mBC 3g = [1 − sin 30° + cos 30°] 2 l g 9.81  ω2 = 7.09 rad/s  = 2.049 = 2.049 = 50.25 l 0.4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1734

PROBLEM 17.24 The 30-kg turbine disk has a centroidal radius of gyration of 175 mm and is rotating clockwise at a constant rate of 60 rpm when a small blade of weight 0.5 N at Point A becomes loose and is thrown off. Neglecting friction, determine the change in the angular velocity of the turbine disk after it has rotated through (a) 90°, (b) 270°.

SOLUTION m A = 51 grams = 0.051 kg

Mass of blade.

m A g = (0.051)(9.81) = 0.5 N

Weight of blade. Moment of inertia about O.

I O = mk 2 − mA r22 = 30(0.175) 2 − 51 × 10−3 (0.3)2 = 0.91416 kg ⋅ m 2

Location of mass center for the position shown. (m − mA ) x = − mA rA

Position 1.

x =−

m A rA m − mA

θ = 0°, ω1 = 60 rpm = 2π rad/s T1 =

Kinetic energy:

1 I Oω12 2

Center of gravity lies at the level of Point O.

h1 = 0

Potential energy:

V1 = (mg − m A g )h1 = 0

(a)

θ = 90°

Position 2.

T2 =

Kinetic energy:

Center of gravity lies a distance

h2 =

Potential energy:

1 I Oω 22 2

m ArA above Point O. m − mA m ArA m − mA

V2 = (mg − m A g )h2 = mA grA = (0.5)(0.3) = 0.150 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1735

PROBLEM 17.24 (Continued) T1 + V1 = T2 + V2 :

Conservation of energy.

1 1 I Oω12 + 0 = I Oω 22 + V2 2 2

ω22 = ω12 −

2V2 (2)(0.15) − (2π ) 2 − IO 0.91416

ω2 = 6.257016 rad/s

Δω = ω2 − ω1 = 6.257016 − 2π = −0.02617 rad/s

Δω = −0.250 rpm 

(b)

Position 3.

θ = 270°

Kinetic energy:

T3 =

Center of gravity lies a distance

1 I Oω32 2

mArA below Point O. m − mA h3 = −

Potential energy: Conservation of energy.

m ArA m − mA

V3 = (mg − mA g )h3 = −mA grA = −(0.5)(0.3) = −0.15 N ⋅ m T1 + V1 = T3 + V3 :

1 1 I Oω12 + 0 = I Oω32 + V3 2 2

ω32 = ω12 −

2V3 (2)(−0.15) = (2π )2 − IO 0.91416

ω3 = 6.309246 rad/s Δω = ω 3 − ω1 = 6.309246 − 2π = 0.026061 rad/s

Δω = 0.249 rpm 

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PROBLEM 17.25 A rope is wrapped around a cylinder of radius r and mass m as shown. Knowing that the cylinder is released from rest, determine the velocity of the center of the cylinder after it has moved downward a distance s.

SOLUTION Point C is the instantaneous center. v = rω

Position 1. At rest.

ω=

v r

T1 = 0

Position 2. Cylinder has fallen through distance s. T2 =

1 1 mv 2 + I ω 2 2 2

1 11  v  mv 2 +  mr 2   2 2 2  r  3 = mv 2 4

2

=

Work.

U1→2 = mgs

Principle of work and energy. 3 mv 2 4 4 gs  v2 = 3

T1 + U1→ 2 = T2: 0 + mgs =

v=

4 gs  3

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1737

PROBLEM 17.26 Solve Problem 17.25, assuming that the cylinder is replaced by a thin-walled pipe of radius r and mass m. PROBLEM 17.25 A rope is wrapped around a cylinder of radius r and mass m as shown. Knowing that the cylinder is released from rest, determine the velocity of the center of the cylinder after it has moved downward a distance s.

SOLUTION Point C is the instantaneous center. v = rω

ω=

v r

T1 = 0

Position 1. At rest.

Position 2. Cylinder has fallen through distance s. T2 = =

1 1 mv 2 + I ω 2 2 2 1 1 v  mv 2 + (mr 2 )   2 2 r

2

= mv 2 U1→2 = mgs

Work. Principle of work and energy.

T1 + U1→2 = T2 : 0 + mgs = mv 2 v 2 = gs

v = gs 

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PROBLEM 17.27 A 45-lb uniform cylindrical roller, initially at rest, is acted upon by a 20-lb force as shown. Knowing that the body rolls without slipping, determine (a) the velocity of its center G after it has moved 5 ft, (b) the friction force required to prevent slipping.

SOLUTION Since the cylinder rolls without slipping, the point of contact with the ground is the instantaneous center. Kinematics:

v = rω

Position 1. At rest.

T1 = 0

Position 2.

s = 5 ft vG = v T2 =

ω=

vG r

1 1 mv 2 + I ω 2 2 2 2

1 11  v  = mvG2 +  mr 2  G  2 22  r  3 3  45  2 = mvG2 =  vG = 1.04815vG2  4 4  32.2 

Work: (a)

U1→2 = Ps = (20)(5) = 100 lb ⋅ ft.

F f does no work.

Principle of work and energy. T1 + U1→ 2 = T2 : 0 + 100 = 1.0481vG2 vG2 = 95.407

(b)

Since the forces are constant,

vG = 9.77 ft/s



aG = a = constant vG2 2s 95.407 = (2)(5)

aG =

= 9.5407 ft/s 2 ΣFx = ma : P − F f = ma F f = P − ma

 45  = 20 −   (9.5407)  32.2 

F f = 6.67 lb



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PROBLEM 17.28 A small sphere of mass m and radius r is released from rest at A and rolls without sliding on the curved surface to Point B where it leaves the surface with a horizontal velocity. Knowing that a = 1.5 m and b = 1.2 m, determine (a) the speed of the sphere as it strikes the ground at C, (b) the corresponding distance c.

SOLUTION

U1→2 = mga

Work: Kinetic energy:

T1 = 0

Rolling motion at position 2.

v2 = rω or 60ω = T2 =

v r

1 1 mv 2 + I ω 2 2 2 2

=

1 12 7  v  mv 2 +  mr 2   = mv 2 2 2 5 10 r  

Principle of work and energy. T1 + U1→ 2 = T2 : 0 + mga =

7 mv 2 10

10 ga (10)(9.81 m/s 2 )(1.5 m) = = 21.021 m/s 2 7 7 v = 4.5849 m/s

v2 =

For path B to C the motion is projectile motion. Let t = 0 at Point B. Let y = 0 at Point C. Vertical motion:

v y = (v y )0 − gt = − gt y = y0 + (v y )0 t −

At Point C,

0=b+0− tC =

1 2 gt 2

1 2 gtC 2

2b (2)(1.2 m) = = 0.49462 s g 9.81 m/s 2

(v y )C = − gtC = −(9.81 m/s 2 )(0.49462 s) = −4.8522 m/s

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PROBLEM 17.28 (Continued)

Horizontal motion: Let the x coordinate point to the left with origin below B. vx = (vx ) B = v = 4.5849 m/s

(a)

Speed at C.

vC = (vx )C2 + (v y )C2 vC = (4.5849) 2 + (4.8522) 2 vC = 6.68 m/s 

(b)

Distance c.

c = vx tC c = (4.5849 m/s)(0.49462 s)

c = 2.27 m 

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PROBLEM 17.29 The mass center G of a 3-kg wheel of radius R = 180 mm is located at a distance r = 60 mm from its geometric center C. The centroidal radius of gyration of the wheel is k = 90 mm. As the wheel rolls without sliding, its angular velocity is observed to vary. Knowing that ω = 8 rad/s in the position shown, determine (a) the angular velocity of the wheel when the mass center G is directly above the geometric center C, (b) the reaction at the horizontal surface at the same instant.

SOLUTION

v1 = ( BG )ω1 = (0.18)2 + (0.06) 2 (8) = 8 0.036 m/s v2 = 0.24ω2 m = 3 kg k = 0.09 m

Position 1.

V1 = 0 1 1 T1 = mv12 + I ω12 2 2 1 1 = (3)(8 0.036) 2 + (3)(0.09) 2 (8)2 2 2 = 4.2336 J

Position 2.

V2 = Wh = mgh = (3)(9.81)(0.06) = 1.7658 J 1 1 T2 = mv22 + I ω22 2 2 1 1 = (3)(0.24ω2 )2 + (3)(0.09) 2 ω22 2 2 = 0.09855ω22

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PROBLEM 17.29 (Continued)

(a)

Conservation of energy. T1 + V1 = T2 + V2

4.2336 J + 0 = 0.09855ω22 + 1.7658 J

ω22 = 25.041 ω2 = 5.004 rad/s (b)

ω2 = 5.00 rad/s 

Reaction at B.

man = m(CG )ω22 = (3 kg)(0.06 m)(5.00 rad/s) 2 = 4.5 N ΣFy = ma y : N − mg = −man N − (3)(9.81) = −4.5

N = 24.9 N 

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PROBLEM 17.30 A half section of pipe of mass m and radius r is released from rest in the position shown. Knowing that the pipe rolls without sliding, determine (a) its angular velocity after it has rolled through 90°, (b) the reaction at the horizontal surface at the same instant. [Hint: Note that GO = 2r/π and that, by the parallel-axis theorem, I = mr 2 − m(GO )2 .]

SOLUTION

ω1 = 0 v1 = 0 T1 = 0

Position 1.

Position 2.

Kinematics:

2  v2 = ( AG )ω2 = r 1 −  ω2  π 2

Moment of inertia: Kinetic energy:

4   2r   I = mr 2 − m(0.6) 2 = mr 2 − m   = mr 2 1 − 2  π   π  T2 =

1 1 mv22 + I ω22 2 2 2

1  2 1 4   = m 1 −  r 2ω22 + mr 2 1 − 2  ω22 2  π 2  π   1 4 4   4  = mr 2 1 − + 2  + 1 − 2   2  π π   π   =

Work: Principle of work and energy:

1 2 4 mr  2 −  π 2  

U1→2 = W (OG ) = mg

2r

π

=

2

π

mgr

T1 + U1→2 = T2 2r 1 2  4 = mr  2 −  ω22 0 + mg π 2 π   g g 2 2 ⋅ = 1.7519 ω2 = 2 r r π (1 − π )

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1744

PROBLEM 17.30 (Continued)

(a)

Angular velocity.

(b)

Reaction at A.

ω2 = 1.324

g r



Kinematics: Since O moves horizontally, (a0 ) y = 0 an = (0.6)ω22

2r  g 1.7519  r π  = 1.1153g

=

Kinetics:

ΣFy = Σ( Fy )eff : A − mg = 1.1153mg

A = 2.12mg 

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PROBLEM 17.31 A sphere of mass m and radius r rolls without slipping inside a curved surface of radius R. Knowing that the sphere is released from rest in the position shown, derive an expression (a) for the linear velocity of the sphere as it passes through B, (b) for the magnitude of the vertical reaction at that instant.

SOLUTION Kinematics: The sphere rolls without slipping. v = rω ω =

v r

Kinetic energy. T=

1 1 mv 2 + I ω 2 2 2

1 12  v  mv 2 +  mr 2   2 25  r  7 T = mv 2 10 7 T1 = 0 T2 = mv22 10

2

=

U1− 2 = mgh = mg ( R − r )(1 − cos β )

Work.

Principle of work and energy.

T1 + U1− 2 = T2 :

0 + mg ( R − r )(1 − cos β ) =

(a)

7 mv22 10 v2 =

Linear velocity at B.

10 g ( R − r )(1 − cos β )  7

Free body diagram when β = 0. ΣF = mat : at = 0 ΣM G = I α :

α =0

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PROBLEM 17.31 (Continued)

The sphere rolls so that its mass center moves on a circle of radius ρ = R − r. a = an =

v22 R−r

ΣFy = Σ( Fy )eff : N − mg = ma

 1  10  N − mg = m  g ( R − r )(1 − cos β )     R − r  7   10  N = mg 1 + (1 − cos β )  7  

(b)

N=

Vertical reaction.

1 mg[17 − 10 cos β ]  7

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PROBLEM 17.32 Two uniform cylinders, each of weight W = 14 lb and radius r = 5 in., are connected by a belt as shown. Knowing that at the instant shown the angular velocity of cylinder B is 30 rad/s clockwise, determine (a) the distance through which cylinder A will rise before the angular velocity of cylinder B is reduced to 5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.

SOLUTION Kinematics. v D = v E = rω B

Point C is the instantaneous center of cylinder A. vD rωB 1 = = ωB 2r 2 cd 1 v A = rω A = rω B 2 vD = 2v A

ωA =

Kinetic energy of the system. T=

1 1 1 mv A2 + I ω A2 + I ωB2 2 2 2 2

2

1 r 11   1  1  m  ωB  +  mr 2  ω B  +  mr 2  ωB2 2 2 2 2 2 2       7 T = mr 2ωB2 16 T=

Position 1:

(ωB )1 = 30 rad/s

Position 2:

(ωB ) 2 = 5 rad/s

(1)

Work. For the system considered, the only force which does work is the weight of disk A. U1− 2 = −Wh = −mgh

where h is the rise of cylinder A.

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PROBLEM 17.32 (Continued)

Principle of work and energy. T1 + U1− 2 = T2 :

7 7 mr 2 (ωB )12 − mgh = mr 2 (ωB ) 22 16 16 h=

7 r2 [(ωB )12 − (ωB ) 22 ] 16 g

(2)

2

h=

7 5  1 ft  [(30 rad/s) 2 − (5 rad/s)2 ] = 2.064 ft  16  12  32.2 ft/s 2 h = 2.06 ft 

(a)

Rise of cylinder A.

(b)

Tension in cord DE. Let Q be its value. Recall that vD = 2v A thus D moves twice the distance that A moves, i.e 2h 1 I (ωB )12 2 1 T2 = I (ωB ) 22 2 U1− 2 = −Q(2h) T1 =

T1 + U1− 2 = T2 1 1 I (ωB )12 − 2Qh = I (ωB )22 2 2 Qh =

1 I [(ω B )12 − (ωB )22 ] 4

(3)

Divide Equation (3) by Equation (2): Q=

1 16 g 1  1 2  16 g 2 2 =  mr  2 = mg = W I 2 4 7r 4 2 7 7  7r

Q=

2 (14 lb) 7

(4) Tension = Q = 4.00 lb 

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PROBLEM 17.33 Two uniform cylinders, each of weight W = 14 lb and radius r = 5 in., are connected by a belt as shown. If the system is released from rest, determine (a) the velocity of the center of cylinder A after it has moved through 3 ft, (b) the tension in the portion of belt connecting the two cylinders.

SOLUTION Kinematics. v D = v E = rω B

Point C is the instantaneous center of cylinder A. vD rω 1 = B = ωB 2r 2 CD 1 v A = rω A = rωB 2 v D = 2v A

ωA =

Kinetic energy of the system. T=

1 1 1 mv A2 + I ω A2 + I ωB2 2 2 2 2

2

1 r 11 11   1   m  ω B  +  mr 2  ωB  +  mr 2  ωB2 2 2 2 2 2 2 2       7 T = mr 2ωB2 16

T=

(1)

T1 = 0

Position 1:

At rest

Position 2:

Center of cylinder C has moved 3 ft .

Work. For the system considered, the only force which does work is the weight of disk A. U1− 2 = Wh = (14 lb)(3 ft) = 42 ft ⋅ lb

where h is the distance that cylinder A falls.

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PROBLEM 17.33 (Continued)

Principle of work and energy: 2

T1 + U1→2 = T2 : 0 + 42 ft ⋅ lb =

7 14 lb  5  ft (ω B ) 22 16 32.2 ft/s 2  12 

(ωB ) 2 = 35.662 rad/s vA =

1 1 5  rωB =  ft  (35.66 rad/s) 2 2  12 

(a)

Velocity of A.

(b)

Tension in cord DE. Let Q be its value.

v A = 7.43 ft/s 

Recall that vD = 2v A thus D moves twice the distance that A moves, i.e 2h T1 = 0 1 I (ωB ) 22 2 = Q (2h)

T2 = U1− 2

T1 + U1− 2 = T2 0 + Qh = Q=

11W 2 2 r  ωB  2 2 g  1 W 2 ω B2 r 4 g h 2

1 14 lb  5  (35.662 rad/s) 2 = ft 4 32.2 ft/s 2  12  6 ft = 4.00 lb Q = 4.00 lb. 

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PROBLEM 17.34 A bar of mass m = 5 kg is held as shown between four disks each of mass m′ = 2 kg and radius r = 75 mm. Knowing that the forces exerted on the disks are sufficient to prevent slipping and that the bar is released from rest, for each of the cases shown determine the velocity of the bar after it has moved through the distance h.

SOLUTION Let v be the velocity of the bar ( v = v ), v′ be the velocity of the mass center G of the upper left disk, ( v′ = v′ ) and ω be its angular velocity.

For all three arrangements, the magnitudes of mass center velocities are the same for all disks. Likewise, the angular speeds are the same for all disks. Moment of inertia of one disk.

I =

1 m′r 2 2

Kinetic energy.

T =

1 2 mv + 2

T =

1  1 11 (5)v 2 + 4  (2)(v′) 2 +   (2)r 2ω 2  2 22 2 

1 1  4  m′(v′) 2 + I ω 2  2 2 

= 2.5v 2 + 4(v′) 2 + 2r 2ω 2

T1 = 0

Position 1.

Initial at rest position.

Position 2.

Bar has moved down a distance h. All the disks move down a distance h′.

Work

U1→ 2 = mgh + 4m′gh′ = 5 gh + 8 gh′

Kinematics and kinetic energy for case (a). The mass center of each disk is not moving. v′ = 0,

h′ = 0

v r

rω = v

ω=

T2a = 2.5v 2 + 0 + 2v 2 = 4.5v 2

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PROBLEM 17.34 (Continued)

Kinematics and kinetic energy for case (b). The instantaneous center C of a typical disk lies at its point of contact with the fixed wall.

ω=

v 2r

v′= rω =

1 v, 2

h′ =

1 h 2

2

2

1  1  T2b = 2.5v 2 + (4)  v  + 2  v  = 4.0 v 2 2  2 

Kinematics and kinetic energy for case (c). The mass center of each disk moves with the bar. v′ = v,

h′′ = h

The instantaneous center C of a typical disk lies at its point of contact with the fixed wall. v′= rω = v, T2c = 2.5v 2 + (4)v 2 + 2v 2 = 8.5 v 2

Principle of Work and Energy.

T1 + U1→ 2 = T2

(a)

0 + 5 gh + 0 = 4.5v 2

v = 1.054 gh 

(b)

0 + 5 gh + 4 gh = 4.0v 2

v = 1.500 gh 

(c)

0 + 5 gh + 8gh = 8.5v 2

v = 1.237 gh 

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PROBLEM 17.35 The 5-kg rod BC is attached by pins to two uniform disks as shown. The mass of the 150-mm-radius disk is 6 kg and that of the 75-mm-radius disk is 1.5 kg. Knowing that the system is released from rest in the position shown, determine the velocity of the rod after disk A has rotated through 90°.

SOLUTION T1 = 0

Position 1.

Position 2. Kinematics.

vB = v AB vC = v AB

Kinetic energy.

T2 =

vB v AB = v A = 2vB = 2v AB BE 0.075 m v v AB ωC = C = ω AB = 0 CF 0.075 m

ωA =

1 1 1 1 1 2 m Av A2 + I Aω A2 + m AB v AB + mB vB2 + I BωB2 2 2 2 2 2

1 1  v  = (6 ft/s)(2v AB ) 2 + (6 kg)(0.15 m) 2  AB  + (5 kg)v 2AB 2  2  0.075  2

+ (1.5 kg)(v AB ) 2 +

2 1  v   (1.5 kg)(0.075 m) 2  AB   2  0.075  

1 2 [ 24 + 12 + 5 + 1.5 + 0.75] vAB 2 2 T2 = 21.625 v AB =

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PROBLEM 17.35 (Continued)

Work:

U1→2 = WAB (0.1125 m − 0.075 m) = (5 kg)(9.81)(0.0375 m)

U1→2 = 1.8394 J

Principle of work and energy:

T1 + U1→2 = T2 2 0 + 1.8394 J = 21.625 v AB 2 v AB = 0.08506

v AB = 0.2916 m/s vAB = 292 mm/s

Velocity of the rod.



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PROBLEM 17.36 The motion of the uniform rod AB is guided by small wheels of negligible mass that roll on the surface shown. If the rod is released from rest when θ = 0, determine the velocities of A and B when θ = 30°.

SOLUTION

θ =0 v A = vB = 0 ω=0 T1 = 0 V1 = 0

Position 1.

θ = 30°

Position 2.

Kinematics. Locate the instantaneous center C. Triangle ABC is equilateral. v A = vB = Lω vG = Lω cos 30° I =

Moment of inertia. T2 =

Kinetic energy.

1 ml 2 12

1 2 1 1 1 1  mvG + I ω 2 : T2 = m( Lω cos 30°) 2 +  mL2  ω 2 2 2 2 2  12  =

5 2 2 ml ω 12

V2 = − mg

Potential energy.

L 1 sin 30° = − mgL 2 4

Conservation of energy. T1 + V1 = T2 + V2 : 0 + 0 =

5 1 mL2ω 2 − mgL 12 4

ω 2 = 0.6

g L

g L v A = 0.775 gL

ω = 0.775

vB = 0.775 gL







vA = 0.775 gL vB = 0.775 gL

 60° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1756

PROBLEM 17.37 A 5-m long ladder has a mass of 15 kg and is placed against a house at an angle θ = 20°. Knowing that the ladder is released from rest, determine the angular velocity of the ladder and the velocity of A when θ = 45°. Assume the ladder can slide freely on the horizontal ground and on the vertical wall.

SOLUTION Kinematics: Let v A = v A

, v B = vB , and ω = ω . Locate the instantaneous

center C by drawing AC perpendicular to vA and BC perpendicular to vB. Triangle GCB is isosceles. GA = GB = GC = L /2. The velocity of the mass center G is v = vG = Lω /2

Kinetic energy:

1 1 mv 2 + I ω 2 2 2 1 1  =  I + mL2  ω 2 2 4 

T=

Since the ladder can slide freely, the friction forces at A and B are zero. Use the principle of conservation of energy. T1 + V1 = T2 + V2 :

Potential energy: Use the ground as the datum. V = mgh

where

h=

L cos θ 2

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PROBLEM 17.37 (Continued)

Position 1.

θ = 20°;

rest (T1 = 0)

Position 2.

θ = 45°;

ω =?

0 + mg

L L 1 1  cos 20° =  I + mL2  ω 2 + mg cos 45° 2 2 4 2  m = 15 kg. L = 5 m

Data:

I =

Assume I +

g = 9.81 m/s 2

1 1 mL2 = (15 kg)(5 m) 2 = 31.25 kg ⋅ m 2 12 12

1 2 1 mL = 31.25 kg ⋅ m 2 + (15 kg)(5 m) 2 = 125 kg ⋅ m 2 4 4

(15 kg)(9.81 m/s 2 )(2.5 m)(cos 20° − cos 45°) =

ω 2 = 1.3690 rad 2 /s2

1 (125 kg ⋅ m 2 )ω 2 2

ω = 1.17004 rad/s ω = 1.170 rad/s

Angular velocity.



Velocity of end A. v A = ω L cos θ = (1.17004 rad/s)(5 m) cos 30°

v A = 5.07 m/s



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PROBLEM 17.38 A long ladder of length l, mass m, and centroidal mass moment of inertia I is placed against a house at an angle θ = θ0. Knowing that the ladder is released from rest, determine the angular velocity of the ladder when θ = θ 2 . Assume the ladder can slide freely on the horizontal ground and on the vertical wall.

SOLUTION Kinematics: Let v A = v A

, v B = vB , and ω = ω . Locate the instantaneous

center C by drawing AC perpendicular to vA and BC perpendicular to vB. Triangle GCB is isosceles. GA = GB = GC = L /2. The velocity of the mass center G is v = vG = Lω /2

Kinetic energy:

1 1 mv 2 + I ω 2 2 2 1 1 2 2 =  I + mL  ω 2 4 

T=

Since the ladder can slide freely, the friction forces at A and B are zero. Use the principle of conservation of energy. T1 + V1 = T2 + V2 :

Potential energy: Use the ground as the datum. V = mgh where Position 1.

h=

L cos θ 2

θ = θ 0 ; rest (T1 = 0)

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PROBLEM 17.38 (Continued)

Position 2.

θ = θ2 ; ω = ? 0 + mg

Assume

L L 1 1  cos θ =  I + mL2  ω 2 + mg cos θ 2 2 2 4 2 

1 mL2 12 1 2 1 2 I + mL = mL 4 3 3 g ω2 = (cos θ0 − cos θ 2 ) L I =

ω = 3g (cos θ0 − cos θ 2 )/L

Angular velocity.



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PROBLEM 17.39 The ends of a 9-lb rod AB are constrained to move along slots cut in a vertical plate as shown. A spring of constant k = 3 lb/in. is attached to end A in such a way that its tension is zero when θ = 0. If the rod is released from rest when θ = 50°, determine the angular velocity of the rod and the velocity of end B when θ = 0.

SOLUTION

L ω2 2 vB = Lω2 v2 =

x1 = L − L cos 50° = (25 in.)(1 − cos 50°) = 8.9303 in. L 1 sin 50° + kx12 2 2 25 in. 1   2 V1 = −(9 lb)   sin 50° + (3 lb/in.)(8.9303 in.) 2  2  = −86.18 + 119.63 = 33.45 in. ⋅ lb = 2.787 ft ⋅ lb. T1 = 0

Position 1.

V1 = −W

Position 2.

V2 = (Vg ) 2 + (Ve ) 2 = 0 T2 =

1 1 mv22 + I ω22 2 2 2

=

1 L  1 1  m ω2 +  mL2  ω22 2  2  2  12 

=

1 2 2 1  9 lb  25 in.  2 2 mL ω2 =   ω2 = 0.2022ω2 6 6  32.2   12 

2

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PROBLEM 17.39 (Continued)

Conservation of energy:

T1 + V1 = T2 + V2 0 + 2.787 ft ⋅ lb = 0.2022ω22

ω22 = 13.7849 ω2 = 3.713 rad/s Velocity of B:

ω 2 = 3.71 rad/s



 25 in.  vB = Lω2 =   (3.713 rad/s)  12  = 7.735 ft/s

vB = 7.74 ft/s 

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PROBLEM 17.40 The ends of a 9-lb rod AB are constrained to move along slots cut in a vertical plane as shown. A spring of constant k = 3 lb/in. is attached to end A in such a way that its tension is zero when θ = 0 . If the rod is released from rest when θ = 0, determine the angular velocity of the rod and the velocity of end B when θ = 30°.

SOLUTION Moment of inertia. Rod. Position 1.

I =

1 mL2 12

θ1 = 0

v1 = 0

ω1 = 0

h1 = elevation above slot.

h1 = 0

e1 = elongation of spring.

e1 = 0

1 1 mv12 + I ω12 = 0 2 2 1 2 V1 = ke1 + Wh1 = 0 2 T1 =

Position 2.

θ = 30° e2 + L cos 30° = L e2 = L(1 − cos 30°) 1 L sin 30° = − L 2 4 1 2 1 2 1 V2 = ke2 + Wh2 = k L (1 − cos 30°) 2 − WL 2 2 4 h2 = −

Kinematics. Velocities at A and B are directed as shown. Point C is the instantaneous center of rotation. From geometry, b = L2 . L ω 2 vB = ( L cos 30°)ω v = bω =

T2 =

1 1 mv 2 + I ω 2 2 2 2

1 L  1 1  m  ω  +  mL2  2 2  2  12  1W 2 2 Lω = 6 g =

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PROBLEM 17.40 (Continued)

Conservation of energy. T1 + V1 = T2 + V2 : 0 + 0 =

ω2 = Data:

1W 2 2 1 2 1 L ω + kL (1 − cos 30°) 2 − WL 6 g 2 4 3g 3 kg (1 − cos 30°)2 − 2L W

W = 9 lb g = 32.2 ft/s 2 L = 25 in. = 2.0833 ft k = 3 lb/in. = 36 lb/ft (3)(32.2) (3)(36)(32.2)(1 − cos 30°) 2 − (2)(2.0833) 9 = 16.2484

ω2 =

vB = (2.0833)(cos30°)(4.03)

ω = 4.03 rad/s



vB = 7.27 ft/s 

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PROBLEM 17.41 The motion of a slender rod of length R is guided by pins at A and B which slide freely in slots cut in a vertical plate as shown. If end B is moved slightly to the left and then released, determine the angular velocity of the rod and the velocity of its mass center (a) at the instant when the velocity of end B is zero, (b) as end B passes through Point D.

SOLUTION The rod AB moves from Position 1, where it is nearly vertical, to Position 2, where vB = 0. In Position 2, vA is perpendicular to both CA and AB, so CAB is a straight line of length 2L and slope angle 30°. In Position 3 the end B passes through Point D.

Position 1:

T1 = 0

V1 = Wh = mg

R 2

Position 2: Since instantaneous center is at B, 1 Rω2 2 1 1 T2 = mv22 + I ω22 2 2 v2 =

2

1 1 1 1   m Rω2  +  mR 2  ω22 2  2 2 12    1 = mR 2ω22 6 R V2 = Wh2 = mg 4 =

Position 3:

V3 = 0

Since both vA and vB are horizontal, ω3 = 0 T3 =

(1)

1 mv22 2

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PROBLEM 17.41 (Continued)

(a)

From 1 to 2: Conservation of energy T1 + V1 = T2 + V2 : 0 +

1 1 1 mgR = mR 2ω22 + mgR 2 6 4 3g 2R 3g ω2 = 2R

ω22 =

v2 =

(b)

ω2 = 1.225

1 1 3 3 Rω2 = gR = gR 2 2 2 8

vR = 0.612 gR

g R



60° 

From 1 to 3: Conservation of energy ω3 = 0 

From Eq. (1) we have T1 + V1 = T3 + V3 : 0 +

1 1 mgR = mv32 2 2 2 v3 = gR

v3 = gR



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PROBLEM 17.42 Each of the two rods shown is of length L = 1 m and has a mass of 5 kg. Point D is connected to a spring of constant k = 20 N/m and is constrained to move along a vertical slot. Knowing that the system is released from rest when rod BD is horizontal and the spring connected to Point D is initially unstretched, determine the velocity of Point D when it is directly to the right of Point A.

SOLUTION I =

Moments of inertia.

1 1 mL2 , I A = mL2 12 3

Use the principle of conservation of energy applied to the system consisting of both rods. Use the level at A as the datum for the potential energy of each rod. Position 1.

(no motion) T1 = 0 1 1  V1 = mg  L  + mgL + kx12 2 2   3 1 = mgL + kx12 2 2

Position 2. L L sin 60° + mg sin 60° 2 2 3 1 = mgL + kx22 2 2

V2 = mg

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PROBLEM 17.42 (Continued)

Kinematics.

ω AB = ω AB vB = Lω AB

v B = Lω AB

30°

v D = vD

Locate the instantaneous center C of rod BD by drawing BC perpendicular to vB and DC perpendicular to vD. Point C coincides with Point A in position 2. Let

ω BD = ωBD

vB = ω AB L L vE = ω AB 2

ωBD =

vG = ( L sin 60°)ωBD =

3 Lω AB 2

vD = LωBD = Lω AB T2 =

(1)

1 1 1 2 2 I Aω AB + I ω BD + mvG2 2 2 2

 11 1 1 1  3  2  2 ω =  mL2  ω AB +  mL2  ω AB + m  AB 23 2  12 2  2   

2

7 1 1 3 2 2 2 = + +  mL ω AB = mL2ω AB 12  6 24 8 

Principle of conservation of energy. T1 + V1 = T2 + V2 : 0 +

Data:

3 1 7 3 1 2 mgL + kx12 = mL2ω AB + mgL + kx22 2 2 12 2 2   7 3 3 1 2 mL2ω AB mgL − k ( x22 − x12 ) = −  2 2  12 2  

(2)

m = 5 kg, L = 1 m, g = 9.81 m/s 2 k = 20 N ⋅ m, x1 = 0, x2 = L = 1 m 3 3 2  −  mgL = (0.63397)(5 kg)(9.81 m/s )(1 m) = 31.096 J 2 2   1 1 − k ( x22 − x12 ) = (20 N/m)(1 m) 2 = −10 J 2 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1768

PROBLEM 17.42 (Continued)

By Eq. (2),

7  35  2 2 mL2ω AB =  kg ⋅ m 2  ω AB = 21.096 J 12 12   2 ω AB = 7.2329 rad 2 /s 2 ω AB = 2.6894 rad/s

By Eq. (1),

vD = (1 m)(2.6894 rad/s)

v D = 2.69 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1769

PROBLEM 17.43 The 4-kg rod AB is attached to a collar of negligible mass at A and to a flywheel at B. The flywheel has a mass of 16 kg and a radius of gyration of 180 mm. Knowing that in the position shown the angular velocity of the flywheel is 60 rpm clockwise, determine the velocity of the flywheel when Point B is directly below C.

SOLUTION Moments of inertia. Rod AB:

Flywheel:

1 m AB L2AB 12 1 = (4 kg)(0.72 m)2 12 = 0.1728 kg ⋅ m 2

I AB =

I C = mk 2 = (16 kg)(0.18 m) 2 = 0.5184 kg ⋅ m 2

Position 1. As shown.

ω = ω1 0.24 β = 19.471° 0.72 1 h1 = (0.72) cos β = 0.33941 m 2 V1 = WAB h1 = (4)(9.81)(0.33941) = 13.3185 J

sin β =

Kinematics. Bar AB is in translation.

vB = rω1 = 0.24ω1

ω AB = 0,

v = vB

1 1 1 2 m AB v 2 + IABω AB + I C ω12 2 2 2 1 1 = (4)(0.24ω1 ) 2 + 0 + (0.5184)ω12 2 2 2 = 0.3744ω1

T1 =

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PROBLEM 17.43 (Continued)

Position 2. Point B is directly below C. 1 LAB − r 2 1 = (0.72) − 0.24 2 = 0.12 m V2 = WAB h2 h2 =

= (4)(9.81)(0.12) = 4.7088 J

Kinematics.

vB = rω2 = 0.24ω2 vB = 0.33333ω2 0.72 1 v = vB = 0.12ω2 2 1 1 1 2 + I C ω22 T2 = mAB v 2 + IABω AB 2 2 2 1 1 1 = (4)(0.12ω2 )2 + (0.1728)(0.33333ω2 ) 2 + (0.5184)ω22 2 2 2 2 = 0.2976ω2

ω AB =

Conservation of energy.

T1 + V1 = T2 + V2 : 0.3744ω12 + 13.3185 = 0.2976ω22 + 4.7088

Angular speed data:

ω1 = 60 rpm = 2π rad/s

Solving Equation (1) for ω2 ,

ω2 = 8.8655 rad/s

ω2 = 84.7 rpm

(1)



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PROBLEM 17.44 If in Problem 17.43 the angular velocity of the flywheel is to be the same in the position shown and when Point B is directly above C, determine the required value of its angular velocity in the position shown. PROBLEM 17.43 The 4-kg rod AB is attached to a collar of negligible mass at A and to a flywheel at B. The flywheel has a mass of 16 kg and a radius of gyration of 180 mm. Knowing that in the position shown the angular velocity of the flywheel is 60 rpm clockwise, determine the velocity of the flywheel when Point B is directly below C.

SOLUTION Moments of inertia. Rod AB:

Flywheel:

1 m AB L2AB 12 1 = (4 kg)(0.72 m)2 12 = 0.1728 kg ⋅ m 2

I AB =

I C = mk 2 = (16 kg)(0.18 m) 2 = 0.5184 kg ⋅ m 2

Position 1. As shown.

ω = ω1 0.24 β = 19.471° 0.72 1 h1 = (0.72) cos β = 0.33941 m 2 V1 = WAB h1 = (4)(9.81)(0.33941) = 13.3185 J

sin β =

Kinematics. Bar AB is in translation.

vB = rω1 = 0.24ω1

ω AB = 0,

v = vB

1 1 1 2 m AB v 2 + IABω AB + I C ω12 2 2 2 1 1 = (4)(0.24ω1 ) 2 + 0 + (0.5184)ω12 2 2 2 = 0.3744ω1

T1 =

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PROBLEM 17.44 (Continued)

Position 2. Point B is directly above C. 1 LAB + r 2 1 = (0.72) + 0.24 2 = 0.6 m V2 = WAB h2 h2 =

= (4)(9.81)(0.6) = 23.544 J

Kinematics.

vB = rω2 = 0.24ω2 vB = 0.33333ω2 0.72 1 v = vB = 0.12ω2 2 1 1 1 2 + I C ω22 T2 = mAB v 2 + IABω AB 2 2 2 1 1 1 = (4)(0.12ω2 )2 + (0.1728)(0.33333ω2 ) 2 + (0.5184)ω22 2 2 2 2 = 0.2976ω2

ω AB =

Conservation of energy. Angular speed data: Then,

T1 + V1 = T2 + V2 : 0.3744ω12 + 13.3135 = 0.2976ω22 + 23.544

ω2 = ω1 0.0760ω12 = + 0.4105

ω1 = 11.602 rad/s

ω1 = 110.8 rpm



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PROBLEM 17.45 The uniform rods AB and BC of masses 2.4 kg and 4 kg, respectively, and the small wheel at C is of negligible mass. If the wheel is moved slightly to the right and then released, determine the velocity of pin B after rod AB has rotated through 90°.

SOLUTION Moments of inertia. Rod AB:

IA =

1 1 mAB L2AB = (2.4)(0.360) 2 = 0.10368 kg ⋅ m 2 3 3

Rod BC:

I =

1 1 mBC L2BC = (4)(0.600) 2 = 0.1200 kg ⋅ m 2 12 12

Position 1. As shown with bar AB vertical. Point G is the midpoint of BC. V1 = mAB ghAB + mBC ghBC = (2.4)(9.81)(0.180) + (4)(9.81)(0.180) = 11.3011 J

ω BC = 0

Rod BC is at rest.

ω AB =

v = vG = vB = vC = 0

vB =0 LAB

T1 = 0

Position 2. Rod AB is horizontal.

V2 = 0

Kinematics.

ω AB = T2 =

vB vB = LAB 0.360

ω BC =

vB vB = LBC 0.600

1 vB 2

1 1 1 2 2 I Aω AB + mBC v 2 + I ωBC 2 2 2 2

=

v =

2

1 1 1  1  v   v  (0.10368)  B  + (4)  vB  + (0.1200)  B  2 0.360 2 2 2      0.600 

2

= 1.06667vB2

Conservation of energy. T1 + V1 = T2 + V2: 0 + 11.3011 = 1.06667vB2 vB = 3.25 m/s

v B = 3.25 m/s 

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PROBLEM 17.46 The uniform rods AB and BC of masses 2.4 kg and 4 kg, respectively, and the small wheel at C is of negligible mass. Knowing that in the position shown the velocity of wheel C is 2 m/s to the right, determine the velocity of pin B after rod AB has rotated through 90°.

SOLUTION Moments of inertia. Rod AB: Rod BC:

1 1 I A = m AB L2AB = (2.4)(0.36) 2 = 0.10368 kg ⋅ m 2 3 3 I =

1 1 mBC LBC 2 = (4)(0.600) 2 = 0.1200 kg ⋅ m 2 12 12

Position 1. As shown with rod AB vertical. Point G is the midpoint of BC. V1 = WAB hAB + WBC hBC = (2.4)(9.81)(0.180) + (4)(9.81)(0.180) = 11.301 J

Kinematics: At the instant shown in Position 1,

ωBC = 0 v = vG = vB = vC = 2 m/s

ω AB =

vB 2 = = 5.5556 rad/s LAB 0.36

1 1 1 2 2 IAω AB + mBC v 2 + IωBC 2 2 2 1 1 = (0.10368)(5.5556) 2 + (4)(2) 2 + 0 2 2 = 9.6 J

T1 =

Position 2. Rod AB is horizontal.

V2 = 0

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1775

PROBLEM 17.46 (Continued)

Kinematics.

ω AB =

vB v = B LAB 0.36

ωBC =

vB v = B LBC 0.60

v= T2 =

1 vB 2 1 1 1 2 2 I Aω AB + mBC v 2 + I ωBC 2 2 2 2

=

2

1 1 1  1  v   v  (0.10368)  B  + (4)  vB  + (0.12)  B  2 2 2  2  0.36   0.60 

2

= 1.0667vB2

Conservation of energy.

T1 + V1 = T2 + V2 : 9.6 + 11.301 = 1.0667vB2 + 0 vB = 4.4266 m/s

vB = 4.43 m/s 

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PROBLEM 17.47 The 80-mm-radius gear shown has a mass of 5 kg and a centroidal radius of gyration of 60 mm. The 4-kg rod AB is attached to the center of the gear and to a pin at B that slides freely in a vertical slot. Knowing that the system is released from rest when θ = 60°, determine the velocity of the center of the gear when θ = 20°.

SOLUTION Kinematics.

vA = v A vB = vB

Point D is the instantaneous center of rod AB. vA L cos θ vB = ( L sin θ )ω AB = v A tan θ

ω AB =

vG =

vA L ω AB = 2 2cos θ

Gear A effectively rolls without slipping, with Point C being the contact point. vC = 0

Angular velocity of gear

ωA =

vA . r

Potential energy: Use the level of the center of gear A as the datum. 1 L  V = −WAB  cos θ  = − m AB gL cos θ 2 2 

Kinetic energy: Masses and moments of inertia:

T=

1 1 1 1 2 m Av A2 + I Aω A2 + mAB vθ2 + I ABω AB 2 2 2 2

m A = 5 kg, m AB = 4 kg I A = m A k 2 = (5)(0.060)2 = 0.018 kg ⋅ m 2 I AB =

1 1 m AB L2 = (4)(0.320)2 = 0.03413 kg ⋅ m 2 12 12

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PROBLEM 17.47 (Continued)

Conservation of energy: Position 1:

T1 + V1 = T2 + V2

θ = 60° v A = 0 T1 = 0 1 V1 = − (4)(9.81)(0.320) cos 60° 2 = −3.1392 J

Position 2:

θ = 20° v A = ? 2

1 1 1  vA  v   T2 = (5)v A2 + (0.018)  A  + (4)  2 2 2  2 cos 20°   0.080  vA 1   + (0.03413)   2 0.320cos 20 °  

2

2

= (2.5 + 1.40625 + 0.56624 + 0.18875)v A2 = 4.66124v 2A 1 V2 = − (4)(9.81)(0.320) cos 20° 2 = −5.8998 J

Conservation of energy:

T1 + V1 = T2 + V2 0 − 3.1392 = 4.66124v 2A − 5.8998 v A2 = 0.59225 m 2 /s 2 v A = 0.770 m/s

vA = 0.770 m/s



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PROBLEM 17.48 Knowing that the maximum allowable couple that can be applied to a shaft is 15.5 kip ⋅ in., determine the maximum horsepower that can be transmitted by the shaft at (a) 180 rpm, (b) 480 rpm.

SOLUTION M = 15.5 kip ⋅ in. = 1.2917 kip ⋅ ft = 1291.7 lb ⋅ ft

(a)

ω = 180 rpm = 6π rad/s Power = M ω = (1291.7 lb ⋅ ft)(6π rad/s) = 24348 ft ⋅ lb/s 24348 550 = 44.3 hp

Horsepower =

(b)



ω = 480 rpm = 16π rad/s Power = Mω = (1291.7 lb ⋅ ft)(16π rad/s) = 64930 ft ⋅ lb/s 64930 550 = 118.1 hp

Horsepower =



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1779

PROBLEM 17.49 Three shafts and four gears are used to form a gear train which will transmit 7.5 kW from the motor at A to a machine tool at F. (Bearings for the shafts are omitted from the sketch.) Knowing that the frequency of the motor is 30 Hz, determine the magnitude of the couple which is applied to shaft (a) AB, (b) CD, (c) EF.

SOLUTION Kinematics.

Gears B and C.

ω AB = 30 Hz = 30(2π )rad/s = 60π rad/s rB = 75 mm rC = 180 mm

rBω AB = rC ωCD : (75 mm)(60π rad/s) = (180 mm)(ωCD )

Gears D and E.

ωCD = 25π rad/s rD = 75 mm rE = 180 mm

rDωCD = rE ωEF : (75 mm)(25π rad/s) = (180 mm)(ωEF )

ωEF = 10.4167π rad/s Power = 7.5 kW

(a)

Shaft AB.

Power = M ABω AB : 7500 W = M AB (60π rad/s)

M AB = 39.8 N ⋅ m 

(b)

Shaft CD.

Power = M CDωCD: 7500 W = M CD (25π rad/s)

M CD = 95.5 N ⋅ m 

(c)

Shaft EF.

Power = M EF ωEF : 7500 W = M EF (10.4167π rad/s)

M EF = 229 N ⋅ m 

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PROBLEM 17.50 The shaft-disk-belt arrangement shown is used to transmit 2.4 kW from Point A to Point D. Knowing that the maximum allowable couples that can be applied to shafts AB and CD are 25 N ⋅ m and 80 N ⋅ m, respectively, determine the required minimum speed of shaft AB.

SOLUTION Power.

2.4 kW = 2400 W M AB < 25 N ⋅ m P = M ABω AB min ω AB =

2400 P = = 96 rad/s max M AB 25

M CD < 80 N ⋅ m P = M CDωCD min ωCD =

Kinematics.

P 2400 = = 30 rad/s max M CD 80

rAω AB = rC ωCD min ω AB =

rC (min ωCD ) rA

 120  =  (30)  30  = 120 rad/s

Choose the larger value for min ω AB .

min ω AB = 120 rad/s

min ω AB = 1146 rpm 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1781

PROBLEM 17.51 The experimental setup shown is used to measure the power output of a small turbine. When the turbine is operating at 200 rpm, the readings of the two spring scales are 10 and 22 lb, respectively. Determine the power being developed by the turbine.

SOLUTION Angular velocity.

ω = 200 rpm = 20.944 rad/s

Moments about the fixed axle.

 9  M = (22 lb − 10 lb)  ft  = 9 lb ⋅ ft  12 

Power = Mω = (9)(20.994) = 188.5 lb ⋅ ft/s 188.5 lb ⋅ ft/s = 550 lb ⋅ ft/s/hp

Power = 0.343 hp 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1782

PROBLEM 17.CQ6 Slender bar A is rigidly connected to a massless rod BC in Case 1 and two massless cords in Case 2 as shown. The vertical thickness of bar A is negligible compared to L. If bullet D strikes A with a speed v0 and becomes embedded in it, how will the speeds of the center of gravity of A immediately after the impact compare for the two cases? (a) (b) (c)

Case 1 will be larger. Case 2 will be larger. The speeds will be the same.

SOLUTION Answer: (b)

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PROBLEM 17.CQ7 A 1-m long uniform slender bar AB has an angular velocity of 12 rad/s and its center of gravity has a velocity of 2 m/s as shown. About which point is the angular momentum of A smallest at this instant? (a) (b) (c) (d ) (e)

P1 P2 P3 P4 It is the same about all the points.

SOLUTION Answer: (a)

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PROBLEM 17.F1 The 350-kg flywheel of a small hoisting engine has a radius of gyration of 600 mm. If the power is cut off when the angular velocity of the flywheel is 100 rpm clockwise, draw an impulse-momentum diagram that can be used to determine the time required for the system to come to rest.

SOLUTION Answer:

Syst. Momenta1

+

Syst. Ext. Imp.1→2

=

Syst. Momenta2

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PROBLEM 17.F2 A sphere of radius r and mass m is placed on a horizontal floor with no linear velocity but with a clockwise angular velocity ω 0 . Denoting by μk the coefficient of kinetic friction between the sphere and the floor, draw the impulse-momentum diagram that can be used to determine the time t1 at which the sphere will start rolling without sliding.

SOLUTION Answer:

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PROBLEM 17.F3 Two panels A and B are attached with hinges to a rectangular plate and held by a wire as shown. The plate and the panels are made of the same material and have the same thickness. The entire assembly is rotating with an angular velocity ω0 when the wire breaks. Draw the impulse-momentum diagram that is needed to determine the angular velocity of the assembly after the panels have come to rest against the plate.

SOLUTION Answer:

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PROBLEM 17.52 The rotor of an electric motor has a mass of 25 kg, and it is observed that 4.2 min is required for the rotor to coast to rest from an angular velocity of 3600 rpm. Knowing that kinetic friction produces a couple of magnitude 1.2 N⋅m, determine the centroidal radius of gyration for the rotor.

SOLUTION Coasting time: Initial angular velocity:

t = 4.2 min = 252 s

ω1 = 3600 rpm = 120π rad/s

Principle of impulse and momentum.

Syst. Momenta1

Moments about axle A:

+

ΣSyst. Impulses. 1→2

=

Syst. Momenta 2

I ω1 − Mt = 0 I = =

Mt

ω1 (1.2 N ⋅ m)(252 s) 120π rad/s

= 0.80214 kg ⋅ m 2 I = mk 2

Radius of gyration: 

k =

I 0.80214 kg ⋅ m 2 = = 0.1791 m m 25 kg

k = 179.1 mm 



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1788

PROBLEM 17.53 A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.

SOLUTION Use the principle of impulse and momentum applied to the grinding wheel and rotor with t1 = 0

t2 = 70 s

ω1 = 3600 rpm = 120π rad/s

ω2 = 0 2

Moment of inertia.

I = mk 2 =

Syst. Momenta1

Moments about A:

+

6 lb  2  2  ft  = 0.00518 lb ⋅ ft ⋅ s 32.2 ft/s 2  12 

Syst. Ext. Imp. 1→2

=

Syst. Momenta 2

I ω1 − Mt = 0 (0.00518)(120π ) − M (70 s) = 0 M = 0.02788 lb ⋅ ft M = 0.33451 lb ⋅ in.



M = 0.335 lb ⋅ in. 



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1789

PROBLEM 17.54 A bolt located 50 mm from the center of an automobile wheel is tightened by applying the couple shown for 0.10 s. Assuming that the wheel is free to rotate and is initially at rest, determine the resulting angular velocity of the wheel. The wheel has a mass of 19 kg and has a radius of gyration of 250 mm.

SOLUTION Moment of inertia. Applied couple.

I = mk 2 = (19 kg)(0.25 m) 2 = 1.1875 kg-m 2 M = (100 N)(0.460 m) = 46 N-m

Syst. Momenta1

Moments about axle:

+

Syst. Ext. Imp.1→ 2

=

Syst. Momenta 2

0 + Mt = I ω 0 + (46 N-m)(0.10 s) = (1.1875 kg-m 2 )ω

ω = 3.87 rad/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1790

PROBLEM 17.55 Two disks of the same thickness and same material are attached to a shaft as shown. The 8-lb disk A has a radius rA = 3 in., and disk B has a radius rB = 4.5 in. Knowing that a couple M of magnitude 20 lb ⋅ in. is applied to disk A when the system is at rest, determine the time required for the angular velocity of the system to reach 960 rpm.

SOLUTION 2

r  WB =  B  WB  rA 

Weight of disk B.

2

 4.5 in.  =  (8 lb)  3 in.  = 18 lb I = I A + IB

Moment of inertia.

2

1 8 lb  3  1 18 lb  4.5  ft  + ft =  2 32.2  12  2 32.2  12 

2

= 0.04707 lb ⋅ ft ⋅ s 2

Angular velocity.

ω2 = 960 rpm = 100.53 rad/s

Moment.

M = 20 lb ⋅ in. = 1.667 lb ⋅ ft

Principle of impulse and momentum.

Syst. Momenta1 + Syst. Ext. Imp.1→2 = Moments about C: Required time.

Syst. Momenta2

0 + Mt = I ω2

I ω2 M (0.04707 lb ⋅ ft ⋅ s2 )(100.53 rad/s) = 1.667 lb ⋅ ft

t=

t = 2.839 s

t = 2.84 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1791

PROBLEM 17.56 Two disks of the same thickness and same material are attached to a shaft as shown. The 3-kg disk A has a radius rA = 100 mm, and disk B has a radius rB = 125 mm. Knowing that the angular velocity of the system is to be increased from 200 rpm to 800 rpm during a 3-s interval, determine the magnitude of the couple M that must be applied to disk A.

SOLUTION 2

Mass of disk B.

r  mB =  B  mA  rA  2

 125 mm  =  3 kg  100 mm  = 4.6875 kg

Moment of inertia.

I = I A + IB 1 1 (3 kg)(0.1 m)2 + (4.6875 kg)(0.125 m) 2 2 2 2 = 0.05162 kg ⋅ m =

Angular velocities.

ω1 = 200 rpm = 20.944 rad/s ω2 = 800 rpm = 83.776 rad/s

Principle of impulse and momentum.

Syst. Momenta1 + Syst. Ext. Imp.1→2 = Moments about B: Couple M.

Syst. Momenta2

I ω1 + Mt = I ω 2

M= =

I (ω2 − ω1) t 0.05162 kg ⋅ m 2 (83.776 rad/s − 20.944 rad/s) 3s

M = 1.081 N ⋅ m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1792

PROBLEM 17.57 A disk of constant thickness, initially at rest, is placed in contact with a belt that moves with a constant velocity v. Denoting by μk the coefficient of kinetic friction between the disk and the belt, derive an expression for the time required for the disk to reach a constant angular velocity.

SOLUTION I =

Moment of inertia.

ω2 =

Final state of constant angular velocity.

Syst. Momenta1 y components: Moments about A:

+

1 2 mr 2 v r

Syst. Ext. Imp.1→2

=

Syst. Momenta2

0 + Nt − mgt = 0 N = mg 0 + μ k Ntr = I ω2 1 mr 2 vr I ω2 v 2 t= = = μk mgr μk mgr 2 μk g

t=

v 2 μk g



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1793

PROBLEM 17.58 Disk A, of weight 5 lb and radius r = 3 in., is at rest when it is placed in contact with a belt which moves at a constant speed v = 50 ft/s. Knowing that μk = 0.20 between the disk and the belt, determine the time required for the disk to reach a constant angular velocity.

SOLUTION I =

Moment of inertia.

ω2 =

Final state of constant angular velocity.

Syst. Momenta1 y components: Moments about A:

Data:

1 2 mr 2

+

v r

Syst. Ext. Imp.1→2

0 + Nt − mgt = 0

=

Syst. Momenta2

N = mg

0 + μk Ntr = I ω2 t=

1 mr 2 v I ω2 v r = 2 = μk mgr μk mgr 2μk g

t=

v 2 μk g

v = 50 ft/s μk = 0.20 t=

50 (2)(0.20)(32.2)

t = 3.88 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1794

PROBLEM 17.59 A cylinder of radius r and weight W with an initial counterclockwise angular velocity ω0 is placed in the corner formed by the floor and a vertical wall. Denoting by μk the coefficient of kinetic friction between the cylinder and the wall and the floor derive an expression for the time required for the cylinder to come to rest.

SOLUTION I =

For the cylinder

1 2 mr , W = mg 2

Principle of impulse and momentum.

Syst. Momenta1 Linear momentum

:

+

Syst. Ext. Imp.1→2

=

Syst. Momenta2

0 + N B t − FAt = 0 N B = FA

Linear momentum

:

0 + N At + FB t − Wt = 0

N A + FB = N A + μk N B = N A + μk FA + N A + μk2 N A = W W NA = 1 + μk2 μW FA = μk N A = k 2 1 + μk μW NB = k 2 1 + μk FB =

Moments about G:

μk2W 1 + μ k2

I ω0 − FA rt − FB rt = 0 t=

I ω0 (1 + μk2 ) I ω0 = ( FA + FB )r μk (1 + μk )Wr

t=

1 + μk2 rω0  2 μk (1 + μk ) g

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1795

PROBLEM 17.60 Two uniform disks and two cylinders are assembled as indicated. Disk A has a mass of 10 kg and disk B has a mass of 6 kg. Knowing that the system is released from rest, determine the time required for cylinder C to have a speed of 0.5 m/s. Disks A and B are bolted together and the cylinders are attached to separate cords wrapped on the disks.

SOLUTION Moments of inertia. Disk A:

IA =

1 1 mA rA2 = (10 kg)(0.200 m) 2 = 0.2 kg ⋅ m 2 2 2

Disk B:

IB =

1 1 mB rB2 = (6 kg)(0.150 m) 2 = 0.0675 kg ⋅ m 2 2 2

Kinematics:

vC = vC

ω A = ωA

=

vC rA

ωB = ω A = ω v D = vD = ω rB =

rB vC rA

Principle of impulse and momentum.

Syst. Momenta1

+ Syst. Ext. Imp.1→2

=

Syst. Momenta2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1796

PROBLEM 17.60 (Continued)

Moments about axle 0. 0 + m0 gtrA − mD gtrB = mC vC rA + mD vD rB + ( I A + I B )ω vC = 0.5 m/s

Data:

(1)

t =?

mC grA = (8 kg)(9.81 m/s 2 )(0.200 m) = 15.696 N ⋅ m mD grB = (10 kg)(9.81 m/s 2 )(0.150 m) = 14.715 N ⋅ m mC grA − mD grB = 0.981 N ⋅ m mC vC rA = (8 kg)(0.5 m/s)(0.200 m) = 0.8 kg ⋅ m 2 /s

 0.150 m  2 mD vD rB = (10 kg)   (0.5 m/s)(0.150 m) = 0.5625 kg ⋅ m /s 0.200 m    0.5 m/s  2 ( I A + I B )ω = (0.2675 kg ⋅ m 2 )   = 0.66875 kg ⋅ m /s 0.200 m   mC vC rA + mD vD rB + ( I A + I B )ω = 2.03125 kg ⋅ m 2 /s

Solving Eq. (1) for t,

t=

2.03125 kg ⋅ m 2 /s 0.981 N ⋅ m t = 2.07 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1797

PROBLEM 17.61 Two uniform disks and two cylinders are assembled as indicated. Disk A has a mass of 10 kg and disk B has a mass of 6 kg. Knowing that the system is released from rest, determine the time required for cylinder C to have a speed of 0.5 m/s. The cylinders are attached to a single cord that passes over the disks. Assume that no slipping occurs between the cord and the disks.

SOLUTION Moments of inertia. Disk A:

IA =

1 1 mA rA2 = (10 kg)(0.200 m) 2 = 0.2 kg ⋅ m 2 2 2

Disk B:

IB =

1 1 mB rB2 = (6 kg)(0.150 m) 2 = 0.0675 kg ⋅ m 2 2 2

Kinematics:

vC = v

ωA =

v rA

vD = v

ωB =

v rB

Principle of impulse and momentum. Disk A and cylinder C

Disk B and cylinder D

Syst. Momenta1

+

Syst. Ext. Imp.1→2 =

Syst. Momenta2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1798

PROBLEM 17.61 (Continued)

Disk A and cylinder C.

Moments about A: QtrA − mC gtrA = mA vrA + I Aω A

Disk B and cylinder D.

(1)

Moments about B: −QtrB − mD gtrB = mD rB v + I BωB

(2)

To eliminate Qt divide Equation (1) by rA and Equation (2) by rB, and then add the resulting equations.  I I  (mD − mC ) gt =  mA + A2 + mB + B2  v  rA rB   v = 0.5 m/s

Data:

(3)

t =?

(mD − mC ) g = (2 kg)(9.81 m/s 2 ) = 19.62 N mC +

IA rA2

+ mD +

Equation (3) becomes

IB rB2

= 8 kg +

0.2 kg ⋅ m 2 0.0675 kg ⋅ m 2 10 kg + + = 26 kg (0.200 m)2 (0.150 m) 2

(19.62 N)t = (26 kg)(0.5 m) t = 0.66259 s

t = 0.663 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1799

PROBLEM 17.62 Disk B has an initial angular velocity ω 0 when it is brought into contact with disk A which is at rest. Show that the final angular velocity of disk B depends only on ω 0 and the ratio of the masses m A and mB of the two disks.

SOLUTION Let Points A and B be the centers of the two disks and Point C be the contact point between the two disks. Let ω A and ωB be the final angular velocities of disks A and B, respectively, and let vC be the final velocity at C common to both disks. vC = rAω A = rBωB Kinematics: No slipping Moments of inertia. Assume that both disks are uniform cylinders. 1 1 I A = mA rA2 I B = mB rB2 2 2 Principle of impulse and momentum. Disk A

Disk B

Syst. Momenta1

Disk A: Moments about A:

+

Syst. Ext. Imp. 1→ 2

=

Syst. Momenta 2

0 + rA Ft = I Aω A I Aω A 1 mA rA2 vC = rA 2 rA2 1 = mA vC 2 1 = mA rBω B 2

Ft =

Disk B: Moments about B:

I Bω0 − rB Ft = I Bω B

1 1  1 mB rB2ω0 − rB  m A rBωB  = mrB2ωB 2 2  2

ωB =

ω0 1+

mA mB



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PROBLEM 17.63 The 7.5-lb disk A has a radius rA = 6 in. and is initially at rest. The 10-lb disk B has a radius rB = 8 in. and an angular velocity ω 0 of 900 rpm when it is brought into contact with disk A. Neglecting friction in the bearings, determine (a) the final angular velocity of each disk, (b) the total impulse of the friction force exerted on disk A.

SOLUTION Let Points A and B be the centers of the two disks and Point C be the contact point between the two disks. Let ω A and ωB be the final angular velocities of disks A and B, respectively, and let vC be the final velocity at C common to both disks. vC = rAω A = rBωB

Kinematics: No slipping

Moments of inertia. Assume that both disks are uniform cylinders. IA =

1 mA rA2 2

IB =

1 mB rB2 2

Principle of impulse and momentum. Disk A

Disk B

Syst. Momenta1

Disk A:

Moments about A:

+

Syst. Ext. Imp. 1→ 2

=

Syst. Momenta 2

0 + rA Ft = I Aω A I ω Ft = A A rA 1 mA rA2 vC 1 = m AvC 2 rA2 2 1 = m A rBωB 2 =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1801

PROBLEM 17.63 (Continued)

I Bω0 − rB Ft = I Bω B

Disk B: Moments about B:

1 1  1 mB rB2ω0 − rB  m A rBωB  = mrB2ω B 2 2   2

ωB =

Data:

ω0 1+

mA mB

7.5 = 0.23292 lb ⋅ s 2 /ft 32.2 mA WA 7.5 = = = 0.75 mB WB 10 mA =

8 ft 12 rB 8 = rA 6 rB =

ω0 = 900 rpm = 30π rad/s (a)

ωB =

ω0

1 + 0.75 30π = 1.75 = 53.856 rad/s

ωA =

rB ωB rA

8 =   (53.856) 6 = 71.808 rad/s

1 (0.23292) ( 12 ) (30π ) 2 1 + 0.75

ω A = 686 rpm



ω B = 514 rpm



8

(b)

Ft =

Ft = 4.18 lb ⋅ s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1802

PROBLEM 17.64 A tape moves over the two drums shown. Drum A weighs 1.4 lb and has a radius of gyration of 0.75 in., while drum B weighs 3.5 lb and has a radius of gyration of 1.25 in. In the lower portion of the tape the tension is constant and equal to TA = 0.75 lb. Knowing that the tape is initially at rest, determine (a) the required constant tension TB if the velocity of the tape is to be v = 10 ft/s after 0.24 s, (b) the corresponding tension in the portion of tape between the drums.

SOLUTION Kinematics.

Drums A and B rotate about fixed axes. Let v be the tape velocity in ft/s. v = rAω A =

0.9 ωA 12

ω A = 13.3333v

v = rBωB =

1.5 ωB 12

ωB = 8v 2

Moments of inertia.

 1.4  0.75  2 −6 I A = m A k A2 =    = 169.837 × 10 lb ⋅ s ⋅ ft  32.2  12  2

 3.5  1.25  I B = mB k B2 =  = 1.17942 × 10−3 lb ⋅ s 2 ⋅ ft    32.2  12 

State 1.

t=0

v=0

State 2.

t = 0.24 s,

v = 10 ft/s

ω A = ωB = 0

ω A = (13.3333)(10) = 133.333 rad/s ωB = (8)(10) = 80 rad/s Drum A.

Syst. Momenta1 +

Syst. Ext. Imp. 1→ 2 =

Syst. Momenta 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1803

PROBLEM 17.64 (Continued)

0 + rATAB t − rATAt = I Aω A

Moments about A:

 0.9   0.9  −6 0+  (TAB t ) −   (0.75)(0.24) = (169.837 × 10 )(133.333)  12   12  TAB t = 0.48193 lb ⋅ s

Drum B.

Syst. Momenta1

+

Syst. Ext. Imp. 1→2 =

Syst. Momenta 2

0 + rBTB t − rBTAB t = I BωB

Moments about B: 0+

1.5  1.5  −3 (TB t ) −   (0.48193) = (1.17942 × 10 )(80) 12  12  TB t = 1.23676 lb ⋅ s

(a)

TB =

TB t 1.23676 = t 0.24

(b)

TAB =

TAB t 0.48193 = t 0.24

TB = 5.15 lb  TAB = 2.01 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1804

PROBLEM 17.65 Show that the system of momenta for a rigid slab in plane motion reduces to a single vector, and express the distance from the mass center G to the line of action of this vector in terms of the centroidal radius of gyration k of the slab, the magnitude v of the velocity of G, and the angular velocity ω.

SOLUTION

=

Syst. Momenta

Components parallel to mv :

mv = X

Moments about G:

I ω = (mv ) d d=

Single Vector

I ω mk 2ω = mv mv

X = mv 

d=

k 2ω  v

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1805

PROBLEM 17.66 Show that, when a rigid slab rotates about a fixed axis through O perpendicular to the slab, the system of the momenta of its particles is equivalent to a single vector of magnitude mr ω , perpendicular to the line OG, and applied to a Point P on this line, called the center of percussion, at a distance GP = k 2/ r from the mass center of the slab.

SOLUTION

Kinematics. Point O is fixed.

v = rω

System momenta. Components parallel to mv : Moments about G:

X = mv = mr ω

X = mr ω 

(GP ) X = I ω (GP ) m r ω = mk 2ω

(GP ) =

k2  r

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PROBLEM 17.67 Show that the sum HA of the moments about a Point A of the momenta of the particles of a rigid slab in plane motion is equal to IAω , where ω is the angular velocity of the slab at the instant considered and IA the moment of inertia of the slab about A, if and only if one of the following conditions is satisfied: (a) A is the mass center of the slab, (b) A is the instantaneous center of rotation, (c) the velocity of A is directed along a line joining Point A and the mass center G.

SOLUTION Kinematics. Let

ω = ωk

and

rG/ A = rG/ A

Then,

vG/ A = ω × rG/ A = (ω rG/ A )

θ

where

β = θ + 90°

Also

v = vA + vG/ A

Define

h = rG/ A × vG/ A

β

h = (rG/ A )(vG/ A )k = (rG/ A )2 ωk = (rG/ A ) 2 ω

System momenta. Moments about A: H A = rG/ A × mv + Iω

= rG/ A × m( vA + vG/ A ) + I ω = rG/ A × mvA + mrG/ A × vG/ A + Iω = rG/ A × mvA + mh + Iω = rG/ A × mvA + mrG2/ Aω + Iω = rG/ A × mvA + (m rG2/ A + I ) ω

The first term on the right hand side is equal to zero if (a)

rG/ A = 0

or

(b)

vA = 0

or

(c)

In the second term, Thus,

(A is the mass center) (A is the instantaneous center of rotation) rG/ A is perpendicular to vA .

mrG2/ A + I = I A

by the parallel axis theorem.

H A = I Aω

when one or more of the conditions (a), (b) or (c) is satisfied.

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PROBLEM 17.68 Consider a rigid slab initially at rest and subjected to an impulsive force F contained in the plane of the slab. We define the center of percussion P as the point of intersection of the line of action of F with the perpendicular drawn from G. (a) Show that the instantaneous center of rotation C of the slab is located on line GP at a distance GC = k 2/GP on the opposite side of G. (b) Show that if the center of percussion were located at C the instantaneous center of rotation would be located at P.

SOLUTION (a)

Locate the instantaneous center C corresponding to center of percussion P. Let d P = GP.

+

Syst. Momenta1

Components parallel to FΔt :

=

Syst. Momenta 2

0 + F Δt = mv 0 + d P ( F Δt ) = I ω

Moments about G:

v

Eliminate F Δt to obtain

ω

=

I md P

=

k2 dP

GC = dC =

Kinematics. Locate Point C. (b)

Syst. Ext. Imp. 1→2

v

ω

=

k2 dP

GC =

k2  GP

Place the center of percussion at P′ = C. Locate the corresponding instantaneous center C ′. Let d P′ = GP′ = GC = dC .

Syst. Momenta1

+

Syst. Ext. Imp. 1→ 2

=

Syst. Momenta 2

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PROBLEM 17.68 (Continued)

Components parallel to FΔt : Moments about G:

0 + F Δt = mv′ 0 + d P′ ( F Δt ) = I ω ′

Eliminate F Δt to obtain Kinematics. Locate Point C ′. Using dC = d P′ =

v′ I k2 = = ω ′ md P′ d P′ GC ′ = dC ′ =

k2 gives dP

v′ k 2 k 2 = = ω ′ d P′ d C dC ′ = d P or GC ′ = GP 

Thus Point C ′ coincides with Point P.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1809

PROBLEM 17.69 A flywheel is rigidly attached to a 1.5-in.-radius shaft that rolls without sliding along parallel rails. Knowing that after being released from rest the system attains a speed of 6 in./s in 30 s, determine the centroidal radius of gyration of the system.

SOLUTION Kinematics. Rolling motion. Instantaneous center at C. v = vG = rω I = mk 2

Moment of inertia. Kinetics.

Syst. Momenta1

Moments about C:

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta 2

0 + (mgt )r sin β = mvr + I ω  k2  mgtr sin β = m  r + v r  

Solving for k 2, Data:

 gt sin β  k 2 = r2  − 1 v   r = 1.5 in. = 0.125 ft g = 32.2 ft/s t = 30 s v = 6 in./s = 0.5 ft/s  (32.2)(30) sin15°  k 2 = (0.125)2  − 1 0.5   = 7.7974 ft 2

k = 2.79 ft 

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PROBLEM 17.70 A wheel of radius r and centroidal radius of gyration k is released from rest on the incline shown at time t = 0. Assuming that the wheel rolls without sliding, determine (a) the velocity of its center at time t, (b) the coefficient of static friction required to prevent slipping.

SOLUTION Kinematics. Rolling motion. Instantaneous center at C. v = vG = rω

ω=

v r

I = mk 2

Moment of inertia. Kinetics.

Syst. Momenta1

moments about C:

+

Syst. Momenta 2

0 + (mgt )r sin β = mvr + I ω (mgr sin β )t = mrv +

(a)

=

Syst. Ext. Imp. 1→ 2

mk 2 v r v=

Velocity of Point G.

r 2 gt sin β r2 + k 2

β 

components parallel to incline: 0 + mgt sin β − Ft = mv Ft = mgt sin β − =

mr 2 gt sin β r2 + k2

k 2 mgt sin β r2 + k 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1811

PROBLEM 17.70 (Continued)

components normal to incline: 0 + Nt − mgt cos β = 0 Nt = mgt cos β

(b)

Required coefficient of static friction. F N Ft = Nt

μs ≥

=

k 2 mgt sin β (r + k 2 )mgt cos β 2

μs ≥

k 2 tan β  r2 + k 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1812

PROBLEM 17.71 The double pulley shown has a mass of 3 kg and a radius of gyration of 100 mm. Knowing that when the pulley is at rest, a force P of magnitude 24 N is applied to cord B, determine (a) the velocity of the center of the pulley after 1.5 s, (b) the tension in cord C.

SOLUTION rC = 0.150 m

For the double pulley,

rB = 0.080 m k = 0.100 m

Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Momenta 2

v = rC ω

Kinematics. Point C is the instantaneous center. Moments about C:

=

Syst. Ext. Imp. 1→ 2

0 + Pt (rC + rB ) − mgtrC = I ω + mvrC = mk 2ω + m(rC ω )rC

ω=

Pt (rC + rB ) − mgtrC

(

m k 2 + rC2

)

(24)(1.5)(0.230) − (3)(9.81)(1.5)(0.150) 3(0.1002 + 0.1502 ) = 17.0077 rad/s =

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PROBLEM 17.71 (Continued)

(a)

v = (0.150) (17.0077) = 2.55115 m/s

Linear components:

v = 2.55 m/s 

0 + Pt − mgt + Qt = mv mv + mg − P t (3)(2.55115) = + (3)(9.81) − 24 1.5

Q=

(b)

Q = 10.53 N 

Tension in cord C.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1814

PROBLEM 17.72 A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 2.5 lb is applied as shown for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine the resulting velocity of (a) the carriage, (b) the center of the cylinder.

SOLUTION I =

Moment of inertia.

=

1 mAr 2 2 1  18 lb  9 in.   2  32.2   12 

2

= 0.15722 slug ⋅ ft 2

Cylinder alone:

Syst. Momenta1 + Syst. Ext. Imp. 1→2 =

Syst. Momenta 2

0 + 0 = I ω − mAv A r

Moments about C:

 18  9  0 = 0.15722ω −    v A  32.2  12 

or

(1)

Cylinder and carriage:

Syst. Momenta1

Horizontal components: or

+

Syst. Ext. Imp. 1→2 =

Syst. Momenta 2

0 + Pt = m Av A + mB vB

 18   6  0 + (2.5)(1.2) =   v A +  32.2  vB 32.2    

(2)

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PROBLEM 17.72 (Continued)

Kinematics.

v A = vB − rω

 9  v A = vB −   ω  12 

Solving Equations (1), (2) and (3) simultaneously gives

(3) ω = 7.16 rad/s

(a)

Velocity of the carriage.

vB = 8.05 ft/s



(b)

Velocity of the center of the cylinder.

vA = 2.68 ft/s



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PROBLEM 17.73 A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 2.5 lb is applied as shown for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine the resulting velocity of (a) the carriage, (b) the center of the cylinder.

SOLUTION I =

Moment of inertia.

=

1 mAr 2 2 1  18 lb  9 in.   2  32.2   12 

2

= 0.15722 slug ⋅ ft 2

Cylinder alone:

Syst. Momenta1

+ Syst. Ext. Imp. 1→2 =

Syst. Momenta 2

0 + Ptr = I ω + mA v A r

Moments about C:

 9   18  9  0 + (2.5)(1.2)   = 0.15722ω +    v A  12   32.2  12 

or

(1)

Cylinder and carriage:

Syst. Momenta1

Horizontal components: or

+

Syst. Ext. Imp. 1→2 =

Syst. Momenta 2

0 + Pt = m Av A + mB vB

 18   6  vA +  0 + (2.5)(1.2) =    vB  32.2   32.2 

(2)

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PROBLEM 17.73 (Continued)

Kinematics.

v A = vB + rω

 9 v A = vB +   ω  12 

Solving Equations (1), (2) and (3) simultaneously gives

(3) ω = 2.39 rad/s

(a)

Velocity of the carriage.

v B = 2.68 ft/s



(b)

Velocity of the center of the cylinder.

v A = 4.47 ft/s



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PROBLEM 17.74 Two uniform cylinders, each of mass m = 6 kg and radius r = 125 mm, are connected by a belt as shown. If the system is released from rest when t = 0, determine (a) the velocity of the center of cylinder B at t = 3 s, (b) the tension in the portion of belt connecting the two cylinders.

SOLUTION v AB = rω B

Kinematics.

Point C is the instantaneous center of cylinder A. v AB 1 = ωB 2r 2 1 v A = rω A = rωB 2

ωA =

I =

Moment of inertia. (a)

1 2 mr 2

Velocity of the center of A. Cylinder B:

Syst. Momenta1

+

Syst. Ext. Imp. 1→ 2

=

Syst. Momenta 2

0 + Ptr = IωB

Moments about B:

(1)

Cylinder. A:

Syst. Momenta1

+

Syst. Ext. Imp. 1→ 2

=

Syst. Momenta 2

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PROBLEM 17.74 (Continued)

Moments about C:

0 − 2 Ptr + mgtr = mv A r + I ωA

1  1  0 − 2 I ωB + mgtr = m  rω B  r + I  ω B  2  2  1 2 5  I + mr  ωB = mgrt 2 2  2  5 mr  1 + mr 2  ωB = mgrt  2 2 2 

7 rω B = gt 4 4 gt ωB = 7 r vA =

(b)

1 2 2 rωB = gt = (9.81)(3) 2 7 7

(2) v A = 8.41 m/s 

Tension in the belt. From Eqs. (1) and (2),

 4 gt  Ptr = I   7 r 

P=

1  1 2  4 gt  2 2 mr   = 7 mg = 7 (6)(9.81) = 16.817 N tr  2 7 r  

P = 16.82 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1820

PROBLEM 17.75 Two uniform cylinders, each of mass m = 6 kg and radius r = 125 mm, are connected by a belt as shown. Knowing that at the instant shown the angular velocity of cylinder A is 30 rad/s counterclockwise, determine (a) the time required for the angular velocity of cylinder A to be reduced to 5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.

SOLUTION v AB = rω B

Kinematics.

Point C is the instantaneous center of cylinder A. v AB 1 = ωB 2r 2 1 v A = rω A = rωB 2

ωA =

I =

Moment of inertia. (a)

1W 2 r 2 g

Required time. Cylinder B:

Syst. Momenta1

Moments about B:

+

Syst. Ext. Imp. 1→ 2 =

Syst. Momenta 2

I (ωB )1 − Ptr = I (ωB ) 2 Ptr = I [(ω B )1 − (ωB )2 ] =

1 2 mr [(ωB )1 − (ωB ) 2 ] 2

(1)

Cylinder A:

Syst. Momenta1

+

Syst. Ext. Imp. 1→ 2 =

Syst. Momenta 2

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PROBLEM 17.75 (Continued)

I (ω A )1 + m(v A )1 r + 2 Ptr − mgtr = I (ω A ) 2 + m(v A ) 2 r

Moments about C:

1 2 mr [(ω A )1 − (ω A ) 2 + mr[(ω A )1 − (ω A ) 2 ]r + 2 Ptr − mgtr = 0 2  3 2  1  1 1  mr  ωB  − (ω B ) 2  + 2  mr 2 [(ωB )1 − (ωB ) 2 ] − mgtr = 0 2 2 2 2 1     7 2 mr [(ω B )1 − (ωB )2 ] − mgtr = 0 4 t=

(2)

m = 6 kg r = 125 mm = 0.125 m

Data:

t=

From Equation (2), (b)

7r[(ωB )1 − (ω B )2 ] 4g

(7)(0.125)(30 − 5) = 0.55747 (4)(9.81)

t = 0.557 s 

Tension in belt between cylinders. From Equation (1),

1 (6)(0.125)2 (30 − 5) 2 = 1.172 N ⋅ m ⋅ s

Ptr =

P=

Ptr 1.172 = = 16.817 tr (0.55747)(0.125)

P = 16.82 N 

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PROBLEM 17.76 In the gear arrangement shown, gears A and C are attached to rod ABC, which is free to rotate about B, while the inner gear B is fixed. Knowing that the system is at rest, determine the magnitude of the couple M which must be applied to rod ABC, if 2.5 s later the angular velocity of the rod is to be 240 rpm clockwise. Gears A and C weigh 2.5 lb each and may be considered as disks of radius 2 in.; rod ABC weighs 4 lb.

SOLUTION Kinematics of motion Let ω ABC = ω

v A = vC = ( BC )ω = 2rω

Since gears A and C roll on the fixed gear B,

ω A = ωC =

vC 2rω = = 2ω r r

Principle of impulse and momentum.

Syst. Momenta1 + Syst. Ext. Imp. 1→2

Moments about D:

=

Syst. Momenta 2

0 + (Qt )r = mC vC r + I C wC

(Qt ) r = mC (2rω )r +

1 mC r 2 (2ω ) 2

Qt = 3mC rω

Syst. Momenta1

+

Syst. Ext. Imp. 1→2

(1)

=

Syst. Momenta 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1823

PROBLEM 17.76 (Continued)

Moments about B:

Mt − Qt (4r ) = I ABC ω

1 m ABC (4r ) 2ω 12 4 Mt − 4(Qt )r = mABC r 2ω 3

Mt − 4(Qt )r =

(2)

Substitute for (Qt) from (1) into (2): 4 m ABC r 2ω 3 4 Mt = r 2ω (m ABC + 9mC ) 3

Mt − 4(3mC rω ) r =

(3)

Couple M. Data:

t = 2.5 s

2 ft 12 4 lb m ABC = 32.2 ft/s 2.5 lb mC = 32.2 ft/s 2 r=

ω = 240 rpm = 8π rad/s 2

Eq. (3):

M (2.5 s) =

 4 4 2   2.5   ft  (8π rad/s)  + 9   3  12   32.2    32.2

2.5 M = 0.76607 M = 0.3064 lb ⋅ ft

M = 0.306 lb ⋅ ft



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1824

PROBLEM 17.77 A sphere of radius r and mass m is projected along a rough horizontal surface with the initial velocities shown. If the final velocity of the sphere is to be zero, express (a) the required magnitude of ω 0 in terms of v0 and r, (b) the time required for the sphere to come to rest in terms of v0 and coefficient of kinetic friction μk .

SOLUTION I =

Moment of inertia. Solid sphere.

Syst. Momenta1

y components: x components: Moments about G:

+

2 2 mr 5

Syst. Ext. Imp. 1→2

Nt − Wt = 0 mv0 − Ft = 0

=

Syst. Momenta 2

N = W = mg

(1)

Ft = mv0

(2)

I ω0 − Ftr = 0

(3)

2 2 mr ω0 − mv0 r = 0 5

(a)

Solving for ω0 ,

(b)

Time to come to rest. From Equation (2),

t=

mv0 mv0 = F μk mg

ω0 =

5 v0  2 r

t=

v0  μk g

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1825

PROBLEM 17.78 A bowler projects an 8.5-in.-diameter ball weighing 16 lb along an alley with a forward velocity v0 of 25 ft/s and a backspin ω 0 of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1.

SOLUTION Radius:

r=

1 1 d = (8.5 in.) = 4.25 in. = 0.35417 ft 2 2

Mass:

m=

W 16 lb = = 0.49689 lb ⋅ s 2 /ft g 32.2 ft/s 2

Moment of inertia:

I =

2 2 2 mr =   (0.49689)(0.35417) 2 = 0.02493 lb ⋅ s 2 ⋅ ft 5 5

Use the principle of impulse and momentum.

Syst. Momenta1

: Nt − Wt = 0

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta 2

N = W = 16 lb

Friction force:

FF = μ k N = (0.10)(16) = 1.6 lb.

: mv0 − FF t = mv2 FF t 1.6t = 25 ft/s − m 0.49689 = 25 − 3.22t

v2 = v0 −

Moments about G:

I ω0 − FF tr = − I ω2 F tr (1.6t )(0.35417) ω2 = F − ω0 = −9 I 0.02493 = 22.731t − 9

Slipping stops when v2 = rω2

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PROBLEM 17.78 (Continued)

(a)

Time t when slipping stops. (25 − 3.22t ) = (0.35417 ft)(22.731t − 9) (25 + 3.1875) = (3.22 + 8.0506)t t = 2.501 s

(b)

t = 2.50 s 

Corresponding velocity. v2 = 25 − 3.22t

v2 = 16.95 ft/s 

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PROBLEM 17.79 Four rectangular panels, each of length b and height 12 b, are attached with hinges to a circular plate of diameter 2b and held by a wire loop in the position shown. The plate and the panels are made of the same material and have the same thickness. The entire assembly is rotating with an angular velocity ω0 when the wire breaks. Determine the angular velocity of the assembly after the panels have come to rest in a horizontal position.

SOLUTION Kinematics:

When the panels are in the up position, the speed of the mass center of each panel is v0 =

1 bω0 2

When the panels are in the down position the speed of the mass center of each panel is v1 =

3 bω1 4

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1828

PROBLEM 17.79 (Continued) Let ρ = mass density of plate and of panels t = thickness of plate and of panels m = ρV = ρπ t (0.707b)2 = ρ tπ b 2 (0.500)

Disk:

I disk = I D =

1 1 m(rdisk ) 2 = ρ tπ b 2 (0.500)(0.707b) 2 2 2

1 I D = ρ tπ b 4 8

Each panel:

1 1  mρ = b  b  ρ t = ρ tb 2 2 2 

In up position

I0 =

In down position

I1 =

1 1 1 1  mρ b 2 =  ρ tb 2  b 2 = ρ tb 4 12 12  2 24  2  1 5 1   1 1 5 mρ  b2 +  b   =  ρ tb 2  b 2 = ρ tb 4   12 2 12 2 4 96      

Principle of impulse and momentum.

+

Syst. Momenta0

Syst. Ext. Imp. 0→1 =

Syst. Momenta1

In the up position, the angular momentum of one panel about the vertical axle is 1  mρ v0  b  + I 0ω0 2 

In the down position it is 3 mρ v1 b + I1ω1 4

Conservation of angular momentum. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1829

PROBLEM 17.79 (Continued)

    1  3  I disk ω0 + 4  mρ vD  b  + I 0ω0  = I diskω1 + 4  mρ v1  b  + I1ω1  2 4         2 2  1    3   I disk ω0 + 4  mρ  b  + I 0  ω0 = I diskω1 + 4  mρ  b  + I1  ω1   2     4   2  1 1   1 4 21  ρ tb 4   ω0  ρ tπ b + 4  ρ tb  b  + 8 24 2   2    2 1   5  1 4 23  ρ tb 4   ω1 =  ρ tπ b + 4  ρ tb  b  + 8  4  96  2   

π 1 1  π 9 5   + +  ω0 =  + +  ω1  8 2 6  8 8 24   {1.059}ω0 = {1.726}ω1 ω1 = 0.614ω0



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PROBLEM 17.80 A 2.5-lb disk of radius 4 in. is attached to the yoke BCD by means of short shafts fitted in bearings at B and D. The 1.5-lb yoke has a radius of gyration of 3 in. about the x axis. Initially the assembly is rotating at 120 rpm with the disk in the plane of the yoke (θ = 0). If the disk is slightly disturbed and rotates with respect to the yoke until θ = 90°, where it is stopped by a small bar at D, determine the final angular velocity of the assembly.

SOLUTION 2

 1.5   3  −3 2 I C = mkC2 =    12  = 2.9115 × 10 lb ⋅ s ⋅ ft 32.2   

Moment of inertia of yoke: Moment of inertia of disk:

1 θ = 0: I A = mr 2 4 =

1  2.5  4   4  32.2   12 

2

= 2.15666 × 10−3 lb ⋅ s 2 ⋅ ft 1 θ = 90°: I A = mr 2 2 =

1  2.5  4   2  32.2   12 

2

= 4.3133 × 10−3 lb ⋅ s 2 ⋅ ft

Total moment of inertia about the x axis:

θ = 0: ( I x )1 = I C + I A = 5.0682 × 10−3 lb ⋅ s 2 ⋅ ft

θ = 90°: ( I x )2 = I C + I A = 7.2248 × 10−3 lb ⋅ s 2 ⋅ ft

Angular momentum about the x axis:

θ = 0: H1 = ( I x )1ω1 = 5.0682 × 10−3ω 1

θ = 90°: H 2 = ( I x ) 2 ω2 = 7.2248 × 10−3ω 2

Conservation of angular momentum. H1 = H 2 : 5.0682 × 10−3 ω1 = 7.2248 × 10−3 ω2

ω2 = 0.7015ω1 = (0.7015)(120 rpm)

ω2 = 84.2 rpm 

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PROBLEM 17.81 Two 10-lb disks and a small motor are mounted on a 15-lb rectangular platform which is free to rotate about a central vertical spindle. The normal operating speed of the motor is 180 rpm. If the motor is started when the system is at rest, determine the angular velocity of all elements of the system after the motor has attained its normal operating speed. Neglect the mass of the motor and of the belt.

SOLUTION Kinematics.

Motor speed:

ω M = 180 rpm = 6π rad/s

Let ω A , ωB , and ωP be the angular velocities, respectfully, of disk A, disk B and the platform. Since the motor speed is the angular velocity of disk B relative to the platform,

ω B = ω P + ω M = ω P + 6π

(1)

ωB = ω A

Since, the disks have the same outer radius,

(2)

Velocity of the center of disk A

vA =

4 ωP 12

(3)

Velocity of the center of disk B

vB =

4 ωP 12

(4)

Moments of inertia. 2

Disks A and B:

I A = IB =

Platform:

IP =

1 W 2 1  10  3  −3 2 r =   = 9.705 × 10 lb ⋅ s ⋅ ft 2 g 2  32.2   12 

2 2 1 W 2 1  15   16   6  2 −3 (a + b 2 ) =  +       = 78.718 × 10 lb ⋅ s ⋅ ft 12 g 12  32.2   12   12  

Principle of impulse and momentum for system.

Syst. Momenta1

+

Syst. Ext. Imp.1→ 2

=

Syst. Momenta 2

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PROBLEM 17.81 (Continued) Moments about O: 0 + 0 = I Pω P + m Av AlOA + I Aω A + mBvBlOB + I Bω B  10  4  4   10  4  4  = (78.718 × 10−3 )ωP +  ωP   + (9.705 × 10−3 )(ω P + 6π ) +  ω P      32.2  12  12   32.2  12  12  + (9.705 × 10−3 )(ωP + 6π ) = 167.141 × 10−3ω P + 365.87 × 10−3

ω P = − 2.189 rad/s ω A = ω B = − 2.189 + 6π = 16.66 rad/s Angular velocities.

ω A = 159.1 rpm



ω B = 159.1 rpm



ω P = 20.9 rpm



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PROBLEM 17.82 A 3-kg rod of length 800 mm can slide freely in the 240-mm cylinder DE, which in turn can rotate freely in a horizontal plane. In the position shown the assembly is rotating with an angular velocity of magnitude ω = 40 rad/s and end B of the rod is moving toward the cylinder at a speed of 75 mm/s relative to the cylinder. Knowing that the centroidal mass moment of inertia of the cylinder about a vertical axis is 0.025 kg ⋅ m 2 and neglecting the effect of friction, determine the angular velocity of the assembly as end B of the rod strikes end E of the cylinder.

SOLUTION Kinematics and geometry.

v1 = (0.04 m)ω1 = (0.4 m)(40 rad/s)

v2 = (0.28 m)ω2

v1 = 1.6 m/s

Initial position

Final position

Conservation of angular momentum about C.

Moments about C:

I AB =

1 (3 kg)(0.8 m)2 = 0.16 kg ⋅ m 2 12

I ABω1 + mv1 (0.04 m) + I DE ω1 = I ABω2 + mv2 (0.028 m) + I DE ω2 (0.16 kg ⋅ m 2 )(40 rad/s) + (3 kg)(1.6 m/s)(0.04 m) + (0.025 kg ⋅ m2 )(40 rad/s) = (0.16 kg ⋅ m 2 )ω2 + (3 kg)(0.28ω2 )(0.28) + (0.025 kg ⋅ m 2 )ω2 (6.4 + 0.192 + 1.00) = (0.16 + 0.2352 + 0.025)ω2 7.592 = 0.4202ω2 ; ω2 = 18.068 rad/s

ω2 = 18.07 rad/s 

Angular velocity.

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PROBLEM 17.83 A 1.6-kg tube AB can slide freely on rod DE, which in turn can rotate freely in a horizontal plane. Initially the assembly is rotating with an angular velocity ω = 5 rad/s and the tube is held in position by a cord. The moment of inertia of the rod and bracket about the vertical axis of rotation is 0.30 kg ⋅ m 2 and the centroidal moment of inertia of the tube about a vertical axis is 0.0025 kg ⋅ m 2 . If the cord suddenly breaks, determine (a) the angular velocity of the assembly after the tube has moved to end E, (b) the energy lost during the plastic impact at E.

SOLUTION Let Point C be the intersection of axle C and rod DE. Let Point G be the mass center of tube AB. Masses and moments of inertia about vertical axes. m AB = 1.6 kg I AB = 0.0025 kg ⋅ m 2 I DCE = 0.30 kg ⋅ m 2

State 1.

1 (125) 2 = 62.5 mm

(rG/A )1 =

ω1 = 5 rad/s State 2.

(rG/A )2 = 500 − 62.5 = 437.5 mm ω = ω2

Kinematics.

(vG )θ = vθ = rG/C ω

Syst. Momenta1

+ Syst. Ext. Imp. 1→2 = Syst. Momenta2

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PROBLEM 17.83 (Continued)

Moments about C: I ABω1 + I DCE ω1 + m AB (vθ )1 (rG/C )1 + 0 = I ABω2 + I DCE ω2 + m AB (vθ ) 2 ( rG/C ) 2  I AB + I DCE + mAB (rG/G )12  ω1 =  I AB + I DCE + mAB (rG/C ) 22  ω2     [0.0025 + 0.30 + (1.6)(0.0625) 2 ](5) = [0.0025 + 0.30 + (1.6)(0.4375) 2 ]ω2 (0.30875)(5) = 0.60875ω2

ω2 = 2.5359 rad/s (a)

2.54 rad/s 

Angular velocity after the plastic impact. Kinetic energy.

T=

1 1 1 I ABω 2 + I DCE ω 2 + mAB v 2 2 2 2

1 1 1 (0.0025)(5) 2 + (0.30)(5)2 + (1.6)(0.0625) 2 (5) 2 2 2 2 = 3.859375 J

T1 =

1 1 1 (0.0025)(2.5359) 2 + (0.30)(2.5359) 2 + (1.6)(0.4375) 2 (2.5359) 2 2 2 2 = 1.9573 J

T2 =

(b)

T1 − T2 = 1.902 J 

Energy lost.

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PROBLEM 17.84 In the helicopter shown, a vertical tail propeller is used to prevent rotation of the cab as the speed of the main blades is changed. Assuming that the tail propeller is not operating, determine the final angular velocity of the cab after the speed of the main blades has been changed from 180 to 240 rpm. (The speed of the main blades is measured relative to the cab, and the cab has a centroidal moment of inertia of 650 lb ⋅ ft ⋅ s 2 . Each of the four main blades is assumed to be a slender 14-ft rod weighting 55 lb.)

SOLUTION Let Ω be the angular velocity of the cab and ω be the angular velocity of the blades relative to the cab. The absolute angular velocity of the blades is Ω + ω .

ω1 = 180 rpm = 6π rad/s ω2 = 240 rpm = 8π rad/s Moments of inertia. Cab: Blades:

I C = 650 lb ⋅ ft ⋅ s 2 1  I B = 4  mL2  3   1  55  2 = ( 4 )    (14)  3  32.2 

= 446.38 lb ⋅ ft ⋅ s 2

Assume Ω1 = 0. Conservation of angular momentum about shaft. I B (ω1 + Ω1 ) + I C Ω1 = I B (ω2 + Ω 2 ) + I C Ω 2 Ω2 = −

I B (ω2 − ω1 ) IC + I B

(446.38)(8π − 6π ) 446.38 + 650 = −2.5581 rad/s  =−

Ω 2 = −24.4 rpm 

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PROBLEM 17.85 Assuming that the tail propeller in Problem 17.84 is operating and that the angular velocity of the cab remains zero, determine the final horizontal velocity of the cab when the speed of the main blades is changed from 180 to 240 rpm. The cab weighs 1250 lb and is initially at rest. Also determine the force exerted by the tail propeller if the change in speed takes place uniformly in 12 s.

SOLUTION Let Ω be the angular velocity of the cab and ω be the angular velocity of the blades relative to the cab. The absolute angular velocity of the blades is Ω + ω .

ω1 = 180 rpm = 6π rad/s ω2 = 240 rpm = 8π rad/s Moments of inertia. Cab:

I C = 650 lb ⋅ ft ⋅ s 2

Blades:

1   1  55  2 I B = 4  mL2  = (4)    (14) 3   3  32.2 

= 446.38 lb ⋅ ft ⋅ s 2

The cab does not rotate.

Ω1 = Ω 2 = 0 Syst. Momenta1 + Syst. Ext. Imp.1→2 = Syst. Momenta 2

Moments about shaft:

I B (ω1 + Ω1 ) + I C Ω1 + Frt = I B (ω2 + Ω 2 ) + I C Ω 2 Frt = I B (ω2 − ω1 ) = (446.38)(8π − 6π ) = 2804.7 lb ⋅ ft ⋅ s Frt 2804.7 = = 175.29 lb ⋅ s Ft = 16 r

Linear components:

mv1 + Ft = mv2 v2 − v1 =

Ft = m

1250 32.2

175.29 55 + (4) ( 32.2 )

= 3.8398 ft/s

(a)

Assume v1 = 0.

(b)

Force.

v2 = 3.84 ft/s  F=

Ft 175.29 = t 12

F = 14.61 lb 

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PROBLEM 17.86 The circular platform A is fitted with a rim of 200-mm inner radius and can rotate freely about the vertical shaft. It is known that the platform-rim unit has a mass of 5 kg and a radius of gyration of 175 mm with respect to the shaft. At a time when the platform is rotating with an angular velocity of 50 rpm, a 3-kg disk B of radius 80 mm is placed on the platform with no velocity. Knowing that disk B then slides until it comes to rest relative to the platform against the rim, determine the final angular velocity of the platform.

SOLUTION I A = mA k 2

Moments of inertia.

= (5 kg)(0.175 m) 2 = 0.153125 kg ⋅ m 2 1 I B = mB rB2 2 1 = (3 kg)(0.08 m) 2 2 = 9.6 × 10−3 kg ⋅ m 2

State 1

Disk B is at rest.

State 2

Disk B moves with platform A.

Kinematics.

In State 2,

Syst. Momenta1

vB = (0.12 m)ω2

Principle of conservation of angular momentum. Moments about D:

I Aω1 = I Aω2 + I Bω2 + mB vB (0.12 m)

(0.153125 kg ⋅ m 2 )ω1 = (0.153125 kg ⋅ m 2 )ω2 −3

Syst. Momenta2

+ (9.6 × 10 kg ⋅ m )ω2 + (3 kg)(0.12 m) ω2 2

2

0.153125ω2 = 0.20593ω1

ω2 = 0.7436ω1 = 0.7436(50 rpm)

ω2 = 37.2 rpm 

Final angular velocity

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PROBLEM 17.87 Two 4-kg disks and a small motor are mounted on a 6-kg rectangular platform which is free to rotate about a central vertical spindle. The normal operating speed of the motor is 240 rpm. If the motor is started when the system is at rest, determine the angular velocity of all elements of the system after the motor has attained its normal operating speed. Neglect the mass of the motor and of the belt.

SOLUTION Moments of inertia. Disks:

I A = IB =

Platform:

IP =

1 2 1 mr = (4 kg)(0.075 m)2 = 0.01125 kg ⋅ m 2 2

1 1 m(a 2 + b 2 ) = (6 kg)[(0.15 m)2 + (0.4 m)2 ] = 0.09125 kg ⋅ m 2 12 12

Kinematics:  240 rev   1 min   2π rad  ωM =     = 8π rad/s  min   60 s   rev 

Let ω A , ωB and ωP be the angular velocities of A, B, and the platform. The motor speed is the angular velocity of B relative to the platform.

ωB = ωP + ωM = ωP + 8π ω A = ωB Velocity of center of disk A.

v A = ωP rA/O = 0.1 ω A

Velocity of center of disk B.

vB = ωP rB /O = 0.1 ω A

Principle of impulse and momentum for system.

Syst. Momenta1 +

Syst. Ext. Imp.1→ 2

=

Syst. Momenta 2

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PROBLEM 17.87 (Continued) Moments about O: 0 + 0 = I PωP + mAv ArA/O + I Aω A + mB vB rB /O + I BωB 0 = I PωP + mA (ω P rA2/O ) + I A (ωP + 8π ) + mB (ωP rB2/O ) + I B (ωP + 8π )

ωP = −

8π ( I A + I B ) I P + mArA/O + I A + mB rB2/O + I B

ωP = −

8π (0.01125 + 0.01125) 0.09125 + (4)(0.1) 2 + 0.01125 + 4(0.1)2 + 0.01125

= −2.9186 rad/s = −27.87 rpm

ωB = ω A = −2.9186 + 8π = 22.214 rad/s  60s  1 rev  = 22.214 rad/s    = 212.13 rpm  min  2π 

Angular velocities.

ω A = 212 rpm



ωB = 212 rpm



ωP = 27.9 rpm



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PROBLEM 17.88 The 4-kg rod AB can slide freely inside the 6-kg tube CD. The rod was entirely within the tube (x = 0) and released with no initial velocity relative to the tube when the angular velocity of the assembly was 5 rad/s. Neglecting the effect of friction, determine the speed of the rod relative to the tube when x = 400 mm.

SOLUTION Let l be the length of the tube and the rod and Point O be the point of intersection of the tube and the axle. IT =

Moments of inertia.

1 mT l 2 , 12

(vθ )T = rT ω =

Kinematics.

IR =

1 mRl 2 12

l ω 2

l  (vθ ) R = rRω =  + x  ω , 2 

(vr )T = vr

Angular momentum about Point O. H O = mT rT (vθ ) + IT ω + mR rR (vθ ) R + I Rω

= mT

ll  1 1 l  l  mT l 2ω + mR  + x  + x ω + mRl 2ω  ω  + 2 2  2 2 12 12

1 1  =  mT l 2 + mR  l 2 + lx + x 2   ω 3  3

Kinetic energy. T =

Potential energy.

1 1 1 1 1 mT (vθ )T2 + IT ω 2 + mR (vθ 2 ) R + mRvr2 + I Rω 2 2 2 2 2 2

=

2 2  1 l  1 1 1 1 l  mT l 2ω 2 + mR  + x ω 2 + mRvr2 + mRl 2ω 2   mT  ω  + 2  2   2  12 2 2 12 

=

1 1 1 1 1 1 2 2 2  2 2 2  mT l + mR  l + lx + x   ω + mRvr = H Oω + mRvr 2 3 2 2 2 3 

All motion is horizontal. V = 0

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PROBLEM 17.88 (Continued)

State 1.

ω = ω1 = 5 rad/s,

x = 0, ( H O )1 = T1 =

vr = 0

1 1 (mT + mR )l 2ω = (6 + 4)(0.800)2 (5) = 10.6667 kg ⋅ m 2 /s 3 3 1 1 1 1  (mT + mR )l 2ω 2  + 0 = ( H O )1ω = (10.6667)(5) = 26.667 J  2 3 2 2 

V1 = 0

State 2.

x =

l = 0.400 m, 2

ω = ω 2 = ?,

vr = ?

1 1  ( H O ) 2 =  (6)(0.800)2 + (4)  (0.800)2 + (0.800)(0.400) + (0.400) 2   ω2 3  3 = 4.05333ω 2 T2 =

1 1 (4.05333ω2 )ω2 + (4)vr2 = 2.026667ω22 + 2vr2 2 2

V2 = 0

Conservation of angular momentum: 10.6667 = 4.05333ω 2

Conservation of energy.

( H O )1 = ( H O )2

ω 2 = 2.6316 rad/s

T1 + V1 = T2 + V2

26.667 + 0 = (2.02667)(2.6316) 2 + 2vr2 + 0 vr2 = 6.3158 m 2 /s 2

vr = 2.51 m/s  

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PROBLEM 17.89 A 1.8-kg collar A and a 0.7-kg collar B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about its vertical axis of symmetry. The two collars are connected by a cord running over a pulley that is attached to the frame at O. At the instant shown, the velocity v A of collar A has a magnitude of 2.1 m/s and a stop prevents collar B from moving. The stop is suddenly removed and collar A moves toward E. As it reaches a distance of 0.12 m from O, the magnitude of its velocity is observed to be 2.5 m/s. Determine at that instant the magnitude of the angular velocity of the frame and the moment of inertia of the frame and pulley system about CD.

SOLUTION Components of velocity of collar A.

v A2 = (v A ) r2 + (v A )θ2

(1)

(v A )θ = rAω

(2)

Constraint of rod OE.

ΔrA = ΔyB ,

Constraint of cable AB.

(v A ) r = vB

(ΔrA ) = 0.1 m, [(v A ) r ]1 = 0,

Position 1.

(2.1) 2 = 0 + [(v A )θ ]12

From Equation (1),

(2.1) = 0.1ω1

From Equation (2),

ω1 = 21 rad/s

vB = 0

Potential energy. Take position 1 as datum.

V1 = 0

( H O )1 = I (21) + (1.8)(2.1)(0.1) T1 =

Kinetic energy. T1 =

From Equation (1),

(4)

( H O )1 = I ω1 + m A[(v A ) r ]1(rA )1:

Angular momentum.

From Equation (2),

(v A )1 = 2.1 m/s

[(v A )θ ]1 = 2.1 m/s

From Equation (3),

Position 2.

(3)

( H O )1 = 21I + 0.378

(5)

1 2 1 1 I ω1 + m Av A2 + mBvB2 : 2 2 2

1 1 I (21) 2 + (1.8)(2.1) 2 2 2 (rA ) 2 = 0.12 m,

T1 = 220.5I + 3.969

(v A )2 = 2.5 m/s

(6)

ω = ω2

[(v A )θ ]2 = 0.12ω2 [(v A ) r ]22 = (v A ) 22 − [(v A )θ ]22 = (2.5) 2 − (0.12) 2ω22 = 6.25 − 0.0144ω 22

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PROBLEM 17.89 (Continued)

vB2 = 6.25 − 0.0144ω 22

From Equation (3),

ΔrA = (rA ) 2 − (rA )1 = 0.02 m

Change in radial position.

ΔyB = 0.02 m

From Equation (3),

V2 = mB g (ΔyB ) = (0.7)(9.81)(0.02)

Potential energy.

V2 = 0.13734 J

(7)

( H O ) 2 = I ω2 + mA[(v A )θ ]2 (rA ) 2:

Angular momentum.

( H O ) 2 = I ω2 + (1.8)(0.12ω2 )(0.12) T2 =

Kinetic energy. T2 =

( H O )2 = ( I + 0.02592)ω2

(8)

1 2 1 1 I ω 2 + mAv A2 + mBvB2 : 2 2 2

1 2 1 1 I ω2 + (1.8)(2.5)2 + (0.7)(6.25 − 0.0144ω22 ) 2 2 2 T2 = (0.5I − 0.00504)ω22 + 7.8125

(9)

( H O )1 = ( H O ) 2:

Conservation of angular momentum.

21I + 0.378 = ( I + 0.02592)ω2

Solving for ω 2 , Conservation of energy.

ω2 =

21I + 0.378 N = I + 0.02592 D

(10)

T1 + V1 = T2 + V2: 220.5I + 3.969 = (0.5I − 0.00504)ω22 + 7.8125 + 0.13734 220.5I − (0.5I − 0.00504)

N2 − 3.98084 = 0 D2

220.5ID 2 − 0.5IN 2 + 0.00504 N 2 − 3.98084 D 2 = 0 220.5I ( I 2 + 0.05184I + 0.0006718464) − 0.5I (441I 2 + 15.876I + 0.142884) + 0.00504(441I 2 + 15.876I + 0.142884) − (3.98084)( I 2 + 0.05184 I + 0.0006718464) = 0 0I 3 + 1.73452I 2 − 0.04965167 I − 0.001954378 = 0

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PROBLEM 17.89 (Continued)

Solving the quadratic equation for I , I =

0.04965167 ± 0.126590 = 0.050804 3.46904

and

− 0.022179

Reject the negative root. From Equation (10),

ω2 =

(21)(0.050804) + 0.378 0.050804 + 0.02592

ω = 18.83 rad/s  I = 0.0508 kg ⋅ m 2 

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PROBLEM 17.90 A 6-lb collar C is attached to a spring and can slide on rod AB, which in turn can rotate in a horizontal plane. The mass moment of inertia of rod AB with respect to end A is 0.35 lb ⋅ ft ⋅ s2. The spring has a constant k = 15 lb/in. and an undeformed length of 10 in. At the instant shown the velocity of the collar relative to the rod is zero, and the assembly is rotating with an angular velocity of 12 rad/s. Neglecting the effect of friction, determine (a) the angular velocity of the assembly as the collar passes through a point located 7.5 in. from end A of the rod, (b) the corresponding velocity of the collar relative to the rod.

SOLUTION Potential energy of spring: undeformed length = 10 in. Position 1:

Position 2:

Δ = 26 in. − 10 in. = 16 in. 1 1 V1 = k Δ 2 = (15 lb/in.)(16 in.) 2 2 2 = 1920 in. ⋅ lb

Δ = 12.5 in. − 10 in. = 2.5 in. 1 1 V2 = k Δ 2 = (15 lb/in.)(2.5 in.) 2 2 2 = 46.875 in. ⋅ lb

V1 = 160 ft ⋅ lb

V2 = 3.91 ft ⋅ lb

Kinematics:

Kinetics: Since moments of all forces about shaft at A are zero, (H A )1 = (H A )2 I Rω1 + mC (v0 )r1 = I RωC + mC (v0 )2 r2

(I

R

)

(

)

+ mC r12 ω1 = I R + mC r23 ω2

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PROBLEM 17.90 (Continued)

I R = 0.35 lb ⋅ ft ⋅ s 2 , mC =

Data:

r1 = 2 ft, r2 =

6 lb 32.2

7.5 ft, ω1 = 12 rad/s 12

2  6 lb 6 lb  7.5    2 2 2 0.35 lb ⋅ ft ⋅ s + 32.2 (2 ft)  (12 rad/s) = 0.35 lb ⋅ ft ⋅ s + 32.2  12 ft   ω2      

13.1441 = 0.42279ω2 ; ω2 = 31.089 rad/s

(a)

ω2 = 31.1 rad/s 

Angular velocity. Kinetic energy.

1 1 1 I Aω12 + mC (vD )12 + mC (vr )12 2 2 2 1 1  6 lb  (2 ft) 2 (12 rad/s) 2 + 0 = (0.35 lb ⋅ ft ⋅ s 2 )(12 rad/s) 2 +  2 2  32.2 

T1 =

T1 = 78.865 ft ⋅ lb 1 1 1 I Rω22 + m(vB ) 22 + m2 (vr )22 2 2 2 1 = (0.35 lb ⋅ ft ⋅ s 2 )(31.089 rad/s) 2 2

T2 =

2

1  6 lb  7.5  1  6 lb  2 +  ft  (31.089 rad/s) 2 +    (vr ) 2 2  32.2  12  2  32.2 

T2 = 204.32 + 0.09317(vr ) 22

Principle of conservation of energy: Recall:

T1 + V1 = T2 + V2

V1 = 160 ft ⋅ lb and V2 = 3.91 ft ⋅ lb 78.865 + 160 = 204.32 + 0.09317(vr ) 22 + 3.91 30.638 = 0.09317(vr ) 22

(b)

(vr ) 2 = 18.13 ft/s 

Velocity of collar relative to rod.

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PROBLEM 17.91 A small 4-lb collar C can slide freely on a thin ring of weight 6 lb and radius 10 in. The ring is welded to a short vertical shaft, which can rotate freely in a fixed bearing. Initially the ring has an angular velocity of 35 rad/s and the collar is at the top of the ring (θ = 0) when it is given a slight nudge. Neglecting the effect of friction, determine (a) the angular velocity of the ring as the collar passes through the position θ = 90°, (b) the corresponding velocity of the collar relative to the ring.

SOLUTION IR =

Moment of inertia of ring.

1 mR R 2 2

Position 2

Position 1 Position 1.

θ =0 vC = 0

Position 2.

θ = 90° (vC ) y = v y = Rω2

Conservation of angular momentum about y axis for system. I Rω1 = I Rω2 + mC v y R 1 1 mR R 2ω1 = mR R 2ω2 + mC R 2ω2 2 2 2 mR R ω1 = ( mR + 2mC ) R 2ω2

ω2 =

mR ω1 mR + 2mC

(1)

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PROBLEM 17.91 (Continued)

Potential energy. Datum is the center of the ring. V1 = mC gR

V2 = 0

1 11  I Rω12 =  mR R 2  ω 12 2 2 2  1 = mR R 2ω 12 4 1 1 T2 = I Rω22 + mC vx2 + v 2y 2 2 1 1 1 = mR R 2ω22 + mC R 2ω22 + mC v 2y 4 2 2

T1 =

Kinetic energy:

(

)

Principle of conservation of energy: T1 + V1 = T2 + V2 1 1 1 1  mR R 2ω12 + mC gR =  mR + mC  R 2ω22 + mC v 2y 4 2 2 4 

(2)

WC = 4 lb

Data:

WR = 6 lb R = 10 in. = 0.83333 ft

ω1 = 35 rad/s (a)

Angular velocity. From Eq. (1),

(b)

ω2 =

6 lb g 6 lb g

+2

( ) 4 lb g

ω2 = 15.00 rad/s 

(35 rad/s)

Velocity of collar relative to ring. 2

From Eq. (2),

1  6 lb  10   10  2   ft  (35 rad/s) + (4 lb)  ft  4  32.2  12   12  2

 1  6 lb  1  4 lb    10  1  4 lb  2 2 =  +     ft  (15 rad/s) +   vy 2  32.2   4  32.2  2  32.2    12  39.629 + 3.3333 = 16.984 + 0.062112v 2y



v y2 = 418.25



v y = 20.5 ft/s 

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PROBLEM 17.92 A uniform rod AB, of mass 7 kg and length 1.2 m, is attached to the 11-kg cart C. Knowing that the system is released from rest in the position shown and neglecting friction, determine (a) the velocity of Point B as rod AB passes through a vertical position (b) the corresponding velocity of cart C.

SOLUTION Kinematics

vC = vA

v AB

= vC

+ (0.6 m)ω

vC = v AB − 0.6ω

(1)

AB = 1.2 m

Weights. Kinetics

Linear momentum : 0 = mC vC + m AB v AB v AB = −

Substitute into Eq. (1):

vC = −

mC (11 kg) 11 vC = − vC , v AB = − vC mAB (7 kg) 7 11 vC − 0.6ω 7

18 vC = −0.6ω 7

Substitute into Eq. (2):

v AB = −

(2)

vC = − 0.23333ω

(3)

11 (−0.23333ω ) 7

v AB = 0.36667ω

(4)

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PROBLEM 17.92 (Continued)

Kinetic and potential energies. T1 = 0 V1 = m AB gb = (7 kg)(9.81)(0.6 m)(1 − cos 30°) = 5.520 N ⋅ m V2 = 0 1 1 1 2 mC vC2 + m AB v AB + I ABω 2 2 2 2 1 1 1 1  = (11)(−0.23333ω ) 2 + (7)(0.36667ω ) 2 +  (7)(1.2) 2  ω 2 2 2 2  12 

T2 =

= (0.29944 + 0.47056 + 0.4200)ω 2 = 1.190ω 2

Conservation of energy:

T1 + V1 = T2 + V2 0 + 5.52 = 1.190ω 2

ω 2 = 4.6387 (b)

Velocity of C:

(a)

Velocity of B:

Eq. (3)

ω = 2.1538 rad/s

vC = −0.23333(2.1538) v B = vC + [(1.2)ω v B = [0.50254

vC = 0.503 m/s

] = [0.50254 m/s ] + [2.5845

]

] + [1.2(2.1538)



]

v B = 2.08 m/s



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PROBLEM 17.93 In Problem 17.82, determine the velocity of rod AB relative to cylinder DE as end B of the rod strikes end E of the cylinder.

SOLUTION Kinematics and geometry.

v1 = (0.04 m)ω1 = (0.4 m)(40 rad/s)

v2 = (0.28 m)ω2

v1 = 1.6 m/s

Final position

Initial position Conservation of angular momentum about C.

Moments about C:

I AB =

1 (3 kg)(0.8 m)2 = 0.16 kg ⋅ m 2 12

I ABω1 + mv1 (0.04 m) + I DE ω1 = I ABω2 + mv2 (0.28 m) + I DE ω2 (0.16 kg ⋅ m 2 )(40 rad/s) + (3 kg)(1.6 m/s)(0.04 m) + (0.025 kg ⋅ m 2 )(40 rad/s) = (0.16 kg ⋅ m 2 )ω2 + (3 kg)(0.28ω2 )(0.28) + (0.025 kg ⋅ m 2 )ω2 (6.4 + 0.192 + 1.00) = (0.16 + 0.2352 + 0.025)ω2 7.592 = 0.4202ω2 ; ω2 = 18.068 rad/s: ω2 = 18.07 rad/s

Conservation of energy

(vr ) = 0.075 m/s

V1 = V2 = 0 1 1 1 1 I DE ω12 + I ABω12 + m AB v12 + mAB (vr )12 2 2 2 2 1 1 = (0.025 kg ⋅ m 2 )(40 rad/s) 2 + (0.16 kg ⋅ m 2 )(40 rad/s)2 2 2 1 1 + (3 kg)(1.6 m/s)2 + (3 kg)(0.075 m/s) 2 2 2

T1 =

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PROBLEM 17.93 (Continued)

T1 = 20 J + 128 J + 3.84 J + 0.008 J = 151.85 J v2 = (0.28 m)ω2 = (0.28 m)(18.068 rad/s) = 5.059 m/s 1 1 1 1 I DE ω22 + I ABω22 + m AB v22 + m AB (vr ) 22 2 2 2 2 1 = (0.025 kg ⋅ m 2 )(18.068 rad/s) 2 2 1 + (0.16 kg ⋅ m 2 )(18.068 rad/s) 2 2 1 1 = (3 kg)(5.059 m/s) 2 + (3 kg)(vr ) 22 2 2

T2 =

T2 = 4.081 J + 26.116 J + 38.391 J + 1.5(vr ) 22 T2 = 68.587 J + 1.5(vr ) 22 T1 + V1 = T2 + V2 : 151.85 J + 0 = 68.587 J + 1.5(vr ) 22 83.263 = 1.5(vr ) 22 (vr ) 2 = 7.45 m/s 

Velocity of rod relative to cylinder.

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PROBLEM 17.94 In Problem 17.83 determine the velocity of the tube relative to the rod as the tube strikes end E of the assembly. PROBLEM 17.83 A 1.6-kg tube AB can slide freely on rod DE which in turn can rotate freely in a horizontal plane. Initially the assembly is rotating with an angular velocity ω = 5 rad/s and the tube is held in position by a cord. The moment of inertia of the rod and bracket about the vertical axis of rotation is 0.30 kg ⋅ m 2 and the centroidal moment of inertia of the tube about a vertical axis is 0.0025 kg ⋅ m 2 . If the cord suddenly breaks, determine (a) the angular velocity of the assembly after the tube has moved to end E, (b) the energy lost during the plastic impact at E.

SOLUTION Let Point C be the intersection of axle C and rod DE. Let Point G be the mass center of tube AB. Masses and moments of inertia about vertical axes. m AB = 1.6 kg,

I AB = 0.0025 kg ⋅ m 2 ,

1 (125) = 62.5 mm, 2

State 1.

(rG/ A )1 =

State 2.

(rG/ A ) 2 = 500 − 62.5 = 437.5 mm,

I DCE = 0.30 kg ⋅ m 2

ω1 = 5 rad/s, (vr )1 = 0 ω = ω2 ,

vr = (vr ) 2 = 0

(vG )θ = vθ = rG/C ω

Kinematics.

Syst. Momenta1 + Syst. Ext. Imp.1→2 = Syst. Momenta2 Moments about C: I ABω1 + I DCE ω1 + m AB (vθ )1 ( rG/C )1 + 0 = I ABω2 + I DCE ω2 + m AB (vθ ) 2 (rG/C ) 2

 I AB + I DCE + mAB (rG/C )12  ω1 =  I AB + I DCE + mAB (rG/C ) 22  ω2     [0.0025 + 0.30 + (1.6)(0.0625) 2 ](5) = [0.0025 + 0.30 + (1.6)(0.4375) 2 ]ω2 (0.30875)(5) = 0.60875ω2

ω2 = 2.5359 rad/s

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PROBLEM 17.94 (Continued)

Kinetic energy.

1 1 1 I ABω 2 + I DCE ω 2 + mAB v 2 2 2 2 1 1 1 = I ABω 2 + I DCE ω 2 + mAB rG2/C ω 2 + vr2 2 2 2 1 1 1 T1 = (0.0025)(5) 2 + (0.3)(5)2 + (1.6)(0.0625) 2 + 0 = 3.859375 J 2 2 2 T=

(

)

1 1 (0.0025)(2.5359)2 + (0.30)(2.5359) 2 2 2 1 1 + (1.6)(0.4375) 2 (2.5359) 2 + (1.6)(vr ) 22 2 2 2 = 1.95737 + 0.8(vr ) 2

T2 =

Work. The work of the bearing reactions at C is zero. Since the sliding contact between the rod and the tube is frictionless, the work of the contact force is zero. U1→2 = 0 T1 + U1→ 2 = T2

Principle of work and energy.

3.859375 + 0 = 1.95737 + 0.8(vr ) 22 (vr ) 2 = 1.542 m/s 

Velocity of the tube relative to the rod.

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PROBLEM 17.95 The 6-lb steel cylinder A and the 10-lb wooden cart B are at rest in the position shown when the cylinder is given a slight nudge, causing it to roll without sliding along the top surface of the cart. Neglecting friction between the cart and the ground, determine the velocity of the cart as the cylinder passes through the lowest point of the surface at C.

SOLUTION Kinematics (when cylinder is passing C) vB = vC = rω − v A

ω=

v A + vB r

Principle of impulse and momentum.

Syst. of Momenta1 x components:

+ Syst. Ext. Imp.1→2

=

Syst. Momenta2

m A v A − mB vB = 0 6 lb 10 lb vA = vB ; vB = 0.6 v A g g

Work: Kinetic energy:

 6  U1→2 = WA (6 in.) = (6 lb)  ft  = 3 ft ⋅ lb; T1 = 0  12  T2 =

1 1 1 m Av A2 + I ω 2 + mB v32 2 2 2

v A + v0 v A + 0.6v A 1.6v A = = r r r 2 1  6 lb  2 1  1 6 lb 2   1.6v A  1 10 lb (0.6v A ) 2 T2 =  r  +  vA +   2 g  2 2 g 2 g  r  3 2 3.84 2 1.8 2 8.64 2 vA + vA = vA = vA + g g g g

vB = 0.6v A ; ω =

Principle of work and energy:

T1 + U1→ 2 = T2 8.64 2 vA 32.2 v A2 = 11.181 v A = 3.344 ft/s

0 + 3 ft ⋅ lb =

vB = 0.6vA = 0.6(3.344)

v B = 2.01 ft/s



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PROBLEM 17.F4 A uniform slender rod AB of mass m is at rest on a frictionless horizontal surface when hook C engages a small pin at A. Knowing that the hook is pulled upward with a constant velocity v0, draw the impulse-momentum diagram that is needed to determine the impulse exerted on the rod at A and B. Assume that the velocity of the hook is unchanged and that the impact is perfectly plastic.

SOLUTION Answer:

Syst. Momenta1

+

Syst. Ext. Imp . 1→2

=

Syst. Momenta 2

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PROBLEM 17.F5 A uniform slender rod AB of length L is falling freely with a velocity v0 when cord AC suddenly becomes taut. Assuming that the impact is perfectly plastic, draw the impulse-momentum diagram that is needed to determine the angular velocity of the rod and the velocity of its mass center immediately after the cord becomes taut.

SOLUTION Answer: Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta 2

Note: For the momentum after the impact a general aGx and aGy can be used. These can be related to ω and vA using kinematics.

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PROBLEM 17.F6 A slender rod CDE of length L and mass m is attached to a pin support at its midpoint D. A second and identical rod AB is rotating about a pin support at A with an angular velocity ω1 when its end B strikes end C of rod CDE. The coefficient of restitution between the rods is e. Draw the impulse-momentum diagrams that are needed to determine the angular velocity of each rod immediately after the impact.

SOLUTION Answer: Rod AB.

Syst. Momenta1

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

Rod CE.

Syst. Momenta1

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

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PROBLEM 17.96 At what height h above its center G should a billiard ball of radius r be struck horizontally by a cue if the ball is to start rolling without sliding?

SOLUTION I =

Moment of inertia.

2 2 mr 5

Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta2

Kinematics. Rolling without sliding. Point C is the instantaneous center of rotation. Linear components:

0 + PΔt = mv2 = mrω2

Moments about G:

0 + hPΔt = I ω2

2  0 + h( mrω2 ) =  mr 2  ω2 5 

h=

2 r  5

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PROBLEM 17.97 A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the lower end of a slender 15-lb bar of length L = 30 in. Knowing that h = 12 in. and that the bar is initially at rest, determine (a) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at C, assuming that the bullet becomes embedded in 0.001 s.

SOLUTION L = 30 in. = 2.5 ft m =

Bar:

I = m0 =

Bullet: Support location:

15 = 0.46584 lb ⋅ s 2 /ft 32.2

1 1 mL2 = (0.46584)(2.5) 2 = 0.24262 lb ⋅ s 2 ⋅ ft 12 12 0.08 = 0.0024845 lb ⋅ s 2 /ft 32.2

h = 12 in. = 1.0 ft vB = ( L − h)ω = (2.5 − 1.0)ω = 1.5ω

Kinematics.

L  vG =  − h  ω = (1.25 − 1.0)ω = 0.25ω 2  

Kinetics.

Syst. Momenta1 + Moments about C:

Syst. Ext. Imp. 1→2 =

Syst. Momenta2

L  m0 v0 ( L − h) = m0 vB ( L − h) + mv0  − h  + I ω 2  

(0.0024845)(1800)(1.5) = (0.0024845)(1.5ω ) + (0.46584)(0.25ω )(0.25) + (0.24262ω )

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PROBLEM 17.97 (Continued)

6.7082 = 0.27546ω

(a)

or

ω = 24.353

ω = 24.4 rad/s



vB = (1.5)(24.353) = 36.53 ft/s vG = (0.25)(24.353) = 6.0881 ft/s

Horizontal components: − m0 v0 + C (Δt ) = − m0 vB − mvG : C ( Δt ) = m0 (v0 − vB ) − mv0 C (Δt ) = (0.0024845)(1800 − 36.53) − (0.46584)(6.0881) = 1.545 lb ⋅ s

(b)

C=

C Δt 1.545 = Δt 0.001

C = 1545 lb



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PROBLEM 17.98 In Problem 17.97, determine (a) the required distance h if the impulsive reaction at C is to be zero, (b) the corresponding angular velocity of the bar immediately after the bullet becomes embedded. PROBLEM 17.97 A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the lower end of a slender 15-lb bar of length L = 30 in. Knowing that h = 12 in. and that the bar is initially at rest, determine (a) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at C, assuming that the bullet becomes embedded in 0.001 s.

SOLUTION Bar:

L = 30 in. = 2.5 ft m = I =

15 = 0.46584 lb ⋅ s 2 /ft 32.2

1 1 mL2 = (0.46584)(2.5) 2 = 2.24262 lb ⋅ s 2 /ft 12 12 0.08 = 0.0024845 lb ⋅ s 2 /ft 32.2

Bullet:

m0 =

Kinematics.

vB = ( L − h)ω = (2.5 − h)ω

L  vG =  − h  ω = (1.25 − h)ω 2 

Kinetics.

Syst. Momenta1 + Syst. Ext. Imp. 1→2 = Syst. Momenta2 moment about B:

L 0 + 0 = I ω − mvG   2 0 + 0 = 0.24262ω − (0.46584)(1.25 − h)ω (1.25)

Divide by ω

0 = 0.24262 − 0.5823(1.25 − h)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1864

PROBLEM 17.98 (Continued)

h = 0.8333 ft

(a)

h = 10.00 in. 

vB = (2.5 − 0.8333)ω = 1.6667ω vG = (1.25 − 0.8333)ω = 0.4167ω

Horizontal components:

m0 v0 + 0 = mvG + m0 vB

(0.0024845)(1800) + 0 = (0.46584)(0.4167ω ) + (0.0024845)(1.6667ω )

(b)

ω = 22.56

ω = 22.6 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1865

PROBLEM 17.99 A 16-lb wooden panel is suspended from a pin support at A and is initially at rest. A 4-lb metal sphere is released from rest at B and falls into a hemispherical cup C attached to the panel at a point located on its top edge. Assuming that the impact is perfectly plastic, determine the velocity of the mass center G of the panel immediately after the impact.

SOLUTION Mass and moment of inertia Ws = 4 lb WP = 16 lb 2

I =

Velocity of sphere at C.

1 1  16  18  2 mP ( L) 2 =   = 0.18634 slug ⋅ ft 6 6  32.2  12  

(vC )1 = 2 gy = 2(32.2 ft/s 2 ) ( 129 ft ) = 6.9498 ft/s

Impact analysis. Kinematics: Immediately after impact in terms of ω2 9 ω2 12 7 (vC )2 = ω2 12 v2 =

Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta2

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PROBLEM 17.99 (Continued)

Moments about A:  7   7   9  ms (vC )1  ft  + 0 = ms (vC ) 2  ft  + I ω2 + mP v2  ft  12 12      12   4 lb   7   4 lb  7  7   16 lb  9  9   32.2  (6.9498 ft/s)  12 ft  =  32.2  12 ω2  12 ft  + 0.18634ω2 +  32.2  12 ω2  12 ft              0.50361 = (0.042271 + 0.18634 + 0.2795)ω2

ω2 = 0.99115 rad/s ω2 = 0.99115 rad/s Velocity of the mass center  9   9  v2 =  ft  ω2 =  ft  (0.99115 rad/s)  12   12 

v2 = 0.74336 ft/s

v 2 = 8.92 in./s



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PROBLEM 17.100 An 16-lb wooden panel is suspended from a pin support at A and is initially at rest. A 4-lb metal sphere is released from rest at B′ and falls into a hemispherical cup C′ attached to the panel at the same level as the mass center G. Assuming that the impact is perfectly plastic, determine the velocity of the mass center G of the panel immediately after the impact.

SOLUTION Ws = 4 lb WP = 16 lb 1 I = mP (0.5 m) 2 6 2 1  16  18  =   6  32.2   12  = 0.18634 slug ⋅ ft 2

Mass and moment of inertia.

(vC ′ )1 = 2 gy

Velocity of sphere at C′.

18 = 2(32.2 ft/s 2 ) ( 12 ft )

= 9.8285 ft/s

Impact analysis. Kinematics: Immediately after impact in terms of ω2 . 2

2

 7   9  AC ′ =   +   = 0.95015 ft  12   12  ( vC ′ ) 2 = AC ′ω2 = 0.95015ω2 v2 =

θ

(perpendicular to AC.)

9 ω2 12

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PROBLEM 17.100 (Continued)

Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Ext. Imp.1− 2

=

Syst. Momenta 2

Moments about A:  7   9  ms (vC )1  ft  + 0 = ms (vC ) 2 (0.95015 ft) + Iω2 + mP v2  ft   12   12   4 lb   7   4 lb    (9.8285 ft/s)  ft  =   (0.95015ω2 )(0.95015 ft) + 0.18634ω2  32.2   12   32.2   16 lb  9  9  + ω2  ft    32.2  12  12  0.71221 = (0.11215 + 0.18634 + 0.2795)ω2

ω2 = 1.2322 Velocity of the mass center.

 9 ft  v2 =   ω2  12   9 ft  =  (1.2322 rad/s)  12  = 0.92418 ft/s

v 2 = 11.09 in./s



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PROBLEM 17.101 A 45-g bullet is fired with a velocity of 400 m/s at θ = 30° into a 9-kg square panel of side b = 200 mm. Knowing that h = 150 mm and that the panel is initially at rest, determine (a) the velocity of the center of the panel immediately after the bullet becomes embedded, (b) the impulsive reaction at A, assuming that the bullet becomes embedded in 2 ms.

SOLUTION mB = 0.045 kg mP = 9 kg

IG =

1 1 mP b2 = (9)(0.200) 2 = 0.06 kg ⋅ m 2 6 6

Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity ω = ω . b vG = ω 2

Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.

+

Syst. Momenta1 (a)

Syst. Ext. Imp. 1→2

=

Syst. Momenta2

Moments about A: b b (mB v0 cos 30°)h + mB v0 sin 30°   + 0 = I G ω + mP vG 2 2   1 b     mB v0  h cos 30° + sin 30°  =  I G + mP b2  ω 2 4     (0.045)(400)(0.150 cos 30° + 0.100sin 30°)

1   = 0.06 + (9)(0.2) 2  ω = 0.15ω 4   ω = 21.588 rad/s vB = (0.100)(21.588) = 2.1588 m/s

vG = 2.16 m/s



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PROBLEM 17.101 (Continued)

(b)

mB v0 cos 30° + Ax ( Δt ) = mP vG

Linear momentum:

(0.045)(400 cos 30°) + Ax (0.002) = (9)(2.1588) Ax = 1920 N

Linear momentum:

A x = 1920 N

− mB v0 sin 30° + Ay ( Δt ) = 0 − (0.045)(400) sin 30° + Ay (0.002) = 0 Ay = 4500 N

A y = 4500 N

A = 4892 N = 4.892 kN

tan β =

4500 1920

β = 66.9° A = 4.87 kN

66.9° 

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PROBLEM 17.102 A 45-g bullet is fired with a velocity of 400 m/s at θ = 5° into a 9-kg square panel of side b = 200 mm. Knowing that the panel is initially at rest, determine (a) the required distance h if the horizontal component of the impulsive reaction at A is to be zero, (b) the corresponding velocity of the center of the panel immediately after the bullet becomes embedded.

SOLUTION mB = 0.045 kg mP = 9 kg

IG =

1 1 mP b2 = (9)(0.200) 2 = 0.06 kg ⋅ m 2 6 6

Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity ω = ω b vG = ω 2

.

.

Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact. AX (Δt ) = 0.

Also

Syst. Momenta1 Linear momentum:

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta2

b  mB v0 cos 5° + 0 = mP vG = mP  ω  2  (0.045)(400 cos 5°) = (9)(0.100)ω ω = 19.9239 rad/s vG = (0.100)(19.9239) = 1.99239 m/s

Moments about A:

b b = I G ω + mP vG 2 2

b 1     mB v0  h cos 5° + sin 5°  =  I G + mP b 2  ω 2 4     1   (0.045)(400)( h cos 5° + 0.100sin 5°) = 0.06 + (9)(0.200) 2  (19.9239) 4   17.9315h + 0.1569 = 2.9886 h = 0.15792 m h = 158 mm 

(a) (b)

(mB v0 cos 5°)h + (mB v0 sin 5°)

(1)

vG = 1.992 m/s

From Eq. (1),



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PROBLEM 17.103 The uniform rods, each of mass m, form the L-shaped rigid body ABC which is initially at rest on the frictionless horizontal surface when hook D of the carriage E engages a small pin at C. Knowing that the carriage is pulled to the right with a constant velocity v0, determine immediately after the impact (a) the angular velocity of the body, (b) the velocity of corner B. Assume that the velocity of the carriage is unchanged and that the impact is perfectly plastic.

SOLUTION Kinematics:

v B = vB

, ω =ω

L ]+  ω 2 L   v BC = v0 − ω  2   v B = [v0 ] + [Lω

  

v BC = [v0

]

v B = [v0 − Lω ]

(v AB ) x = vB = [v0 − Lω ] L L (v AB ) y = (vB ) y + ω = ω 2 2

Let m be the mass of each rod. I =

Moment of inertia of each rod. (a)

1 mL2 12

Principle of impulse and momentum.

Syst. Momenta0

+

Syst. Ext. Imp. 0−1

=

Syst. Momenta1

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PROBLEM 17.103 (Continued)

Moments about C: 1  1  0 + 0 = m(v AB ) y  L  − m(v AB ) x L + I ω − m(vBC )  L  + I ω 2  2  1 L  1  1  L  1   0 = m  ω  L  − m(v0 − Lω ) L + mL2ω − m  v0 − ω  L  + mL2ω 12 2  2  12  2  2   3 5 2 9 v0 0 = − mLv0 + mL ω ω= 2 3 10 L

(a)

Angular velocity

(b)

Velocity of B.

vB = v0 − Lω =

1 v0 10

ω = 0.900 v0 /L



v B = 0.100 v0



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PROBLEM 17.104 The uniform slender rod AB of weight 5 lb and length 30 in. forms an angle β = 30° with the vertical as it strikes the smooth corner shown with a vertical velocity v1 of magnitude 8 ft/s and no angular velocity. Assuming that the impact is perfectly plastic, determine the angular velocity of the rod immediately after the impact.

SOLUTION 2

I =

Moment of inertia.

1 1  5  30  −3 2 mL2 =  = 80.875 × 10 lb ⋅ s ⋅ ft 12 12  32.2   12 

β = 30°

Kinematics. (Rotation about A)

vG =

L 15 ω = ω 2 12

Kinetics.

Syst. Momenta1

moments about A:

+

Syst. Ext. Imp.1→ 2

mv1

=

Syst. Momenta 2

L L sin β + 0 = I ω + mvG 2 2

 5   15   5  15  15  −3  32.2  (8)  12  sin 30° + 0 = 80.875 × 10 ω +  32.2  12 ω  12         

ω = 2.4 rad/s ω = 2.40 rad/s



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PROBLEM 17.105 A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the 15-lb wooden rod AB of length L = 30 in. The rod, which is initially at rest, is suspended by a cord of length L = 30 in. Determine the distance h for which, immediately after the bullet becomes embedded, the instantaneous center of rotation of the rod is Point C.

SOLUTION Let mB be the mass of the bullet and m the mass of the rod. The moment of inertia I of the rod is I =

1 mL2 12

Principle of impulse and momentum.

Syst. Momenta1

Moments about G: x components:

+ Syst. Ext. Imp.1→ 2 =

Syst. Momenta 2

mB v0 h = I ω2

(1)

mB v0 = mv2

(2)

mB W v0 = B v0 m W

From Eq. (2).

v2 =

From Eq. (3).

m vh w2 = B 0 = I

For the instantaneous center to lie at Point C,

v2 =

WB g

(3)

v0 h

1 W 12 g

2

L

= 12

WB v0 h W L2

(4)

3 Lω2 2

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PROBLEM 17.105 (Continued)

Substitute for v2 and ω2 from Equations (3) and (4). WB 3  W v h v0 = L 12 B ⋅ 02  W 2  W L  1 30 in. h= L= 18 18

h = 1.667 in. 

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PROBLEM 17.106 A uniform sphere of radius r rolls down the incline shown without slipping. It hits a horizontal surface and, after slipping for a while, it starts rolling again. Assuming that the sphere does not bounce as it hits the horizontal surface, determine its angular velocity and the velocity of its mass center after it has resumed rolling.

SOLUTION I =

Moment of inertia. Solid sphere.

2 2 mr 5

Kinematics.

Before

After

Before impact (rolling). v1 = rω1

After slipping has stopped. v2 = rω2

Kinetics.

Syst. Momenta1

Moments about C:

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

I ω1 + mv1r cos β + 0 = I ω2 + mv2 r I ω1 + mr 2ω1 cos β = I ω2 + mr 2ω2 2 mr 2 + mr 2 cos β I + mr 2 cos β 5 ω2 = ω1 = 2 2 ω1 I + mr 2 mr + mr 2 5

v2 = rω2 =

2 + 5cos β rω1 7

ω2 =

1 (2 + 5cos β )v1 /r 7



v2 =

1 (2 + 5cos β )v1 7



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PROBLEM 17.107 A uniformly loaded rectangular crate is released from rest in the position shown. Assuming that the floor is sufficiently rough to prevent slipping and that the impact at B is perfectly plastic, determine the smallest value of the ratio a/b for which corner A will remain in contact with the floor.

SOLUTION We consider the limiting case when the crate is just ready to rotate about B. At that instant the velocities must be zero and the reaction at corner A must be zero. Use the principle of impulse and momentum.

Syst. Momenta1 + Syst. Ext. Imp.1→2 = Syst. Momenta 2

Moments about B: I ω1 + (mv1 ) 2

Note:

sin φ =

b a − (mv1 ) y + 0 = 0 2 2 b a +b 2

2

v1 = ( AG )v1 =

, cos φ =

(1) a

a + b2 2

1 2 a + b 2 ω1 2

Thus:

(mv1 ) x = (mv1 )sin φ =

m 2 b 1 a + b2 ω 1 = mbω 1 2 2 2 2 a +b

Also,

(mv1 ) y = (mv1 ) cos φ =

1 maω 2

I =

From Eq. (1)

1 m( a 2 + b 2 ) 12

1 1 b 1 a m( a 2 + b 2 )ω1 + (mbω1 ) − (maω1 ) = 0 12 2 2 2 2 1 2 1 mb ω1 − ma 2 v1 = 0 3 6

a2 =2 b2

a = 2 b

a = 1.414  b

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PROBLEM 17.108 A bullet of mass m is fired with a horizontal velocity v0 and at a height h = 12 R into a wooden disk of much larger mass M and radius R. The disk rests on a horizontal plane and the coefficient of friction between the disk and the plane is finite. (a) Determine the linear velocity v1 and the angular velocity ω1 of the disk immediately after the bullet has penetrated the disk. (b) Describe the ensuing motion of the disk and determine its linear velocity after the motion has become uniform.

SOLUTION (a)

Conditions immediately after the bullet has penetrated the disk. Principle of impulse and momentum:

Syst. Momenta0

+

Syst. Ext. Imp. 0→1

y components:

0 + N Δt − W Δt = 0 N = W

x components:

mv0 − F Δt = M v1

=

Syst. Momenta1

mv0 − μW Δt = M v1

Since Δt ≈ 0, Moments about G: Since Δt ≈ 0,

v1 =

mv0 = mv1

(1) 

mv0 = ( R − h) − R ( μW Δt ) = I ω1 mv0 = ( R − h) =

1 MR 2ω1 2

ω1 = 2 But

mv0 M

h=

m R−h v0 M R2

(2)

1 R 2

ω1 = 2

m R − 12 R M R2

ω1 =

mv0 MR



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1880

PROBLEM 17.108 (Continued)

(b)

After the motion becomes uniform, the disk rolls without slipping. Kinematics.

Syst. Momenta0

Moments about A:

v2 = Rω2

+

Syst. Ext. Imp. 0→ 2

=

Syst. Momenta 2

mv0 h + 0 = M v2 R + I ω2

Since h = 12 R is given: 1 1  mv0  R  = ( MRω2 ) R + MR 2ω2 2 2  1 3 mv0 = MRω2 2 2

ω2 =

mv0 3MR

v2 = Rω2

At first the disk slides

v2 =

mv0 3M



and rotates , it latter rolls with a constant linear velocity v 2 and a constant

angular velocity ω 2 . 

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PROBLEM 17.109 Determine the height h at which the bullet of Problem 17.108 should be fired (a) if the disk is to roll without sliding immediately after impact, (b) if the disk is to slide without rolling immediately after impact.

SOLUTION Principle of impulse and momentum:

Syst. Momenta0

y components: x components:

+

Syst. Ext. Imp.

0 + N Δt − W Δt = 0

=

Syst. Momenta1

N =W

mv0 − F Δt = M v1 mv0 − μW Δt = M v1

Since Δt ≈ 0; Moments about G: Since Δt ≈ 0;

mv0 M

(1)

mv0 = ( R − h) − R ( μW Δt ) = I ω1 mv0 = ( R − h) =

ω1 = 2 (a)

v1 =

mv0 = mv1

1 MR 2ω1 2

m R−h v0 M R2

(2)

If disk is to roll without sliding immediately after impact, we must have

ω1

and v1 = Rω1 mv0  2m R − h  = −R  ⋅ 2 v0  M R M  R−h 1 = −2 R

(b)

If disk is to slide without rotating,





ω1 =

2m R − h ⋅ 2 v0 = 0 M R

h=

3 R  2

h = R 

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PROBLEM 17.110 A uniform slender bar of length L = 200 mm and mass m = 0.5 kg is supported by a frictionless horizontal table. Initially the bar is spinning about its mass center G with a constant angular velocity ω 1 = 6 rad/s. Suddenly latch D is moved to the right and is struck by end A of the bar. Knowing that the coefficient of restitution between A and D is e = 0.6, determine the angular velocity of the bar and the velocity of its mass center immediately after the impact.

SOLUTION Moment of inertia. Before impact.

1 mL2 12 L ( vA )1 = ω1 2 I =

1 eLω1 2 L 1 1 v2 = (v A ) 2 + ω2 = eLω1 + Lω2 Kinematics after impact. 2 2 2 Principle of impulse-momentum at impact.

Impact condition.

Syst. Momenta1

Moments about D:

Coefficient of restitution,

( vA ) 2 = −e( vA )1 =

+

Syst. Ext. Imp . 1→2

=

Syst. Momenta 2

L 2 1 1  L I ω1 = I ω2 + m  eLω1 + Lω2  2 2 2

I ω1 + 0 = I ω2 + mv2

1 1 1 1 mL2ω1 = mL2ω2 + mL2 eω1 + mL2ω2 12 12 4 4 1 ω2 = (1 − 3e)ω1 4 1 1 1 1 v2 = Leω1 + L   (1 − 3e)ω1 = (1 + e)ω1 L 2 2 4 8 e = 0.6 1 ω2 = (1 − (3)(0.6))(6 rad/s) = −1.200 ω2 = 1.200 rad/s 4 1 v2 = (1 + 0.6)(6 rad/s)(0.2 m) = 0.240 m/s 8



v 2 = 0.240 m/s 

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PROBLEM 17.111 A uniform slender rod of length L is dropped onto rigid supports at A and B. Since support B is slightly lower than support A, the rod strikes A with a velocity v1 before it strikes B. Assuming perfectly elastic impact at both A and B, determine the angular velocity of the rod and the velocity of its mass center immediately after the rod (a) strikes support A, (b) strikes support B, (c) again strikes support A.

SOLUTION I =

Moment of inertia. (a)

1 mL2 12

First impact at A.

Syst. Momenta1

+

Syst. Ext. Imp. 1→2

Syst. Momenta 2

e = 1: ( v A ) 2 = v1

Condition of impact:

v2 =

Kinematics: Moments about A:

=

mv1

L L ω − (v A ) 2 = ω − v1 2 2

L L + 0 = mv2 + I ω2 2 2 L L  1  = m  ω − v1  +  mL2  ω2 2  2  12  v2 =

L  3v1  1 − v1 = v1   2 L  2

ω2 =

3v1 L

v2 =

1 v1  2



(vB )2 = Lω − (v A ) 2 = 3v1 − v1 = 2v1

(b)

Impact at B.

Syst. Momenta1

Condition of impact. Kinematics:

+

Syst. Ext. Imp . 1→2

=

Syst. Momenta3

e = 1: ( v B )3 = 2v1 v3 = (vB ) 2 −

L L ω = 2v1 − ω 2 2

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PROBLEM 17.111 (Continued)

Moments about B:

− mv2

L L + I ω2 + 0 = mv3 − I ω3 2 2

L L  1 1 L  1   3v    − m  v1  +  mL2   1  + 0 = m  2v1 − ω3  −  mL2  ω3 2  2  12  2  2  12  L    v3 = 2v1 −

L  3v1  1 = v1 2  L  2

ω3 =

3v1 L

v3 =

1 v1  2



(v A )3 = Lω − (vB )3 = 3v1 − 2v1 = v1

(c)

Second impact at A.

Syst. Momenta3

+

Syst. Ext. Imp. 3→4

Syst. Momenta 4

e = 1: ( v A ) 4 = v1

Condition of impact.

v4 = (v A ) 4 +

Kinematics:

Moments about A:

=

mv3

L L ω4 = v1 + ω4 2 2

L L + I ω3 + 0 = mv4 + I ω4 2 2

L L  1 1 L  1   3v    m  v1  +  mL2   1  + 0 = m  v1 + ω4  +  mL2  ω4 2  2  12  2  2  12  L    v4 = v1 + 0

ω4 = 0  v 4 = v1 

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PROBLEM 17.112 The slender rod AB of length L forms an angle β with the vertical as it strikes the frictionless surface shown with a vertical velocity v1 and no angular velocity. Assuming that the impact is perfectly plastic, derive an expression for the angular velocity of the rod immediately after the impact.

SOLUTION 1 mL2 12

Moment of inertia.

I =

Perfectly plastic impact.

e=0

[(v A ) y ]2 = −e(v A ) y1 = 0

v A = (v A ) x i + (v A ) y j = (v A ) x i

Kinetics.

Syst. Momenta1

+

0 + 0 = mvx

horizontal components: Kinematics.

Syst. Ext. Imp.1→2 = Syst. Momenta 2

vG = v A + vG/ A

[v y ] = [(v A ) x

L ] + ω 2



β 

L v y = − ω sin β 2

Velocity components : Moments about A:

mvx = 0

mv1

L L sin β + 0 = −mv y sin β + I ω 2 2 1 L L L mv1 sin β = m  ω sin β  sin β + mL2ω 2 12 2 2

1 2 2  1 2  12 mL + 4 mL sin β 

1   ω = 2 mv1 L sin β 

ω=

6sin β v1  1 + 3sin 2 β L

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PROBLEM 17.113 The slender rod AB of length L = 1 m forms an angle β = 30° with the vertical as it strikes the frictionless surface shown with a vertical velocity v1 = 2 m/s and no angular velocity. Knowing that the coefficient of restitution between the rod and the ground is e = 0.8, determine the angular velocity of the rod immediately after the impact.

SOLUTION I =

Moment of inertia.

1 mL2 12

[(v A ) y ] 2 = −e[(v A ) y ]1 = ev1

Apply coefficient of restitution.

v A = (v A ) x i + (v A ) y j = (v A ) x i + ev1 j

Kinetics.

Syst. Momenta1 + Syst. Ext. Imp.1→2 =

0 + 0 = mvx

horizontal components: Kinematics.

vG = vA + vG/ A

Moments about A:

mv1

mvx = 0

[v y ] = [ev1 ] + [(v A ) x v y = ev1 −

Velocity components :

Syst. Momenta 2

L ] + ω 2



β 

L ω sin β 2

L L sin β + 0 = −mv y sin β + I ω 2 2 1 L L  L mv1 sin β = m  ω sin β − ev1  sin β + mL2ω 2 12 2 2

1 2 2  1+ e  1 2  12 mL + 4 mL sin β  ω = 2 mv1 L sin β   6(1 + e)sin β v1 ω= 1 + 3sin 2 β 1L PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1887

PROBLEM 17.113 (Continued)

Data:

L = 1 m, β = 30°, v1 = 2 m/s, e = 0.8

ω=

(6)(1.8)sin 30° 2 m/s ⋅ 1 + 3sin 2 30° 1 m

ω = 6.17 rad/s



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PROBLEM 17.114 The trapeze/lanyard air drop (t/LAD) launch is a proposed innovative method for airborne launch of a payload-carrying rocket. The release sequence involves several steps as shown in (1) where the payload rocket is shown at various instances during the launch. To investigate the first step of this process where the rocket body drops freely from the carrier aircraft until the 2-m lanyard stops the vertical motion of B, a trial rocket is tested as shown in (2). The rocket can be considered a uniform 1-m by 7-m rectangle with a mass of 4000 kg. Knowing that the rocket is released from rest and falls vertically 2 m before the lanyard becomes taut, determine the angular velocity of the rocket immediately after the lanyard is taut.

SOLUTION While the lanyard is slack, the rocket falls freely without rotation. Considering its motion relative to the airplane (a Newtonian frame of reference), its vertical velocity is v12 = v02 + 2 gy = 2 gy v1 = 2 gy = (2)(9.81 m/s)(2 m)

Moment of inertia:

v1 = 6.2642 m/s

(1)

1 m(a 2 + b 2 ) 12 1 = (4000 kg)[(7 m)2 + (1 m) 2 ] = 16.667 kg ⋅ m 2 12

I =

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PROBLEM 17.114 (Continued)

For impact use the principle of impulse and momentum.

Syst. Momenta1

x-components

:

Moments about B: Kinematics.

+

Syst. Ext. Imp.1→2

0 + 0 = mv′x

Syst. Momenta 2

vx′ = 0

mv1 (3.5) + 0 = I ω ′ + mv′y (3.5)

(2)

v B = vB v = [v x

y-components :

=

] + [v y ] = [vB

] + [3.5ω ′ ] + [0.5ω

]

v y = 3.5ω ′

(3)

Substitute from Eqs. (1) and (3) into Eq. (2). mv1 (3.5) = [ I + m(3.5) 2 ]ω ′ (4000 kg)(6.2642 m/s)(3.5 m) = [16667 kg ⋅ m 2 + (4000 kg)(3.5 m) 2 ]ω ′ ω′ = 1.336 rad/s

Angular velocity.



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PROBLEM 17.115 The uniform rectangular block shown is moving along a frictionless surface with a velocity v1 when it strikes a small obstruction at B. Assuming that the impact between corner A and obstruction B is perfectly plastic, determine the magnitude of the velocity v1 for which the maximum angle θ through which the block will rotate is 30°.

SOLUTION Let m be the mass of the block. a = 0.200 m b = 0.100 m

Dimensions:

Moment of inertia about the mass center. I =

1 m( a 2 + b 2 ) 12

Let d be one half the diagonal.

d=

1 2 a + b 2 = 0.1118 m 2

Kinematics. Before impact

v1 = v1

, ω1 = 0

After impact, the block is rotating about corner at B. ω 2 = ω2

v 2 = d ω2

Principle of impulse and momentum.

Syst. Momenta1

Moments about B:

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta 2

mv1b + 0 = I ω2 + mdv2 2 1 1 mv1b = m(a 2 + b 2 )ω2 + md 2ω2 2 12 1 = m(a 2 + b 2 )ω2 3

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PROBLEM 17.115 (Continued)

ω2 =

Angular velocity after impact

3v1b

(1)

2( a 2 + b 2 )

The motion after impact is a rotation about corner B. Position 2 (immediately after impact). Position 3

v2 = d ω2 b = tan −1 0.5 = 26.565° a h = d sin( β + 30°) = 0.11180sin 56.565° = 0.093301 m v3 = 0 ω3 = 0

β = tan −1

(θ = 30°).

Potential energy:

V2 =

Kinetic energy:

T2 =

mgb 2

V3 = mgh

1 2 1 1 I ω2 + mv22 = ( I + md 2 )ω22 2 2 2 1 T3 = 0 = m( a 2 + b 2 )ω22 6

Principle of conservation of energy: T2 + V3 = T3 + V3 1 mgb = 0 + mgh m(a 2 + b 2 )ω22 + 6 2

ω22 =

3g (2h − b) (3)(9.81)(0.18660 − 0.100) = (a 2 + b 2 ) (0.200) 2 + (0.100)2

= 50.974 (rad/s)2

ω2 = 7.1396 rad/s

Magnitude of initial velocity. Solving Eq. (1) for v1

v1 =

2(a 2 + b 2 )ω2 3b



v1 =

(2)[(0.200) 2 + (0.100) 2 ](7.1396)  (3)(0.100)

v1 = 2.38 m/s 

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PROBLEM 17.116 A slender rod of length L and mass m is released from rest in the position shown. It is observed that after the rod strikes the vertical surface it rebounds to form an angle of 30° with the vertical. (a) Determine the coefficient of restitution between knob K and the surface. (b) Show that the same rebound can be expected for any position of knob K.

SOLUTION For analysis of the downward swing of the rod before impact and for the upward swing after impact use the principle of conservation of energy. Before impact.

V1 = 0 V2 = −W

L L = − mg 2 2

T1 = 0 2

T2 =

1 1 1 1 1 1  1  I ω22 + mv22 =  mL2  v22 + m  ω2  = mL2ω22 2 2 2  12 2 2 6   

T1 + V1 = T2 + V2 : 0 =

L g 1 2 2 mL ω2 = −mg ; ω22 = 3 L 6 2

ω 2 = 1.73205

g L

After impact.

V3 = −W

L L = − mg 2 2

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PROBLEM 17.116 (Continued) L cos 30° 2 1 1 T3 = I ω32 + mv32 2 2

V4 = −W

1 1 1 1  2 2  mL  ω3 + m  ω3  2  12 2 2   1 = mL2ω32 6 T4 = 0

2

=

T3 + V3 = T4 + V4 :

L L 1 2 2 mL ω3 − mg = 0 − mg cos 30° 6 2 2

ω32 = 3(1 − cos 30°)

g L

ω3 = 0.63397

g L

Analysis of impact. Let r be the distance BK. Before impact,

( v k )3 = rω3

= 1.73205r

g L

After impact,

( v k ) 4 = rω4

= 0.63397r

g L

Coefficient of restitution.

e=

|(vk ) 4 n | |(vk )3n |

e=

0.63397 1.73205

e = 0.366 

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PROBLEM 17.117 A slender rod of mass m and length L is released from rest in the position shown and hits edge D. Assuming perfectly plastic impact at D, determine for b = 0.6 L, (a) the angular velocity of the rod immediately after the impact, (b) the maximum angle through which the rod will rotate after the impact.

SOLUTION For analysis of the falling motion before impact use the principle of conservation of energy. Position 1:

T1 = 0, V1 = mg

Position 2:

V2 = 0

L 4

2

1 L  1 1  T2 = m  ω2  +  mL2  ω22 2 2  2  12  T2 =

1 2 2 mL ω2 6

T1 + V1 = T2 + V2 : 0 + mg

L 1 2 2 = mL ω2 4 6

ω2 =

3g 2L

Analysis of impact. Kinematics Before impact, rotation is about Point A.

v2 =

L ω2 2

After impact, rotation is about Point D.

v3 =

L ω3 10

Principle of impulse-momentum.

Syst. Momenta 2

Moments about D:

+

Syst. Ext. Imp. 2→3

=

Syst. Momenta3

 L  L I ω2 − mv2   = I ω3 + mv3    10   10 

(1)

1 1 L  L  L L mL2ω2 − m  ω2  = mL2ω3 + m  ω3  12  2  10 12  10  10 1  1   1  1  12 − 20  ω2 =  12 + 100  ω3     PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1895

PROBLEM 17.117 (Continued)

(a)

ω3 =

Angular velocity.

5 5 3g ω2 = 14 14 2 L

ω3 = 0.437

g L



For analysis of the rotation about Point D after the impact use the principle of conservation of energy.

(just after impact)

Position 3

v3 =

L ω3 V3 = 0 10

2

T3 =

1  L 1 1 14   m  ω3  +  mL2  ω32 = mL2ω32 2  10  2  12 300  2

 5 3g  14 mgL = mL2   =  300 112  14 2 L 

Position 4.

θ = maximum rotation angle. h′ =

L sin θ 10

mgL sin θ 10 v4 = 0, ω4 = 0, T4 = 0

V4 = mgh′ =

T3 + V3 = T4 + V4 ;

(b)

Maximum rotation angle.

mgL mgL +0=0+ sin θ 112 10 sin θ =

10 112

θ = 5.12° 

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PROBLEM 17.118 A uniformly loaded square crate is released from rest with its corner D directly above A; it rotates about A until its corner B strikes the floor, and then rotates about B. The floor is sufficiently rough to prevent slipping and the impact at B is perfectly plastic. Denoting by ω 0 the angular velocity of the crate immediately before B strikes the floor, determine (a) the angular velocity of the crate immediately after B strikes the floor, (b) the fraction of the kinetic energy of the crate lost during the impact, (c) the angle θ through which the crate will rotate after B strikes the floor.

SOLUTION Let m be the mass of the crate and c be the length of an edge. Moment of inertia

I =

Syst. Momenta1

Kinematics:

Moments about B:

1 1 m(c 2 + c 2 ) = mc 2 12 6

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta 2

1 2cω0 2 1 2cω v = rG/Bω = 2

v0 = rG/ Aω0 =

I ω0 + 0 = I ω + rG/B mv 1 2 1 1  1  2 mc ω0 + 0 = mc 2ω +  2c  m  2cω  = mc 2ω 6 6 2  2  3

(a)

1 ω = ω0 4

Solving for ω ,



Kinetic Energy. Before impact:

1 2 1 I ω0 + mv02 2 2 2 11 2 2 1 1  2cω0  =  mc  ω0 + m  26 2 2   1 2 2 = mc ω0 3

T1 =

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PROBLEM 17.118 (Continued)

After impact:

T2 =

1 2 1 11 1 1   I ω + mv 2 =  mc 2  ω 2 + m  2cω  2 2 26 2 2  

2

2

1 1 1 1  mc 2ω0 = mc 2ω 2 = mc 2  ω0  = 3 3 4 48  

(b)

T1 − T2 = T1

Conservation of energy during falling.

T0 + V0 = T1 + V1

(1)

Conservation of energy during rising.

T3 + V3 = T2 + V2

(2)

T0 = 0,

Conditions:

T3 = 0

From Equation (2),

T2 = V3 − V2 = mgh3 −

1 2

2 −1

1 16

=1−

1 16

15  16

1 T1 16 V3 = mgh3

1 mgc 2

1 1  h3 =  + ( 2 − 1)  c 2 16   h3 =

From geometry,

1 3

1 ( 2 − 1)mgc 2

T1 = V0 − V1 =

=

1 48

1  V1 = V2 = mg  c  2 

From Equation (1),

h3 − 12 c

1 3

T2 =

1  V0 = mg  2c  2 

(c)



Fraction of energy lost:

1 2c sin (θ + 45°) 2

Equating the two expressions for h3 , sin (45° + θ ) =

1 2

+ 161 ( 2 − 1) 1 2

45° + θ = 46.503°

2

θ = 1.50° 

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PROBLEM 17.119 A 1-oz bullet is fired with a horizontal velocity of 750 mi/h into the 18-lb wooden beam AB. The beam is suspended from a collar of negligible mass that can slide along a horizontal rod. Neglecting friction between the collar and the rod, determine the maximum angle of rotation of the beam during its subsequent motion.

SOLUTION Mass of bullet.

W ′ = 1 ounce = 0.0625 lb

Mass of beam AB.

W = 18 lb

Mass ratio.

β=

W′ = 0.0034722 W

W ′ = β W and m′ = β m

Since β is so small, the mass of the bullet will be neglected in comparison with that of the beam in determining the motion after the impact. Moment of inertia.

I =

1 mL2 12

Impact kinetics.

Syst. Momenta1 + Syst. Ext. Imp. 1→2 =

linear components: Moments about B:

− β mv0 + 0 = mv2

Syst. Momenta 2

v2 = β v0

0 + 0 = I ω − mv2

L 2

mv2 L 12mβ v0 L = 2I 2mL2 6β v0 ω= L

ω=

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PROBLEM 17.119 (Continued)

Motion during rising. Position 2. Just after the impact. V2 = − mg T2 = =

L 2

(datum at level A)

1 1 mv22 + I ω22 2 2 1 1 1   6β v0  m( β v0 ) 2 +  mL2   2 2  12  L 

2

= 2β 2 mv02

ω = 0,

Position 3.

Syst. Momenta 2

+

θ = θm .

Syst. Ext. Imp. 2→3

V3 = − mg T3 =

linear components: Conservation of energy.

=

Syst. Momenta3

L cos θ m 2

1 mv32 2

mv2 + 0 = mv3

v3 = v2 = β v0 where v0 = 750 mi/h = 1100 ft/s

T2 + V2 = T3 + V3 : 2 β 2 mv02 − mg

L 1 L = m( β v0 ) 2 − mg cos θ m 2 2 2

3β 2 v02 = 1 − cos θ m gL cos θ m = 1 −

3β 2 v02 gL

(3)(0.0034722) 2 (1100) 2 (32.2)(4) = 0.66021 =1−

θ m = 48.7° 

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PROBLEM 17.120 For the beam of Problem 17.119, determine the velocity of the 1-oz bullet for which the maximum angle of rotation of the beam will be 90°. PROBLEM 17.119 A 1-oz bullet is fired with a horizontal velocity of 350 m/s into the 18-lb wooden beam AB. The beam is suspended from a collar of negligible weight that can slide along a horizontal rod. Neglecting friction between the collar and the rod, determine the maximum angle of rotation of the beam during its subsequent motion.

SOLUTION Mass of bullet.

W ′ = 1 ounce = 0.0625 lb

Mass of beam AB.

W = 18 lb

β=

Mass ratio.

W′ = 0.0034722 W

W ′ = β W and m′ = β m

Since β is so small, the mass of the bullet will be neglected in comparison with that of the beam in determining the motion after the impact. I =

Moment of inertia.

1 mL2 12

Impact Kinetics.

Syst. Momenta1 + Syst. Ext. Imp. 1→2 =

linear components: Moments about B:

− β mv0 + 0 = mv2

Syst. Momenta 2

v2 = β v0 L 2 12mβ v0 L

0 + 0 = I ω − mv2 mv2 L = 2I 6β v0 ω= L

ω=

2mL2

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PROBLEM 17.120 (Continued)

Motion during rising. Position 2. Just after the impact. V2 = − mg T2 =

L 2

(datum at level A)

1 1 mv22 + I ω22 2 2

1 1 1   6β v0  = m( β v0 ) 2 +  mL2   2 2  12  L 

2

= 2β 2 mv02

ω = 0,

Position 3.

Syst. Momenta 2

+

θ = θ m .

Syst. Ext. Imp. 2→3

V3 = − mg T3 =

linear components:

=

Syst. Momenta3

L cos θ m 2

1 mv32 2

mv2 + 0 = mv3

v3 = v2 = β v0

Conservation of energy. T2 + V2 = T3 + V3 :

2 β 2 mv02 − mg

L 1 L = m( β v0 ) 2 − mg cos θ m 2 2 2

β v0 =

1 gL(1 − cos θ m ) 3

1 =   (32.2)(4)(1 − cos 90°) 3 = 6.5524 ft/s



v0 =

6.5524  0.0034722

v0 = 1887 ft/s 

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PROBLEM 17.121 The plank CDE has a mass of 15 kg and rests on a small pivot at D. The 55-kg gymnast A is standing on the plank at C when the 70-kg gymnast B jumps from a height of 2.5 m and strikes the plank at E. Assuming perfectly plastic impact and that gymnast A is standing absolutely straight, determine the height to which gymnast A will rise.

SOLUTION I =

Moment of inertia.

1 1 mP (2 L)2 = mP L2 12 3

(v)1 = 2 gh1

Velocity of jumper at E.

(1)

Principle of impulse-momentum.

Syst. Momenta1

Kinematics: Moments about D:

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

vC = Lω vD = Lω mE v1 L + 0 = mE vE L + mC vC L + I ω 1 = mE L2ω + mC L2ω + mP L2ω 3 mE v1 ω= mE + mC + 13 mP L vC = Lω =

mE v1 mE + mC + 13 mP

vC2 2g

Gymnast (flier) rising.

hC =

Data:

mE = mB = 70 kg

(2)

(3)

mC = mA = 55 kg mP = 15 kg h1 = 2.5 m

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PROBLEM 17.121 (Continued)

From Equation (1)

v1 = (2)(9.81)(2.5) = 7.0036 m/s (70)(7.0036) 70 + 55 + 5 = 3.7712 m/s

From Equation (2)

vC =

From Equation (3)

h2 =

(3.7712) 2 (2)(9.81) = 0.725 m

h2 = 725 mm 

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PROBLEM 17.122 Solve Problem 17.121, assuming that the gymnasts change places so that gymnast A jumps onto the plank while gymnast B stands at C. PROBLEM 17.121 The plank CDE has a mass of 15 kg and rests on a small pivot at D. The 55-kg gymnast A is standing on the plank at C when the 70-kg gymnast B jumps from a height of 2.5 m and strikes the plank at E. Assuming perfectly plastic impact and that gymnast A is standing absolutely straight, determine the height to which gymnast A will rise.

SOLUTION I =

Moment of inertia.

1 1 mP (2 L)2 = mP L2 12 3

(v)1 = 2 gh1

Velocity of jumper at E.

(1)

Principle of impulse-momentum.

Syst. Momenta1

Kinematics: Moments about D:

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

vC = Lω vD = Lω mE v1 L + 0 = mE vE L + mC vC L + I ω 1 = mE L2ω + mC L2ω + mP L2ω 3 mE v1 ω= mE + mC + 13 mP L vC = Lω =

Gymnast (flier) rising.

hC =

mE v1 mE + mC + 13 mP

vC2 2g

(2)

(3)

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PROBLEM 17.122 (Continued)

Data:

mE = mA = 55 kg mC = mB = 70 kg mP = 15 kg h1 = 2.5 m

From Equation (1)

v1 = (2)(9.81)(2.5) = 7.0036 m/s (55)(7.0036) 55 + 70 + 5 = 2.9631 m/s

From Equation (2)

vC =

From Equation (3)

h2 =

(2.9631) 2 (2)(9.81)

= 0.447 m

h2 = 447 mm 

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PROBLEM 17.123 A small plate B is attached to a cord that is wrapped around a uniform 8-lb disk of radius R = 9 in. A 3-lb collar A is released from rest and falls through a distance h = 15 in. before hitting plate B. Assuming that the impact is perfectly plastic and neglecting the weight of the plate, determine immediately after the impact (a) the velocity of the collar, (b) the angular velocity of the disk.

SOLUTION The collar A falls a distance h. From the principle of conservation of energy. v1 = 2 gh

Impact analysis:

e=0

Kinematics. Collar A and plate B move together. The cord is inextensible. v2 = R ω

or

ω2 =

v2 R

Let m = mass of collar A and M = mass of disk. Moment of inertia of disk:

I =

1 MR 2 2

Principle of impulse and momentum. I ω1 = 0

Syst. Momenta1 +

Moments about C:

Syst. Ext. Imp.1→2

=

m v1 R = I ω2 + m v2 R

Syst. Momenta 2

(1)

1 v  MR 2  2  + m v2 R 2 R 1 m v1 = Mv2 + m v2 2 2m v2 = v1 2m + M

m v1 R =

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PROBLEM 17.123 (Continued)

m = 3 lb/g

Data:

M = 8 lb/g h = 15 in. = 1.25 ft R = 9 in. = 0.75 ft v1 = (2)(32.2)(1.25) = 8.972 ft/s

(a)

(b)

Velocity of A.

Angular velocity.

v2 =

(2)(3) 3 v1 = v1 (2)(3) + 8 7

v2 =

3 (8.972) = 3.8452 ft/s 7

ω2 =

3.8452 0.75

v 2 = 3.85 ft/s  ω 2 = 5.13 rad/s



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PROBLEM 17.124 Solve Problem 17.123, assuming that the coefficient of restitution between A and B is 0.8. PROBLEM 17.123 A small plate B is attached to a cord that is wrapped around a uniform 8-lb disk of radius R = 9 in. A 3-lb collar A is released from rest and falls through a distance h = 15 in. before hitting plate B. Assuming that the impact is perfectly plastic and neglecting the weight of the plate, determine immediately after the impact (a) the velocity of the collar, (b) the angular velocity of the disk.

SOLUTION WD = 8 lb WD 8 = = 0.2484 lb ⋅ s 2 /ft g 32.2 R = 9 in. = 0.75 ft 1 I D = mD R 2 = 0.06988 lb ⋅ s 2 ⋅ ft 2 WA = 3 lb mD =

WA 3 = = 0.09317 lb ⋅ s 2 /ft g 32.2 h = 15 in. = 1.25 ft

mA =

Collar A falls through distance h. Use conservation of energy. T1 = 0 V1 = WA h 1 mA v A2 2 V2 = 0 T2 =

T1 + V1 = T2 + V2 : 0 + WA h = v A2 =

1 mA v A2 + 0 2 2m A h = 2 gh WA

= (2)(32.2)(1.25) = 80.5 ft 2 /s 2 vA = 8.972 ft/s

Impact. Neglect the mass of plate B. Neglect the effect of weight over the duration of the impact.

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PROBLEM 17.124 (Continued)

ω′ = ω

Kinematics.

v′B = R ω = 0.75ω ′

Conservation of momentum.

m A v A R + 0 = m Av′A R + I Dω ′ + mB vB′ R

Moments about D:

(0.09317)(8.972)(0.75) = (0.09317)(0.75)v′A + 0.06988 ω ′

Coefficient of restitution.

(1)

vB′ − v′A = e(v A − vB ) 0.75ω ′ − v′A = 0.8(8.972 − 0)

(2)

Solving Eqs. (1) and (2) simultaneously (a)

Velocity of A.

v′A = −0.25648 ft/s

(b)

Angular velocity.

ω ′ = 9.228 rad/s

vA′ = 0.256 ft/s 

ω′ = 9.23 rad/s



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PROBLEM 17.125 Two identical slender rods may swing freely from the pivots shown. Rod A is released from rest in a horizontal position and swings to a vertical position, at which time the small knob K strikes rod B which was at rest. If h = 12 l and e = 12 , determine (a) the angle through which rod B will swing, (b) the angle through which rod A will rebound.

SOLUTION m = mAC = mBD

Let

I AC = I =

Moment of inertia.

I BD =

1 mL2 12

1 mL2 12

Rod AB falls to vertical position.

Position 0.

V1 = 0

T1 = 0

Position 1.

V2 = − mg

L 2

L (ω AC )1 2 1 1 T1 = m(v AC )12 + I (ω AC )12 2 2 1 2 = mL (ω AC )12 6 1 1 T0 + V0 = T1 + V1: 0 + 0 = mL2 (ω AC )12 − mgL 6 2 3g (ω AC )12 = L (v AC )1 =

Conservation of energy.

(1)

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PROBLEM 17.125 (Continued)

Impact.

+

Syst. Momenta1

Syst. Ext. Imp.1→ 2 = Syst. Momenta2

L (ω AC )1 2 L (v AC ) 2 = (ω AC ) 2 2 (v AC )1 =

Kinematics

Moments about C:

m(v AC )1

L L + I (ω AC )1 − L  Kdt = m(v AC ) 2   + I (ω AC ) 2 2 2 1 2 1 mL (ω AC )1 − L  Kdt = mL2 (ω AC ) 2 3 3

(2)

Syst. Momenta1 + Syst. Ext. Imp.1→ 2 = Syst. Momenta2 Kinematics Moments about D:

(vBD ) 2 =

L (ωBD ) 2 2

0 + ( L − h)  Kdt = m(vBD ) 2

L + I (ω BD )2 2

1 ( L − h)  Kdt = mL2 (ωBD ) 2 3

(3)

Multiply Eq. (2) by (L − h) and Eq. (3) by L and then add to eliminate  Kdt. 1 2 1 1 mL ( L − h)(ω AC )1 = mL2 ( L − h)(ω AC ) 2 + mL3 (ωBD ) 2 3 3 3

(1)

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PROBLEM 17.125 (Continued)

L(ω AC )2 − ( L − h)(ωBD ) 2 = −eL(ω AC )1

Condition of impact:

(2)

For h = 12 L and e = 0.5 Eqs. (1) and (2) become 1 3 1 1 mL (ω AC )1 = mL3 (ω AC ) 2 + mL3 (ω BD ) 2 6 6 3 −

Dividing Eq. (3) by

1 6

1 = 0.5L(ω AB )1 2

(3) (4)

mL3 and transposing terms gives (ω AC ) 2 + 2(ωBD )2 = (ω AC )1

(5)

Dividing Eq. (4) by L /2 and transposing terms gives 2(ω AC ) 2 − (ωBD ) 2 = −(ω AC )1

(6)

Solving Eqs. (5) and (6) simultaneously for (ω AC ) 2 and (ωBD ) 2 gives

(a)

(ω AC ) 2 = −0.2(ω AC )1

(7)

(ωBD ) 2 = 0.6(ω AC )1

(8)

Angle of swing θ B for rod B.

Apply the principle of conservation of energy to rod B. T2 + V2 = T3 + V3

Position (2): Just after impact. Position (3): At maximum angle of swing.

Potential energy. Use the pivot point D as the datum. L 2 L V3 = − mg cos θ B 2

V2 = − mg

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PROBLEM 17.125 (Continued)

Kinetic energy. 1 1 I D (ωBD ) 22 = mL3 (ω BD ) 22 2 6 T3 = 0

T2 =

1 2 L L mL (ω BD ) 22 − mg = 0 − mg cos θ B 6 2 2 1L 1 − cos θ B = (ω BD ) 22 3g 1L = [0.6(ω AB )1 ]2 3g  L   3g  1 =   (0.36)     = 0.36 3  g  L 

cos θ B = 0.64

(b)

θ B = 50.2° 

Angle of rebound θA for rod A.

Apply the principle of conservation of energy to rod A. T2 + V2 = T4 + V4

Position (2): Just after impact. Position (4): At maximum angle of rebound.

Potential energy. Use the pivot Point C as the datum. V2 = − mg

L 2

V4 = −mg

L cos θ A 2

Kinetic energy. T2 =

1 1 I C (ω AB ) 22 = mL2 (ω AC )22 2 6

T4 = 0

1 2 L L mL (ω AC )22 − mg = 0 − mg cos θ A 6 2 2 1L 1 − cos θ A = (ω AC )22 3g 1L = [−0.2(ω AB )1 ]2 3g  L   3g  1 =   (0.04)     = 0.04 3    g  L 

cos θ A = 0.96

θ A = 16.26° 

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PROBLEM 17.126 A 2-kg solid sphere of radius r = 40 mm is dropped from a height h = 200 mm and lands on a uniform slender plank AB of mass 4 kg and length L = 500 mm which is held by two inextensible cords. Knowing that the impact is perfectly plastic and that the sphere remains attached to the plank at a distance a = 40 mm from the left end, determine the velocity of the sphere immediately after impact. Neglect the thickness of the plank.

SOLUTION Masses and moments of inertia. Sphere:

mS = 2 kg, r = 40 mm = 0.040 m IS =

Plank AB:

2 2 mS r 2 =   (2 kg)(0.04 m)2 = 1.28 × 10−3 kg ⋅ m 2 5 5

m AB = 4 kg, L = 500 mm = 0.5 m I AB =

1  1  mAB L2 =   (4 kg)(0.5 m)2 = 83.333 × 10−3 kg ⋅ m 2 12  12 

Velocity of sphere at impact. vS = 2 gh = (2)(9.81 m/s)(0.200 m) = 1.9809 m/s

Before impact. Linear momentum:

mS v S = (4 kg)(1.9809 m/s) = 7.9236 kg ⋅ m/s

with its line of action lying at distance

L 2

− a from the midpoint of the plank.

L − a = 0.25 m − 0.04 m = 0.21 m. 2

After impact. Assume that both cables are taut so vA is perpendicular to the cable at A and vB is perpendicular to the cable at B.

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PROBLEM 17.126 (Continued)

Kinematics. To locate the instantaneous center C draw line AC perpendicular to vA and line BC perpendicular to vB. Let point G be the mass center of the plank AB and Point S be that of the sphere. CH = L cos 30° + r = (0.500 m) cos 30° + 0.040 m = 0.47301 m HS =

L − a = 0.21 m 2 2

2

CS = CH + HS = 0.51753 m HS = 0.44397 β = 23.94° CH vS = (CS )ω = 0.51753ω

tan β =

vG = ( L cos 30°)ω = 0.43301ω

Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Ext. Imp.1→ 2

=

Syst. Momenta2

Moments about C: L  mS vS  − a  + 0 = mS v1S (CS ) + m AB vG1 (CH ) + I S ω + I ABω 2  2 L  mS vS  − a  = [mS (CS ) 2 + m AB CG + I S + I AB ]ω = I C ω 2  

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PROBLEM 17.126 (Continued)

L  mS vS  − a  = (2 kg)(1.9809 m/s)(0.21 m) = 0.83198 kg ⋅ m 2 /s 2  

where and

I C = (2 kg)(0.51753 m) 2 + (4 kg)(0.43301 m) 2 + 1.28 × 10−3 kg ⋅ m 2 + 83.333 × 10 −3 kg ⋅ m 2 = (0.53567 + 0.75 + 0.00128 + 0.08333) kg ⋅ m 2 = 1.37028 kg ⋅ m 2 0.83198 kg ⋅ m 2 /s = (1.37028 kg ⋅ m 2 )ω

ω = 0.60716 rad/s



v1S = (0.51753 m)(0.60716 rad/s) = 0.31422 m/s vG1 = (0.43301 m)(0.60716 rad/s) = 0.26291 m/s

To check that neither cable becomes slack during the impact, we show that  Adt and  B dt are positive quantities. components:

− mS vS + (  Adt +  Bdt ) cos 30° = − mv1S sin β 3 (  Adt +  Bdt ) = [mS vS − mS v11 sin β ]/ cos 30° 2 = 7.9236 − (2)(0.31422) sin 23.94° = 7.6686 N ⋅ s

components:

0 + (  Adt −  Bdt ) sin 30° = mAB vG1 + mS vS cos β 1 (  Adt −  Bdt ) = (4)(0.26291) + (2)(0.31422) cos 23.94° 2 = 1.6260 N ⋅ s

Solving the simultaneous equation gives  Adt = 6.05 N ⋅ s

 Bdt = 2.80 N ⋅ s

The cables remain taut as assumed. v1S = 0.314 m/s

Velocity of sphere:

23.9° 

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PROBLEM 17.127 Member ABC has a mass of 2.4 kg and is attached to a pin support at B. An 800-g sphere D strikes the end of member ABC with a vertical velocity v1 of 3 m/s. Knowing that L = 750 mm and that the coefficient of restitution between the sphere and member ABC is 0.5, determine immediately after the impact (a) the angular velocity of member ABC, (b) the velocity of the sphere.

SOLUTION mD = 0.800 kg L = 0.750 m 1 L = 0.1875 m 4 m AC = 2.4 kg

Let Point G be the mass center of member ABC. 1 mAC L2 12 1 = (2.4)(0.750) 2 12 = 0.1125 kg ⋅ m 2

IG =

ω′ = ω ′

Kinematics after impact.

v′G =

,

L ω′ , 4

v′A =

L ω′ 4

Conservation of momentum.

Moments about B:

L L L + 0 = mD vD′ + I Gω ′ + mAC vG′ 2 2 4 2 L L  L  ′ mD vD = mD vD +  I G + m AD    ω ′ 4 4   4  

mD vD

(0.800)(3)(0.1875) = (0.800)(0.1875)vD′ + [0.1125 + (2.4)(0.1875) 2 ]ω ′ 0.45 = 0.15vD′ + 0.196875 ω ′

(1)

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PROBLEM 17.127 (Continued)

L ω′ 4 = − e( v D − v A )

vD′ − v′A = vD′ −

Coefficient of restitution.

vD′ − 0.1875ω ′ = −(0.5)(3 − 0)

(2)

Solving Eqs. (1) and (2) simultaneously. (a)

Angular velocity.

ω′ = 3

(b)

Velocity of D.

vD′ = −0.9375

ω′ = 3.00 rad/s



vD′ = 0.938 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1919

PROBLEM 17.128 Member ABC has a mass of 2.4 kg and is attached to a pin support at B. An 800-g sphere D strikes the end of member ABC with a vertical velocity v1 of 3 m/s. Knowing that L = 750 mm and that the coefficient of restitution between the sphere and member ABC is 0.5, determine immediately after the impact (a) the angular velocity of member ABC, (b) the velocity of the sphere.

SOLUTION Let M be the mass of member ABC and I its moment of inertia about B. M = 2.4 kg

I =

1 M (2 L) 2 12

where

L = 750 mm = 0.75 m

Let m be the mass of sphere D.

m = 800 g = 0.8 kg

Impact kinematics and coefficient of restitution.

(v1 sin θ )e = Lω2 − (vD ) n : (vD ) n = Lω2 − (v1 sin θ ) e

(1)

Principle of impulse and momentum.

Syst. Momenta1

Moments about B:

+

Syst. Ext. Imp.1→2 =

Syst. Momenta 2

mv1 L sin θ = I ω2 + m(vD ) n L mv1 L sin θ =

1 M (2 L) 2 ω2 + m[ Lω2 − (v1 sin θ )e]L 12

1 mv1 sin θ = MLω2 − mLω2 − m(v1 sin θ )e 3 v1 1  m(1 + e) sin θ =  M + m  ω2 L 3   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1920

PROBLEM 17.128 (Continued)

(a)

Angular velocity.

ω2 =

(3)(1 + e) mv1 sin θ ( M + 3 m) L

(3)(1.5)(0.8)(3) sin 60° (2.4 + 2.4)(0.75) = 2.5981

ω2 =

(b)

ω 2 = 2.60 rad/s



Velocity of D. From Eq. (1),

(vD ) n = (0.75)(2.5981) − (3sin 60°)(0.5) = 0.64976 m/s (vD )t = v1 cos 60° = 3cos 60° = 1.5 m/s ( v D ) n = 0.64976 m/s ( v D )t = 1.5 m/s

30°

30°

vD = (0.64976) 2 + (1.5) 2 = 1.63468 m/s 0.64976 tan θ = 1.5 θ = 23.4° θ + 30° = 53.4°

vD = 1.635 m/s

53.4° 

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PROBLEM 17.129 Sphere A of mass mA = 2 kg and radius r = 40 mm rolls without slipping with a velocity v1 = 2 m/s on a horizontal surface when it hits squarely a uniform slender bar B of mass mB = 0.5 kg and length L = 100 mm that is standing on end and at rest. Denoting by μk the coefficient of kinetic friction between the sphere and the horizontal surface, neglecting friction between the sphere and the bar, and knowing the coefficient of restitution between A and B is 0.1, determine the angular velocities of the sphere and the bar immediately after the impact.

SOLUTION Before impact sphere A rolls without slipping so that its instantaneous center of rotation is its contact point with the floor.

ω1 =

v1 2 m/s = = 50 rad/s r 0.040 m

ω1 = 50 rad/s



Analysis of impact. Use the principle of impulse and momentum. Let point A be the center of sphere A, point B be the mass center of bar B, and Points P and Q the contact point between the sphere and the bar, Point P being on sphere A and Point Q being on the bar B.

Syst. Momenta1

+

Syst. Ext. Imp.1→2

ω A = ωA

ω B = ωB

v A = vA

v B = vB

=

Syst. Momenta 2

,

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PROBLEM 17.129 (Continued)

Sphere A alone. Moments about A:

I Aω1 + 0 = I Aω A b=

Kinematics:

ω A = ω1

ω A = 50.0 rad/s



L − r = 50 mm − 40 mm = 10 mm = 0.010 m 2

vQ = (vB + bωB ) v A − v Q = −ev1

Condition of impact.

v A − vB − bωB = −ev1

(1)

Bar B alone: Moments about Q:

0 + 0 = I BωB − bmB vB 1 mB L2ω B − bmB vB = 0 12 1 bvB − L2ω B = 0 12

(2)

Sphere A and bar B together. components: mA v1 + 0 = mA v A + mB vB m A v A + mB vB = mA v1

Data:

m A = 2 kg,

mB = 0.5 kg,

v1 = 2 m/s,

L = 0.100 m,

(3) e = 0.1 b = 0.010 m

v A − vB − (0.010 m)ωB = −(0.1)(2 m/s) (0.010 m)vB −

(1)′

1 (0.100 m) 2 ωB = 0 12

(2)′

(2 kg)v A + (0.5 kg)vB = (2 kg)(2 m/s)

(3)′

Solving Eqs. (1)′, (2)′, and (3)′ simultaneously, v A = 1.599 m/s

vB = 1.606 m/s

ωB = 19.27 rad/s ω B = 19.27 rad/s



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PROBLEM 17.130 A large 3-lb sphere with a radius r = 3 in. is thrown into a light basket at the end of a thin, uniform rod weighing 2 lb and length L = 10 in. as shown. Immediately before the impact the angular velocity of the rod is 3 rad/s counterclockwise and the velocity of the sphere is 2 ft/s down. Assume the sphere sticks in the basket. Determine after the impact (a) the angular velocity of the bar and sphere, (b) the components of the reactions at A.

SOLUTION Let Point G be the mass center of the sphere and Point C be that of the rod AB. Rod AB:

WAB = 2 lb. m AB =

2 = 0.06211 lb ⋅ s 2 /ft 32.2 2

I AB =

Sphere:

1 1  10  mP L2 = (0.06211)   = 0.003594 lb ⋅ s 2 ⋅ ft 12 12  12 

WS = 3 lb mS =

3 = 0.09317 lb ⋅ s 2 /ft 32.2 2

IG =

2 2  3 mS r 2 = (0.09317)   = 0.002329 lb ⋅ s 2 ⋅ ft 5 5  12 

Impact. Before impact, bar AB is rotating about A with angular velocity ω0 = ω0

(ω0 = 3 rad/s) and the

sphere is falling with velocity v 0 = v0 (v0 = 2 ft/s). After impact, the rod and the sphere move together, rotating about A with angular velocity ω = ω . Geometry.

R = L2 + r 2 = 102 + 32 = 10.44 in. = 0.8700 ft r 3 tan θ = = θ = 16.7° L 10

Kinematics: Before impact,

vC =

L  5 ω0 =   (3) = 1.25 ft/s 2  12 

After impact,

vC =

L ω′ , 2

vG = Rω ′

θ

Principle of impulse and momentum. Neglect weights of the rod and sphere over the duration of the impact.

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PROBLEM 17.130 (Continued)

(a)

Moments about A: mS v0 L − I ABω0 − m AB vC

or

L L + 0 = I G ω ′ + mS vG′ R + I ABω ′ + m AB vC 2 2

mS v0 L − I ABω02 − m AB vC

L  1  =  I G + mS R 2 + I AB + m AB L2  ω ′ 2  4 

(1)

 10   5 (0.09317)(2)   − (0.003594)(3) − (0.06211)(1.25)    12   12  2  1  10   = 0.002329 + (0.09317)(0.87) 2 + 0.003594 + (0.06211)    ω ′ 4  12    0.112152 = 0.087226ω ′ ω ′ = 1.2858 ω′ = 1.286 rad/s



Normal accelerations at C and G. (aC ) n =

L  5 (ω ′)2 =   (1.2858) 2 = 0.6889 ft/s 2 2  12 

(aG ) n = R(ω ′)2 = (0.87)(1.2858) 2 = 1.4384 ft/s 2

Tangential accelerations at C and G.

α =α

L 5 α= α (aG )t = Rα = 0.87α 2 12 Kinetics. Use bar AB and the sphere as a free body.

(aC )t =

(b)

16.7°

16.7°

ΣM A = Σ( M A )eff : WAB

L L + WS L = I ABα + m AB (aC )t + I Gα + mS (aG )t R 2 2

1   =  I AB + mAB L2 + I G + mS R 2  α 4   2  1  5  10    10  (2)   + (3)   = 0.003594 + (0.06211)   + 0.002329 + (0.09317)(0.87) 2  α 4  12   12    12  

3.3333 = 0.087226α

α = 38.214 rad/s 2

 5 (aC )t =   (38.214) = 15.923 ft/s 2 ,  12 

(aG )t = (0.87)(38.214) = 33.247 ft/s 2

16.7°

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PROBLEM 17.130 (Continued)

ΣFx = Σ( Fx )eff : Ax = − mAB (aC ) n − mS (aG ) n cos16.7° + mS ( aG )t sin16.70 Ax = − (0.06211)(0.6889) − (0.09317)(1.4384) cos16.7° + (0.09317)(33.247) sin16.7° Ax = 0.719 lb

A x = 0.719 lb



ΣFy = Σ( Fy )eff : Ay − WAB − WS = −m AB (aC )t − mS (aG )t cos16.7° − mS (aG ) n sin16.7° Ay − 2 − 3 = −(0.06211)(15.923) − (0.09317)(33.247) cos16.7° − (0.09317)(1.4384) sin16.7° Ay = 1.006 lb

A y = 1.006 lb 

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PROBLEM 17.131 A small rubber ball of radius r is thrown against a rough floor with a velocity v A of magnitude v0 and a backspin ω A of magnitude ω 0 . It is observed that the ball bounces from A to B, then from B to A, then from A to B, etc. Assuming perfectly elastic impact, determine the required magnitude ω 0 of the backspin in terms of v0 and r.

SOLUTION I =

Moment of inertia.

2 2 mr Ball is assumed to be a solid sphere. 5

Impact at A.

Syst. Momenta1

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

For the velocity of the ball to be reversed on each impact, v′A = v A = v0 ω A′ = ω A = ω0

This is consistent with the assumption of perfectly elastic impact. Moments about C:

mv A r cos 60° − I ω A + 0 = I ω ′A − mv′A r cos 60° mv0 r cos 60° −

2 2 2 mr ω0 + 0 = mr 2ω0 − mv0 r cos 60° 5 5 2 rω0 = v0 cos 60° 5

ω0 =

5 v0  4 r

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PROBLEM 17.132 Sphere A of mass m and radius r rolls without slipping with a velocity v1 on a horizontal surface when it hits squarely an identical sphere B that is at rest. Denoting by μk the coefficient of kinetic friction between the spheres and the surface, neglecting friction between the spheres, and assuming perfectly elastic impact, determine (a) the linear and angular velocities of each sphere immediately after the impact, (b) the velocity of each sphere after it has started rolling uniformly.

SOLUTION I =

Moment of inertia.

2 2 mr 5

Analysis of impact. Sphere A.

+

Syst. Momenta1

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

Kinematics: Rolling without slipping in Position 1.

ωA =

v1 r

I ω1 + 0 = I ω A

Moments about G:

ω A = ω1 = Linear components:

v1 r



mv1 − Pdt = mv A

(1)

Analysis of impact. Sphere B.

Syst. Momenta1

Linear components:

+

Syst. Ext. Imp.1→2

0 +  Pdt = mvB

=

Syst. Momenta 2

(2)

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PROBLEM 17.132 (Continued)

Add Equations (1) and (2) to eliminate  Pdt. mv1 = mv A + mvB

Condition of impact. e = 1.

or vB + v A = v1

(3)

vB − v A = ev1 = v1

(4)

Solving Equations (3) and (4) simultaneously,

Moments about G: (a)

v A = 0,

vB = v1

0 + 0 = I ωB

ωB = 0 vA = 0; ω A =

Velocities after impact.

v1 r

;

vB = v1

; ωB = 0 

Motion after Impact. Sphere A.

Syst. Momenta1 +

Syst. Ext. Imp. 1→ 2 =

Condition of rolling without slipping:

Syst. Momenta 2

v′A = ω ′A r

I ω A + 0 + I ω A′ + mv′A r

Moments about C:

 2 2   v1  2 2  mr    + 0 =  mr  ω ′A + m(rω A′ )r 3  r  5  2v ω A′ = 1 7 r 2 v′A = v1 7

Motion after impact. Sphere B.

Syst. Momenta1 +

Syst. Ext. Imp.1→2 =

Syst. Momenta 2

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PROBLEM 17.132 (Continued)

Condition of rolling without slipping: Moments about C:

vB′ = rω B′

mvB r + 0 = I ωB′ + mvB′ r 2  mv1r + 0 =  mr 2  ωB′ + m(rω B′ )r 5  5v ωB′ = 1 7 r 5 vB′ = v1 7

(b)

v′A =

Final Rolling Velocities.

2 v1 7

; v′B =

5 v1 7



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PROBLEM 17.133 In a game of pool, ball A is rolling without slipping with a velocity v 0 as it hits obliquely ball B, which is at rest. Denoting by r the radius of each ball and by μk the coefficient of kinetic friction between a ball and the table and assuming perfectly elastic impact, determine (a) the linear and angular velocity of each ball immediately after the impact, (b) the velocity of ball B after it has started rolling uniformly.

SOLUTION

(a)

2 2 mr 5

I =

Moment of inertia. Impact analysis. Ball A:

+

Syst. Momenta1



Syst. Ext. Imp.1→2

ω0 =

Kinematics of rolling:

=

Syst. Momenta 2

v0  r



mv0 cos θ − Pdt = m(v A ) x

(1)

mv0 sin θ + 0 = m(v A ) y

(2)

Moments about y axis:

I ω0 cos θ + 0 = I ω A cos β

(3)

Moments about x axis:

− I ω0 sin θ + 0 = − I ω A sin β

(4)

Linear components: Linear components:

Ball B:

Syst. Momenta1

Linear components: Linear components:

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

0 + Pdt = m(vB ) x



(5)

0 + 0 = m ( vB ) y

(6)

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PROBLEM 17.133 (Continued)

Moments about y axis:

0 + 0 = I ω B cos γ

(7)

Moments about x axis:

0 + 0 = I ωB sin γ

(8)



Adding Equations (1) and (5) to eliminate Pdt , mv0 cos θ + 0 = m(v A ) x + m(vB ) x (vB ) x + (v A ) x = v0 cos θ

or Condition of impact.

(9)

(vB ) x − (v A ) x = ev0 cos θ = v0 cos θ

e = 1:

(10)

Solving Equations (9) and (10) simultaneously, (v A ) x = 0, (vB ) x = v0 cos θ (v A ) y = v0 sin θ , (vB ) y = 0

From Equations (2) and (6),

v A = (v0 sin θ ) j  v B = (v0 cos θ )i 

From Equations (3) and (4) simultaneously,

β = θ , ω A = ω0 =

v0 r

ωA =

v0 (− sin θ i + cos θ j)  r

From Equations (7) and (8) simultaneously,

ωB = 0 (b)

ωB = 0 

Subsequent motion of ball B.

Syst. Momenta1

+

Syst. Ext. Imp.1→2 =

Kinematics of rolling without slipping. Moments about C:

Syst. Momenta 2

vB′ = rω B′

mvB r + 0 = I ωB′ + mvB′ r 2 2 mr ωB′ + m(rω B′ )r 5 5 v′ 5 v cos θ ωB′ = B = 1 7 r 7 r 5 vB′ = v1 cos θ  7 =

v′B =

5 (v0 cos θ )i  7

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PROBLEM 17.134 Each of the bars AB and BC is of length L = 400 mm and mass m = 1.2 kg. Determine the angular velocity of each bar immediately after the impulse QΔt = (1.5 N ⋅ s) i is applied at C.

SOLUTION Principle of impulse and momentum. Bar BC:

Syst. Momenta1 + Syst. Ext. Imp.1→2 =

vBC = vB +

Kinematics Moments about B:

Syst. Momenta 2

L L ωBC = Lω AB + ωBC 2 2

0 + (QΔt ) L = I ωBC + mvBC

L 2

L 1  L mL2ωBC + m  Lω AB + ω BC  12 2  2 1 1 QΔt = mLω AB + mLωBC 2 3

(QΔt ) L =

x components:

L   QΔt − Bx Δt = m  Lω AB + ωBC  2  

(1) (2)

Bar AB:

Syst. Momenta1 + Syst. Ext. Imp.1→2 =

Syst. Momenta 2

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PROBLEM 17.134 (Continued)

Moments about A:

0 + ( Bx Δt ) L = I ω AB + mv AB ( Bx Δt ) L =

L 2

1 L L mL2ω AB + m  ω AB  12 2  2

1 Bx Δt = mLω AB 3

(3)

We now have 3 unknowns (Bx Δ t1, ωAB, and ωBC) and 3 equations. Add Eqs. (2) and (3): Subtract Eq. (1) from Eq. (4):

QΔt =

4 1 mLω AB + mLω BC 3 2

0=

5 1 mLω AB + mLωBC 6 6

(4)

ωBC = −5ω AB Substitute for ωBC in Eq. (1):

1 1 mLω AB + mL(−5ω AB ) 2 3 7 = − mLω AB 6

QΔt =

ω AB = − Substituting into Eq. (5):

6 QΔt 7 mL

(6)

 6 QΔt    7 mL 

ωBC = −5  − ωBC =

Given data:

(5)

30 QΔt 7 mL

(7)

L = 0.400 m QΔt = 1.5 N ⋅ s m = 1.2 kg

Angular velocity of bar AB.

ω AB = −

Angular velocity of bar BC.

ωBC =

6 Q Δt (6)(1.5) =− 7 mL (7)(1.2)(0.4)

30 QΔt (30)(1.5) = 7 mL (7)(1.2)(0.4)

ω AB = 2.68 rad/s

ω BC = 13.39 rad/s

 

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PROBLEM 17.135 A uniform disk of constant thickness and initially at rest is placed in contact with the belt shown, which moves at a constant speed v = 80 ft/s. Knowing that the coefficient of kinetic friction between the disk and the belt is 0.15, determine (a) the number of revolutions executed by the disk before it reaches a constant angular velocity, (b) the time required for the disk to reach that constant angular velocity.

SOLUTION F f = μ k N = 0.15 N

Kinetic friction.

ΣFy = N cos 25° − F f sin 25° − mg = 0 (cos 25° − μ k sin 25°)N = mg mg cos 25° − 0.15sin 25° = 1.18636 mg

N=

F f = (0.15)(1.18636)mg = 0.177954 mg

ω2 =

Final angular velocity.

I =

Moment of inertia. (a)

v r 1 2 mr 2

Principle of work and energy. T1 + W1→ 2 = T2 : T1 = 0

W1→2 = F f rθ = 0.177954mgrθ 2

1 11 1  v  I ω22 =  mr 2   = mv 2 2 22 4  r  1 0 + 0.177954mgrθ = mv 2 4 v2 θ = 1.40486 gr T2 =

=

(1.40486)(80) 2 (32.2) ( 125 )

= 670.14 radians

θ = 106.7 rev 

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PROBLEM 17.135 (Continued)

(b)

Principle of impulse-momentum.

Syst. Momenta1

Moments about A:

+

Syst. Ext. Imp.1→2 =

Syst. Momenta 2

0 + F f tr = I ω2 t= =

I ω2 Ff r

(

1 2

mr 2

)( ) v r

0.177954mgr v = 2.8097 g (2.8097)(80) =  32.2

t = 6.98 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1936

PROBLEM 17.136 The 8-in.-radius brake drum is attached to a larger flywheel that is not shown. The total mass moment of inertia of the flywheel and drum is 14 lb ⋅ ft ⋅ s 2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.35. Knowing that the initial angular velocity of the flywheel is 360 rpm counterclockwise, determine the vertical force P that must be applied to the pedal C if the system is to stop in 100 revolutions.

SOLUTION

ω1 = 360 rpm = 12π rad/s ω2 = 0

Kinetic energy.

1 2 I ω1 2 1 = (14)(12π ) 2 2 = 9.9486 × 103 ft ⋅ lb T2 = 0 T1 =

θ = (100)(2π ) = 628.32 rad

Work.

 8  M D = Ff r = Ff    12   8  U1→2 = − M Dθ = − F f   (628.32)  12  = −418.88F f

Principle of work and energy. T1 + U1→ 2 = T2 : 9.9486 × 103 − 418.88F f = 0 F f = 23.75 lb

Kinetic friction force.

F f = μk N N=

Statics.

Ff

μk

=

23.75 = 67.859 lb 0.35

ΣM A = 0: (9 in.) P + (2 in.) F f − (10 in.) N = 0 9P + (2)(23.75) − (10)(67.859) = 0 P = 70.12

P = 70.1 lb 

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PROBLEM 17.137 A 6 × 8 − in. rectangular plate is suspended by pins at A and B. The pin at B is removed and the plate swings freely about pin A. Determine (a) the angular velocity of the plate after it has rotated through 90°, (b) the maximum angular velocity attained by the plate as it swings freely.

SOLUTION Let m be the mass of the plate. a = 8 in. = 0.66667 ft b = 6 in. = 0.5 ft

Dimensions: Moment of inertia about A

1 I A = m(a 2 + b2 ) 3

Position 1. Initial position.

ω1 = 0

Position 2. Plate has rotated about A through 90°. Position 3. Mass center is directly below pivot A.

Potential energy. Use level A as datum. V1 = −

V2 = −

mga 2

V3 = −mgd

1 2 a + b 2 = 0.41667 ft 2

Where

d=

Kinetic energy.

T1 = 0

(a)

mab 2

T2 =

1 I Aω22 2

T3 =

1 I Aω32 2

90° rotation. Conservation of energy. mgb 1 1 mga = ⋅ m(a 2 + b 2 )ω22 + 2 2 3 2 3 ( ) (3)(32.2)(0.66667 0.5) − − g a b ω22 = 2 = = 23.184(rad/s)2 (0.66667) 2 + (0.5) 2 a + b2

T1 + V1 = T2 + V2 : 0 −

ω 2 = 4.81 rad/s



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PROBLEM 17.137 (Continued)

(b)

ω is maximum. Conservation of energy. mgb 1 1 = ⋅ m(a 2 + b 2 )ω32 − mgd 2 2 3 (32.2)(2.5 − 1.5) g (6d − 3b) 2 ω3 = 2 2 = = 46.386 (rad/s)2 2 2 (0.66667) + (0.5) a +b

T1 + V1 = T3 + V3 : 0 −

ω3 = 6.81 rad/s



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PROBLEM 17.138 The gear shown has a radius R = 150 mm and a radius of gyration k = 125 mm. The gear is rolling without sliding with a velocity v1 of magnitude 3 m/s when it strikes a step of height h = 75 mm. Because the edge of the step engages the gear teeth, no slipping occurs between the gear and the step. Assuming perfectly plastic impact, determine (a) the angular velocity of the gear immediately after the impact, (b) the angular velocity of the gear after it has rotated to the top of the step.

SOLUTION Part (a) Conditions just after impact. Kinematics. Just before impact, the contact Point C with the floor the instantaneous center of rotation of the gear. v1 = Rω1

Just after impact, Point S is the instantaneous center of rotation. v 2 = Rω2

θ

(perpendicular to GS )

Principle of impulse and momentum.

Moments about S:

mv2 ( R − h) + I ω1 = mv2 R + I ω2

m( Rω1 )( R − h) + mk 2ω1 = m( Rω2 ) R + mk 2ω2 [ R( R − h) + k 2 ]ω1 = ( R 2 + k 2 )ω2

ω2 = Data:

R 2 + k 2 − Rh ω1 R2 + k 2



Rh



ω2 = 1 − 2 ω 2 1  R +k 

(1)

R = 150 mm, k = 125 mm, v1 = 3 m/s, h = 75 mm

ω1 =

v1 3 m/s = = 20 rad/s R 0.150 m

Angular velocity. From (1),



ω2 = 1 − 

(150)(75)   (20 rad/s) = 0.7049(20) (1502 + 1252 ) 

ω2 = 14.10 rad/s



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PROBLEM 17.138 (Continued)

Part (b) Conditions at the top of the step. The gear pivots about the edge of the step. Use the principle of conservation of energy. Position (2): The gear has just broken contact with the floor. Position (3): The center of the gear is above the edge of the step.

Kinematics: (Rotation about S) v = ω R Kinetic energy:

Position (2):

Position (3):

1 1 I ω 2 + mv 2 2 2 1 1 = mk 2ω 2 + mR 2ω 2 2 2 1 2 2 = m(k + R )ω 2 2

T=

1 m(k 2 + R 2 )ω22 2 V2 = mgR T2 =

1 m( k 2 + R 2 )ω32 2 V3 = mg ( R + h) T3 =

T2 + V2 = T3 + V3

Principle of conservation of energy:

1 1 m( k 2 + R 2 )ω22 + mgR = m( k 2 + R 2 )ω32 + mg ( R + h) 2 2

Angular velocity:

ω32 = ω32 −

2 gh k + R2 2

= (14.10 rad/s)2 −

(2)(9.81 m/s 2 )(0.075 m) (0.125 m) 2 + (0.150 m) 2

= 160.21 rad 2 /s 2 ω3 = 12.66 rad/s



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PROBLEM 17.139 A uniform slender rod is placed at corner B and is given a slight clockwise motion. Assuming that the corner is sharp and becomes slightly embedded in the end of the rod, so that the coefficient of static friction at B is very large, determine (a) the angle β through which the rod will have rotated when it loses contact with the corner, (b) the corresponding velocity of end A.

SOLUTION Position 1

T1 = 0

Position 2

mgL 2 mgL cos β V2 = mgh2 = 2 1 2 11 2 2 T2 = Iω2 =  mL  ω2 2 23  V1 = mgh1 =

Principle of conservation of energy. T1 + V1 = T2 + V2 0+

mgL 1  1 2  2 mgL cos β =  mL  ω2 + 2 23 2  3g ω22 = (1 − cos β ) L

(1)

Normal acceleration of mass center. an =

L 2 3 ω2 = g (1 − cos β ) 2 2

ΣF = +ΣFeff = man mg cos β =

(a)

(b)

Angle β .

3 mg (1 − cos β ) 2

5 3 cos β = 2 2

cos β = 0.6

3g g (1 − 0.6) = 1.2 L L

From (1)

ω22 =

Velocity of end A

v A = Lω2

β = 53.1°  ω2 = 1.09545

g L vA = 1.095 gL

53.1° 

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PROBLEM 17.140 The motion of the slender 250-mm rod AB is guided by pins at A and B that slide freely in slots cut in a vertical plate as shown. Knowing that the rod has a mass of 2 kg and is released from rest when θ = 0, determine the reactions at A and B when θ = 90°.

SOLUTION Let Point G be the mass center of rod AB. m = 2 kg L = 0.25 m 1 I G = mL2 = 0.0104667 kg ⋅ m 2 12

Kinematics.

θ = 90° AD = R = 0.125 m AB = L = 0.25 m R 1 = β = 30° L 2 L AG = = 0.125 m 2 BG = 0.125 m

sin β =

Point E is the instantaneous center of rotation of bar AB. L ω = 0.125 ω 2 v A = ( L cos 30°)ω = 0.21651ω

vG =

vB = ( L sin 30°)ω = 0.125ω

Use principle of conservation of energy to obtain the velocities when θ = 90°: Use level A as the datum for potential energy. Position 1

θ =0 T1 = 0 V1 = − mg

L = −(2)(9.81)(0.125) = −2.4525 J 2

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PROBLEM 17.140 (Continued)

Position 2

θ = 90° 1 1 I G ω 2 + mvG2 2 2 1 1 = (0.0104667)ω 2 + (2)(0.125ω )2 2 2 2 = 0.0208583ω

T2 =

L   V2 = − mg  R + cos β  2   = −(2)(9.81)(0.125 + 0.125cos30°) = −4.5764 J T1 + V1 = T2 + V2 : 0 − 2.4525 = 0.0208583ω 2 − 4.5764

ω 2 = 101.826 rad 2 /s 2 ω = 10.091 rad/s v A = (0.21651)(10.091) = 2.1848 m/s vG = (0.125)(10.091) = 1.2614 m/s

More kinematics:

For Point A moving in the curved slot, v A2 j R (2.1847) 2 = ( aC ) x i + j 0.125 = ( aC ) x i + 38.1833j

a A = ( aC ) x i +

For the rod AB,

α = α k , v B = vB j rA/B = − L sin 30°i + L cos 30° j

rG/B

= −0.125i + 0.21651j 1 = rA/B 2 = −0.0625i + 0.108253j

a A = a B + α + rA/B − ω 2 rA/B = aB j + α k × (−0.125i + 0.21651j) − (10.091) 2 ( −0.125i + 0.21651j) = aB j − 0.125α j − 0.21651α i + 12.7285i − 22.0468 j

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PROBLEM 17.140 (Continued)

Matching vertical components of a A 38.1833 = aB − 0.125α − 22.0468 aB = 0.125α + 60.2301 aG = a B + aG/B = a B + α k × rG/B − ω 2 rG/B = (0.125α + 60.2324) j + α k × (−0.0625i + 0.108253j) − (10.091) 2 (−0.0625i + 0.108253j) = 0.125α j + 60.2301j − 0.0625α j − 0.108253α i + 6.3643i − 11.0232 j aG = ( −0.108253α + 6.3643)i + (0.0625α + 49.2069) j

Kinetics:

Use rod AB as a free body. ΣM E = Σ(M E )eff : L sin β k = I G α + rG /E × ( maG ) 2 −(2)(9.81)(0.125)sin 30°k − mg

= 0.0104667α + (0.0625i + 0.10825 j) × (maG ) −1.22625 = 0.0104667α + 0.03125α + 4.7730 0.0417167α = −5.99925

α = −143.808 rad/s 2 aG = ( −21.933 m/s 2 )i + (40.2189 m/s 2 ) j

ΣFx = Σ( Fx )eff = m(aG ) x : −B = (2)(−21.932) = 43.864 N

B = 43.9 N



ΣFy = Σ( Fy )eff = m(aG ) y : A − mg = (2)(40.4289) = 80.4378 A = (2)(9.81) + 80.4378 = 100.058

A = 100.1 N 

Check by considering ΣM G = ΣM G eff :

ΣM G = (0.0625) A − 0.108253B = 1.5052 N ⋅ m Σ( M G )eff = I G (−α ) = (0.0104667)(143.808) = 1.5052 N ⋅ m

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PROBLEM 17.141 A 35-g bullet B is fired horizontally with a velocity of 400 m/s into the side of a 3-kg square panel suspended from a pin at A. Knowing that the panel is initially at rest, determine the components of the reaction at A after the panel has rotated 45°.

SOLUTION mB = 0.035 kg mP = 3 kg b = 500 mm = 0.5 m 1 1 I G = mP b 2 = (3)(0.5) 2 = 0.125 kg ⋅ m 2 6 6 Note: The mass of the bullet is neglected in comparison with that of the plate after impact. Analysis of impact: Use principle of impulse and momentum.

Masses and moment of inertia.

Kinematics: After impact the plate rotates about the pin at A. vG =

Moments about A:

b 2

ω=

0.5 2

ω

 b b  b + mB v0   = I G ω + m p vG 2  2 2 2 3

1 mP b 2 )ω 2 2 2 3 1   (0.035)(400)(0.5) = 0.125 + (3)(0.5) 2  ω 2 2 2   ω = 14.8492 rad/s 0.5 (14.8492) = 5.25 m/s vG = 2 mG v0b = ( I G +

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PROBLEM 17.141 (Continued)

Corresponding kinetic energy.

1 1 I G ω 2 + mP vG2 2 2 1 1 T1 = (0.125)(14.8492)2 + (3)(5.25) 2 2 2 = 55.125 J T1 =

Plate rotates through 45°.

θ = 0°

Position 1:

Use Point A as the datum for potential energy. V1 = −mP g

b 2

= −(3)(9.81)

0.5 2

= −10.4051 J

θ = 90°

Position 2:

V2 = 0 since G is at level A. T2 =

1 1 I G ω22 + mP (vG ) 22 2 2

1 1  0.5  = (0.125)ω22 + (3)  ω2  2 2  2 

2

= 0.25ω22

Principle of conservation of energy: T1 + V1 = T2 + V2 55.125 J − 10.4051 J = 0.25ω22 + 0

ω22 = 178.879 (rad/s 2 ) ω2 = 13.3746 rad/s Analysis at 90° rotation. Kinematics:

α =α (aG )t =

b

(aG ) n =

b

=

2

α=

0.5 2

α

(aG )t = 0.35355α

ω2

2 (0.5)(178.879) 2

(aG ) n = 63.2434 m/s 2

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PROBLEM 17.141 (Continued)

Kinematics: Use the free body diagram of the plate. ΣM A = Σ( M A )eff : mP g

b 2

= I Gα + mP (aG )t

b 2

1   =  I G + mP b 2  α 2   (3)(9.81)(0.5)  1  = 0.125 + (3)(0.5) 2  α 2 2   α = 20.810 rad/s 2 (aG )t = 7.3574 m/s 2 ΣFx = Σ( Fx )eff : Ax = mP (aG ) n = (3)(63.2434)

A x = 189.7 N



ΣFy = Σ( Fy )eff : Ay − mP g = −mP ( aG ) y Ay − (3)(9.81) = (3)(−7.3574)

A y = 7.36 N 

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PROBLEM 17.142 Two panels A and B are attached with hinges to a rectangular plate and held by a wire as shown. The plate and the panels are made of the same material and have the same thickness. The entire assembly is rotating with an angular velocity ω0 when the wire breaks. Determine the angular velocity of the assembly after the panels have come to rest against the plate.

SOLUTION Geometry and kinematics:

Panels in up position

Panels in down position

v0 = bω0

v2 =

3 b ω0 2

Let ρ = mass density, t = thickness Plate:

mplate = ρ t (2b)(4b) = 8 ρ tb 2 1 (8 ρ tb 2 )[(2b) 2 + (4b) 2 ] 12 160 = ρ tb 4 12 40 = ρ tb 4 3

I plate =

Each panel: Panel in up position

mpanel = ρ t (b)(2b) = 2 ρ tb 2 1 (2 ρ t b 2 )(2b) 2 12 8 2 = ρ t b4 = ρ tb4 12 3

( I panel )0 =

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PROBLEM 17.142 (Continued)

Panel in down position

1 (2 ρ tb2 )[b 2 + (2b)2 ] 12 10 = ρ tb 4 12 5 = ρ tb 4 6

( I panel )1 =

Conservation of angular momentum about the vertical spindle.

Final momenta

Initial momenta Moments about C:

  3b   I plateω0 + 2[( I panel )0 ω0 + mpanel v0 (b)] = I plateω1 + 2 ( I panel )1 ω1 + mpanel v1     2   5 40 2  40 3  3   ρ tb 4ω0 + 2  ρ tb 4ω0 + (2 ρ tb2 )(bω0 )b  = ρ tb 4ω0 + 2  ρ tb 4ω1 + 2 ρ tb2  bω0  b   3 3 3 6 2     2     40 10   40 4  4 4  3 + 3 + 4  ρ tb ω0 =  3 + 6 + 9  ρ tb ω1     56 ω0 = 24ω1 3 56 ω1 = (3)(24)

7 9

ω1 = ω 2 

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PROBLEM 17.143 Disks A and B are made of the same material and are of the same thickness; they can rotate freely about the vertical shaft. Disk B is at rest when it is dropped onto disk A, which is rotating with an angular velocity of 500 rpm. Knowing that disk A has a mass of 8 kg, determine (a) the final angular velocity of the disks, (b) the change is kinetic energy of the system.

SOLUTION m A = 8 kg rA = 0.15 m

Disk A:

IA =

1 1 mA rA2 = (8)(0.15)2 = 0.0900 kg ⋅ m 2 2 2

rB = 0.100 m

Disk B:

2

r   0.10  mB = mA  B  = (8)   = 3.5556 kg r  0.15   A 1 1 I B = mB rB2 = (3.5556)(0.10)2 = 0.017778 kg ⋅ m 2 2 2 2

Principle of impulse and momentum.

+

Syst. Momenta1

Syst. Ext. Imp. 1→2 =

Syst. Momenta2

I Aω0 + 0 + 0 = I Aω2 + I Bω2

Moments about B:

ω2 = Initial angular velocity of disk A:

IA 0.09 ω1 = ω1 = 0.83505ω1 0.10778 IA + IB

ω1 = 500 rpm = 52.36 rad/s

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PROBLEM 17.143 (Continued)

(a)

Final angular velocity of system:

ω2 = (0.83505)(52.36) ω2 = 43.723 rad/s

Initial kinetic energy:

Final kinetic energy:

(b)

Change in energy:

T1 =

1 I Aω12 2

T1 =

1 (0.09)(52.36)2 = 123.37 J 2

T2 =

1 ( I A + I B )ω22 2

T2 =

1 (0.10778)(43.723) 2 = 103.02 J 2

T2 − T1 = −20.35 J

ω2 = 418 rpm 

ΔT = −20.4 J 

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PROBLEM 17.144 A square block of mass m is falling with a velocity v1 when it strikes a small obstruction at B. Knowing that the coefficient of restitution for the impact between corner A and the obstruction B is e = 0.5, determine immediately after the impact (a) the angular velocity of the block, (b) the velocity of its mass center G.

SOLUTION 1 2 mb 6 Before impact, block is translating. I =

Moments of inertia. Kinematics.

v1 = v1

ω1 = 0

vA = ev1

After impact,

v 2 = vA + vG/ A

 45° 

 b = [ev1 ] +  ω2  2

(1)

Principle of impulse and momentum.

Syst. Momenta1

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta2

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PROBLEM 17.144 (Continued) Moments about A: mv1

b b b b = Iω1 − mev1 + m ω2 2 2 2 2 1 I = mb 2 6 2

 b  b 1 b mv1 = mb 2ω2 − mev1 + m   ω2 2 6 2  2 (1 + e)v1b 2 2 = b ω2 2 3

(a)

Angular velocity.

(b)

Velocity of the mass center. From Eq. (1),

ω2 =

3 (1 + e)v1 4b

ω 2 = 1.125

 45°  

 b 3 v 2 = ev1 +  (1 + e)v1  2 4b

  3  3 = ev1 +  (1 + e)v1 sin 45°  +  (1 + e)v1 cos 45°  4 2 4 2  3 3 = ev1 +  (1 + e)v1  +  (1 + e)v1  8 8

 5 3   3 =  ev1 − v1   +  (1 + e)v1 8   8  8  5 3  3 =  (0.5) −  v1  + (1 + 0.5)v1 8  8  8 = [−0.0625 ] + [0.5625

v1  b

  

        

] v 2 = 0.566 m/s

6.34°

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PROBLEM 17.145 A 3-kg bar AB is attached by a pin at D to a 4-kg square plate, which can rotate freely about a vertical axis. Knowing that the angular velocity of the plate is 120 rpm when the bar is vertical, determine (a) the angular velocity of the plate after the bar has swung into a horizontal position and has come to rest against pin C, (b) the energy lost during the plastic impact at C.

SOLUTION Moments of inertia about the vertical centroidal axis. 1 1 mL2 = (4)(0.500)2 = 0.083333 kg ⋅ m 2 12 12

Square plate.

I =

Bar AB vertical.

I = approximately zero

Bar AB horizontal.

I =

Position 1. Bar AB is vertical.

I1 = 0.083333 kg ⋅ m 2

Angular velocity.

ω1 = 120 rpm = 4π rad/s

1 1 mL2 = (3)(0.500) 2 = 0.0625 kg ⋅ m 2 12 12

Angular momentum about the vertical axis. ( H O )1 = I1ω1 = (0.083333)(4π ) = 1.04720 kg ⋅ m 2 /s 1 1 I1ω12 = (0.083333)(4π ) 2 = 6.5797 J 2 2

Kinetic energy.

T1 =

Position 2. Bar AB is horizontal.

I 2 = 0.145833 kg ⋅ m 2

( H O ) 2 = I 2ω2 = 0.145833ω2

Conservation of angular momentum. ( H O )1 = ( H O ) 2 : 1.04720 = 0.145833ω2 ω2 = 7.1808 rad/s

(a)

Final angular velocity.

(b)

Loss of energy.

ω2 = 68.6 rpm 

T1 − T2 = T1 −

1 1 I 2ω22 = 6.5797 − (0.145833)(7.1808) 2 2 2 T1 − T2 = 2.82 J 

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PROBLEM 17.146 A 1.8-lb javelin DE impacts a 10-lb slender rod ABC with a horizontal velocity of v0 = 30 ft/s as shown. Knowing that the javelin becomes embedded into the end of the rod at Point C and does not penetrate very far into it, determine immediately after the impact (a) the angular velocity of the rod ABC after the impact, (b) the components of the reaction at B. Assume that the javelin and the rod move as a single body after the impact.

SOLUTION Masses and moments of inertia. m AC =

WAC 10 lb = = 0.31056 lb ⋅ s 2 /ft g 32.2 ft/s 2

WDE 1.8 lb = = 0.05590 lb ⋅ s 2 /ft g 32.2 ft/s 2 1 1 = m AC L2AC = (0.31056)(10) 2 = 2.5880 lb ⋅ s 2 ⋅ ft 12 12 1 1 = mDE L2DE = (0.05590)(8.5) 2 = 0.3365 lb ⋅ s 2 ⋅ ft 12 12

mDE = I AC I DE

(a)

Angular velocity immediately after the impact.

Principle of impulse and momentum. ω DE = ω AB = ω = ω

Syst. Momenta1 Moments about B: where

+

Syst. Ext. Imp. 1→2

=

Syst. Momenta2

mDE v0 r0 + 0 = I AC ω + m AC v AC r1 + I DE ω + mDE vDE r2 v0 = 30 ft/s r0 = 10 ft − 2 ft = 8 ft r1 = 5 ft − 2 ft = 3 ft r2 = (4.25 ft) 2 + (8 ft) 2 = 9.0588 ft

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PROBLEM 17.146 (Continued)

Kinematics: (Rotation about B) v AC = r1ω vDE = r2ω tan β =

4.25 ft 8 ft

β = 27.98°

mDE v0 r0 = I AC ω + m AC r12ω + I DE ω + mDE r22ω = I Bω

where

I B = I AC + mAC r12 + I DE + mDE r22 = 2.5880 + (0.31056)(3) 2 + 0.3365 + (0.05590)(9.0588) 2 = 10.3068 lb ⋅ s 2 ⋅ ft mDE v0 r0 (0.05590 lb ⋅ s 2 /ft)(30 ft/s)(8 ft) = IB 10.3068 lb ⋅ s 2 ⋅ ft = 1.30167 rad/s

ω=

ω = 1.302 rad/s



α =α

Accelerations:

a AD = [r1α

] + [r1ω 2 ]

aDE = [r2α

β ] + [r2ω 2

β]

Free body and kinetic diagrams.

Moments about B: WDE r4 = I AC α + mAC ( a AC )t r1 + I DEα + mDE ( aDE )t r2 = I AC α + mAC r12α + I DEα + mDE r22α = I Bα

α=

WDE r4 (1.8 lb)(4.25) = IB 10.3068 lb ⋅ s 2 ⋅ ft

= 0.74223 rad/s 2

α = 0.742 rad/s 2

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PROBLEM 17.146 (Continued)

(b)

Components of reaction at B. ΣF = Σma :

B + [WAC ] + [WDE ] = [m AC r1α

Component

] + [mAC r1ω 2 ] + [mDE r2α

β ] + [ mDE r2ω 2

β]

: Bx = m AC r1α + mDE ( r2 cos β )α + mDE (r2 sin β )ω 2 = (0.31056)(3)(0.74223) + (0.05590)(8)(0.74223) + (0.05590)(4.25)(1.30167) 2 = 0.6915 + 0.3319 + 0.4025

B x = 1.426 lb

Component



:

By − WAC − WDC = m AC r1ω 2 − mDE (r1 sin β )α + mDE (r2 cos β )ω 2 B y − 10 − 1.8 = (0.31056)(3)(1.30167) 2 − (0.05590)(4.25)(0.74223) + (0.05590)(8)(1.30167) 2 B y = 11.8 + 1.5785 − 0.1763 + 0.7577

B y = 13.96 lb 

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CHAPTER 18

PROBLEM 18.1 A thin, homogeneous disk of mass m and radius r spins at the constant rate ω1 about an axle held by a fork-ended vertical rod, which rotates at the constant rate ω 2 . Determine the angular momentum H G of the disk about its mass center G.

SOLUTION Angular velocity:

ω = ω2 j + ω1k

Moments of inertia:

Ix =

Products of inertia: by symmetry,

I xy = I yz = I zx = 0

Angular momentum:

H G = I xω x i + I yω y j + I zω z k

1 2 mr , 4

HG = 0 +

Iy =

1 2 1 mr , I z = mr 2 4 2

1 2 1 mr ω2 j + mr 2ω1k 4 2

HG =

1 2 1 mr ω2 j + mr 2ω1k  4 2

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PROBLEM 18.2 A thin rectangular plate of weight 15 lb rotates about its vertical diagonal AB with an angular velocity ω. Knowing that the z axis is perpendicular to the plate and that ω is constant and equal to 5 rad/s, determine the angular momentum of the plate about its mass center G.

SOLUTION

h = (9 in.)2 + (12 in.) 2 = 15 in.

Resolving ω along the principal axes x′, y′, z: 12 ω = 0.8(5 rad/s) = 4 rad/s 15 9 ω y′ = ω = 0.6(5 rad/s) = 3 rad/s 15 ωz = 0

ω x′ =

2

Moments of inertia:

I x′ =

1  15 lb  9  ft  = 0.021836 slug ⋅ ft 2 12  32.2   12 

I y′ =

1  15 lb  12  ft  = 0.038820 slug ⋅ ft 2 12  32.2  12  

2

From Eqs. (18.10):

H x′ = I x′ω x′ = (0.021836)(4) = 0.087345 slug ⋅ ft 2 /s H y′ = I y′ω y ′ = (0.038820)(3) = 0.11646 slug ⋅ ft 2 /s H z′ = I z ωz = 0

H G = (0.087345 slug ⋅ ft 2 /s)i ′ + (0.11646 slug ⋅ ft 2 /s) j′

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PROBLEM 18.2 (Continued)

Components along x and y axes: 3 4 3 4 H x′ − H y′ =   (0.087345) −   (0.11646) 5 5 5 5 = −0.040761 4 3 H y = H x′ + H y ′ 5 5 4 3 =   (0.087345) +   (0.11646) = 0.13975 5 5 Hx =

H G = −(0.0408 slug ⋅ ft 2 /s)i + (0.1398 slug ⋅ ft 2 /s) j 

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PROBLEM 18.3 Two uniform rods AB and CE, each of weight 3 lb and length 2 ft, are welded to each other at their midpoints. Knowing that this assembly has an angular velocity of constant magnitude ω = 12 rad/s, determine the magnitude and direction of the angular momentum HD of the assembly about D.

SOLUTION m=

W 3 = = 0.093168 lb ⋅ s2 /ft, g 32.2

l = 2 ft,

ω = (12 rad/s)i

For rod ADB,

H D = I xω i ≈ 0,

since I x ≈ 0.

For rod CDE, use principal axes x′, y′ as shown. cosθ =

9 , 12

θ = 41.410°

ω x′ = ω cosθ = 9 rad/s 2 ω y′ = ω sin θ = 7.93725 rad/s 2 ω z′ = 0 I x′ ≈ 0 I y′ =

1 2 1 (0.093168)(2)2 ml = 12 12

= 0.0310559 lb ⋅ s 2 ⋅ ft

H D = I x′ω x′i′ + I y′ω y′ j′ + I z′ω z′k′ = 0 + (0.0310559)(7.93725) j′ + 0 = 0.246498 j′ H D = 0.246 lb ⋅ s ⋅ ft 

H D = 0.246498(sin θ i + cos θ j) = 0.163045i + 0.184874 j cosθ x =

0.163045 0.246498

θ x = 48.6° 

cosθ y =

0.184874 0.246498

θ y = 41.4° 

cosθ z = 0

θ z = 90° 

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PROBLEM 18.4 A homogeneous disk of weight W = 6 lb rotates at the constant rate ω 1 = 16 rad/s with respect to arm ABC, which is welded to a shaft DCE rotating at the constant rate ω 2 = 8 rad/s. Determine the angular momentum H A of the disk about its center A.

SOLUTION ω = ω1 j + ω2 i = (8 rad/s)i + (16 rad/s) j

For axes x′, y ′, z ′ parallel to x, y, z with origin at A, m=

W 6 = = 0.186335 lb ⋅ s 2 /ft g 32.2

I x′ =

1 2 1  8 mr = (0.186335)   4 4  12 

2

= 0.020704 lb ⋅ s 2 ⋅ ft I z ′ = I x′ = 0.020704 lb ⋅ s 2 ⋅ ft I y ′ = I x′ + I z′ = 0.041408 lb ⋅ s 2 ⋅ ft H A = I x′ω x′i + I y′ω y′ j + I z ′ω z′k = (0.020704)(8)i + (0.041408)(16) j = 0.1656i + 0.6625 j H A = (0.1656 lb ⋅ ft ⋅ s)i + (0.663 lb ⋅ ft ⋅ s) j 

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PROBLEM 18.5 A thin disk of mass m = 4 kg rotates at the constant rate ω2 = 15 rad/s with respect to arm ABC, which itself rotates at the constant rate ω1 = 5 rad/s about the y axis. Determine the angular momentum of the disk about its center C.

SOLUTION r = 150 mm

Angular velocity of disk: ω = ω1 j + ω2k = (5 rad/s) j + (15 rad/s)k

Centroidal moments of inertia: I x′ = I y ′ =

1 2 mr 4

1 (4)(0.150 m) 2 = 0.0225 kg ⋅ m 2 4 1 I z′ = mr 2 = 0.045 kg ⋅ m 2 2 =

Angular momentum about Point C. H C = I x′ω x′ i + I y′ω y′ j + I z′ω z′k = 0 + (0.0225)(5) j + (0.045)(15)k = (0.1125 kg ⋅ m 2 /s) j + (0.6750 kg ⋅ m 2 /s)k H C = (0.1125 kg ⋅ m 2 /s) j + (0.675 kg ⋅ m 2 /s)k 



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PROBLEM 18.6 A solid rectangular parallelepiped of mass m has a square base of side a and a length 2a. Knowing that it rotates at the constant rate ω about its diagonal AC ′ and that its rotation is observed from A as counterclockwise, determine (a) the magnitude of the angular momentum H G of the parallelepiped about its mass center G, (b) the angle that H G forms with the diagonal AC ′.

SOLUTION Body diagonal:

d = a 2 + (2a) 2 + a 2 = 6a ω= Ix Iy Iz

(a)

HG

ω

( − ai + 2 aj − ak ) = −

ω

d 6 1 5 = m[(2a) 2 + a 2 ] = ma 2 12 12 1 1 2 2 2 = m[a + a ] = ma 12 6 1 5 2 2 = m[a + (2a) ] = ma 2 12 12 = I xω x i + I y ω y j + I z ω z k

i+

2ω 6

j−

ω 6

k

ω   5   ω   1 2   2ω   5 2  =  ma 2   −  i +  ma    j +  ma   − k 12 6 12 6  6    6    ma 2ω = (−5i + 4 j − 5k ) 12 6 HG =

(b)

HG ⋅ ω = =

ma 2ω 2 11ma 2ω 5 + 4 2 + 52 = 12 12 6

H G = 0.276 ma 2ω 

ma 2ω 2 ( −5i + 4 j − 5k ) ⋅ (−i + 2 j − k ) (12)(6) 18ma 2ω 2 1 = ma 2ω (12)(6) 4

11 ma 2ω 2 12 H ⋅ω 12 cos θ = G = = 0.90453 H Gω 4 11

H Gω =

θ = 25.2° 

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PROBLEM 18.7 Solve Problem 18.6, assuming that the solid rectangular parallelepiped has been replaced by a hollow one consisting of six thin metal plates welded together. PROBLEM 18.6 A solid rectangular parallelepiped of mass m has a square base of side a and a length 2a. Knowing that it rotates at the constant rate ω about its diagonal AC ′ and that its rotation is observed from A as counterclockwise, determine (a) the magnitude of the angular momentum H G of the parallelepiped about its mass center G, (b) the angle that H G forms with the diagonal AC ′.

SOLUTION d = a 2 + ( 2a ) + a 2 = 6 a 2

Body diagonal:

ω=

ω d

( −ai + 2aj − ak ) = −

ω 6

i+

2ω 6

j−

ω 6

k

Total area = 2( a 2 + 2a 2 + 2a 2 ) = 10a 2 1 m 10 1 13 13 I x = m′a 2 + m′a 2 = m′a 2 = ma 2 12 12 120 1 1 I y = m′a 2 = ma 2 6 60 13 Iz = Ix = ma 2 120

For each square plate:

m′ =

For each plate parallel to the yz plane:

m′ =

1 m 5

Ix =

1 5 1 m′[a 2 + (2a )2 ] = m′a 2 = ma 2 12 12 12

Iy =

1 1 1 a m′a 2 + m′   = m′a 2 = ma 2 12 3 15 2

Iz =

1 7 7 a m(2a) 2 + m′   = m′a 2 = ma 2 12 2 12 60  

2

2

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PROBLEM 18.7 (Continued)

For each plate parallel to the xy plane:

1 m′ = m 5 2

Ix =

1 7 7 a m′(2a )2 + m′   = m′a 2 = ma 2 12 2 12 60   2

1 1 1 a m′a 2 + m′   = m′a 2 = ma 2 12 3 15 2 1 5 1 I z = m′[ a 2 + (2a) 2 ] = m′a 2 = ma 2 12 12 12

Iy =

Total moments of inertia: 1 7  37 2  13 Ix = 2 ma + +  ma 2 = 60  120 12 60  1 1  3  1 I y = 2 + +  ma 2 = ma 2 10  60 15 15  7 1  37 2  13 Iz = 2 ma + +  ma 2 = 60  120 60 12 

(a)

H G = I xω x i + I y ω y j + I z ω z k = HG =

ma 2ω 60 6

ma 2ω 60 6

(−37i + 36 j − 37k )

(37) 2 + (36)2 + (37)2 = 0.43216 ma 2ω H G = 0.432 ma 2ω 

(b)

HG ⋅ ω = cos θ =

ma 2ω (−37i + 36 j − 37k ) ⋅ (−i + 2 j − k ) = 0.40556 ma 2ω 2  (60)(6) H G ⋅ ω 0.40556 = = 0.93844 0.43216 H Gω

θ = 20.2° 

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PROBLEM 18.8 A homogeneous disk of mass m and radius r is mounted on the vertical shaft AB. The normal to the disk at G forms an angle β = 25° with the shaft. Knowing that the shaft has a constant angular velocity ω, determine the angle θ formed by the shaft AB and the angular momentum HG of the disk about its mass center G.

SOLUTION Use the principal centroidal axes Gx′, y′ z. Moments of inertia: I x′ = I z = I y′ =

1 2 mr 4

1 2 mr 2

Angular velocities:

ω x′ = − ω sin β ω y′ = ω cos β ωz = 0 Using Eq. (18.10): 1 H x′ = I x′ω x′ = − mr 2ω sin β 4 1 2 H y′ = I y′ω y ′ = mr ω cos β 2 H z = I zω z = 0

We have H G = H x′ i ′ + H y′ j′ + H z k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1970

PROBLEM 18.8 (Continued)

where i′, j′, k are the unit vectors along the x′, y′, z axes. 1 1 H G = − mr 2ω sin β i ′ + mr 2ω cos β j′ 4 2

HG =

(1)

1 2 mr ω (− sin β i ′ + 2cos β j′ ) 4

Forming the scalar product, H G ⋅ ω = |H G | ω cos θ cos θ =

But or observing that

HG ⋅ ω =

HG ⋅ ω |H G | ω

(2)

1 2 mr ω (− sin β i ′ + 2 cos β j′) ⋅ ω j 4

i ′ ⋅ i = − sin β

and

j′ ⋅ j = cos β

1 2 2 mr ω (sin 2 β + 2cos 2 β ) 4 1 = mr 2ω 2 (1 + cos 2 β ) 4

HG ⋅ ω =

Also,

|H G | ω = =

(3)

1 2 2 mr ω sin 2 β + 4cos 2 β 4 1 2 2 mr ω 1 + 3cos 2 β 4

(4)

Substituting from Eqs. (3) and (4) into Eq. (2), cos θ =

For β = 25°,

1 + cos 2 β 1 + 3cos 2 β

cos θ = 0.9786

θ = 11.88° 

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PROBLEM 18.9 Determine the angular momentum HD of the disk of Problem 18.4 about Point D. PROBLEM 18.4 A homogeneous disk of weight W = 6 lb rotates at the constant rate ω 1 = 16 rad/s with respect to arm ABC, which is welded to a shaft DCE rotating at the constant rate ω 2 = 8 rad/s. Determine the angular momentum H A of the disk about its center A.

SOLUTION ω = ω2 i + ω1 j = (8 rad/s)i + (16 rad/s) j

For axes x′, y ′, z ′ parallel to x, y, z with origin at A, m=

W 6 = = 0.186335 lb ⋅ s 2 /ft g 32.2

I x′ =

1 2 1  8  mr = (0.186335)   4 4  12 

2

= 0.020704 lb ⋅ s 2 ⋅ ft

I z ′ = I x′ = 0.020704 lb ⋅ s 2 ⋅ ft, I y ′ = I x′ + I z′ = 0.041408 lb ⋅ s 2 ⋅ ft

H A = I x′ω x′i + I y′ω y ′ j + I z′ω z ′k = (0.020704)(8)i + (0.041408)(16) j = (0.1656 lb ⋅ ft ⋅ s)i + (0.6625 lb ⋅ ft ⋅ s) j

Point A is the mass center of the disk. rA/D = (1.0 ft)i + (0.75 ft) j − (0.75 ft)k v = v A = ω2 i × rA/D = 8i × (1.0i + 0.75 j − 0.75k ) = (6 ft/s) j + (6 ft/s)k

mv = (1.118 lb ⋅ s) j + (1.118 lb ⋅ s)k

i j k rA/D × mv = 1.0 0.75 −0.75 0 1.118 1.118 = (1.677 lb ⋅ ft ⋅ s)i − (1.118 lb ⋅ ft ⋅ s) j + (1.118 lb ⋅ ft ⋅ s)k H D = H A + rA/D × mv

H D = (1.843 lb ⋅ ft ⋅ s)i − (0.455 lb ⋅ ft ⋅ s) j + (1.118 lb ⋅ ft ⋅ s)k 

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PROBLEM 18.10 Determine the angular momentum of the disk of Problem 18.5 about Point A. PROBLEM 18.5 A thin disk of mass m = 4 kg rotates at the constant rate ω2 = 15 rad/s with respect to arm ABC, which itself rotates at the constant rate ω1 = 5 rad/s about the y axis. Determine the angular momentum of the disk about its center C.

SOLUTION r = 150 in.

Angular velocity of disk: ω = ω1 j + ω2 k = (5 rad/s) j + (15 rad/s)k

Centroidal moments of inertia: I x′ = I y ′ =

1 2 mr 4

1 (4)(0.150 m) 2 = 0.0225 kg ⋅ m 2 4 1 2 I z′ = mr = 0.045 kg ⋅ m 2 2 =

Angular momentum about Point C. H C = I x′ω x′ i + I y′ω y′ j + I z ′ω z ′k = 0 + (0.0225)(5) j + (0.045)(15)k = (0.1125 kg ⋅ m 2 /s) j + (0.6750 kg ⋅ m 2 /s)k

Location of mass center. Velocity of mass center.

rC/ A = (0.450 m)i + (0.225 m) j v = ω1 × rC/ A = 5 j × (0.45i + 0.225 j)

= −(2.25 m/s)k

Angular momentum about Point A. H A = H C + rC/ A × (mv ) H A = 0.1125 j + 0.675k + (0.45i + 0.225 j) × [−(4)(2.25)k ]

= 0.1125 j + 0.675k + 4.05 j − 2.025i



H A = −(2.03 kg ⋅ m 2 /s)i + (4.16 kg ⋅ m 2 /s) j + (0.675 kg ⋅ m 2 /s)k 

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PROBLEM 18.11 Determine the angular momentum H O of the disk of Sample Problem 18.2 from the expressions obtained for its linear momentum mv and its angular momentum HG using Eq. (18.11). Verify that the result obtained is the same as that obtained by direct computation. PROBLEM 18.2 A homogeneous disk of radius r and mass m is mounted on an axle OG of length L and negligible mass. The axle is pivoted at the fixed Point O, and the disk is constrained to roll on a horizontal floor. Knowing that the disk rotates counterclockwise at the rate ω1 about the axle OG, determine (a) the angular velocity of the disk, (b) its angular momentum about O, (c) its kinetic energy, (d) the vector and couple at G equivalent to the momenta of the particles of the disk.

SOLUTION Using Equation (18.11), H O = r × mv + H G 1 r   = ( Li ) × ( mrω1k ) + mr 2ω1  i − j 2  2L  = − mrLω1 j + HO =

1 2 1 r3 mr ω1i − m ω1 j 2 4 L

1 2 1   rω   mr ω1i − m  L2 + r 2  1  j 2 4  L  

which is the answer obtained in Part b of Sample Problem 18.2.

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PROBLEM 18.12 The 100-kg projectile shown has a radius of gyration of 100 mm about its axis of symmetry Gx and a radius of gyration of 250 mm about the transverse axis Gy. Its angular velocity ω can be resolved into two components; one component, directed along Gx, measures the rate of spin of the projectile, while the other component, directed along GD, measures its rate of precession. Knowing that θ = 6° and that the angular momentum of the projectile about its mass center G is H G = (500 g ⋅ m 2 /s)i − (10 g ⋅ m 2 /s)j, determine (a) the rate of spin, (b) the rate of precession.

SOLUTION m = 100 kg, k x = 100 mm = 0.1 m, k y = 250 mm = 0.25 m I x = mk x2 = (100)(0.1) 2 = 1 kg ⋅ m 2

I y = I z = mk y2 = (100)(0.25)2 = 6.25 kg ⋅ m 2 H G = ( H G ) x i + ( H G ) y j + ( H G ) z k = I xω x i + I y ω y j + I z ω z k

ωx = ωy =

( H G ) x 0.500 kg ⋅ m 2 /s = = 0.5 rad/s Ix 1 kg ⋅ m 2 (HG ) y

Iy

=

−0.01 kg ⋅ m 2 /s = −0.0016 rad/s 6.25 kg ⋅ m 2

ωz = 0 ωP sin θ = −ω y ωP = (a)

−ω y sin θ

=

0.0016 = 0.015307 rad/s sin 6°

Rate of spin.

ω x = ωs + ωP cos θ ωs = ω x − ω P cos θ = 0.5 − 0.15307 cos 6° = 0.4847

(b)

ωs = 0.485 rad/s  ωP = 0.01531 rad/s 

Rate of precession.

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PROBLEM 18.13 Determine the angular momentum HA of the projectile of Problem 18.12 about the center A of its base, knowing that its mass center G has a velocity v of 750 m/s. Give your answer in terms of components respectively parallel to the x and y axes shown and to a third axis z pointing toward you. PROBLEM 18.12 The 100-kg projectile shown has a radius of gyration of 100 mm about its axis of symmetry Gx and a radius of gyration of 250 mm about the transverse axis Gy. Its angular velocity ω can be resolved into two components; one component, directed along Gx, measures the rate of spin of the projectile, while the other component, directed along GD, measures its rate of precession. Knowing that θ = 6° and that the angular momentum of the projectile about its mass center G is H G = (500 g ⋅ m 2 /s)i − (10 g ⋅ m 2 /s)j, determine (a) the rate of spin, (b) the rate of precession.

SOLUTION m = 100 kg, rG /A = (0.300 m)i v = v cos θ i − v sin θ j H G = (0.50 kg ⋅ m 2 /s)i − (0.10 kg ⋅ m 2 /s) j mv = (100)(750)(cos 6°i − sin 6° j) = (74589 kg ⋅ m/s)i − (7839 kg ⋅ m/s)j rG /A × mv = 0.3i × (74589i − 7839.6 j)

= −(2351.9 kg ⋅ m 2 /s)k H A = H G + rG /A × mv = 0.5i − 0.1j + (−2351.9k ) H A = (0.500 kg ⋅ m 2 /s)i − (0.100 kg ⋅ m 2 /s) j − (2350 kg ⋅ m 2 /s)k 

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PROBLEM 18.14 (a) Show that the angular momentum H B of a rigid body about Point B can be obtained by adding to the angular momentum HA of that body about Point A the vector product of the vector rA/B drawn from B to A and the linear momentum mv of the body: H B = HA + rA/B × mv

(b) Further show that when a rigid body rotates about a fixed axis, its angular momentum is the same about any two Points A and B located on the fixed axis (HA = H B ) if, and only if, the mass center G of the body is located on the fixed axis.

SOLUTION (a)

Angular momenta H A and H B are related to H G and mv by HA = rG/ A × mv + H G

and H B = rG/B × mv + H G

H B − H A = rG/B × mv − rG/ A × mv

Subtracting,

H B = H A + (rG/B − rG/ A ) × mv = H A + (rG/B + rA/G ) × mv H B = H A + rA/B × mv

(b)

It follows that H A = H B if, and only if rA/B × mv = 0

With Points A and B located on the fixed axis, ω = ωλ

where λ is a unit vector along the fixed axis, and v = ω× rG/ A = ωλ × rG/ A

Then

rA/B × (mωλ × rG /A ) = 0

but rA/B is parallel to λ, hence, λ × (λ × rG /A ) = 0

Let u = λ × rG/ A , so that λ × u = 0. Note that u must be either perpendicular to λ or equal to zero. But if u is perpendicular to λ, λ × u cannot be equal to zero. Hence,

u = λ × rG/ A = 0

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PROBLEM 18.15 A 5-kg rod of uniform cross section is used to form the shaft shown. Knowing that the shaft rotates with a constant angular velocity ω of magnitude 12 rad/s, determine (a) the angular momentum H G of the shaft about its mass center G, (b) the angle formed by H G and the axis AB.

SOLUTION

ω = (12 rad/s)i,

ω y = ωz = 0

( H G ) x = I xω ( H G ) y = − I xyω ( H G ) z = − I xzω

The shaft is comprised of 8 sections, each of length m a = 0.25 m and of mass m′ = = 0.625 kg. 8 10 10 1  I x = (4)  m′a 2  + (2)(m′a 2 ) = m′a 2 = (0.625)(0.25)2 = 0.130208 kg ⋅ m 2 3 3 3  I xy = 0 a  I xz = (4)  m′a  = 2m′a 2 = (2)(0.625)(0.25)2 = 0.078125 kg ⋅ m 2 2  ( H G ) x = (0.130208)(12) = 1.5625 kg ⋅ m 2 /s (HG ) y = 0 ( H G ) z = −(0.078125)(12) = −0.9375 kg ⋅ m 2 /s H G = (1.563 kg ⋅ m 2 /s)i − (0.938 kg ⋅ m 2 /s)k 

(a) HG =

(b)

(1.5625) 2 + (0.9375) 2 = 1.82217 kg ⋅ m 2 /s

H G ⋅ ω = (1.5625i − 0.9375k ) ⋅ 12i = 18.75 kg ⋅ m 2 /s 2

cos θ =

HG ⋅ ω 18.75 = = 0.85749 (1.82217)(12) H Gω

θ = 31.0° 

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PROBLEM 18.16 Determine the angular momentum of the shaft of Problem 18.15 about (a) Point A, (b) Point B.

SOLUTION

ω = (12 rad/s)i,

ω y = ω z = 0,

m = 5 kg

( H G ) x = I xω ( H G ) y = − I xyω ( H G ) z = − I xzω

The shaft is comprised of 8 sections, each of length m a = 0.25 m and of mass m′ = = 0.625 kg. 8 10 10 1  I x = (4)  m′a 2  + (2)(m′a 2 ) = m′a 2 = (0.625)(0.25)2 = 0.130208 kg ⋅ m 2 3 3 3  I xy = 0 a  I xz = (4)  m′a  = 2m′a 2 = (2)(0.625)(0.25) 2 = 0.078125 kg ⋅ m 2 2  ( H G ) x = (0.130208)(12) = 1.5625 kg ⋅ m 2 /s (HG ) y = 0 ( H G ) z = −(0.078125)(12) = −0.9375 kg ⋅ m 2 /s H G = (1.5625 kg ⋅ m 2 /s)i − (0.9375 kg ⋅ m 2 /s)k

Since Point G lies on the axis of rotation, its velocity is zero. v = vG = 0

(a)

H A = H G + rG /A × mv = H G

H A = (1.563 kg ⋅ m 2 /s)i − (0.938 kg ⋅ m 2 /s)k 

(b)

H B = H G + rG /B × mv = H G

H B = (1.563 kg ⋅ m 2 /s)i − (0.938 kg ⋅ m 2 /s)k 

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PROBLEM 18.17 Two L-shaped arms, each weighing 4 lb, are welded at the third points of the 2-ft shaft AB. Knowing that shaft AB rotates at the constant rate ω = 240 rpm, determine (a) the angular momentum of the body about A, (b) the angle formed by the angular momentum and shaft AB.

SOLUTION W = 4 lb, m =

ω=

4 = 0.12422 lb ⋅ s 2/ft , a = 8 in. = 0.66667 ft 32.2

(2π )(240) = 8π rad/s, ω x = 0, ω y = 0, ω z = 8π rad/s 60

Use parallel axes x′, y ′, z ′ with origin at Point A as shown. ( H A ) x′ = − I x′z′ω ( H A ) y ′ = − I y′z′ω ( H A ) z′ = I z′ω

Segments 1, 2, 3, and 4, each of mass m′ = 0.06211 lb ⋅ s 2/ft , contribute to I x′z′ , I y′z′ , and I z′ . Part

Σ

I x′z′

I y′z ′

I z′

2m′a 2

− m′a 2

 1 1  2  12 + 4 + 1 m′a  

m′a 2

0

1 m′a 2 3

1 − m′a 2 2

0

1 m′a 2 3

− m′a 2

1 − m′a 2 2

 1 1  2  12 + 4 + 1 m′a  

3 m′a 2 2

3 − m′a 2 2

10 m′a 2 3

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PROBLEM 18.17 (Continued)

(a)

Angular momentum about A. 3 ( HA ) x′ = − I x′z′ω = − m′a 2ω 2 3 = − (0.06211)(0.66667) 2 (8π ) 2 = −1.04067 lb ⋅ s ⋅ ft  3  ( HA ) y′ = − I y′z ′ω = −  − m′a 2  ω  2  3 = (0.06211)(0.66667) 2 (8π ) 2 = 1.04067 lb ⋅ ft ⋅ s ( HA ) z ′ = I z′ω =

10 m′a 2ω 3

10 (0.06211)(0.66667)2 (8π ) 3 = 2.3126 lb ⋅ ft ⋅ s

=

HA = −(1.041 lb ⋅ ft ⋅ s)i + (1.041 lb ⋅ ft ⋅ s)j + (2.31 lb ⋅ ft ⋅ s)k 

HA = (1.04067)2 + (1.04067)2 + (2.3126) 2 = 2.7412 lb ⋅ ft ⋅ s

(b)

Angle formed by HA and shaft AB.

Unit vector along shaft AB:

λ = −k cos θ =

H A ⋅ λ −2.3126 = = −0.84365 2.7412 HA

θ = 147.5° 

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PROBLEM 18.18 For the body of Problem 18.17, determine (a) the angular momentum about B, (b) the angle formed by the angular momentum about shaft BA.

SOLUTION W = 4 lb. m =

4 = 0.12422 lb ⋅ s 2 /ft 32.2

a = 8 in. = 0.66667 ft

ω=

(2π )(240) = 8π rad/s, ω x = 0, ω y = 0, ω z = 8π rad/s 60

Use parallel axes x′, y ′, z with origin at Point B as shown. ( H B ) x = − I xzω ( H B ) y = − I yz ω ( H B ) z = I zω

Segments 1, 2, 3, and 4, each of mass m′ = 0.06211 lb ⋅ s 2/ft, contribute to I xz , I yz , and I z . Part

Σ

I xz

I yz

Iz

− m′a 2

1 m′a 2 2

 1 1  2  12 + 4 + 1 m′a  

1 − m′a 2 2

0

1 m′a 2 3

m′a 2

0

1 m′a 2 3

2m′a 2

m′a 2

 1 1  2  12 + 4 + 1 m′a  

3 m′a 2 2

3 m′a 2 2

10 m′a 2 3

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PROBLEM 18.18 (Continued)

(a)

Angular momentum about B. 3 ( HB ) x = − I xz ω = − m′a 2ω 2 3 = − (0.06211)(0.66667)2 (8π ) 2 = −1.04067 lb ⋅ s ⋅ ft 3 ( HB ) y = − I yz ω = − m′a 2ω 2 3 = − (0.06211)(0.66667) 2 (8π ) 2 = −1.04067 lb ⋅ s ⋅ ft ( HB ) z = I zω =

10 m′a 2ω 3

10 (0.06211)(0.66667)2 (8π ) 3 = 2.3126 lb ⋅ s ⋅ ft

=

H B = −(1.041 lb ⋅ ft ⋅ s)i − (1.041 lb ⋅ ft ⋅ s)j + (2.31 lb ⋅ ft ⋅ s)k 

HB = (1.04067)2 + (1.04067)2 + (2.3126) 2 = 2.7412 lb ⋅ ft ⋅ s

(b)

Angle formed by H B and shaft BA.

Unit vector along shaft BA:

λ =k cos θ =

H B ⋅ λ 2.3126 = = 0.84365 2.7412 HB

θ = 32.5° 

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PROBLEM 18.19 The triangular plate shown has a mass of 7.5 kg and is welded to a vertical shaft AB. Knowing that the plate rotates at the constant rate ω = 12 rad/s, determine its angular momentum about (a) Point C, (b) Point A. (Hint: To solve part b find v and use the property indicated in part a of Problem 18.14.)

SOLUTION ω = (12 rad/s)j, ω x = 0, ω y = 12 rad/s, ω z = 0

( H C ) x = − I xyω , ( H C ) y = I yω , (H C ) z = − I yzω

(a)

Use axes with origin at C as shown. Divide the plate ABD into right triangles ACD and CBD. For plate ACD, the product of inertia of the area is ( I xy )area = −

1 2 2 a c 24

For plate BCD, it is ( I xy )area =

1 2 2 a b 24

For both areas together, ( I xy )area = −

Area:

A= ( I xy ) mass =

For both areas together,

1 2 (c − b 2 ) a 2 24

1 (c + b)a 2 m m(c − b ) a ( I xy )area = − A 12

( I y )area =

1 (c + b) a 3 12

( I y ) mass =

m 1 ( I y )area = ma 2 A 6

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PROBLEM 18.19 (Continued)

m = 7.5 kg a = 120 mm = 0.12 m

Data:

b = 90 mm = 0.09 m c = 160 mm = 0.16 m ( I xy ) mass = − ( I y ) mass =

(7.5)(0.07)(0.12) = −0.00525 kg ⋅ m 2 12

(7.5)(0.12) 2 = 0.018 kg ⋅ m 2 6

( H C ) x = −(−0.00525)(12) = 0.063 kg ⋅ m 2/s ( H C ) y = (0.018)(12) = 0.216 kg ⋅ m 2/s (HC )z = 0

Locate the mass center. rG/C = Velocity of mass center:

H C = (0.063 kg ⋅ m 2/s)i + (0.216 kg ⋅ m 2/s)j 

a i + yj 3

v = ω × rG/C

1 a  1 v = ω j ×  i + y j  = − ω ak = −   (12)(0.12)k = −(0.48 m/s)k 3 3   3 rC/ A = cj = (0.16 m)j

rC/ A × mv = (0.16 j) × [(7.5)(−0.48k )] = −(0.576 kg ⋅ m 2/s)i

(b)

HA = H C + rC/ A × mv = (0.063 − 0.576)i + 0.216 j HA = −(0.513 kg ⋅ m 2/s)i + (0.216 kg ⋅ m 2/s)j 

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PROBLEM 18.20 The triangular plate shown has a mass of 7.5 kg and is welded to a vertical shaft AB. Knowing that the plate rotates at the constant rate ω = 12 rad/s, determine its angular momentum about (a) Point C, (b) Point B. (See hint of Problem 18.19.)

SOLUTION ω = (12 rad/s)j, ω x = 0, ω y = 12 rad/s, ω z = 0

( H C ) x = − I xyω , ( H C ) y = I yω ,

(a)

(H C ) z = − I yzω

Use axes with origin at C as shown. Divide the plate ABD into right triangles ACD and CBD. For plate ACD, the product of inertia of the area is ( I xy )area = −

1 2 2 a c 24

For plate BCD, it is ( I xy )area =

1 2 2 a b 24

For both areas together, ( I xy )area = −

Area:

A= ( I xy ) mass =

1 2 (c − b 2 ) a 2 24

1 (c + b)a 2 m m(c − b ) a ( I xy )area = − A 12

For both areas together, 1 (c + b ) a 3 12 1 m = ( I y )area = ma 2 6 A ≈0

( I y )area = ( I y ) mass ( I xz ) mass

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PROBLEM 18.20 (Continued)

m = 7.5 kg a = 120 mm = 0.12 m

Data:

b = 90 mm = 0.09 m c = 160 mm = 0.16 m ( I xy ) mass = − ( I y ) mass =

(7.5)(0.07)(0.12) = −0.00525 kg ⋅ m 2 12

(7.5)(0.12) 2 = 0.018 kg ⋅ m 2 6

( H C ) x = −( −0.00525)(12) = 0.063 kg ⋅ m 2/s ( H C ) y = (0.018)(12) = 0.216 kg ⋅ m 2/s (HC )z = 0

H C = (0.063 kg ⋅ m 2/s)i + (0.216 kg ⋅ m 2/s)j 

Locate the mass center. rG/C = Velocity of mass center:

a i + yj 3

v = ω × rG/C 1 a  1 v = ω j ×  i + y j  = − ω ak = −   (12)(0.12)k = −(0.48 m/s)k 3 3   3

rC/B = −bj = −(0.09 m)j rC/B × mv = (−0.09 j) × [(7.5)(−0.48k )] = (0.324 kg ⋅ m 2/s)i

(b)

H B = H C + rC/B × mv = (0.063 + 0.324)i + 0.216 j H B = (0.387 kg ⋅ m 2/s)i + (0.216 kg ⋅ m 2/s)j 



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PROBLEM 18.21 One of the sculptures displayed on a university campus consists of a hollow cube made of six aluminum sheets, each 1.5 × 1.5 m, welded together and reinforced with internal braces of negligible weight. The cube is mounted on a fixed base at A and can rotate freely about its vertical diagonal AB. As she passes by this display on the way to a class in mechanics, an engineering student grabs corner C of the cube and pushes it for 1.2 s in a direction perpendicular to the plane ABC with an average force of 50 N. Having observed that it takes 5 s for the cube to complete one full revolution, she flips out her calculator and proceeds to determine the mass of the cube. What is the result of her calculation? (Hint: The perpendicular distance from the diagonal joining two vertices of a cube to any of its other six vertices can be obtained by multiplying the side of the cube by 2/3.)

SOLUTION Let m′ = 16 m be the mass of one side of the cube. Choose x, y, and z axes perpendicular to the face of the cube. Let a be the side of the cube. For a side perpendicular to the x axis, ( I x )1 = 16 m′a 2 . For a side perpendicular to the y or z axis, Total moment of inertia:

1  1 1 ( I x ) 2 =  +  m′a 2 = m′a 2 12 4 3   5 5 I x = 2( I x )1 + 4( I x ) 2 = m′a 2 = ma 2 3 18

By symmetry, I y = I x and I z = I x . Since all three moments of inertia are equal, the ellipsoid of inertia is a sphere. All centroidal axes are principal axes. 5 I v = ma 2 Moment of inertia about the vertical axis: 18 Let b =

2 a 3

be the moment arm of the impulse applied to the corner.

Using the impulse-momentum principle and taking moments about the vertical axis, bF (Δt ) = H v = I vω = a = 1.5 m,

Data:

ω = Solving Equation (1) for m,

b=

2π = 1.25664 rad/s, 5 m=

5 ma 2ω 18

(1)

2 (1.5) = 1.22474 m 3

F = 50 N,

Δt = 1.2 s.

18 bF (Δt ) 18 (1.22474)(50)(1.2) = = 93.563 kg 5 a 2ω 5 (1.5) 2 (1.25664)

m = 93.6 kg 

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PROBLEM 18.22 If the aluminum cube of Problem 18.21 were replaced by a cube of the same size, made of six plywood sheets with mass 8 kg each, how long would it take for that cube to complete one full revolution if the student pushed its corner C in the same way that she pushed the corner of the aluminum cube?

SOLUTION Let m′ = 16 m be the mass of one side of the cube. Choose x, y, and z axes perpendicular to the face of the cube. Let a be the side of the cube. For a side perpendicular to the x axis, ( I x )1 = 16 m′a 2 . 1  1 1 ( I x ) 2 =  +  m′a 2 = m′a 2 12 4 3   5 5 I x = 2( I x )1 + 4( I x ) 2 = m′a 2 = ma 2 3 18

For a side perpendicular to the y or z axis, Total moment of inertia: By symmetry, I y = I x and I z = I x .

Since all three moments of inertia are equal, the ellipsoid of inertia is a sphere. All centroidal axes are principal axes. 5 Moment of inertia about the vertical axis: I v = ma 2 18 Let b =

2 a 3

be the moment arm of the impulse applied to the corner.

Using the impulse-momentum principle and taking moments about the vertical axis, bF (Δt ) = H v = I vω = m′ = 8 kg,

Data: b=

5 ma 2ω 18

m = 6m′ = 48 kg,

2 a = 1.22474 m, 3

F = 50 N,

(1) a = 1.5 m, Δt = 1.2 s

Solving (1) for ω ,

ω =

18 bF (Δt ) 18 (1.22474)(50)(1.2) = = 2.4495 rad/s 5 ma 2 5 (48)(1.5) 2 t =



ω

=

2π 2.4495

t = 2.57 s 

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PROBLEM 18.23 A uniform rod of total mass m is bent into the shape shown and is suspended by a wire attached at B. The bent rod is hit at D in a direction perpendicular to the plane containing the rod (in the negative z direction). Denoting the corresponding impulse by FΔt, determine (a) the velocity of the mass center of the rod, (b) the angular velocity of the rod.

SOLUTION We apply the principle of impulse and momentum, considering only the impulsive forces.

(a)

Velocity of mass center vy = 0

From constraints:

0 = mvx

x components:

T Δt = mv y = 0

y components: z components:

vx = 0 T Δt = 0 vz = −

− F Δt = mvt

F Δt m

v=−

(b)

F Δt k  m

Angular velocity Equating moments about G: (− ai − a j) × (− F Δt k ) = H x i + H y j + H z k aF Δt i − aF Δt j = H x i + H y j + H z k

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PROBLEM 18.23 (Continued)

Thus:

H x = aF Δt , H y = −aF Δt , H z = 0

(1)

To determine angular velocity, we shall use Eqs. (18.7). First, we determine the moments & products of inertia: Ix =

1 m m m 2 (2a )2 + a 2 + a 2 = ma 2   12  2  4 4 3

(2)

1  m   1 I y = 2    a 2  = ma 2 3 4   6 I xy =

(3)

ma m a 1 (a ) +  −  (− a) = + ma 2   4 2 4  2 4

I xz = 0

(4)

I yz = 0

(5)

We substitute the expressions (1) through (5) into Eqs. (18.7): aF Δt =

2 2 1 ma ω x − ma 2ω y + 0 3 4

(6)

1 1 − aF Δt = − ma 2ω x + ma 2ω y + 0 4 6

(7)

0 = 0 + 0 + I zωz

(8)

Multiplying Eq. (7) by 3/2 and adding to Eq. (6): 1 7 − aF Δt = ma 2ω x 2 24

ωx = −

12 F Δt 7 ma

Substituting for ω into (7): − aF Δt +

From Eq. (8):

1  122 − 4  7

 1 2 60 F Δt  aF Δt = 6 ma ω y , ω y = − 7 ma 

I zω z = 0 ω z = 0 ω=

Thus:

12 F Δt (−i − 5 j)  7 ma

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PROBLEM 18.24 Solve Problem 18.23, assuming that the bent rod is hit at C. PROBLEM 18.23 A uniform rod of total mass m is bent into the shape shown and is suspended by a wire attached at B. The bent rod is hit at D in a direction perpendicular to the plane containing the rod (in the negative z direction). Denoting the corresponding impulse by FΔt, determine (a) the velocity of the mass center of the rod, (b) the angular velocity of the rod.

SOLUTION We apply the principle of impulse and momentum, consider only impulsive forces.

(a)

Velocity of mass center vy = 0

From constraints:

0 = mvx

x components:

T Δt = mv y = 0

y components: z components:

vx = 0 T Δt = 0 vz = −

− F Δt = mvz

F Δt m

v=

(b)

F Δt k  m

Angular velocity Equating moments about G: − a j × (− F Δt k ) = H x i + H y j + H z k

aF Δt i = H x i + H y j + H z k

Thus:

H x = aF Δt , H y = 0, H z = 0

(1)

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PROBLEM 18.24 (Continued)

To determine angular velocity, we shall use Eqs. (18.7) first, we determine the moments & products of inertia: Ix =

1 m m m 2 (2a )2 + a 2 + a 2 = ma 2   12  2  4 4 3

(2)

1  m   1 I y = 2    a 2  = ma 2 3 4   6 I xy =

ma m a 1 (a ) +  −  (− a) = + ma 2   4 2 4  2 4

I xz = 0 aF Δt =

(3)

(4)

I yz = 0

(5)

2 2 1 ma ω x − ma 2ω y + 0 3 4

(6)

1 1 0 = − ma 2ω x + ma 2ω y + 0 4 6

(7)

0 = 0 + 0 + I zωz

(8)

Multiplying Eq. (7) by 3/2 and adding to Eq. (6):  2 3 aF Δt =  −  ma 2ω x,  3 8

ωx =

24 F Δt 7 ma

Substituting for ωx into (7): 3 3 24 36 F Δt ω y = ωx = aF Δt , ω y = 2 2 7 7 ma

From Eq. (8):

I zω z = 0, ω z = 0 ω=

Thus: 

12 F Δt (2i + 3j)  7 ma

Note that ω y ≠ 0, even though Point C where impulse is applied is on the y axis.

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PROBLEM 18.25 Three slender rods, each of mass m and length 2a, are welded together to form the assembly shown. The assembly is hit at A in a vertical downward direction. Denoting the corresponding impulse by F Δt , determine immediately after the impact (a) the velocity of the mass center G, (b) the angular velocity of the rod.

SOLUTION Computation of moments and products of inertia.  1 a2  8 2 I x = ( I x )1 + ( I x )2 + ( I x )3 = ma 2 + ma 2 + m  a 2 +  = ma 3 3  3   a2  1 2 8 2 2 I y = ( I y )1 + ( I y )2 + ( I y )3 = m  a 2 +  + ma + ma = ma 3  3 3 

(1)

1 1 2 I z = ( I z )1 + ( I z ) 2 + ( I z )3 = ma 2 + 0 + ma 2 = ma 2 3 3 3 I xy = 0, I yz = 0, I zx = 0

Impulse-momentum principle.

The impulses consist of −( F Δt ) j applied at A and (T Δt ) j applied at G. Because of constraints, v y = 0. (a)

Velocity of mass center. Equate sums of vectors: (T Δt ) j − ( F Δt ) j = mvx i + mvz k Thus,

T Δt = F Δt , vx = 0, vz = 0. Since v y = 0 from above,

v=0 

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PROBLEM 18.25 (Continued) (b)

Angular velocity. Equate moments about G: (ai + ak ) × (− F Δt ) j = H x i + H y j + H z k −(aF Δt )k + (aF Δt )i = H x i + H y j + H z k

Thus,

H x = aF Δt , H y = 0, H z = − aF Δt

(2)

Since the three products of inertia are zero, the x, y, and z axes are principal centroidal axes and we can use Eqs. (18.10). Substituting from Eqs. (1) and (2) into these equations, we have H x = I xω x :

8 aF Δt = ma 2ω x 3

H y = I yω y :

8 0 = ma 2ω y 3

H z = I zωz :

−aF Δt =

ω x = 3F Δt/8ma

(3)

ωy = 0

(4)

2 2 ma ω z = −3F Δt/2ma 3

(5)

Therefore, ω = (3F Δt/8ma )(i − 4k ) 

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PROBLEM 18.26 Solve Problem 18.25, assuming that the assembly is hit at B in a direction opposite to that of the x axis. PROBLEM 18.25 Three slender rods, each of mass m and length 2a, are welded together to form the assembly shown. The assembly is hit at A in a vertical downward direction. Denoting the corresponding impulse by F Δt , determine immediately after the impact (a) the velocity of the mass center G, (b) the angular velocity of the rod.

SOLUTION Computation of moments and products of inertia.  1 a2  8 2 I x = ( I x )1 + ( I x )2 + ( I x )3 = ma 2 + ma 2 + m  a 2 +  = ma 3 3  3   a2  1 2 8 2 2 I y = ( I y )1 + ( I y )2 + ( I y )3 = m  a 2 +  + ma + ma = ma 3  3 3 

(1)

1 1 2 I z = ( I z )1 + ( I z ) 2 + ( I z )3 = ma 2 + 0 + ma 2 = ma 2 3 3 3 I xy = I yz = I zx = 0

Impulse-momentum principle.

The only impulse is F Δt = −( F Δt )i. (a)

Velocity of mass center. Equate sums of vectors: Thus,

−( F Δt )i = mvx i + mv y j + mvz k

vx = − F Δt/m, v y = 0, vz = 0

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PROBLEM 18.26 (Continued)

(b)

Angular velocity. Equate moments about G: (aj − ak ) × (− F Δt )i = H x i + H y j + H z k (aF Δt )k + ( aF Δt ) j = H x i + H y j + H z k

Thus,

H x = 0, H y = aF Δt , H z = aF Δt

(2)

Since the three products of inertia are zero, the x, y, and z axes are principal centroidal axes and we can use Eqs. (18.10). Substituting from Eqs. (1) and (2) into these equations, we have H x = I xω x :

8 0 = ma 2ω x 3

H y = I yω y : H z = I zωz :

aF Δt =

ωx = 0

(3)

8 aF Δt = ma 2ω y 3

ω y = 3F Δt/8 ma

(4)

2 2 ma ω z 3

ω z = 3F Δt/2 ma

(5)

Therefore, ω = (3F Δt/8ma)( j + 4k ) 

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PROBLEM 18.27 Two circular plates, each of mass 4 kg, are rigidly connected by a rod AB of negligible mass and are suspended from Point A as shown. Knowing that an impulse F Δt = −(2.4 N ⋅ s)k is applied at Point D, determine (a) the velocity of the mass center G of the assembly, (b) the angular velocity of the assembly.

SOLUTION Moments and products of inertia: 1  1  I x = 2  mr 2 + mb 2  = 2  (4)(0.18) 2 + (4)(0.15) 2  = 0.2448 kg ⋅ m 2 4  4  1  I y = 2  mr 2 + mr 2  = 3mr 2 = (3)(4)(0.18)2 = 0.3888 kg ⋅ m 2 2  1  I z = 2  mr 2 + m(b 2 + r 2 )  4   1  = 2  (4)(0.18) 2 + (4)(0.152 + 0.182 )  = 0.504 kg ⋅ m 2 4  I xy = mb( −r ) + m( −b)( r ) = −2mbr = −(2)(4)(0.15)(0.18) = −0.216 kg ⋅ m 2 I xz = 0,

Constraint of cable:

I yz = 0,

total mass = 2m

(vG ) y = 0 v = v G

Principle of impulse-momentum:

2mv = 0, H G = 0 initially.

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PROBLEM 18.27 (Continued)

(a)

Direct components: FΔt = 2mv

0 = 2mvx

vx = 0

T Δt = 2mv y = 0 − F Δt = 2mvz vz = −

(b)

Moments about G.

F Δt 2.4 =− = −0.3 m/s 2m (2)(4)

v = −(0.300 m/s)k 

(bj − ri ) × ( − F Δtk ) = H G −bF (Δt )i − rF Δtj = ( I xω x − I xyω y )i + (− I xyω x + I y ω y ) j + I z ω z k

Resolve into components and apply the numerical data. i : − (0.15)(2.4) = 0.2448ω x − (−0.216)ω y

(1)

j: − (0.18)(2.4) = −(−0.216)ω x + 0.3888ω y

(2)

k: Solving Eqs. (1) and (2),

0 = 0.504ω z

ωz = 0

ω x = −0.962 rad/s, ω y = −0.577 rad/s ω = −(0.962 rad/s)i − (0.577 rad/s) j 

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PROBLEM 18.28 Two circular plates, each of mass 4 kg, are rigidly connected by a rod AB of negligible mass and are suspended from Point A as shown. Knowing that an impulse FΔt = (2.4 N ⋅ s) j is applied at Point D, determine (a) the velocity of the mass center G of the assembly, (b) the angular velocity of the assembly.

SOLUTION Moments and products of inertia: 1  1  I x = 2  mr 2 + mb 2  = 2  (4)(0.18) 2 + (4)(0.15) 2  = 0.2448 kg ⋅ m 2 4  4  1  I y = 2  mr 2 + mr 2  = 3mr 2 = (3)(4)(0.18)2 = 0.3888 kg ⋅ m 2 2  1  I z = 2  mr 2 + m(b 2 + r 2 )  4   1  = 2  (4)(0.18)2 + (4)(0.152 + 0.182 )  = 0.504 kg ⋅ m 2 4  I xy = mb( −r ) + m( −b)( r ) = −2mbr = −(2)(4)(0.15)(0.18) = −0.216 kg ⋅ m 2 I xz = 0,

Constraint of cable:

I yz = 0,

total mass = 2m

(vG ) y = 0 v = v G

Principle of impulse-momentum:

2mv = 0, H G = 0 initially.

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PROBLEM 18.28 (Continued)

(a)

Direct components: FΔt = 2mv 0 = 2mvx vy =

F (Δt ) + T Δt = 2mv y = 0

F (Δt ) + T (Δt ) 2m

0 = 2mvz

vz = 0

T (Δt ) = 0

Point A moves upward. The cord becomes slack. vy =

(b)

vx = 0

2.4 = 0.300 m/s (2)(4)

v = (0.300 m/s) j 

(bj − ri ) × (− F Δtj) = H G

Moments about G.

− rF (Δt )i − rF ( Δt )k = ( I xω x − I xyω y )i + (− I xyω x + I y ω y ) j + I z ω z k

Resolve into components and apply the numerical data. i : − (0.18)(2.4) = 0.2448ω x − (−0.216)ω y j:

k: Solving Eqs. (1) and (2),

0 = −( −0.216)ω x + 0.3888ω y −(0.18)(2.4) = 0.504ω z

(1) (2)

ω z = −0.857 rad/s

ω x = −3.46 rad/s, ω y = −1.923 rad/s ω = −(3.46 rad/s)i + (1.923 rad/s) j − (0.857 rad/s)k 

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PROBLEM 18.29 A circular plate of mass m is falling with a velocity v 0 and no angular velocity when its edge C strikes an obstruction. Assuming the impact to be perfectly plastic (e = 0), determine the angular velocity of the plate immediately after the impact.

SOLUTION

+

Syst. Momenta1

1 1 mR 2 , I x = I z = mR 2 2 4

Iy =

Principal moments of inertia. Principle of impulse and momentum.

=

Syst. Ext. Imp.1→2

Syst. Momenta2

− mv0 j + C Δt j = mvx i + mv y j + mvz k

Linear momentum: i:

0 = mvx

j:

− mv 0 + C Δt = mv y

k:

0 = mvz

rC /G =

Geometry:

1 2

vx = 0 C Δt = m(v0 + v y ) vz = 0 R (i − k )

(e = 0)

Condition of impact:

(1)

(vC ) y = 0

vC = v + ω × rC /G

Kinematics:

(vC ) x i + (vC ) z k = v y j + (ω x i + ω y j + ω z k ) × = vy j +

j : 0 = vy +

R 2

R 2

ωx j −

(ω x + ω z )

R 2

R 2

(i − k ) R

ω y (k + i ) + vy = −

2

R 2

ωz j

(ω x + ω z )

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PROBLEM 18.29 (Continued)

Moments about G: 0 + rC/G × C Δt j = ( H G ) x i + ( H G ) y j + ( H G ) z k R 2

(i − k ) × (C Δt j) = I xω x i + I yω y j + I z ω z k R 2

i:

R 2

j: k:

From Eqs. (1) and (2),

1 1 1 mR 2ω x i + mR 2ω y j + mR 2ω z k 4 2 4

(2)

C Δt =

1 mR 2ω x 4

(3)

0=

1 mR 2ω y 2

C Δt =

1 mR 2ω z 4

(C Δt )(k + i) =

R 2

ωy = 0 (4)

  R C Δt = m v0 − (ω x + ω z )  2  

ωx = 2 2

v0 − 2(ω x + ω z ) R

3ω x + 2ω z = 2 2

v0 R

(5)

ωz = 2 2

v0 − 2(ω x + ω z ) R

2ω x + 3ω z = 2 2

v0 R

(6)

Solving Eqs. (5) and (6) simultaneously,

ωx = ωz = Angular velocity.

2 2 v0 5 R

ω = ωx i + ω y j + ωz k

ω=

2 2 v0 (i + k )  5 R

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PROBLEM 18.30 For the plate of Problem 18.29, determine (a) the velocity of its mass center G immediately after the impact, (b) the impulse exerted on the plate by the obstruction during the impact.

SOLUTION 1 1 mR 2 , I x = I z = mR 2 2 4

Iy =

Principal moments of inertia. Principle of impulse and momentum.

+

Syst. Momenta1

Syst. Ext. Imp.1→2

=

Syst. Momenta2

− mv0 j + C Δt j = mvx i + mv y j + mvz k

Linear momentum: i:

0 = mvx

j:

− mv0 + C Δt = mv y

k:

0 = mvz

rC /G =

Geometry:

1 2

vx = 0 C Δt = m(v0 + v y ) vz = 0 R (i − k )

(e = 0)

Condition of impact:

(1)

(vC ) y = 0

vC = v + ω × rC /G

Kinematics:

(vC ) x i + (vC ) z k = v y j + (ω x i + ω y j + ω z k ) × = vy j +

j : 0 = vy +

R 2

R 2

ωx j −

(ω x + ω z )

R 2

R 2

(i − k ) R

ω y (k + i ) + vy = −

2

R 2

ωz j

(ω x + ω z )

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PROBLEM 18.30 (Continued)

Moments about G: 0 + rC /G × C Δt j = ( H G ) x i + ( H G ) y j + ( H G ) z k R 2

(i − k ) × (C Δt j) = I xω x i + I yω y j + I z ω z k R 2

1 1 1 mR 2ω x i + mR 2ω y j + mR 2ω z k 4 2 4

(2)

C Δt =

1 mR 2ω x 4

(3)

0=

1 mR 2ω y 2

C Δt =

1 mR 2ω z 4

(C Δt )(k + i ) =

i:

R 2

j: k:

From Eqs. (1) and (2),

R 2

ωy = 0 (4)

  R C Δt = m v0 − (ω x + ω z )  2  

ωx = 2 2

v0 − 2(ω x + ω z ) R

3ω x + 2ω z = 2 2

v0 R

(5)

ωz = 2 2

v0 − 2(ω x + ω z ) R

2ω x + 3ω z = 2 2

v0 R

(6)

Solving Eqs. (5) and (6) simultaneously,

ωx = ωz = (a)

2 2 v0 5 R

Velocity of the mass center. From Eq. (2),

vy = −

R

4 (ω x + ω z ) = − v0 5 2

v = vx i + v y j + vz k

(b)

4 v = − v0 j  5

Impulse at C. From Eq. (1),

4  1  C Δt = m  v0 − v0  = mv0 5  5 

CΔt =

1 mv0 j  5

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PROBLEM 18.31 A square plate of side a and mass m supported by a ball-and-socket joint at A is rotating about the y axis with a constant angular velocity ω = ω 0 j when an obstruction is suddenly introduced at B in the xy plane. Assuming the impact at B to be perfectly plastic (e = 0), determine immediately after impact (a) the angular velocity of the plate, (b) the velocity of its mass center G.

SOLUTION For the x′ and y ′ axes shown, the initial angular velocity ω0 j has components

ω x′ =

2 ω 0, 2

ω y′ =

2 ω 0, 2

Initial angular momentum about the mass center: (H G )0 = I x′ω x′ i ′ + I y′ω y ′ j′ =

1 2 ma 2 ω0 (i′ + j′) 12 2

v0 = 0

Initial velocity of the mass center:

Let ω be the angular velocity and v be the velocity of the mass center immediately after impact. Let ( F Δt )k be the impulse at B. Kinematics:

v B = ω × rB/ A = (ω x′ i ′ + ω y′ j′ + ω z′k ′) × (−aj′) v B = a(ω z ′ i ′ + ω x′ k ′)

Since the corner B does not rebound, (vB ) z′ = 0 or ω x′ = 0 1  v = ω × rG/ A = (ω y′ j′ + ω z′k ′) ×  a  ( −i ′ − j′) 2  1 = a(ω z ′i ′ − ω z′ j + ω y k ′) 2

Also,

and

rG/ A × mv =

1 2 ma (−ω y′ i ′ + ω y′ j′ + 2ω z′k ′) 4

H G = I x′ω x′i ′ + I y′ω y′ j′ + I z′ω z′k ′ =

1 1 ma 2ω y′ j′ + ma 2ω z ′k ′ 12 6

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PROBLEM 18.31 (Continued)

Principle of impulse-momentum.

(H A )0 + (−aj) × ( F Δt )k = H A

Moments about A:

(H G )0 + rG/ A × mv 0 − (aF Δt )i = H G + rG/ A × mv

Resolve into components. i′:

1 1 2ma 2ω0 − aF ( Δt ) = − ma 2ω y′ 24 4

j′:

1 1 1 2 2ma 2ω0 = ma 2ω y′ + ma 2ω y′ ω y′ = ω0 24 12 4 8

k ′: 0 =

(a)

ω=

(b)

v=

1 2 1 ma ω z ′ + ma 2ω z ′ ω z′ = 0 6 2 2 1 2 2ω0 ( j − i) ω0 j′ = 8 8 2 1 2 aω y k ′ = aω0 k 2 16

1 ω = ω0 (−i + j)  8 v = 0.0884aω0 k 

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PROBLEM 18.32 Determine the impulse exerted on the plate of Problem 18.31 during the impact by (a) the obstruction at B, (b) the support at A. PROBLEM 18.31 A square plate of side a and mass m supported by a ball-and-socket joint at A is rotating about the y axis with a constant angular velocity ω = ω 0 j when an obstruction is suddenly introduced at B in the xy plane. Assuming the impact at B to be perfectly plastic (e = 0), determine immediately after impact (a) the angular velocity of the plate, (b) the velocity of its mass center G.

SOLUTION For the simpler x′ and y′ axes, the initial angular velocity ω0 j has components

ω x′ =

2 ω0 , 2

ω y′ =

2 ω0 , 2

Initial angular momentum about the mass center: (H G )0 = I x′ω x′ i ′ + I y′ω y ′ j′ =

Initial velocity of the mass center:

1 2 ma 2 ω0 (i′ + j′) 12 2

v0 = 0

Let ω be the angular velocity and v be the velocity of the mass center immediately after impact. Let ( F Δt )k be the impulse at B. Kinematics:

vB = ω × rB/ A = (ω x′ i ′ + ω y′ j′ + ω z′k ′) × (−aj′) vB = a(ω z′ i′ + ω x′k ′)

Since the corner B does not rebound, (vB ) z′ = 0 or ω x′ = 0 1  v = ω × rG/ A = (ω y′ j′ + ω z′k ′) ×  a  ( −i ′ − j′) 2  1 = a(ω z ′ i ′ − ω z′ j + ω y k ′) 2

Also, and

rG/ A × mv =

1 2 ma (−ω y′ i ′ + ω y′ j′ + 2ω z′k ′) 4

H G = I x′ω x′i ′ + I y′ω y′ j′ + I z′ω z′k ′ =

1 1 ma 2ω y′ j′ + ma 2ω z′k ′ 12 6

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PROBLEM 18.32 (Continued)

Principle of impulse-momentum.

Moments about A:

(H A )0 + (−aj′) × ( F Δt )k = H A (H G )0 + rG/ A × mv0 + aF Δti ′ = H G + rG/ A × mv

Resolve into components. i′:

1 1 2ma 2ω0 − aF ( Δt ) = − ma 2ω y′ 24 4

j′:

1 1 1 1 2ma 2ω0 = ma 2ω y′ + ma 2ω y ′ ω y ′ = 2ω0 24 12 4 8

k ′: 0 =

(a)

From Eq. (1),

F Δt =

(1)

1 2 1 ma ω z ′ + ma 2ω z ′ ω z′ = 0 6 2 1 1 7 2 maω0 + 2 maω0 = 2 maω0 24 32 96 ( F Δt )k = 0.1031maω0 k 

v=

Linear momentum:

mv 0 + AΔt + F Δtk = mv 0 + AΔt +

(b)

1 1 2 aω0k ′ aω y k ′ = 2 16

7 1 2maω0 k ′ = 2maω0 k ′ 96 16

AΔt = −

1 2maω0 96

AΔt = −0.01473maω0 k 

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PROBLEM 18.33 The coordinate axes shown represent the principal centroidal axes of inertia of a 3000-lb space probe whose radii of gyration are k x = 1.375 ft, k y = 1.425 ft, and k z = 1.250 ft. The probe has no angular velocity when a 5-oz meteorite strikes one of its solar panels at Point A with a velocity v 0 = (2400 ft/s)i − (3000 ft/s) j + (3200 ft/s)k relative to the probe. Knowing that the meteorite emerges on the other side of the panel with no change in the direction of its velocity, but with a speed reduced by 20 percent, determine the final angular velocity of the probe.

SOLUTION Masses:

Space probe:

m′ =

3000 = 93.17 lb ⋅ s 2 /ft 32.2

Meteorite:

m=

5 = 0.009705 lb ⋅ s 2 /ft (16)(32.2)

Point of impact:

rA = (9 ft)i + (0.75 ft)k

Initial linear momentum of the meteorite, (lb ⋅ s): mv 0 = (0.009705)(2400i + 3000 j + 3200k ) = 23.292i − 29.115 j + 31.056k

Its moment about the origin, (lb ⋅ ft ⋅ s): i j k rA × mv 0 = 9 0 0.75 = 21.836i − 262.04 j − 262.04k 23.292 −29.115 31.056

Final linear momentum of meteorite and its moment about the origin, (lb ⋅ s) and (lb ⋅ s ⋅ ft): 0.8mv 0 = 18.634i − 23.292 j + 24.845k rA × (0.8mv 0 ) = 17.469i − 209.63j − 209.63k

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PROBLEM 18.33 (Continued) Let H A be the angular momentum of the probe and m′ be its mass. Conservation of angular momentum about the origin for a system of particles consisting of the probe plus the meteorite: rA × mv 0 = H A + rA × (0.8mv 0 ) H A = (4.367 lb ⋅ s ⋅ ft)i − (52.41 lb ⋅ s ⋅ ft) j − (52.41 lb ⋅ s ⋅ ft)k I xω x = ( H A ) x

ωx =

(H A )x 4.367 = = 0.02479 rad/s 2 m′k x (93.17)(1.375) 2

I yω y = ( H A ) y

ωy =

(H A ) y

I zωz = ( H A ) z

ωz =

m′k y2

=

−52.41 = −0.2770 rad/s (93.17)(1.425) 2

(H A )z −52.41 = = −0.3600 rad/s 2 ′ (93.17)(1.250)2 m kz ω = (0.0248 rad/s)i − (0.277 rad/s) j − (0.360 rad/s)k 

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PROBLEM 18.34 The coordinate axes shown represent the principal centroidal axes of inertia of a 3000-lb space probe whose radii of gyration are k x = 1.375 ft, k y = 1.425 ft, and k z = 1.250 ft. The probe has no angular velocity when a 5-oz meteorite strikes one of its solar panels at Point A and emerges on the other side of the panel with no change in the direction of its velocity, but with a speed reduced by 25 percent. Knowing that the final angular velocity of the probe is ω = (0.05 rad/s)i − (0.12 rad/s) j + ω z k and that the x component of the resulting change in the velocity of the mass center of the probe is −0.675 in./s, determine (a) the component ω z of the final angular velocity of the probe, (b) the relative velocity v 0 with which the meteorite strikes the panel.

SOLUTION Masses:

Space probe:

m′ =

3000 = 93.17 lb ⋅ s 2 /ft 32.2

Meteorite:

m=

5 = 0.009705 lb ⋅ s 2 /ft (16)(32.2)

rA = (9 ft)i + (0.75 ft)k

Point of impact:

Initial linear momentum of the meteorite, (lb ⋅ s): mv 0 = (0.009705)(vx i + v y j + vz k )

Its moment about the origin, (lb ⋅ ft ⋅ s): i (H A )0 = rA × mv 0 = 0.009705 9 vx

j 0 vy

k 0.75 vz

= 0.009705[−0.75v y i + (0.75vx − 9vz ) j + 9v y k ]

Final linear momentum of the meteorite, (lb ⋅ s): 0.75mv 0 = 0.007279(vx i + v y j + vz k )

Its moment about the origin, (lb ⋅ ft ⋅ s): rA × (0.75mv 0 ) = 0.007279[−0.75v y i + (0.75vx − 9vz ) j + 9v y k ]

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PROBLEM 18.34 (Continued)

m′v′0 = 0

Initial linear momentum of the space probe, (lb ⋅ s): Final linear momentum of the space probe, (lb ⋅ s):

 0.675  m′(vx′ i + v′y j + vz′ k ) = 93.17  − i + v′y j + vz′ k   12 

Final angular momentum of space probe, (lb ⋅ ft ⋅ s):

(

HA = m′ k x2ω x i + k y2ω y j + k z2ω z k

)

= 93.17[(1.375) 2 (0.05)i + (1.425) 2 (−0.12) j + (1.250) 2 ω z k ] = 8.8075i − 22.703j + 145.58ω z k

Conservation of linear momentum of the probe plus the meteorite, (lb ⋅ s): 0.009705(vx i + v y j + vz k ) = 0.007279(vx i + v y j + vz k ) + 93.17(−0.05625i + v′y j + vz′ k )

i:

0.002426vx = −5.2408

j:

0.002426v y = 93.17v′y

k:

0.002426vz = 93.17v′z

vx = −2160 ft/s

Conservation of angular momentum about the origin, (lb ⋅ ft ⋅ s): (0.009705)[ −0.75v y i + (0.75vx − 9vz ) j + 9v y k ] = (0.007279)[−0.75v y i + (0.75vx − 9vz ) j + 9v y k ] + 8.8075i − 22.703j + 145.58ω z k

i: j: k: (a)

− 0.0018195v y = 8.8075 − 0.021834vz + 0.0018195vx = −22.703

v y = 4840.5 ft/s vz = 0.08333vx + 1039.8 = 859.8 ft/s

− 0.021834v y = 145.58ω z

ω z = −149.98 × 10−6 v y

ω z = −0.726 rad/s 

v 0 = −(2160 ft/s)i − (4840 ft/s) j + (860 ft/s)k 

(b)

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PROBLEM 18.35 A 2500-kg probe in orbit about the moon is 2.4 m high and has octagonal bases of sides 1.2 m. The coordinate axes shown are the principal centroidal axes of inertia of the probe, and its radii of gyration are k x = 0.98 m, k y = 1.06 m, and k z = 1.02 m. The probe is equipped with a main 500-N thruster E and with four 20-N thrusters A, B, C, and D which can expel fuel in the positive y direction. The probe has an angular velocity ω = (0.040 rad/s)i + (0.060 rad/s)k when two of the 20-N thrusters are used to reduce the angular velocity to zero. Determine (a) which of the thrusters should be used, (b) the operating time of each of these thrusters, (c) for how long the main thruster E should be activated if the velocity of the mass center of the probe is to remain unchanged.

SOLUTION

(

H G = I xω x i + I yω y j + I zω z k = m k x2ω x i + k y2ω y2 j + k22ω z k

)

= (2500)[(0.98) 2 (0.040)i + (1.06)2 (0) j + (1.02) 2 (0.060)k ] = (96.04 kg ⋅ m 2 /s)i + (156.06 kg ⋅ m 2 /s)k

Let − Aj, − Bj, − Cj, and − Dj be the impulses provided by the 20 N thrusters at A, B, C, and D, respectively. Let Ej be that provided by the 500 N main thruster. Position vectors for intersections of the lines of action of the thruster impulses with the xz plane: a=

1 (1.2) = 0.6 m, 2

b = 0.6 + 0.6 2 = 1.4485 m

rA = − ai + bk ,

rB = ai + bk

rC = ai − bk ,

rD = − ai − bk

The final linear and angular momenta are zero. Principle of impulse-momentum. Moments about G: H G + rA × (− Aj) + rB × (− Bj) + rC × (−Cj) + rD × (− Dj) = 0  H G + b( A + B − C − D)i + a( A − B − C + D )k = 0 

Resolve into components. i:

A+ B−C − D = −

(HG )x 96.04 =− = −66.30 N ⋅ s 1.4485 b

(1)

k:

A− B−C + D = −

(HG )z 156.06 =− = −260.1 N ⋅ s 0.6 a

(2)

Of A, B, C, and D, two must be zero or positive, the other two zero. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2014

PROBLEM 18.35 (Continued)

Set A = 0 and B − D = N . Solve the simultaneous equations (1) and (2). C = 163.2 N ⋅ s and N = 96.9.

Set D = 0 and B = 96.9 N ⋅ s Use thrusters C and B. 

(a) (b)

(c)

Linear momentum:

FC (ΔtC ) = C ,

ΔtC =

C 163.2 = 20 FC

ΔtC = 8.16 s 

FB (Δt B ) = B,

Δt B =

B 96.9 = 20 FB

Δt B = 4.84 s 

Ej − Bj − Cj = 0, FE (Δt E ) = E

E = 30.291 lb ⋅ s Δt E =

E 260.1 = 500 FE

Δt E = 0.520 s 

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PROBLEM 18.36 Solve Problem 18.35, assuming that the angular velocity of the probe is ω = (0.060 rad/s)i − (0.040 rad/s)k. PROBLEM 18.35 A 2500-kg probe in orbit about the moon is 2.4 m high and has octagonal bases of sides 1.2 m. The coordinate axes shown are the principal centroidal axes of inertia of the probe, and its radii of gyration are k x = 0.98 m, k y = 1.06 m, and k z = 1.02 m. The probe is equipped with a main 500-N thruster E and with four 20-N thrusters A, B, C, and D which can expel fuel in the positive y direction. The probe has an angular velocity ω = (0.040 rad/s)i + (0.060 rad/s)k when two of the 20-N thrusters are used to reduce the angular velocity to zero. Determine (a) which of the thrusters should be used, (b) the operating time of each of these thrusters, (c) for how long the main thruster E should be activated if the velocity of the mass center of the probe is to remain unchanged.

SOLUTION H G = I xω x i + I yω y j + I zω z k

(

= m k x2ω x i + k y2ω y2 j + k22ω z k

)

= (2500)[(0.98) 2 (0.060) + (1.06) 2 (0) + (1.02)2 (−0.040)k ] = (144.06 kg ⋅ m 2 /s)i − (104.04 kg ⋅ m 2 /s)k

Let − Aj, − Bj, − Cj, and − Dj be the impulses provided by the 20 N thrusters at A, B, C, and D, respectively. Let Ej be that provided by the 500 N main thruster. Position vectors for intersections of the lines of action of the thruster impulses with the xz plane: a=

1 (1.2) = 0.6 m, 2

b = 0.6 + 0.6 2 = 1.4485 m

rA = − ai + bk ,

rB = ai + bk

rC = ai − bk ,

rD = − ai − bk

The final linear and angular momenta are zero. Principle of impulse-momentum. Moments about G: H G + rA × (− Aj) + rB × (− Bj) + rC × (−Cj) + rD × (− Dj) = 0  H G + b( A + B − C − D)i + a( A − B − C + D )k = 0

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PROBLEM 18.36 (Continued)

Resolve into components. i:

A+ B−C − D = −

(HG )x 144.06 =− = −99.455 N ⋅ s 1.4485 b

(1)

k:

A− B−C + D = −

(HG )z −104.04 =− = 173.4 N ⋅ s 0.6 a

(2)

Of A, B, C, and D, two must be zero or positive, the other two zero. Set B = 0 and A − C = N . Solve the simultaneous equations (1) and (2). D = 136.43 N ⋅ s and N = 36.97 N ⋅ s. Set C = 0 and A = 36.97 N ⋅ s

Use thrusters D and A. 

(a) (b)

(c)

Linear momentum:

FD (Δt D ) = D,

Δt D =

D 136.43 = 20 FD

Δt D = 6.82 s 

FA (Δt A ) = A,

Δt A =

A 36.97 = 20 FA

Δt A = 1.848 s 

Ej − Dj − Aj = 0 FE (Δt E ) = E

E = 173.4 N ⋅ s Δt E =

E 173.4 = 500 FE

Δt E = 0.347 s 

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PROBLEM 18.37 Denoting, respectively, by ω, H O , and T the angular velocity, the angular momentum, and the kinetic energy of a rigid body with a fixed Point O, (a) prove that H O ⋅ ω = 2T ; (b) show that the angle θ between ω and HO will always be acute.

SOLUTION H 0 = ( I xω x − I xyω y − I xzω z )i + (− I xy ω x + I yω y − I yzω z ) j + (− I xzω x − I yzω y + I zω z )k

(a)

ω = ωx i + ω y j + ωz k H 0 ⋅ ω = I xω x2 − I xyω yω x − I xz ω zω x − I xyω xω y + I yω y2 − I yzω z ω y − I xzω xω z − I yz ω y ω z + I zω z2

1 = (2)   I xω x2 + I yω y2 + I zω z2 − 2 I xyω xω y − 2 I yzω yω z − 2 I xzω xω z 2

(

)

= 2T H 0 ⋅ ω = H 0ω cos θ

(b)

2T = H 0ω cos θ cos θ =

But

2T H 0ω T > 0, H 0 > 0, ω > 0

cos θ > 0

θ < 90°

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PROBLEM 18.38 Show that the kinetic energy of a rigid body with a fixed Point O can be expressed as T = 12 I OLω 2 , where ω is the instantaneous angular velocity of the body and I OL is its moment of inertia about the line of action OL of ω. Derive this expression (a) from Eqs. (9.46) and (18.19), (b) by considering T as the sum of the kinetic energies of particles Pi describing circles of radius ρi about line OL.

SOLUTION T=

(a) Let

(

1 I xω x2 + I yω y2 + I zω z2 − 2 I xyω xω y − 2 I yz ω yω z − 2 I xzω xω z 2

)

ω x = ω cos θ x = ωλx ω y = ω cos θ y = ωλ y ω z = ω cos θ z = ωλz T= =

(b)

(

)

1 I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I xz λx λz ω 2 2 1 I OLω 2 2

T=

1 I OLω 2  2

T=

1 I OLω 2  2

Each particle of mass (Δm)i describes a circle of radius ρi . The speed of the particle is vi = ρiω. Its kinetic energy is

(ΔT )i =

1 1 (Δmi )vi2 = (Δmi ) ρi2ω 2 2 2

The kinetic energy of the entire body is T = Σ( ΔT )i =

but

1 Σ(Δmi ) ρi2ω 2 2

I OL = Σ( Δmi ) ρi2

Hence,

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PROBLEM 18.39 Determine the kinetic energy of the disk of Problem 18.1. PROBLEM 18.1 A thin, homogeneous disk of mass m and radius r spins at the constant rate ω1 about an axle held by a fork-ended vertical rod, which rotates at the constant rate ω 2 . Determine the angular momentum H G of the disk about its mass center G.

SOLUTION Angular velocity:

ω = ω2 j + ω1k

Moments of inertia:

Ix =

Products of inertia: by symmetry,

I xy = I yz = I zx = 0

Kinetic energy:

1 2 mr , 4

(

Iy =

1 2 mr , 4

Iz =

1 2 mr 2

)

T=

1 I xω x2 + I y ω y2 + I zω z 2

T=

1   1 2  2  1 2  2 mr  ω2 +  mr  ω1  0 + 2   4  2  

(

)

1 T = mr 2 ω22 + 2ω12  8

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PROBLEM 18.40 Determine the kinetic energy of the plate of Problem 18.2. PROBLEM 18.2 A thin rectangular plate of weight 15 lb rotates about its vertical diagonal AB with an angular velocity ω. Knowing that the z axis is perpendicular to the plate and that ω is constant and equal to 5 rad/s, determine the angular momentum of the plate about its mass center G.

SOLUTION h = (9 in.)2 + (12 in.) 2 = 15 in.

Resolving ω along the principal axes x′, y′, z: 12 ω = 0.8(5 rad/s) = 4 rad/s 15 9 ω y′ = ω = 0.6(5 rad/s) = 3 rad/s 15 ωz = 0

ω x′ =

Moments of inertia: I x′ =

1  15 lb  9  ft  = 0.021836 slug ⋅ ft 2 12  32.2  12  

I y′ =

1  15 lb  12  ft  = 0.038820 slug ⋅ ft 2 12  32.2   12 

2

From Eqs. (18.20): 1 ( I x′ω x2′ + I y′ω y2′ + I z′ω z2′ ) 2 1 = [(0.021836)(4) 2 + (0.038820)(3) 2 + 0] 2 T = 0.34938 ft ⋅ lb T=

T = 0.349 ft ⋅ lb 

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PROBLEM 18.41 Determine the kinetic energy of the assembly of Problem 18.3. PROBLEM 18.3 Two uniform rods AB and CE, each of weight 3 lb and length 2 ft, are welded to each other at their midpoints. Knowing that this assembly has an angular velocity of constant magnitude ω = 12 rad/s, determine the magnitude and direction of the angular momentum HD of the assembly about D.

SOLUTION m=

W 3 = = 0.093168 lb ⋅ s 2 /ft, g 32.2

l = 24 in. = 2 ft,

ω = (12 rad/s)i

For rod ADB, T =

1 I xω 2 ≈ 0, since I x ≈ 0. 2

For rod CDE, use principal axes x′, y′ as shown. cosθ =

9 , 12

θ = 41.410°

ω x′ = ω cosθ = 9 rad/s 2 ω y′ = ω sin θ = 7.93725 rad/s 2 ω z′ = 0 I x′ ≈ 0 I y′ =

1 2 1 ml = (0.093168)(2)2 12 12

= 0.0310559 lb ⋅ s 2 ⋅ ft T =

1 1 1 1 mv 2 + I x′ω 2x′ + I y′ω 2y′ + I z ′ω 2z′ 2 2 2 2

=0+0+

1 (0.0310559)(7.93725) 2 + 0 2

= 0.97826 ft ⋅ lb

T = 0.978 ft ⋅ lb 

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PROBLEM 18.42 Determine the kinetic energy of the disk of Problem 18.4. PROBLEM 18.4 A homogeneous disk of weight W = 6 lb rotates at the constant rate ω 1 = 16 rad/s with respect to arm ABC, which is welded to a shaft DCE rotating at the constant rate ω 2 = 8 rad/s. Determine the angular momentum H A of the disk about its center A.

SOLUTION m=

6 W = = 0.1863 lb ⋅ s 2 /ft g 32.2

ω = ω2 i + ω1 j = (8 rad/s)i + (16 rad/s) j

For axes x′, y ′, z ′ parallel to x, y, z with origin at A, 2

I x′ =

1 2 1  8  mr = (0.1863)   = 0.0207 lb ⋅ s 2 ⋅ ft 4 4  12 

I z′ = I x′ = 0.0207 lb ⋅ s 2 ⋅ ft, I y ′ = I x′ + I z′ = 0.0414 lb ⋅ s 2 ⋅ ft

Point A is the mass center of the disk. rA/C = (9 in.)i − (9 in.)k = (0.75 ft)i − (0.75 ft)k v = v A = ω2 i × rA/C = 8i × (0.75 j − 0.75k )

= (6 ft/s) j + (6 ft/s)k v = (6)2 + (6) 2 = 8.4853 ft/s

Kinetic energy:

1 1 1 1 mv 2 + I x′ω x2 + I y′ω y2 + I z′ω z2 2 2 2 2 1 1 1 T = (0.1863)(8.4853)2 + (0.0207)(8) 2 + (0.0414)(16) 2 + 0 2 2 2 = 6.7068 + 0.6624 + 5.2992 = 12.6684 ft ⋅ lb T = 12.67 ft ⋅ lb  T=

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PROBLEM 18.43 Determine the kinetic energy of the disk of Problem 18.5. PROBLEM 18.5 A thin disk of mass m = 4 kg rotates at the constant rate ω2 = 15 rad/s with respect to arm ABC, which itself rotates at the constant rate ω1 = 5 rad/s about the y axis. Determine the angular momentum of the disk about its center C.

SOLUTION r = 150 mm

Angular velocity of disk: ω = ω1 j + ω2 k

= (5 rad/s) j + (15 rad/s)k

Centroidal moments of inertia: I x′ = I y ′ =

1 2 mr 4

1 (4)(0.150 m) 2 = 0.0225 kg ⋅ m 2 4 1 2 I z′ = mr = 0.045 kg ⋅ m 2 2 =

Location of mass center. Velocity of mass center.

rC/ A = (0.450 m)i + (0.225 m) j v = ω1 × rC/ A = 5 j × (0.45i + 0.225 j)

= − (2.25 m/s)k

Kinetic energy:

1 1 1 1 mv 2 + I xω x2 + I y ω y2 + I zω z2 2 2 2 2 1 1 1 = (4)(2.25) 2 + 0 + (0.0225)(5)2 + (0.0450)(15) 2 2 2 2

T=

= 10.125 + 0 + 0.28125 + 5.0625

T = 15.47 J 

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PROBLEM 18.44 Determine the kinetic energy of the solid parallelepiped of Problem 18.6. PROBLEM 18.6 A solid rectangular parallelepiped of mass m has a square base of side a and a length 2a. Knowing that it rotates at the constant rate ω about its diagonal AC ′ and that its rotation is observed from A as counterclockwise, determine (a) the magnitude of the angular momentum H G of the parallelepiped about its mass center G, (b) the angle that H G forms with the diagonal AC ′.

SOLUTION d = a 2 + (2a) 2 + a 2 = 6a

Body diagonal:

ω=

ω

( − ai + 2 aj − ak ) = −

ω

d 6 1 5 I x = m[(2a) 2 + a 2 ] = ma 2 12 12 1 1 I y = m[a 2 + a 2 ] = ma 2 12 6 1 5 I z = m[a 2 + (2a) 2 ] = ma 2 12 12

i+

2ω 6

j−

ω 6

k

Axis of rotation passes through the mass center, hence v = 0. Kinetic energy:

T=

1 1 1 1 mv 2 + I xω x2 + I yω y2 + I zω z2 2 2 2 2 2

2

2

1 5 1  1 2   2ω  1 5 1 2 2  ω  2  ω  T = 0 +  ma 2    +  ma    +  ma    = ma ω 2  12 26 2  12 8  6   6   6  T = 0.1250 ma 2ω 2 

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PROBLEM 18.45 Determine the kinetic energy of the hollow parallelepiped of Problem 18.7. PROBLEM 18.7 Solve Problem 18.6, assuming that the solid rectangular parallelepiped has been replaced by a hollow one consisting of six thin metal plates welded together.

SOLUTION d = a 2 + (2a) 2 + a 2 = 6a

Body diagonal:

ω=

ω d

( −ai + 2aj − ak ) = −

ω 6

i+

2ω 6

j−

ω 6

k

Total area = 2( a 2 + 2a 2 + 2a 2 ) = 10a 2 1 m 10 1 13 13 I x = m′a 2 + m′a 2 = m′a 2 = ma 2 12 12 120 1 1 I y = m′a 2 = ma 2 6 60 13 Iz = Ix = ma 2 120

For each square plate,

m′ =

For each plate parallel to the yz plane,

m′ =

1 m 5

1 5 1 m′[ a 2 + (2a) 2 ] = m′a 2 = ma 2 12 12 12 2 1 1 1 a I y = m′a 2 + m′   = m′a 2 = ma 2 12 3 15 2 Ix =

2

Iz =

1 7 7 a m′(2a )2 + m′   = m′a 2 = ma 2 12 60  2  12

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PROBLEM 18.45 (Continued)

For each plate parallel to the xy plane,

m′ =

1 m 5

Ix =

1 7 7 a m′(2a )2 + m′   = m′a 2 = ma 2 12 2 12 60  

2

2

1 1 1 a m′a 2 + m′   = m′a 2 = ma 2 12 3 15 2 1 5 1 I z = m′[ a 2 + (2a) 2 ] = m′a 2 = ma 2 12 12 12

Iy =

Total moments of inertia: 1 7  37 2  13 Ix = 2 + +  ma 2 = ma 60  120 12 60  1 1  3  1 I y = 2 + +  ma 2 = ma 2 10  60 15 15  7 1  37 2  13 Iz = 2 ma + +  ma 2 = 60  120 60 12 

Axis of rotation passes through the mass center, hence v = 0. Kinetic energy: T=

1 1 1 1 mv 2 + I xω x2 + I y ω y2 + I zω z2 2 2 2 2 2

2

2

1  37 1 3 1  37 2   ω  73  ω  2   2ω  T = 0 +  ma 2   ma 2ω 2  +  ma    +  ma    = 2  60 2  10 2  60 360  6   6   6  T = 0.203 ma 2ω 2 

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PROBLEM 18.46 Determine the kinetic energy of the disk of Problem 18.8. PROBLEM 18.8 A homogeneous disk of mass m and radius r is mounted on the vertical shaft AB. The normal to the disk at G forms an angle β = 25° with the shaft. Knowing that the shaft has a constant angular velocity ω, determine the angle θ formed by the shaft AB and the angular momentum HG of the disk about its mass center G.

SOLUTION Use the principal centroidal axes Gx′y ′z. Moments of inertia. I x′ = I z = I y′ =

1 2 mr 4

1 2 mr 2

Angular velocities.

ω x′ = −ω sin β ω y′ = ω cos β ωz = 0 Kinetic energy:

1 1 1 1 mv 2 + I x2′ω x2′ + I y′ω y2′ + I z′ω z2′ 2 2 2 2 1 1 1 1 = 0 + ⋅ mr 2 (−ω sin β ) 2 + ⋅ mr 2 (ω cos β ) 2 + 0 2 4 2 2 1 = mr 2ω 2 (sin 2 β + 2 cos 2 β ) 8 1 = mr 2ω 2 (sin 2 25° + 2 cos 2 25°) 8

T=

T = 0.228 mr 2ω 2 

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PROBLEM 18.47 Determine the kinetic energy of the shaft of Problem 18.15. PROBLEM 18.15 A 5-kg rod of uniform cross section is used to form the shaft shown. Knowing that the shaft rotates with a constant angular velocity ω of magnitude 12 rad/s, determine (a) the angular momentum H G of the shaft about its mass center G, (b) the angle formed by H G and the axis AB.

SOLUTION

ω = (12 rad/s)i, T = =

ωy = ωz = 0

1 1 1 I xω 2x + I yω 2y + I zω 2z − I xyω xω y − I yzω yω z − I xzω xω z 2 2 2 1 I xω 2 2

The shaft is comprised of eight sections, each of length m a = 0.25 m and of mass m′ = = 0.625 kg. 8 10 10 1  (0.625)(0.25) 2 = 0.130208 kg ⋅ m 2 I x = (4)  m′a 2  + (2)( m′a 2 ) = m′a 2 = 3 3 3  T =

1 (0.130208)(12) 2 = 9.38 kg ⋅ m 2/s 2 = 9.38 N ⋅ m = 9.38 J 2 T = 9.38 J 

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PROBLEM 18.48 Determine the kinetic energy of the body of Problem 18.17. PROBLEM 18.17 Two L-shaped arms, each weighing 4 lb, are welded at the third points of the 2-ft shaft AB. Knowing that shaft AB rotates at the constant rate ω = 240 rpm, determine (a) the angular momentum of the body about A, (b) the angle formed by the angular momentum and shaft AB.

SOLUTION 4 = 0.12422 lb ⋅ s 2 /ft 32.2 a = 8 in. = 0.66667 ft (2π )(240) = 8π rad/s, ω x = 0, ω y = 0, ω z = 8π rad/s ω= 60

W = 4 lb m =

For rotation about the fixed Point B, the kinetic energy is T=

Calculation of Iz:

1 1 1 1 I xω x2 + I y ω y2 + I zω z2 − I xyω xω y − I xzω xω z − I yz ω yω z = I z ω 2 2 2 2 2

Segments 1, 2, 3, and 4, each of mass m′ = 0.06211 lb ⋅ s 2 /ft contribute to Part

Iz  1 1  2  12 + 4 + 1 m′a  

1 m′a 2 3 1 m′a 2 3  1 1  2  12 + 4 + 1 m′a  

10 m′a 2 3

Σ

Kinetic energy:

T=

1  10 1  10   m′a 2  ω 2 =   (0.06211)(0.66667)2 (8π ) 2  2 3 2 3  

T = 29.1 ft ⋅ lb 

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PROBLEM 18.49 Determine the kinetic energy of the triangular plate of Problem 18.19. PROBLEM 18.19 The triangular plate shown has a mass of 7.5 kg and is welded to a vertical shaft AB. Knowing that the plate rotates at the constant rate ω = 12 rad/s, determine its angular momentum about (a) Point C, (b) Point A. (Hint: To solve part b find v and use the property indicated in part a of Problem 18.13.)

SOLUTION ω = (12 rad/s) j, ω x = 0, ω y = 12 rad/s, ω z = 0

1 1 1 I xω x2 + I yω y2 + I z ω z2 − I yz ω y ω z − I zx ω zω x − I xy ω xω y 2 2 2 1 = I yω 2 2

Kinetic energy:

T=

For the plate:

b = AB = 90 + 160 = 250 mm = 0.25 m h = CD = 120 mm = 0.12 m

Area:

1 bh = 15 × 10−3 m 2 2 1 ( I y )area = bh3 = 36 × 10−6 m 4 12 m = 7.5 kg A=

( I y ) mass = T=

m (7.5)(36 × 10−6 ) = 18 × 10−3 kg ⋅ m 2 ( I y )area = −3 A 15 × 10 1 (18 × 10−3 )(12)2 = 1.296 J 2

T = 1.296 J 

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PROBLEM 18.50 Determine the kinetic energy imparted to the cube of Problem 18.21. PROBLEM 18.21 One of the sculptures displayed on a university campus consists of a hollow cube made of six aluminum sheets, each 1.5 × 1.5 m, welded together and reinforced with internal braces of negligible weight. The cube is mounted on a fixed base at A and can rotate freely about its vertical diagonal AB. As she passes by this display on the way to a class in mechanics, an engineering student grabs corner C of the cube and pushes it for 1.2 s in a direction perpendicular to the plane ABC with an average force of 50 N. Having observed that it takes 5 s for the cube to complete one full revolution, she flips out her calculator and proceeds to determine the mass of the cube. What is the result of her calculation? (Hint: The perpendicular distance from the diagonal joining two vertices of a cube to any of its other six vertices can be obtained by multiplying the side of the cube by 2/3.)

SOLUTION Let m′ = 16 m be the mass of one side of the cube. Choose x, y, and z axes perpendicular to the face of the cube. Let a be the side of the cube. For a side perpendicular to the x axis, ( I x )1 = 16 m′a 2 . 1  1 1 For a side perpendicular to the y or z axis, ( I x ) 2 =  +  m′a 2 = m′a 2 3  12 4  5 5 I x = 2( I x )1 + 4( I x ) 2 = m′a 2 = ma 2 3 18

Total moment of inertia: By symmetry, I y = I x and I z = I x .

Since all three moments of inertia are equal, the ellipsoid of inertia is a sphere. All centroidal axes are principal axes. Moment of inertia about the vertical axis: Let b =

2 a 3

Iv =

5 ma 2 18

be the moment arm of the impulse applied to the corner.

Using the impulse-momentum principle and taking moments about the vertical axis, bF ( Δt ) = H v = I vω =

Data:

a = 1.5 m, b =

ω=

5 ma 2ω 18

(1)

2 (1.5) = 1.22474 m 3

2π = 1.25664 rad/s, F = 50 N, Δt = 1.2 s. 5

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PROBLEM 18.50 (Continued)

Solving Equation (1) for m, m=

18 bF (Δt ) 18 (1.22474)(50)(1.2) = = 93.563 kg 5 a 2ω 5 (1.5) 2 (1.25664) m = 93.6 kg

For principal axes,

1 1 1 I xω x2 + I yω y2 + I z ω z2 2 2 2 1 1 5  T = I vω 2 =  ma 2  ω 2 2 2  18  1 5 2 = (93.563)(1.5) (1.2566)2 2 18 T=

T = 46.2 J 

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PROBLEM 18.51 Determine the kinetic energy lost when edge C of the plate of Problem 18.29 hits the obstruction.

SOLUTION For principal moments of inertia, the kinetic energy is T=

Before impact:

1 1 1 1 mv 2 + I xω x2 + I yω y2 + I zω z2 2 2 2 2

v = v0

ωx = ω y = ωz = 0 T0 =

1 mv02 2

After impact: From Problem 18.30,

vx = vz = 0 4 v y = − v0 5 4 v = v0 5

From Problem 18.29,

ωx = ωz =

2 2 v0 5 R

ωy = 0 Ix = Iz =

1 mR 2 4 2

2 1  4   1  1   2 2 v0   1  1   2 2 v0  + 0 +   mR 2   T = m  v0  +   mR 2     2  5   2  4   5 R   2  4   5 R  1  16 2 2 2 =  + + 0 +  mv02 = mv02 2  25 25 25  5

Energy loss:

T0 − T =

1 2 mv02 − mv02 2 5

T0 − T =

2

1 mv02  10

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PROBLEM 18.52 Determine the kinetic energy lost when the plate of Problem 18.31 hits the obstruction at B.

SOLUTION For the x′ and y ′ axes shown, the initial angular velocity ω0 j has components

ω x′ =

2 2 ω0 , ω y′ = ω0 2 2

Initial angular momentum about the mass center: (H G )0 = I x′ω x′ i ′ + I y ′ω y′ j′ =

1 2 ma 2 ω0 (i′ + j′) 12 2

v0 = 0

Initial velocity of the mass center:

Let ω be the angular velocity and v be the velocity of the mass center immediately after impact. Let ( F Δ t )k be the impulse at B. v B = ω × rB/ A = (ω x′ i ′ + ω y′ j′ + ω z′k ′) × (−aj′)

Kinematics:

v B = a(ω z′i ′ + ω x′k ′)

Since the corner B does not rebound,

(vB ) z′ = 0 or ω x′ = 0

1  v = ω × rG/ A = (ω y′ j′ + ω z′k ′) ×  a  ( −i ′ − j′) 2  1 = a(ω z ′ i ′ − ω z′ j + ω y k ′) 2

Also, and

rG/ A × mv =

1 2 ma (−ω y′ i ′ + ω y′ j′ + 2ω z′k ′) 4

H G = I x′ω x′i ′ + I y′ω y′ j′ + I z′ω z′k ′ =

1 1 ma 2ω y′ j′ + ma 2ω z′k ′ 12 6

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PROBLEM 18.52 (Continued) Principle of impulse-momentum.

(H A )0 + ( −aj) × ( F Δt )k = H A

Moments about A:

(H G )0 + rG/ A × mv 0 − (aF Δt )i = H G + rG/ A × mv

Resolve into components. i′:

1 1 2 ma 2ω0 − aF (Δt ) = − ma 2ω y′ 24 4

1 1 1 2 2ma 2ω0 = ma 2ω y′ + ma 2ω y ′ ω y ′ = ω0 24 12 4 8 1 1 k ′: 0 = ma 2ω z ′ + ma 2ω z ′ ω z′ = 0 6 2 j′:

2 1 2 ω0 j′ = 2ω0 ( j − i) 8 8 2 1 2 v = aω y k ′ = aω0 k 2 16

ω=

1 1 1 1 mv 2 + I xω x2′ + I yω y2′ + I z′ω z2′ 2 2 2 2

Kinetic energy:

T=

Before impact:

v0 = 0, ω x′ =

2 2 ω0 , ω y ′ = ω0 , ω z′ = 0 2 2

2

2

1 1 1 1 2  1  2   T0 = 0 +  ma 2   ω ω0  + 0 = ma 2ω02 +  ma 2   −  0    2  12 2  12 24  2  2  

After impact:

v=

2 2 aω0 , ω x′ = 0, ω y ′ = ω0 , ω z′ = 0 16 8 2

2

 1  2 1 1 1  2  T1 = m  aω0  + 0 +  ma 2   ma 2ω02 ω0  + 0 =    2  16 2 12 8 192      T0 − T1 =

Kinetic energy lost.

7 ma 2ω02  192

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PROBLEM 18.53 Determine the kinetic energy of the space probe of Problem 18.33 in its motion about its mass center after its collision with the meteorite.

SOLUTION Masses:

Space probe:

m′ =

3000 = 93.17 lb ⋅ s 2 /ft 32.2

Meteorite:

m=

5 = 0.009705 lb ⋅ s 2 /ft (16)(32.2)

Point of impact:

rA = (9 ft)i + (0.75 ft)k

Initial linear momentum of the meteorite, (lb ⋅ s): mv 0 = (0.009705)(2400i + 3000 j + 3200k ) = 23.292i − 29.115 j + 31.056k

Its moment about the origin, (lb ⋅ ft ⋅ s): i j k rA × mv 0 = 9 0 0.75 = 21.836i − 262.04 j − 262.04k 23.292 −29.115 31.056

Final linear momentum of meteorite and its moment about the origin, (lb ⋅ s) and (lb ⋅ s ⋅ ft): 0.8mv 0 = 18.634i − 23.292 j + 24.845k rA × (0.8mv 0 ) = 17.469i − 209.63j − 209.63k

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PROBLEM 18.53 (Continued)

Let H A be the angular momentum of the probe and m′ be its mass. Conservation of angular momentum about the origin for a system of particles consisting of the probe plus the meteorite: rA × mv 0 = H A + rA × (0.8mv 0 ) H A = (4.367 lb ⋅ s ⋅ ft)i − (52.41 lb ⋅ s ⋅ ft) j − (52.41 lb ⋅ s ⋅ ft)k I xω x = ( H A ) x

ωx =

(H A )x 4.367 = = 0.02479 rad/s 2 m′k x (93.17)(1.375) 2

I yω y = ( H A ) y

ωy =

(H A ) y

I zωz = ( H A ) z

ωz =

m′k y2

=

−52.41 = −0.2770 rad/s (93.17)(1.425) 2

(H A )z −52.41 = = −0.3600 rad/s 2 ′ (93.17)(1.250) 2 m kz

Kinetic energy of motion of the probe about its mass center:

(

)

(

)

1 m 2 2 I xω x2 + I y ω y2 + I z ω z2 = k x ω x + k y2ω y2 + k z2ω z2 2 2 93.17 [(1.375)2 (0.02479)2 + (1.425)2 (−0.2770)2 + (1.250)2 (−0.3600)2 ] = 2

T′ =

T ′ = 16.75 ft ⋅ lb 

= 16.75 ft ⋅ lb

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PROBLEM 18.54 Determine the kinetic energy of the space probe of Problem 18.34 in its motion about its mass center after its collision with the meteorite.

SOLUTION Masses:

Point of impact:

Space probe:

m′ =

3000 = 93.17 lb ⋅ s 2/ft 32.2

Meteorite:

m=

5 = 0.009705 lb ⋅ s 2/ft (16)(32.2)

rA = (9 ft)i + (0.75 ft)k

Initial linear momentum of the meteorite, ( lb ⋅ s ) : mv 0 = (0.009705)(vx i + v y j + vz k )

Its moment about the origin, (lb ⋅ ft ⋅ s): i H r v ( A )0 = A × m 0 = 0.009705 9 vx

j 0 vy

k 0.75 vz

= 0.009705[− 0.75v y i + (0.75vx − 9vz ) j + 9v y k ]

Final linear momentum of the meteorite, (lb ⋅ s): 0.75mv 0 = 0.007279(vx i + v y j + vz k )

Its moment about the origin, (lb ⋅ ft ⋅ s): rA × (0.75mv 0 ) = 0.007279[− 0.75v y i + (0.75vx − 9vz ) j + 9v y k ]

m′v′0 = 0

Initial linear momentum of the space probe, (lb ⋅ s):

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PROBLEM 18.54 (Continued)

Final linear momentum of the space probe, (lb ⋅ s):  0.675  i + v′y j + vz′ k  m′(vx′ i + v′y j + vz′ k ) = 93.17  −  12 

Final angular momentum of space probe, (lb ⋅ ft ⋅ s): H A = m′(k x2ω x i + k y2ω y j + k z2ω z k ) = 93.17[(1.375) 2 (0.05)i + (1.425) 2 (−0.12) j + (1.250) 2 ω z k ] = 8.8075i − 22.703j + 145.58ω z k

Conservation of linear momentum of the probe plus the meteorite (lb ⋅ s): 0.009705(vx i + v y j + vz k ) = 0.007279(vx i + v y j + vz k ) + 93.17(−0.05625i + v′y j + vz′ k ) i:

0.002426vx = −5.2408

j:

0.002426v y = 93.17v′y

k:

0.002426vz = 93.17vz′

vx = −2160 ft/s

Conservation of angular momentum about the origin (lb ⋅ ft ⋅ s): (0.009705)[− 0.75v y i + (0.75vx − 9vz ) j + 9v y k ] = (0.007279)[ − 0.75v y i + (0.75vx − 9vz ) j + 9v y k ] + 8.8075i − 22.703j + 145.58ω z k i: − 0.0018195v y = 8.8075 k:

v y = − 4840.5 ft/s

− 0.021834v y = 145.58ω z

ω z = −149.98 × 10−6 v y

ω z = − 0.726 rad/s

Kinetic energy of motion of probe relative to its mass center:

(

)

(

)

1 1 I xω x2 + I y ω y2 + I z ω z2 = m k x2ω x2 + k y2ω z2ω z2 2 2 1 = (93.17)[(1.375)2 (0.05) 2 + (1.425) 2 (−0.12) 2 + (1.250) 2 (−0.726) 2 ] 2

T′ =

T ′ = 39.9 ft ⋅ lb 

= 39.9 ft ⋅ lb

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PROBLEM 18.55  of the angular momentum H of the disk Determine the rate of change H G G of Problem 18.1.

PROBLEM 18.1 A thin, homogeneous disk of mass m and radius r spins at the constant rate ω1 about an axle held by a fork-ended vertical rod, which rotates at the constant rate ω 2 . Determine the angular momentum H G of the disk about its mass center G.

SOLUTION Angular velocity:

ω = ω2 j + ω1k

Moments of inertia:

Ix =

Products of inertia: by symmetry,

I xy = I yz = I zx = 0

Angular momentum:

H G = I xω x i + I y ω y j + I z ω z k

1 2 mr , 4

Iy =

1 2 mr , 4

Iz =

1 2 mr 2

1 1  HG = 0 +  mr 2  ω2 j + mr 2ω1k 2 4  1 1 H G = mr 2ω2 j + mr 2ω1k 4 2

Rate of change of angular momentum. Let the frame of reference Gxyz be rotating with angular velocity Ω = ω1 j

Then

 = (H  ) H G G Gxyz + Ω × H G 1 1  = 0 + ω2 j ×  mr 2ω2 j + mr 2ω1k  2 4  =

1 2 mr ω1ω2 i 2

 = 1 mr 2ω ω i  H G 1 2 2

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PROBLEM 18.56  of the angular momentum H of Determine the rate of change H G G the plate of Problem 18.2.

PROBLEM 18.2 A thin rectangular plate of weight 15 lb rotates about its vertical diagonal AB with an angular velocity ω. Knowing that the z axis is perpendicular to the plate and that ω is constant and equal to 5 rad/s, determine the angular momentum of the plate about its mass center G.

SOLUTION h = (9 in.)2 + (12 in.) 2 = 15 in.

Resolving ω along the principal axes x′, y′, z: 12 ω = 0.8(5 rad/s) = 4 rad/s 15 9 ω y′ = ω = 0.6(5 rad/s) = 3 rad/s 15 ωz = 0

ω x′ =

Moments of inertia: 2

I x′ =

1  15 lb  9  2   ft  = 0.021836 slug ⋅ ft 12  32.2  12 

I y′ =

1  15 lb  12  2   ft  = 0.038820 slug ⋅ ft 12  32.2  12 

2

From Eqs. (18.10): H x′ = I x′ω x′ = (0.021836)(4) = 0.087345 slug ft 2 /s H y′ = I y′ω y ′ = (0.038820)(3) = 0.11646 slug ft 2 /s H z = I zωz = 0 H G = (0.087345 slug ft 2 /s)i ′ + (0.11646 slug ft 2 /s) j′

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PROBLEM 18.56 (Continued)

Components along x and y axes: 3 4 H x′ − H y ′ 5 5 3 4 = (0.087345) − (0.11646) 5 5 = −0.040761 4 3 H y = H x′ + H y ′ 5 5 4 3 = (0.087345) + (0.11646) 5 5 = 0.13975 Hx =

H G = (−0.040761 slug ⋅ ft 2 /s)i + (0.13975 slug ⋅ ft 2 /s) j

where the frame Axyz rotates with the plate with the angular velocity. Ω = ω = (5 rod/s) j  ) We have (H G Axyz = 0. Substituting into Eq. (18.22):

 = (H  ) H G G Axyz + Ω × H G = 0 + 5 j × ( −0.040761i + 0.13975 j) = (0.20380 ft/lb)k

 = (0.204 ft/lb)k  H G

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PROBLEM 18.57  of the angular momentum Determine the rate of change H D H D of the assembly of Problem 18.3.

PROBLEM 18.3 Two uniform rods AB and CE, each of weight 3 lb and length 2 ft, are welded to each other at their midpoints. Knowing that this assembly has an angular velocity of constant magnitude ω = 12 rad/s, determine the magnitude and direction of the angular momentum HD of the assembly about D.

SOLUTION m=

W 3 = = 0.093168 lb ⋅ s 2 /ft, 32.2 g

l = 24 in. = 2 ft,

ω = (12 rad/s)i

For rod ADB,

H D = I xω i ≈ 0,

since I x ≈ 0.

For rod CDE, use principal axes x′, y′ as shown. cosθ =

9 , 12

θ = 41.410°

ω x′ = ω cos θ = 9 rad/s 2 ω y′ = ω sin θ = 7.93725 rad/s 2 ω z′ = 0 I x′ ≈ 0 I y′ =

1 2 1 ml = (0.093168)(2)2 12 12

= 0.0310559 lb ⋅ s 2 ⋅ ft  H D = I x′ω x′i′ + I y′ω y′ j′ + I z′ω z′k′

= 0 + (0.0310559)(7.93725) j′ + 0 = 0.246498 j′ H D = 0.246498(sin θ i + cos θ j) = 0.163045i + 0.184874 j

Let the frame of reference Dxyz be rotating with angular velocity Ω = ω = (12 rad/s)i

Then,

 = (H  ) H D D Dxyz + Ω × Η D = 0 + ω × Η D  = 12i × (0.163045i + 0.184874 j) H D

 = (2.22 lb ⋅ ft)k  H D

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PROBLEM 18.58  of the angular momentum HA of Determine the rate of change H A the disk of Problem 18.4.

PROBLEM 18.4 A homogeneous disk of weight W = 6 lb rotates at the constant rate ω 1 = 16 rad/s with respect to arm ABC, which is welded to a shaft DCE rotating at the constant rate ω 2 = 8 rad/s. Determine the angular momentum H A of the disk about its center A.

SOLUTION ω = ω2 i + ω1 j = (8 rad/s)i + (16 rad/s) j

For axes x′, y ′, z ′ parallel to x, y, z with origin at A, m=

W 6 = = 0.186335 lb ⋅ s 2 /ft g 32.2

I x′ =

1 2 1  8  mr = (0.186335)   = 0.020704 lb ⋅ s 2 ⋅ ft 4 4  12 

2

I z ′ = I x′ = 0.020704 lb ⋅ s 2 ⋅ ft I y ′ = I x′ + I z′ = 0.041408 lb ⋅ s 2 ⋅ ft HA = I x′ω x′i ′ + I y′ ω y′ j + I z′ ω z′k = (0.020704)(8)i + (0.041408)(16) j = (0.1656 lb ⋅ s ⋅ ft)i + (0.6625 lb ⋅ s ⋅ ft) j

Let the frame of reference Axyz be rotating with angular velocity Ω = ω2 i = (8 rad/s)i

Then

 = (H  ) H A A Axyz + Ω × Η A = 0 + ω2 i × ΗA  = 8i × (0.1656i + 0.6625 j) = (5.30 lb ⋅ ft)k H A

 = (5.30 lb ⋅ ft)k  H A

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PROBLEM 18.59  of the angular momentum H of Determine the rate of change H C C the disk of Problem 18.5.

PROBLEM 18.5 A thin disk of mass m = 4 kg rotates at the constant rate ω2 = 15 rad/s with respect to arm ABC, which itself rotates at the constant rate ω1 = 5 rad/s about the y axis. Determine the angular momentum of the disk about its center C.

SOLUTION r = 150 mm

Angular velocity of disk: ω = ω1 j + ω2 k

= (5 rad/s) j + (15 rad/s)k

Centroidal moments of inertia: I x′ = I y ′ =

1 2 mr 4

1 (4)(0.150 m) 2 = 0.0225 kg ⋅ m 2 4 1 2 I z′ = mr = 0.045 kg ⋅ m 2 2

=

Angular momentum about C. H C = I x′ ω x′ i + I y ′ ω y ′ j + I z ′ ω z ′ k = 0 + (0.0225)(5) j + (0.045)(15)k = (0.1125 kg ⋅ m 2 /s) j + (0.6750 kg ⋅ m 2 /s)k

Rate of change of angular momentum. Let the reference frame Oxyz be rotating with angular velocity. Ω = ω1 = (5 rad/s) j

Then

 = (H  ) H C C Oxyz + Ω × H C

= 0 + 5 j × (0.1125 j + 0.6750k ) = (3.3750 N ⋅ m)i

 = (3.38 N ⋅ m)i  H C

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PROBLEM 18.60  of the angular momentum HG of Determine the rate of change H G the disk of Problem 18.8.

PROBLEM 18.8 A homogeneous disk of mass m and radius r is mounted on the vertical shaft AB. The normal to the disk at G forms an angle β = 25° with the shaft. Knowing that the shaft has a constant angular velocity ω, determine the angle θ formed by the shaft AB and the angular momentum HG of the disk about its mass center G.

SOLUTION Use the principal centroidal axes Gx′y ′z ′, Moments of inertia. I x′ = I z = I y′ =

1 2 mr 4

1 2 mr 2

Angular velocity.

ω x′ = − ω sin β ω y′ = ω cos β ωz = 0 Angular momentum about G. Using Eqs. (18.10), 1 H x′ = I x′ω x′ = − mr 2ω sin β 4 1 H y′ = I y′ω y′ = mr 2ω cos β 2 H z = I zω z = 0 H G = H x′ i ′ + H y′ j′ + H z k

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PROBLEM 18.60 (Continued)

where i′, j′, k are the unit vectors along the x′y ′z axes. 1 1 H G = − mr 2ω sin β i ′ + mr 2ω cos β j′ 4 2 Rate of change of angular momentum. Let the frame of reference Gx′y ′z ′ be rotating with angular velocity Ω = ω j = ω (− sin β i ′ + cos β j′)

Then

 = (H  ) H G G Gxyz + Ω × H G 1  1  = 0 + ω (− sin β i′ + cos β j′) ×  − mr 2ω sin β i ′ + mr 2ω cos β j′  4 2   1 2 2 1 2 2 = mr ω cos β sin β k − mr ω sin β cos β k 4 2 1 2 2 = − mr ω sin β cos β k 4

With β = 25°,

 = − 1 mr 2ω 2 sin 25° cos 25°k H G 4

 = −0.0958mr 2ω 2 k  H G

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PROBLEM 18.61  of the angular momentum HD of Determine the rate of change H D the assembly of Problem 18.3, assuming that at the instant considered the assembly has an angular velocity ω = (12 rad/s)i and an angular acceleration α = −(96 rad/s 2 )i.

SOLUTION m=

W 3 = = 0.093168 lb ⋅ s 2 /ft 2 , l = 24 in. = 2 ft, ω = (12 rad/s)i g 32.2

For rod ADB, H D = I xω i ≈ 0, since I x ≈ 0. For rod CDE, use principal axes x′, y ′ as shown. 9 , θ = 41.410° 12 ω x′ = ω cos θ = 9 rad/s 2

cos θ =

ω y′ = ω sin θ = 7.93725 rad/s 2 ωz′ = 0 α = −(96 rad/s 2 )i α x′ = α cos θ = −72 rad/s α y′ = α sin θ = −63.498 rad/s 2 I x′ ≈ 0 1 1 I y′ = ml 2 = (0.93168)(2) 2 = 0.0310559 kg ⋅ m 2 12 12 I z′ = I y′ H D = I x′ω x′i ′ + I y′ω y′ j′ + I z′ω z′k ′ = 0 + (0.0310559)(7.93725) j′ + 0 = (0.24698 lb ⋅ s/ft)j′

Let the reference frame Dx′y ′z ′ be rotating with angular velocity Ω = ω i = ω x′ i ′ + ω y′ j′ = (9 rad/s)i ′ + (7.93725 rad/s)j′  = (H  ) H D D Dx′y ′z ′ + Ω × H D = I x′α x′i′ + I y′α y′ j′ + I z′α z ′k ′ + Ω × H D = 0 + (0.0310559)(−63.498) j + 0 + (9i ′ + 7.93725 j′) × (0.246498 j′) = −1.97199 j′ + 2.21848k = −(1.97199 kg ⋅ m 2 /s 2 )(sin θ i + cos θ j) + (2.21848 kg ⋅ m 2 /s 2 )k



 = −(1.304 N ⋅ m)i − (1.479 N ⋅ m)j + (2.22 N ⋅ m)k  H D

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PROBLEM 18.62  of the angular momentum HD of the Determine the rate of change H D assembly of Problem 18.3, assuming that at the instant considered the assembly has an angular velocity ω = (12 rad/s)i and an angular acceleration α = (96 rad/s 2 )i.

SOLUTION m=

W 3 = = 0.093168 lb ⋅ s 2 /ft, l = 24 in. = 2 ft, ω = (12 rad/s)i g 32.2

For rod ADB, H D = I xω i ≈ 0, since I x ≈ 0. For rod CDE, use principal axes x′, y ′ as shown. 9 , θ = 41.410° 12 ω x′ = ω cos θ = 9 rad/s 2

cos θ =

ω y′ = ω sin θ = 7.93725 rad/s 2 ω z′ = 0 α = (96 rad/s 2 )i α x′ = α cos θ = 72 rad/s α y′ = α sin θ = 63.498 rad/s 2 I x′ ≈ 0 1 1 I y′ = ml 2 = (0.93168)(2) 2 = 0.0310559 kg ⋅ m 2 12 12 I z′ = I y′ H D = I x′ω x′ i ′ + I y′ω y′ j′ + I z′ω z′k ′ = 0 + (0.0310559)(7.93725) j + 0 = (0.246498 lb ⋅ s/ft)j′

Let the reference frame Dx′y ′z ′ be rotating with angular velocity Ω = ω i = ω x′ i′ + ω y′ j′ = (9 rad/s)i′ + (7.93725 rad/s)j′  = (H  ) H D D Dx′y ′z ′ + Ω × H D = I x′α x′ i ′ + I y′α y ′ j′ + I z′α z ′k ′ + Ω × H D = 0 + (0.310559)(63.498) j + 0 + (9i ′ + 7.93725 j′) × (0.246498 j′) = 1.97199 j′ + 2.21848k = (1.97199 kg ⋅ m 2 /s 2 )(sin θ i + cos θ j) + (2.21848 kg ⋅ m 2 /s 2 )k



 = (1.304 N ⋅ m)i + (1.479 N ⋅ m)j + (2.22 N ⋅ m)k  H D

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PROBLEM 18.63 A thin homogeneous square of mass m and side a is welded to a vertical shaft AB with which it forms an angle of 45°. Knowing that the shaft rotates with an angular velocity ω = ω j and an angular acceleration  of the angular momentum HA of α = α j, determine the rate of change H A the plate assembly.

SOLUTION Use principal axes y ′, z ′ as shown.

ω y ′ = ω cos 45°, ω z′ = ω sin 45° ω x′ = 0 1 1 I x′ = ma 2 , I y′ = ma 2 3 12 5 I z ′ = I x′ + I y ′ = ma 2 12 H A = I x′ω x′i ′ + I y′ω y′ j′ + I z′ω z′k ′

 1   5  = 0 +  ma 2  (ω cos 45°) j′ +  ma 2  (ω sin 45°)k ′  12   12   1  H A =  ma 2  (ω cos 45°)(cos 45° j − sin 45°k )  12   5  +  ma 2  (ω sin 45°)(sin 45° j + cos 45°k ) 12   2 ma ω HA = (3j + 2k ) 12

Angular acceleration.

α = α j; ω y′ = α cos 45°, ω z′ = α sin 45°

Let the reference frame Axyz be rotating with angular velocity Ω = ωj

With respect to this frame,  )    (H A Axyz = I x′ω x′ i ′ + I y ′ω y ′ j′ + I z ′ω z k ′  1   5  = 0 +  ma 2  (α cos 45°)i ′ +  ma 2  (α sin 45°)k ′  12   12  =

ma 2α (3j + 2k ) 12

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PROBLEM 18.63 (Continued)

With respect to the fixed reference frame,  = (H  ) H A A Axyz + Ω × H A =

 ma 2ω  ma 2α (3j + 2k ) + ω j ×  (3j + 2k )  12  12  2  = ma (2ω 2 i + 3α j + 2α k )  H A 12

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PROBLEM 18.64  of the angular momentum H of the Determine the rate of change H G G disk of Problem 18.8, assuming that at the instant considered the assembly has an angular velocity ω = ω j and an angular acceleration α = α j.

PROBLEM 18.8 A homogeneous disk of mass m and radius r is mounted on the vertical shaft AB. The normal to the disk at G forms an angle β = 25° with the shaft. Knowing that the shaft has a constant angular velocity ω, determine the angle θ formed by the shaft AB and the angular momentum HG of the disk about its mass center G.

SOLUTION Use the principal centroidal axes Gx′y ′z ′ Moments of inertia. I x′ = I z ′ = I y′ =

1 2 mr 4

1 2 mr 2

Angular velocity.

ω x′ = −ω sin β ω y′ = ω cos β ωz = 0 Angular momentum about G. Using Eqs. (18.10), 1 H x′ = I x′ω x′ = − mr 2ω sin β 4 1 2 H y′ = I y′ω y′ = mr ω cos β 2 H z = I zω z = 0 H G = H x′ i ′ + H y ′ j + H z k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2053

PROBLEM 18.64 (Continued) where i, j, k are the unit vectors along the x′, y ′, z ′ axes. 1 1 H G = − mr 2ω sin β i ′ + mr 2ω cos β j′ 4 2

Angular acceleration.

α = α j, ω x′ = −α sin β , ω y′ = α cos β , ω z′ = 0

Rate of change of angular momentum. Let the reference frame Gxyz be rotating with angular velocity Ω = ω j = −(ω sin β )i ′ + (ω cos β ) j′

With respect to this frame,  )    (H G Gxyz = I x′ω x ′ i ′ + I y ′ω y ′ j′ + I z ′ω z ′k ′ 1  1  =  mr 2  (−α sin β )i ′ +  mr 2  (α cos β ) j′ + 0 4  2  1 1 = − mr 2α sin β i ′ + mr 2α cos β j′ 4 2 1 1 = − mr 2α sin β (i cos β − j sin β ) + mr 2α cos β (i sin β + j cos β ) 4 2 1 2 = mr α [i sin β cos β + j(2 cos 2 β + sin 2 β )] 4 1 2 = mr α [i sin 25° cos 25° + j(2 cos 2 25° + sin 2 25°)] 4 = mr 2α (0.0957556i + 0.45535 j)

With respect to the fixed reference frame,  = (H  ) H G G Gxyz + Ω × H G

where

Ω × H G = + (−ω sin β i ′ + ω cos β j′)

1  1  ×  − mr 2ω sin β + mr 2ω cos β  4 2   1 2 2 1 2 2 = mr ω cos β sin β k − mr ω sin β cos β k 4 2 1 2 2 = − mr ω sin β cos β k 4 1 2 2 = − mr ω sin 25° cos 25°k = −0.095756mr 2ω 2k 4  = mr 2 (0.0958α i + 0.455α j − 0.0958ω 2 k )  H G

Then

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PROBLEM 18.65 A slender, uniform rod AB of mass m and a vertical shaft CD, each of length 2b, are welded together at their midpoints G. Knowing that the shaft rotates at the constant rate ω, determine the dynamic reactions at C and D.

SOLUTION Using the principal axes Gx′y ′z: 1 I x′ = 0, I y ′ = I z = mb 2 3

ω x′ = −ω sin β , ω y′ = ω cos β , ω z = 0 Angular momentum about G.

H G = I x′ω x′ i ′ + I y′ω y′ j′ + I zω z k 1 H G = mb 2ω cos β j′ 3

or, since j′ = i sin β + j cos β :

1 H G = mb 2ω cos β (sin β i + cos β j) 3

(1)

Rate of change of angular momentum.  = (H  ) H G G Gxyz + Ω × H G = 0 + ω × H B 1  = ω j ×  mb 2ω cos β (sin β i + cos β j)  3   1 2 2 = − mb ω sin β cos β k 3

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PROBLEM 18.65 (Continued)

Equations of motion. We equate the systems of external and effective forces  ΣM D = Σ(M D )eff : 2bj × (C x i + C z k ) = H G 1 − 2bC x k + 2bC z i = − mb 2ω 2 sin β cos β k 3

1 mbω 2 sin β cos β , C z = 0 6

Thus,

Cx =



ΣF = ΣFeff : C + D = 0 

C=

1 mbω 2 sin β cos β i  6

1 D = − mbω 2 sin β cos β i  6

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PROBLEM 18.66 A thin homogeneous triangular plate of weight 10 pounds is welded to a light vertical axle supported by bearings at A and B. Knowing that the plate rotates at the constant rate ω = 8 rad/s, determine the dynamic reactions at A and B.

SOLUTION We shall use Eqs. (18.1) and (18.28):  ) ΣF = m a ΣM A = (H A Axyz + Ω × H A

Computation of H A : ω = ωj H A = − I xyω i + I yω j − I yzω k

The moment of inertia of the triangle is ( I y )area =

1 3 b h, 12

( I y ) mass =

m 1 ( I y )area = mb 2 A 6

A=

1 bh 2

The product of inertia of the triangle is ( I xy )area =

1 2 2 b h 24

( I xy ) mass =

m 1 ( I xy )area = mbh A 12

Since the z-coordinate is negligible, ( I yz ) mass = 0

Thus,

HA = −

1 1 mbhω i + mb2ω j. 12 6

(1)

where the frame of reference Axyz rotates with the plate with the angular velocity Ω = ω =ωj PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2057

PROBLEM 18.66 (Continued)

Equations of motion. (Weight is omitted for dynamic reactions.) Eq. (18.28),  ) ΣM A = ( H A Axyz + Ω × H A .

Since ω = constant, it follows from Eq. (1) that  ) (H A Axyz = 0

Thus,  mbh mb 2  ωi + ω j  − hBx k + hB z i hj × ( Bx i + Bz k ) = 0 + ω j ×  − 6  12  1 = + mbhω 2 k 12

Equating coefficients of unit vectors:

Bx = − B=−

1 mbω 2 , Bz = 0 12 1 mbω 2 i 12

The dynamic reactions must also satisfy Eq. (18.1): b ΣF = m a: A + B = − mxω 2 i = − m   ω 2 i 3 1 1  1  A = − mbω 2 i −  − mbω 2 i  , A = − mbω 2 i 3 4  12 

Given data:

W = 10 lb, b = 12 in. = 1 ft, h = 24 in. = 2 ft, ω = 8 rad/s 1  10 lbs  A=−  (1 ft)(8 rad/s)2 i 4  32.2 



B=−

1  10 lbs  (1 ft)(8 rad/s) 2 i  12  32.2 

A = −(4.97 lb)i  B = −(1.656 lb)i 

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PROBLEM 18.67 The assembly shown consists of pieces of sheet aluminum of uniform thickness and of total weight 2.7 lb welded to a light axle supported by bearings at A and B. Knowing that the assembly rotates at the constant rate ω = 240 rpm, determine the dynamic reactions at A and B.

SOLUTION Mass of sheet metal: Sheet metal dimension: Area of sheet metal: Let Moments and products of inertia:

2.7 = 0.08385 lb ⋅ s 2 /ft 32.2 b = 6 in. = 0.5 ft 1 1 A = b 2 + b 2 + b 2 + b 2 = 3b 2 = 0.75 ft 2 2 2 m 0.08385 ρ= = = 0.1118 lb ⋅ s 2 /ft 3 = mass per unit area. A 0.75 I mass = ρ I area m=

xy plane (rectangles) 1 1 2 I x = b4 + b4 = b4 3 3 3 2 4 I x = ρb 3 2 = (0.1118)(0.5)4 3 = 4.658 × 10−3 lb ⋅ s 2 ⋅ ft

 3  1   5  1  I xy = (b 2 )  b  b  + (b 2 )  b  − b   2  2   2  2  1 = − b4 2 1 1 I xy = − ρ b4 = − (0.1118)(0.5) 4 2 2 = −3.4938 × 10−3 lb ⋅ s 2 ⋅ ft

xz plane (triangles) 1 1 1 I x = b4 + b4 = b4 12 12 6 1 1 I x = ρ b4 = (0.1118)(0.5)4 6 6 = 1.1646 × 10 −3 lb ⋅ s 2 ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2059

PROBLEM 18.67 (Continued)

For calculation of I xz , use pairs of elements dA1 and dA2 : dA2 = dA1. z b  z I xz =  x dA1 +  (4b − x)  −  dA2 = −  (2b − x) zdA1 = −  0 (2b − x) z 2 dx 2  2 z = x.

but

a 1  5 2 I xz = −  0 (2bx 2 − x3 )dx = −  b 4 − b 4  = − b 4 4  12 3

Hence,

I xz = −

5 5 ρ b 4 = − (0.1118)(0.5)4 = −2.9115 × 10 −3 lb ⋅ s 2 ⋅ ft 12 12

I x = 4.658 × 10−3 + 1.1646 × 10−3 = 5.823 × 10 −3 lb ⋅ s 2 ⋅ ft

Total for I x :

The mass center lies on the rotation axis, therefore a=0

ΣF = A + B = m a = 0 A = − B H A = I xω i − I xyω j − I xzω k ω = ω i, α = α i

Let the frame of reference Axyz be rotating with angular velocity Ω = ω = ωi  = (H  ) ΣM A = H A A Axyz + Ω × H A M 0 i + 4bi × ( B y j + Bz k ) = I xα i − I xyα j − I xzα k + ω i × ( I xω i − I xyω j − I xzω k ) M 0 i − 4bBz j + 4bBy k = I xα i − ( I xyα − I xz ω 2 ) j − ( I xzα + I xyω 2 )k

Resolve into components and solve for By and Bz . i:

j: k: Data:

M 0 = I xα Bz =

( I xyα − I xzω 2 ) 4b

By = −

( I xzα + I xyω 2 )

4b 2π (240) α = 0, ω = = 25.133 rad/s, b = 0.5 ft 60

Bz =

M0 = 0

0 − (−2.9115 × 10 −3 )(25.133) 2 = 0.91955 lb. (4)(0.5)

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PROBLEM 18.67 (Continued)

By =

0 − (−3.4938 × 10−3 )(25.133) 2 = 1.10346 lb. (4)(0.5)

Ay = − B y = −1.10346 lb. Az = − Bz = −0.91955 lb.

A = −(1.103 lb)j − (0.920lb)k  B = (1.103 lb)j + (0.920 lb)k 



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PROBLEM 18.68 The 8-kg shaft shown has a uniform cross section. Knowing that the shaft rotates at the constant rate ω = 12 rad/s, determine the dynamic reactions at A and B.

SOLUTION ω = ωi

Angular velocity:

Angular momentum about the mass center G: H G = I xω i − I xyω j − I xzω k

Let the reference frame Axyz be rotating with angular velocity Ω = ω i.  = (H  ) H +Ω×H G

G Axyz

G

= 0 + ω i × ( I xω i − I xyω j − I xzω k ) = I xzω 2 j − I xyω 2 k

Since the shaft lies in the xz plane, I xy = 0. By symmetry, the mass center lies on line AB. ma = 0 ΣF = ΣFeff : A + B = m a = 0 A and B form a couple.

A = −B

 bi × ( B y j + Bz k ) = H G −bBz j + bB y k = I xzω j Bz = −

I xzω 2 , By = 0 b

Calculation of I xz . Divide the shaft into eight segments, each of length a = 200 mm = 0.2 m Let m′ be the mass of one segment. m′ =

For

,

,

, and

8 kg = 1 kg 8

, I xz = 0

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PROBLEM 18.68 (Continued)

Then

 a  a a a I xz = m′( −a)  −  + m′  −  (− a) + m′   ( a) + m′(a )   = 2m′a 2  2  2 2 2 2 2 2 2 2m′a ω (2)(1)(0.2) (12) Bz = − =− = −14.4 N 4a (4)(0.2) Ay = 0, Az = − Bz = 14.4 N

A = (14.4 N)k  B = −(14.4 N)k 

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PROBLEM 18.69 After attaching the 18-kg wheel shown to a balancing machine and making it spin at the rate of 15 rev/s, a mechanic has found that to balance the wheel both statically and dynamically, he should use two corrective masses, a 170-g mass placed at B and a 56-g mass placed at D. Using a right-handed frame of reference rotating with the wheel (with the z axis perpendicular to the plane of the figure), determine before the corrective masses have been attached (a) the distance from the axis of rotation to the mass center of the wheel and the products of inertia I xy and I zx , (b) the force-couple system at C equivalent to the forces exerted by the wheel on the machine.

SOLUTION m = 18 kg, ω = 2π (15)i = (94.248 rad/s)i mB = 170 g = 0.17 kg mD = 56 g = 0.056 kg xB = 75 mm = 0.075 m, xD = −0.075 m,

(a)

yB = 182 mm = 0.182 m

yD = −0.182 m

Balance masses are added to move the mass center to Point C and to reduce the products of inertia to zero. mB yB + mD yD + my = 0 (0.17)(0.182) + (0.056)(−0.182) + 18 y = 0

y = −1.15267 × 10−3 m z =0

y = −1.153 mm  z =0 

mB xB yB + mD xD yD + I xy = 0 (0.17)(0.075)(0.182) + (0.056)(−0.075)(−0.182) + I xy = 0

I xy = −3.0848 × 10−3

I xy = −3.08 × 10−3 kg ⋅ m 2  I zx = 0 

 

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PROBLEM 18.69 (Continued)

(b)

Force-couple system exerted on the wheel: F′ = m an = − m rω 2 = −(18)(−1.15267 × 10−3 j)(94.248)2 = (184.3 N)j

H C = I xω i − I xy ω j − I xz ω k  = ω × H = I ω 2 j − I ω 2k M ′C = H C C xz xy = 0 − ( −3.0848 × 10−3 )(94.248)2 k = (27.4 N ⋅ m)k

Force-couple system exerted by the wheel on the machine: F = − F′

F = −(184.3 N) j 

M C = −M ′C

M C = −(27.4 N ⋅ m)k 

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PROBLEM 18.70 When the 18-kg wheel shown is attached to a balancing machine and made to spin at a rate of 12.5 rev/s, it is found that the forces exerted by the wheel on the machine are equivalent to a force-couple system consisting of a force F = (160 N)j applied at C and a couple M C = (14.7 N ⋅ m)k , where the unit vectors form a triad which rotates with the wheel. (a) Determine the distance from the axis of rotation to the mass center of the wheel and the products of inertia I xy and I zx . (b) If only two corrective masses are to be used to balance the wheel statically and dynamically, what should these masses be and at which of the Points A, B, D, or E should they be placed?

SOLUTION m = 18 kg, ω = 2π (12.5)i = (78.54 rad/s)i

(a)

The force-couple system acting on the wheel is F′ = −(160 N)j, M ′C = −(14.7 N ⋅ m)k

F′ = m an = − m rω 2 y= z=

− Fy′ mω

2

=

r = yj + zk =

−F ′ mω 2

160 = 1.441 × 10−3 m (18)(78.54) 2

y = 1.441 mm

− Fz′ =0 mω 2

r = 1.441 mm 

H C = I xω i − I xy ω j − I xz ω k  = ω × H = ω i × (I ω i − I ω j − I ωk) M ′C = H C C x xy xz

( M C′ ) y j + ( M C′ ) z k = I xzω 2 j − I xyω 2 k I xy = −

( M C′ ) z

ω2

=

−(−14.7) = 2.3831 × 10−3 kg ⋅ m 2 (78.54) 2 I xy = 2.38 × 10−3 kg ⋅ m 2 

I xz =

( M C′ ) y

ω2

=0

I xz = 0 

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PROBLEM 18.70 (Continued)

(b)

Positions for the balance masses. y A = yB = − yE = − yD = 182 mm = 0.182 m − x A = xB = xE = − xD = 75 mm = 0.075 m

Balance masses must be added to move the mass center to Point C and to reduce the product of inertia to zero. m A y A + mB yB + mE yE + mD yD + my = 0 (0.182)(mA + mB − mE − mD ) + (18)(1.441 × 10−3 ) m A + mB − mE − mD = −0.1425 kg

(1)

m A x A y A + mB xB yB + mE xE yE + mD xD yD + I xy = 0 (0.075)(0.182)(− mA + mB − mE + mD ) + 2.3831 × 10−3 = 0 − mA + mB − mE + mD = −0.17459 kg

(2)

To solve Eqs. (1) and (2), set mB = 0 and let m AD = m A − mD . Then and

m AD − mE = −0.1425 − m AD − mE = −0.17459

Solving,

m AD = 0.01605 kg, mE = 0.1585 kg

Set mD = 0, so that m A = 0.01605 kg

mE = 158.5 g  m A = 16.05 g 



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PROBLEM 18.71 Knowing that the assembly of Problem 18.65 is initially at rest (ω = 0) when a couple of moment M 0 = M 0 j is applied to shaft CD, determine (a) the resulting angular acceleration of the assembly, (b) the dynamic reactions at C and D immediately after the couple is applied.

SOLUTION Using the principal axes G x′ y ′ z: 1 I x′ = 0, I y ′ = I z = mb 2 3

ω x′ = −ω sin β , ω y′ = ω cos β , ω z = 0 Angular momentum about G. H G = I x′ω x′ i′ + I y′ω y′ j′ + I zω z k 1 H G = mb 2ω cos β j′ 3 1 j′ = i sin β + j cos β : H G = mb 2ω cos β (sin β i + cos β j) 3

or, since

(1)

Rate of change of angular momentum. Eq. (18.22):

 = (H  )  H G G Gxyz + Ω × H G = ( H G )Gxyz + 0

Since Ω = ω = 0 when couple is applied, thus, 1 2  = (H  ) H G G Gxyz = mb α cos β (sin β i + cos β j) 3

(2)

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PROBLEM 18.71 (Continued)

Equations of motion: Equivalence of applied and effective forces.

1 1 ΣM D = Σ(M D )eff : 2bj × (C x i + Cz k ) + M 0 j = mb 2α sin β cos β i + mb 2α cos 2 β j 3 3 1 1 −2bCx k + 2bC z i + M 0 j = mb 2α sin β cos β i + mb 2α cos 2 β j 3 3

Equating the coefficients of i, j, k: i:

(a)

1 2bCz = mb 2α sin β cos β 3

j:

1 M 0 = mb 2α cos 2 β 3

(4)

k:

Cx = 0

(5)

Angular acceleration.

α=

From Eq. (4), (b)

(3)

3M 0 mb cos 2 β 2



Initial dynamic reactions. From Eq. (3), Cz =

 3M 1 1 mbα sin β cos β = mb sin β cos β  2 0 2 6 6  mb cos β

  

M  Cz =  0  tan β  2b 

Recalling Eq. (5),

Cx = 0 M C= 0  2b

ΣF = Σ(F)eff :

C + D = 0, D = −C

  tan β k  

M  D = −  0  tan β k   2b 

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PROBLEM 18.72 Knowing that the plate of Problem 18.66 is initially at rest (ω = 0) when a couple of moment M 0 = (0.75 ft ⋅ lb)j is applied to it, determine (a) the resulting angular acceleration of the plate, (b) the dynamic reactions A and B immediately after the couple has been applied. PROBLEM 18.66 A thin homogeneous triangular plate of weight 10 pounds is welded to a light vertical axle supported by bearings at A and B. Knowing that the plate rotates at the constant rate ω = 8 rad/s, determine the dynamic reactions at A and B.

SOLUTION We shall use Eqs. (18.1) and (18.28):  ) ΣF = m a ΣM A = (H A Axyz + Ω × HA

Computation of H A : ω = ωj

H A = − I xyω i + I yω j − I yzω k

The moment of inertia of the triangle is ( I y )area =

1 3 b h, 12

( I y ) mass =

m 1 ( I y )area = mb 2 A 6

A=

1 bh 2

The product of inertia of the triangle is ( I xy )area =

1 2 2 b h 24

( I xy ) mass =

m 1 ( I xy )area = mbh A 12

Since the z-coordinate is negligible, ( I yz ) mass = 0 1 1 mbhω i + mb2ω j 12 6 where the frame of reference Axyz rotates with the plate with the angular velocity Ω = ω = ωj

Thus,

HA = −

(1)

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PROBLEM 18.72 (Continued)

Equations of motion. We first use Eq. (18.28):  ) Σ M = (H A

A Axyz

+ Ω × HA

Where H A was obtained in Eq. (1) of Problem 18.66: 1 1 mbhω i + mb 2ωj 12 6 Differentiating with respect to the rotating frame: HA = −

 ) (H A Axyz = −

1 1 mbhα i + mb 2α j 12 6

 ) Substituting for H A and (H A Axyz into Eq. (18.28), noting that ω = 0, and computing ΣMA from diagram: 1 1 mbhα i + mb2α j + 0 12 6

M 0 j + hj × ( Bx i + Bz k ) = − M 0 j − hBx k + hBz i = −

1 1 mbhα i + mb 2α j 12 6

Equating the coefficients of the unit vectors: M0 =

j:

α=

(a)

1 2 mb α 6 6(0.75 lb ⋅ ft) lb ( 10 )(1 ft) 2 32.2

k:

− hBx = 0

i:

hBz = −

α=

6M 0

mb 2

= 14.49 rad/s 2

Bx = 0 1 1 1  10 lb  mbhα , Bz = − mbα = −  (1)(14.49) 12 12 12  32.2 

Bz = −0.375 lb

(b)

α = (14.49 rad/s 2 ) j 

B = −(0.375 lb)k 

We shall now apply Eq. (18.1): ΣF = m a : b 1 a = at = α j × i = − bα k 3 3

Since ω = 0 : Substituting in Eq. (12.1):

1 A + B = − mbα k 3 1 A = − mbα k − B 3 1 1 = − mbα k + mbα k 3 12 1 1  10 lb  A = − mbα k = −  (1 ft)(14.49 rad/s 2 )k 4 4  32.2 

A = −(1.125 lb)k 

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PROBLEM 18.73 The assembly of Problem 18.67 is initially at rest (ω = 0) when a couple M 0 is applied to axle AB. Knowing that the resulting angular acceleration of the assembly is α = (150 rad/s 2 )i, determine (a) the couple M 0 , (b) the dynamic reactions at A and B immediately after the couple is applied.

SOLUTION 2.7 = 0.08385 lb ⋅ s 2 /ft 32.2

Mass of sheet metal:

m=

Sheet metal dimension:

b = 6 in. = 0.5 ft

Area of sheet metal:

A=

1 2 1 b + b2 + b 2 + b 2 = 3b 2 = 0.75 ft 2 2 2

Let

ρ=

m 0.08385 = = 0.1118 lb ⋅ s 2 /ft 3 = mass per unit area 0.75 A

Moments and products of inertia:

( I )mass = ρ I (area)

xy plane (rectangles) 1 1 2 I x = b4 + b 4 = b 4 3 3 3 2 4 2 I x = ρ b = (0.1118)(0.5) 4 3 3 = 4.658 × 10−3 lb ⋅ s 2 ⋅ ft 1  3  1   5  1  I xy = (b 2 )  b  b  + (b 2 )  b  − b  = − b 4 2 2 2 2 2       1 1 I xy = − ρ b 4 = − (0.1118)(0.5) 4 4 4 = −3.4938 × 10−3 lb ⋅ s 2 ⋅ ft

xz plane (triangles) Ix =

1 4 1 4 1 4 b + b = b 12 12 6

1 4 1 ρ b = (0.1118)(0.5)4 6 6 = 1.1646 × 10−3 lb ⋅ s 2 ⋅ ft

Ix =

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PROBLEM 18.73 (Continued) For calculation of I xz , use pairs of elements dA1 and dA2 :

dA2 = dA1

z b  z I xz =  x dA1 +  (4b − x)  −  dA2 = −  (2b − x) zdA1 = −  0 (2b − x) z 2 dx 2  2

But z = x. a 1  5 2 I xz = −  0 (2bx 2 − x3 )dx = −  b 4 − b 4  = − b 4 3 4 12  

Hence,

I xz = −

5 5 ρ b 4 = − (0.1118)(0.5) 4 = −2.9115 × 10−3 lb ⋅ s 2 ⋅ ft 12 12

I x = 4.658 × 10−3 + 1.1646 × 10−3 = 5.823 × 10−3 kg ⋅ m 2

Total for I x :

The mass center lies on the rotation axis, therefore, a = 0 ΣF = A + B = m a = 0

A = −B

H A = I xω i − I xyω j − I xzω k

ω = ω i,

α =αi

Let the frame of reference Axyz be rotating with angular velocity Ω = ω = ω i.  = (H  ) ΣM A = H A A Axyz + Ω × H A

M 0 i + 4bi × ( By j + Bz k ) = I xα i − I xyα j − I xzα k + ω i × ( I xω i − I xyω j − I xzωk ) M 0 i − 4bBz j + 4bBy k = I xα i − ( I xyα − I xz ω 2 ) j − ( I xzα + I xyω 2 )k

Resolve into components and solve for By and Bz . i:

Data: (a) (b)

M 0 = I xα ( I xyα − I xzω 2 )

j:

Bz =

k:

By = −

4b ( I xzα + I xyω 2 ) 4b

α = 150 rad/s 2 ,

ω = 0,

b = 0.5 ft

M 0 = (5.823 × 10−3 )(150) = 0.87345 lb ⋅ ft Bz =

M 0 = (0.873 lb ⋅ ft)i 

(−3.4938 × 10−3 )(150) − 0 = −0.262 lb (4)(0.5)

By = −

(−2.9115 × 10−3 )(150) + 0 = 0.218 lb (4)(0.5)

Ay = − By = −0.218 lb Az = − Bz = 0.262 lb

A = −(0.218 lb) j + (0.262 lb)k  B = (0.218 lb)j − (0.262 lb)k 

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PROBLEM 18.74 The shaft of Problem 18.68 is initially at rest (ω = 0) when a couple M0 is applied to it. Knowing that the resulting angular acceleration of the shaft is α = (20 rad/s 2 )i, determine (a) the couple M0, (b) the dynamic reactions at A and B immediately after the couple is applied. PROBLEM 18.68 The 8-kg shaft shown has a uniform cross section. Knowing that the shaft rotates at the constant rate ω = 12 rad/s, determine the dynamic reactions at A and B.

SOLUTION  = ω i = α i ω

Angular velocity and angular acceleration:

H G = I xω i − I xyω j − I xzω k

Angular momentum about the mass center G:

Ω = ωi = 0

Let the reference frame Axyz be rotating with angular velocity  = (H  ) H G G Axyz + Ω × H G = I xα i − I xyα j − I xzα k + 0 = I xα i − I xyα j − I xzα k

Since the shaft lies in the xz plane, I xy = 0. By symmetry, the mass center lies on line AB.

ma = 0

ΣF = ΣFeff : A + B = m a = 0

A and B form a couple. A = −B

 M 0 i + bi × ( By j + Bz k ) = H G

M 0 i + bBy k − bBz j = I xα i − I xzα k M 0 = I xα , B y = −

I xzα , Bz = 0 b

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PROBLEM 18.74 (Continued)

Calculation of I x and I xz . Divide the shaft into eight segments, each of length a = 200 mm = 0.2 m

Let m′ be the mass of one segment. m′ =

8 kg = 1 kg 8

For  and ,

Ix = 0

For , , , and ,

1 I x = ma 2 3

For  and ,

I x = m′a 2

Total:

Ix =

For , , , and ,

I xz = 0

For , , , and ,

 a  a a a I xz = m′( −a)  −  + m′  −  (− a) + m′   ( a) + m′(a )   = 2m′a 2  2  2 2 2

(a)

M0 =

(b)

By = −

10 m′a 2 3

10 10 m′a 2α = (1)(0.2) 2 (20) 3 3 2m′a 2α (2)(1)(0.2)2 (20) =− (4)(0.2) b

Az = 0,

Ay = − By = 2 N

M 0 = (2.67 N ⋅ m)i  B = −(2.00 N) j  A = (2.00 N) j 

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PROBLEM 18.75 The assembly shown weighs 12 lb and consists of 4 thin 16-in.diameter semicircular aluminum plates welded to a light 40-in.-long shaft AB. The assembly is at rest (ω = 0) at time t = 0 when a couple M 0 is applied to it as shown, causing the assembly to complete one full revolution in 2 s. Determine (a) the couple M0, (b) the dynamic reactions at A and B at t = 0.

SOLUTION α = α i,

Fixed axis rotation with constant angular acceleration. 1 2

1 2

θ = θ0 + ω0 t + α t 2 = α t 2

α=

ω = ωi

2θ 2(2π ) = = 3.1416 rad/s 2 t2 22

ω=0 H C = I xω i − I xy ω j − I xzω k

Use centroidal axes x, y, z with origin at C.

Ω = ω = ωi

Let the reference frame Cxyz rotate with angular velocity

2 2  = (H  ) H C C Cxyz + Ω × H C = I xα i − I xyα j − I xzα k + I xz ω j − I xy ω k

Required moments and products of inertia. Let ρ = I mass = ρ I area

m = mass per unit area. A

r = 8 in. = 0.66667 ft m =

Part

Σ

12 = 0.37267 lb ⋅ s 2 /ft 32.2

A

Ix

I xy

I xz

1 2 πr 2

1 4 πr 8

2 − r4 3

0

1 2 πr 2

1 4 πr 8

0

2 − r4 3

1 2 πr 2

1 4 πr 8

0

2 − r4 3

1 2 πr 2

1 4 πr 8

2 − r4 3

0

2π r 2

1 4 πr 2

4 − r4 3

4 − r4 3

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PROBLEM 18.75 (Continued)

ρ=

m 0.37267 = = 0.13345 lb ⋅ s 2 /ft 3 2 2π r 2π (0.66667) 2

1  I x = (0.13345)  π  (0.66667)4 = 0.041407 lb ⋅ s 2 ⋅ ft 2   4 I xy = (0.13345)  −  (0.66667) 4 = −0.035147 lb ⋅ s 2 ⋅ ft  3  4 I xz = (0.13345)  −  (0.66667) 4 = −0.035147 lb ⋅ s 2 ⋅ ft  3

Since the mass center lies on the rotation axis,

a=0

ΣF = A + B = ΣFeff = m a = 0

A = −B

ΣM C = M 0 i + (−bi ) × A + bi × B = M 0 i + 2bi × ( By j + Bz k ) = M 0 i − 2bBz j + 2bBy k

 ΣM C = H C

(a)

i:

(b)

j:

2b = 40 in. = 3.3333 ft

Resolve into components.

M 0 = I xα = (0.041407)(3.1416)

M 0 = (0.1301 lb ⋅ ft)i 

− 2b Bz = − I xyα + I xzω 2 Bz = −

k:

where

−( −0.035147)(3.1416) + 0 = −0.0331 lb 3.3333

Az = 0.0331 lb

2bBy = − I xzα − I xyω 2 By = −

( −0.035147)(3.1416) + 0 = 0.0331 lb 3.3333

Ay = −0.0331 lb A = −(0.0331 lb) j + (0.0331 lb)k  B = (0.0331 lb) j − (0.0331 lb)k 

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PROBLEM 18.76 For the assembly of Problem 18.75, determine the dynamic reactions at A and B at t = 2 s. PROBLEM 18.75 The assembly shown weighs 12 lb and consists of four thin 16-in.-diameter semicircular aluminum plates welded to a light 40-in.-long shaft AB. The assembly is at rest (ω = 0) at time t = 0 when a couple M 0 is applied to it as shown, causing the assembly to complete one full revolution in 2 s. Determine (a) the couple M0, (b) the dynamic reactions at A and B at t = 0.

SOLUTION

α = α i,

Fixed axis rotation with constant angular acceleration. 1 2

1 2

θ = θ0 + ω0 t + α t 2 = α t 2

α=

ω = ωi

2θ 2(2π ) = = 3.1416 rad/s 2 2 2 t 2

ω = ω0 + α t = α t = (3.1416)(2) = 6.2832 rad/s H C = I xω i − I xy ω j − I xzω k

Use centroidal axes x, y, z with origin at C.

Ω = ω = ωi

Let the reference frame Cxyz rotate with angular velocity

2 2  = (H  ) H C C Cxyz + Ω × H C = I xα i − I xyα j − I xzα k + I xz ω j − I xy ω k

Required moments and products of inertia. Let ρ = I mass = ρ I area

m = mass per unit area. A

r = 8 in. = 0.66667 ft m =

Part



12 = 0.37267 lb ⋅ s 2 /ft 32.2

A

Ix

I xy

I xz

1 2 πr 2

1 4 πr 8

2 − r4 3

0

1 2 πr 2

1 4 πr 8

0

2 − r4 3

1 2 πr 2

1 4 πr 8

0

2 − r4 3

1 2 πr 2

1 4 πr 8

2 − r4 3

0

2π r 2

1 4 πr 2

4 − r4 3

4 − r4 3

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PROBLEM 18.76 (Continued)

ρ=

m 0.37267 = = 0.13345 lb ⋅ s 2 /ft 3 2 2π r 2π (0.66667)2

1  I x = (0.13345)  π  (0.66667) 4 = 0.041407 lb ⋅ s 2 ⋅ ft 2   4 I xy = (0.13345)  −  (0.66667) 4 = −0.035147 lb ⋅ s 2 ⋅ ft  3  4 I xz = (0.13345)  −  (0.66667) 4 = −0.035147 lb ⋅ s 2 ⋅ ft  3

Since the mass center lies on the rotation axis,

a=0

ΣF = A + B = ΣFeff = m a = 0

A = −B

ΣM C = M 0 i + (−bi ) × A + bi × B = M 0 i + 2bi × ( By j + Bz k ) = M 0 i − 2bBz j + 2bBy k

 ΣM C = H C

i: j: Bz = −

Resolve into components.

M 0 = I xα = (0.041407)(3.1416)

M 0 = 0.1301 lb ⋅ ft

− 2b Bz = − I xyα + I xzω 2

−(−0.035147)(3.1416) + (−0.035147)(6.2832) 2 = 0.383 lb, 3.3333

k: By = −

2b = 40 in. = 3.3333 ft

where

Az = −0.383 lb

2bBy = − I xzα − I xyω 2

( −0.035147)(3.1416) + (−0.035147)(6.2832) 2 = 0.449 lb, 3.3333

Ay = −0.449lb

A = −(0.449 lb) j − (0.383 lb)k  B = (0.449 lb) j + (0.383 lb)k 

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PROBLEM 18.77 The sheet-metal component shown is of uniform thickness and has a mass of 600 g. It is attached to a light axle supported by bearings at A and B located 150 mm apart. The component is at rest when it is subjected to a couple M0 as shown. If the resulting angular acceleration is α = (12 rad/s 2 )k , determine (a) the couple M0, (b) the dynamic reactions at A and B immediately after the couple has been applied.

SOLUTION The sheet metal component rotates about the fixed z axis, so that Equations (18.29) of the textbook apply. These are relisted below as equations (1), (2), and (3). ΣM x = − I xzα + I yzω 2

(1)

ΣM y = − I yzα − I xzω 2

(2)

ΣM z = I z α

(3)

Calculation of the required moment and products of inertia. Total mass:

m = 600 g = 0.6 kg

Total area:

A=

1 1 (0.075) 2 + (0.150)(0.075) + (0.075)2 = 16.875 × 10−3 m 2 2 2

Let ρ be the mass per unit area.

ρ=

m = 35.556 kg/m 2 A

The component is comprised of 3 parts: triangle , triangle , and rectangle  as shown. Let the lengths of 75 mm be labeled b. Triangle . The equation of the upper edge is y=

b −z 2

Use elements of width dz and height y. dm = ρ ydz (dI xz )el = 0

(dI z )el =

1 2 y dm 12

(dI yz )el = 0

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PROBLEM 18.77 (Continued)

Coordinates of the element mass center: xel = −b,

yel =

1 y, 2

zel = z

dI xz = ( dI xz )el + xel zel dm b  = 0 + (−b) z ( ρ yd z ) = − ρ bz  − z  dz 2  

I xz = − ρ b



b /2 − b /2

1 b  z  − z dz = ρ b 4 12 2 

dI yz = ( dI yz )el + yel zel dm 1 =0+ 2

I yz =

1 ρ 2



2

1 b   y  z ( ρ ydz ) = ρ  − z  zdz 2 2   2

1 b  4  − z  zdz = − 24 ρ b − b /2  2  b/ 2

dI z = (dI z )el + ( xel2 + yel2 )dm 1  1   1  =  y 2 + b 2 + y 2  ρ ydz = ρ  b 2 y + y 3  dz 4  3   12  5 1 1   13 = ρ  b3 − b 2 z + bz 2 − z 3  dz 24 4 2 3   Iz = ρ



1 1  7  13 3 5 2 b − b z + bz 2 − z 3  dz = ρ b 4  − b / 2  24 4 2 3  12 b /2

Applying the data, I xz =

1 (35.556)(0.075)4 = 93.75 × 10−6 kg ⋅ m 2 12

I yz = − Iz =

1 (35.556)(0.075)4 = −46.875 × 10−6 kg ⋅ m 2 24

7 (35.556)(0.075) 4 = 656.25 × 10−6 kg ⋅ m 2 12

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PROBLEM 18.77 (Continued)

b  Triangle . The equation of the lower edge is y = −  − z  . 2  

Use elements of width dz and height − y. dm = − ρ ydz ,

( dI z )el =

1 2 y dm, 12

(dI zx )el = 0,

( dI yz )el = 0

Coordinates of the element mass center: xel = b,

yel =

1 y, 2

zel = z

The integrals for I xz , I yz , and I z turn out to be the same as those of triangle . I xz = 93.75 × 10−6 kg ⋅ m 2 , I yz = −46.875 × 10−6 kg ⋅ m 2 , I z = 656.25 × 10−6 kg ⋅ m 2

Rectangle .

Area:

A = (0.150)(0.075) = 11.25 × 10−3 m 2

Mass:

m = ρ A = 400 × 10−3 kg I xz = 0, Iz =

Totals.

I yz = 0

1 1 m(2b) 2 = (400 × 10−3 )(0.150) 2 = 750 × 10−6 kg ⋅ m 2 12 12

I xz = ΣI xz = 187.5 × 10−6 kg ⋅ m 2 I yz = ΣI yz = −93.75 × 10−6 kg ⋅ m 2 I z = ΣI z = 2062.5 × 10−6 kg ⋅ m 2

Since the mass center lies on the fixed axis, the acceleration a of the mass center is zero. ΣF = m a = 0

The reactions at A and B form a couple. B = −A

Let

A = Ax i + Ay j

Resultant couple acting on the body: M = ck × ( Ax i + Ay j) + M 0 k

= −cAy i + cAx j + M 0 k

(a)

Moment M 0 . Using Equation (3), ΣM z = M 0 = I zα = (2062.5 × 10−6 )(12)

M 0 = 24.8 × 10−3 N ⋅ m 

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PROBLEM 18.77 (Continued)

(b)

Reactions at A and B for the case ω = 0. ΣM x = −cAy = − I xzα − I yzω 2 = − I xzα Ay =

I xzα (187.5 × 10−6 )(12) = = 15 × 10−3 N 0.150 c

ΣM y = cAx = − I yzα − I xzω 2 = − I yzα Ax = −

I yzα c

=−

(−93.75 × 10−6 )(12) = 7.5 × 10−3 N 0.150 A = (7.50 × 10−3 N)i + (15.00 × 10−3 N) j  B = −(7.50 × 10−3 N)i − (15.00 × 10−3 N) j 

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PROBLEM 18.78 For the sheet-metal component of Problem 18.77, determine (a) the angular velocity of the component 0.6 s after the couple M0 has been applied to it, (b) the magnitude of the dynamic reactions at A and B at that time.

SOLUTION The sheet metal component rotates about the fixed z axis with angular acceleration α = (12 rad/s 2 )k. (a)

Angular velocity at

t = 0.6 s.

ω = ω0 + α t = 0 + (12)(0.6) = 7.2 rad/s (b)

ω = (7.20 rad/s)k 

Dynamic reactions. Equations (18.29) of the textbook apply. These are relisted below as equations (1), (2), and (3). ΣM x = − I xzα + I yzω 2

(1)

ΣM y = − I yzα − I xzω 2

(2)

ΣM z = I z α

(3)

Calculation of the required moment and products of inertia. m = 600 g = 0.6 kg Total mass: 1 1 (0.075) 2 + (0.150)(0.075) + (0.075) 2 = 16.875 × 10−3 m 2 2 2 m ρ = = 35.556 kg/m 2 Let ρ be the mass per unit area. A The component is comprised of 3 parts: triangle , triangle , and rectangle  as shown. Let the lengths of 75 mm be labeled b. Triangle . The equation of the upper edge is b y= −z 2 Use elements of width dz and height y. 1 2 dm = ρ ydz (dI z )el = y dm 12

Total area:

A=

(dI xz )el = 0

(dI yz )el = 0

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PROBLEM 18.78 (Continued)

Coordinates of the element mass center: xel = −b, yel =

1 y, 2

zel = z

dI xz = (dI xz )el + xel zel dm b  = 0 + ( −b) z ( ρ ydz ) = − ρ bz  − z  dz 2 

I xz = − ρ b



1 b  z  − z  dz = ρ b 4 − b/2  2 12  b/ 2

dI yz = (dI yz )el + yel zel dm 2

1 b 1   = 0 +  y  z ( ρ ydz ) = ρ  − z  zdz 2 2 2   I yz

1 = ρ 2



2

1 b  4  − z  zdz = − 24 ρ b − b/2  2  b/ 2

(

)

dI z = (dI z )el + xel2 + yel2 dm 1  1   1  =  y 2 + b 2 + y 2  ρ yi dz = ρ  b2 y + y 3  dz 12 4 3     5 1 1   13 = ρ  b3 − b 2 z + bz 2 − z 3  dz 24 4 2 3   Iz = ρ



b /2

1 2 1 3 7  13 3 5 2 4  24 b − 4 b z + 2 bz − 3 z  dz = 12 ρ b 

−b / 2 

Applying the data, I xz =

1 (35.556)(0.075)4 = 93.75 × 10−6 kg ⋅ m 2 12

I yz = − Iz =

1 (35.556)(0.075)4 = −46.875 × 10−6 kg ⋅ m 2 24

7 (35.556)(0.075) 4 = 656.25 × 106 kg ⋅ m 2 12

Triangle . The equation of the lower edge is

b  y = −  − z . 2  

Use elements of width dz and height − y. dm = − ρ ydz , (dI z )el =

1 2 y dm, ( dI zx )el = 0, (dI yz )el = 0 12

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PROBLEM 18.78 (Continued)

Coordinates of the element mass center: xel = b,

yel =

1 y, 2

zel = z

The integrals for I xz , I yz , and I z turn out to be the same as those of triangle . I xz = 93.75 × 10−3 kg ⋅ m 2 , I yz = −46.875 × 10−6 kg ⋅ m 2 , I z = 656.25 × 10−6 kg ⋅ m 2

Rectangle .

Area:

A = (0.150)(0.075) = 11.25 × 10−3 m 2

Mass:

m = ρ A = 400 × 10−3 kg I xz = 0, Iz =

I yz = 0

1 1 m(2b) 2 = (400 × 10−3 )(0.150) 2 = 750 × 10−6 kg ⋅ m 2 12 12

I xz = ΣI xz = 187.5 × 10−6 kg ⋅ m 2

Totals.

I yz = ΣI yz = −93.75 × 10−6 kg ⋅ m 2 I z = ΣI z = 2062.5 × 10−6 kg ⋅ m 2

Since the mass center lies on the fixed axis, the acceleration a of the mass center is zero. ΣF = m a = 0

The reactions at A and B form a couple. B = −A A = Ax i + Ay j

Let

Resultant couple acting on the body: M = ck × ( Ax i + Ay j) + M 0 k

= −cAy i + cAx j + M 0 k −cAy = − I xzα + I yz ω 2

From Eq. (1), Ay =

2 I xzα I yzω (187.5 × 10−6 )(12) (−93.75 × 10 −6 )(7.2) 2 − = − = 47.4 × 10−3 N 0.150 0.150 c c

cAx = − I yzα − I xzω 2

From Eq. (2), Ax = −

I yzα c



I xz ω 2 (−93.75 × 10−6 )(12) (187.5 × 10−6 )(7.2) 2 =− − = −57.3 × 10−3 N 0.150 0.150 c A = −(57.3 × 10−3 N)i + (47.3 × 10−3 N) j  B = (57.3 × 10−3 N)i − (47.3 × 10−3 N) j 

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PROBLEM 18.79 The blade of an oscillating fan and the rotor of its motor have a total weight of 300 g and a combined radius of gyration of 75 mm. They are supported by bearings at A and B, 125 mm apart, and rotate at the rate ω1 = 1800 rpm. Determine the dynamic reactions at A and B when the motor casing has an angular velocity ω 2 = (0.6 rad/s) j.

SOLUTION ω1 = ω1i 2π (1800) 60 = 188.5 rad/s

ω1 =

ω = ω1i + ω2 j

Angular velocity:

H G = I xω1i + I yω2 j

Angular momentum:

Let the reference frame be rotating with angular velocity Ω = ω2 j.  = (H  ) H G G Gxyz + Ω × H G = 0 + ω2 j × ( I xω1i + I yω2 j) = − I xω1ω2 k

Assume that the acceleration of the mass center is negligible. Then the dynamic reactions at A and B reduce to a couple. A = −B M = bi × B = bi × ( By j + Bz k ) = −b Bz j + b By k

 M=H G j:

k:

− b Bz = 0

Resolve into components. Bz = 0,

b By = − I xω1ω2

Az = 0 By = −

I xω1ω2 mk 2ω ω =− x 1 2 b b

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PROBLEM 18.79 (Continued)

Data:

m = 0.300 kg k x = 0.075 m b = 0.125 m (0.300)(75 × 10−3 ) 2 (188.5)(0.6) = −1.527 N 0.125 Ay = 1.527 N

By = −

A = (1.527 N) j  B = −(1.527 N) j 

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PROBLEM 18.80 The blade of a portable saw and the rotor of its motor have a total weight of 2.5 lb and a combined radius of gyration of 1.5 in. Knowing that the blade rotates as shown at the rate ω1 = 1500 rpm, determine the magnitude and direction of the couple M that a worker must exert on the handle of the saw to rotate it with a constant angular velocity ω 2 = −(2.4 rad/s) j.

SOLUTION (2π )(1500) 60 = 157.08 rad/s ω1 = ω1k ω1 =

Angular velocity: Angular momentum of rotor:

ω = ω2 j + ω1k H G = I yω2 j + I zω1k

Let the reference frame Gxyz be rotating with angular velocity Ω = ω2 j.  = (H  ) H G G Gxyz + Ω × H G = 0 + ω2 j × ( I yω2 j + I zω1k ) = I z ω1ω2 i

Couple exerted on the saw:

 M=H G = I zω1ω2 i = mk z2ω1ω2 i

Data:

W = 2.5 lb 2.5 m= 32.2 = 0.07764 lb ⋅ s 2 /ft k z = 1.5 in. = 0.125 ft M = (0.07764)(0.125) 2 (157.08)(−2.4) i

M = −(0.457 lb ⋅ ft) i 

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PROBLEM 18.81 The flywheel of an automobile engine, which is rigidly attached to the crankshaft, is equivalent to a 400-mmdiameter, 15-mm-thick steel plate. Determine the magnitude of the couple exerted by the flywheel on the horizontal crankshaft as the automobile travels around an unbanked curve of 200-m radius at a speed of 90 km/h, with the flywheel rotating at 2700 rpm. Assume the automobile to have (a) a rear-wheel drive with the engine mounted longitudinally, (b) a front-wheel drive with the engine mounted transversely. (Density of steel = 7860 kg/m3 . )

SOLUTION Let the x axis be a horizontal axis directed along the engine mounting, i.e., longitudinally for rear-wheel drive and transversely for front-wheel drive. Let the y axis be vertical. The angular velocity of the automobile, ω2 , is equal to (v/ρ ) j, where v = 90 km/h = 25 m/s and ρ = 200 m. ω2 =

25 j = (0.125 rad/s) j 200 ω1 = ω1i

Angular velocity of the fly wheel relative to the automobile:

ω1 =

where

2π (2700) = 282.74 rad/s 60 H G = I xω1i + I yω y j

Angular momentum of fly wheel: Let the reference frame Gxyz be rotating with angular velocity

Ω = ω2 j

 = (H  ) H G G Gxyz + Ω × H G = 0 + ω2 j × ( I xω1i + I yω2 j) = − I xω1ω2 k

Couple exerted by the shaft on the fly wheel:

M = − I xω1ω2 k

Couple exerted by the fly wheel on the shaft:

M ′ = −M = I xω1ω2k

(1)

Data for fly wheel:

π π  m = ρ  d 2  t = (7860) (0.4)2 (0.015) = 14.816 kg 4 4 

For a circular plate,

Ix =

Using Equation (1),

1 2 1 mr = (14.816)(0.2) 2 = 0.29632 kg ⋅ m 2 2 2

M ′ = (0.29632)(282.74)(0.125)k = (10.47 N ⋅ m)k

(a)

Magnitude of couple for rear-wheel drive:

M ′ = 10.47 N ⋅ m 

(b)

For front-wheel drive:

M ′ = 10.47 N ⋅ m 

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PROBLEM 18.82 Each wheel of an automobile has a mass of 22 kg, a diameter of 575 mm, and a radius of gyration of 225 mm. The automobile travels around an unbanked curve of radius 150 m at a speed of 95 km/h. Knowing that the transverse distance between the wheels is 1.5 m, determine the additional normal force exerted by the ground on each outside wheel due to the motion of the car.

SOLUTION For each wheel,

v = 95 km/h = 26.389 m/s ωx = 0

ωy =

v

ρ

=

26.389 = 0.17593 rad/s 150

d 575 = = 287.5 mm = 0.2875 m 2 2 I z = mk 2 = (22)(0.225)2 = 1.11375 kg ⋅ m 2 r=

26.389 = −91.787 rad/s 0.2875 H G = I xω x i + I y ω y j + I z ω z k v r

ωz = − = −

= I yω y j + I z ω z k

Let reference frame Gxyz be rotating with angular velocity Ω = ω y j.  = (H  ) H G G Gxyz + Ω × H G = 0 + ω y j × ( I yω y j + I zω z k ) = I zωzω y i

 = (1.11375)( −91.787)(0.17593)i H G = −(17.985 N ⋅ m)i

Let O be the point at the center of the axle. For the two wheels plus the axle,  =H  +H  = −(35.97 N ⋅ m)i H O G G

The distance between the wheels is 1.5 m. M O = 1.5k × Fy j = −1.5Fy i

 and solve for F . Set M O = H y O Fy =

−35.97 = 23.98 N −1.5

Fy = 24.0 N 

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PROBLEM 18.83 The uniform thin 5-lb disk spins at a constant rate ω 2 = 6 rad/s about an axis held by a housing attached to a horizontal rod that rotates at the constant rate ω1 = 3 rad/s. Determine the couple which represents the dynamic reaction at the support A.

SOLUTION Angular velocity:

ω x = ω1,

ω y = 0,

ω z = ω 2.

ω = ω1i + ω 2k

Angular momentum:

H O = I xω x i + I yω y j + I zω zk = I xω1i + I zω 2k

Let frame Oxyz be rotating with angular velocity Ω = ω1 i. Rate of change of angular momentum.  = (H  ) H O O Oxyz + Ω × Η O = I xω1i + I zω 2 k + ω1i × ( I xω1i + I zω2 k ) 1 = 0 + 0 + 0 − I zω1ω2 j = − mr 2ω1ω2 j 2

Dynamic reaction couple:

 M=H O 2

1 1  5  4  M = − mr 2ω1ω2 j = −   (3)(6) j 2 2  32.2   12  M = − (0.1553 lb ⋅ ft) j 

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PROBLEM 18.84 The essential structure of a certain type of aircraft turn indicator is shown. Each spring has a constant of 500 N/m, and the 200-g uniform disk of 40-mm radius spins at the rate of 10,000 rpm. The springs are stretched and exert equal vertical forces on yoke AB when the airplane is traveling in a straight path. Determine the angle through which the yoke will rotate when the pilot executes a horizontal turn of 750-m radius to the right at a speed of 800 km/h. Indicate whether Point A will move up or down.

SOLUTION Let the x axis lie along the axle AB and the y axis be vertical. 2π (10, 000) = 1047.2 rad/s 60 v = 800 km/h = 222.22 m/s ρ = 750 m 22.222 = −0.2963 rad/s ω y = − ρv = − 750 ωz = 0

ωx =

Angular momentum:

H G = I xω x i + I yω y j + I zω z k = I xω x i + I yω y j

Let the reference frame Gxyz be turning about the y axis with angular velocity Ω = ω y j .  = (H  ) H G G Gxyz + Ω × Η G = ω y j × ( I xω x i + I y ω y j) = − I xω x ω y k

Data for the disk:

m = 200 g = 0.2 kg 1 2 mr 2 1 = (0.2)(0.040) 2 2 = 160 × 10−6 kg ⋅ m 2  MG = H G Ix =

= −(160 × 10−6 ) (1047.2) (−0.2963)k = (0.049646 N ⋅ m)k

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PROBLEM 18.84 (Continued)

The spring forces FA and FB exerted on the yoke provide the couple M G . The force exerted by spring B is upward. Let

FB = Fj

Then

FA = − Fj M G = rB/ A i × Fj = 0.100i × Fj = 0.1Fk

From

 , MG = H G

0.1F = 0.049646 F = 0.49646 N.

Compression of spring B:

F k 0.49646 = 500 = 0.99291 × 10−3 m = 0.9929 mm

δB =

Point B moves 0.9929 mm down. Point A moves 0.9929 mm up. Turning angle for yoke:

θ=

0.9929 + 0.9929 100

= 0.019858 rad

θ = 1.138°



Point A moves up. 

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PROBLEM 18.85 A slender rod is bent to form a square frame of side 6 in. The frame is attached by a collar at A to a vertical shaft which rotates with a constant angular velocity ω. Determine the value of ω for which line AB forms an angle β = 48° with the horizontal x axis.

SOLUTION Choose principal axes x′, y ′, z ′ with origin at the fixed Point A. 2

Angular velocity:

1  m  1   m  a  I x′ = 2   a 2  + 2    = ma 2 6  4  12   4  2  5  m  1   m  I z′ = 2   a 2  +   (a) 2 = ma 2 4 3 4 12      7 I y′ = I x′ + I z′ = ma 2 12 ω = −ω sin β i ′ + ω cos β j′

Angular momentum about A:

HA = − I x′ω sin β i ′ + I y′ω cos β j′

Let the reference from Axyz be rotating with angular velocity Ω = ω.  = (H  ) H + Ω ×H = 0 + ω× H A

A Axyz

A

A

= (−ω sin β i ′ + ω cos β j′) × (− I x′ ω sin β i ′ + I y′ ω cos β j′) = −( I y − I z ) ω 2 sin β cos β k 5 ma 2ω 2 sin β cos β k 12 a  Σ MA = − mg cos β k = H A 2 5 a − mg cos β = − ma 2ω 2 sin β cos β 2 12 6 g (6)(32.2) ω2 = = = 103.99 (rad/s)3 5 a sin β (5)(0.5)sin 48° =−

ω = 10.20 rad/s 

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PROBLEM 18.86 A uniform semicircular plate of radius 120 mm is hinged at A and B to a clevis which rotates with a constant angular velocity ω about a vertical axis. Determine (a) the angle β that the plate forms with the horizontal x axis when ω = 15 rad/s, (b) the largest value of ω for which the plate remains vertical ( β = 90°).

SOLUTION Moments and products of inertia. We use the axes Cx′y ′z shown. We note that I x′ and I y′ are half those for a circular plate, and so is the mass m. Thus, 1 2 mr 4 1 I y′ = mr 2 2 I x′ =

Because of symmetry, all products of inertia are equal to zero: I x′y′ = I y′z = I zx′ = 0

Angular momentum about C. H C = I x′ ω x′ i ′ + I y′ω y ′ j′ 1 2 1 mx (−ω sin β ) i′ + m r 2 (ω cos β ) j′ 4 2 1 = mx 2ω (− sin β i′ + 2 cos β j′) 4 =

Since C is a fixed point, we can use Equation (18.28):  ) ΣM C = ( H C C x′ y ′ z + Ω × H C = 0 + ω j × H C

Or, since j = −i ′ sin β + j′ cos β :

1 ΣM C = ω (−i ′ sin β + j′ cos β ) × mx 2ω (− sin β i ′ + 2 cos β j′) 4 1 2 2 = mr ω (−2 sin β cos β k + cos β sin β k ) 4 1 ΣM C = − m r 2ω 2 sin β cos β k 4

(1)

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PROBLEM 18.86 (Continued)

ΣM C = − mgx′ cos β k

But

= − mg

Equating (1) and (2):

4r cos β k 3π

(2)

1 2 2 4 mgr mr ω sin β cos β = cos β 4 3π

ω 2 sin β = (a)

Let ω = 15 rad/s in Eq. (3):

(b)

Let β = 90° in Eq. (3):

sin β =

16 g 16 9.81 m/s 2 = 3π r 3π 0.12 m 138.78 = 0.61681 (15)2

ω 2 = 138.78 s −2

ω 2 sin β = 138.78 s −2 (3) β = 38.1°  ω = 11.78 rad/s 

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PROBLEM 18.87 A uniform semicircular plate of radius 120 mm is hinged at A and B to a clevis which rotates with a constant angular velocity ω about a vertical axis. Determine the value of ω for which the plate forms an angle β = 50° with the horizontal x axis.

SOLUTION Moments and products of inertia. We use the axes Cx′y ′z shown. We note that I x′ and I y′ are half those for a circular plate, and so is the mass m. Thus, 1 2 mr 4 1 I y′ = mr 2 2 Because of symmetry, all products of inertia are equal to zero: I x′ =

I x′y′ = I y′z = I zx′ = 0

Angular momentum about C. H C = I x′ω x′i ′ + I y ′ω y′ j′ 1 2 1 mr (−ω sin β )i ′ + mr 2 (ω cos β ) j′ 4 2 1 = mr 2ω (− sin β i ′ + 2 cos β j′) 4 =

Since C is a fixed point, we can use Equation (18.28):  ) Σ M C = (H C C x′y ′z + Ω × H C = 0 + ω j × HC

Or, since j = −i ′ sin β + j′ cos β :

1 ΣM C = ω (−i ′ sin β + j′ cos β ) × mr 2ω (− sin β i ′ + 2 cos β j′) 4 1 2 2 = mr ω (−2sin β cos β k + cos β sin β k ) 4 1 ΣM C = − mr 2ω 2 sin β cos β k 4

(1)

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PROBLEM 18.87 (Continued)

But

ΣM C = − mgx′ cos β k = − mg

Equating (1) and (2):

(2)

1 2 2 4 mgr mr ω sin β cos β = cos β 4 3π

ω 2 sin β = Let β = 50° in Equation (3):

4r cos β k 3π

ω2 =

16 g 16 9.81 m/s 2 = 3π r 3π 0.12 m 138.78 s −2 = 181.17 s −2 sin 50°

ω 2 sin β = 138.78 s −2 (3) ω = 13.46 rad/s 

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PROBLEM 18.88 The slender rod AB is attached by a clevis to arm BCD which rotates with a constant angular velocity ω about the centerline of its vertical portion CD. Determine the magnitude of the angular velocity ω.

SOLUTION Let

AB = L = 15 in. = 1.25 ft, and

BC = b = 5 in. = 0.41667 ft

1 mL2 12

Choose x, y, z axes as shown.

I x ≈ 0, I y = I z =

Angular velocity:

ω = ω sin 30°i + ω cos 30° j

Angular momentum of rod AB about its mass center G: H G = I xω x i + I yω y j + I zω z k = I yω cos 30° j

Let the reference frame Gxyz be rotating with angular velocity Ω = ω.  = (H  ) H G G Gxyz + Ω × Η G

= 0 + (ω sin 30°i + ω cos 30° j) × I yω cos 30° j = I yω 2 sin 30° cos 30°k =

3 2 2 mL ω k 48

L cos 30° + b = 0.95793 ft 2

Radius of circular path of Point G:

r=

Acceleration of the mass center:

a = rω 2

Equations of motion:

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PROBLEM 18.88 (Continued)

ΣM B = mg

L L   cos 30°k =  sin 30°  mrω 2 k + H G 2 2  

1 3 3 2 2 mgLk =  mLr + mL  ω k   4 48 4  1 3 3  2 g = r+ L ω  4 48  4 1  3 3 (32.2) =  (0.95793) + (1.25)  ω 2 4 48  4 

ω 2 = 48.994

ω = 7.00 rad/s 

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PROBLEM 18.89 The slender rod AB is attached by a clevis to arm BCD, which rotates with a constant angular velocity ω about the centerline of its vertical portion CD. Determine the magnitude of the angular velocity ω.

SOLUTION Let AB = L = 15 in. = 1.25 ft, and BC = b = 5 in. = 0.41667 ft Choose x, y, z axes as shown. Angular velocity:

I x ≈ 0, I y = I z =

1 mL2 12

ω = −ω sin 30°i + ω cos 30° j

Angular momentum of rod AB about its mass center G: H G = I xω x i + I yω y j + I zω z k = I yω cos 30° j

Let the reference frame Gxyz be rotating with angular velocity Ω = ω.  = (H  ) H G G Gxyz + Ω × Η G

= 0 + (−ω sin 30°i + ω cos 30° j) × I yω cos 30° j = − I yω 2 sin 30° cos 30°k = −

3 2 2 mL ω k 48

L cos 30° − b = 0.1246 ft 2

Radius of circular path of Point G:

r=

Acceleration of the mass center:

a = rω 2

Equations of motion:

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PROBLEM 18.89 (Continued)

ΣM B = − mg −

L L   cos 30°k = −  sin 30°  mrω 2k + H G 2 2  

1 3 3 2 2 mgLk = −  mLr + mL  ω k   4 48 4  1 3 3  2 g = r+ L ω  4 48  4 1  3 3 (32.3) =  (0.1246) + (1.25)  ω 2 4 48  4 

ω 2 = 182.85

ω = 13.52 rad/s 

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PROBLEM 18.90 The 950-g gear A is constrained to roll on the fixed gear B, but is free to rotate about axle AD. Axle AD, of length 400 mm and negligible mass, is connected by a clevis to the vertical shaft DE, which rotates as shown with a constant angular velocity ω1. Assuming that gear A can be approximated by a thin disk of radius 80 mm, determine the largest allowable value of ω1 if gear A is not to lose contact with gear B.

SOLUTION

β = 30° L = 400 mm = 0.4 m r = 80 mm = 0.08 m

Choose principal axes x, y, z as shown. Kinematics:

ω1 = ω1 sin β i + ω1 cos β j ω 2 = ω2 j ω = ω1 + ω2 = ω1 sin β i + (ω1 cos β + ω2 ) j

ω x = ω1 sin β ω y = ω1 cos β + ω2 vG = ω1 × rG/D = ω1 j × rG/D = −ω1 L sin β k

aG = ω1 × v G = ω12 L sin β vC = vG + ω × rC/G 0 = −ω1 L sin β k + (ω x i + ω y j) × (− r i ) = −ω1 L sin β k + ω y r k

ω y = ω1 Angular momentum:

L sin β r

H G = I xω x i + I y ω y j

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PROBLEM 18.90 (Continued) Let the reference frame Dxyz be rotating with angular velocity Ω = ω1.  = (H  ) H G G Gxyz + Ω × Η G

= 0 + (ω1 sin β i + ω1 cos β j) × ( I xω x i + I yω y j) = ( I yω yω1 sin β − I xω xω1 cos β )k 1 1  =  mr 2ω y sin β − mr 2ω x cos β  ω1k 4 2 

Moments about D: where

M D = Fd k − mgL sin β k d = ( DC ) = L2 + r 2  + r × ma (M D )eff = H G G/D G  + mω 2 L2 sin β cos β k =H 1 G

Equating M D = (M D )eff and taking the z component, 1 1  FC d − mgL sin β =  mr 2 L sin 2 β − mr 2 sin β cos β  ω12 + mω12 L2 sin β cos β r 4 2  1 1  = mω12 sin β  rL sin β − r 2 cos β − L2 cos β  2 4   gL 1 1 L2 cos β + r 2 cos β − rL sin β 4 2 (9.81)(0.4) = (0.4) 2 cos 30° + 14 (0.08)2 cos 30° − 12 (0.08)(0.4) sin 30°

Set FC = 0 and solve for ω12 : ω12 =

ω2 = 5.45 rad/s 

= 29.739

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PROBLEM 18.91 Determine the force F exerted by gear B on gear A of Problem 18.90 when shaft DE rotates with the constant angular velocity ω1 = 4 rad/s. (Hint: The force F must be perpendicular to the line drawn from D to C.)

SOLUTION

β = 30° L = 400 mm = 0.4 m r = 80 mm = 0.08 m

Choose principal axes x, y, z as shown. Kinematics:

ω1 = ω1 sin β i + ω1 cos β j ω 2 = ω2 j ω = ω1 + ω2 = ω1 sin β i + (ω1 cos β + ω2 ) j

ω x = ω1 sin β ω y = ω1 cos β + ω2 vG = ω1 × rG/D = ω1 j × rG/D = −ω1 L sin β k

aG = ω1 × v G = ω12 L sin β vC = vG + ω × rC/G 0 = −ω1 L sin β k + (ω x i + ω y j) × (− r i) = −ω1 L sin β k + ω y r k

ω y = ω1 Angular momentum:

L sin β r

H G = I xω x i + I y ω y j

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PROBLEM 18.91 (Continued)

Let the reference frame Dxyz be rotating with angular velocity Ω = ω1.  = (H  ) H G G Gxyz + Ω × Η G = 0 + (ω1 sin β i + ω1 cos β j) × ( I xω x i + I yω y j) = ( I yω yω1 sin β − I xω xω1 cos β )k 1 1  =  mr 2ω y sin β − mr 2ω x cos β  ω1k 4 2 

M D = Fd k − mgL sin β k

Moments about D:

d = ( DC ) = L2 + r 2  + r × ma = H  + mω 2 L2 sin β cos β k (M D )eff = H G G/ D G G 1

where

Equating M D = (M D )eff and taking the z component, 1 1  FC d − mgL sin β =  mr 2 L sin 2 β − mr 2 sin β cos β  ω12 + mω12 L2 sin β cos β r 4 2  1 1  = mω12 sin β  rL sin β − r 2 cos β − L2 cos β  4 2  FC =

Solving for FC ,

m sin β  1 2 1  2 2  gL − ω1  L cos β + r cos β − rL sin β   d  4 2  

m = 950 g = 0.95 kg, ω1 = 4 rad/s

Additional data:

d = (0.4) 2 + (0.08) 2 = 0.40792 m FC =

(0.95) sin 30°  1 1  2 2 2 (9.81)(0.4) − (4) (0.4) cos 30° + (0.08) cos 30° − (0.4)(0.08) sin 30°  0.40792  4 2  

= 2.11 N

α = β − tan −1

r  4  = 30° − tan −1   L  20 

α = 18.7°

2.11 N

18.7° 

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PROBLEM 18.92 The essential structure of a certain type of aircraft turn indicator is shown. Springs AC and BD are initially stretched and exert equal vertical forces at A and B when the airplane is traveling in a straight path. Each spring has a constant of 600 N/m and the uniform disk has a mass of 250 g and spins at the rate of 12,000 rpm. Determine the angle through which the yoke will rotate when the pilot executes a horizontal turn of 800-m radius to the right at a speed of 720 km/h. Indicate whether point A will move up or down.

SOLUTION Aircraft speed:

v = 720 km/h = 200 m/s

Radius of turn:

ρ = 800 m ω x = 12000 rpm = 1256.6 rad/s

Angular velocity:

ω y is negative since the aircraft is turning to the right. ωy = −

v

ρ

=−

200 m/s = −0.25 rad/s 800 m

ωz = 0 H G = I xω x i + I y ω y j

Angular momentum:

where the reference frame Gxyz is turning with the aircraft with an angular velocity Ω = ωy j

Rate of change of angular momentum:  ) Since ω x and ω y are constant, (H G Gxyz = 0 and Eq. (18.22) yields

 = (H  ) H G G Gxyz + Ω × H G = 0 + ω y j × ( I xω x i + I y ω y j) 1  = − I xω xω y k = −  mr 2  ω xω y k 2  1 = − (0.25 kg)(0.05 m)2 (1256.6 rad/s)(−0.250 rad/s)k 2  H G = + (98.172 × 10−3 N ⋅ m)k

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PROBLEM 18.92 (Continued)

Free Body and kinetic diagrams

 . They must therefore be The forces A and B exerted by the springs must be equivalent to the couple H G directed as shown, which means that the spring at A will stretch and A will move up. 

We have A = B,

(0.12 m) A = 98.172 × 10−3 N ⋅ m F = A = 0.81810 N

Deflection of spring = x = Angle of rotation =

F 0.81810 N = = 1.3635 × 10−3 m k 600 N/m

x 1.3635 × 10−3 m = = 0.022725 rad = 1.30° GA 0.06 m



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PROBLEM 18.93 The 10-oz disk shown spins at the rate ω1 = 750 rpm, while axle AB rotates as shown with an angular velocity ω 2 of 6 rad/s. Determine the dynamic reactions at A and B.

SOLUTION ω = ω2 i − ω1k

Angular velocity:

H C = I xω x i + I yω y j + I z ω z k = I xω2 i − I zω1k

Angular momentum:

Let the reference frame Cxyz be rotating with angular velocity Ω = ω2 i.  = (H  ) H C C Cxyz + Ω × H C = 0 + ω2 i × ( I xω2 i − I z ω1k ) = I z ω2ω1 j

Acceleration of mass center:

a=0 ΣF = m a Ak − Bk = 0 A= B  M =H C

C

LBj = I zω2ω1 j B =

Data:

I zω2ω1 L

W 10 = = 0.01941 lb ⋅ s 2 /ft r = 2 in. = 0.16667 ft g (16)(32.2) 1 1 I Z = mr 2 = (0.01941)(0.16667) 2 = 269.6 × 10−6 lb ⋅ s 2 ⋅ ft 2 2 2π (750) = 25π rad/s, ω2 = 6 rad/s, L = 8 in. = 0.66667 ft ω1 = 60 m=

A=B=

(269.6 × 10−6 )(6)(25π ) = 0.1906 lb 0.66667

A = (0.1906 lb)k  B = −(0.1906 lb)k 



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PROBLEM 18.94 The 10-oz disk shown spins at the rate ω1 = 750 rpm, while axle AB rotates as shown with an angular velocity ω 2. Determine the maximum allowable magnitude of ω2 if the dynamic reactions at A and B are not to exceed 0.25 lb each.

SOLUTION ω = ω2 i − ω1k

Angular velocity:

H C = I xω x i + I yω y j + I z ω z k = I xω2 i − I zω1k

Angular momentum:

Let the reference frame Cxyz be rotating with angular velocity Ω = ω2 i.  = (H  ) H C C Cxyz + Ω × H C = 0 + ω2 i × ( I xω2 i − I z ω1k ) = I z ω2ω1 j

Acceleration of mass center: a=0

ΣF = m a Ak − Bk = 0 A= B  M =H C

C

LBj = I zω2ω1 j B =

Data:

I zω2ω1 L

W 10 = = 0.01941 lb ⋅ s 2 /ft r = 2 in. = 0.16667 ft g (16)(32.2) 1 1 I z = mr 2 = (0.01941)(0.16667) 2 = 269.6 × 10−6 lb ⋅ s 2 ⋅ ft 2 2 2π (750) = 25π rad/s, L = 8 in. = 0.66667 ft ω1 = 60 m=

A = B = 0.25 lb



ω2 =

LB (0.66667)(0.25)  = I z ω1 (269.6 × 10−6 )(25π )

ω2 = 7.87 rad/s 

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PROBLEM 18.95 Two disks, each of mass 5 kg and radius 100 mm, spin as shown at the rate ω1 = 1500 rpm about a rod AB of negligible mass which rotates about a vertical axis at the rate ω2 = 45 rpm. (a) Determine the dynamic reactions at C and D. (b) Solve part a, assuming that the direction of spin of disk B is reversed.

SOLUTION Angular momentum of each disk about its mass center. 1 1 H G = I xω x i + I yω y j = − mr 2ω1i + mr 2ω2 j 2 4 1 H G = mr 2 (−2ω1i + ω2 j) 4

(1)

Eq. (18.22): 1 2  = (H  ) H G G Gxyz + Ω × H G = 0 + ω2 j × mr (−2ω1i + ω2 j) 4  = + 1 mr 2ω ω k H B 1 2 2

(2)

Equations of motion.

 = mr 2ω ω k. Since m a A and m aB cancel out, effective forces reduce to couple 2H G 1 2

It follows that the reactions form an equivalent couple with  mr 2ω1ω2 −C = D =   0.3 m 

  i 

(3)

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PROBLEM 18.95 (Continued)

(a)

With m = 5 kg, r = 0.1 m, ω1 = 1500 rpm = 50π rad/s, and ω2 = 45 rpm = 1.5π rad/s, Eq. (3) yields  1.5π rad/s  C = D = (5 kg)(0.1 m) 2 (50π rad/s)   = 123.37 N  0.3 m  C = −(123.4 N)i; D = (123.4 N)i 

(b)

With direction of spin of B reversed, its angular momentum will also be reversed and the effective forces (and thus, the applied forces) reduce to zero: C = D = 0 

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PROBLEM 18.96 Two disks, each of mass 5 kg and radius 100 mm, spin as shown at the rate ω1 = 1500 rpm about a rod AB of negligible mass which rotates about a vertical axis at a rate ω2 . Determine the maximum allowable value of ω2 if the dynamic reactions at C and D are not to exceed 250 N each.

SOLUTION Angular momentum of each disk about its mass center. 1 1 H G = I xω x i + I yω y j = − mr 2ω1i + mr 2ω2 j 2 4 1 H G = mr 2 (−2ω1i + ω2 j) 4

(1)

Eq. (18.22): 1 2  = (H  ) H G G Gxyz + Ω × H G = 0 + ω2 j × mr (−2ω1i + ω2 j) 4  = + 1 mr 2ω ω k H B 1 2 2

(2)

Equations of motion.

 = mr 2ω ω k. It follows that Since m a A and m aB cancel out, effective forces reduce to couple 2H G 1 2 the reactions form an equivalent couple with  mr 2ω1ω2 −C = D =   0.3 m 

  i 

(3)

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PROBLEM 18.96 (Continued)

Making C = D − 250 N in Eq. (3) yields mr 2ω1ω2 = 250 N 0.3 m

With m = 5 kg, r = 0.1 m, ω1 = 1500 rpm = 50π rad/s We have

ω2 =

(250 N)(0.3 m) = 9.5493 rad/s (5 kg)(0.1 m) 2 (50π rad/s)

ω2 = 91.2 rpm 

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PROBLEM 18.97 A stationary horizontal plate is attached to the ceiling by means of a fixed vertical tube. A wheel of radius a and mass m is mounted on a light axle AC which is attached by means of a clevis at A to a rod AB fitted inside the vertical tube. The rod AB is made to rotate with a constant angular velocity Ω causing the wheel to roll on the lower face of the stationary plate. Determine the minimum angular velocity Ω for which contact is maintained between the wheel and the plate. Consider the particular cases (a) when the mass of the wheel is concentrated in the rim, (b) when the wheel is equivalent to a thin disk of radius a.

SOLUTION Angular momentum H G of wheel:

ω y = Ω, ω x =

R Ω, ω z = 0 a

H G = I xω x i + I yω y j + I zω z k HG = I x

R Ωi + I y Ωj a

Rate of change of angular momentum: Since ω x and ω y are constant, and observing that the frame Gxyz rotates with the angular velocity Ω = Ωi:  ) (H G Gxyz = 0

Eq. (18.22):

 = (H  ) H G G Gxyz + Ω × H G  = 0 + Ωj ×  I R ai + I Ωj  H G y  x a     = − I R Ω 2k H G z a

(1)

The free body and kinetic diagrams

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PROBLEM 18.97 (Continued)

Equating moments about A:  R i × (−W j) = H G R 2 Ωk a mga Ix

− Rmg k = − I z Ω=

(a)

mass in rim: I x = ma 2

(b)

thin disk: I x =

Ω = g/a 

1 2 ma 2

Ω = 2 g /a 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2117

PROBLEM 18.98 Assuming that the wheel of Problem 18.97 weights 8 lb, has a radius a = 4 in. and a radius of gyration of 3 in., and that R = 20 in., determine the force exerted by the plate on the wheel when Ω = 25 rad/s. PROBLEM 18.97 A stationary horizontal plate is attached to the ceiling by means of a fixed vertical tube. A wheel of radius a and mass m is mounted on a light axle AC which is attached by means of a clevis at A to a rod AB fitted inside the vertical tube. The rod AB is made to rotate with a constant angular velocity Ω causing the wheel to roll on the lower face of the stationary plate. Determine the minimum angular velocity Ω for which contact is maintained between the wheel and the plate. Consider the particular cases (a) when the mass of the wheel is concentrated in the rim, (b) when the wheel is equivalent to a thin disk of radius a.

SOLUTION Angular momentum H G of wheel: ω y = Ω, ω x =

R Ω, ω z = 0 a

H G = I xω x i + I yω y j + I zω z k HG = I x

R Ωi + I y Ωj a

Rate of change of angular momentum: Since ω x and ω y are constant, and observing that the frame Gxyz rotates with the angular velocity Ω = Ωj:  ) (H G Gxyz = 0

Eq. (18.22):

 = (H  ) H G G Gxyz + Ω × H G  = 0 + Ωj ×  I R ai + I Ωj  H G y  x a     = − I R Ω 2k H G z a

(1)

With

R = 20 in., a = 4 in., Ω = 25 rad/s, m =

and

I x = mk x2 =

8 g

2

8 3  −3 2  ft  = 15.53 × 10 lb ⋅ ft ⋅ s g  12 

 = −15.53 × 10−3  20  (25) 2 k = 48.52k H G    4 

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PROBLEM 18.98 (Continued)

Taking moments about A:  Ri × (− Dj − W j) = H G 20 i × (− Dj − 8 j) = −48.52k 12 5 − ( D + 8)k = −48.52k 3 3 D = (48.52) − 8 = 21.1 lb 5

D = 21.1 lb 

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PROBLEM 18.99 A thin disk of mass m = 4 kg rotates with an angular velocity ω2 with respect to arm ABC, which itself rotates with an angular velocity ω1 about the y axis. Knowing that ω 1 = 5 rad/s and ω2 = 15 rad/s and that both are constant, determine the force-couple system representing the dynamic reaction at the support at A.

SOLUTION Angular velocity of the disk.

ω = ω1 j + ω2 k = (5 rad/s)j + (15 rad/s)k

Moments of inertia about principal axes passing through the mass center. I x′ = I y ′ =

1 2 mr 4

1 (4)(0.150 m) 2 = 0.0225 kg ⋅ m 2 4 1 I z′ = mr 2 = 0.045 kg ⋅ m 2 2 =

Angular momentum about mass center C. H C = I x′ω x′i + I y′ω y′ j + I z ′ω z ′k

= 0 + (0.0225)5 j + (0.045)15k H C = (0.1125 kg ⋅ m 2 /s)j + (0.6750 kg ⋅ m 2 /s)k

Rate of change of H C . Let the frame Axyz be turning with angular velocity Ω = ω1 j.  = (H  ) H C C Axyz + Ω × H C = 0 + Ω × H C = 5 j × (0.1125 j + 0.675k ) = (3.375 N ⋅ m)i

Position vector of Point C. rC /A = (0.450 m)i + (0.225 m)j

Velocity of Point C, the mass center of the disk. vC = ω1 × rC /A = 5 j × (0.45i + 0.225 j)

= −(2.25 m/s)k

Acceleration of Point C. aC = α1 j × rC /A + ω1 j × v C = 0 + 5 j × (−2.25k )= − (11.25 m/s 2 )i

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PROBLEM 18.99 (Continued)

maC = (4)(−11.25i ) = −(45 N)i

Free body and kinetic diagrams

Linear components: Moments about A.

A = maC

A = −(45 N)i 

 MA = rC /A × maC + H C MA = (0.450i + 0.225 j) × (−45i) + 3.375i MA = (3.38 N ⋅ m)i + (10.13 N ⋅ m)k 



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PROBLEM 18.100 An experimental Fresnel-lens solar-energy concentrator can rotate about the horizontal axis AB, which passes through its mass center G. It is supported at A and B by a steel framework, which can rotate about the vertical y axis. The concentrator has a mass of 30 Mg, a radius of gyration of 12 m about its axis of symmetry CD, and a radius of gyration of 10 m about any transverse axis through G. Knowing that the angular velocities ω1 and ω2 have constant magnitudes equal to 0.20 rad/s and 0.25 rad/s, respectively, determine for the position θ = 60° (a) the forces exerted on the concentrator at A and B, (b) the couple M 2 k applied to the concentrator at that instant.

SOLUTION Let the y axis be vertical and the y ′ axis be the symmetry axis. Let the z axis be directed along axle BA as shown and the x′ axis be the transverse axis perpendicular to BA. Unit vectors.

β = 90° − θ i ′ = i cos β − j sin β j′ = i sin β + j cos β i = i ′ cos β + j′ sin β j = −i ′ sin β + j cos β

Angular velocity.

ω = ω2 j + ω1k

ω = −(ω 2 sin β )i′ + (ω2 cos β ) j′ + ω 1k = −ω 2 cos θ i + ω 2 sin θ j + ω 1k ω x′ = ω 2 cos θ

ω y′ = ω 2 sinθ

ω z = ω1

θ = ω1 ω1 and ω 2 are constant. ω x′ = ω2ω 1 sin θ ω y′ = ω2ω1 cos θ , ω 2 = 0 Radii of gyration:

k x′ = 10 m k y′ = 12 m k z = 10 m

Moments of inertia:

I = mk 2

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PROBLEM 18.100 (Continued)

H G = I x′ω x′ i ′ + I yω y′ j + I zω z k

Angular momentum about Point G.

H G = m[(10) 2 ω x′i ′ + (12) 2 ω y′ j′ + (10) 2 ω z k ]

= m(100ω z cos θ i ′ + 144ω2 sin θ j′ + 100ω1k ) m = 30,000 kg

where m is the mass. Acceleration of the mass center.

Since the mass center lies at the center of the axle BA,

a=0

Rate of change of angular momentum. Let the reference frame Gx′y′z be turning with angular velocity Ω = ω  =H  H G Gx′y ′z + Ω × H G     H Gx ′y ′z = I x′ω x′ i ′ + I y ′ω y j′ + I z ω z k

where

= m[(10)2 ω 2ω1 sin θ i ′ + (12) 2 ω 2ω1 cos θ j′ + 0] = m[(100)(0.25)(0.20)sin 60°i ′ + (144)(0.25)(0.20) cos 60° j′] = m(4.3301i ′ + 3.6 j′)

and

i′ j′ k Ω × H G = m −ω 2 cos θ ω 2 sin θ ω1 −100ω 2 cos θ 144ω 2 sin θ 100ω1

= m[−44ω1ω2 sin θ i ′ − 44ω 22 sin θ cos θ k ] = m[−(44)(0.20)(0.25)sin 60°i ′ + (44)(0.25) 2 sin 60° cos 60°k ] = m(−1.9053i ′ + 1.1908k )  = m[2.4248i′ + 3.6 j′ − 1.1908k ] H G = (30, 000)[2.4248(i cos 30° − j sin 30°) + 3.6(i sin 30° + j cos 30°) − 1.1908k ] = (117 × 103 N ⋅ m)i + (57.158 × 103 N ⋅ m) j − (35.724 × 103 N ⋅ m)k W = − mgj = −(30 × 103 )(9.81) j

Weight:

= −(294.3 × 103 N) j

Equations of motion.  ΣM B = Σ(M B )eff : rA /B × ( Ax i + By j) + rG /B × W ) + M 2 k = H G  32k × ( Ax i + Ay j) + 16k × ( −294.3 × 103 ) j + M 2 k = H G  (−32 Ay + 4.7088 × 106 )i + 32 Ax j + M 2 k = H G

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PROBLEM 18.100 (Continued)

Equate like components: i : − 32 Ay + 4.7088 × 106 = 117 × 103

Ay = 143.5 × 103 N

j: 32 Ax = 57.158 × 103

Ax = 1.786 × 103 N

k : M 2 = −35.724 × 103

(a)

A = (1.786 kN)i + (143.5 kN) j 

Reaction at A. ΣF = ΣFeff : B + A + W = 0 B = − W − A = 294.3 × 103 j − A

B = −(1.786 kN)i + (150.8 kN) j 

Reaction at B. (b)

M 2k = −(35.7 kN ⋅ m)k 

Couple M 2k :

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PROBLEM 18.101 A 6-lb homogeneous disk of radius 3 in. spins as shown at the constant rate ω1 = 60 rad/s. The disk is supported by the forkended rod AB, which is welded to the vertical shaft CBD. The system is at rest when a couple M 0 = (0.25 ft ⋅ lb) j is applied to the shaft for 2 s and then removed. Determine the dynamic reactions at C and D after the couple has been removed.

SOLUTION Angular velocity of shaft CBD and arm AB:

Ω = ω2 j

Angular velocity of disk A:

ω = ω2 j + ω1k

H A = I xω x i + I yω y j + I zω z k = I yω2 j + I z ω1k

Its angular momentum about A:

Let the reference frame Bxyz be rotating with angular velocity Ω.  = (H  ) H A A Axyz + Ω × H A = I yω 2 j + I zω1k + ω2 j × ( I yω2 j + I zω1k ) = I zω1ω2 i + I yω 2 j + I zω1k 1 2 1 1 mr ω1ω2 i + mr 2ω 2 j + mr 2ω1k 2 4 2 Velocity and acceleration of the mass center A of the disk: =

v = ω2 j × ci = −cω2 k , a = ω 2 j × ci + ω2 j × v = −cω 2k − cω22 i

ΣF = ΣFeff = m a C x i + C y k + Dx i + Dz k = m a

Resolve into components. C x + Dx = − mcω22 C z + Dz = − mcω 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2125

PROBLEM 18.101 (Continued)

 =H  + r × ma ΣM D = H D A A/ D  + (ci + bj) × (−cω k − cω 2 i ) =H A 2 2 2  M 0 j + 2bj × (C x i + C z k ) = H A − mbcω 2 i + mc ω 2 j + mbcω22 k 1  1  1  2bC z i + M 0 j − 2bC x k = m  r 2ω1ω2 − bcω 2  i + m  r 2 + c 2  ω 2 j + m  r 2ω1 + bcω22  k 2 4 2       1  j: M 0 = m  r 2 + c 2  ω 2 4  m 1 2 m 1   r ω1 + bcω22  Dx = −  − r 2ω1 + bcω22   2b  2 2b  2  

(2)

m 1 2 m 1   r ω1ω2 − bcω 2  Dz = −  r 2ω1ω2 + bcω 2   2b  2 2 b 2   

(3)

k : Cx = − i:

Data:

Cz =

(1)

W = 6 lb, r = 3 in. = 0.25 ft, b = 4 in. = 0.33333 ft, c = 5 in. = 0.41667 ft, ω1 = 60 rad/s, ω1 = 0

While the couple is applied,

M 0 = 0.25 ft ⋅ lb

Rearranging Equation (1)

ω 2 =

m

(

M0 1 4

r +c 2

2

=

) ( ) 6 32.2

0.25  14 

(0.25)2 + (0.41667)2 

At t = 2s,

ω2 = (ω2 )0 + ω 2 t = 0 + (7.0899)(2) = 14.18 rad/s

For t > 2s,

M 0 = 0, ω 2 = 0

From Equations (2), (3)

Cx = Dx =

6 32.2

(2)(0.33333) 6 32.2

(2)(0.33333)

= 7.0899 rad/s 2

[0 + (0.33333)(0.41667)(14.1798) 2 ] = −7.8054 lb [0 + (0.33333)(0.41667)(14.1798) 2 ] = −7.8054 lb

Cz =

1  (0.25) 2 (60)(14.1798) − 0  = 7.4312 lb  (2)(0.33333)  2 

Dz =

 1  − (0.25) 2 (60)(14.1798) − 0  = −7.4312 lb (2)(0.33333)  2 

6 32.2

6 32.2

C = −(7.81 lb)i + (7.43 lb)k  D = −(7.81 lb)i − (7.43 lb)k 

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PROBLEM 18.102 A 6-lb homogeneous disk of radius 3 in. spins as shown at the constant rate ω1 = 60 rad/s. The disk is supported by the forkended rod AB, which is welded to the vertical shaft CBD. The system is at rest when a couple M 0 is applied as shown to the shaft for 3 s and then removed. Knowing that the maximum angular velocity reached by the shaft is 18 rad/s, determine (a) the couple M 0 , (b) the dynamic reactions at C and D after the couple has been removed.

SOLUTION Angular velocity of shaft CBD and arm AB:

Ω = ω2 j

Angular velocity of disk A:

ω = ω2 j + ω1k H A = I xω x i + I yω y j + I zω z k = I yω2 j + I z ω1k

Its angular momentum about A:

Let the reference frame Bxyz be rotating with angular velocity Ω.  = (H  ) H A A Axyz + Ω × H A

= I yω 2 j + I zω1k + ω2 j × ( I yω2 j + I zω1k ) = I zω1ω2 i + I yω 2 j + I zω1k 1 2 1 1 mr ω1ω2 i + mr 2ω 2 j + mr 2ω1k 2 4 2 Velocity and acceleration of the mass center A of the disk: =

v = ω2 j × ci = −cω2 k , a = ω 2 j × ci + ω2 j × v = −cω 2k − cω22 i

ΣF = ΣFeff = m a C x i + C y k + Dx i + Dz k = m a

Resolve into components. C x + Dx = − mcω22 C z + Dz = − mcω 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2127

PROBLEM 18.102 (Continued)

 =H  + r × ma ΣM D = H D A A/ D  + (ci + bj) × (−cω k − cω 2 i ) =H A 2 2 2  M 0 j + 2bj × (C x i + C z k ) = H A − mbcω 2 i + mc ω 2 j + mbcω22 k

1  1  1  2bC z i + M 0 j − 2bC x k = m  r 2ω1ω2 − bcω 2  i + m  r 2 + c 2  ω 2 j + m  r 2ω1 + bcω22  k 2 4 2       1  j: M 0 = m  r 2 + c 2  ω 2 4  m 1 2 m 1   r ω1 + bcω22  Dx = −  − r 2ω1 + bcω22   2b  2 2b  2  

(2)

m 1 2 m 1   r ω1ω2 − bcω 2  Dz = −  r 2ω1ω2 + bcω 2   2b  2 2 b 2   

(3)

k : Cx = − i:

W = 6 lb, r = 3 in. = 0.25 ft, b = 4 in. = 0.33333 ft, c = 5 in. = 0.41667 ft, ω1 = 60 rad/s, ω1 = 0

Data:

(a)

Cz =

While the couple is applied ω 2 = From Equation (1)

ω2 t

=

18 = 6 rad/s 2 3

1  M 0 = m  r 2 + c 2  ω 2 4    6 1 2 2 =   4 (0.25) + (0.41667)  (6) 32.2   

(b)

(1)

For t > 3 s,

From Equations (2), (3)

M 0 = (0.212 ft ⋅ lb) j 

M 0 = 0, ω 2 = 0 Cx = Dx =

6 32.2

(2)(0.33333) 6 32.2

(2)(0.33333)

[0 + (0.33333)(0.41667)(18) 2 ] = −12.578 lb [0 + (0.33333)(0.41667)(18) 2 ] = −12.578 lb

Cz =

1  (0.25) 2 (60)(18) − 0  = 9.4332 lb  (2)(0.33333)  2 

Dz =

 1  − (0.25) 2 (60)(18) − 0  = −9.4332 lb  (2)(0.33333)  2 

6 32.2

6 32.2

C = −(12.58 lb)i + (9.43 lb)k  D = −(12.58 lb)i − (9.43 lb)k 

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PROBLEM 18.103 A 2.5 kg homogeneous disk of radius 80 mm rotates with an angular velocity ω 1 with respect to arm ABC, which is welded to a shaft DCE rotating as shown at the constant rate ω 2 = 12 rad/s. Friction in the bearing at A causes ω1 to decrease at the rate of 15 rad/s 2. Determine the dynamic reactions at D and E at a time when ω1 has decreased to 50 rad/s.

SOLUTION Angular velocity of shaft DCE and arm CBA:

Ω = ω2 k

Angular velocity of disk A:

ω = ω1 j + ω2 k

Its angular momentum about A:

H A = I xω x i + I yω y j + I zω z k = I yω1 j + I z ω2 k

Let the reference frame Cxyz be rotating with angular velocity Ω.  = (H  ) H A A Axyz + Ω × H A = I yω1 j + I zω 2 k + ω2 k × ( I yω1 j + I zω2k ) = − I yω2ω1i + I yω1 j + I zω 2 k 1 1 1 = − mr 2ω2ω1i + mr 2ω1 j + mr 2ω 2 k 2 2 4

Velocity and acceleration of the mass center A of the disk: v A = ω2 k × rA/C = ω2 k × (bi − cj) = cω2 i + bω2 j a A = ω 2 k × rA/C + ω2k × vA = (cω 2 − bω22 )i + (bω 2 + cω22 ) j

ΣF = ΣFeff Dx i + D y j + E x i + E y j = ma A

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2129

PROBLEM 18.103 (Continued)

Resolve into components. Dx + Ex = m(cω 2 − bω22 ) D y + E y = m(bω 2 + cω22 )  =H  + r × ma = H  + (bi − cj + lk ) × ma ΣM E = H E A A/E A A A  + m(blω − clω 2 )i + m(clω + blω 2 ) j + m(b 2 + c 2 )ω k 2lk × ( Dx i + Dy j) + M 0 k = H 2 2 2 2 2 A

 1  −2lDy i + 2lDx j + M 0 k = m  − r 2ω2ω1 + blω 2 − clω22  i  2  1  1  + m  r 2ω1 + clω 2 − blω22  j + m  r 2 + b 2 + c 2  ω 2 k 2 4     k: j: i:

Data:

1  M 0 = m  r 2 + b 2 + c 2  ω 2 4  m1 m 1   Dx =  r 2ω1 + clω 2 − blω22  Ex =  − r 2ω1 + clω 2 − blω22  2l  2 2l  2   Dy =

m1 2 m 1 2 2 2  r ω2ω1 + blω 2 + clω2  E y =  − r ω2ω1 + blω 2 + clω2  2l  2 2l  2  

m = 2.5 kg, r = 80 mm = 0.08 m b = 120 mm = 0.12 m, c = 60 mm = 0.06 m, l = 150 mm = 0.15 m

ω1 = 50 rad/s, ω1 = −15 rad/s 2 , ω2 = 12 rad/s, ω 2 = 0 Dx =

2.5  1  (0.08) 2 (−15) + 0 − (0.12)(0.15)(12) 2  = −22.0 N  (2)(0.15)  2 

Dy =

2.5  1  (0.08) 2 (12)(50) + 0 + (0.06)(0.15)(12) 2  = 26.8 N (2)(0.15)  2 

D = − (22.0 N)i + (26.8 N) j  Ex =

2.5  1  − (0.08) 2 (−15) + 0 − (0.12)(0.15)(12) 2  = − 21.2 N  (2)(0.15)  2 

Ey =

2.5  1  − (0.08) 2 (12)(50) + 0 + (0.06)(0.15)(12)2  = −5.20 N  (2)(0.15)  2 

E = − (21.2 N)i − (5.20 N) j 

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PROBLEM 18.104 A 2.5-kg homogeneous disk of radius 80 mm rotates at the constant rate ω 1 = 50 rad/s with respect to arm ABC, which is welded to a shaft DCE. Knowing that at the instant shown shaft DCE has an angular velocity ω 2 = (12 rad/s)k and an angular acceleration α 2 = (8 rad/s 2 )k , determine (a) the couple which must be applied to shaft DCE to produce that acceleration, (b) the corresponding dynamic reactions at D and E.

SOLUTION Angular velocity of shaft DCE and arm CBA:

Ω = ω2 k

Angular velocity of disk A:

ω = ω1 j + ω2 k

Its angular momentum about A:

H A = I xω x i + I yω y j + I zω z k = I yω1 j + I z ω2 k

Let the reference frame Cxyz be rotating with angular velocity Ω.  = (H  ) H A A Axyz + Ω × H A = I yω1 j + I zω 2 k + ω2 k × ( I yω1 j + I zω2k ) = − I yω2ω1i + I yω1 j + I zω 2 k 1 1 1 = − mr 2ω2ω1i + mr 2ω1 j + mr 2ω 2 k 2 2 4

Velocity and acceleration of the mass center A of the disk: v A = ω2 k × rA/C = ω2 k × (bi − cj) = cω2 i + bω2 j a A = ω 2 k × rA/C + ω2k × vA = (cω 2 − bω22 )i + (bω 2 + cω22 ) j

ΣF = ΣFeff Dx i + D y j + Ex i + E y j = ma A

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PROBLEM 18.104 (Continued)

Resolve into components. Dx + Ex = m(cω 2 − bω22 ) D y + E y = m(bω 2 + cω22 )  =H  + r × ma = H  + (bi − cj + lk ) × ma ΣM E = H E A A/E A A A  + m(blω − clω 2 )i + m(clω + blω 2 ) j + m(b 2 + c 2 )ω k 2lk × ( Dx i + Dy j) + M 0 k = H A 2 2 2 2 2

 1  −2lDy i + 2lDx j + M 0 k = m  − r 2ω2ω1 + blω 2 − clω22  i  2  1  1  + m  r 2ω1 + clω 2 − blω22  j + m  r 2 + b 2 + c 2  ω 2 k 2 4     k: j: i:

Data:

1  M 0 = m  r 2 + b 2 + c 2  ω 2 4  m1 m 1   Dx =  r 2ω1 + clω 2 − blω22  E x =  − r 2ω1 + clω 2 − blω22  2l  2 2l  2   Dy =

m1 2 m 1 2 2 2  r ω2ω1 + blω 2 + clω2  E y =  − r ω2ω1 + blω 2 + clω2  2l  2 2l  2  

m = 2.5 kg, r = 80 mm = 0.08 m b = 120 mm = 0.12 m, c = 60 mm = 0.06 m, l = 150 mm = 0.15 m

ω1 = 50 rad/s, ω1 = 0, ω2 = 12 rad/s, ω 2 = α 2 = 8 rad/s 2 (a)

1  M 0 = (2.5)  (0.08)2 + (0.12) 2 + (0.06) 2  (8) 4  

(b)

Dx =

2.5 [0 + (0.06)(0.15)(8) − (0.12)(0.15)(12)2 ] = −21.0 N (2)(0.15)

Dy =

2.5  1  (0.08) 2 (12)(50) + (0.12)(0.15)(8) + (0.06)(0.15)(12)2  = 28.0 N (2)(0.15)  2 

M 0 = (0.392 N ⋅ m)k 

D = − (21.0 N)i + (28.0 N) j 

Ex =

2.5 [−0 + (0.06)(0.15)(8) − (0.12)(0.15)(12)2 ] = − 21.0 N (2)(0.15)

2.5  1  − (0.08) 2 (12)(50) + (0.12)(0.15)(8) + (0.06)(0.15)(12) 2   (2)(0.15)  2  = −4.00 N

Ey =

E = − (21.0 N)i − (4.00 N) j 

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PROBLEM 18.105 For the disk of Problem 18.99, determine (a) the couple M1j which should be applied to arm ABC to give it an angular acceleration α1 = −(7.5 rad/s 2 ) j when ω1 = 5 rad/s, knowing that the disk rotates at the constant rate ω2 = 15 rad/s, (b) the force-couple system representing the dynamic reaction at A at that instant. Assume that ABC has a negligible mass. PROBLEM 18.99 A thin disk of mass m = 4 kg rotates with an angular velocity ω2 with respect to arm ABC, which itself rotates with an angular velocity ω1 about the y axis. Knowing that ω 1 = 5 rad/s and ω2 = 15 rad/s and that both are constant, determine the force-couple system representing the dynamic reaction at the support at A.

SOLUTION Angular velocity of the disk.

ω = ω1 j + ω2 k = (5 rad/s) j + (15 rad/s)k

Moments of inertia about principal axes passing through the mass center. Ix = Iy =

1 2 mr 4

1 (4)(0.150 m) 2 = 0.0225 kg ⋅ m 2 4 1 I z = mr 2 = 0.045 kg ⋅ m 2 2 =

Angular momentum about mass center C. H C = I x′ω x′ i + I y′ω y′ j + I z ′ω z ′k = 0 + (0.0225)5 j + (0.045)15k = (0.1125 kg ⋅ m 2 /s) j + (0.6750 kg ⋅ m 2 /s)k

Rate of change of H C . Let the frame Axyz be turning with angular velocity Ω = ω1 j.  = (H  )  x i + I yω y j + I z ω z k + Ω × H C H C C Axyz + Ω × H C = I xω

= 0 + (0.0225)(−7.5) j + 0 + 5 j × (0.1125 j + 0.675k ) = −(0.16875 N ⋅ m) j + (3.375 N ⋅ m)i

Position vector of Point C.

rC/ A = (0.450 m)i + (0.225 m) j

Velocity of Point C, the mass center of the disk. vC = ω1 × rC/ A = 5 j × (0.45i + 0.225 j) = −(2.25 m/s)k

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PROBLEM 18.105 (Continued)

Acceleration of Point C. aC = α1 j × rC/ A + ω1 j × v C = ( −7.5 j) × (0.45i + 0.225 j) + 5 j × (−2.25k ) = (3.3750 m/s 2 )k − (11.25 m/s 2 )i maC = (4)(−11.25i + 3.3750k ) = −(45 N)i + (13.5 N)k

Free body and kinetic diagrams

Linear components: Moments about A.

A = maC

A = −(45 N)i + (13.5 N)k

 M A = rC/ A × maC + H C

M A = (0.450i + 0.225 j) × ( −45i + 13.5k ) − 0.16875 j + 3.375i = −6.0750 j + 10.125k + 3.0375i − 0.16875 j + 3.375i = 6.4125i − 6.2438 j + 10.125k

(a)

Required couple.

(b)

Dynamic reaction.

M1 j = −(6.24 N ⋅ m) j  A = −(45.0 N)i + (13.50 N)k  M A = (6.41 N ⋅ m)i + (10.13 N ⋅ m)k 

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PROBLEM 18.106* A slender homogeneous rod AB of mass m and length L is made to rotate at the constant rate ω2 about the horizontal z axis, while frame CD is made to rotate at the constant rate ω1 about the y axis. Express as a function of the angle θ (a) the couple M1 required to maintain the rotation of the frame, (b) the couple M2 required to maintain the rotation of the rod, (c) the dynamic reactions at the supports C and D.

SOLUTION Angular momentum H G : We resolve the angular velocity ω = ω1 j + ω2 k into components along the principal axes Gx′y ′z :

ω x′ = − ω1 cos θ ω y′ = ω1 sin θ ω z = ω2 Moments of inertia: I x′ = I z =

We have

1 mL2 12

H x′ = I x′ ω x′ = −

I y′ = 0

1 mL2 ω1 cos θ 12

H y′ = I y′ ω y′ = 0 H z = I z ωz =

1 mL2 ω2 12

Computing the components of H G along the x, y, z axes: 1 1 mL2 ω1 cos θ sin θ = − mL2 ω1 sin 2θ 12 24 1 H y = − H x′ cos θ = + mL2 ω1 cos 2 θ 12 1 H z = mL2 ω2 12 H x = H x′ sin θ = −

The angular momentum is therefore 1 1 1 mL2ω1 sin 2θ i + mL2ω1 cos 2 θ j + mL2ω2 k 24 12 12 where the reference frame Gxyz rotates with the angular velocity HG = −

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PROBLEM 18.106* (Continued) Rate of change of H G . We note that ω1 and ω2 are constant, while θ varies with t, with θ = ω2 .

Eq. (18.22) yields 1 1  = (H  ) H mL2ω1 (2 cos 2θθ)i + mL2ω1 ( −2 cos θ sin θθ) j G G Gxyz + Ω × H G = − 24 12 1  1  + ω1 j ×  − mL2ω1 sin 2θ i + mL2ω2 k  12  24  1 1 = − mL2ω1ω2 cos 2θ i − mL2ω1ω2 sin 2θ j 12 12 1 1 + mL2ω12 sin 2θ k + mL2ω1ω2 i 24 12 1  = mL2ω ω (1 − cos 2θ )i − 1 mL2ω ω sin 2θ j H G 1 2 1 2 12 12 1 mL2ω12 sin 2θ k + (1) 24

Equivalence of external and effective forces.

Equating the moments of the variable couples:  Lj × C k + M 1 j + M 2 k = H G LCi + M 1 j + M 2 k =

1 1 mL2ω1ω2 (1 − cos 2θ )i − mL2ω1ω2 sin 2θ j 12 12 1 2 2 + mL ω1 sin 2θ k 24

Equating the coefficients of the unit vectors: j: M 1 = −

(a)

1 mL2ω1ω2 sin 2θ 12 M1 = −

Couple M1: k: M2 =

1 mL2ω1ω2 sin 2θ j  12

1 mL2ω12 sin 2θ 24

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PROBLEM 18.106* (Continued)

(b)

M2 =

Couple M 2 : i : LC = C=

1 mL2ω12 sin 2θ k  24

1 mL2ω1ω2 (1 − cos 2θ ) 12 1 1 mLω1ω2 (1 − cos 2θ ) = mLω1ω2 sin 2 θ 12 6

Dynamic reactions. C=

1 1 mLω1ω2 sin 2 θ k; D = − mLω1ω2 sin 2 θ k  6 6

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PROBLEM 18.107 A solid cone of height 9 in. with a circular base of radius 3 in. is supported by a ball-and-socket joint at A. Knowing that the cone is observed to precess about the vertical axis AC at the constant rate of 40 rpm in the sense indicated and that its axis of symmetry AB forms an angle β = 40° with AC, determine the rate at which the cone spins about the axis AB.

SOLUTION Use principal axes xyz with origin at A as shown. r = 3 in. = 0.25 ft

For the solid cone,

3h = 0.5625 ft 4 3 3  1  I = mr 2 I ′ = m  h 2 + r 2  10 5  4  3  2 1 2 I′ − I = m h − r  5  4  h = 9 in. = 0.75 ft c =

Angular velocity. spin: ψ about negative z axis precession: φ about positive Z axis ω = φ K − ψ k = φ (cos β k + sin β j) − ψ k ω x = 0, ω y = φ sin β , ω z = φ cos β − ψ

Angular momentum about fixed Point A. H A = I ′ω x i + I ′ω y j + I ω z k = I ′φ sin β j + I (φ cos β − ψ )k

Let frame Axyz be rotating with angular velocity Ω . Ω = φ K = φ cos β j + φ sin β k Rate of change of H A .

i j k    φ cos β H A = Ω × H A = 0 φ sin β 0 I ′φ sin β I (φ cos β − ψ )   sin β ]i = − [( I ′ − I ) φ2 cos β sin β + I φψ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2138

PROBLEM 18.107 (Continued)

Moment about A.

M A = − mgc sin β i  leads to MA = H A

I ′ − I 2 I   φ cos β + φψ m m 3 1  3   =  h 2 − r 2  φ2 cos β + r 2φψ 5 4  10  20 gc = (12h 2 − 3r 2 )φ2 cos β + 6r 2φψ gc =

ψ = Data:

20 gc − (12h 2 − 3r 2 )φ2 cos β 6r 2φ

β = 40° φ = 40 rpm = 4.1888 rad/s ψ =

(20)(32.2)(0.5625) − [(12)(0.75)2 − (3)(0.25)2 ]4.18882 cos 40° (6)(0.25) 2 (4.1888)

= 174.46 rad/s

ψ = 1666 rpm 

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PROBLEM 18.108 A solid cone of height 9 in. with a circular base of radius 3 in. is supported by a ball-and-socket joint at A. Knowing that the cone is spinning about its axis of symmetry AB at the rate of 3000 rpm and that AB forms an angle β = 60° with the vertical axis AC, determine the two possible rates of steady precession of the cone about the axis AC.

SOLUTION Use principal axes xyz with origin at A as shown. For the solid cone,

r = 3 in. = 0.25 ft 3h = 0.5625 ft 4 3 3  1  I = mr 2 I ′ = m  h 2 + r 2  10 5  4  3  1  I ′ − I = m  h2 − r 2  5  4  h = 9 in. = 0.75 ft c =

Angular velocity. spin: ψ about negative z axis precession: φ about positive Z axis ω = φ K − ψ k = φ (cos β k + sin β j) − ψ k

ω x = 0, ω y = φ sin β , ω z = φ cos β − ψ Angular momentum about fixed Point A. H A = I ′ω x i + I ′ω y j + I ω z k = I ′φ sin β j + I (φ cos β − ψ )k

Let frame Axyz be rotating with angular velocity Ω . Ω = φK = φ cos β j + φ sin β k Rate of change of H A . i j   H A = Ω × H A = 0 φ sin β 0 I ′φ sin β

k  φ cos β  I (φ cos β − ψ )

  sin β ]i = − [( I ′ − I ) φ2 cos β sin β + I φψ

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PROBLEM 18.108 (Continued)

Moment about A.

M A = − mgc sin β i  leads to MA = H A I ′ − I 2 I   φ cos β + φψ m m 3 1  3   =  h 2 − r 2  φ2 cos β + r 2φψ 5 4  10  20 gc = (12h 2 − 3r 2 )φ2 cos β + 6r 2φψ gc =

(12h 2 − 3r 2 )φ2 cos β + 6r 2ψ φ − 20 gc = 0

Data:

ψ = 3000 rpm = 314.16 rad/s β = 60° [(12)(0.75) 2 − (3)(0.25) 2 ](cos 60°) φ2 + (6)(0.25) 2 (314.16) φ − (20)(32.2)(0.5625) = 0 3.28125φ2 + 117.81φ − 362.25 = 0

Solving the quadratic equation, φ = 2.8488 rad/s, −38.753 rad/s

φ = 27.2 rpm, −370 rpm 

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PROBLEM 18.109 The 85-g top shown is supported at the fixed Point O. The radii of gyration of the top with respect to its axis of symmetry and with respect to a transverse axis through O are 21 mm and 45 mm, respectively. Knowing that c = 37.5 mm and that the rate of spin of the top about its axis of symmetry is 1800 rpm, determine the two possible rates of steady precession corresponding to θ = 30°.

SOLUTION Use principal axes x, y, z with origin at O. Angular velocity:

ω = ϕ sin θ i + (ψ + ϕ cos θ )k

ω x = ϕ sin θ , ω y = 0, ω z = ψ + ϕ cos θ Angular momentum about O: H O = I xω x i + I yω y j + I zω z k = I ′ϕ sin θ i + I ω z k

Let the reference frame Oxyz be rotating with angular velocity Ω = ϕ sin θ i + ϕ cos θ k.  = Ω×H H O Ο  = (ϕ sin θ i + ϕ cos θ k ) × ( I ′ϕ sin θ + I ω z k ) = ( I ′ϕ sin θ cos θ − I ω z sin θ )ϕ j  ΣM O = H O −Wc sin θ j = ( I ′ϕ sin θ cos θ − I ω z sin θ ) j Wc = ( I ω z − I ′ϕ cos θ )ϕ Wc = [ Iψ − ( I ′ − I )ϕ cos θ ]ϕ

Data:

(1)

m = 85 g = 0.085 kg W = mg = (0.085)(9.81) = 0.83385 N I = mk z2 = (0.085)(0.021) 2 = 37.485 × 10−6 kg ⋅ m 2 I ′ = mk x2 = (0.085)(0.045) 2 = 172.125 × 10−6 kg ⋅ m 2 c = 37.5 mm = 0.0375 m, ψ = 1800 rpm = 188.496 rad/s

θ = 30°

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PROBLEM 18.109 (Continued)

Substituting into Eq. (1), (0.83385)(0.0375) = [(37.485 × 10−6 )(188.496) − (134.64 × 10−6 )ϕ cos 30°]ϕ 116.602 × 10−6 ϕ 2 − 7.0658 × 10−3 ϕ + 31.269 × 10−3 = 0

ϕ = 30.299 ± 25.492

ϕ = 4.807 rad/s, 55.791 rad/s ϕ = 45.9 rpm, 533 rpm 

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PROBLEM 18.110 The top shown is supported at the fixed Point O and its moments of inertia about its axis of symmetry and about a transverse axis through O are denoted, respectively, by I and I ′. (a) Show that the condition for steady precession of the top is ( I ω z − I ′φ cos θ )φ = Wc

where φ is the rate of precession and ω z is the rectangular component of the angular velocity along the axis of symmetry of the top. (b) Show that if the rate of spin ψ of the top is very large compared with its rate of precession φ, the   ≈ Wc. (c) Determine the percentage error condition for steady precession is Iψφ introduced when this last relation is used to approximate the slower of the two rates of precession obtained for the top of Problem 18.109.

SOLUTION Use principal axes x, y, z with origin at O. ω = ϕ sin θ i + (ψ + ϕ cos θ )k

Angular velocity:

ω x = ϕ sin θ , ω y = 0, ω z = ψ + ϕ cos θ Angular momentum about O: H O = I xω x i + I yω y j + I zω z k = I ′ϕ sin θ i + I ω z k

Let the reference frame Oxyz be rotating with angular velocity Ω = ϕ sin θ i + ϕ cos θ k.  = Ω×H H O Ο  = (ϕ sin θ i + ϕ cos θ k ) × ( I ′ϕ sin θ + I ω z k ) = ( I ′ϕ sin θ cos θ − I ω z sin θ )ϕ j  ΣM O = H O −Wc sin θ j = ( I ′ϕ sin θ cos θ − I ω z sin θ ) j

Wc = ( I ω z − I ′ϕ cos θ )ϕ 

(a)

(b)

For |ϕ |

1 I 2

I − I′ <

1 I 2

Hence, Also, sec θ > 1:

I >2 I − I′

ϕ > 2ψ tan γ =

(b)

and

I tan θ I′

or tan θ =

I′ 1 tan γ > tan γ I 2

Angle of surface of space cone is γ − θ . Angle of axis is θ . tan θ > [1 + tan θ tan(γ − θ )]tan θ >

1 1 tan θ + tan(γ − θ ) tan[θ + (γ − θ )] = 2 2 1 + tan θ tan(γ − θ ) 1 1 tan θ + tan(γ − θ ) 2 2

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PROBLEM 18.122 (Continued)

tan θ + tan 2 θ tan(γ − θ ) >

1 1 tan θ + tan(γ − θ ) 2 2

1 1 tan θ > tan(γ − θ ) − tan 2 θ tan(γ − θ ) 2 2 1 1 tan θ > tan(γ − θ ) 2 2 tan θ > tan(γ − θ )

θ > (γ − θ ) 

The axis lies outside the space cone.

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PROBLEM 18.123 Using the relation given in Problem 18.121, determine the period of precession of the north pole of the earth about the axis of symmetry of the earth. The earth may be approximated by an oblate spheroid of axial moment of inertia I and of transverse moment of inertia I ′ = 0.9967 I . (Note: Actual observations show a period of precession of the north pole of about 432.5 mean solar days; the difference between the observed and computed periods is due to the fact that the earth is not a perfectly rigid body. The free precession considered here should not be confused with the much slower precession of the equinoxes, which is a forced precession. See Problem 18.118.)

SOLUTION ω = −ϕ sin θ i + (ϕ cos θ + ψ )k

Angular velocity of the body:

Let the reference frame Gxyz be rotating with angular velocity Ω, where Ω = −ϕ sin θ i + ϕ cos θ j α = ω Gxyz + Ω × ω

Angular acceleration of the body:

The rate of change of angular velocity as observed from the body is −α. Assume that −α may be represented as the angular velocity vector rotating with angular velocity li + mj + nk. −α = (li + mj + nk) × (ω x i + ω z k ) = − mω z i + (nω x − lω z j) − mω x k

Matching components:

i : 0 = mω z

  sin θ = nω x − lω z −ϕψ

j: k:

From Equation (1), From which Using Equation (18.44) with or

m=0

0 = − mω x

(1)

m=0

  sin θ = − nϕ sin θ − l (ϕ cos θ + ψ ) −ϕψ l = 0 and n = ψ . ΣM 0 = 0 yields I ω z − I ′ϕ cos θ = 0

ϕ cos θ =

I ωz I′

But

ω z = ψ + ϕ cos θ = ψ +

Solving for ψ ,

ψ =

I ωz I′

I′ − I ωz I

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PROBLEM 18.123 (Continued)

Using Data for Earth:

ω z = ω2 and n = ψ

yields n =

I′ − I ω2 I′

I ′ = 0.9967 I I ′ − I = −0.0033I , 0.0033 ω 2 = −0.003311ω 2 n=− 0.9967 2π 1 2π 2π period = = = 302.03 | n | 0.003311 ω2 ω2 = (302.03)(24 h) = 7248 h

period = 302 days 

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PROBLEM 18.124 A coin is tossed into the air. It is observed to spin at the rate of 600 rpm about an axis GC perpendicular to the coin and to precess about the vertical direction GD. Knowing that GC forms an angle of 15° with GD, determine (a) the angle that the angular velocity ω of the coin forms with GD, (b) the rate of precession of the coin about GD.

SOLUTION 1 2 mr 2 1 I ′ = mr 2 4 I=

Moments of inertia:

Euler angle θ for steady precession:

θ = 15°

For axisymmetric body under no force, Equation (18.49) gives for the body cone angle: tan γ =

(a)

Angular velocity:

I tan θ = 2 tan15° γ = 28.187° I′

ω = ω ( −i sin γ + k cosγ )

Its projection onto the vertical direction is

ω cos β = ω ⋅ k = ω (−i sinγ + k cosγ ) ⋅ (− sin θ i + cosθ k ) = ω (sin γ sin θ + cosγ cos θ ) = ω cos(γ − θ ) cos β = cos(γ − θ )

β = |γ − θ | = 28.187° − 15° = 13.187° β = 13.19° 

Angle between ω and vertical direction GD: tan γ = −

(b)

ωx ϕ sin θ = ω z ϕ cos θ + ψ

ψ sin γ ψ sin 28.187° = = −2.0705ψ sin(θ − γ ) − sin13.187°

from which

ϕ =

With

ψ = 600 rpm, ϕ = (−2.0705)(600)

|ϕ | = 1242 rpm (retrograde) 

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PROBLEM 18.125 The angular velocity vector of a football which has just been kicked is horizontal, and its axis of symmetry OC is oriented as shown. Knowing that the magnitude of the angular velocity is 200 rpm and that the ratio of the axis and transverse moments of inertia is I/I ′ = 13, determine (a) the orientation of the axis of precession OA, (b) the rates of precession and spin.

SOLUTION

ωx ωz γ = 15°

tan γ = −

For steady precession with no force, I′ tan γ I = 3 tan15° θ = 38.794°

tan θ =

(a) (b)

β = θ − γ = 38.794 − 15° ω x = −ϕ sin θ = −ω sin γ ω sin γ (200 rpm)sin15° ϕ = = sin θ sin(38.794°) = 82.621 rpm ω z = ψ + ϕ cos θ = ω cos γ ψ = ω cos γ − ϕ cos θ = 200cos15° − 82.621cos 38.794°

β = 23.8° 

precession: ϕ = 82.6 rpm 

spin: ψ = 128.8 rpm 

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PROBLEM 18.126 The space capsule has no angular velocity when the jet at A is activated for 1 s in a direction parallel to the x axis. Knowing that the capsule has a mass of 1000 kg, that its radii of gyration are k z = k y = 1.00 m and k z = 1.25 m, and that the jet at A produces a thrust of 50 N, determine the axis of precession and the rates of precession and spin after the jet has stopped.

SOLUTION Initial angular momentum about the mass center: H G = I xω x i + I y ω y j + I z ω z k = 0

Applied impulse at A:

A Δt = (50 N)(1 s)i = (50 N ⋅ s)i

Its moment about the mass center G: rA/G × A Δt = [(2 m) j − (1.25 m)k ] × (50 N ⋅ s)i = −(100 N ⋅ m ⋅ s)k − (62.5 N ⋅ m ⋅ s) j

Principle of impulse and momentum. (Moments about G) (H G )0 = rA/G × A Δt = H G

where H G is the final angular momentum about G. 0 + rG /A × A Δt = ( H G ) x i + ( H G ) y j + ( H G ) z k

Angular momentum vector components:

i:

0 = (HG )x

j:

−62.5 N ⋅ m ⋅ s = ( H G ) y = −62.5 kg ⋅ m 2 /s

k:

−100 N ⋅ m ⋅ s = ( H G ) z = −100 kg ⋅ m 2 /s tan θ =

(HG ) y (HG )z

= 0.625

θ = 32.005° 

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PROBLEM 18.126 (Continued)

Moments of inertia: I x = mkk2 = (1000 kg)(1 m) 2 = 1000 kg ⋅ m 2 I y = mk y2 = (1000 kg)(1 m) 2 = 1000 kg ⋅ m 2 I z = mk z2 = (1000 kg)(1.25 m) 2 = 1562.5 kg ⋅ m 2

Angular velocity vector components:

ωx =

(HG )x =0 Ix

ωy =

(HG ) y

ωz = tan γ =

Iy

=

−62.5 = −0.0625 rad/s 1000

(HG )z −100 = = −0.0640 rad/s Iz 1562.5

ωy = 0.97656 ωz

γ = 44.321°

ω = ω y2 + ω z2 = 0.089455 rad/s. Rates of precession and spin. The angular velocity is resolved into a component ϕ (rate of precession) parallel to HG and ψ (rate of spin) parallel to the axis of symmetry.

γ − θ = 12.316° Law of sines:

ϕ

ψ ω = sin γ sin(γ − θ ) sin θ ω sin γ 0.089455 sin 44.321° ϕ = = sin θ sin 32.005° ω sin(γ − θ ) 0.089455 sin12.316° ψ = = sin θ sin 32.005° =

Rate of precession:

ϕ = 0.1179 rad/s

ϕ = 1.126 rpm 

Rate of spin:

ψ = 0.0360 rad/s

ψ = 0.344 rpm 

Since γ > θ , the precession is retrograde.

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PROBLEM 18.127 The space capsule has an angular velocity ω = (0.02 rad/s)j + (0.10 rad/s)k when the jet at B is activated for 1 s in a direction parallel to the x axis. Knowing that the capsule has a mass of 1000 kg, that its radii of gyration are k z = k y = 1.00 m and k z = 1.25 m, and that the jet at B produces a thrust of 50 N, determine the axis of precession and the rates of precession and spin after the jet has stopped.

SOLUTION Moments of inertia: I x = mkk2 = (1000 kg)(1 m) 2 = 1000 kg ⋅ m 2 I y = mk y2 = (1000 kg)(1 m) 2 = 1000 kg ⋅ m 2 I z = mk z2 = (1000 kg)(1.25 m) 2 = 1562.5 kg ⋅ m 2

Initial angular velocity:

(ω x )0 = 0 (ω y )0 = 0.02 rad/s (ω z )0 = 0.10 rad/s

Initial angular momentum about G: (H G )0 = I x (ω x )0 i + I y (ω y )0 j + I z (ω z )0 k = (1000)(0)i + (1000)(0.02) j + (1562.5)(0.10)k = (20 kg ⋅ m 2 /s) j + (156.25 kg ⋅ m 2 /s)k

Applied impulse at B:

B Δt = (50 N)(1 s)i = (50 N ⋅ s)i

Its moment about the mass center G: rB /G × B Δt = [(1.25 m) j + (2 m)k ] × (50 N ⋅ s)i = (100 N ⋅ m ⋅ s) j − (62.5 N ⋅ m ⋅ s)k

Principle of impulse and momentum. (Moments about G): (H G )0 + rB /G × B Δt = H G

where H G is the final angular momentum about G.

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PROBLEM 18.127 (Continued)

Angular momentum vector components: 0 + 0 = (HG )x = 0

i: j:

20 kg ⋅ m 2 /s + 100 N ⋅ m ⋅ s = ( H G ) y = 120 kg ⋅ m 2 /s

k:

156.25 kg ⋅ m 2 /s − 62.5 N ⋅ m ⋅ s = ( H G ) z = 93.75 kg ⋅ m 2 /s

tan θ =

(HG ) y

θ = 52.001° 

= 1.28

(HG )z

Angular velocity vector components:

ωx =

(HG )x =0 Ix

ωy =

(HG ) y

ωz =

(HG )z 93.75 = = 0.060 rad/s Iz 1562.5

tan γ =

Iy

=

120 = 0.12 rad/s 1000

ωy = 2.0000 γ = 63.435° ωz

ω = ω y2 + ω z2 = 0.134164 rad/s Rates of precession and spin. The angular velocity is resolved into a component ϕ (rate of precession) parallel to HG and ψ (rate of spin) parallel to the axis of symmetry.

γ − θ = 11.434° Law of sines:

ϕ

ψ ω = sin γ sin(γ − θ ) sin θ ω sin γ 0.134164sin 63.435° ϕ = = sin θ sin 52.001° ω sin(γ − θ ) 0.134164sin11.434° ψ = = sin θ sin 52.001° =

Rate of precession:

ϕ = 0.1523 rad/s 

Rate of spin:

ψ = 0.0338 rad/s 

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PROBLEM 18.128 Solve Sample Problem 18.6, assuming that the meteorite strikes the satellite at C with a velocity v 0 = (2000 m/s)i. PROBLEM 18.6 A space satellite of mass m is known to be dynamically equivalent to two thin disks of equal mass. The disks are of radius a = 800 mm and are rigidly connected by a light rod of length 2a. Initially the satellite is spinning freely about its axis of symmetry at the rate ω0 = 60 rpm. A meteorite, of mass m0 = m /1000 and traveling with a velocity v0 of 2000 m/s relative to the satellite, strikes the satellite and becomes embedded at C. Determine (a) the angular velocity of the satellite immediately after impact, (b) the precession axis of the ensuing motion, (c) the rates of precession and spin of the ensuing motion.

SOLUTION (a)

Angular velocity after impact. From Sample Problem 18.6: 1 5 ma 2 I ′ = I x = I y = ma 2 2 4 Conservation of angular momentum: Angular momentum after impact: I = Iz =

H G = rC × m0 v 0 + I ω0 k = (− aj − ak ) × m0 v0 i + I ω0 k = − am0 v0 j + ( I ω0 + am0 v0 )k

Data:

a = 800 mm = 0.8 m ω0 = 60 rpm = 2π rad/s v0 = 2000 m/s, m0 =

m 1000

ωx =

(HG )x =0 Ix

ωy =

(HG ) y

ωz =

am v m v (HG )z = ω0 + 0 20 = ω0 + 2 0 0 1 Iz m a ma 2

Iy

=−

am0 v0 5 4

ma

2

=−

4 m0 v0 4  1  2000 =−  = −2 rad/s 5 m a 5  1000  0.8

 1  2000 = 2π + 2  = 11.2832 rad/s   1000  0.8 ω = −(2.00 rad/s)j + (11.28 rad/s)k

ω = (2) 2 + (11.2832)2 = 11.4591 rad/s = 109.426 rpm

ω = 109.4 rpm 

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PROBLEM 18.128 (Continued)

tan γ =

(b)

ωy ωz

=

2 11.2832

γ = 10.0515°

γ x = 90°, γ y = 100.05°, γ z = 10.05° 

Precession axis. I′ tan γ I 2 5 =   (2)  4  11.2832 θ = 23.900°

tan θ =

(c)

θ x = 90°, θ y = 113.9°, θ z = 23.9° 

Rates of precession and spin.

θ − γ = 13.8484° Law of sines.

ϕ

ψ ω = sin γ sin(θ − γ ) sin θ ω sin γ 109.4sin10.05° ϕ = = sin θ sin 23.9° =

precession: ϕ = 47.1 rpm 

ψ =

ω sin(θ − γ ) 109.4sin13.85° = sin θ sin 23.9° spin: ψ = 64.6 rpm 

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PROBLEM 18.129 An 800-lb geostationary satellite is spinning with an angular velocity ω 0 = (1.5 rad/s)j when it is hit at B by a 6-oz meteorite traveling with a velocity v 0 = −(1600 ft/s)i + (1300 ft/s)j + (4000 ft/s)k relative to the satellite. Knowing that b = 20 in. and that the radii of gyration of the satellite are k x = k z = 28.8 in. and k y = 32.4 in. , determine the precession axis and the rates of precession and spin of the satellite after the impact.

SOLUTION Mass of satellite:

m=

W 800 = = 24.845 lb ⋅ s 2 /ft g 32.2

Principal moments of inertia:

Ix =

mk x2

 28.8  = (24.845)    12 

2

= 143.106 lb ⋅ s 2 ⋅ ft Iy =

mk y2

 32.4  = (24.845)    12 

2

= 181.118 lb ⋅ s 2 ⋅ ft I z = I x = 143.106 lb ⋅ s 2 ⋅ ft

Mass of meteorite:

m′ =

6 = 0.011649 lb ⋅ s 2 /ft (16)(32.2)

Initial momentum of meteorite: m′v 0 = (0.011649)(−1600i + 1300 j + 4000k ) = −(18.633 lb ⋅ s)i + (15.140 lb ⋅ s)j + (46.584 lb ⋅ s)k

Assume that the position of the mass center of the satellite plus the meteorite is essentially that of the satellite alone. Position of Point B relative to the mass center: 20   42 rB/G =  i − j 12 12   = (3.5 ft)i − (1.66667 ft)j Angular velocity of satellite before impact: ω0 = (1.5 rad/s)j

(ω 0 ) x = (ω 0 ) z = 0, (ω 0 ) y = 1.5 rad/s

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PROBLEM 18.129 (Continued)

Angular momentum of satellite-meteorite system before impact: (H G )0 = I yω0 j + rB/G × (m′v 0 ) i j k = (181.118)(1.5) j + 3.5 −1.66667 0 −18.633 15.140 46.584

= −(77.64 lb ⋅ s ⋅ ft)i + (108.637 lb ⋅ s ⋅ ft)j + (21.935 lb ⋅ s ⋅ ft)k

Principle of impulse and momentum for satellite-meteorite system. Moments about G: (H G )1 = (H G )0 = H G

Angular velocity immediately after impact. ω = ωx i + ω y j + ω z k

Neglect the mass of the meteorite. H G = I xω x i + I y ω y j + I z ω z k

ωx =

(HG )x −77.64 = = −0.54253 rad/s 143.106 Ix

ωy =

(HG ) y

ωz =

(HG )z 21.934 = = 0.15327 rad/s 143.106 Iz

Iy

=

108.637 = 0.59981 rad/s 181.118

ω = −(0.54253 rad/s)i + (0.59981 rad/s)j + (0.15327 rad/s)k

ω = (0.54253)2 + (0.59981) 2 + (0.15327) 2 = 0.82317 rad/s H G = (77.64) 2 + (108.637) 2 + (21.935) 2 = 135.319 lb ⋅ s ⋅ ft

Motion after impact. Since the moments of inertia I x and I z are equal, the body moves as an axisymmetrical body with the y axis as the symmetry axis. Moment of inertia about the symmetry axis:

I = I y = 181.118 lb ⋅ s2 ⋅ ft

Moment of inertia about a transverse axis through G:

I ′ = I x = I z = 143.106 lb ⋅ s 2 ⋅ ft

The motion is a steady precession φ about the precession axis together with a steady spin ψ about the spin or symmetry axis. Since I > I ′, the precession is retrograde.

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PROBLEM 18.129 (Continued) Precession axis. The precession axis is directed along the angular momentum vector H G , which remains fixed. Immediately after impact, its direction cosines relative to the body axes x, y , z are: cos θ x =

(HG )x −77.64 = = −0.57376 HG 135.319

θ x = 125.0° 

cos θ y =

(HG ) y

108.637 = 0.80282 135.319

θ y = 36.6° 

cos θ z =

(H G ) z 21.935 = = 0.16210 HG 135.319

θ z = 80.7° 

HG

=

The angle θ between the spin axis (y axis) and the precession axis remains constant.

θ = θ y = 36.600° The angle γ between the angular velocity vector and the spin axis (y axis) is

ω y 0.59981 = ω 0.82317 The angle γ could also have been calculated from cos γ =

γ = 43.226°

tan γ =

I 181.118 tan β = tan 36.600° I′ 143.106

The angle between the precession axis and the angular velocity vector is

γ − θ = 6.626° Rates of precession and spin. Set up the triangle of vector addition for the components of angular velocity. Apply the law of sines.

ω ψ ϕ = = sin θ sin(γ − θ ) sin γ 0.82317 ψ ϕ sin 36.600°

Rate of precession:

=

sin 6.626°

=

sin 43.226°

ϕ = 0.946 rad/s  ψ = 0.1593 rad/s 

Rate of spin:

Since γ > θ , precession is retrograde. 

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PROBLEM 18.130 Solve Problem 18.129, assuming that the meteorite hits the satellite at A instead of B. PROBLEM 18.129 An 800-lb geostationary satellite is spinning with an angular velocity ω0 = (1.5 rad/s)j when it is hit at B by a 6-oz meteorite traveling with a velocity v 0 = −(1600 ft/s)i + (1300 ft/s)j + (4000 ft/s)k relative to the satellite. Knowing that b = 20 in. and that the radii of gyration of the satellite are k x = k z = 28.8 in. and k y = 32.4 in., determine the precession axis and the rates of precession and spin of the satellite after the impact.

SOLUTION W 800 = = 24.845 lb ⋅ s 2 /ft g 32.2

Mass of satellite:

m=

Principal moments of inertia:

 28.8  2 I x = mk x2 = (24.845)   = 143.106 lb ⋅ s ⋅ ft 12  

2

2

Iy =

mk y2

 32.4  2 = (24.845)   = 181.118 lb ⋅ s ⋅ ft  12 

I z = I x = 143.106 lb ⋅ s 2 ⋅ ft

Mass of meteorite:

m′ =

6 = 0.011649 lb ⋅ s 2 /ft (16)(32.2)

Initial momentum of meteorite: m′v 0 = (0.011649)( −1600i + 1300 j + 4000k ) = −(18.633 lb ⋅ s)i + (15.140 lb ⋅ s)j + (46.584 lb ⋅ s)k

Assume that the position of the mass center of the satellite plus the meteorite is essentially that of the satellite alone. Position of Point A relative to the mass center: rA/G =

42 i = (3.5 ft)i 12

Angular velocity of satellite before impact: ω0 = (1.5 rad/s)j,

(ω 0 ) x = (ω 0 ) z = 0, (ω 0) y = 1.5 rad/s

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PROBLEM 18.130 (Continued)

Angular momentum of satellite-meteorite system before impact: (H G )0 = I yω0 j + rA/G × (m′v 0 ) i

= (181.118)(1.5) j +

j

k

3.5 0 0 −18.633 15.140 46.584

= (108.637 lb ⋅ s ⋅ ft)j + (52.99 lb ⋅ s ⋅ ft)k

Principle of impulse and momentum for satellite-meteorite system. Moments about G: (H G )1 = (H G )0 = H G

Angular velocity immediately after impact. ω = ωx i + ω y j + ωz k

Neglect the mass of the meteorite. H G = I xω x i + I yω y j + I zω z k

ωx =

(HG )x =0 Ix

ωy =

(HG ) y

ωz =

(HG )z 52.99 = = 0.37028 rad/s Iz 143.106

Iy

=

108.637 = 0.59981 rad/s 181.118

ω = (0.59981 rad/s)j + (0.37028 rad/s)k

ω = (0.59981)2 + (0.37028)2 = 0.70490 rad/s H G = (108.637) 2 + (52.99) 2 = 120.872 lb ⋅ s ⋅ ft

Motion after impact. Since the moments of inertia I x and I z are equal, the body moves as an axisymmetrical body with the y axis as the symmetry axis. Moment of inertia about the symmetry axis: I = I y = 181.118 lb ⋅ s 2 ⋅ ft

Moment of inertia about a transverse axis through G: I ′ = I x = I z = 143.106 lb ⋅ s 2 ⋅ ft

The motion is a steady precession φ about the precession axis together with a steady spin ψ about the spin or symmetry axis. Since I > I ′, the precession is retrograde.

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PROBLEM 18.130 (Continued) Precession axis. The precession axis is directed along the angular momentum vector H G , which remains fixed. Immediately after impact, its direction cosines relative to the body axes x, y , z are: cos θ x =

(HG )x =0 HG

θ x = 90.0° 

cos θ y =

(HG ) y

108.637 = 0.89878 120.872

θ y = 26.0° 

cos θ z =

(HG )z 52.99 = = 0.43840 HG 120.872

θ z = 64.0° 

HG

=

The angle θ between the spin axis (y axis) and the precession axis remains constant.

θ = θ y = 26.002° The angle γ between the angular velocity vector and the spin axis (y axis) is

ω y 0.59981 γ = 31.689° = ω 0.70490 The angle γ could also have been calculated from cos γ =

tan γ =

I 181.118 tan θ = tan 26.002° I′ 143.106

The angle between the precession axis and the angular velocity vector is

γ − θ = 5.687° Rates of precession and spin. Set up the triangle of vector addition for the components of angular velocity. Apply the law of sines.

ω ψ ϕ = = sin θ sin(γ − θ ) sin γ ψ ϕ 0.70490 sin 26.002°

Rate of precession:

=

sin 5.687°

=

sin 31.689°

ϕ = 0.844 rad/s  ψ = 0.1593 rad/s 

Rate of spin:

Since γ > θ , the precession is retrograde. 



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PROBLEM 18.131 A homogeneous rectangular plate of mass m and sides c and 2c is held at A and B by a fork-ended shaft of negligible mass, which is supported by a bearing at C. The plate is free to rotate about AB, and the shaft is free to rotate about a horizontal axis through C. Knowing that, initially, θ 0 = 40°, θ0 = 0, and φ0 = 10 rad/s, determine for the ensuring motion (a) the range of values of θ , (b) the minimum value of φ, (c) the maximum value of θ.

SOLUTION Let the fixed Z axis lie along the axle of the fork-ended shaft. Let the axes Gxyz be attached at the mass center with x perpendicular to the plate, y along the axle AB and z parallel to the long edges of the plate.  = ϕ k + θ j Angular velocity vector: ω = ϕ cos θ i + θ j + ϕ sin θ k

Conservation of angular momentum. Since plate is free to rotate about Z axis, H Z = constant

But

(1)

H Z = H x cos θ + H z sin θ

H Z = I xω x cos θ + I zω z sin θ 1 5 = mc 2φ cos 2 θ + mc 2φ sin 2 θ 12 12 1 = mc 2φ(cos 2 θ + 5sin 2 θ ) 12 1 = mc 2φ(1 + 4sin 2 θ ) 12 Using the initial conditions, Eq. (1) yields φ(1 + 4sin 2 θ ) = φ0 (1 + 4sin 2 θ0 )

(2)

Conservation of energy. Since no work is done, we have T = constant

(3)

where

T=

1 ( I xω x2 + I yω y2 + I zω z2 ) 2

1 1 1 5  mc 2φ2 cos 2 θ + mc 2θ 2 + mc 2φ2 sin 2 θ  2  12 3 12  1 2 2 2 2 2 mc [4θ + φ (cos θ + 5sin θ )] = 24 1 mc 2 [4θ 2 + φ2 (1 + 4sin 2 θ )] = 24

T=

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PROBLEM 18.131 (Continued) Using the initial conditions, including θ0 = 0, Eq. (3) yields 4θ 2 + φ2 (1 + 4sin 2 θ ) = φ02 (1 + 4sin 2 θ 0 )

(a)

(4)

With θ 0 = 40° and φ0 = 10 rad/s in. Eqs. (2) and (4),

φ(1 + 4sin 2 θ ) = 26.527 26.527 1 + 4sin 2 θ 4θ 2 + φ2 (1 + 4sin 2 θ ) = 265.27

φ =

(2′)

Eliminate φ and solve for θ 2 : 4θ 2 = 265.27 −

(b)

(5)

(26.527)2 = 2.6527 265.27 sin 2 θ = 0.4132 sin θ = 0.6428

For

θ 2 = 0, 1 + 4sin 2 θ =

From which

θ = 40° and 140°

40° < θ < 140° 

From Eq. (2′), φ is minimum for θ = 90°.

φmin = 5.3054 (c)

(26.527) 2 1 + 4sin 2 θ

φmin = 5.31 rad/s 

From Eq. (5),

θ 2 = 66.318 −

175.92 1 + 4sin 2 θ

θ 2 is maximum for θ = 90°. 2 θmax = 31.134 rad 2 /s 2

θmax = 5.58 rad/s 

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PROBLEM 18.132 A homogeneous rectangular plate of mass m and sides c and 2c is held at A and B by a fork-ended shaft of negligible mass which is supported by a bearing at C. The plate is free to rotate about AB, and the shaft is free to rotate about a horizontal axis through C. Initially the plate lies in the plane of the fork (θ0 = 0) and the shaft has an angular velocity φ0 = 10 rad/s. If the plate is slightly disturbed, determine for the ensuring motion (a) the minimum value of φ, (b) the maximum value of θ.

SOLUTION Let the fixed Z axis lie along the axle of the fork-ended shaft. Let the axes Gxyz be attached at the mass center with x perpendicular to the plate, y along the axle AB and z parallel to the long edges of the plate.  = ϕ k + θ j ω Angular velocity vector: = ϕ cos θ i + θ j + ϕ sin θ k

Conservation of angular momentum. Since plate is free to rotate about Z axis, H Z = constant

But

(1)

H Z = H x cos θ + H z sin θ

H Z = I xω x cos θ + I zω z sin θ 1 5 = mc 2φ cos 2 θ + mc 2φ sin 2 θ 12 12 1 2 2 = mc φ (cos θ + 5sin 2 θ ) 12 1 = mc 2φ(1 + 4sin 2 θ ) 12 Using the initial conditions, Eq. (1) yields φ(1 + 4sin 2 θ ) = φ0 (1 + 4sin 2 θ0 )

(2)

Conservation of energy. Since no work is done, we have T = constant

(3)

where

T=

1 ( I xω x2 + I yω y2 + I zω z2 ) 2

1 1 1 5  mc 2φ2 cos 2 θ + mc 2θ 2 + mc 2φ2 sin 2 θ   2  12 3 12  1 mc 2 [4θ 2 + φ2 (cos 2 θ + 5sin 2 θ )] = 24 1 mc 2 [4θ 2 + φ2 (1 + 4sin 2 θ )] = 24

T=

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PROBLEM 18.132 (Continued) Using the initial conditions, including θ0 = 0, Eq. (3) yields 4θ 2 + φ2 (1 + 4sin 2 θ ) = φ02 (1 + 4sin 2 θ 0 )

(a)

With

θ 0 = 0, φ0 = 10 rad/s

Eq. (2) yields

φ =

10 1 + 4sin 2 θ

φ is minimum for θ = 90° : (b)

(4)

φmin = 2.00 rad/s 

1   4θ 2 = 100 − φ2 (1 + 4sin 2 θ ) = 100 1 − 2  θ 1 4sin +  

Eq. (4) yields

θ 2 is largest for θ = 90°:  1 2 4θmax = 100 1 −   5 θ 2 = 20

θmax = 4.47 rad/s 

max

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PROBLEM 18.133 A homogeneous disk of radius 180 mm is welded to a rod AG of length 360 mm and of negligible mass which is connected by a clevis to a vertical shaft AB. The rod and disk can rotate freely about a horizontal axis AC, and shaft AB can rotate freely about a vertical axis. Initially, rod AG is horizontal (θ0 = 90°) and has no angular velocity about AC. Knowing that the maximum value φm of the angular velocity of shaft AB in the ensuing motion is twice its initial value φ0 , determine (a) the minimum value of θ , (b) the initial angular velocity φ0 of shaft AB.

SOLUTION Let the Z axis be vertical. For principal axes xyz with origin at A, the principal moments of inertia are I′ = Ix = Iy 1  17 = m  a 2 + (2a )2  = ma 2 4  4 1 I = I z = ma 2 2

Angular velocity components:

ω x = ϕ sin θ ω y = −θ ω z = ψ + ϕ cos θ

Angular momentum about A:

H A = I xω x i + I y ω y j + I z ω z k

= I ′ϕ sin θ i − I ′θ j + I ω z k 1 1 1 I xω x2 + I yω y2 + I z ω z2 2 2 2 1 1 T = I ′(ϕ 2 sin 2 θ + θ 2 ) + I ω z2 2 2

Kinetic energy:

T=

Potential energy:

V = −2mga cos θ

Conservation of angular momentum about fixed Z axis: H A ⋅ K = H A ⋅ (i sin θ + k cos θ ) = I ′ϕ sin 2 θ + I ω z cos θ =

where α is a constant.

17 2 1 ma ϕ sin 2 θ + ma 2ω z cos θ = α 4 2

(1)

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PROBLEM 18.133 (Continued)

Conservation of energy:

T + V = E , where E is a constant. 17 2 2 2 1 ma (ϕ sin θ + θ 2 ) + ma 2ω z2 − 2mga cos θ = E 8 4

Constraint of clevis: (a)

From Eq. (1),

ψ = 0

(2)

ω z = ϕ cos θ

17 2 1 17 1 ma ϕm sin 2 θ m + ma 2ϕm cos 2 θ m = ma 2ϕ0 sin 2 90° + ma 2ϕ0 cos 2 90° 4 2 4 2 17 2 1 17 ϕ0  17  1  17 =    = sin θ m + cos 2 θ m = 4 2 4 ϕm  4  2  8 17 2 1 17 sin θ m + (1 − sin 2 θ m ) = 4 2 8 13 sin 2 θ m = 30 sin θ m = 0.65828

θ m = 41.169° (b)

θ = 0

At the minimum value of θ , From Eq. (2),

θ m = 41.2° 

17 2 2 1 ma (ϕm sin 2 θ m + 0) + ma 2ϕm2 cos 2 θ m − 2mga cos θ m 8 4 =

17 2 2 2 1 ma (ϕ0 sin θ0 + 0) + ma 2ϕ02 cos 2 90° − 2mga cos 90° 8 4

 17  2 2 17  2 2 1 2 2   (2) sin θ m +   (2) cos θ m −  ma ϕ0 = 2mga cos θ m 8 4  8  2.1250 ma 2ϕ02 = 1.5055mga

ϕ02 = 0.70849

g 9.81 = 0.70849 = 38.613 a 0.18

ϕ0 = 6.21 rad/s 

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PROBLEM 18.134 A homogeneous disk of radius 180 mm is welded to a rod AG of length 360 mm and of negligible mass which is connected by a clevis to a vertical shaft AB. The rod and disk can rotate freely about a horizontal, axis AC, and shaft AB can rotate freely about a vertical axis. Initially, rod AG is horizontal (θ 0 = 90°) and has no angular velocity about AC. Knowing that the smallest value of θ in the ensuing motion is 30°, determine (a) the initial angular velocity of shaft AB, (b) its maximum angular velocity.

SOLUTION Let the Z axis be vertical. For principal axes x, y, z with origin at A, the principal moments of inertia are I′ = Ix = Iy

Angular velocity components:

1  17 = m  a 2 + (2a )2  = ma 2 4  4 1 2 I = I z = ma 2 ω x = ϕ sin θ ω = −θ y

ω z = ψ + ϕ cos θ Angular momentum about A:

H A = I xω x i + I y ω y j + I z ω z k = I ′ϕ sin θ i − I ′θ j + I ω z k

Kinetic energy:

Potential energy:

1 1 1 I xω x2 + I yω y2 + I z ω z2 2 2 2 1 1 T = I ′(ϕ 2 sin 2 θ + θ 2 ) + I ω z2 2 2 V = −2mga cos θ T=

Conservation of angular momentum about fixed Z axis: H A ⋅ K = H A ⋅ (i sin θ + k cos θ ) = I ′ϕ sin 2 θ + I ω z cos θ =

17 2 1 ma ϕ sin 2 θ + ma 2ω z cos θ = α 4 2

(1)

where α is a constant PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2188

PROBLEM 18.134 (Continued)

Conservation of energy:

T + V = E , where E is a constant. 17 2 2 2 1 ma (ϕ sin θ + θ 2 ) + ma 2ω z2 − 2mga cos θ = E 8 4

Constraint of clevis: (a)

ψ = 0

ω z = ϕ cos θ θ = 0.

At θ = θ m = 30° and at θ = θ0 = 90°, From Eq. (1),

17 2 1 17 ma ϕm sin 2 θ m + ma 2ϕm cos 2 θ m = ma 2ϕ0 sin 2 θ0 + 0 4 2 4 2  17 1 2 17 2   1  3   2    +   ma ϕm = ma ϕ0   4  2   2   2   4   68 ϕm = ϕ0 23 2

From Eq. (2),

(2)

2

17 2  68  1  68  ma  ϕ0  sin 2 30° + ma 2  ϕ0  cos 2 30° − 2mga cos 30° 8 4  23   23  =

17 1 ma 2ϕ02 sin 2 90° + ma 2ϕ02 cos 2 90° − 2mga cos 90° 32 4

 17 3  68 2 17  2 2  +   −  ma ϕ0 = 2mga cos 30° 8   32 16  23  g ϕ02 = 0.41660 a 9.81 = 0.41660 0.18 = 22.705

ϕ0 = 4.7649 rad/s (b)

ϕm =

68 (4.7649) 23

ϕ0 = 4.76 rad/s  ϕm = 14.09 rad/s 

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PROBLEM 18.135 The slender homogeneous rod AB of mass m and length L is free to rotate about a horizontal axle through its mass center G. The axle is supported by a frame of negligible mass which is free to rotate about the vertical CD. Knowing that, initially, θ = θ0 , θ = 0, and φ = φ0 , show that the rod will oscillate about the horizontal axle and determine (a) the range of values of angle θ during this motion, (b) the maximum value of θ, (c) the minimum value of φ.

SOLUTION Angular velocity. Using the coordinate axes x, y, z shown (with y running into the paper), we have ω = −φ sin θ i + θ j + φ cos θ k

(1)

Moments of inertia. For slender rod of length L are mass m: Ix = Iy 1 mL2 12 Iz = 0 =

(2)

Conservation of energy. Since the x, y , z axes are principal axis, we use

(

)

1 I xω x2 + I y ω y2 + I z ω z2 2 1 1 1  =  mL2φ2 sin 2 θ + mL2θ 2 + 0  2  12 12  1 T= mL2 (φ2 sin 2 θ + θ 2 ) 24 T=

(3)

Using a datum through G, we have V = 0. T + V = constant:

1 mL2 (φ2 sin 2 θ + θ 2 ) = constant 24

Recalling the initial conditions θ = θ0 , θ = 0, ϕ = φ0 , we determine the constant and write

φ2 sin 2 θ + θ 2 = φ02 sin 2 θ 0

(4)

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PROBLEM 18.135 (Continued)

Conservation of angular momentum. Since the only forces exerted on the rod are its weight and the reaction at G, we have Σ M G = 0. Using a fixed reference frame GXYZ , with Z directed vertically upward, we have from Eq. (18.2)  = ΣM = 0 H G G

or, integrating with respect to the frame GXYZ , H G = constant

Considering the vertical component of H G , H Z = constant H Z = H z cos θ − H x sin θ = I z ω z cos θ − I xω x sin θ

But

=0− HZ =

Thus,

1 mL2 (−φ sin θ )sin θ 12

1 m L2φ sin 2 θ = constant 12

Recalling the initial conditions, we obtain

φ sin 2 θ = φ0 sin 2 θ0

(5)

Solving Eq. (5) for φ and substituting into Eq. (4):  φ0 sin 2 θ0  2  sin θ

2

 2 2 2  sin θ + θ = φ0 sin θ0  

θ 2 = φ02 sin 2 θ0 1 − 

(a)

sin 2 θ0   sin 2 θ 

(6)

Range of values of θ. Since θ ≥ 0, we must have 1−

sin 2 θ0 sin 2 θ

≥0

sin 2 θ ≥ sin 2 θ0 |sin θ | ≥ |sin θ0 |

From trigonometric circle, we conclude that the range is

θ 0 ≤ θ ≤ 180° − θ0  Rod oscillates about axle and about horizontal line (dashed line). PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2191

PROBLEM 18.135 (Continued)

(b)

Maximum value of θ.

Referring to Eq. (6), we note that this occurs when sin θ = 1, that is, when θ = 90°. We have 2 θmax = φ02 sin 2 θ 0 (1 − sin 2 θ 0 )

= φ02 sin 2 θ 0 cos 2 θ0

(c)

θmax = φ0 sin θ0 cos θ0 

Minimum value of φ0 .

Referring to Eq. (5), we note that φ is minimum when sin θ = 1, that is, when θ = 90°. We have

φmin = φ0 sin 2 θ0 

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PROBLEM 18.136 The gimbal ABA′B′, is of negligible mass and may rotate freely about the vertical AA′. The uniform disk of radius a and mass m may rotate freely about its diameter BB′, which is also the horizontal diameter of the gimbal. (a) Applying the principle of conservation of energy, and observing that, since ΣM AA′ = 0, the component of the angular momentum of the disk along the fixed axis AA′ must be constant, write two first-order differential equations defining the motion of the disk. (b) Given the initial conditions θ 0 ≠ 0, φ0 = 0, and θ0 = 0, express the rate of nutation θ as a function of θ . (c) Show that the angle θ will never be larger than θ 0 during the ensuing motion.

SOLUTION We use a reference frame Cxyz attached to the disk as shown. The angular velocity of the disk is

ω = φk + θ j ω = −φ sin θ i + θ j + φ cos θ k (a)

Conservation of energy:

T + V = constant

Since V = constant, we have T = constant. For principal centroidal axes and v = 0, the kinetic energy is given by T=

But

1 ( I z ω z2 + I yω y2 + I z ω z2 ) 2

Ix = Iy =

1 2 ma , 4

Iz =

1 2 ma 2

Using the components of ω computed above: 1 1 1 T = ma 2φ2 sin 2 θ + ma 2θ 2 + ma 2ω s 2θ 8 8 4 1 T = ma 2 [(1 + cos 2 θ )φ2 + θ 2 ] 8

We have therefore

(1 + cos 2 θ )φ2 + θ 2 = constant

(1) 

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PROBLEM 18.136 (Continued)

We now determine the angular momentum HC: H C = I zω x i + I yω y j + I z ω z k 1 1 1 = − ma 2φ sin θ i + ma 2θ j + ma 2φ cos θ k 4 4 2

Since H z = constant, we write H z = H C ⋅ K = constant 1 2  ma (−φ sin θ i + θ j + 2φ/ cos θ k ) ⋅ K = constant 4

Since i ⋅ K = − sin θ , j ⋅ K = 0, k ⋅ K = cos θ , we have

φ sin 2 θ + 2φ cos2 θ = constant φ(1 + cos 2 θ ) = constant (b)

(2) 

We determine the constants in (1) and (2) from the initial conditions θ 0 , φ0 , θ0 = 0, and write

Solving (2′) for φ :

(1 + cos 2 θ )φ2 + θ 2 = (1 + cos 2 θ0 )φ02

(1′)

φ(1 + cos 2 θ ) = φ0 (1 + cos 2 θ0 )

(2′)

φ = φ0

1 + cos 2 θ 0 1 + cos 2 θ

Substituting into (1′): (1 + cos 2 θ0 ) 2 2  2 φ0 + θ = (1 + cos 2 θ 0 )φ02 1 + cos 2 θ



θ 2 = φ02 (1 + cos 2 θ0 ) 1 − 

θ = φ0 (c)

1 + cos 2 θ 0   1 + cos 2 θ 0 

(1 + cos θ0 )(cos 2 θ − cos 2 θ0 ) 2

1 + cos 2 θ



For θ to be real, we need cos 2 θ 0 − cos 2 θ 0 ≥ 0 Thus

|cos θ | ≥ |cos θ 0 |

Assuming that the axes have been chosen so that θ 0 ≤ 90°, we must have cos θ ≥ cos θ 0

θ ≤ θ0 

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PROBLEM 18.137* The top shown is supported at the fixed Point O. Denoting by φ , θ , and ψ the Eulerian angles defining the position of the top with respect to a fixed frame of reference, consider the general motion of the top in which all Eulerian angles vary. (a)

Observing that ΣM Z = 0 and ΣM z = 0, and denoting by I and I ′, respectively, the moments of inertia of the top about its axis of symmetry and about a transverse axis through O, derive the two first-order differential equations of motion I ′φ sin 2 θ + I (ψ + φ cos θ ) cos θ = α I (ψ + φ cos θ ) = β

where α and β are constants depending upon the initial conditions. These equations express that the angular momentum of the top is conserved about both the Z and z axes, i.e., that the rectangular component of H O along each of these axes is constant. (b)

Use Eqs. (1) and (2) to show that the rectangular component ω z of the angular velocity of the top is constant and that the rate of precession φ depends upon the value of the angle of nutation θ .

SOLUTION Use a rotating frame of reference with the y axis pointing into the paper. Angular velocity of the frame: Ω = − φ sin θ i + θ j + φ cos θ k Angular velocity of the top: ω = − φ sin θ i + θ j + (ψ + φ cos θ )k Its angular momentum about O: H O = I xω x i + I y ω y j + I z ω z k = − I ′φ sin θ i + I ′θ j + I (ψ + φ cos θ )k

where

I x = I y = I ′ and I z = I .

The moment M O about O is due to the weight mg. M 0 = mgc sin θ j

(a)

Since the fixed Z axis refers to a Newtonian frame of reference and ( M O ) Z = 0, it follows that ( H O ) Z is constant. Thus, ( H O ) Z = H O ⋅ K = H O ⋅ (−i sin θ + k cos θ ) = I ′φ sin 2 θ + I (ψ + φ cos θ ) cos θ = α

(1) 

where α is a constant. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2195

PROBLEM 18.137* (Continued)

 = (H  ) MO = H O O Oxyz + Ω × H O d  d (φ sin θ ) i + I ′θj + I (ψ + φ cos θ ) k dt dt      ′ + [ I (ψ + φ cos θ ) θ − I φθ cos θ ]i + [ I (ψ + φ cos θ ) φ sin θ − I ′φ2 sin θ cos θ ]j

mgc sin θ j = − I ′

0=

z-components:

d (ψ + φ cos θ ) dt I (ψ + φ cos θ ) = β (2) 

Integrating, where β is a constant. (b)

Since

ω z = ψ + φ cos θ

From Eqs. (1) and (2), I ′φ sin 2 θ + β cos θ = α

ωz =

β I

φ =

= constant (3) 

α − β cos θ (4)  I ′ sin 2 θ

which is a function of θ .

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PROBLEM 18.138* (a) (b)

Applying the principle of conservation of energy, derive a third differential equation for the general motion of the top of Problem 18.137. Eliminating the derivatives φ and ψ from the equation obtained and from the two equations of Problem 18.139, show that the rate of nutation θ is defined by the differential equation θ 2 = f (θ ), where f (θ ) =

(c)

  α − β cos θ  1 β2 − 2mgc cos θ  −   2 E −  I′  I   I ′ sin θ 

2

Further show, by introducing the auxiliary variable x = cos θ , that the maximum and minimum values of θ can be obtained by solving for x the cubic equation   β2 1 − 2mgcx  (1 − x 2 ) − (α − β x) 2 = 0  2 E − I I′  

SOLUTION (a)

Angular velocity of the top: ω = −ϕ sin θ i + θ j + (ψ + ϕ cos θ )k

Kinetic energy: 1 1 1 I xω x2 + I yω y2 + I zω z2 2 2 2 1 1 1 = I ′(ϕ sin θ ) 2 + I ′θ 2 + I zω z2 2 2 2

T=

V = mgc cos θ

Potential energy:

Principle of conservation of energy:

T +V = E

1 1 1 I ′(ϕ sin θ ) 2 + I ′θ 2 + I ω z2 + mgc cos θ = E  2 2 2

(b)

Solving for θ 2 ,

θ 2 =

1 (2 E − I zω z2 − 2mgc cos θ ) − (ϕ sin θ ) 2 I′

(A)

Equation (2) of Problem 18.137, with ω z = ψ + ϕ cos θ gives I zω z2 =

β2

(B)

I

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PROBLEM 18.138* (Continued)

Equation (1) of Problem 18.137 gives I ′ϕ sin 2 θ + β cos θ = α

ϕ =

α − β cos θ I ′ sin 2 θ

(C) (D)

Substituting Equations (D) and (B) into Equation (A),

θ 2 = f (θ ) where (c)

  α − β cos θ 2 1 β2 f (θ ) =  2 E − − 2mgc cos θ  −   I′  I   I ′ sin θ 

(1) 

Maximum and minimum values of θ occur when f (θ ) = 0. Setting cos θ = x and sin 2 θ = 1 − x 2 in Equation (1), and letting f (θ ) = 0 gives  (α − β x)2 1 β2 + 2mgcx  − =0  2 E − 2 2 I′  I  ( I ′) (1 − x )

Multiplying by I ′(1 − x 2 ) gives the cubic equation F ( x) = 0:   β2 1 − 2mgcx  (1 − x 2 ) − (α − β x) 2 = 0  2 E − I I′  

(2) 

Solving this equation will yield three values of x. The two values lying between −1 and +1 correspond to the maximum and minimum values of θ .

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PROBLEM 18.139 A solid cone of height 180 mm with a circular base of radius 60 mm is supported by a ball and socket at A. The cone is released from the position θ 0 = 30° with a rate of spin ψ 0 = 300 rad/s, a rate of precession φ0 = 20 rad/s, and a zero rate of nutation. Determine (a) the maximum value of θ in the ensuing motion, (b) the corresponding values of the rates of spin and precession. [Hint: Use Eq. (2) of Prob. 18.138; you can either solve this equation numerically or reduce it to a quadratic equation, since one of its roots is known.]

SOLUTION Data:

r = 60 mm = 0.06 m,

θ 0 = 30°,

h = 180 mm = 0.18 m,

ψ 0 = 300 rad/s,

c=

φ0 = 20 rad/s,

3 h = 0.135 m, 4

θ0 = 0

Calculate the following:

Initially,

I 3 2 3 r = (0.06)2 = 1.08 × 10−3 m 2 = m 10 10 I′ 3  1 2  3 1  =  r + h 2  =  (0.06) 2 + (0.18)2  = 19.98 × 10−3 m 2 m 5 4 5 4     ω = −φ sin θ = −20sin 30° = −10 rad/s, ω = θ = 0 0

x

0

y

0

ω z = ψ 0 + φ0 cosθ 0 = 300 + 20 cos 30° = 317.32 rad/s T 1 I′ 2 1 I 2 (ω x + ω y2 ) + ωz = m 2m 2m 1 1 = (19.98 × 10−3 )(102 + 0) + (1.08 × 10−3 )(317.32) 2 = 55.3728 m 2 /s 2 2 2 V = gc cos θ0 = (9.81)(0.135) cos 30° = 1.1469 m 2 /s 2 m 2E = 2(55.3728 + 1.1469) = 113.0394 m 2 /s 2 m I β = ω z = (1.08 × 10−3 )(317.32) = 0.342706 m 2 /s m m

α m

=

I′  φ0 sin 2 θ 0 + β cos θ 0 m

= (19.98 × 10−3 )(20) sin 2 30° + 0.342706cos 30° = 0.396692 m 2 /s PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2199

PROBLEM 18.139 (Continued)

After dividing by m, Equation (2) of Problem 18.138 becomes  2E F ( x) =  −  m 

β2 m2 I m

 2 mα βx − 2 gcx  (1 − x 2 ) −  − =0  I′  m m   

  (0.342706)2 (0.396692 − 0.342706 x) 2 − (2)(9.81)(0.135) x  (1 − x 2 ) − =0 113.0394 − −3 1.08 × 10 19.98 × 10−3  

(4.29198 − 2.6487 x)(1 − x 2 ) − 5.87825(1.157529 − x) 2 = 0

(a)

x = cosθ = 0.68170,

Roots are:

0.86603,

2.2919

θ max = cos −1(0.68170) = 47.023° (b)

θ max = 47.0° 

By Equation (4) of Problem 18.137,

φ =

α − β cos θ 0.396692 − (0.342706)(0.68170) = = 15.2474 I ′ sin 2 θ (19.98 × 10−3 ) sin 2 47.023°

ω z = 317.32 rad/s ψ = ω z − φ cos θ = 317.32 − (15.2474)(0.68170) 

spin: ψ = 307 rad/s  precession: φ = 15.25 rad/s 



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PROBLEM 18.140 A solid cone of height 180 mm with a circular base of radius 60 mm is supported by a ball and socket at A. The cone is released from the position θ 0 = 30° with a rate of spin ψ 0 = 300 rad/s, a rate of precession φ0 = −4 rad/s, and a zero rate of nutation. Determine (a) the maximum value of θ in the ensuing motion, (b) the corresponding values of the rates of spin and precession, (c) the value of θ for which the sense of the precession is reversed. (See hint of Problem 18.139.)

SOLUTION Data:

r = 60 mm = 0.06 m,

θ0 = 30°,

h = 180 mm = 0.18 m,

ψ 0 = 300 rad/s,

φ0 = −4 rad/s,

c=

3 h = 0.135 m, 4

θ0 = 0

Calculate the following:

Initially,

I 3 2 3 r = (0.06)2 = 1.08 × 10−3 m 2 = m 10 10 I′ 3  1 2  3 1  =  r + h 2  =  (0.06) 2 + (0.18)2  = 19.98 × 10−3 m 2 m 5 4  5 4  ω x = −φ0 sin θ0 = −(−4) sin 30° = 2 rad/s, ω y = θ0 = 0

ω z = 300 + (−4) cos 30° = 296.536 rad/s T 1 I′ 2 1 I 2 (ω x + ω y2 ) + ωz = m 2m 2m 1 1 = (19.98 × 10−3 )((2) 2 + 0) + (1.08 × 10−3 )(296.536)2 = 47.5241 m 2 /s 2 2 2 V = gc cos θ = (9.81)(0.135) cos 30° = 1.1469 m 2 /s 2 m 2E = 2(47.5241 + 1.1469) = 97.3419 m 2 /s 2 m I β = ω z = (1.08 × 10−3 )(296.536) = 0.320259 m 2 /s m m

α m

=

I′  φ0 sin 2 θ 0 + β cos θ 0 m

= (19.98 × 10−3 )(−4) sin 2 30° + 0.320259 cos 30° = 0.257372 m 2 /s PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2201

PROBLEM 18.140 (Continued)

After dividing by m, Equation (2) of Problem 18.138 becomes  2E F ( x) =  −  m 

β2 m2 I m

 2 mα βx − 2 gcx  (1 − x 2 ) −  − =0  I′  m m   

  (0.320259) 2 (0.257372 − 0.320259 x) 2 − (2)(9.81)(0.135) x  (1 − x 2 ) − =0 97.3419 − −3 1.08 × 10 19.98 × 10−3  

(2.37324 − 2.6487 x)(1 − x 2 ) − 5.13342(0.80364 − x) 2 = 0

(a)

x = cos θ = 0.23732, 0.86603, 1.73

Roots are:

θ max = cos −1(0.23732) = 76.272° (b)

θ max = 76.3° 

By Equation (4) of Problem 18.137,

φ =

α − β cos θ 0.257372 − (0.320259)(0.23732) = = 9.6192 rad/s 2 I ′ sin θ (19.98 × 10−3 )sin 2 76.272°

ω z = 296.536 rad/s ψ = ω z − φ cos θ = 296.536 − (9.6192)(0.23732) 

(c)

spin: ψ = 294 rad/s  precession: φ = 9.62 rad/s 



φ = cos θ =

α − β cos θ =0 I ′ sin 2 θ α 0.257372 = β 0.320259

θ = 36.5° 

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PROBLEM 18.141* A homogeneous sphere of mass m and radius a is welded to a rod AB of negligible mass, which is held by a ball-and-socket support at A. The sphere is released in the position β = 0 with a rate of precession φ0 = 17 g /11a with no spin or nutation. Determine the largest value of β in the ensuing motion.

SOLUTION Conservation of angular momentum about the Z and z axes. Since the only external forces are the weight of the sphere and the reaction at A, a reasoning similar to that used in Problem 18.109 shows that the angular momentum is conserved about the Z and z axes. Choosing the principal axes Axyz shown (with y horizontal and pointing into the paper), we have ω = −φ cos β i + β j + (ψ − φ sin β )k

The moments of inertia are Iz =

2 2 ma 5

Ix = Iy =

2 2 22 2 ma + m(2a) 2 = ma 5 5

Angular momentum about A: H A = I xω x i + I y ω y j + I z ω z k HA = −

22 2  22 2  2 2 ma φ cos β i + ma β j + ma (ψ − φ sin β )k 5 5 5

We write H z = constant, or H A ⋅ Κ = constant Since i ⋅ K = − cos β ,

j ⋅ K = 0,

k ⋅ K = − sin β

22 2  ma φ cos β (− cos β ) 5 2 + ma 2 (ψ − φ sin β )(− sin β ) = constant 5

H0 ⋅ K = −

With the initial conditions φ = φ0 , ψ = β = 0, β = 0, we find that the constant is 11φ cos 2 β − (ψ − φ sin β ) sin β = 11φ0

22 5

ma 2φ0 . Thus,

(1)

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PROBLEM 18.141* (Continued)

We now write H z = constant: Hz =

2 2 ma (ψ − φ sin β ) = constant 5

and, from the initial conditions, we find that the constant is zero. Thus, ψ − φ sin β = 0

(2)

Conservation of energy. We have 1 ( I xω x2 + I yω y2 + I zω z2 ) 2 1  22 22 2  2 2 2  T =  ma 2φ2 cos 2 β + ma β + ma (ψ − φ sin β )2  2 5 5 5  T=

and selecting the datum at β = 0: V = −2mga sin β T + V = constant:

1  22 2 2 22 2  2 2 2  ma φ cos2 β + ma β + ma (ψ − φ sin β ) 2  − 2mga sin β = constant  2 5 5 5 

From the initial conditions φ = φ0 , ψ = β = 0, β = 0, we find that the constant is g 11φ2 cos 2 β + 11β 2 + (ψ − φ sin β )2 − 10 sin β = 11φ02 a

11 ma 2φ02 . 5

Thus, (3)

Substituting for ψ − φ sin β from Eq. (2) into Eqs. (1) and (3): Eq. (1):

11φ cos 2 β = 11φ0

(1′)

Eq. (3):

g 11φ2 cos 2 β + 11β 2 − 10 sin β = 11φ02 a

(3′)

Solving (1′) for φ,

φ = φ0 sec2 β

(4)

Substituting for φ from Eq. (4) into Eq. (3′): g 11(φ0 sec 2 θ ) 2 cos 2 β + 11β 2 − 10 sin β = 11φ02 a For the maximum value of β , we have β = 0 and Eq. (5) yields

(5)

 1  10 g − 1 = sin β 2 θ cos   11 a

φ 02 

φ02 =

10 g cos 2 β 11 a sin β

(6)

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PROBLEM 18.141* (Continued)

Given data:

g φ02 = 17 ⋅ Substituting into Eq. (6), 11 a

17 g 10 g cos 2 β = 11 a 11 a sin β

cos 2 β = 1.7sin β

Letting cos 2 β = 1 − sin 2 β , we have sin 2 β + 1.7sin β − 1 = 0

Solving the quadratic, sin β = −2.162 (impossible) and sinβ = 0.46244

β max = 27.5° 

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PROBLEM 18.142* A homogeneous sphere of mass m and radius a is welded to a rod AB of negligible mass, which is held by a ball-and-socket support at A. The sphere is released in the position β = 0 with a rate of precession φ = φ0 with no spin or nutation. Knowing that the largest value of β in the ensuing motion is 30°, determine (a) the rate of precession φ0 of the sphere in its initial position, (b) the rates of precession and spin when β = 30°.

SOLUTION Conservation of angular momentum about the Z and z axes. Since the only external forces are the weight of the sphere and the reaction at A, a reasoning similar to that used in Problem 18.109 shows that the angular momentum is conserved about the Z and z axes. Choosing the principal axes Axyz shown (with y horizontal and pointing into the paper), we have ω = −φ cos β i + β j + (ψ − φ sin β )k

The moments of inertia are 2 2 ma 5 2 22 2 I x = I y = ma 2 + m(2a) 2 = ma 5 5 Iz =

Angular momentum about A: H A = I xω x i + I y ω y j + I z ω z k HA = −

22 2  22 2  2 2 ma φ cos β i + ma β j + ma (ψ − φ sin β )k 5 5 5

We write H Z = constant, or H A ⋅ Κ = constant Since i ⋅ K = − cos β ,

j ⋅ K = 0,

k ⋅ K = − sin β 22 2  ma φ cos β (− cos β ) 5 2 + ma 2 (ψ − φ sin β )(− sin β ) = constant 5

H0 ⋅ K = −

With the initial conditions φ = φ0 , ψ = β = 0, β = 0, we find that the constant is 11φ cos 2 β − (ψ − φ sin β ) sin β = 11φ0

22 5

ma 2φ0 . Thus,

(1)

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PROBLEM 18.142* (Continued) We now write H z = constant: Hz =

2 2 ma (ψ − φ sin β ) = constant 5

and, from the initial conditions, we find that the constant is zero. Thus, ψ − φ sin β = 0

(2)

Conservation of energy. We have 1 ( I xω x2 + I yω y2 + I zω z2 ) 2 1  22 22 2  2 2 2  T =  ma 2φ2 cos 2 β + ma β + ma (ψ − φ sin β )2  2 5 5 5  T=

and, selecting the datum at β = 0: V = −2mga sin β T + V = constant:

1  22 2 2 22 2  2 2 2  ma φ cos 2 β + ma β + ma (ψ − φ sin β ) 2  − 2mga sin β = constant  2 5 5 5 

From the initial conditions φ = φ0 , ψ = β = 0, β = 0, we find that the constant is g 11φ2 cos 2 β + 11β 2 + (ψ − φ sin β )2 − 10 sin β = 11φ02 a

11 ma 2φ02 . 5

Thus, (3)

Substituting for ψ − φ sin β from Eq. (2) into Eqs. (1) and (3), Eq. (1):

11φ cos 2 β = 11φ0

(1′)

Eq. (3):

g 11φ2 cos 2 β + 11β 2 − 10 sin β = 11φ02 a

(3′)

Solving (1′) for φ,

φ = φ0 sec 2 β

(4)

Substituting for φ from Eq. (4) into Eq. (3′), g 11(φ0 sec2 θ )2 cos 2 β + 11β 2 − 10 sin β = 11φ02 a For the maximum value of β , we have β = 0 and Eq. (5) yields

(5)

 1  10 g − 1 = sin β 2 θ cos   11 a

φ 02 

φ02 =

10 g cos 2 β 11 a sin β

(6)

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PROBLEM 18.142* (Continued)

(a)

Making β = 30° in Eq. (6), we have

φ02 = (b)

10 g 0.75 15 g = 11 a 0.5 11 a

φ0 =

15 g  11 a

φ = 2

20 g  33 a

Substituting for φ0 in Eq. (4), and making β = 30°:

φ = φ0 sec2 30° =

15 g  4  15(16) g =   11 a  3  11(9) a

Substituting for in Eq. (2), and making β = 30° : 1 2

ψ = φ sin 30° = φ

ψ =

20 g  33 a

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PROBLEM 18.143* Consider a rigid body of arbitrary shape which is attached at its mass center O and subjected to no force other than its weight and the reaction of the support at O. (a) Prove that the angular momentum H O of the body about the fixed Point O is constant in magnitude and direction, that the kinetic energy T of the body is constant, and that the projection along H O of the angular velocity ω of the body is constant. (b) Show that the tip of the vector ω describes a curve on a fixed plane in space (called the invariable plane), which is perpendicular to HO and at a distance 2T/H O from O. (c) Show that with respect to a frame of reference attached to the body and coinciding with its principal axes of inertia, the tip of the vector ω appears to describe a curve on an ellipsoid of equation I xω x2 + I y ω y2 + I zω z2 = 2T = constant

The ellipsoid (called the Poinsot ellipsoid ) is rigidly attached to the body and is of the same shape as the ellipsoid of inertia, but of a different size.

SOLUTION (a)

From Equation (18.27),

 ΣM O = H O

Since

 = 0. ΣM O = 0, H O

Conservation of energy:

T + V = constant

Since

V = 0,

H O = constant

(1) 

T = constant

(2) 

For a rigid body rotating about Point O,

(

1 I xω x2 + I yω y2 + I zω z2 2 H O = I xω x i + I y ω y j + I z ω z k T=

)

ω = ωx i + ω y j + ω z k H O ⋅ω = I xωx2 + I yω y2 + I zω z2 = 2T

Let β be the angle between the vectors H O and ω. H O ⋅ ω = H Oω cos β

The projection of ω along H O is ω cos β

ω cos β =

2T = constant HO

ω cos β = constant (3) 

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PROBLEM 18.143* (Continued)

(b)

ω cos β is the perpendicular distance from the invariable plane. This distance is equal to

(c)

For a frame of reference attached to the body, the moments of inertia with respect of orthogonal axes of the frame do not change. I xω x2 + I yω y2 + I zω z2 = 2T

a1 =

Let

Then

ω x2 a12

+

ω y2 b12

+

ω z2 c12

2T HO

.

(4)

2T , Ix

=1

b1 =

2T , Iy

c1 =

2T Iz

(5)

(6)

which is the equation of an ellipsoid.

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PROBLEM 18.144* Referring to Problem 18.143, (a) prove that the Poinsot ellipsoid is tangent to the invariable plane, (b) show that the motion of the rigid body must be such that the Poinsot ellipsoid appears to roll on the invariable plane. [Hint: In part a, show that the normal to the Poinsot ellipsoid at the tip of ω is parallel to H O . It is recalled that the direction of the normal to a surface of equation F ( x, y, z ) = constant at a Point P is the same as that of grad F at Point P.]

SOLUTION (a)

From Problem 18.143, the equation of the Poinsot ellipsoid is I xω x2 + I y ω y2 + I zω z2 = 2T = constant

Let

F (ω x , ω y , ω z ) = I xω x2 + I yω y2 + I zω z2

grad F =

∂F ∂F ∂F i+ j+ k ∂ω x ∂ω y ∂ω z

= 2 I xω x i + 2 I y ω y j + 2 I z ω z k = 2H O

The direction of the normal at a point on the surface of the ellipsoid is parallel to grad F, which in turn is parallel to H O . Since H O is normal also to the invariable plane, it follows that the Poinsot ellipsoid is tangent to the invariable plane at the point common to the plane and the ellipsoid. (b)

The Poinsot elipsoid moves with the body. Thus, its angular velocity is ω, the angular velocity of the body. Since Point O is regarded as fixed, the angular velocity vector lies along the axis of rotation, or the locus of points of zero velocity. Thus, the velocity of the point of contact of the Poinsot ellipsoid with the invariable plane is zero. The Poinsot ellipsoid rolls without slipping on the invariable plane.

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PROBLEM 18.145* Using the results obtained in Problems 18.143 and 18.144, show that for an axisymmetrical body attached at its mass center O and under no force other than its weight and the reaction at O, the Poinsot ellipsoid is an ellipsoid of revolution and the space and body cones are both circular and are tangent to each other. Further show that (a) the two cones are tangent externally, and the precession is direct, when I < I ′, where I and I ′ denote, respectively, the axial and transverse moment of inertia of the body, (b) the space cone is inside the body cone, and the precession is retrograde, when I > I ′.

SOLUTION Let I x = I y = I ′ and I z = I so that the z axis is the symmetry axis. Then, the equation of the Poinsot ellipsoid (Equation (4) of Problem 18.143) becomes

(

)

I ′ ω x2 + ω y2 + I ω z2 = 2T = constant

which is the equation of an ellipsoid of revolution. It follows that the tip of ω describes circles on both the Poinsot ellipsoid and on the invariable plane, and that the vector ω itself describes circular body and space cones. The Poinsot ellipsoid, the invariable plane, and the body and space cones are shown below for cases a and b. (a)

I < I′

(b) I > I ′

Retrograde Precession

Direct Precession

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PROBLEM 18.146* Refer to Problems 18.143 and 18.144. (a) Show that the curve (called polhode) described by the tip of the vector ω with respect to a frame of reference coinciding with the principal axes of inertia of the rigid body is defined by the equations I xω x2 + I y ω y2 + I zω z2 = 2T = constant I x2ω x2

+

I y2ω y2

+

I z2ω z2

=

H O2

= constant

(1) (2)

and that this curve can, therefore, be obtained by intersecting the Poinsot ellipsoid with the ellipsoid defined by Eq. (2). (b) Further show, assuming I x > I y > I z , that the polhodes obtained for various values of H O have the shapes indicated in the figure. (c) Using the result obtained in part b, show that a rigid body under no force can rotate about a fixed centroidal axis if, and only if, that axis coincides with one of the principal axes of inertia of the body, and that the motion will be stable if the axis of rotation coincides with the major or minor axis of the Poinsot ellipsoid (z or x axis in the figure) and unstable if it coincides with the intermediate axis ( y axis).

SOLUTION (a)

Equation (1) expresses conservation of energy as shown in the solution to Problem 18.143. It is the equation of the Poinsot ellipsoid. Let

a1 =

Then

ω x2 a12

2T , b1 = Ix +

ω y2 b12

+

ω z2 c12

2T , c1 = Iy

2T Iz

=1

(3)

which is the equation of an ellipsoid. Equation (2) expresses the constancy of H O2 = ( H O ) 2x + ( H O ) 2y + ( H O ) 2z , the square of the magnitude of the angular momentum vector. a2 =

Let

Then

ω x2 a22

+

ω y2 b22

+

ω z2 c22

HO H H , b2 = O , c2 = O Ix Iy Iz

=1

(4)

which is the equation of a second ellipsoid. Since the coordinates ω x , ω y , ω z of the tip of the vector ω must satisfy both Equations (1) and (2), the curve described by the tip of ω is the intersection of the two ellipsoids. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2213

PROBLEM 18.146* (Continued)

(b)

Assume I x > I y > I z . Then

a1 < b1 < c1 and a2 < b2 < c2 .

Thus, for both ellipsoids, the minor axis is directed along the x axis, the intermediate axis along the y axis, and the major axis along the z axis. However, because the ratio of the major to minor semiaxis is

Ix Iz

for the Poinsot ellipsoid and is

Ix Iz

for the second ellipsoid, the deviation from a spherical shape

is more pronounced in the second ellipsoid. The largest ellipsoid of the second type to be in contact with the Poinsot ellipsoid will lie outside that ellipsoid and touch it at its points of intersection with the x axis, and the smallest will lie inside the Poinsot ellipsoid and touch it at its points of intersection with the z axis (see left-hand sketch). All ellipsoids of the second type comprised between these two will intersect the Poinsot ellipsoid along the curves called polhodes, as shown in the right-hand figure.

Note that the ellipsoid of the second type, which has the same intermediate axis as the Poinsot ellipsoid, intersects that ellipsoid along two ellipses whose planes contain the y axis. These curves are not polhodes, since the tip of ω will not describe them, but they separate the polhodes into four groups. Two groups loop around the minor axis (x axis) and the other two around the major axis (z axis). (c)

If the body is set to spin about one of the principal axes, the Poinsot ellipsoid will remain in contact with the invariable plane at the same point (on the x, y, or z axis); the rotation is steady. In any other case, the point of contact will be located on one of the polhodes, and the tip of ω will start describing that polhode, while the Poinsot ellipsoid rolls on the invariable plane. A rotation about the minor or the major axis (x or z axis) is stable. If that motion is disturbed, the tip of ω will move to a very small polhode surrounding that axis and stay close to its original position. On the other hand, a rotation about the intermediate axis (y axis) is unstable. If that motion is disturbed, the tip of ω will move to one of the polhodes located near that axis and start describing it, departing completely from its original position and causing the body to tumble.

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PROBLEM 18.147 Three 25-lb rotor disks are attached to a shaft which rotates at 720 rpm. Disk A is attached eccentrically so that its mass center is 14 in. from the axis of rotation, while disks B and C are attached so that their mass centers coincide with the axis of rotation. Where should 2-lb weights be bolted to disks B and C to balance the system dynamically?

SOLUTION The system is dynamically balanced if the effective forces are equivalent to zero. Let Points A, B, and C be the points where the disks are attached to the shaft and let Point G be the mass center of disk A. Let m be the mass of each rotor and m′ the magnitude of each added mass.

Since

ω = constant, H G = 0

for each disk. We treat the added masses as particles so that their moments of inertia about their mass centers are negligible. The effective force of disk A is. ( FA )eff = −mrAω 2 j

The effective forces of the added masses are m′rBω 2 j and m′rC ω 2 j,

respectively for the masses added to disks B and C. ΣFeff = 0: − mrAω 2 j + m′rBω 2 j + m′rC ω 2 j = 0 m rB + rC = rA m′ Σ( M 0 )eff = 0 4i × ( −mrAω 2 j) + 10i × (m′rBω 2 j) + 16i × (m′rC ω 2 j) m 10rB + 16r2 = 4 rA m′

(1)

(2)

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PROBLEM 18.147 (Continued)

Data:

rA =

1 in. = 0.25 in. 4

m 25 lb = = 12.5 m′ 2 lb

rA + rB = 3.125 in.

(1)′

10rA + 16rB = 12.5 in.

(2)′

Solving the simultaneous equations. rB = 6.25 in.

rC = −3.125

Placement of added masses. On B : 6 14 in. below shaft. On C : 3 81 in. above shaft 

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PROBLEM 18.148 A homogeneous disk of mass m = 5 kg rotates at the constant rate ω1 = 8 rad/s with respect to the bent axle ABC, which itself rotates at the constant rate ω2 = 3 rad/s about the y axis. Determine the angular momentum H C of the disk about its center C.

SOLUTION Using frame Cx′y ′z ′: I x′ = I y ′ =

1 2 mr 4

1 2 mr 2 H C = I y′ω2 j + I z′ω1k I z′ =

= HC =

1 2 mr (ω2 j + 2ω1k ) 4 1 (5 kg)(0.25 m) 2 [(3 rad/s)j + 2(8 rad/s)k ] 4 H C = (0.234 kg ⋅ m 2 /s)j + (1.250 kg ⋅ m 2 /s)k 

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PROBLEM 18.149 A rod of uniform cross section is used to form the shaft shown. Denoting by m the total mass of the shaft and knowing that the shaft rotates with a constant angular velocity ω, determine (a) the angular momentum H G of the shaft about its mass center G, (b) the angle formed by HG and the axis AB, (c) the angular momentum of the shaft about Point A.

SOLUTION Length of rod:

L = 2r + 2π r = 8.2832r

Angular velocity:

ω = ωi H G = I xω i − I xzω k

Angular momentum: Calculation of I x and I xz : Let ρ = mass per unit length =

m . L

For portions AC and DB, I x = 0, I xz = 0

For portion CG, use polar coordinate θ . x = − r (1 − cos θ ), z = −r sin θ



I x = z 2 dm =



I xz = xzdm =

π

ρ r 3,

Likewise, for portion GD,

Ix =

Total:

I x = πρ r 3 =

(a)

Angular momentum H G .

HG =

2

 

π 0

π 0

r 2 sin 2 θρ rdθ =

dm = ρ rdθ

π 2

ρr3

r 2 (1 − cos θ ) sin θρ rdθ = 2 ρ r 3

I xz = 2 ρ r 3

π mr 3 , L

I xz = 4 ρ r 3 =

4mr 3 L

π mr 3 4mr 3 ωi − ω k = mr 2ω (0.37927i − 0.48291k ) L L H G = mr 2ω (0.379i − 0.483k ) 

H G = mr 2ω (0.37927 2 + 0.482912 )1/ 2 = 0.61404 mr 2ω

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PROBLEM 18.149 (Continued)

(b)

Angle formed by H G and the axis AB . H G ⋅ i = 0.37927mr 2ω

cos θ =

(c)

H G ⋅ i 0.37927mr 2ω = = 0.61766 HG 0.61404mr 2ω

θ = 51.9° 

Angular momentum about Point A : H A = H G + rG /A × (mv ) H A = mr 2ω (0.379i − 0.483k ) 

H A = HG + 0

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PROBLEM 18.150 A uniform rod of mass m and length 5a is bent into the shape shown and is suspended from a wire attached at Point B. Knowing that the rod is hit at Point A in the negative y direction and denoting the corresponding impulse by −( F Δt ) j, determine immediately after the impact (a) the velocity of the mass center G, (b) the angular velocity of the rod.

SOLUTION Moments and products of inertia: Part OA

1 5

m

AB

1 5

m

BC

1 5

CD

1 5

DE

1 5

Σ

Iy

Ix

m

( 15 m ) ×

(

1 12

a + 2

+ a

1 2 a 4

2

)

×

(

1 12

a +a + a 2

2

1 4

)

( m)(

m

( 15 m ) ( 121 a 2 )

0

m

( 15 m ) ( 14 a 2 )

( 15 m ) ( 13 a 2 )

m

m

( m)(

1 4

( 15 m )

1 5

1 4

a

2

( 15 m ) ×

(

1 12

a 2 + 14 a 2 + 14 a 2

1 5

)

0.35ma 2

1 2 a 3

2

(

1 12

a 2 + a 2 + 14 a 2

0.66667ma 2

I xy

( 15 m ) ( a 2 + 14 a 2 )

( 15 m ) ( − 12 a 2 )

( 15 m )

)

( 15 m ) ×

)

Iz

×

(

1 12

a + 2

1 2 a 4

+ a 1 4

2

( 15 m ) ( 121 a 2 ) ( 15 m ) ×

)

(

1 12

a + 2

1 2 a 4

)

( 15 m ) ( − 14 a 2 ) 0

+ a 1 4

2

)

( 15 m ) ( − 14 a 2 )

( 15 m ) ( a 2 + 14 a 2 )

( 15 m ) ( − 12 a 2 )

0.75ma 2

−0.3ma 2

 1   1   1  1  I xz =  m   a 2  +  m  a 2  = 0.2ma 2  5   2   5  2   1  1   1  1  I yz =  m   − a 2  +  m   − a 2  = −0.1ma 2  5  4   5  4 

Angular momentum about the mass center.

( H G ) x = I xωx − I xyω y − I xzωz = 0.35ma 2ωx + 0.3ma 2ω y − 0.2ma 2ωz ( H G ) y = − I xyω x + I yω y − I yzω z = 0.3ma 2ωx + 0.66667ma 2ω y + 0.1ma 2ωz ( H G ) z = − I xzω x − I yzω z + I zω z = − 0.2ma 2ωx + 0.1ma 2ω y + 0.75ma 2ωz PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2220

PROBLEM 18.150 (Continued)

vy = 0

Constraint of the supporting cable: Impulse-momentum principle:

(a)

Before impact,

F(Δt ) + T Δtj = mv

Linear momentum:

0 = mvx ,

H G = 0.

Resolve into components.

−F Δt + T Δtj = 0,

T Δt = F Δt ,

vx = 0,

(b)

v = 0,

Angular momentum, moments about G:

0 = mvz v = 0

vz = 0

rA/G × FΔt = HG

a   j − ai  × [−( F Δt ) j] = (aF Δt )k = ( H G ) x i + ( H G ) y j + ( H G ) z k 2  

Using expressions for ( H G ) x , (H G ) y , and ( H G ) z and resolving into components, i:

0 = 0.35ma 2ω x + 0.3ma 2ω y − 0.2ma 2ω z

j:

0 = 0.3ma 2ω x + 0.66667ma 2ω y + 0.1ma 2ω z

k : aF Δt = −0.2ma 2ω x + 0.1ma 2ω y + 0.75ma 2ω z Solving,

ω x = 2.5

( F Δt ) , ma

ω y = −1.454

( F Δt ) , ma

ω z = 2.19

( F Δt ) ma

 F Δt  ω=  (2.50i − 1.454 j + 2.19k )   ma 

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PROBLEM 18.151 A four-bladed airplane propeller has a mass of 160 kg and a radius of gyration of 800 mm. Knowing that the propeller rotates at 1600 rpm as the airplane is traveling in a circular path of 600-m radius at 540 km/h, determine the magnitude of the couple exerted by the propeller on its shaft due to the rotation of the airplane.

SOLUTION We assume senses shown for ω x, ω y , and v. v = 540 km/h = 150 m/s  2π rad    60 s 

ω x = 1600 rpm 

= 167.55 rad/s

ωy =

v

ρ

=

150 m/s = 0.25 rad/s 600 m

I x = mk 2 = (160 kg)(0.8 m) 2 = 102.4 kg ⋅ m 2

Angular momentum about G: H G = I xω x i + I y ω y j

Eq. (18.22),

 = (H  ) H G G Gxyz + Ω × H G = 0 + ω y j × ( I xω x i + I y ω y j)

 = − I ω ω k = −(102.4 kg ⋅ m 2 )(167.55 rad/s)(0.25 rad/s)k H G x x y  = −(4289 N ⋅ m)k = − (4.29 kN ⋅ m)k H G  = −(4.29 kN ⋅ m)k , and the couple exerted by The couple exerted on the propeller, therefore, must be M = H G the propeller on its shaft is −M = (4.29 kN ⋅ m)k. 4.29 kN ⋅ m 

Magnitude of couple.

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PROBLEM 18.152 A 2.4-kg piece of sheet steel with dimensions 160 × 640 mm was bent to form the component shown. The component is at rest (ω = 0) when a couple M 0 = (0.8 N ⋅ m)k is applied to it. Determine (a) the angular acceleration of the component, (b) the dynamic reactions at A and B immediately after the couple is applied.

SOLUTION m = 2.4 kg, b = 160 mm = 0.16 m

Area of sheet metal:

A = b 2 + (2b)b + b2 = 4b 2 = 0.1024 m 2

Let

ρ=

m 2.4 = A 0.1024 = 23.4375 kg/m 2 = mass per unit area

Angular velocity and angular acceleration:

ω =ωk α =αk

H G = − I xzω i − I yzω j + I zω k

Angular momentum about G:

= − I yzω j + I zω k

Let the frame of reference Gxyz be rotating with angular velocity

Ω = ω =ωk

 = (H  ) H G G Gxyz + Ω × H G = − I yzα j + I zα k + I yz ω 2 i

Required moment and product of inertia: I mass = ρ I area Part plate A

plate AB

A

Iz

2

1 2 1  b + b2  b  6 2 

b

2b 2 2

plate B

b

Σ

4b 2

I yz 2

1 (2b)b3 12 1 2 1  b + b2  b  6 2  b4

1  b 2 (b)  b  2  0

2

 1  b 2 (−b)  − b   2 

I z = ρ b4

= (23.4375)(0.16) 4 = 0.01536 kg ⋅ m 2 I yz = ρ b 4

= 0.01536 kg ⋅ m 2

b4

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PROBLEM 18.152 (Continued)

a=0

Since the mass center lies on the rotation axis, ΣF = A + B = m a B = −A

ΣM G = M 0 k + bk × A + ( −bk ) × B = M 0 k + 2bk × ( Ax i + Ay j) = M 0 k + 2bAx j − 2bAy i  ΣM G = H G

(a)

k:

(b)

j: i:

Resolve into components.

M 0 = I zα α = 2bAx = − I yzα

M0 0.8 = = 52.083 rad/s 2 0.01536 Iz

Ax = −

− 2bAy = I yzω 2 = 0

I yzα 2b

=−

Ay = 0

α = 52.1 rad/s 2 

(0.01536)(52.083) = −2.50 N (2)(0.16)

A = −(2.50 N)i  B = (2.50 N)i 

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PROBLEM 18.153 A homogeneous disk of weight W = 6 lb rotates at the constant rate ω1 = 16 rad/s with respect to arm ABC, which is welded to a shaft DCE rotating at the constant rate ω2 = 8 rad/s. Determine the dynamic reactions at D and E.

SOLUTION Angular velocity of shaft DE and arm CBA: Ω = ω2 i ω = ω2 i + ω1 j

Angular velocity of disk A: Angular velocity about its mass center A:

HA = I xω x i + I y ω y j + I zω z k = I xω2 i + I yω1 j =

1 2 1 mr ω2 i + mr 2ω1 j 4 2

Let the reference frame Oxyz be rotating with angular velocity Ω = ω2 i.  = (H  ) H A A Oxyz + Ω × HA 1 2 1 mr ω 2 i + mr 2ω1 j + ω2 i × HA 4 2 1 2 1 2 1 = mr ω 2 i + mr ω1 j + mr 2ω1ω2 k 4 2 2 vA = ω2 i × rA/O =

Velocity of Point A:

= ω2 i × (−bk + cj) = bω2 j + cω2 k

Acceleration of Point A:

a A = ω 2 j × rA/O + ω2 j × vA

a A = (bω 2 − cω22 )j + (cω 2 + bω2 )k Consider the system of particles consisting of the shaft, the arm, and the disk. Neglect the mass of the arm. ΣF = ma A Dy j + Dz k + E y j + Ez k = maA

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PROBLEM 18.153 (Continued)

Resolve into components. Dy + E y = m(bω 2 − cω22 ) Dz + Ez = m(cω 2 + bω22 )  =H  +r ΣM = H D

D

A

A/D

× maA

 + (li + cj − bk ) × ma ( M 0 )i + 2li × ( E y j + E z k ) = H A A

1  1  ( M 0 )i − 2lEz j + 2lE y k = m  r 2 + b 2 + c 2  ω 2 i + m  r 2ω1 − lcω 2 − lbω22  j 4 2     1  + m  r 2ω1ω2 + lbω 2 − lcω22  k 2  

i:

1  M 0 = m  r 2 + b2 + c 2  ω 2 4 

k:

Ey =

m1 2 2  r ω1ω2 + lbω 2 − lcω2  2l  2 

Dy =

m 1 2 2  − r ω1ω2 + lbω 2 − lcω2  2l  2 

Ez =

m 1 2  2  lcω 2 + lbω2 − r ω1  2l  2 

Dz =

m 1 2  2  lcω 2 + lbω2 + r ω1  2l  2 

j:

Data:

6 = 0.186335 r = 8 in. = 0.66667 ft 32.2 b = c = 9 in. = 0.75 ft l = 12 in. = 1.0 ft

W = 6 lb. m =

ω1 = 16 rad/s, ω1 = 0, ω2 = 8 rad/s, ω 2 = 0 Dy =

0.186335   1  2 2  −   (0.66667) (16)(8) + 0 − (1.0)(0.75)(8)  = −7.12 lb (2)(1.0)   2  

Dz =

0.186335 [0 + (1.0)(0.75)(8)2 + 0] = 4.47 lb (2)(1.0) D = −(7.12 lb)j + (4.47 lb)k 

Ey =

 0.186335  1  (0.66667) 2 (16)(8) + 0 − (1.0)(0.75)(8) 2  = −1.822 lb  (2)(1.0)  2  

Ez =

0.186335 [0 + (1.0)(0.75)(8) 2 + 0] = 4.47 lb (2)(1.0) E = −(1.822 lb)j + (4.47 lb)k 

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PROBLEM 18.154 A 48-kg advertising panel of length 2a = 2.4 m and width 2b = 1.6 m is kept rotating at a constant rate ω1 about its horizontal axis by a small electric motor attached at A to frame ACB. This frame itself is kept rotating at a constant rate ω 2 about a vertical axis by a second motor attached at C to the column CD. Knowing that the panel and the frame complete a full revolution in 6 s and 12 s, respectively, express, as a function of the angle θ , the dynamic reaction exerted on column CD by its support at D.

SOLUTION Use principal axes x′, y ′, z ′ as shown. Moments of inertia:

Kinematics:

1 I x ′ = m( a 2 + b 2 ) 3 1 1 I y′ = ma 2 , I z′ = mb 2 3 3  θ = ω1

ω x′ = ω2 sin θ , ω y ′ = ω2 cos θ , ω z ′ = ω1 ω x′ = ω1ω2 cos θ , ω y′ = −ω1ω2 sin θ , ω z′ = 0 Since the acceleration of the mass center is zero, the resultant force acting on the column CD is zero. R=0 

Euler’s equations of motion for the plate: ΣM x′ = I x′ω x′ − ( I y′ − I z′ )ω y′ω z′ = I x′ω1ω2 cos θ − ( I y′ − I z ′ )ω1ω2 cos θ = ( I x′ + I z′ − I y′ )ω1ω2 cos θ =

2 2 mb ω1ω2 cos θ 3

ΣM y′ = I y′ω y ′ − ( I z ′ − I x′ )ω x′ω z′ = − I y ω1ω2 sin θ − ( I z − I x )ω1ω2 sin θ = ( I x′ − I y′ − I z′ )ω1ω2 sin θ = 0 ΣM z′ = I z′ω z′ − ( I x′ − I y )ω x′ω y′ = 0 − ( I x′ − I y )ω22 sin θ cos θ 1 = − mb 2ω22 sin θ cos θ 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2227

PROBLEM 18.154 (Continued)

ΣM = =

Resolve into components:

2 2 1 mb ω1ω2 cos θ i ′ − mb2ω22 sin θ cos θ k 3 3 2 2 1 mb ω1ω2 cos θ (cos θ i + sin θ j) − mb 2ω22 sin θ cos θ k 3 3

ΣM x =

2 2 mb ω1ω2 cos 2 θ 3

2 2 mb ω1ω2 sin θ cos θ 3 1 ΣM z = − mb 2ω22 sin θ cos θ 3

ΣM y =

Data:

m = 48 kg,

ω1 =

b = 0.8 m

2π 2π = 1.0472 rad/s, ω2 = = 0.5236 rad/s 6 12

2 (48)(0.8) 2 (1.0472)(0.5236) cos 2 θ 3 = 11.23cos 2 θ N ⋅ m

ΣM x =

2 (48)(0.8) 2 (1.0472)(0.5236)sin θ cos θ 3 = 11.23sin θ cos θ N ⋅ m

ΣM y =

1 ΣM z = − (48)(0.8) 2 (0.5236)2 sin θ cos θ 3 = −2.81sin θ cos θ N ⋅ m M D = (11.23 N ⋅ m) cos 2 θ i + (11.23 N ⋅ m) sin θ cos θ j − (2.81 N ⋅ m) sin θ cos θ k 

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PROBLEM 18.155 A 2500-kg satellite is 2.4 m high and has octagonal bases of sides 1.2 m. The coordinate axes shown are the principal centroidal axes of inertia of the satellite, and its radii of gyration are k x = k z = 0.90 m and k y = 0.98 m. The satellite is equipped with a main 500-N thruster E and four 20-N thrusters A, B, C, and D, which can expel fuel in the positive y direction. The satellite is spinning at the rate of 36 rev/h about its axis of symmetry G y , which maintains a fixed direction in space, when thrusters A and B are activated for 2 s. Determine (a) the precession axis of the satellite, (b) its rate of precession, (c) its rate of spin.

SOLUTION  2π rad  1 h     rev  3600 s  = 0.062832 rad/s m = 2500 kg

ω 0 = (36 rev/h) 

I x = mk x2 = (2500)(0.90) 2 = 2025 kg ⋅ m 2



I z = I x = 2025 kg ⋅ m 2 I y = mk y2 = (2500)(0.98) 2 = 2401 kg ⋅ m 2 (H G )0 = I yω0 j = (2401)(0.062832) j = (150.86 kg ⋅ m 2 /s) j a = 0.6 m, b = 0.6 + 1.2 cos 45° = 1.4485

When thrusters A and B are activated, M G = −b( FA + FB )i

= −(1.4485)(40)i  = −(57.941 N ⋅ m)i

Angular momentum after 2 s: H G = (H G )0 + M G (Δt ) = 150.86 j + ( −57.941)(2)i = −(115.88 kg ⋅ m 2 /s)i + (150.86 kg ⋅ m 2 /s) j

ωx =

Hx 115.88 =− = −0.057225 rad/s = −32.788 rev/h 2025 Ix

ωy =

Hy

ωz =

Hz =0 Iz

Iy

= 36 rev/h

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PROBLEM 18.155 (Continued)

(a)

Precession axis: tan θ = −

H x 115.88 = θ = 37.529° H y 150.86

tan γ = −

ωx ωy

32.788 36 γ = 42.327°

θ x = 52.5°, θ y = 37.5°, θ z = 90° 

=





γ − θ = 4.798° ω = ω x2 + ω y2 = 48.693 rev/h



Law of sines.

ϕ sin γ

(b)

=

ϕ =

ψ ω = sin(γ − θ ) sin θ 48.693sin 42.327° sin 37.529°

ϕ = 53.8 rev/h  (c)

ψ =

48.693sin 4.798° sin 37.529°

ψ = 6.68 rev/h 

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PROBLEM 18.156 A thin disk of weight W = 8 lb rotates with an angular velocity ω 2 with respect to arm OA, which itself rotates with an angular velocity ω1 about the y axis. Determine (a) the couple M1 j which should be applied to arm OA to give it an angular acceleration α1 = (6 rad/s 2 ) j with ω1 = 4 rad/s, knowing that the disk rotates at the constant rate ω 2 = 12 rad/s, (b) the force-couple system representing the dynamic reaction at O at that instant. Assume that arm OA has a negligible mass.

SOLUTION Angular velocity of arm OA:

Ω = ω1 j

Angular velocity of disk A:

ω = ω1 j + ω 2 k

Angular momentum about its mass center A: H A = I xω x i + I y ω y j + I z ω z k = I yω1 j + I zω2 k =

1 2 1 mr ω1 j + mr 2ω2k 4 2

Let the reference Oxyz be rotating with angular velocity Ω = ω2 j.  = (H  ) H A A Oxyz + Ω × H A

1 2 1 mr ω 1 j + mr 2ω 2 k + ω1 j × H A 4 2 1 1 1 = mr 2ω1ω 2 i + mr 2ω 1 j + mr 2ω 2 k 2 4 2 =

Velocity of Point A:

v A = ω1 j × rA/O = ω1 j × (bi − cj) = −bω1k

Acceleration of Point A:

a A = ω 1 j × rA/O + ω1 j × v A = −bω 12 i − bω 1k

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PROBLEM 18.156 (Continued)

Consider the system of particles consisting of the arm OA and disk A. Neglect the mass of the arm.

ΣF = ma A : Rx i + Rz k = −mbω 12 i − mbω 1k Rx = − mbω 12 Rz = − mbω 1  =H  + r × ma ΣM O = H O A A/O A

1  ( M O ) x i + ( M O ) y j + ( M O ) z k = m  r 2ω 1ω 2 + bcω 1  i 2   1  1  + m  r 2 + b 2  ω 1 j + m  r 2ω 2 − bcω 12  k 4 2    

Data:

W = 8 lb 8 = 0.24845 lb ⋅ s 2 /ft 32.2 r = 5 in. = 0.41667 ft b = 16 in. = 1.33333 ft c = 10 in. = 0.83333 ft

m=

ω 1 = 4 rad/s ω 1 = 6 rad/s 2 ω 2 = 12 rad/s ω 2 = 0 (a)

Required couple: The required couple is the y-component of the couple at Point O. 1  (M O ) y = (0.24845)  (0.41667) 2 + (1.33333) 2  (6) j 4  

(M O ) y = (2.71 lb ⋅ ft) j 

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PROBLEM 18.156 (Continued)

(b)

Dynamic reaction at Point O. Rx = −(0.24845)(1.33333)(4) 2 = −5.30 lb,

Ry = 0

Rz = −(0.24845)(1.33333)(6) = −1.988 lb R = −(5.30 lb)i − (1.988 lb)k 

 1   ( M O ) x = (0.24845)   (0.41667) 2 (4)(12) + (1.33333)(0.83333)(6)   2   = 2.69 lb ⋅ ft ( M O ) z = (0.24845)[0 − (1.33333)(0.83333)(4) 2 ] = −4.42 lb ⋅ ft M O = (2.69 lb ⋅ ft)i − (4.42 lb ⋅ ft)k 

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PROBLEM 18.157 A homogeneous disk of mass m connected at A and B to a fork-ended shaft of negligible mass which is supported by a bearing at C. The disk is free to rotate about its horizontal diameter AB and the shaft is free to rotate about a vertical axis through C. Initially, the disk lies in a vertical plane (θ0 = 90°) and the shaft has an angular velocity φ0 = 8 rad/s. If the disk is slightly disturbed, determine for the ensuring motion (a) the minimum value of φ, (b) the maximum value of θ.

SOLUTION Place the origin at the center of mass and let Oxyz be a principal axis frame of reference with the y axis directed along the moving axle AB. Let the Z axis lie along the fixed axle. Useful unit vectors are i, j and k along the x, y, z axes and K along the Z axis. K = −i sin θ + k cos θ

Angular velocity:

ω = ϕ k + θ j ω = −ϕ sin θ i + θ j + ϕ cos θ k

Moments of inertia: Angular momentum about O.

I x = e1 I y ,

I z = e2 I y

H O = I xω x i + I y ω y j + I z ω z k = I y (−e1 ϕ sin θ i + θ j + e2ϕ cos θ k )

The moment about the fixed Z axis is zero, hence, H O ⋅ K = constant. H O ⋅ K = I y (e1 sin 2 θ + e2 cos 2 θ )ϕ = I y C1

ϕ =

C1 e1 sin θ + e2 cos 2 θ 2

C1 = (e1 sin 2 θ 0 + e2 cos 2 θ0 )ϕ0

Twice the kinetic energy:

2T = I xω x2 + I y ω y2 + I zω z2 = constant 2T = I y [(e1 sin 2 θ + e2 cos 2 θ )ϕ 2 + θ 2 ] = I y (C1ϕ + θ 2 ) = I y C2 2

θ = C2 − C1ϕ C2 = θ02 + C1ϕ0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2234

PROBLEM 18.157 (Continued)

Data:

e1 = 1 e2 = 2 e1 sin 2 θ + e2 cos 2 θ = 1 + cos 2 θ

θ 0 = 90° θ0 = 0 ϕ0 = 16 rad/s C1 = (1 + cos 2 90°)(8) = 8 rad/s

ϕ = (a)

ϕmin =

8 1 + cos θ 8 =4 1 + cos θ

ϕmin = 4.00 rad/s 

C2 = 0 + (8)(8) = 64 (rad/s)2 θ 2 = C2 − C1ϕ

(b)

(θ 2 ) max = C2 − C1ϕmin = 64 − (8)(4) = 32 (rad/s) 2

θmax = 5.66 rad/s 

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PROBLEM 18.158 The essential features of the gyrocompass are shown. The rotor spins at the rate ψ about an axis mounted in a single gimbal, which may rotate freely about the vertical axis AB. The angle formed by the axis of the rotor and the plane of the meridian is denoted by θ , and the latitude of the position on the earth is denoted by λ . We note that line OC is parallel to the axis of the earth, and we denote by ωe the angular velocity of the earth about its axis. (a) Show that the equations of motion of the gyrocompass are I ′θ + I ω z ωe cos λ sin θ − I ′ω e2 cos 2 λ sin θ cos θ = 0 I ω z = 0

where ω z is the rectangular component of the total angular velocity ω along the axis of the rotor, and I and I′ are the moments of inertia of the rotor with respect to its axis of symmetry and a transverse axis through O, respectively. (b)

Neglecting the term containing ωe2 , show that for small values of θ, we have

θ +

I ω zωe cos λ θ =0 I′

and that the axis of the gyrocompass oscillates about the north-south direction.

SOLUTION 

(a)

Angular momentum about O. We select a frame of reference Oxyz attached to the gimbal. The angular velocity of Oxyz with respect to a Newtonian frame is Ω = ωe K + θ j

 

where

K = − cos λ sin θ i + sin λ j + cos λ cos θ k

Thus,

Ω = −ωe cos λ sin θ i + (θ + ωe sin λ ) j + ωe cos λ cos θ k

(1)

The angular velocity ω of the rotor is obtained by adding its spin ψ k to Ω. Setting

ψ + ωe cos λ cos θ = ω z .



We have

ω = −ωe cos λ sin θ i + (θ + ωe sin λ ) j + ω z k

(2)

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PROBLEM 18.158 (Continued)

The angular momentum H O of the rotor is H O = I xω z i + I yω y j + I zω z k

where I x = I y = I ′ and I z = I . Recalling Eq. (2), we write H O = − I ′ωe cos λ sin θ i + I ′(θ + ωe sin λ ) j + I ω z k

(3)

Equations of motion. Eq. (18.28):

 ) ΣM O = ( H O Oxyz + Ω × H O or, from Eqs. (1) and (3):

ΣM O = − I ′ωe cos λ cos θθi + I ′θj + I ω z k i + −ωe cos λ sin θ − I ′ωe cos λ sin θ

j k  θ + ωe sin λ ωe cos λ cos θ  ′ I (θ + ω sin λ ) Iω e

(4)

z

We observe that the rotor is free to spin about the z axis and free to rotate about the y axis. Therefore, the y and z components of ΣM O must be zero. It follows that the coefficients of j and k at the righthand member of Eq. (4) must also be zero. Setting the coefficient of j in the right-hand member of Eq. (4) equal to zero, I ′θ + (− I ′ωe cos λ sin θ )(ωe cos λ cos θ ) − (−ωe cos λ sin θ ) I ω z = 0 I ′θ + I ω zωe cos λ sin θ − I ′ωe2 cos 2 λ sin θ cos θ = 0

(5)  Q. E. D

Setting the coefficient of k equal to zero, I ω z + (−ωe cos λ sin θ ) I ′(θ + ωe sin λ ) − (− I ′ωe cos λ sin θ )(θ + ωe sin λ ) = 0

Observing that the last two terms cancel out, we have I ω z = 0

(b)

Q. E. D.

It follows from Eq. (6) that

ω z = constant

Rewrite Eq. (5) as follows:

I ′θ + ( I ω z − I ′ωe cos λ cos θ )ωe cos λ sin θ = 0

(6)  (7)

It is evident that ω z >>> ωe . We can therefore neglect the second term in the parenthesis and write I ′θ + I ω zωe cos λ sin θ = 0

or

θ +

I ω zωe cos λ sin θ = 0 I′

(8)

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PROBLEM 18.158 (Continued) where the coefficient of sin θ is a constant. The rotor, therefore, oscillates about the line NS as a simple pendulum. For small oscillations, sin θ ≈ θ , and Eq. (8) yields

θ +

I ω zωe cos λ θ =0 I′

Q. E. D.

(9) 

Eq. (9) is the equation of simple harmonic motion with period

τ = 2π

I′ I ω zωe cos λ

(10)

Since its rotor oscillates about the line NS, the gyrocompass can be used to determine the direction of that line. We should note, however, that for values of λ close to 90° or –90°, the period of oscillation becomes very large and the line about which the rotor oscillates cannot be determined. The gyrocompass, therefore, cannot be used in the polar regions.

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CHAPTER 19

PROBLEM 19.1 Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 3 mm and a frequency of 20 Hz.

SOLUTION Frequency:

f = 20 Hz

ωn = 2π f = (2π )(20) = 125.66 rad/s Amplitude: Simple harmonic motion:

xm = 3 mm x = xm sin(ωn t + φ )

v = x = ωn xm cos(ωn t + φ ) a = v =  x = − ωn2 xm sin(ωn t + φ )

Maximum velocity:

vm = ωn xm = (125.66 rad/s)(3 mm) = 377 mm/s vm = 0.377 m/s 

Maximum acceleration:

am = ωn2 xm = (125.66 rad/s)2 (3 mm) = 47.3 × 103 mm/s 2 am = 47.3 m/s 2 

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PROBLEM 19.2 A particle moves in simple harmonic motion. Knowing that the amplitude is 15 in. and the maximum acceleration is 15 ft/s2, determine the maximum velocity of the particle and the frequency of its motion.

SOLUTION Simple harmonic motion.

x = xm sin(ωn t + φ ) xm = 15 in. = 1.25 ft x = v = xmωn cos(ωn t + φ )  x = a = − xmωn2 sin(ωn t + φ ) am = − xmωn2 | am | = 15 ft/s 2 = (1.25 ft)ωn2

Natural frequency

ωn = 3.4641 rad/s ω f n = n = 0.55133 Hz 2π f n = 0.551 Hz 

Maximum velocity

vm = xmωn = (1.25 ft)(3.4641 rad/s) = 4.3301 ft/s vm = 4.33 ft/s 

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PROBLEM 19.3 Determine the amplitude and maximum velocity of a particle which moves in simple harmonic motion with a maximum acceleration of 15 ft/s2 and a frequency of 8 Hz.

SOLUTION Simple harmonic motion

x = xm sin(ωn t + φ )

ωn = 2π f n = 2π (8 Hz) = 16π rad/s x = v = xmωn cos(ωn t + φ ) vm = xmωn

 x = a = − xmωn2 sin(ωn t + φ ) am = xmωn2 15 ft/s 2 = xm (16π rad/s)2

Maximum displacement.

xm = 0.005937 ft = 0.0712 in. xm = 0.0712 in. 

Maximum velocity.

vm = xmωn = (0.005937 ft)(16π rad/s) = 0.2984 ft/s = 3.58 in./s vm = 3.58 in./s 

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PROBLEM 19.4 A 32-kg block is attached to a spring and can move without friction in a slot as shown. The block is in its equilibrium position when it is struck by a hammer, which imparts to the block an initial velocity of 250 mm/s. Determine (a) the period and frequency of the resulting motion, (b) the amplitude of the motion and the maximum acceleration of the block.

SOLUTION x = xm sin(ωn t + φ )

(a)

k 12 × 103 N/m = m 32 kg

ωn =

ωn = 19.365 rad/s 2π τn = ωn τn =

2π 19.365

τ n = 0.324 s  fn =

(b)

At t = 0, x0 = 0,

x0 = v0 = 250 mm/s

Thus,

x0 = 0 = xm sin(ωn (0) + φ )

and

1

τn

=

1 = 3.08 Hz  0.324

φ =0 x0 = v0 = xmωn cos(ωn (0) + 0) = xmωn v0 = 0.250 m/s = xm (19.365 rad/s) xm =

(0.250 m/s) (19.365 rad/s)

xm = 12.91 × 10−3 m

xm = 12.91 mm 

am = xmωn2 = (12.91 × 10−3 m)(19.365 rad/s) 2

am = 4.84 m/s 2 

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PROBLEM 19.5 A 12-kg block is supported by the spring shown. If the block is moved vertically downward from its equilibrium position and released, determine (a) the period and frequency of the resulting motion, (b) the maximum velocity and acceleration of the block if the amplitude of its motion is 50 mm.

SOLUTION (a)

Simple harmonic motion. Natural frequency.

x = xm sin(ωn t + φ )

ωn =

k m

ωn =

(5000 N/m) 12 kg

k = 5 kN/m = 5000 N/m

ωn = 20.412 rad/s 2π τn = ωn τn =

2π = 0.30781 s 20.412

τ n = 0.308s  fn =



1

τn

=

1 = 3.25 Hz  0.30781

xm = 50 mm = 0.05 m

(b)

x = 0.05 sin (20.412t + φ )

Maximum velocity.

vm = xmωn = (0.05 m)(20.412 rad/s)

vm = 1.021 m/s 

Maximum acceleration.

am = xmωn2 = (0.05 m)(20.412 rad/s)2

am = 20.8 m/s 2 

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PROBLEM 19.6 An instrument package A is bolted to a shaker table as shown. The table moves vertically in simple harmonic motion at the same frequency as the variable-speed motor which drives it. The package is to be tested at a peak acceleration of 150 ft/s2. Knowing that the amplitude of the shaker table is 2.3 in., determine (a) the required speed of the motor in rpm, (b) the maximum velocity of the table.

SOLUTION In simple harmonic motion, amax = xmax ωn2

 2.3  2 150 ft/s 2 =  ft  ωn  12 

ωn2 = (782.6 rad/s) 2 ωn = 27.98 rad/s ω fn = n 2π 27.98 2π = 4.452 Hz (cycles per second) =

(a)

Motor speed.

(b)

Maximum velocity.

(4.452 rev/s)(60 s/min)  2.3  ft  (27.98 rad/s) vmax = xmaxωn =   12 

speed = 267 rpm  vmax = 5.36 ft/s 

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PROBLEM 19.7 A simple pendulum consisting of a bob attached to a cord oscillates in a vertical plane with a period of 1.3 s. Assuming simple harmonic motion and knowing that the maximum velocity of the bob is 0.4 m/s, determine (a) the amplitude of the motion in degrees, (b) the maximum tangential acceleration of the bob.

SOLUTION (a)

θ = θ m sin(ωn t + φ ) 2π 2π = ωn = τ n (1.3 s)

Simple harmonic motion

= 4.8332 rad/s  θ = θ mωn cos(ωn t + φ ) θ = θ ω m

m

n

vm = lθm = lθ mωn

Thus,

θm =

vm lωn

For a simple pendulum,

ωn =

g l

l=

Thus,

g

ωn2

(1)

9.81 m/s 2 (4.8332 rad/s) 2

=

= 0.41995 m

Amplitude from (1),

θm =

vm 0.4 m/s = lωn (0.42 m)(4.833rad/s)

= 0.19707 rad = 11.291°

(b)

Maximum tangential acceleration

θ m = 11.29° 

at = lθ

The maximum tangential acceleration occurs when θ is maximum.

θ = −θ mωn2 sin(ωn t + φ ) θm = θ mωn2 (at ) m = lθ mωn2 (at ) m = (0.41995 m)(0.19707 rad)(4.8332 rad/s) 2

= 1.933 m/s 2

(at )m = 1.933 m/s 2 

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PROBLEM 19.8 A simple pendulum consisting of a bob attached to a cord of length l = 800 mm oscillates in a vertical plane. Assuming simple harmonic motion and knowing that the bob is released from rest when θ = 6°, determine (a) the frequency of oscillation, (b) the maximum velocity of the bob.

SOLUTION (a)

Frequency.

ωn =

g (9.81 m/s 2 ) = l (0.8 m)

ωn = 3.502 rad/s ω (3.502 rad/s) fn = n = 2π 2π (b)

Simple harmonic motion. where Maximum velocity.

f n = 0.557 Hz 

θ = θ m sin(ωn t + φ ) θ m = 6° = 0.10472 rad θ = θ mωn cos(ωn t + φ ) θm = θ mωn vm = lθm = lθ mωn = (0.8 m)(0.10472)(3.502) vm = 293.4 × 10−3 m/s

vm = 293 mm/s 

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PROBLEM 19.9 An instrument package B is placed on the shaking table C as shown. The table is made to move horizontally in simple harmonic motion with a frequency of 3 Hz. Knowing that the coefficient of static friction is μs = 0.40 between the package and the table, determine the largest allowable amplitude of the motion if the package is not to slip on the table. Given the answers in both SI and U.S. customary units.

SOLUTION Maximum allowable acceleration of B.

μs = 0.40 ΣF = ma :

Fm = mam

μs mg = mam am = μ s g Simple harmonic motion.

f n = 3 Hz =

am = 0.40 g

ωn 2π

ωn = 6π rad/s am = xmωn2

0.40 g = xm (6π rad/s) 2 xm = 1.1258 × 10−3g

Largest allowable amplitude. SI:

xm = 1.1258 × 10−3 (9.81) = 11.044 × 10−3 m

U.S.:

xm = 1.1258 × 10−3 (32.2) = 0.03625 ft

xm = 11.04 mm  xm = 0.435 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2249

PROBLEM 19.10 A 5-kg fragile glass vase is surrounded by packing material in a cardboard box of negligible weight. The packing material has negligible damping and a force-deflection relationship as shown. Knowing that the box is dropped from a height of 1 m and the impact with the ground is perfectly plastic, determine (a) the amplitude of vibration for the vase, (b) the maximum acceleration the vase experiences in g’s.

SOLUTION v = 2 gh

Velocity at end of free fall:

v = (2)(9.81 m/s 2 )(1 m) = 4.4294 m/s

Assume that the spring is unstretched during the free fall. Use a simple spring-mass model for the motion of the vase and the packing material. m = 5 kg 100 N (slope from graph) k= 10 mm k = 10 N/m = 10000 N/m

k 10000 N/m = = 44.721 rad/s m 5 kg

ωn =

Natural frequency:

x = xm sin(ωn t + φ ) v = x = ωn xm cos(ωn t + φ )

Simple harmonic motion:

Let t = 0 at the instant when the box bottom hits the ground. Then, at t = 0, from which and

x=0

and

v = 4.4294 m/s

φ =0 ωn xm = 4.4294 m/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2250

PROBLEM 19.10 (Continued)

(a)

Amplitude:

xm =

4.4294 m/s = 0.099045 m 44.721 rad/s xm = 99.0 mm 

(b)

Maximum acceleration: am = ωn2 xm = (44.721 rad/s) 2 (0.099045 m) = 198.087 m/s 2 = (20.192)(9.81 m/s2 ) am = 20.2 g 

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PROBLEM 19.11 A 3-lb block is supported as shown by a spring of constant k = 2 lb/in. which can act in tension or compression. The block is in its equilibrium position when it is struck from below by a hammer which imparts to the block an upward velocity of 90 in./s. Determine (a) the time required for the block to move 3 in. upward, (b) the corresponding velocity and acceleration of the block.

SOLUTION x = xm sin(ωn t + φ )

Simple harmonic motion. Natural frequency.

ωn =

k , k = 2 lb/in. = 24 lb/ft m

ωn =

24 lb/ft

(

3 lb 32.2 ft/s2

)

ωn = 16.05 rad/s x(0) = 0 = xm sin(0 + φ )

φ =0 x (0) = xmωn cos(0 + 0) x (0) =

90 = 7.5 ft/s 12

7.5 = xm (16.05)

xm = 0.4673 ft

x = (0.4673)sin(16.05t )(ft/s)

(a)

(1)

Time at x = 3 in. (x = 0.25 ft) 0.25 = 0.4673sin(16.05t ) t=

(b)

0.25 sin −1 ( 0.4673 )

16.05

t = 0.0352 s 

Velocity and acceleration. x = xmωn cos(ωn t )  x = − xmωn2 sin ωn t t = 0.0352 x = (0.4673)(16.05) cos[(16.05)(0.0352)] x = 6.34 ft/s

v = 6.34 ft/s 

 x = −(0.4673)(16.05) 2 sin[(16.05)(0.0352)] = −64.4 ft/s 2

a = 64.4 ft/s 2 

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PROBLEM 19.12 In Problem 19.11, determine the position, velocity, and acceleration of the block 0.90 s after it has been struck by the hammer.

SOLUTION Simple harmonic motion. Natural frequency.

x = xm sin(ωn t + φ )

ωn =

k , k = 2 lb/in. = 24 lb/ft m

ωn =

24 lb/ft

(

3 lb 32.2 ft/s 2

)

ωn = 16.05 rad/s x(0) = 0 = xm sin(0 + φ )

φ =0 90 = 7.5 ft/s 12 xm = 0.4673 ft

x (0) = xmωn cos(0 + 0) 7.5 = xm (16.05)

x (0) =

x = (0.4673)sin(16.05t )(ft/s)

Simple harmonic motion.

x = xm sin(ωn t + φ ) x = xmωn cos(ωn t + φ )

 x = − xmωn2 sin(ωn t + φ )

At 0.90 s:



x = (0.4673)sin[(16.05)(0.90)] = 0.445 ft

x = 0.445 ft 

x = (0.4673)(16.05) cos[(16.05)(0.90)] = −2.27 ft/s

v = 2.27 ft/s 

 x = −(0.4673)(16.05) 2 sin[(16.05)(0.90)] = −114.7 ft/s 2  a = 114.7 ft/s 2 

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PROBLEM 19.13 The bob of a simple pendulum of length l = 40 in. is released from rest when θ = +5°. Assuming simple harmonic motion, determine 1.6 s after release (a) the angle θ , (b) the magnitudes of the velocity and acceleration of the bob.

SOLUTION For simple harmonic motion and l = 40 in. = 3.333 ft: g 32.2 ft/s 2 = = 3.1082 rad/s l 3.333 ft

ωn =

θ = θ m sin(ωn t + φ )

Angular displacement:

θ (0) = 5° = 0.08727 rad, and θ(0) = 0:

Initial conditions:

θ(0) = 0 = θ mωn cos(0 + φ )

φ=

π 2

5π θ m = θ (0) = = 0.08727 rad 180

θ=

5π π  sin  (3.1082 rad/s)t +  180 2 

π  = (0.08727 rad) sin (3.1082 rad/s)t +  2  (a)

At t = 1.6 s. 5π π  sin (3.1082 rad/s)(1.6 s) +  180 2  = 0.022496 rad = 1.288°

θ=

θ = 1.288°  (b)

Velocity:

θ = θ mωn cos(ωn t + φ ) 5π π  (3.1082 rad/s) cos (3.1082 rad/s)(1.6 s) +  180 2  = 0.262074 rad/s v = lθ = (3.3333 ft)(0.262074 rad/s) = 0.874 ft/s =

v = 0.874 ft/s  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2254

PROBLEM 19.13 (Continued)

Angular acceleration:

θ = −θ mωn2 sin(ωn t + φ ) = −

5π π  (3.1082 rad/s) 2 cos (3.1082 rad/s)(1.6 s) +  180 2 

= −0.21733 rad/s 2

Acceleration: a = (an ) 2 + (at ) 2 v2 = lθ 2 = (3.3333 ft)(0.26207 rad/s)2 = 0.22894 ft/s 2 l at = lθ = (3.333 ft)(−0.21733 rad/s 2 ) 2 = −0.72443 m/s 2

an =

a = 0.75974 ft/s 2 

a = 0.760 ft/s 2 

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PROBLEM 19.14 A 150-kg electromagnet is at rest and is holding 100 kg of scrap steel when the current is turned off and the steel is dropped. Knowing that the cable and the supporting crane have a total stiffness equivalent to a spring of constant 200 kN/m, determine (a) the frequency, the amplitude, and the maximum velocity of the resulting motion, (b) the minimum tension which will occur in the cable during the motion, (c) the velocity of the magnet 0.03 s after the current is turned off.

SOLUTION

m1 = 150 kg m2 = 100 kg k = 200 × 103 N/m

Data: From the first two sketches,

T0 + kxm = ( m1 + m2 ) g

(1)

T0 = m1 g

(2)

Subtracting Eq. (2) from Eq. (1),

kxm = m2 g

xm =

m2 g (100)(9.81) = = 4.905 × 10−3 m = 4.91 mm 3 k 200 × 10 k = m1

200 × 103 = 36.515 rad/s 150

Natural circular frequency:

ωn =

Natural frequency:

fn =

Maximum velocity:

vm = ωn xm = (36.515)(4.905 × 10−3 ) = 0.1791 m/s

(a)

ωn 36.515 = 2π 2π

f n = 5.81 Hz

amplitude xm = 4.91 mm 

Resulting motion:

frequency

f n = 5.81 Hz 

maximum velocity vm = 0.1791 m/s 

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PROBLEM 19.14 (Continued)

(b)

Minimum value of tension occurs when x = − xm . Tmin = T0 − kxm = m1 g − m2 g = ( m1 − m2 ) g = (50)(9.81)

The motion is given by

Tmin = 491 N 

x = xm sin(ωn t + ϕ ) x = ωn xm cos(ωn t + ϕ ) x0 = − xm or sinϕ = −1 x0 = 0 or cosϕ = 0

Initally,

ϕ=−

π 2

π  x = ωn xm cos  ωn t −  2  (c)

Velocity at t = 0.03 s.

ωn t = (36.515)(0.03) = 1.09545 rad ωn t − ϕ = −0.47535 rad cos(ωn t − ϕ ) = 0.88913 

x = (36.515)(4.905 × 10−3 )(0.88913) 

x = 0.1592 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2257

PROBLEM 19.15 A variable-speed motor is rigidly attached to beam BC. The rotor is slightly unbalanced and causes the beam to vibrate with a frequency equal to the motor speed. When the speed of the motor is less than 600 rpm or more than 1200 rpm, a small object placed at A is observed to remain in contact with the beam. For speeds between 600 rpm and 1200 rpm, the object is observed to “dance” and actually to lose contact with the beam. Determine the amplitude of the motion of A when the speed of the motor is (a) 600 rpm, (b) 1200 rpm. Give answers in both SI and U.S. customary units.

SOLUTION At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g. (a)

ω = 600 rpm = 62.832 rad/s am = xmω 2

Eq. (19.15):

(b)

SI:

xm =

9.81 = 2.4849 × 10−3 m (62.832)2

US:

xm =

32.2 = 0.008156 ft (62.832)2

xm =

g (62.832) 2 xm = 2.48 mm  xm = 0.0979 in. 

ω = 1200 rpm = 125.664 rad/s Eq. (19.15):

am = xmω 2

xm =

g (125.664) 2

SI:

xm =

9.81 = 621.2 × 10−6 m (125.664) 2

xm = 0.621 mm 

US:

xm =

32.2 = 0.002039 ft (125.664) 2

xm = 0.0245 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2258

PROBLEM 19.16 A small bob is attached to a cord of length 1.2 m and is released from rest when θ A = 5°. Knowing that d = 0.6 m, determine (a) the time required for the bob to return to Point A, (b) the amplitude θC .

SOLUTION As the pendulum moves between Points A and B, the length of the pendulum is l = l AB = 1.2 m.

ωn = ωn1 = τ1 =



ωn1

=

g l AB

=

9.81 m/s 2 = 2.8592 rad/s 1.2 m

2π = 2.1975 s 2.8592 rad/s

The falling from A to B is one quarter period.

τ AB =

1 τ1 = 0.54938 s. 4

As the pendulum moves between Points B and C, the length of the pendulum is l = lBC = 1.2 m − 0.6 m = 0.6 m.

ωn = ωn 2 = τ2 =



ωn 2

=

g lBC

=

9.81 m/s 2 = 4.0435 rad/s 0.6 m

2π = 1.55389 s 4.0435 rad/s

The motion from B to C and back to B is one half period

τ BCB =

1 τ 2 = 0.77695 s 2

As the pendulum moves from B to A, the length is again 1.2 meters.

τ BA = (a)

1 τ1 = 0.54938 s 4

Time required to return to A.

τ = τ AB + τ BCB + τ BA τ = 1.87571 s

τ = 1.876 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2259

PROBLEM 19.16 (Continued)

For falling from A to B,

θm = θ A

At B,

θB = θm = ωn1θ A vB = l ABθB = l ABωn1θ A

For rising from B to C, vB l AB = ωn1θ A = θmax lBC lBC θ l ω θC = θ max = max = AB n1 θ A ωn 2 lBC ωn 2

θB =

θC = (b)

(1.2 m)(2.8592 rad/s) θ A = 1.4142 θ A (0.6 m)(4.0435 rad/s)

Amplitude θC :

With θ A = 5°, 

θC = 7.07° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2260

PROBLEM 19.17 A 5-kg block, attached to the lower end of a spring whose upper end is fixed, vibrates with a period of 6.8 s. Knowing that the constant k of a spring is inversely proportional to its length, determine the period of a 3-kg block which is attached to the center of the same spring if the upper and lower ends of the spring are fixed.

SOLUTION Equivalent spring constant. k ′ = 2k + 2k = 4k

For case

,

(Deflection of each spring is the same.)

τ n1 = 6.8 s 2π 2π ωn1 = = = 0.924 rad/s τ n1 6.8 ωn2 =

k m1

k = m1ωn21 = (5)(0.924)2 = 4.2689 N/m

For case

,

ωn22 =

4k (4)(4.2689) = = 5.6918 (rad/s) 2 3 m2

ωn 2 = 2.3857 rad/s 2π 2π = τn = ωn 2 2.3857

τ n 2 = 2.63 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2261

PROBLEM 19.18 A 75-lb block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. Knowing that the amplitude of the resulting motion is 2 in., determine (a) the period and frequency of the motion, (b) the maximum velocity and maximum acceleration of the block.

SOLUTION (a)

Determine the constant k of a single spring equivalent to the three springs P = kδ kδ = 90δ + 45δ + 45δ k = 180 lb/in. = 2160 lb/ft

Natural frequency.

ωn = =

k m 2160 lb/ft 75 lb 32.2 ft/s 2

ωn = 30.453 rad/s τn = fn =

(b)



ωn

=

2π = 0.20633 s 30.453

1

τ n = 0.206 s  f n = 4.85 Hz 

τn

x = xm sin(ωn t + φ )

x0 = 2 in. = 0.16667 ft = xm

ωn = 30.453 rad/s x = 0.16667 sin(30.453t + φ ) x = (0.16667)(30.453) cos(30.453t + φ )  x = −(0.16667)(30.453) sin(30.453t + φ ) 2

vmax = 5.08 ft/s  amax = 154.6 ft/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2262

PROBLEM 19.19 A 75-lb block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. Knowing that the amplitude of the resulting motion is 2 in., determine (a) the period and frequency of the motion, (b) the maximum velocity and maximum acceleration of the block.

SOLUTION (a)

Determine the constant k of a single spring equivalent to the two springs shown.

δ = δ1 + δ 2 = 1 1 1 = + k 90 90

P P P + = 90 lb/in. 90 lb/in. k k = 45 lb/in. = 540 lb/ft

Period of the motion.

τn =

2π k m

=

2π 540 75/32.2

τ n = 0.413 s 

= 0.41265 s

fn =

(b)

1

τn

=

1 = 2.42 Hz  0.41265

x = xm sin(ωn t + φ ) x0 = 2 in. = 0.16667 ft = xm

ωn = 2π f n = 2π (2.4233) = 15.226 rad/s x = 0.16667 sin(15.226t + φ ) x = (0.16667)(15.226) cos(15.226t + φ ) vmax = 2.54 ft/s 

 x = −(0.16667)(15.226) 2 sin(15.226t + φ ) 

amax = 38.6 ft/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2263

PROBLEM 19.20 A 13.6-kg block is supported by the spring arrangement shown. If the block is moved from its equilibrium position 44 mm vertically downward and released, determine (a) the period and frequency of the resulting motion, (b) the maximum velocity and acceleration of the block.

SOLUTION

Determine the constant k of a single spring equivalent to the three springs shown. Springs 1 and 2:

δ = δ1 + δ 2 , and

Hence,

k′ =

P1 P1 P1 = + k ′ k1 k2

k1k2 k1 + k2

where k ′ is the spring constant of a single spring equivalent of springs 1 and 2. Springs k ′ and 3: (Deflection in each spring is the same). P = P1 + P2 , and P = kδ , P1 = k ′δ , P2 = k3δ

So

kδ = k ′δ + k3δ

Now

k = k ′ + k3 = k=

or (a)

Period and frequency:

τn =

fn =

k1k2 + k3 k1 + k2

(3.5 kN/m)(2.1 kN/m) + 2.8 kN/m = 4.11 kN/m = 4.11 × 103 N/m (3.5 kN/m) + (2.1 kN/m) 2π k m

1

τn

=

=

2π 4.11×103 N/m 13.6 kg

1 0.3614 s

tn = 0.361 s 

f n = 2.77 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2264

PROBLEM 19.20 (Continued)

(b)

Displacement: x = xm sin(ωn t + φ ) xm = 44 mm = 0.044 m

ωn = 2π f n = (2π )(2.77 Hz) = 17.384 rad/s x = (0.044 m) sin[(17.384 rad/s)t + φ ] x = (0.044 m)(17.384 rad/s) cos[(17.384 rad/s)t + φ ]

Velocity:

vmax = (0.044 m)(17.384 rad/s) = 0.765 m/s

 x = −(0.044 m)(17.384 rad/s) 2 sin[(17.384 rad/s)t + φ ]

Acceleration:

amax = (0.044 m)(17.384 rad/s) 2 = 13.30 m/s 2 vmax = 0.765 m/s  amax = 13.30 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2265

PROBLEM 19.21 A 11-lb block, attached to the lower end of a spring whose upper end is fixed, vibrates with a period of 7.2 s. Knowing that the constant k of a spring is inversely proportional to its length (e.g., if you cut a 10 lb/in. spring in half, the remaining two springs each have a spring constant of 20 lb/in.), determine the period of a 7-lb block which is attached to the center of the same spring if the upper and lower ends of the spring are fixed.

SOLUTION Equivalent spring constant. k ′ = 2k + 2k = 4k

For case

,

(Deflection of each spring is the same.)

τ n1 = 7.2 s 2π 2π ωn1 = = = 0.87266 rad/s τ n1 7.2 ωn2 =

k m1

 11  2 k = m1ωn21 =   (0.87266) = 0.26015 lb/ft 32.2  

For case

,

ωn22 =

4k (4)(0.26015) = = 4.7868 (rad/s) 2 7 m2 32.2

ωn 2 = 2.1879 rad/s 2π 2π τn = = ωn 2 2.1879

τ n 2 = 2.87 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2266

PROBLEM 19.22 Block A of mass m is supported by the spring arrangement as shown. Knowing that the mass of the pulley is negligible and that the block is moved vertically downward from its equilibrium position and released, determine the frequency of the motion.

SOLUTION We first determine the constant keq of a single spring equivalent to the spring and pulley system supporting the block by finding the total displacement δ A of the end of the cable under a given static load P. Owing to the force 2P in the upper spring the pulley moves down a distance

δ1 =

2P 2k

Owing to the force P in the lower spring, Point A moves down an additional distance

δ2 =

P k

The total displacement is

δ A = δ1 + δ 2 = But

δA =

P so that keq

keq =

k 2

ωn =

Natural frequency:

2 P P 2P + = 2k k k

keq m

ωn =

k  2m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2267

PROBLEM 19.23 The period of vibration of the system shown is observed to be 0.2 s. After the spring of constant k2 = 20 lb/in. is removed and block A is connected to the spring of constant k1, the period is observed to be 0.12 s. Determine (a) the constant k1 of the remaining spring, (b) the weight of block A.

SOLUTION Equivalent spring constant for springs in series. ke =

k1k2 ( k1 + k2 )





For k1 and k2 ,

τ=

For k1 alone,

τ′ =

(a)

(k1 + k2 )(k1 ) k +k τ = = 1 2 τ′ k2 (k1k2 )

ke mA

=

( k1k 2 ) ( m A )( k1 + k2 )

2π k1 mA

2

τ  k2   = k1 + k2 τ′  τ 0.2 s = = 1.6667 τ ′ 0.12 s k2 = 20 lb/in. (20 lb/in.)(1.6667) 2 = k1 + 20 lb/in.

(b)

τ′ =

2π k1 mA

k1 = 35.6 lb/in. 

mA =

WA g

(τ ′) 2 k1 (2π )2 k1 = 35.6 lb/in. = 426.7 lb/ft

mA =

WA =

(32.2 ft/s 2 )(0.12 s)2 (426.7 lb/ft) (2π )2

WA = 5.01 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2268

PROBLEM 19.24 The period of vibration of the system shown is observed to be 0.8 s. If block A is removed, the period is observed to be 0.7 s. Determine (a) the mass of block C, (b) the period of vibration when both blocks A and B have been removed.

SOLUTION

τ1 = 0.8 s 2π

m1 = mC + 6 kg

ω1 = ω12



τ1

=

2π rad/s = 0.8 s 0.8

k  2π  ; k = m1ω12 = (mC + 6)  =  m1  0.8 

2

(1)

m2 = mC + 3 kg τ 2 = 0.7 s

ω2 = ω22 =



τ2

=

2π 2π rad/s = 0.7 s 0.7

k  2π  ; k = m2ω22 = (mC + 3)   m2  0.7 

2

(2)

Equating the expressions found for k in Eqs. (1) and (2): 2

 2π   2π  (mC + 6)   = (mC + 3)  0.7  0.8    

2

2

mC + 6  0.8  ; Solve for mC : = mC + 3  0.7 

ω3 = ω32

mC = 6.80 kg 



τ3

 2π  k ; k = mC ω32 = mC  =  mC  τ3 

2

(3)

Equating expressions for k from Eqs. (2) and (3), 2  2π   2π  (mC + 3)    = mC   0.7   τ3 

Recall mC = 6.8 kg:

2  2π   2π  (6.8 + 3)    = 6.8   0.7   τ3  2

6.8  τ3   0.7  = 9.8 ;  

2

2

τ3 0.7

= 0.833

τ 3 = 0.583 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2269

PROBLEM 19.25 The 100-lb platform A is attached to springs B and D, each of which has a constant k = 120 lb/ft. Knowing that the frequency of vibration of the platform is to remain unchanged when an 80-lb block is placed on it and a third spring C is added between springs B and D, determine the required constant of spring C.

SOLUTION Frequency of the original system. Springs B and D are in parallel.

ke = k B + k D = 2(120 lb/ft) = 240 lb/ft

ωn2 = ωn2

ke 240 lb/ft = 100 lb mA 2

(

32.2 ft/s

= 77.28(rad/s)

)

2

Frequency of new system. Springs A, B, and C are in parallel.

ke′ = k B + k D + kC = (2)(120) + ke (ωn′ ) 2 =

ke′ (240 + kC )(32.2 ft/s 2 ) = mA + mB (100 lb + 80 lb)

(ωn′ ) 2 = (0.1789)(240 + kC )

ωn2 = (ωn′ ) 2 77.28 = (0.1789)(240 + kC ) kC = 191.97 lb/ft

kC = 192.0 lb/ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2270

PROBLEM 19.26 The period of vibration for a barrel floating in salt water is found to be 0.58 s when the barrel is empty and 1.8 s when it is filled with 55 gallons of crude oil. Knowing that the density of the oil is 900 kg/m3, determine (a) the mass of the empty barrel, (b) the density of the salt water, ρsw. [Hint: the force of the water on the bottom of the barrel can be modeled as a spring with constant k = ρswgA.]

SOLUTION Area of bottom of barrel:

Mass of oil: Barrel empty:

A=

π D2

4

= 0.2570 m 2

τ1 = 0.58 s 2π 2π = = 10.833 rad/s ωn1 = τ1 0.58 s k mb

(1)

τ 2 = 1.8 s 2π 2π = = 3.4907 rad/s ωn 2 = τ 2 1.8 s ωn 2 =

(a)

π (0.572 m)2

  1 m3 3 moil = (55 gal)   (900 kg/m ) = 187.378 kg 264.172 gal  

ωn1 = Barrel full:

4

=

k = m

k moil + mb

(2)

Mass mb of empty barrel. Divide Eq. (1) by Eq. (2) and square both sides.

ωn21 (10.833)2 m + mb = = 9.6310 = oil 2 2 mb ωn 2 (3.4907) 9.6310 mb = moil + mb mb =

moil 187.378 kg = = 21.710 kg 9.6310 − 1 8.6310 mb = 21.7 kg 

Spring constant:

k = mbωn21 = (21.710)(10.833) 2 = 2.5477 × 103 N/m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2271

PROBLEM 19.26 (Continued)

(b)

Density of the salt water. k = ρsw gA

ρsw =

k 2.5477 × 103 N/m = gA (9.81 m/s 2 )(0.2570 m 2 )

ρsw = 1011 kg/m3 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2272

PROBLEM 19.27 From mechanics of materials it is known that for a cantilever beam of constant cross section, a static load P applied at end B will cause a deflection δ B = PL3 /3EI , where L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia of the cross-sectional area of the beam. Knowing that L = 10 ft, E = 29 × 106 lb/in.2 , and I = 12.4 in.4, determine (a) the equivalent spring constant of the beam, (b) the frequency of vibration of a 520-lb block attached to end B of the same beam.

SOLUTION (a)

Equivalent spring constant.

ke =

P

δB P = k eδ B PL3 3EI  3EI P= 3  L

δB =

 δ B 

3EI L3 (3)(29 × 106 lb/in.2 )(12.4 in.4 ) = (10 × 12 in.)3 ke = 624.3 lb/in. ke =

(b)

Natural frequency.

fn =

ke = 624.3 lb/in. 

ke m

2π ke = 624.3 lb/in. = 7.492 × 103 lb/ft

fn =

(7.492 × 103 lb/ft)  (520 lb)   2    (32.2 ft/s ) 

2π f n = 3.428 Hz

f n = 3.43 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2273

PROBLEM 19.28 From mechanics of materials it is known that when a static load P is applied at the end B of a uniform metal rod fixed at end A, the length of the rod will increase by an amount δ = PL/AE , where L is the length of the undeformed rod. A is its cross-sectional area, and E is the modulus of elasticity of the metal. Knowing that L = 450 mm and E = 200 GPa and that the diameter of the rod is 8 mm, and neglecting the mass of the rod, determine (a) the equivalent spring constant of the rod, (b) the frequency of the vertical vibrations of a block of mass m = 8 kg attached to end B of the same rod.

SOLUTION (a)

P = k eδ PL AE  AE  P= δ  L  AE ke = L

δ=

A=

πd2

=

π (8 × 10−3 m)2

4 4 A = 5.027 × 10−5 m 2 L = 0.450 m

E = 200 × 109 N/m 2 ke =

(5.027 × 10−5 m 2 )(200 × 109 N/m 2 ) (0.450 m)

ke = 22.34 × 106 N/m

(b)

fn = =

ke = 22.3 MN/m 

ke m

2π 22.3×106 8



= 265.96 Hz

f n = 266 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2274

PROBLEM 19.29 Denoting by δ st the static deflection of a beam under a given load, show that the frequency of vibration of the load is f =

g

1 2π

δ st

Neglect the mass of the beam, and assume that the load remains in contact with the beam.

SOLUTION

k= m=

ωn2 =

W

δ st W g k = m

W

δ st W g

=

g

δ st



fn =

ωn 1 = 2π 2π

g

δ st



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2275

PROBLEM 19.30 A 40-mm deflection of the second floor of a building is measured directly under a newly installed 3500-kg piece of rotating machinery, which has a slightly unbalanced rotor. Assuming that the deflection of the floor is proportional to the load it supports, determine (a) the equivalent spring constant of the floor system, (b) the speed in rpm of the rotating machinery that should be avoided if it is not to coincide with the natural frequency of the floor-machinery system.

SOLUTION

(a)

Equivalent spring constant. W = ke δ s ke = =

(b)

Natural frequency.

fn = =

mg

δ 3500(9.81) N 40 mm

ke = 858 N/mm 

ke m

2π (858.38 × 1000 N/m) (3500 kg)

2π f n = 2.4924 Hz 1 Hz = 1 cycle/s

= 60 rpm Speed = (2.424 Hz)

(60 rpm) Hz

Speed = 149.5 rpm 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2276

PROBLEM 19.31 If h = 700 mm and d = 500 mm and each spring has a constant k = 600 N/m, determine the mass m for which the period of small oscillations is (a) 0.50 s, (b) infinite. Neglect the mass of the rod and assume that each spring can act in either tension or compression.

SOLUTION

xs = x

d h

2 F = 2kxs = 2k

d x h

ΣM A = Σ( M A )eff : 2 Fd − mgx = −( mx)h

d   2k  x  d − mgx = −mxh h   2kd 2 g   x+ − x=0 2 h  mh  2kd 2

ωn2 = 

 mh

2



g  h

2

ωn2 Data:

2k  d  g = −   m h h

(1)

d = 0.5 m h = 0.7 m k = 600 N/m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2277

PROBLEM 19.31 (Continued)

(a)

For τ = 0.5 s:

τ=



ωn

; 0.5 =



ωn

ωn = 4π

2

Eq. (1):

(4π ) 2 =

2(600)  0.5  9.81 − m  0.7  0.7

m = 3.561 kg

(b)

2π ωn

m = 3.56 kg 

For τ = infinite:

τ=

Eq. (1):

2(600)  0.5  9.81 − 0= m  0.7  0.7

ωn = 0 2

m = 43.69 kg

m = 43.7 kg 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2278

PROBLEM 19.32 The force-deflection equation for a nonlinear spring fixed at one end is F = 1.5 x1/ 2 where F is the force, expressed in newtons, applied at the other end, and x is the deflection expressed in meters. (a) Determine the deflection x0 if a 4-oz block is suspended from the spring and is at rest. (b) Assuming that the slope of the force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant, determine the frequency of vibration of the block if it is given a very small downward displacement from its equilibrium position and released.

SOLUTION (a)

Deflection x0 .

W = 4 oz = 0.25 lb F =W

= 1.5 x1/2 0 2

 0.25  x0 =    1.5  = 0.027778 ft

x0 = 0.333 in. 

Equivalent spring constant. At x0 ,

1.5 1.5  dF  ( x0 ) −1/ 2 = (0.027778) −1/ 2   = 2 2  dx  x0

 dF   dx  = 4.5 lb/ft   x0 ke = 4.5 lb/ft

(b)

Natural frequency.

fn =

=

ke m

2π (4.5 lb/ft ) (0.25/32.2)



f n = 3.8316 Hz

f n = 3.83 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2279

PROBLEM 19.33* Expanding the integrand in Equation (19.19) of Section 19.4 into a series of even powers of sin φ and integrating, show that the period of a simple pendulum of length l may be approximated by the formula

τ = 2π

l g

 1 2 θm  1 + 4 sin 2   

where θ m is the amplitude of the oscillations.

SOLUTION Using the Binomial Theorem, we write 1 1 − sin 2

( ) θm 2

 θ = 1 − sin 2  m  2 sin 2 φ 

   sin φ   

−1/ 2

θ 1 = 1 + sin 2 m sin 2 φ + ⋅ ⋅ ⋅ ⋅ 2 2 Neglecting terms of order higher than 2 and setting sin 2 φ = 12 (1 − cos 2φ ), we have π /2 

1 2 θm 1 + sin 2  2

1   2 (1 − cos 2φ )   dφ  

τn = 4

l g



=4

l g



=4

l g

 1  2 θm φ +  sin 4 2 

 1 2 θm   φ − 8 sin 2 sin 2φ   0

=4

l g

 π 1  2 θm  +  sin 2 2 4

 π  2 + 0  

0

1 2 θm 1 2 θm  cos 2φ  dφ − sin 1 + sin 2 4 2  4 

π /2  0

π /2

τ n = 2π

l g

 1 2 θm  1 + 4 sin 2    

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2280

PROBLEM 19.34* Using the formula given in Problem 19.33, determine the amplitude θ m for which the period of a simple pendulum is 12 percent longer than the period of the same pendulum for small oscillations.

SOLUTION For small oscillations,

(τ n )0 = 2π

l g

τ n = 1.005(τ n )0

We want

l g

= 1.005 2π

Using the formula of Problem 19.33, we write 

1 4

τ n = (τ n )0 1 + sin 2  = 1.005(τ n )0

sin 2

θm

sin

θm

2 2

θm 2

θm  2 

= 4[1.005 − 1] = 0.02 = 0.02

θ m = 16.26° 

= 8.130°

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2281

PROBLEM 19.35* Using the data of Table 19.1, determine the period of a simple pendulum of length l = 750 mm (a) for small oscillations, (b) for oscillations of amplitude θ m = 60°, (c) for oscillations of amplitude θ m = 90°.

SOLUTION (a)

τ n = 2π

l (Equation 19.18 for small oscillations): g

τ n = 2π

0.750 m 9.81 m/s 2

τ n = 1.737 s 

= 1.737 s

(b)

For large oscillations (Eq. 19.20),  2K  

l 

τn =    2π g   π    =

For θ m = 60°,

2K

π

(1.737 s)

K = 1.686 (Table 19.1)

τ n (60°) =

2(1.686)(1.737 s)

= 1.864 s

(c)

For θ m = 90°,

π

τ n = 1.864 s  τn =

K = 1.854

2(1.854)(1.737 s)

π

= 2.05 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2282

PROBLEM 19.36* Using the data of Table 19.1, determine the length in inches of a simple pendulum which oscillates with a period of 2 s and an amplitude of 90°.

SOLUTION For large oscillations (Eq. 19.20),  2K  

l 

τn =     2π g   π   for

θ m = 90° K = 1.854 (Table 19.1)

(2 s) = (2)(1.854)(2)

l 32.2 ft/s 2

(2 s) 2 (32.2 ft/s 2 ) [(4)(1.854)]2 = 2.342 ft

l=

l = 28.1 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2283

PROBLEM 19.37 The uniform rod shown has mass 6 kg and is attached to a spring of constant k = 700 N/m. If end B of the rod is depressed 10 mm and released, determine (a) the period of vibration, (b) the maximum velocity of end B.

SOLUTION k = 700 N/m W = mg

F = k ( x + δ st )

where

= k (0.5θ + δ st ) ma = mr α = 6(0.1 m)θ = 0.6θ 1 I α = (6)(0.8 m) 2 θ 12 = 0.32θ

(a)

Equation of motion. ΣM C = I α + mad : W (0.1 m) − F (0.5 m) = I α + ma (0.1 m) W (0.1) − k (0.5θ + δ st )(0.5 m) = 0.32θ + 0.6θ(0.1)

But in equilibrium, we have Thus,

W (0.1 m) − kδ st (0.5 m) = 0

−k (0.5) 2 θ = [0.32 + 0.06]θ −(700 N/m)(0.5) 2 θ = 0.38θ

θ + (460.53)θ = 0

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PROBLEM 19.37 (Continued)

Natural frequency and period.

ωn2 = 460.53 ωn = 21.46 rad/s 2π 2π τ= = ωn 21.46 rad/s (b)

At end B.

τ = 0.293 s 

xm = 0.010 m vm = xmωn

= (10 mm)(21.46 rad/s) = 214.6 mm/s

vm = 0.215 m/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2285

PROBLEM 19.38 A belt is placed around the rim of a 500-lb flywheel and attached as shown to two springs, each of constant k = 85 lb/in. If end C of the belt is pulled 1.5 in. down and released, the period of vibration of the flywheel is observed to be 0.5 s. Knowing that the initial tension in the belt is sufficient to prevent slipping, determine (a) the maximum angular velocity of the flywheel, (b) the centroidal radius of gyration of the flywheel.

SOLUTION Denote the initial tension by T0.

Equation of motion.

ΣM 0 = I θ: − TA r + TB r = I θ − (T0 + krθ ) r + (T0 − krθ )θ = I θ

θ +

Data:

W 500 lb = g 32.2 τ = 0.5 s 2π τ= ;

m=

ωn

(a)

2kr 2 θ =0 I 2 kr 2 ωn2 = I

(1)

k = 85 lb/in. = 1020 lb/ft r = 18 in. = 1.5 ft 2π 2π ωn = = = 4π rad/s τ 0.5

Maximum angular velocity. If Point C is pulled down 1.5 in. and released,  1.5 in. 

−3 θ m = θ max =   = 83.333 × 10 rad 18 in.  

θm = θ mωn = (83.333 × 10−3 rad)(4π rad/s)

θm = 1.047 rad/s 

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PROBLEM 19.38 (Continued)

(b)

Centroidal radius of gyration. 2kr 2 I 2(1020 lb/ft)(1.5 ft) 2 (4π rad/s)2 = I I = 29.067 slug ⋅ ft 2

ωn2 =

I = mk 2

or since  500 lb  2  32.2 ft/s

 2 2  k = 29.067 slug ⋅ ft  k = 1.3682 ft

k = 16.42 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2287

PROBLEM 19.39 An 8-kg uniform rod AB is hinged to a fixed support at A and is attached by means of pins B and C to a 12-kg disk of radius 400 mm. A spring attached at D holds the rod at rest in the position shown. If Point B is moved down 25 mm and released, determine (a) the period of vibration, (b) the maximum velocity of Point B.

SOLUTION (a)

ΣM A = (ΣM A )eff : FS = k (0.6θ + δ st )

Equation of motion.

0.6( mR g − FS ) + 1.2mD g = ( I R + I D )α + 0.6( mR )(at ) D + 1.2( mD )(at ) B

At equilibrium (θ = 0),

FS = kδ st ΣM A = 0 = 0.6(mR g − k (δ st )) + 1.2mD g

Substituting Eq. (2) into Eq. (1),

(1)

(2)

( I R + I D )α + 0.6 mR (at ) D + 1.2mD ( at ) B + (0.6)2 kθ = 0

α = θ ( at ) B = 0.6θ (a ) = 1.2θ t D

1 1 mR l 2 = (8)(1.2) 2 12 12 = 0.960 kg ⋅ m 1 1 I D = mD R 2 = (12)(0.4) 2 = 0.960 kg ⋅ m 2 2 IR =

[0.960 + 0.960 + (0.6)2 (8) + (1.2)2 (12)]θ + (0.6) 2 (800)θ = 0 288 N ⋅ m θ + θ =0 (22.08 kg ⋅ m 2 ) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2288

PROBLEM 19.39 (Continued)

(a)

Natural frequency and period. 288 22.08 = 3.6116 rad/s 2π 2π τn = = ωn 3.6116

ωn =

(b)

τ n = 1.740 s 

Maximum velocity at B. (vB )max = (1.2)(θmax ) yB 1.2 0.025 θm = = 0.02083 rad 1.2 θ = θ m sin(ωn t + φ ) θ = θ ω cos(ω t + φ )

θm =

m

θ

max

n

n

= θ mωn = (0.02083)(3.612) = 0.07524 rad/s

(vB )max = (1.2)(θmax ) = (1.2 m)(0.07524) rad/s (vB )max = 0.09029 m/s

(vB )max = 90.3 mm/s 

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PROBLEM 19.40 Solve Problem 19.39, assuming that pin C is removed and that the disk can rotate freely about pin B. PROBLEM 19.39 An 8-kg uniform rod AB is hinged to a fixed support at A and is attached by means of pins B and C to a 12-kg disk of radius 400 mm. A spring attached at D holds the rod at rest in the position shown. If Point B is moved down 25 mm and released, determine (a) the period of vibration, (b) the maximum velocity of Point B.

SOLUTION (a)

Note: This problem is the same as Problem 19.39, except that the disk does not rotate, so that the effective moment I Dα = 0. Equation of motion.

ΣM A = (ΣM A )eff : FS = k (0.60 + δ st ) (0.6)(mR g − FS ) + 1.2mD g = I Rα + (0.6)(mR )(at ) D + 1.2(mD )(at ) B

At equilibrium (θ = 0),

(1)

FS = kδ st ΣM A = 0 = 0.6(mR g − δ st ) + 1.2 mD g

Substituting Eq. (2) into Eq. (1),

(2)

I Rα + 0.6 mR (at ) D + 1.2mD (at ) B + (0.6) 2 kθ = 0

α = θ ( at ) B = 0.6θ (a ) = 1.2θ t D

1 1 mR l 2 = (8)(1.2) 2 12 12 = 0.960 kg ⋅ m

IR =

[0.960 + (0.6)2 (8) + (1.2) 2 (12)]θ + (0.6)2 (800)θ = 0 (288 N ⋅ m) θ + θ =0 21.12 kg ⋅ m 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2290

PROBLEM 19.40 (Continued)

(a)

Natural frequency and period. 288 21.12 = 3.693 rad/s 2π 2π τn = = ωn 3.693

ωn =

(b)

τ n = 1.701 s 

Maximum velocity at B. (vB ) max = (1.2)(θ) max yB 0.025 = = 0.02083 rad 1.2 1.20 θ = θ m sin(ωn t + φ ) θ = θ ω cos(ω t + φ )

θm =

m

n

n

θmax = θ mωn

(vB )max

= (0.02083)(3.693) = 0.07694 rad/s = (1.2)(θ ) max

= (1.2)(0.07694) = 0.09233 m/s

(vB )max = 92.3 mm/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2291

PROBLEM 19.41 A 15-lb slender rod AB is riveted to a 12-lb uniform disk as shown. A belt is attached to the rim of the disk and to a spring which holds the rod at rest in the position shown. If end A of the rod is moved 0.75 in. down and released, determine (a) the period of vibration, (b) the maximum velocity of end A.

SOLUTION

Equation of motion.

ΣM B = Σ( M B )eff : mg

where x = rθ + δ st and from statics, mg

L  L  L  cos θ − kxr = I ABα + m  α   + I diskα 2  2  2 

(1)

L = kδ st r 2

Assuming small angles (cos θ ≈ 1), Equation (1) becomes 2   L L θ − kr 2θ − krδ st =  I AB + m   + I disk  α   2 2   2  mL + + I disk  θ + kr 2θ = 0 4 

mg

  I AB 

Data:

15 32.2 = 0.46584 lb ⋅ s 2 /ft

m=

12 32.2 = 0.37267 lb ⋅ s 2 /ft L = 36 in. = 3.0 ft r = 10 in. = 0.83333 ft k = 30 lb/in. = 360 lb/ft

mdisk =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2292

PROBLEM 19.41 (Continued)

1 mL2 12 1 = (0.46584)(3.0) 2 12 = 0.34938 lb ⋅ s 2 ⋅ ft

I AB =

1 mdisk r 2 2 1 = (0.37267)(0.83333)2 2 = 0.1294 lb ⋅ s 2 ⋅ ft

I disk =

1    2 2 0.34935 + 4 (0.46584)(3.0) + 0.1294  θ + (360)(0.83333) θ = 0   1.5269θ + 250θ = 0 or θ + 163.73θ = 0

(a)

Natural frequency and period.

ωn2 = 163.73 (rad/s)2

ωn = 12.796 rad/s 2π 2π τ= = ωn 12.796 (b)

Maximum velocity.

vm = ωn xm = (12.796)(0.75)

τ = 0.491 s  vm = 9.60 in./s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2293

PROBLEM 19.42 A 30-lb uniform cylinder can roll without sliding on a 15°-incline. A belt is attached to the rim of the cylinder, and a spring holds the cylinder at rest in the position shown. If the center of the cylinder is moved 2 in. down the incline and released, determine (a) the period of vibration, (b) the maximum acceleration of the center of the cylinder.

SOLUTION

x A = x0 + x A/0

Spring deflection.

x A/0 = rθ x0 r x A = 2 x0

θ=

FS = k ( x A + δ st ) = k (2 x0 + δ st )

ΣM C = (ΣM )eff : − 2rk (2 x0 + δ st ) + rW sin15° = rmx0 + I θ

(1)

x0 = 0

But in equilibrium,

ΣM C = 0 = −2rkδ st + rW sin15°

Substituting Eq. (2) into Eq. (1) and noting that rmx0 + I

θ=

(2)

 x0 x , θ = 0 r r

 x0 + 4rkx0 = 0 r 1 I = mr 2 2

3 mrx0 + 4rkx0 = 0 2 8 k   x0 +   x0 = 0 3 m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2294

PROBLEM 19.42 (Continued)

Natural frequency.

8k (8)(30 × 12 lb/ft) = = 32.1 s −1 (30 lb) 3m (3) 2

ωn =

(32.2 ft/s )

(a)

Period.

τn =

2π 2π = = 0.1957 s ωn 32.1

τ n = 0.1957 s 

x0 = ( x0 )m sin(ωn t + φ )

(b) At t = 0,

2 ft x0 = 0 12 x0 = ( x0 )m ωn cos(ωn t + φ ) x0 =

t=0 0 = ( x0 )m ωn cos φ

Thus,

φ=

π

2 t =0

x0 (0) =

1 ft = ( x0 ) m sin φ = ( x0 ) m (1) 6

1 ft 6  x0 = −( x0 ) m ωn2 sin(ωn t + φ ) x0 )max (a0 ) max = (  ( x0 )m =

= −( x0 ) m ωn2

1  = −  ft  (32.1 s −1 ) 2 6  = 171.7 ft/s 2

(a0 ) max = 171.7 ft/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2295

PROBLEM 19.43 A square plate of mass m is held by eight springs, each of constant k. Knowing that each spring can act in either tension or compression, determine the frequency of the resulting vibration (a) if the plate is given a small vertical displacement and released, (b) if the plate is rotated through a small angle about G and released.

SOLUTION (a)

Small vertical displacement. Let the plate be displaced downward a distance x from the equilibrium position. Each corner moves downward a distance x and the four vertical springs exert additional forces k x for each spring. The horizontal springs exert negligible change. ΣF = ma : − 4kx = mx

 x+

4k x=0 m

ωn2 =

Frequency: (b)

f =

ωn 1 = 2π 2π

4k m

4k m

f = 0.318

k  m

Small rotation about G. Let the plate be rotated through a small counterclockwise angle θ from the equilibrium position. The corners A, B, C, and D move as indicated below: A:

(l /2)θ + (l /2)θ

B:

(l /2)θ

C:

(l /2)θ + (l /2)θ

D:

(l /2)θ

+ (l /2)θ

+ (l /2)θ

The additional force exerted by each of the eight springs is F = (kl /2)θ and directed as shown on the free body diagram. The eight forces reduce to four clockwise couples, each of magnitude Fl. For a square plate I =

1 2 ml 6

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2296

PROBLEM 19.43 (Continued)

M G = I α : 4 Fl = I θ − 4( kl 2 /2)θ

=

θ +

Frequency:

f =

ωn 1 12k = 2π 2π m

1 2  ml θ 6

12k θ =0 m 12k ωn2 = m

f = 0.551

k  m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2297

PROBLEM 19.44 Two small weights w are attached at A and B to the rim of a uniform disk of radius r and weight W. Denoting by τ 0 the period of small oscillations when β = 0, determine the angle β for which the period of small oscillations is 2τ 0 .

SOLUTION

α = θ at = rα = rθ ID =

1W 2 r 2 g

Equation of motion. ΣM C = (ΣM C )eff : wr sin( β − θ ) − wr sin( β + θ ) =

2w rat + I α g

wr[sin( β − θ ) − sin( β + θ )] = −2wr sin θ cos β sin θ ≈ θ  2w 2 W 2   r + r  θ + (2 wr cos β ) θ = 0  2g   g

Natural frequency.

ωn =

2 wg cos β 4 g cos β = W ( 2w + 2 ) r (4 + Ww )r

β =0 τn =

τ0 = 2π cos β (4 + Ww ) r

2π = ω0

= 2τ 0 =

(1)

2π 4g (4 + Ww ) r

4π 4g (4 + Ww ) r

2

1 1 cos β =   = 4 2

β = 75.5° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2298

PROBLEM 19.45 Two 40-g weights are attached at A and B to the rim of a 1.5-kg uniform disk of radius r = 100 mm. Determine the frequency of small oscillations when β = 60°.

SOLUTION

α = θ at = rα = rθ ID =

1 mD r 2 2

Equation of motion. ΣM C = I α + mad : wr sin( β − θ ) − wr sin( β + θ ) =

2w rat + I α g

wr[sin( β − θ ) − sin( β + θ )] = −2wr sin θ cos β sin θ ≈ θ  2w 2 W 2   r + r  θ + (2 wr cos β ) θ = 0  2g   g

Natural frequency.

Data:

ωn =

2 wg cos β 4 g cos β = W + Ww )r (4 ( 2w + 2 ) r

w = mg = 0.04 g r = 0.100 m

ωn = fn =

W = mD g = 1.5 g

(1) W 1.5 g = = 37.5 w 0.04 g

β = 60°

(4)(9.81) cos 60° = 2.1743 rad/s (4 + 37.5)(0.10)

ωn 2.1743 = 2π 2π

f n = 0.346 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2299

PROBLEM 19.46 A three-bladed wind turbine used for research is supported on a shaft so that it is free to rotate about O. One technique to determine the centroidal mass moment of inertia of an object is to place a known weight at a known distance from the axis of rotation and to measure the frequency of oscillations after releasing it from rest with a small initial angle. In this case, a weight of Wadd = 50 lb is attached to one of the blades at a distance R = 20 ft from the axis of rotation. Knowing that when the blade with the added weight is displaced slightly from the vertical axis, the system is found to have a period of 7.6 s, determine the centroidal mass moment of inertia of the 3-bladed rotor.

SOLUTION Let the turbine rotor be turned counterclockwise through a small angle θ. The moment of the added weight about Point O is

M = −Wadd R sin θ ΣM 0 = Σ( M 0 )eff : − Wadd R sin θ = I α + madd Raa

= ( I + madd R 2 )α = ( I + m R 2 )θ add

θ + Using sin θ ≈ θ

Wadd R I + madd R 2

sin θ = 0

gives

θ +

Wadd R I + madd R 2

θ =0

θ + ωn2θ = 0

ωn2 =

Wadd R I + madd R 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2300

PROBLEM 19.46 (Continued)

Solving for I , I =

Data:

ωn2

R = 20 ft, madd =

Period and frequency:

Wadd R

− madd R 2

(1)

Wadd = 50 lb

Wadd 50 lb = = 1.5528 lb ⋅ s 2 /ft 2 g 32.2 ft/s

τ = 7.6 s 1 Hz τ 7.6 2π = 0.82673 rad/s ωn = 2π f = 7.6

From Eq. (1),

f =

1

I =

(50 lb)(20 ft) − (1.5528 lb ⋅ s 2 /ft)(20 ft) 2 (0.82673 rad/s) 2

=

= 1463.10 − 621.12 I = 842 lb ⋅ s 2 ⋅ ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2301

PROBLEM 19.47 A connecting rod is supported by a knife-edge at Point A; the period of its small oscillations is observed to be 0.87 s. The rod is then inverted and supported by a knifeedge at Point B and the period of small oscillations is observed to be 0.78 s. Knowing that ra + rb = 10 in. determine (a) the location of the mass center G, (b) the centroidal radius of gyration k .

SOLUTION Consider general pendulum of centroidal radius of gyration k .

Equation of motion.

ΣM 0 = Σ( M 0 )eff : − mgr sin θ = ( mr θ)r + mk 2θ  gr  sin θ = 0 2 2 r + k 

θ +  For small oscillations, sin θ ≈ θ , we have  gr  θ =0 2 2 r + k 

θ + 

ωn2 = τ= For rod suspended at A,

gr r +k2 2



ωn

= 2π

r2 +k2 gr

ra2 + k 2 gra

τ A = 2π

(

gτ A2 ra = 4π 2 ra2 + k 2

For rod suspended at B,

)

(1)

)

(2)

rb2 + k 2 grb

τ B = 2π

(

gτ B2 rb = 4π 2 rb2 + k 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2302

PROBLEM 19.47 (Continued)

(a)

Value of ra .

Subtracting Eq. (2) from Eq. (1),

(

gτ A2 ra − gτ B2 rb = 4π 2 ra2 − rb2 gτ A2 ra



gτ B2 rb

)

= 4π ( ra + rb )(ra − rb ) 2

Applying the numerical data with ra + rb = 10 in. = 0.83333 ft (32.2)(0.87) 2 ra − (32.2)(0.78) 2 rb = 4π 2 (0.83333)(ra − rb ) 24.372ra − 19.590rb = 32.899(ra − rb ) 13.309rb = 8.527ra

rb = 0.6407ra

0.83333 = ra + 0.6407 ra

ra = 0.5079 ft

ra = 6.09 in. 

rb = 0.32543 ft

rb = 3.91 in. 

rb = 0.83333 − 0.5079

(b)

Centroidal radius of gyration. From Eq. (1),

4π 2 k 2 = g τ A2 ra − 4π 2 ra2 = (32.2)(0.87)2 (0.5079) − 4π 2 (0.5079) 2 = 2.1947 ft 2 k = 0.2398 ft

k = 2.83 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2303

PROBLEM 19.48 A 75-mm-radius hole is cut in a 200-mm-radius uniform disk, which is attached to a frictionless pin at its geometric center O. Determine (a) the period of small oscillations of the disk, (b) the length of a simple pendulum which has the same period.

SOLUTION

Equation of motion. ΣM 0 = (ΣM 0 )eff :

− mH g (0.1) sin θ = I Dθ − I H θ − (0.1) 2 mH θ mD = ρ tπ R 2

= ( ρ tπ )(0.2) 2 = (0.04) πρ t mH = ρ tπ r 2

= ( ρ tπ ) (0.075) 2 = (0.005625) πρ t 1 1 I D = mD R 2 = (0.04 πρ t )(.2) 2 2 2 −6 = 800 × 10 πρ t 1 I H = mH r 2 2 1 = (0.005625 πρ t )(0.75)2 2 = 15.82 × 10−6 πρ t

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2304

PROBLEM 19.48 (Continued)

Small angles.

sin θ ≈ θ [800 × 10−6πρ t − 15.82 × 10−6πρ t − (0.1)(0.005625) πρ t ]θ + (0.005625 πρ t ) (9.81)(0.1)θ = 0 727.9 × 10−6 θ + 5.518 × 10−3 θ = 0

(a)

Natural frequency and period. 5.518 × 10−3 727.9 × 10−6 = 7.581

ωn2 =

ωn = 2.753 rad/s 2π τn = ωn (b)

τn =

2π = 2.28 s  2.753

Length and period of a simple pendulum.

τ n = 2π

l g 2

τ  l= n  g  2π  2

 (2.753)  2 l=  (9.81 m/s ) π 2  

l = 1.294 m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2305

PROBLEM 19.49 A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of negligible mass, which can rotate freely in a vertical plane about B. Determine the period of small oscillations (a) if the disk is free to rotate in a bearing at A, (b) if the rod is riveted to the disk at A.

SOLUTION

1 2 mr 2 1 m = (0.250)2 m = 2 32 at = lα = 0.650α α = θ I =

(a)

The disk is free to rotate and is in curvilinear translation. Iα = 0

Thus, ΣM B = Σ( M B )eff :

− mgl sin θ = lmat

sin θ ≈ θ

ml 2θ − mglθ = 0 g 9.81 m/s 2 = I 0.650 m = 15.092

ωn2 =

ωn = 3.885 rad/s 2π 2π τn = = ωn 3.885

τ n = 1.617 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2306

PROBLEM 19.49 (Continued)

(b)

When the disk is riveted at A, it rotates at an angular acceleration α . ΣM B = Σ( M B )eff :

− mgl sin θ = I α + lmat

I=

1 2 mr 2

1 2 2    2 mr + ml θ + mglθ = 0  

ωn2 = =

(

gl 2

r 2

+ l2

)

(9.81 m/s 2 )(0.650 m)  0.2502 + (0.650) 2   2 

(

)

= 14.053 ωn = 3.749 rad/s

τn = =

2π ωn 2π 3.749

τ n = 1.676 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2307

PROBLEM 19.50 A small collar of mass 1 kg is rigidly attached to a 3-kg uniform rod of length L = 750 mm. Determine (a) the distance d to maximize the frequency of oscillation when the rod is given a small initial displacement, (b) the corresponding period of oscillation.

SOLUTION

Equation of motion. ΣM A = (ΣM A )eff : sin θ ≈ θ

−WR

L L sin θ − WC d sin θ = I Rα + mR (at ) R + mC d (at )C 2 2

α = θ, (at ) R =

L L α = θ, (at )C = dα = dθ 2 2

2   L L    I R + mR   + mC d 2 θ +  mR g + mC gd θ = 0   2 2    

IR =

1 mR L2 12

2

m L2 L I R + mR   = R 3 2  mR L2  L   + mC d 2 θ +  mR g + mC gd θ = 0  3 2    

( θ + (

L 2 2

L 3

+ +

mC mR mC mR

)

d g d2

)

θ =0

mC 1 = mR 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2308

PROBLEM 19.50 (Continued)

Natural frequency. (a)

ωn2

( =

L 2 L2 3

mC mR

+ +

mC mR

) =(

d g d

3L + 2 2

d)g

(L + d 2 )

2

To maximize the frequency, we need to take the derivative with respect to d and set it equal to zero.

( )

2 ( L2 + d 2 )(1) − ( 32L + d ) (2d ) 1 d ωn = =0 g d (d ) ( L2 + d 2 ) 2

d 2 + L2 − 3Ld − 2d 2 = 0 d 2 + 3Ld − L2 = 0

Solve for d knowing that L = 0.75 m

d = 0.22708 or − 2.4771 d = 0.22708 m

ωn2

(b)

Period of oscillation.

( =

3(0.75) 2

d = 227 mm 

)

+ 0.22708 9.81

(0.75 + 0.227082 ) = 21.6

τn =

2



ωn

=

2π 4.6476

ωn = 4.6476 rad/s τ n = 1.352 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2309

PROBLEM 19.51 For the uniform square plate of side b = 12 in, determine (a) the period of small oscillations if the plate is suspended as shown, (b) the distance c from O to a Point A from which the plate should be suspended for the period to be a minimum.

SOLUTION

(a)

Equation of motion. ΣM 0 = I α + mad : α = θ 1 2 mb 6 at = (OG )(α ) I =

OG = b

2 2

 2   θ at =  b  2    (OG )(sin θ )( mg ) = −(OG )mat − I α

sin θ ≈ θ

  2  2   1 2 2  b  m  b θ + mb θ +  b  mg θ = 0 6  2   2   2   2 1 1 (b)  +  m θ +  mg θ = 0  2  2 6   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2310

PROBLEM 19.51 (Continued)

θ +

( )gθ = 0 2 2

3 2 g or θ + =0 4 b

( )b 2 3

Natural frequency and period. 3 2 g 3 2(32.2) = = 34.153 4 b (4)(1) = 5.8441 rad/s

ωn20 = ωn 0

τ n0 = (b)

Suspended about A.

2π ωn 0

τ n 0 = 1.075 s 

Let e = (OG − c) at = eα

Equation of motion. ΣM A = I α + mad :

mge sin θ = −emat − I α = −( me 2 + I )α 1   m  e 2 + b 2  θ + mgeθ = 0 6  

ωn2 =

Frequency and period.

τ n2 = τ n2 = For τ n to be minimum,

eg e + 16 b 2 2

4π 2

ωn2 4π 2 g

=

4π 2 (e2 + 16 b 2 ) eg

 b2   e +  6e  

d  b2   e +  = 0 de  6e  1−

b2 =0 6e 2

b2 =6 e2

e=

b 6

2 b b− = 0.29886b 2 6 c = (0.29886)(12 in.) c = OG − e =

c = 3.59 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2311

PROBLEM 19.52 A compound pendulum is defined as a rigid slab which oscillates about a fixed Point O, called the center of suspension. Show that the period of oscillation of a compound pendulum is equal to the period of a simple pendulum of length OA, where the distance from A to the mass center G is GA = k 2 / r . Point A is defined as the center of oscillation and coincides with the center of percussion defined in Problem 17.66.

SOLUTION

ΣM 0 = Σ( M 0 )eff : − W r sin θ = I α + mat r − mgr sin θ = mk 2θ + mr 2θ

θ +

gr sin θ = 0 r +k2

(1)

2

For a simple pendulum of length OA = l ,

g l

θ + θ = 0 Comparing Equations (1) and (2),

l=

(2)

r2 +k2 r

GA = l − r =

k2 Q.E.D.  r

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2312

PROBLEM 19.53 A rigid slab oscillates about a fixed Point O. Show that the smallest period of oscillation occurs when the distance r from Point O to the mass center G is equal to k .

SOLUTION See Solution to Problem 19.52 for derivation of

θ +

gr sin θ = 0 r +k2 2

For small oscillations, sin θ ≈ θ and

τn = For smallest τ n , we must have r +

k2 r



ωn

r2 +k2 2π = gr g

= 2π

r+

k2 r

as a minimum:

(

d r+ dr

k2 r

) =1− k r

2 2

r2 =k2

=0

r = k Q.E.D. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2313

PROBLEM 19.54 Show that if the compound pendulum of Problem 19.52 is suspended from A instead of O, the period of oscillation is the same as before and the new center of oscillation is located at O.

SOLUTION Same derivation as in Problem 19.52 with r replaced by R. Thus,

θ +

gR θ =0 R +k 2

Length of the equivalent simple pendulum is R2 + k 2 k2 =R+ R R 2 k L = (l − r ) + 2 = l

L=

k r

Thus, the length of the equivalent simple pendulum is the same as in Problem 19.52. It follows that the period is the same and that the new center of oscillation is at O. Q.E.D.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2314

PROBLEM 19.55 The 8-kg uniform bar AB is hinged at C and is attached at A to a spring of constant k = 500 N/m. If end A is given a small displacement and released, determine (a) the frequency of small oscillations, (b) the smallest value of the spring constant k for which oscillations will occur.

SOLUTION I =

1 2  1  ml =   (8)(0.250) 2 12  12 

I = 0.04167 kg ⋅ m 2 α = θ at = 0.04α = 0.04θ sin θ ≈ θ

Equation of motion. ΣMC = Σ( MC )eff :

−(0.165) 2 kθ + 0.04mgθ = I θ + (0.04)2 mθ (0.04167 + 0.01280)θ + (0.02722k − 0.32 g )θ = 0

(a)

(1)

Frequency if k = 500 N ⋅ m. 0.05447θ + (10.47)θ = 0 fn =

(b)

For τ n

∞ or ωn

From Equation (1),

ωn = 2π

(

10.47 0.05447



)

f n = 2.21 Hz 

0, oscillations will not occur.

ωn2 = k=

0.02722k − 0.32 g =0 (0.05447) 0.32 g (0.32)(9.81) = 0.02722 (0.02722)

k = 115.3 N/m 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2315

PROBLEM 19.56 Two uniform rods, each of mass m = 12 kg and length L = 800 mm, are welded together to form the assembly shown. Knowing that the constant of each spring is k = 500 N/m and that end A is given a small displacement and released, determine the frequency of the resulting motion.

SOLUTION

Equation of motion.

ΣM 0 = Σ( M 0 )eff :

2 2  L L  L  m AC g − 2k    θ = ( I AC + I BD )α + mAC   α 2  2   2 

mBD = m AC = m I BD = I AC = I =

1 2 mL 2

2 L  1 1  2    L  + θ + − 2 mL k mg  θ = 0 6 4   2      2 

10 2   kL2 mgL  6(kL − mg ) 2 mL θ +  −  θ = 0  ωn = 24 2  5mL  2

Data: Frequency.

L = 800 mm = 0.8m, m = 12 kg, k = 500 N/m

ωn2 =

6[(500)(0.8) − (12)(9.81)] = 35.285 (5)(12)(0.8)

ωn = 5.9401 rad/s

fn =

ωn 2π

f n = 0.945 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2316

PROBLEM 19.57 A 45-lb uniform square plate is suspended from a pin located at the midpoint A of one of its 1.2-ft edges and is attached to springs, each of constant k = 8 lb/in. If corner B is given a small displacement and released, determine the frequency of the resulting vibration. Assume that each spring can act in either tension or compression.

SOLUTION

α = θ b b at = α − θ 2 2 sin θ ≈ θ 2

Equation of motion.

b b ΣM 0 = Σ( M 0 )eff : − mg θ + 2kb 2θ = I α +   mα 2 2 2

1 b2 5 b = mb 2 I + m   = mb 2 + m 2 6 4 12   5 b   mb 2θ +  mg + 2kb θ = 0 12 2  

 12 g 24k  + θ = 0  10 b 5mb 

θ + 

45 = 1.3975 lb ⋅ s 2 /ft 32.2 k = 8 lb/in. = 96 lb/ft

b = 1.2 ft; m =

Data:

 (12)(32.2)  (24)(96) + θ = 0  (10)(1.2) (5)(1.3975)(1.2) 

θ + 

θ + 306.98θ = 0 ωn2 = 306.98 ωn = 17.521 rad/s Frequency.

fn =

ωn 17.521 = 2π 2π

f n = 2.79 Hz 

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PROBLEM 19.58 A 1300-kg sports car has a center of gravity G located a distance h above a line connecting the front and rear axles. The car is suspended from cables that are attached to the front and rear axles as shown. Knowing that the periods of oscillation are 4.04 s when L = 4 m and 3.54 s when L = 3 m, determine h and the centroidal radius of gyration.

SOLUTION Let the mass center of the car be displaced a small distance x to the right. The mass center is moves on a circular arc of radius L − h, so that x = (L − h) sin θ, where θ is the counterclockwise rotation of the car. From kinematics α = θ a = ( L − h)θ t

The moment of the weight force about O is

M 0 = −mg ( L − h) sin θ M 0 = I α + ( L − h) mat − mg ( L − h)sin θ = I θ + m( L − h) 2 θ

Dividing by m and transposing terms yields [k 2 + ( L − h) 2 ]θ + g ( L − h)sin θ = 0

For small angle θ,

sin θ ≈ θ

θ +

g ( L − h) θ =0 k + ( L − h) 2 2

θ + ωn2θ = 0 k 2 + ( L − h) 2 =

ωn2 = g

ωn2

g ( L − h) k + ( L − h) 2 2

( L − h)

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PROBLEM 19.58 (Continued) Using the two different values (L1 and L2) for L, k 2 + ( L1 − h) 2 = k 2 + ( L 2 − h) 2 =

g

ωn21

( L1 − h)

(1)

( L 2 − h)

(2)

g

ωn22

Subtracting Eq. (2) from Eq. (1) to eliminate k 2 , ( L1 − h) 2 − ( L 2 − h) 2 =

(L

2 1



ωn21

gL 2

ωn22

)

 g g − 2 − 2 ω  n1 ωn 2

  h 

− L22 − 2( L1 − L 2 )h = A − Bh

where

A=

and

B= h=

Data:

gL1

L1 = 4 m,



ωn21 g

ωn21



gL 2

ωn22 g

ωn22

L12 − L22 − A 2( L1 − L2 ) − B

L 2 = 3 m,

g = 9.81 m/s 2

2π = = 1.55524 rad/s τ1 4.04 s 2π 2π = = = 1.77491 rad/s τ 2 3.54 s

ωn1 = ωn 2

gL1



(9.81)(4) (9.81)(3) − = 6.8812 m 2 2 (1.55524) (1.77491)2 9.81 9.81 − = 0.94190 m B= 2 (1.55524) (1.77491)2 A=

h=

(4) 2 − (3) 2 − 6.8812 = 0.11228 m 2(4 − 3) − 0.94190 h = 0.1123 m 

L1 − h = 3.88772 m

From Eq. (1),

k 2 + (3.88772) 2 =

L 2 − h = 2.88772 m

(9.81)(3.88772) (1.55524) 2

k 2 = 0.65336 m 2

Checking, using Eq. (2),

k 2 + (2.88772) 2 =

k = 0.808 m 

(9.81)(2.88772) (1.77491) 2

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PROBLEM 19.59 A 6-lb slender rod is suspended from a steel wire which is known to have a torsional spring constant K = 1.5 ft ⋅ lb/rad. If the rod is rotated through 180° about the vertical and released, determine (a) the period of oscillation, (b) the maximum velocity of end A of the rod.

SOLUTION Equation of motion.

K θ =0 I K ωn2 = I

ΣM G = Σ( M G )eff : − Kθ = I θ

θ +

θ + ωn2θ = 0 W = 6 lb.

Data:

m=

W 6 = = 0.186335 lb ⋅ s 2 /ft g 32.2

l = 8 in. = I =

2 ft 3

1 1 2 ml 2 = (0.186335)   12 12 3

2

= 0.006901 lb ⋅ s 2 ⋅ ft K = 1.5 lb ⋅ ft/rad 1.5 = 217.35 ωn2 = 0.006901 ωn = 14.743 rad/s

(a)



2π 14.743

Period of oscillation.

τn =

Simple harmonic motion:

θ = θ m sin (ωn t + ϕ )

ωn

=

τ n = 0.426 s 

θ = ωnθ m cos (ωn t + ϕ ) θm = ωnθ m l 1 (v A )m = θm = lω Aθ m 2 2 θ m = 180° = π radians

(b)

Maximum velocity at end A.

 1  2  (v A )m =    (14.743)(π )  2  3 

(v A )m = 15.44 ft/s 

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PROBLEM 19.60 A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of negligible mass which can rotate freely in a vertical plane about B. If the rod is displaced 2° from the position shown and released, determine the magnitude of the maximum velocity of Point A, assuming that the disk (a) is free to rotate in a bearing at A, (b) is riveted to the rod at A.

SOLUTION

1 2 mr 2 α = θ I =

Kinematics:

at = lα = lθ

(a)

The disk is free to rotate and is in curvilinear translation. Iα = 0

Thus, Equation of motion.

ΣM B = (ΣM B )eff : − mgl sin θ = lmat , sin θ ≈ θ ml 2θ + mglθ = 0 g l 9.81 = 0.650 = 15.092 ωn = 3.8849 rad/s θ m = 2° = 0.034906 rad θ = ω θ = (3.8849)(0.034906) = 0.13561 rad/s 

ωn2 =

Frequency.

m

n m

(v A ) m = lθm = (0.650)(0.13561)



(v A ) m = 0.0881 m/s 



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PROBLEM 19.60 (Continued)

(b)

When the disk is riveted at A, it rotates at an angular acceleration α . Equation of motion.

ΣM B = (ΣM B )eff : − mgl sin θ = I α + lmat , I =

1 2 mr , sin θ ≈ θ 2

1 2 2    2 mr + ml θ + mglθ = 0  

ωn2 =

Frequency.

=

( 1 2

gl r2 2

+ l2

)

(9.81)(0.650) (0.250) 2 + (0.650)2

= 14.053 ωn = 3.7487 rad/s

θ m = 2° = 0.034906 rad θm = ωnθ m = (3.7487)(0.034906) = 0.13085 rad/s  (v A ) m = lθm = (0.650)(0.13085)



(v A ) m = 0.0851 m/s 



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PROBLEM 19.61 Two uniform rods, each of mass m and length l, are welded together to form the T-shaped assembly shown. Determine the frequency of small oscillations of the assembly.

SOLUTION Let the assembly be rotated counterclockwise through the small angle θ about the fixed Point A.

Equation of motion.

l l ΣM A = Σ( M A )eff : − mg sin θ − mgl sin θ = I ABα + ma AB + I CDα + maCD l 2 2 2

3 1 1 l − mgl sin θ = ml 2θ + m   θ + ml 2θ + ml 2θ 2 12 2 12   3 17 2  − mgl sin θ = ml θ 2 12 18 g θ + sin θ = 0 17 l

For small oscillations,

sin θ ≈ θ

θ +

Frequency.

f =

18 g θ =0 17 l 18 g ωn2 = 17 l

ωn 2π

f =

1 2π

ωn = 18 g 17l

18 g 17l f = 0.1638

g  l

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PROBLEM 19.62 A homogeneous wire bent to form the figure shown is attached to a pin support at A. Knowing that r = 220 mm and that Point B is pushed down 20 mm and released, determine the magnitude of the velocity of B, 8 s later.

SOLUTION

Determine location of the centroid G. Let

ρ = mass per unit length

Then total mass

m = ρ (2r + π r ) = ρ r (2 + π )

About C:

 2r  mgc = 0 + π r ρ   g = 2r 2 ρ g π  y=

2r

π

for a semicircle

ρ r (2 + π )c = 2r 2 ρ , c = Equation of motion.

ΣM 0 = Σ( M 0 )eff : α = θ at = cα = cθ − mgc sin θ = I α + mcan ( I + mc 2 )θ + mgcθ = 0

Moment of inertia.

2r (2 + π )

sinθ ≈ θ

I 0θ + mgcθ = 0

I + mc 2 = I 0 I 0 = ( I 0 )semicirc. + ( I 0 )line = msemicirc.r 2 + mline msemicirc. = ρπ r

mline = 2 ρ r

ρ=

(2r ) 2 12

m (2 + π )r

 r2  mr 2  2 I 0 = ρ π r 2 ⋅ r + 2r ⋅  = π+   3  (2 + π )  3  mr 2  2 2r π +  θ + mg θ =0  (2 + π )  3 (2 + π )

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PROBLEM 19.62 (Continued)

Frequency.

ωn2 =

2g (2)(9.81) = 2 (π + 3 ) r (π + 23 ) (0.220)

ωn2 = 23.418 s −2 ωn = 4.8392 rad/s θ = θ m sin(ωn t + φ ) yB = rθ At t = 0,

yB = 20 mm, y B = 0 y B = 0 = ( yB ) m ωn cos(0 + φ ), φ =

π 2

π  yB = 20 mm = ( yB ) m sin  0 +  , (yB ) m = 20 mm 2  π  yB = (20 mm) sin  ωn t +  ωn = 4.8392 rad/s 2  π  y B = 20ω cos  ωn t +  = −(20 mm)ωn sin ωn t 2  At t = 8 s,

y B = −(20)(4.8392) sin[(4.8392)(8)] = −(96.78)(0.8492) = −82.2 mm/s

vB = 82.2 mm/s 

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PROBLEM 19.63 A horizontal platform P is held by several rigid bars which are connected to a vertical wire. The period of oscillation of the platform is found to be 2.2 s when the platform is empty and 3.8 s when an object A of unknown moment of inertia is placed on the platform with its mass center directly above the center of the plate. Knowing that the wire has a torsional constant K = 27 N ⋅ m/rad, determine the centroidal moment of inertia of object A.

SOLUTION

Equation of motion.

ΣM G = I α : −Kθ = I θ K

θ +  θ = 0 I Case 1. The platform is empty.

ωn1 = I1 =

Case 2. Object A is on the platform.

ωn 2 = I2 =

Moment of inertia of object A.



τ1 K

ωn21 2π

τ2 K

ωn22

ωn2 =

K I

=

2π = 2.856 rad/s 2.2

=

27 = 3.31 kg ⋅ m 2 (2.856)2

=

2π = 1.653 rad/s 3.8

=

27 = 9.8814 kg ⋅ m 2 2 (1.653)

I A = I 2 − I1

I A = 6.57 kg ⋅ m 2 

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PROBLEM 19.64 A uniform disk of radius r = 120 mm is welded at its center to two elastic rods of equal length with fixed ends at A and B. Knowing that the disk rotates through an 8° angle when a 500-mN ⋅ m couple is applied to the disk and that it oscillates with a period of 1.3 s when the couple is removed, determine (a) the mass of the disk, (b) the period of vibration if one of the rods is removed.

SOLUTION Torsional spring constant. k=

T

θ

=

0.5 N ⋅ m π (8) ( 180 )

k = 3.581 N ⋅ m/rad

Equation of motion. K ΣM 0 = Σ( M 0 )eff : −Kθ = Iθ θ + θ = 0 I

Natural frequency and period.

ωn2 =

Period.

τn =

Mass moment of inertia.

I=

K I 2π

ωn

= 2π

I K

τ n2 K (1.35) 2 (3581 N ⋅ m/r) = 2 (2π ) (2π ) 2

I = 0.1533 N ⋅ m ⋅ s 2 =

1 2 1 mr = m(0.120 m) 2 2 2

(a)

Mass of the disk.

m=

(0.1533 N ⋅ m ⋅ s 2 )(2) (0.120 m) 2

(b)

With one rod removed:

K′ =

K 3.581 = = 1.791 N ⋅ m/rad 2 2

Period.

τ = 2π

I (0.1533 N ⋅ m ⋅ s 2 ) = 2π K′ 1.791 N ⋅ m/rad

m = 21.3 kg 

τ = 1.838 s 

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PROBLEM 19.65 A 5-kg uniform rod CD of length l = 0.7 m is welded at C to two elastic rods, which have fixed ends at A and B and are known to have a combined torsional spring constant K = 24 N ⋅ m/rad. Determine the period of small oscillation, if the equilibrium position of CD is (a) vertical as shown, (b) horizontal.

SOLUTION (a)

Equation of motion.

α = θ

l l at = α = θ 2 2

l l ΣM C = I α + mad : −Kθ − ( mg ) sin θ = I α + (mat ) 2 2 − Kθ −

1 1 mgl sin θ = I θ + ml 2θ 2 4

1 2   1   I + ml θ + mgl sin θ + Kθ = 0 4 2   1 2   1  1 2  12 ml + 4 ml θ + Kθ + 2 mglθ = 0   1 2   1  ml θ +  K + mgl θ = 0 3 2    3K 3 g  θ =0 + 2 2l   ml

θ + 

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PROBLEM 19.65 (Continued)

K = 24 N ⋅ m/rad, m = 5 kg, l = 0.7 m

Data:

 (3)(24) (3)(9.81)  + θ = 0 2 (2)(0.7)   (5)(0.7) θ + 50.409θ = 0

θ + 

ωn2 = 50.409

Frequency.

τ=

Period. (b)



ωn

=

ωn = 7.1 rad/s

2π 7.1

τ n = 0.885 s 

If the rod is horizontal, the gravity term is not present and the equation of motion is

θ +

3K θ =0 ml 2

ωn2 =

3K (3)(24) = = 29.388 2 ml (5)(0.7) 2

ωn = 5.4210 rad/s

τ=



ωn

=

2π 5.4210

τ n = 1.159 s 

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PROBLEM 19.66 A 1.8-kg uniform plate in the shape of an equilateral triangle is suspended at its center of gravity from a steel wire which is known to have a torsional constant K = 35 mN ⋅ m/rad. If the plate is rotated 360° about the vertical and then released, determine (a) the period of oscillation, (b) the maximum velocity of one of the vertices of the triangle.

SOLUTION Mass moment of inertia of plate about a vertical axis: 3 b 2

h=

For area,

Ix = Iy =

A=

1 3 3b 4 bh = 36 96

Iz = Ix + I y =

For mass,

3 4 b 48

m ( I z )area A  4 m  3 4  1 = b  = mb2    12 48 3 b   

I =

I =

Equation of motion.

1 3 2 bh = b 2 4

1 (1.8)(0.150) 2 = 3.375 × 10−3 kg ⋅ m 2 12

+ ΣM G = Σ( M G )eff : − Kθ = I θ

θ + Frequency.

K θ =0 I

K 35 × 10−3 = = 10.37 I 3.375 × 10−3 ωn = 3.2203 rad/s

ωn2 =

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PROBLEM 19.66 (Continued)

(a)

(b)

Period.

τ=



ωn

=

2π 3.2203

Maximum rotation.

θ m = 360° = 2π rad

Maximum angular velocity.

θm = ωnθ m = (3.2203)(2π ) = 20.234 rad/s

τ = 1.951 s 

Maximum velocity at a vertex. vm = rθm =

2  2 3  2  3  hθ m = b =    (0.150)(20.234) 3 3 2  3   2  vm = 1.752 m/s 

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PROBLEM 19.67 A period of 6.00 s is observed for the angular oscillations of a 4-oz gyroscope rotor suspended from a wire as shown. Knowing that a period of 3.80 s is obtained when a 1.25-in.-diameter steel sphere is suspended in the same fashion, determine the centroidal radius of gyration of the rotor. (Specific weight of steel = 490 lb/ft3.)

SOLUTION ΣM = Σ( M )eff : − Kθ = I θ

θ +

K θ =0 I K ωn2 = I

τ = 2π K=

I K

(1)

4π 2 I

(2)

τ2

Kτ 2 4π 2 d r = = 0.625 in. = 52.083 × 10−3 ft 2 4 4 Vs = π r 3 = π (52.083 × 10−3 )3 3 3 = 591.81 × 10 −6 ft 3 I =

For the sphere, Volume:

Weight:

(3)

Ws = γ Vs

= (490 lb/ft 3 )(591.81 × 10−6 ft 3 ) = 0.28999 lb

Mass:

ms =

Ws 0.28999 = 32.2 g

= 9.0059 × 10−3 lb ⋅ s 2 /ft

Moment of inertia:

Period:

2 2 ms r 2 = (9.0059 × 10−3 )(52.083 × 10−3 )2 5 5 = 9.7719 × 10−6 lb ⋅ s 2 ⋅ ft

I =

τ s = 3.80 s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2332

PROBLEM 19.67 (Continued)

From Eq. (2):

K=

4π 2 (9.7719 × 10−6 ) (3.80) 2

= 26.716 × 10−6 lb ⋅ ft/rad

For the rotor,

m=

W  4  1  =    g  16  32.2 

= 7.764 × 10−3 lb ⋅ s 2 /ft τ = 6.00 s (6.00) 2 4π 2 = 24.362 × 10−6 lb ⋅ s 2 ⋅ ft

From Eq. (3):

I = (26.716 × 10−6 )

Radius of gyration.

I = mk 2 k =

I m

24.362 × 10−6 7.764 × 10−3 = 0.056016 ft =

k = 0.672 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2333

PROBLEM 19.68 The centroidal radius of gyration k y of an airplane is determined by suspending the airplane by two 12-ft-long cables as shown. The airplane is rotated through a small angle about the vertical through G and then released. Knowing that the observed period of oscillation is 3.3 s, determine the centroidal radius of gyration k y .

SOLUTION Let the airplane rotate through the small angle θ about a vertical axis. Suspension Points C and D on the airplane each move horizontally a distance (10 ft) sin θ ≈ (10 ft) θ. Let ϕ be the angle between a cable and the vertical direction. Then, sin ϕ = (10 ft)θ /(12 ft) = 56 θ . ΣF = 0: 2T cos ϕ − W = 0 T=

W W ≈ 2 cos ϕ 2

Let F be the horizontal component of T. F = T sin ϕ ≈

W 5 5 ⋅ θ = Wθ 2 6 12

The two forces F form a couple of moment  5 M = −(20 ft) F = −(20)   W θ  12 

Equation of motion:

 5W  W 2  ΣM y = I yα : − 20  θ  = ky θ  12  g (20)(5) g θ + θ =0 12k y2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2334

PROBLEM 19.68 (Continued)

Natural frequency:

ωn2 =

(20)(5) g (20)(5)(32.2) 268.33 = = 12 k y2 12 k y2 k y2

k y2 =

268.33

ky =

16.381

=

With τ = 3.3 s,

ωn2 ωn

=

16.381 2π f

2.607 = 2.607τ f

k y = (2.607)(3.3) k y = 8.60 ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2335

PROBLEM 19.69 A 1.8-kg collar A is attached to a spring of constant 800 N/m and can slide without friction on a horizontal rod. If the collar is moved 70 mm to the left from its equilibrium position and released, determine the maximum velocity and maximum acceleration of the collar during the resulting motion.

SOLUTION Datum at : Position  T1 = 0

V1 =

1 2 kxm 2

Position 

1 2 mv2 2 xm = ωn xm T2 =

T1 + V1 = T2 + V2 1 2 1 kxm = mωn2 xm2 2 2

ωn2 = 444.4 s −2

V2 = 0

v2 = xm

1 2 1 2 kxm = mxm + 0 2 2 k 800 N/m ωn2 = = m 1.8 kg

0+

ωn = 21.08 rad/s

xm = ωn xm = (21.08 s −1 )(0.070 m)

xm = 1.476 m/s 

 xm = ωn2 xm = (21.08 s −2 )(0.070 m)

xm = 31.1 m/s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2336

PROBLEM 19.70 Two blocks, each of weight 3 lb, are attached to links which are pin-connected to bar BC as shown. The weights of the links and bar are negligible, and the blocks can slide without friction. Block D is attached to a spring of constant k = 4 lb/in. Knowing that block A is moved 0.5 in. from its equilibrium position and released, determine the magnitude of the maximum velocity of block D during the resulting motion.

SOLUTION 1 m(b 2 + c 2 )θ 2 2 1 V = kc 2θ 2 2 kc 2 ωn2 = m(b2 + c 2 ) k = 48 lb/ft 3 lb m= = 0.093167 lb ⋅ s 2 /ft 2 32.2 ft/s (48)(2) 2 = 329.73 ωn2 = (0.093167)(1.52 + 22 ) ωn = 18.158 rad/s T=

θ0 =

0.5 in. 12 in./ft

1.5 ft

= 0.0277 rad

|vD |m = cωnθ0 = (2)(18.158)(0.02778) = 1.009 ft

|vD |m = 12.11 in./s 

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PROBLEM 19.71 A 14-oz sphere A and a 10-oz sphere C are attached to the ends of a rod AC of negligible weight which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.

SOLUTION Datum at : Position  T1 = 0 V1 = WC hC − WA hA hC = BC (1 − cos θ m ) hA = BA(1 − cos θ m )

Small angles. 1 − cos θ m ≈

θ m2 2

V1 = [(WC )( BC ) − (WA )( BA)]

θ m2 2

 10  8   14  5   θ 2 V1 =  lb  ft  −  lb  ft   m  16  12   16  12   2 V1 = (0.4167 − 0.3646)

θ m2 2

= 0.05208

θ m2 2

Position  V2 = 0 8  5 (v A ) m = θm θm 12 12 W 10 mC = C = = 0.019410 lb ⋅ s 2 /ft (16)(32.2) g W 14 mA = A = = 0.027174 lb ⋅ s 2 /ft (16)(32.2) g

(vC )m =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2338

PROBLEM 19.71 (Continued)

1 1 mC (vC ) m2 + m A (v A ) m2 2 2 2 2  1 1  8   5   2 = (0.019410)   + (0.027174)    θ m 2  2  12   12   θ 2 = 0.013344 m 2

T2 =

= 0.013344

Conservation of energy.

ωn2θ m2 2

T1 + V1 = T2 + V2 : 0 + 0.05208

θ m2 2

= 0.013344

ωn2θ m2 2

+0

0.05208 = 3.902 0.013344 ωn = 1.9755 rad/s

ωn2 =

Period of oscillations.

τn =

2π ωn

τ n = 3.18 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2339

PROBLEM 19.72 Determine the period of small oscillations of a small particle which moves without friction inside a cylindrical surface of radius R.

SOLUTION Datum at

:

Position  T1 = 0 V1 = WR(1 − cos θ m )

Small oscillations: (1 − cos θ m ) = 2sin 2 V1 =

θm 2



θ m2 2

WRθ m2 2

Position  vm = Rθm

T2 =

1 2 1 mvm = mR 2θm2 2 2

V2 = 0

Conservation of energy.

T1 + Y1 = T2 + V2

0 + WR

θ m2 2

=

1 mR 2θm2 + 0 2

θm = ωnθ m W = mg

mgR

θ m2 2

=

ωn =

1 mR 2ωn2θ m2 2 g R

τn =

Period of oscillations.



ωn

= 2π

R  g

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2340

PROBLEM 19.73 The inner rim of an 85-lb flywheel is placed on a knife edge, and the period of its small oscillations is found to be 1.26 s. Determine the centroidal moment of inertia of the flywheel.

SOLUTION Datum at : Position  T1 =

1 2 I 0θ m V1 = 0 2

Position  T2 = 0 V2 = mgh

θ   h = r (1 − cos θ m ) = r  2sin 2 m  2   ≈r

θ m2 2

V2 = mgr

θ m2 2

Conservation of energy. T1 + V1 = T2 + V2 :

θm = ωnθ m

For simple harmonic motion, I 0ωn2θ m2 = mgr θ m2

Moment of inertia.

I 0 = I + mr 2 I =

θ2 1 2 I 0 θ m + 0 = 0 + mgr m 2 2

ωn2 =

mgr I0

τ n2 =

I + mr 2 =

(τ ) (mgr ) − mr 2 n

4π 2

2

=

4π 2

ωn2

=

(4π 2 ) I 0 mgr

(τ ) (mgr ) 2 n

4π 2

(1.26 s) 2 (85 lb)  7  (85 lb)  7   ft  −  ft  12 4π 2 (32.2 ft/s 2 )  12   

I = 1.994 − 0.8983

2

I = 1.096 lb ⋅ ft ⋅ s 2 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2341

PROBLEM 19.74 A connecting rod is supported by a knife edge at Point A; the period of its small oscillations is observed to be 1.03 s. Knowing that the distance ra is 6 in. determine the centroidal radius of gyration of the connecting rod.

SOLUTION Position  Displacement is maximum. T1 = 0, V1 = mgra (1 − cos θ m ) ≈

1 mgraθ m2 2

Position  Velocity is maximum. (vG ) m = raθm 1 2 1 2 1 2 2 1 mvG + I θ m = mra θ m + mk 2θm2 2 2 2 2 1 = m ra2 + k 2 θm2 2 V2 = 0 T2 =

(

θm = ωnθ m

For simple harmonic motion,

T1 + V1 = T2 + V2

Conservation of energy. 0+

Data:

)

(

τ n = 1.03 s

ωn =

ra = 6 in. = 0.5 ft k2 =

)

1 1 mgraθ m2 = m ra2 + k 2 ωn2 θ m2 + 0 2 2 gr gr ωn2 = 2 a 2 or k 2 = 2a − ra2 ra + k ωn



τn

=

2π = 6.1002 rad/s 1.03

g = 32.2 ft/s 2

(32.2)(0.5) − (0.5) 2 = 0.43265 − 0.25 = 0.18265 ft 2 2 (6.1002)

k = 0.42738 ft

k = 5.13 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2342

PROBLEM 19.75 A uniform rod AB can rotate in a vertical plane about a horizontal axis at C located at a distance c above the mass center G of the rod. For small oscillations, determine the value of c for which the frequency of the motion will be maximum.

SOLUTION Find ωn as a function of c. Datum at : Position 

T1 = 0 V1 = mgh V1 = mgc(1 − cos θ m ) 1 − cos θ m = 2sin 2 V1 = mgc

Position 

T2 =

θm



2

θ m2 2

θ m2 2

1 2 I Cθ m 2

1 2 ml + mc 2 12  1  l2 T2 = m  + c 2 θm2 V2 = 0 2  12 

I C = I + mc 2 =

T1 + V1 = T2 + V2

0 + mgc

θm = ωnθ m

 l2  θ 2 = m  + c 2  m + 0 2  12  2

θ m2

 l2  gc = m  + c 2  ωn2  12  gc ωn2 = 2 l + c2 12

(

Maximum c, when



g d ωn2 =0= dc

(

)

l2 12

)

+ c 2 − 2c 2 g

(

2

l 12

+ c2

)

l2 − c2 = 0  12

=0

c=

l 12



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2343

PROBLEM 19.76 A homogeneous wire of length 2l is bent as shown and allowed to oscillate about a frictionless pin at B. Denoting by τ 0 the period of small oscillations when β = 0, determine the angle β for which the period of small oscillations is 2τ 0 .

SOLUTION We denote by m the mass of half the wire.

Position  Maximum deflections: l l T1 = 0, V1 = − mg cos(θ m − β ) − mg cos(θ m + β ) 2 2 l = −mg (cosθ m cos β + sin θ m sin β + cosθ m cos β − sin θ m sin β ) 2 V1 = − mgl cos β cos θ m

For small oscillations,

1 cos θ m ≈ 1 − θ m2 2 V1 = −mgl cos β +

1 mgl cos β θ m2 2

Position  Maximum velocity:

Thus,

T2 =

1 2 I Bθ m 2

T2 =

1  2 2  2  ml θ m 23 

but

1  I B = 2  ml 2  3  

l  V2 = −2mg  cos β  = −mgl cos β 2  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2344

PROBLEM 19.76 (Continued)

T1 + V1 = T2 + V2

Conservation of energy. − mgl cos β +

Setting θm = θ mωn ,

1 1 mgl cos β θ m2 = ml 2θm2 − mgl cos β 2 3 1 1 mgl cos β θ m2 = ml 2 θ m2 ωn2 2 3

ωn2 =

3g cos β 2 l

But for β = 0, For τ = 2τ 0 ,

τ=



ωn

τ 0 = 2π 2π

= 2π

2l 3g cos β

(1)

2l 3g

 2l 2l  = 2  2π   3 g cos β 3g  

Squaring and reducing, 1 =4 cos β

cos β =

1 4

β = 75.5° 

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PROBLEM 19.77 A uniform disk of radius r and mass m can roll without slipping on a cylindrical surface and is attached to bar ABC of length L and negligible mass. The bar is attached to a spring of constant k and can rotate freely in the vertical plane about Point B. Knowing that end A is given a small displacement and released, determine the frequency of the resulting oscillations in terms of m, L, k, and g.

SOLUTION 2

V =

1  Lθ  Lθ 2 k mg + 2  2  4

T =

1  L2θ 2  1 mr 2  L2θ 2  m +   2  4  2 2  4r 2 

=

ωn2 = =

3mL2θ 2 16 kL2 8

+

mgL 4

3mL2 16

2 k 2g  +  3 m L  fn =

1 2π

2k 4g +  3m 3L

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2346

PROBLEM 19.78 Two uniform rods, each of weight W = 1.2 lb and length l = 8 in., are welded together to form the assembly shown. Knowing that the constant of each spring is k = 0.6 lb/in. and that end A is given a small displacement and released, determine the frequency of the resulting motion.

SOLUTION Mass and moment of inertia of one rod.

m=

W 1.2 = = 0.037267 lb ⋅ s 2 /ft g 32.2 2

I =

1 1  8  ml 2 = (0.037267)   = 1.38026 × 10−3 lb ⋅ s 2 ⋅ ft 12 12  12 

Approximation. sin θ m ≈ tan θ m ≈ θ m 1 − cos θ m = 2sin 2

Spring constant: Position 

θm 2

1 ≈ θ m2 2

k = 0.6 lb/in. = 7.2 lb/ft 1  1 l  T1 = 2  I θm2  + m  θm  2  2 2 

2

1 1  4  = (2)   (1.38026 × 10−3 )θm2 +   (0.037267)  θm  2 2  12  = 3.4506 × 10−3θ 2

2

m

V1 = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2347

PROBLEM 19.78 (Continued)

T2 = 0

Position 

V2 = − ≈−

Wl 1  l  (1 − cos θ m ) + 2   k  θ m  2 2 2 

2

(1.2)(0.66667)  1 2  1  0.66667  θm   2 θ m  + (2)  2  (7.2)  2 2      

2

≈ 0.6θ m2

Conservation of energy.

T1 + V1 = T2 + V2

3.4506 × 10−3θm2 + 0 = 0 + 0.6θ m2 θ = 13.186θ m

Simple harmonic motion.

m

θm = ωnθ m ωn = 13.186 rad/s

Frequency.

fn =

ωn 2π

f n = 2.10 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2348

PROBLEM 19.79 A 15-lb uniform cylinder can roll without sliding on an incline and is attached to a spring AB as shown. If the center of the cylinder is moved 0.4 in. down the incline and released, determine (a) the period of vibration, (b) the maximum velocity of the center of the cylinder.

SOLUTION

(a)

Position  T1 = 0 V1 =

1 k (δ st + rθ m ) 2 2

Position  1 2 1 I θ m + mvm2 2 2 1 V2 = mgh + k (δ st ) 2 2 T2 =

Conservation of energy. T1 + V1 = T2 + V2 : 0 +

1 1 1 1 k (δ st + rθ m ) 2 = I θm2 + mvm2 + mgh + k (δ st ) 2 2 2 2 2

kδ st2 + 2kδ st rθ m + kr 2θ m2 = I θm2 + mvm2 + 2mgh + kδ st2

(1)

When the disk is in equilibrium, ΣM c = 0 = mg sin β r − kδ st r

Also,

h = r sin β θ m

Thus, mgh − kδ st r = 0

(2)

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PROBLEM 19.79 (Continued)

Substituting Eq. (2) into Eq. (1) kr 2θ m2 = I θm2 + mvm2 θm = ωnθ m vm = rθm = rωnθ m kr 2θ m2 = ( I + mr 2 )θ m2 ωn2

ωn2 = ωn2 =

τn =

(b)

kr 2 J + mr 2

I =

kr 2 1 2 mr + mr 2 2 2π = ωn

=

2 k 3m

2π 2 (4.5 × 12 lb/ft) 3 15 lb ft/s 2

(

1 2 mr 2

τ n = 0.715 s 

)

32.2

Maximum velocity. vm = rθm θm = θ mωn vm = rθ mωn

rθ m =

0.4 ft 12

 0.4   2π  ft   vm =    12   0.715 s 

vm = 0.293 ft/s 

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PROBLEM 19.80 A 3-kg slender rod AB is bolted to a 5-kg uniform disk. A spring of constant 280 N/m is attached to the disk and is unstretched in the position shown. If end B of the rod is given a small displacement and released, determine the period of vibration of the system.

SOLUTION

r = 0.08 m l = 0.3 m Position 

1 1 I diskθm2 + ( IA ) rod θm2 2 2 V1 = 0 T1 =

1 m0 r 2 2 1 I A rod = m AB l 2 3 I disk =

Position 

T2 = 0 V2 =

1 l k (rθ m ) 2 + mAB g (1 − cos θ m ) 2 2

1 − cos θ m = 2sin 2 V2 =

θm 2



θ m2 2

l 1 2 2 m AB g ( 2 ) 2 θm kr θ m + 2 2

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PROBLEM 19.80 (Continued)

Conservation of energy. T1 + V1 = T2 + V2 :

11 1 1 1 l  m0 r 2 + m AB l 2 θm2 + 0 = 0 + kr 2θ m2 + m AB g θ m2  2 2 3 2 2 2 

For simple harmonic motion,

θm = ωnθ m 1 l 2 1  2 2 2 2 2  2 m0 t + 3 mAB l  ωn θ m =  kr + m AB g 2  θ m    

ωn2 = ωn2 = ωn2 = Period of vibration.

τn =

kr 2 + m AB gl 1 2

m0 r 2 + 13 m AB l 2

(280 N/m)(0.08 m) 2 + (3 kg)(9.81 m/s 2 ) ( 0.3 m) 2 1 2

(5 kg)(0.08 m) 2 + 13 (3 kg)(0.300 m) 2

6.207 = 58.55 0.106 2π

ωn

=

2π 58.55

τ n = 0.821 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2352

PROBLEM 19.81 A slender rod AB of mass m and length l is connected to two collars of negligible mass in a horizontal plane as shown. Collar A is attached to a spring of constant k. Knowing that the collars can slide freely on their respective rods and the system is in equilibrium in the position shown, determine the period of vibration if collar A is given a small displacement and released.

SOLUTION Moment of inertia:

I =

1 2 ml 12

Position  Maximum deflection: Let collar A be moved a small distance xm as shown. Since the movement is

horizontal, there is no change in gravitational potential energy. xm = (l cos β )θ m 1 2 1 kxm = k (l cos βθ m ) 2 2 2 1 2 V1 = kl cos 2 βθ m2 2 T1 = 0 V1 =

Position  Maximum velocity: The instantaneous center of rotation lies at Point C, the intersection of lines

perpendicular, respectively, to vA and vB. vm = (GC )ωm = T2 =

1 lωm 2

1 1 mvm2 + I ωm2 2 2 2

1 1 1 1   = m  lωm  +  ml 2  ωm2 2 2 2  12   1 2 2 T2 = ml ωm 6

But,

ωm = −θm

so that

1 2 2 ml θ m 6 V2 = 0 T2 =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2353

PROBLEM 19.81 (Continued)

For simple harmonic motion, Conservation of energy:

θm = ωnθ m T1 + V1 = T2 + V2 : 0 +

Natural frequency:

ωn2 =

Period of vibration:

τ=

1 2 1 kl cos 2 βθ m2 = ml 2ωn2θ m2 + 0 2 6

3k cos 2 β m 2π

τ = 2π m /3k cos 2 β 

ωn

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2354

PROBLEM 19.82 A slender rod AB of mass m and length l is connected to two collars of mass mC in a horizontal plane as shown. Collar A is attached to a spring of constant k. Knowing that the collars can slide freely on their respective rods and the system is in equilibrium in the position shown, determine the period of vibration if collar A is given a small displacement and released.

SOLUTION Moment of inertia of rod:

I =

1 2 ml 12

Position  Maximum deflection: Let collar A be moved a small distance xm as shown. Since the movement is

horizontal, there is no change in gravitational potential energy. xm = (l cos β )θ m 1 2 1 kxm = k (l cos βθ m ) 2 2 2 1 2 V1 = kl cos 2 βθ m2 2 T1 = 0 V1 =

Position  Maximum velocity: The instantaneous center of rotation lies at Point C, the intersection of lines

perpendicular, respectively, to vA and vB. v A = ( AC )ωm = l cos βωm vB = ( BC )ωm = l sin βωm vm = (CG )ωm = T2 =

1 lωm 2

1 1 1 1 m v 2 + I ωm2 + mC v A2 + mC vB2 2 2 2 2 2

1 1 1 1   m  lvm  +  ml 2  ωm2 2 2 2  12   1 1 + mC (l sin βωm ) 2 + mC (l cos βωm ) 2 2 2 11 2 2 1 = ml ωm + mC l 2 (sin 2 β + cos 2 β )ωm2 23 2 11  T2 =  m + mC  l 2ωm2 23  =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2355

PROBLEM 19.82 (Continued)

But,

ωm = −θm

so that

11  2 2  m + mC  l θ m 23  V2 = 0

Conservation of energy:

T2 =

T1 + V1 = T2 + V2 : 0 +

Natural frequency:

ωn2 =

Period of vibration:

τ=

1 2 11  kl cos 2 βθ m2 =  m + mC  l 2ωn2θ m2 2 23 

k cos 2 β m + mC 3 2π

m



τ = 2π  + mC  /k cos 2 β  3 

ωn

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PROBLEM 19.83 An 800-g rod AB is bolted to a 1.2-kg disk. A spring of constant k = 12 N/m is attached to the center of the disk at A and to the wall at C. Knowing that the disk rolls without sliding, determine the period of small oscillations of the system.

SOLUTION

Position  2

1 1 1 1 l  T1 = ( I G ) AB θm2 + mAB  − r  θm + ( I G )disk θm2 + m disk r 2θm2 2 2 2 2 2  1 1 ( I G ) AB = ml 2 = (0.8)(0.6)2 = 0.024 kg ⋅ m 2 12 12 2

l  m AB  − r  = (0.8)(0.3 − 0.25)2 = 0.002 kg ⋅ m 2 2   1 1 ( I G )disk = mdisk r 2 = (1.2)(0.25)2 = 0.0375 kg ⋅ m 2 2 2 2 2 mdisk r = 1.2(0.25) = 0.0750 kg ⋅ m 2

1 [0.024 + 0.002 + 0.0375 + 0.0750]θm2 2 1 T1 = [0.1385]θm2 2 V1 = 0 T1 =

Position 

T2 = 0 V2 =

1 l k (rθ m ) 2 + mAB g (1 − cos θ m ) 2 2

1 − cos θ m = 2sin 2 V2 =

θm 2



θ m2 2

(small angles)

2 1  0.6 m  θ m (12 N/m)(0.25 m)2 θ m2 + (0.8 kg)(9.81 m/s 2 )   2  2  2

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PROBLEM 19.83 (Continued)

1 V2 = [0.750 + 2.354]θ m2 2 1 = (3.104) θ m2 N ⋅ m 2 T1 + V1 = T2 + V2 θ 2 = ω 2θ 2 m

n m

1 1 (0.1385)θ m2 ωn2 + 0 = 0 + (3.104)θ m2 2 2 (3.104 N ⋅ m) ωn2 = (0.1385 kg ⋅ m 2 ) = 22.41 s −2

τn =



ωn

=

2π 22.41

τ n = 1.327 s 

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PROBLEM 19.84 Three identical rods are connected as shown. If b = 34 l , determine the frequency of small oscillations of the system.

SOLUTION l = length of each rod m = mass of each rod

Kinematics:

Position 

l vm = θm 2 (vBE ) m = bθm T1 = 0 l V1 = −2mg cos θ m − mgb cos θ m 2 V1 = −mg (l + b) cos θ m

Position 

l − mgb 2 = − mg (l + b)

V2 = −2mg

1 1  1 T2 = 2  I θm2 + mvm2  + m(vBE ) 2m 2 2  2 2

=

1 2 2 1 l  ml θ m + m  θm  + m(bθm ) 2 12 2 2  

1  1 T2 =  l 2 + b 2  mθm2 3 2  

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2359

PROBLEM 19.84 (Continued)

Conservation of energy.

1  1 T1 + V1 = T2 + V2 : 0 − mg (l + b) cos θ m =  l 2 + b 2  mθm2 − mg (l + b) 2  3 1  1 mg (l + b)(1 − cos θ m ) =  l 2 + b 2  mθm2 3 2  

For small oscillations,

But for simple harmonic motion,

1 (1 − cos θ m ) = θ m2 2 1 1  1 mg (l + b)θ m2 =  l 2 + b 2  mθm2 2 2  3

θm = ωnθ m :

1 1  1 mg (l + b)θ m2 =  l 2 + b 2  m(ωnθ m )2 2 2  3

ωn2 =

ωn2 = 3g

or For b =

1 l+b g 2 1 2 3 l + 12 b 2

3 l , we have 4

ωn2 = 3g = 3g

2l 2 + 3 ( 34 l )

2

7 l 4 59 2 l 16

g l g l

ωn = 1.1932

=

(1)

l + 34 l

= 1.4237

fn =

l +b 2l + 3b 2 2

ωn 2π 1.1932 2π

g l

f n = 0.1899

g  l

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2360

PROBLEM 19.85 A 14-oz sphere A and a 10-oz sphere C are attached to the ends of a 20-oz rod AC which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.

SOLUTION

Position  2

T1 =

I AC =

1 WA  5   1 WC θm + 2 g  12  2 g

1 WAC 12 g

 13     12 

2

1 WAC  8    12 θ m  + 2 g  

2

1 1   2  8 θ m  + 2 I ACθ m  

2

14  5 2 10  8 2 20  1 2 1  20  13 2  2    +   +   +     θm 16  12  16  12  16  8  12  16  12  

T1 =

1 2g

T1 =

1 [0.1519 + 0.2778 + 0.01953 + 0.1223]θm2 2 2(32.2 ft/s )

T1 =

1  0.5715 lb ⋅ ft 2  2  32.2 ft/s 2

 2 1 2 θ m = (0.01775)θm (lb ⋅ ft) 2 

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PROBLEM 19.85 (Continued)

Position  T2 = 0 V2 = −WA

5 8 1 (1 − cos θ m ) + WC (1 − cos θ m ) + WAC (1 − cos θ m ) 12 12 8

1 − cos θ m = 2sin 2

θm



θ m2

2 2   14  5   10  8   20  1   θ m2 V2 =  −    +    +     (lb ⋅ ft)   16  12   16  12   16  8   2 V2 = [ −0.3646 + 0.4167 + 0.1563]

θ m2 2

0.2084θ m2 V2 = 2

Conservation of energy.

T1 + V1 = T2 + V2:

1 0.2084 2 (0.01775)θm2 + 0 = 0 + θm 2 2

Simple harmonic motion.

θm = ωnθ m 0.2084 = 11.738 0.01775 2π 2π τn = = ωn 11.738

ωn2 =

τ n = 1.834 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2362

PROBLEM 19.86 A 10-lb uniform rod CD is welded at C to a shaft of negligible mass which is welded to the centers of two 20-lb uniform disks A and B. Knowing that the disks roll without sliding, determine the period of small oscillations of the system.

SOLUTION WA = WB = Wdisk

Position 

T1 =

1 1  2W 2( I A )disk θm2 +  disk 2 2 g 1 1 WCD + I CDθm2 + 2 2 g

  2  ( rθ m )  2

l  2  2 − r  θm  

1 Wdisk 2 1 (20) 2 10 (1) = r = 2 g 2 g g 1 WCD 2 1 (10) 2 15 = (3) = l = 12 g 12 g 2g

( I A )disk = I CD

1  15 5  20 + 40 + +  θm2  2g  2 2 1 T1 = g (70)θm2 V1 = 0 2 T1 =

Position 

T2 = 0 V2 = WCD

l (1 − cos θ m ) 2

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PROBLEM 19.86 (Continued)

Small angles:

1 − cos θ m = 2sin 2

θm



2

θ m2 2

θ m2

1 V2 = WCD l 2 2 1 = (10) (1.5) θ m2 2 1 = (15)θ m2 2

Conservation of energy and simple harmonic motion. T1 + V1 = T2 + V2 θ = ω θ m

n m

1 1 (70)ωn2θ m2 + 0 = 0 + (15)θ m2 2g 2 15 g ωn2 = 70

Period of oscillations.

τn =



ωn

= 2π

70 (15)(32.2)

τ n = 2.39 s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2364

PROBLEM 19.87 Two uniform rods AB and CD, each of length l and mass m, are attached to gears as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4m, determine the period of small oscillations of the system.

SOLUTION

Kinematics:

2rθ A = rθC 2θ A = θC 2θA = θC

Let

θ A = θm 2θ m = (θC )m 2θm = (θC )m

Position 

T1 =

1 2 1 1 1 I Aθ m + I C (2θm ) 2 + I ABθm2 + I CD (2θm ) 2 2 2 2 2 2

1 1 l  l  + mAB  θm  + mCD  2θm  2 2 2  2 

2

1 (4m)(2r ) 2 = 8mr 2 2 1 1 I C = (m)(r ) 2 = mr 2 2 2 1 1 2 I AB = ml I CD = ml 2 12 12   r2  1  l2 l2 l2 T1 = m 8r 2 +   4 + + + + l 2  2  12 3 4  2   1  5  T1 = m 10r 2 + l 2  θm2 2  3  V1 = 0 IA =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2365

PROBLEM 19.87 (Continued)

Position 

T1 = 0 l mgl V1 = mg (1 − cos θ m ) + (1 − cos 2θ m ) 2 2

For small angles,

1 − cos θ m = 2sin 2

θm 2



θ m2 2

1 − cos 2θ m = 2sin θ m ≈ 2θ m2 2

V1 =

θ2  1 5θ 2 1 mgl  m + 2θ m2  = mgl m  2  2 2 2  

T1 + V1 = T2 + V2

θm2 = ωn2θ m2

5θ 2 1  2 5 2 2 2 1 m 10r + l  ωn θ m + 0 = 0 + mgl m 2  3  2 2

ωn2 = =

5 2 2

gl

10r + 53 l 2 3gl  12r 2 + 2l 2

τn =



ωn

= 2π

12r 2 + 2l 2  3gl

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2366

PROBLEM 19.88 Two uniform rods AB and CD, each of length l and mass m, are attached to gears as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4m, determine the period of small oscillations of the system.

SOLUTION

Kinematics:

Let

2rθ A = rθC 2θA = θC

2θ A = θC

θ A = θm

2θ m = (θC ) m

2θm = (θC )m 2

Position 

T1 =

1 2 1 1 1 1 1 l  l  I Aθ m + I C (2θm ) 2 + I ABθm2 + I CD (2θm ) 2 + m AB  θm  + mCD  2θm  2 2 2 2 2 2 2 2    

2

1 (4m)(2r )2 = 8mr 2 2 1 1 I C = (m)(r 2 ) = mr 2 2 2 1 1 2 I AB = ml I CD = ml 2 12 12   r2  1  l2 l2 l2 T1 = m 8r 2 +   4 + + + + l 2  θm2 2  12 3 4  2   IA =

T1 =

1  2 5 2  2 m 10r + l  θ m 2  3 

V1 = 0

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PROBLEM 19.88 (Continued)

Position 

T2 = 0 l mgl V2 = − mg (1 − cos θ m ) + (1 − cos 2θ m ) 2 2

For small angles,

1 − cos θ m = 2sin 2

θm 2



θ m2 2

1 − cos 2θ m = 2sin θ m ≈ 2θ m2 2

l θ m2 mgl 2 2θ m + 2 2 2 1 3 = mgl θ m2 2 2 θm = ωnθ m T1 + V1 = T2 + V2 V2 = − mg

1  2 5 2 2 2 1 3 m 10r + l  θ mωn + 0 = 0 + mgl θ m2 2  3  2 2

ωn2 = =

3 2 2

gl

10r + 53 l 2 9 gl 60r + 10l 2 2



τn =



ωn

= 2π

60r 2 + 10l 2  9 gl

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2368

PROBLEM 19.89 An inverted pendulum consisting of a rigid bar ABC of length l and mass m is supported by a pin and bracket at C. A spring of constant k is attached to the bar at B and is undeformed when the bar is in the vertical position shown. Determine (a) the frequency of small oscillations, (b) the smallest value of a for which these oscillations will occur.

SOLUTION Moment of inertia:

I =

1 2 ml 12

Position  Maximum deflection. Let rod AC rotate through angle θ m . The spring stretches an amount xm = a sin θ m

and the center of gravity moves down an amount l (1 − cos θ m ) 2 1 V1 = kxm2 + mgym 2 l 1 = k (a sin θ m ) 2 − mg (1 − cos θ m ) 2 2 1  l  1  ≈ ka 2θ m2 − mg   θ m2  2  2  2 

− ym =

1 2 1  ka − mgl  θ m2 2  2  T1 = 0 =

Position  Maximum velocity:

For simple harmonic motion,

θ = −ωnθ m

Velocity of the mass center of the rod:

l v = θ 2

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PROBLEM 19.89 (Continued)

Kinetic energy:

T2 =

1 1 mv 2 + I θ 2 2 2

2  1   lθ  1 =  m   + ml 2θ 2  2   2  12    1  1 2 2  =  ml θ  2 3  11  =  ml 2ωn2θ m2  23  V2 = 0

T1 + V1 = T2 + V2

Conservation of energy:

1 1 11   0 +  ka 2 − mgl θ m2 =  ml 2ωn2θ m2  2 2 2 3   

ωn2 = (a)

Frequency:

6 ka 2 − 3mgl 2ml 2

f = 2πωn f = 2π (6ka 2 − 3mgl )/2ml 2 

(b)

Smallest value of a for oscillations. f is real for 6 ka 2 > 3mgl a>

mgl 2k

amin =

mgl  2k

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2370

PROBLEM 19.90 Two 12-lb uniform disks are attached to the 20-lb rod AB as shown. Knowing that the constant of the spring is 30 lb/in. and that the disks roll without sliding, determine the frequency of vibration of the system.

SOLUTION

Position 1:

T1 = 0,

V1 =

1 2 kxm 2

Position 2:

V2 = 0,

T2 =

1 1 1  m AB vm2 + 2  I ωm2 + mdisk vm2  2 2 2   2

T2 =

1 1  v  m AB vm2 +  mdisk r 2  m  + mdisk vm2 2 2  r 

T2 =

1 (mAB + 3mdisk )vm2 2

Conservation of energy T1 + V1 = T2 + V2:

0+

1 2 1 kxm = (m AB + 3mdisk )vm2 2 2

vm = ω n xm :

But for simple harmonic motion,

1 2 1 kxm = (m AB + 3mdisk )(ωn xm ) 2 2 2

ωn2 = Data:

k = 30 lb/in.

ωn2 = f =

m AB

k + 3mdisk

WAB = 20 lb

Note: Result is independent of r Wdisk = 12 lb

30(12) lb/ft = 207 20 lb/32.2 + 3(12/32.2)

mAB =

WAB g

mdisk =

Wdisk g

ωn = 14.387 rad/s

ωn 14.387 rad/s = 2π 2π

f = 2.29 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2371

PROBLEM 19.91 The 20-lb rod AB is attached to two 8-lb disks as shown. Knowing that the disks roll without sliding, determine the frequency of small oscillations of the system.

SOLUTION Position 

Position 

r = 6 in.

mA = mB =

Masses and moments of inertia.

8 = 0.24845 lb ⋅ s 2 /ft 32.2

1 1  6 m A rA2 = (0.24845)   2 2  12  = 0.031056 lb ⋅ s ⋅ ft 20 = = 0.62112 lb ⋅ s 2 /ft 32.2

2

IA = IB =

m AB

6 vm = rAθm = θm = 0.5θm 12 1  2 v AB =  θ m = θm 6  12 

Kinematics:

Position  (Maximum displacement)

T1 = 0 80  4  V1 = −WAB  cos θ m  = − cos θ m 12  12 

Position  (Maximum speed)

T2 =

1 1 1 1 1 2 mA vm2 + I Aθm2 + mB vm2 + I Bθm2 + mAB v AB 2 2 2 2 2

1 1  1 1  = 2  (0.24845)(0.5θm ) 2 + (0.031056)θm2  + (0.62112)  θm  2 2  2 6  2 = 0.101795θ

2

m

80  4 V2 = −WAB   = − 12 12   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2372

PROBLEM 19.91 (Continued)

T1 + V1 = T2 + V2

Conservation of energy. 0−

80 80 cos θ m = 0.101795θm2 − 12 12 2  θ m = 65.491(1 − cos θ m ) 1  ≈ 65.491 θ m2  2  = 32.745θ m2

θ m = 5.7224 θ m Simple harmonic motion.

θm = ωnθ m

Frequency.

fn =

ωn 5.7224 = 2π 2π

ωn = 5.7224 rad/s f n = 0.911 Hz 

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PROBLEM 19.92 A half section of a uniform cylinder of radius r and mass m rests on two casters A and B, each of which is a uniform cylinder of radius r/4 and mass m/8. Knowing that the half cylinder is rotated through a small angle and released and that no slipping occurs, determine the frequency of small oscillations.

SOLUTION  4r V1 = mgh = mg   3π 1 − cosθ ≈ V1 = 2mgr T2 =

  (1 − cos θ ) 

θ m2 2

θ m2 3π

1 1 11  I Aω A2 + I Bω B2 +  mr 2  ω 2 2 2 22 

2

Where and

I A = IB =

1  m  r  mr 2 =  2  8  256  4 

ω A = ω B = 4ω  mr 2 mr 2  2 5mr 2ω 2 ∴ T2 =  + ω = 4  16  16 V1 = T2 ,

ω n2 =

2 m gr θ m2 3π

=

5 m r 2ω n2 θ m2 16

32 g , 15π r

 1  32 g fn =    2π  15π r

f n = 0.1312

g  r

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2374

PROBLEM 19.93 The motion of the uniform rod AB is guided by the cord BC and by the small roller at A. Determine the frequency of oscillation when the end B of the rod is given a small horizontal displacement and released.

SOLUTION Position . (Maximum deflection): Let θ m be the small angle between the cord CB and the vertical. As the rod is moved from the equilibrium position the center of gravity G moves up an amount ym . 1  1  yB = l (1 − cos θ m ) ≈ l 1 − 1 + θ m2  = lθ m2 2  2  1 1 ym = yG = yB = lθ m2 2 4 1 V1 = mgym = mglθ m2 4 T1 = 0,

Position . (Maximum velocity): At the equilibrium position the motion of the rod is a translation. vm = lω = lθm 1 1 mvm2 = ml 2θm2 2 2 V2 = 0 T2 =

Conservation of energy:

T1 + V1 = T2 + V2 :

For simple harmonic motion;

θm = ωnθ m so that

1 1 mglθ m2 = ml 2θm2 4 2

1 1 mglθ m2 = ml 2ωn2θ m2 4 2

Natural frequency:

ωn2 = f =

g 2l

ωn 1 = 2π 2π

g 2l

f = 0.1125

g  l

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2375

PROBLEM 19.94 A uniform rod of length L is supported by a ball-and-socket joint at A and by a vertical wire CD. Derive an expression for the period of oscillation of the rod if end B is given a small horizontal displacement and then released.

SOLUTION Position  (Maximum deflection)

Looking from above: Horizontal displacement of C:

xC = bθ m

Looking from right:

φm =

xC b = θm h h 1 2 hφm 2

yC = h(1 − cos φm ) ≈ 2

yC =

1 b  1 b2 2 h  θm  = θm 2 h  2 h

1 L  1 b2 2  AG − yC = 2  θ m  AC b 2 h  1 bL ym = ⋅ θ m2 4 h

ym = yG =

We have

T1 = 0 V1 = mgym =

1 mgbL 2 θm 4 h

Position  (Maximum velocity)

Looking from above:

T2 =

1 2 1 I θ m + mvm2 2 2

1 1 1 L   mL2 θm2 + m  θ   2  12 2 2   1 2 2 T2 = mL θ m 6 V2 = 0

2

=

Conservation of energy.

T1 + V1 = T2 + V2 : 0 +

1 mgbL 2 1 2  2 θ m = mL θ m 4 h 6

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PROBLEM 19.94 (Continued)

But for simple harmonic motion,

θm = ωnθ m 1 mgbL 2 1 2 θ m = mL (ωnθ m ) 2 4 h 6 3 bg ωn2 = 2 hL

Period of vibration.

τn =

2π ωn

τ n = 2π

2hL  3bg

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2377

PROBLEM 19.95 A section of uniform pipe is suspended from two vertical cables attached at A and B. Determine the frequency of oscillation when the pipe is given a small rotation about the centroidal axis OO′ and released.

SOLUTION

AA′ = BB′ =

a θ m = lα m 2

αm =

a θm 2l

Position  T1 = 0

For small angles

V1 = mgyc = mgl (1 − cos α )

1 − cos α m = 2sin

αm 2



α m2 2

=

a2 2 θm 8l 2

 a2  V1 = mgl  2 θ m2  8l 

Position 

T2 =

1 2 1 1  I θ m =  ma 2 θm2 2 2  12 

V2 = 0

θm = ω nθ m T1 + V1 = T2 + V2  a2  1 mgl  2  + 0 + ma 2ω n2θ m2 24 8 l  

ω n2 =

3g l fn =

1 2π

3g  l

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2378

PROBLEM 19.96 A 0.6-kg uniform arm ABC is supported by a pin at B and is attached to a spring at A. It is connected at C to a 1.4-kg mass M which is attached to a spring. Knowing that each spring can act in tension or compression, determine the frequency of small oscillations of the system when the weight is given a small vertical displacement and released.

SOLUTION Data:

k A = 260 N/m l AB = 0.200 m

kC = 350 N/m lBC = 0.300 m

l ABC = lBA + lBA = 0.500 m m ABC = 0.6 kg

mC = 1.4 kg

0.200 2 m ABC = (0.6 kg) = 0.24 kg 0.500 5 0.300 3 = m ABC = (0.6 kg) = 0.36 kg 0.500 5 = mBA g = (0.24 kg)(9.81 m/s 2 ) = 2.3544 N

mBA = mBC WBA

WBC = mBC g = (0.36 kg)(9.81 m/s 2 ) = 3.5316 N WC = mC g = (1.4 kg)(9.81 m/s 2 ) = 13.734 N

Let x A and xC be the amounts of stretch from their zero force lengths of the springs at locations A and C, respectively. Let θ be the small clockwise rotation of arm ABC about the fixed Point B, measured from the equilibrium position. Let yBA and yBC be the upward movement of the mass centers of portions BA and BC of the arm ABC.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2379

PROBLEM 19.96 (Continued)

Potential energy: 1 1 k A x 2A + k B xB2 2 2 1  1  = WC lBC sin θ + WBC  lBC sin θ  − WBA  lBA (1 − cos θ )  2  2  1 1 + k A (lBA sin θ + δ A )2 + kC (lBC sin θ + δ C )2 2 2 1 1 = WC lBC sin θ + WBC lBC sin θ − WBAlBA (1 − cos θ ) 2 2 1 1 2 2 + k AlBA sin θ + k AlBAδ A sin θ + k Aδ A2 2 2 1 1 2 + kC lBC sin 2 θ + kC lBC δ C sin θ + kC δ C2 2 2

V = WC yC + WBC yBC + WBA yBA +

(1)

where δ A and δ C are the spring elongations at the equilibrium position. In the static equilibrium position, 1 ΣM B = 0: WC lBC + WBC lBC + (k Aδ A )lBA + (kC δ C )lBC = 0 2

(2)

Substiting Eq. (2) into Eq. (1) gives 1 1 2 V = − WBAlBA (1 − cos θ ) + k AlBA sin 2 θ 2 2 1 1 1 2 2 + kC lBC sin θ + k Aδ A2 + kC δ C2 2 2 2

(3)

Kinematics for position with θ = 0. vC = lBCθ 1 lBCθ 2 1 = lBAθ 2

vBC = vBA

Kinetic energy:

T=

1 1 1 1 1 2 2 mC vC2 + mBC vBC + I BCθ 2 + mBA vBA + I BAθ 2 2 2 2 2 2 2

=

1 1 1 1 1  2 2 2  2 mC lBC θ + mBC  lBCθ  +  mBC lBC θ 2 2 2 2 12     2

1 1 1 1  2   + mBA  lBAθ  +  mBAlBA θ 2 2  12 2   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2380

PROBLEM 19.96 (Continued)

=

1 1 1 2 2 2  2 mC lBC + mBC lBC + mB lBA  θ 2 3 3 

(4)

T1 + V1 = T2 + V2

(5)

Conservation of energy:

Position . (Maximum deflection) θ = θ m T1 = 0

(6)

1 1 2 sin 2 θ m V1 = − WAB l AB (1 − cos θ m ) + k AlBA 2 2 1 1 1 2 sin 2 θ m + k Aδ A2 + kC δ C2 + kC lBC 2 2 2

For small angle θ m ,

sin θ m ≈ θ m 1 − cos θ m = 2sin 2 V1 ≈

θm

1 ≈ θ m2 2 2

1 1 2 2  2  − WBAlBA + k AlBA + kC lBC θ m 2 2  1 1 + k Aδ A2 + kC δ C2 2 2

(7)

Position : Maximum velocity. θ = 0 For simple harmonic motion T2 =

θ = ωnθ m

(8)

1 1 1 2 2 2  2 2 mBC lBC + mBC lBC + mBAlBA   ωn θ m 2 3 3 

(9)

Substituting Eqs. (6), (7), (8), and (9) into Eq. (5) and noting that the terms containing δ A and δ C cancel, 1 1 2 2  2 0 +  − WBAlBC + k AlBA + kC lBC θm 2 2  1 1 1 2 2 2  2 2 =  mC lBC + mBC lBC + mBAlBA  ωn θ m + 0 2 3 3  1 2 2 + kC lBC Applying the numerical data: − WBAlBA + k AlBA 2 1 = − (2.3544)(0.2) + (350)(0.2) 2 + (260)(0.3) 2 2 = −0.23544 + 14.0 + 23.4 = 37.165 N ⋅ m

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PROBLEM 19.96 (Continued)

1 1 2 2 2 mC lBC + mBC lBC + mBAlBA 3 3 1 1 = (1.4)(0.3) 2 + (0.36)(0.3)2 + (0.24)(0.2) 2 3 3 = 0.126 + 0.0108 + 0.0032 = 0.1400 kg ⋅ m 2

Then, Natural frequency:

1 1 (37.165) θ m2 = (0.1400) ωn2θ m2 2 2

ωn2 =

37.165 = 265.46 0.1400

ωn = 16.293 rad/s f =

ωn 2π

f = 2.59 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2382

PROBLEM 19.97* A thin plate of length l rests on a half cylinder of radius r. Derive an expression for the period of small oscillations of the plate.

SOLUTION (r sin θ m ) sin θ m ≈ rθ m2 r (1 − cos θ m ) ≈ r

θ m2 2

T1 = 0

Position  (Maximum deflection)

V1 = Wym = mgr

θ m2 2

1 2 I θm 2 1 1  =   ml 2θm2 2  12 

Position  (θ = 0):

T2 =

θm = ωnθ m 1 1  2 2 2   ml ωn θ m 2  12  T1 + V1 = T2 + V2 T2 =

0+

1 1 1  mgrθ m2 =   ml 2ωn2θ m2 2 2  12  12 gr ωn2 = 2 l

τn =



ωn

= 2π

l2 12 gr

τn =

πl 3gr



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PROBLEM 19.98* As a submerged body moves through a fluid, the particles of the fluid flow around the body and thus acquire kinetic energy. In the case of a sphere moving in an ideal fluid, the total kinetic energy acquired by the fluid is 14 ρVv 2 , where ρ is the mass density of the fluid, V is the volume of the sphere, and v is the velocity of the sphere. Consider a 500-g hollow spherical shell of radius 80 mm, which is held submerged in a tank of water by a spring of constant 500 N/m. (a) Neglecting fluid friction, determine the period of vibration of the shell when it is displaced vertically and then released. (b) Solve Part a, assuming that the tank is accelerated upward at the constant rate of 8 m/s2.

SOLUTION This is not a damped vibration. However, the kinetic energy of the fluid must be included. (a)

Position 

T2 = 0 1 V2 = kxm2 2

Position 

T1 = Tspere + Tfluid = V1 = 0

1 1 ms vm2 + ρVvm2 2 4

Conservation of energy and simple harmonic motion. T1 + V1 = T2 + V2 :

1 1 1 ms vm2 + ρVvm2 + 0 = 0 + kxm2 2 4 2  vm = xm = xmωn

1 1  2 2 1 2  ms + ρ V  xmωn = kxm 2 2 2  k ωn2 = ms + 12 ρV

ωn2 =

500 N/m (0.5 kg) + ( 12 ρ V )

1 1 4  ρ V = (1000 kg/m3 )  π (0.08 m)3  2 2 3  1 ρ V = 1.0723 kg 2 500 N/m = 318 s −2 ωn2 = (0.5 kg) + (1.0723 kg)

Period of vibration. (b)

τn =



ωn

=

2π 318

τ n = 0.352 s  τ n = 0.352 s 

Acceleration does not change mass.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2384

PROBLEM 19.99 A 20-kg block is attached to a spring of constant k = 8 N/m and can move without friction in a vertical slot as shown. The block is acted upon by a periodic force of magnitude P = Pm sin ω f t , where Pm = 100 N. Determine the amplitude of the motion of the block if (a) ω f = 10 rad/s, (b) ω f = 19 rad/s, (c) ω f = 30 rad/s.

SOLUTION Equation of motion:

mx + kx = Pm sin ω f t

The steady state response is

xm =

Pm /k 1−

( )

ωf 2 ωn

where

ωn =

k 8000 N/m = = 20 rad/s m 20 kg

and

Pm /k =

100 N = 0.0125 m 8000 N/m

(a)

ω f = 10 rad/s:

ω f 10 = = 0.5 ωn 20 xm =

(b)

ω f = 19 rad/s:

ω f = 30 rad/s:

xm = 166.7 mm  (in-phase)

ω f 19 = = 0.95 ωn 20 xm =

(c)

0.0125 = 0.01667 m 1 − (0.5) 2

0.0125 = 0.1282 m 1 − (0.95) 2

xm = 128.2 mm  (in-phase)

ω f 30 = =1.5 ωn 20 xm =

0.0125 = − 0.0100 m 1 − (1.5) 2

xm = 10.00 mm  (out-of-phase)

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PROBLEM 19.100 A 20-kg block is attached to a spring of constant k = 8 kN/m and can move without friction in a vertical slot as shown. The block is acted upon by a periodic force of magnitude P = Pm sin ω f t , where Pm = 10 N. Knowing that the amplitude of the motion is 3 mm, determine the value of ω f .

SOLUTION Equation of motion:

mx + kx = Pm sin ω f t

The steady state response is

xm =

Pm /k 1−

( )

ωf 2 ωn

where

ωn =

k 8000 N/m = = 20 rad/s m 20 kg

and

Pm /k =

10 N = 0.00125 m 8000 N/m

Solve for ω f /ωn :

1/2

ωf

 P  = 1 − m  ωn  kxm 

The amplitude is 3 mm so that

xm = ± 0.003 m.

so that

Pm 0.00125 m = = ± 0.41667 kxm ± 0.003

For the in-phase motion,

ωf = (1 − 0.41667)1/2 = 0.76376 ωn ω f = (0.76376)(20 rad/s) For the out-of-phase motion,

ω f = 15.28 rad/s 

ωf = (1 + 0.41667)1/2 = 1.19024 ωn ω f = (1.19024)(20 rad/s)

ω f = 23.8 rad/s 

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PROBLEM 19.101 A 9-lb collar can slide on a frictionless horizontal rod and is attached to a spring of constant k. It is acted upon by a periodic force of magnitude P = Pm sin ω f t , where Pm = 2 lb and ω f = 5 rad/s. Determine the value of the spring constant k knowing that the motion of the collar has an amplitude of 6 in. and is (a) in phase with the applied force, (b) out of phase with the applied force.

SOLUTION Eq. (19.33):

Pm k

xm = 1−

xm = k=

( ) ωf ωn

k m

Pm k − mω 2f Pm + mω 2f xm

Pm = 2 lb,

Data:

ωn2 =

2

m=

W 9 = = 0.2795 lb ⋅ s 2 /ft g 32.2

ω f = 5 rad/s k= =

(a)

(In phase)

(Out of phase)

Pm + 6.9876 xm

xm = 6 in. = 0.5 ft k=

(b)

Pm + (0.2795)(5)2 xm

2 + 6.9876 0.5

k = 10.99 lb/ft 

xm = −6 in. = −0.5 ft k=

2 + 6.9876 −0.5

k = 2.99 lb/ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2387

PROBLEM 19.102 A collar of mass m which slides on a frictionless horizontal rod is attached to a spring of constant k and is acted upon by a periodic force of magnitude P = Pm sin ω f t. Determine the range of values of ω f for which the amplitude of the vibration exceeds two times the static deflection caused by a constant force of magnitude Pm .

SOLUTION

ωn =

Circular natural frequency.

k m

For forced vibration, the equation of motion is mx + kx = Pm sin (ω f t + ϕ )

The amplitude of vibration is Pm k

xm = 1−

=

δ st

( ) (1 − ) 2

ωf ωn

ωf 2 ωn

For ω f < ωn and xm = 2 δ st , we have 2 δ st =

2

δ st 1−

( )

ωf 2 ωn

1 2

ω 2f = ωn2 = For

1 k 2m

k < ω f ≤ ωn , 2m

 ωf  1 or 1 −   = 2  ωn  k 2m

ωf =

(1)

| xm | exceeds 2δ st

For ω f > ωn and xm = 2δ st , we have 2δ st =

ω 2f δ st 1 −1 = or 2 2 2 ωn (ω f − ωn ) − 1 3 2

For

ω 2f = ωn2 =

3 k 2m

ωn ≤ ω f ≤

3k 2m

From Eqs. (1) and (2),

ωf =

3k 2m

(2)

| xm | exceeds 2δ st

Range:

k ωn . Then

|θ m | =

θst

( )

ωf 2 ωn

−1

< θst

2

 ωf    −1 > 1  ωn  2

 ωf    >2  ωn 

ω f > 2ωn = ( 2)(77.254) ω f > 109.3 rad/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2390

PROBLEM 19.105 An 18-lb block A slides in a vertical frictionless slot and is connected to a moving support B by means of a spring AB of constant k = 10 lb/in. Knowing that the displacement of the support is δ = δ m sin ω f t , where δ m = 6 in., determine the range of values of ω f for which the amplitude of the fluctuating force exerted by the spring on the block is less than 30 lb.

SOLUTION Natural circular frequency:

ωn =

Eq. (19.33′):

xm =

k 10 × 12 = = 14.652 rad/s 18 lb m 32.2

δm 1−

( )

ωf 2 ωn

  1 Fm = − k ( xm − δ m ) = − kδ m 1 −  1− ωf ωn 

Spring force:

( )

= kδ m

  2  

( ) 1− ( ) ωf 2 ωn

ωf 2 ωn

( ) = (120)(0.50) 1− ( ) ωf 2 ωn

ωf 2 ωn

( ) = 60 1− ( ) ωf 2 ωn

ωf 2 ωn

| Fm | < 30 lb

Limit on spring force:

60

( ) 1− ( ) ωf 2 ωn ωf ωn

2

< 30 or

( ) 1− ( ) ωf 2 ωn ωf ωn

2

<

1 2

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PROBLEM 19.105 (Continued)

In phase motion.

( ) 1− ( ) ωf 2 ωn ωf ωn

2

<

1 2

2

 ωf  1 1  ωf    < −   2 2  ωn   ωn 

2

2

ωf 1 3  ωf  1 >   < 2  ωn  2 ωn 3 1 ωf < ωn 3

Out of phase motion.

ω f < 8.46 rad/s 

( ) xm  ωn  1−

 ωf  >  xm  ωn 

δm

2

  

δm 



2

ωf <  1−

 ωn xm 

ω f <  1 −  (35.018) 4 

ω f < 24.8 rad/s 

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PROBLEM 19.106 (Continued)

Out of phase motion.

δm

< xm

( ) −1 ( ) −1 > 1 ωf 2 ωn

ωf 2 ωn

δm

2

 ωf  δm   −1 > xm  ωn 

xm

2

ωf    > 1.5  ωn 

ω f > 1.5ωn

ω f > 42.9 rad/s 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2394

PROBLEM 19.107 Rod AB is rigidly attached to the frame of a motor running at a constant speed. When a collar of mass m is placed on the spring, it is observed to vibrate with an amplitude of 15 mm. When two collars, each of mass m, are placed on the spring, the amplitude is observed to be 18 mm. What amplitude of vibration should be expected when three collars, each of mass m, are placed on the spring? (Obtain two answers.)

SOLUTION (a)

One collar:

( xm )1 = 15 mm

(ωn )12 =

k m

(b)

Two collars:

( xm )2 = 18 mm

(ωn ) 22 =

k 1 = (ωn )12 2m 2

ω  ω    = 2   ωn  2  ωn 1

(c)

Three collars: ( xm )3 = unknown, (ωn )32 =

k 1 = (ωn )12 , 3m 3

ω  ω    = 3   ωn 3  ωn 1

We also note that the amplitude δ m of the displacement of the base remains constant. Referring to Section 19.7, Figure 19.9, we note that, since ( xm )2 > ( xm )1 and have

ω

(ωn )1

ω

< 1 and ( xm )1 > 0. However,

(ωn )2

ω

(ωn )2

>

ω

(ωn )1

, we must

may be either < 1 or > 1, with ( xm ) 2 being

correspondingly either > 0 or < 0. 1.

Assuming ( xm )2 > 0: For one collar, ( xm )1 =

δm 1−

( )

+ 15 mm =

2

ω ωn 1

δm 1−

( )

(1)

2

ω ωn 1

For two collars, ( xm )2 =

δm 1−

( )

+ 18 mm =

2

ω ωn 2

δm 1− 2

( )

2

(2)

ω ωn 1

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PROBLEM 19.107 (Continued)

Dividing Eq. (2) by Eq. (1), member by member:

( ) 1.2 = 1 − 2( )

2

ω ωn 1

1−

2

ω  1 ; we find   = 2  ωn 1 7

ω ωn 1

Substituting into Eq. (1),

 

δ m = (15 mm) 1 −

1  90 = mm 7  7

For three collars, ( xm )3 =

2.

δm 1− 3

( )

2

ω ωn 1

=

( 907 ) mm = 90 mm, 4 1 − 3 ( 17 )

( xm )3 = 22.5 mm 

Assuming ( xm )2 < 0: For two collars, we have

−18 mm =

δm 1− 2

( )

(3)

2

ω ωn 1

Dividing Eq. (3) by Eq. (1), member by member:

( ) −1.2 = 1 − 2( )

2

ω ωn 1

1−

2

ω ωn 1

2

2

ω  ω  −1.2 + 2.4   =1−    ωn 1  ωn 1 2

ω  2.2 1.1 =   = ω 3.4 1.7  n 1

Substitute into Eq. (1),

 

δ m = (15 mm) 1 −

For three collars, ( xm )3 =

δm 1− 3

( )

2

ω ωn 1

=

1.1  9 = mm 1.7  1.7

( 1.79 ) = 9 mm , 1.1 1 − 3 ( 1.7 ) −1.6

( xm )3 = − 5.63 mm  (out of phase)

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PROBLEM 19.107 (Continued)

Points corresponding to the two solutions are indicated below:

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PROBLEM 19.108 The crude-oil-pumping rig shown in the accompanying figure is driven at 20 rpm. The inside diameter of the well pipe is 2 in., and the diameter of the pump rod is 0.75 in. The length of the pump rod and the length of the column of oil lifted during the stroke are essentially the same, and equal to 6000 ft. During the downward stroke, a valve at the lower end of the pump rod opens to let a quantity of oil into the well pipe, and the column of oil is then lifted to obtain a discharge into the connecting pipeline. Thus, the amount of oil pumped in a given time depends upon the stroke of the lower end of the pump rod. Knowing that the upper end of the rod at D is essentially sinusoidal with a stroke of 45 in. and the specific weight of crude oil is 56.2 lb/ft3, determine (a) the output of the well in ft3/min if the shaft is rigid, (b) the output of the well in ft3/min if the stiffness of the rod is 2210 N/m, the equivalent mass of the oil and shaft is 290 kg and damping is negligible.

SOLUTION Forcing frequency:

ω f = 20 rpm = 2.0944 rad/s

Cross sectional area of the flow chamber Aoil =

π 2 2 2 2 (2 in.) − (0.75 in.)  = 2.6998 in = 0.018749 ft 4

Let s be the stroke at the lower end of the pump in feet. Stroke is twice the amplitude. s = 2 xm Volume of oil pumped per revolution: Voil = Aoil s = 0.018749 s

Amplitude of motion at top of shaft: 1 2

δ m = (45 in.) = 22.5 in. = 1.875 ft Amplitude of motion at bottom of shaft:

δm

xm = 1−

(a)

Rigid shaft:

( )

ωf 2 ωn

ωn = ∞ xm = δ m = 1.875 ft s = (2)(1.875) = 3.75 ft Voil = (0.018749 ft 2 )(3.75 ft) = 0.070309 ft 3 /rev

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PROBLEM 19.108 (Continued)

output rate: (b)

(0.070309 ft 3 /rev)(20 rev/min)

1.406 ft 3 /min 

Flexible shaft. k = 2210 N/m

ωn =

k = meq

meq = 290 kg 2210 N/m = 2.7606 rad/s 290 kg

ω f 2.0944 = = 0.75869 ωn 2.7606 1.875 = 4.4178 ft 1 − (0.75869)2 s = (2)(4.4178) = 8.8358 ft

xm =

Voil = (0.018749 ft 2 )(8.8358 ft) = 0.16566 ft 3 /rev

output rate:

(0.16566 ft 3 /rev)(20 rev/min)

3.31 ft 3 /min 

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PROBLEM 19.109 A simple pendulum of length l is suspended from a collar C which is forced to move horizontally according to the relation xC = δ m sin ωft. Determine the range of values of ω f for which the amplitude of the motion of the bob is less than δ m . (assume that δ m is small compared with the length l of the pendulum).

SOLUTION x = xC + l sin θ

Geometry.

sin θ =

x − xC l

ΣFy = ma y ≈ 0: T cos θ − mg = 0

T ≈ mg

−T sin θ = mx

ΣFx = max : m  x+

mg ( x − xC ) =0 l g g  x + x = xC l l

Using the given motion of xC,  x+

Circular natural frequency.

g g x = δ m sin ω f t l l

ωn =

g l

 x + ωn2 x = ωn2δ m sin ω f t

The steady state response is

xm =

xm2 =

Consider

δm 1−

( )

ωf 2 ωn

δ m2  1 − 

( )  ωf ωn

2

2

≤ δ m2

xm2 = δ m2 .

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PROBLEM 19.109 (Continued)

2

Then

2 4   ω 2   ωf   ωf  f 1 −    = 1− 2  +  =1 ωn   ωn    ωn      2

2

 ωf   ωf    = 0 and   =2  ωn   ωn  ωf ωf  = 2   = 0 and ωn  ωn 

For

0<

ωf < 2, | xm | > δ m ωn

For

ωf > 2, | xm | < δ n ωn

Then

ω f > 2 ωn =

2g l

ωf >

2g  l

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PROBLEM 19.110 The 2.75-lb bob of a simple pendulum of length l = 24 in. is suspended from a 3-lb collar C. The collar is forced to move according to the relation xC = δ m sin ω f t , with an amplitude δm = 0.4 in. and a frequency ff = 0.5 Hz. Determine (a) the amplitude of the motion of the bob, (b) the force that must be applied to collar C to maintain the motion.

SOLUTION (a) ΣFx = max −T sin θ = mx ΣFy = T cos θ − mg = 0

For small angles cos θ ≈ 1. Acceleration in the y direction is second order and is neglected. T = mg mx = − mg sin θ x − xc sin θ = l mg g mg δ m sin ω f t mx + x = xc = l l l g ωn2 = l 2  x + ωn x = ωn2δ m sin ω f t

From Equation (19.33′): xm =

So

δm 1−

ω 2f ωn2

ω 2f = (2π f f ) 2 = 4π 2 (0.5) 2 = π 2 s −2 ωn2 = xm =

g 32.2 ft/s 2 = = 16.1 s −2 l 2 ft 0.4 12

ft

π 1 − 16.1 2

= 0.086137 ft

xm = 1.034 in. 

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PROBLEM 19.110 (Continued)

(b) ac =  xc = −δ mω 2f sin ω f t ΣFx = mc ac F − T sin θ = mc ac

From Part (a): Thus,

T = mg , sin θ =

x − xc l

 x − xc  F = − mg  xc  + mc   l 

= − mωn2 x + mωn2 xc + mc  xc = − mωn2 xm sin ω f t + mωn2δ m sin ω f t − mcω 2f δ m sin ω f t   2.75 lb   2.75 lb   3 lb  2 0.4  (16.1)(0.086137) +  (16.1) ( 0.4 = −  −    π ( 12 )  sin π t 12 )   32.2   32.2    32.2   = −0.10326sin π t

F = − 0.1033sin π t (lb) 

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PROBLEM 19.111 An 18-lb block A slides in a vertical frictionless slot and is connected to a moving support B by means of a spring AB of constant k = 8 lb/ft. Knowing that the acceleration of the support is a = am sin ω f t , where am = 5 ft/s 2 and ω f = 6 rad/s, determine (a) the maximum displacement of block A, (b) the amplitude of the fluctuating force exerted by the spring on the block.

SOLUTION (a)

Support motion. a = δ = am sin ω f t a  m  sin ω f t  ω 2f   

δ = − δm =

− am

ω 2f

=−

5 ft/s 2 = −0.13889 ft (6 rad/s)2

From Equations (19.31 and 19.33′): xm =

xm =

(b)

δm 1−

ω 2f ωn2

ωn2 =

k 8 lb/ft = 18 = 14.311 (rad/s) 2 m 32.2

−0.13889 = 0.091643 ft 36 1 − ( 14.311 )

xm = 1.100 in. 

x is out of phase with δ for ω f = 6 rad/s. Thus, Fm = k ( xm + δ m ) = 8 lb/ft (0.091643 ft + 0.13889 ft) = 1.8443 lb

Fm = 1.844 lb 

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PROBLEM 19.112 A variable-speed motor is rigidly attached to a beam BC. When the speed of the motor is less than 600 rpm or more than 1200 rpm, a small object placed at A is observed to remain in contact with the beam. For speeds between 600 and 1200 rpm the object is observed to “dance” and actually to lose contact with the beam. Determine the speed at which resonance will occur.

SOLUTION Let m be the unbalanced mass and r the eccentricity of the unbalanced mass. The vertical force exerted on the beam due to the rotating unbalanced mass is P = mr ω 2f sin ω f t = Pm sin ω f t

Then from Eq. 19.33,

Pm k

xm = 1−

( )

ωf 2 ωn

mr ω 2f k

=

1−

( )

ωf 2 ωn

For simple harmonic motion, the acceleration is am =

−ω 2f xm

mr ω 4f k

=

1−

( )

ωf 2 ωn

When the object loses contact with the beam, the acceleration | am | is greater than g. Let ω1 = 600 rpm = 62.832 rad/s. | am |1 =

mr ω14 k

1−

( )

ω1 2 ωn

=

mrωn4U 4 k 2

1−U

ω1 . ωn

where

U=

Let

ω2 = 1200 rpm = 125.664 rad/s = 2ω1 | am | 2 =

(1)

mr ω24 k 2

( ) ω2 ωn

−1

=

mrωn4 (2U )4 k 2

4U − 1

(2)

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PROBLEM 19.112 (Continued)

Dividing Eq. (1) by Eq. (2), 1=

4U 2 − 1 16(1 − U 2 )

20U 2 = 17

ω1 17 = ωn 20

U=

or 16 − 16U 2 = 4U 2 − 1 17 20

ωn =

20 ω1 = 1.08465 ω1 17

ωn = (1.08465)(600 rpm)

ωn = 651 rpm 

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PROBLEM 19.113 A motor of mass M is supported by springs with an equivalent spring constant k. The unbalance of its rotor is equivalent to a mass m located at a distance r from the axis of rotation. Show that when the angular velocity of the motor is ω f , the amplitude xm of the motion of the motor is

xm =

( ) 1− ( )

r ( Mm )

ωf 2 ωn

ωf 2 ωn

k . M

where ωn =

SOLUTION Rotor

Motor ΣF = ma Pm sin ω f t − kx = Mx Mx + kx = Pm sin ω f t  x+

P k x = m sin ω f t M M k ωn2 = M

From Equation (19.33): xm =

But

Pm k

1−

( )

ωf 2 ωn

2 Pm mrω f = k k

k = M ωn2

Pm  m  ω f  = r    k  M   ωn 

2

xm =

Thus,

( ) 1− ( )

r ( Mm )

ωf 2 ωn

ωf 2 ωn

Q.E.D. 

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PROBLEM 19.114 As the rotational speed of a spring-supported 100-kg motor is increased, the amplitude of the vibration due to the unbalance of its 15-kg rotor first increases and then decreases. It is observed that as very high speeds are reached, the amplitude of the vibration approaches 3.3 mm. Determine the distance between the mass center of the rotor and its axis of rotation. (Hint: Use the formula derived in Problem 19.113.)

SOLUTION Use the equation derived in Problem 19.113. xm =

For very high speeds,

thus,

( ) 1− ( )

r ( Mm )

ωf 2 ωn

1

( ) ωf ωn

ωf 2 ωn

2

r ( Mm )

=

1

( )

0 and

ωf 2 ωn

−1 rm , M

xm

 15  3.3 mm = r    100 

r = 22 mm 

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PROBLEM 19.115 A motor of weight 40 lb is supported by four springs, each of constant 225 lb/in. The motor is constrained to move vertically, and the amplitude of its motion is observed to be 0.05 in. at a speed of 1200 rpm. Knowing that the weight of the rotor is 9 lb, determine the distance between the mass center of the rotor and the axis of the shaft.

SOLUTION W = 40 lb

Four springs each of constant 225 lb/in. We note that the motor is constrained to move vertically. 4(225 lb/in.) = 900 lb/in. = 10800 lb/ft k = m

ωn =

10800 lb/ft = 93.242 rad/s (40 lb/32.2)

For ω = 1200 rpm = 125.664 rad/s we have xm = 0.05 in. = 4.1667 × 10−3 ft

xm =

Eq. (19.33):

Pm k

1−

( ) ω ωn

2

  ω 2  Pm = xm 1 −    k   ωn    

Thus:

= (4.1667 × 10

−3

  125.664 2  −3 ft) 1 −    = −3.4015 × 10 ft (out of phase)   93.242  

Pm = (10800 lb/ft)(3.4015 × 10−3 ft) = 36.736 lb

We have found: Pm = 36.736 lb For an unbalanced rotor of weight WR = 9 lb, rotating at ω = 1200 rpm = 125.664 rad/s, with the mass center at a distance r from the axis of rotation, we have, Pm = mR r ω 2 r =

Pm 36.736 lb = = 8.3231 × 10 −3 ft 2 mRω (9 lb/32.2)(125.664 rad/s) 2 r = 0.0999in. 

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PROBLEM 19.116 A motor weighing 400 lb is supported by springs having a total constant of 1200 lb/in. The unbalance of the rotor is equivalent to a 1-oz weight located 8 in. from the axis of rotation. Determine the range of allowable values of the motor speed if the amplitude of the vibration is not to exceed 0.06 in.

SOLUTION Let M = mass of motor, m = unbalance mass, r = eccentricity 400 = 12.4224 lb ⋅ s 2 /ft 32.2  1  1  2 m =    = 0.001941 lb ⋅ s /ft  16  32.2  r = 8 in. = 0.66667 ft k = 1200 lb/in. = 14, 400 lb/ft

M=

Natural circular frequency:

k 14, 400 = = 34.047 rad/s 12.4224 M rm (0.66667)(0.001941) = 12.4224 M = 0.00014017 ft = 0.00125 in.

ωn =

From the derivation given in Problem 19.113,

xm

) = 1− ( ) ( rmM ) ( ωω

( ) 1− ( )

2

f

=

n

ωf 2 ωn

0.00125

ωf 2 ωn

ωf 2 ωn

in.

In phase motion with | xm | < 0.06 in.

( ) 1− ( )

0.00125

ωf 2 ωn

ωf 2 ωn

2

< 0.06

2

 ωf   ωf   ωf  0.00125     < 0.06 − 0.06    ωn   ωn   ωn 

2

2

 ωf  0.06125   < 0.06  ωn 

ωf 0.06 < = 0.98974 0.06125 ωn ω f < (0.98974)(34.047) = 33.698 rad/s

ω f < 322 rpm 

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PROBLEM 19.116 (Continued)

Out of phase motion with | xm | = 0.06 in. 0.00125

( )

ωf 2 ωn

( )

ωf 2 ωn

< 0.06

−1 2

2

 ωf  ωf  0.00125   < 0.06   − 0.06  ωn   ωn  ωf  0.06 < 0.05875    ωn 

2

ωf 0.06 > = 1.01058 0.05875 ωn ω f > (1.01058)(34.047) = 34.407 rad/s

ω f > 329 rpm 

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PROBLEM 19.117 A 180-kg motor is bolted to a light horizontal beam. The unbalance of its rotor is equivalent to a 28-g mass located 150 mm from the axis of rotation, and the static deflection of the beam due to the weight of the motor is 12 mm. The amplitude of the vibration due to the unbalance can be decreased by adding a plate to the base of the motor. If the amplitude of vibration is to be less than 60 μ m for motor speeds above 300 rpm, determine the required mass of the plate.

SOLUTION M1 = 180 kg, m = 28 × 10−3 kg

Before the plate is added,

r = 150 mm = 0.150 m

Equivalent spring constant:

k=

W1

k=

(180)(9.81) = 147.15 × 103 N/m −3 12 × 10

δ st

=

M1 g

δ st

Let M2 be the mass of motor plus the plate. k M2

Natural circular frequency.

ωn =

Forcing frequency:

ω f = 300 rpm = 31.416 rad/s 2

ω 2f M 2 (31.416) 2 M 2 ωf  = = = 0.006707 M 2   k 147.15 × 103  ωn  From the derivation in Problem 19.113,

xm

For out of phase motion with

( )( ) = rm M2

1−

ωf 2 ωn

( )

ωf 2 ωn

xm = −60 × 10−6 m,

−60 × 10

−6

 (0.150)(28×10−3 )  (0.006707 M ) 2 M2   = 1 − 0.006707 M 2

−60 × 10−6 + (60 × 10−6 )(0.006707)M 2 = 28.170 × 10−6 402.49 × 10−9 M 2 = 88.170 × 10−6 M 2 = 219.10 kg

Added mass:

ΔM = M 2 − M1 = 219.10 − 180

ΔM = 39.1 kg 

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PROBLEM 19.118 The unbalance of the rotor of a 400-lb motor is equivalent to a 3-oz weight located 6 in. from the axis of rotation. In order to limit to 0.2 lb the amplitude of the fluctuating force exerted on the foundation when the motor is run at speeds of 100 rpm and above, a pad is to be placed between the motor and the foundation. Determine (a) the maximum allowable spring constant k of the pad, (b) the corresponding amplitude of the fluctuating force exerted on the foundation when the motor is run at 200 rpm.

SOLUTION M=

Mass of motor.

400 = 12.422 lb ⋅ s 2 /ft 32.2

Unbalance mass.

 3  1  2 m =    = 0.005823 lb ⋅ s /ft  16  32.2 

Eccentricity.

r = 6 in. = 0.5 ft

Equation of motion:

Mx + kx = Pm sin ω f t = mrω 2f sin ω f t (− M ω 2f + k ) xm = mrω 2f

xm =

Fm = kxm =

Transmitted force.

For out of phase motion, (a)

| Fm | =

mrω 2f k − M ω 2f

kmrω 2f k − M ω 2f

kmrω 2f

(1)

M ω 2f − k

Required value of k. Solve Eq. (1) for k.

| Fm | ( M ω 2f − k ) = kmrω 2f

k (mrω 2f + | Fm |) = | Fm | M ω 2f k=

Data:

| Fm | = 0.2 lb

| Fm | M ω 2f

mrω 2f + | Fm |

ω f = 100 rpm = 10.472 rad/s k=

(0.2)(12.422)(10.472) 2 = 524.65 (0.005823)(0.5)(10.472)2 + 0.2

k = 525 lb/ft 

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PROBLEM 19.118 (Continued)

(b)

Force amplitude at 200 rpm. From Eq. (1),

ω f = 20.944 rad/s | Fm | =

(524.65)(0.005823)(0.5)(20.944) 2 (12.422)(20.944) 2 − 524.65 | Fm | = 0.1361 lb 

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PROBLEM 19.119 A counter-rotating eccentric mass exciter consisting of two rotating 100-g masses describing circles of radius r at the same speed but in opposite senses is placed on a machine element to induce a steady-state vibration of the element. The total mass of the system is 300 kg, the constant of each spring is k = 600 kN/m, and the rotational speed of the exciter is 1200 rpm. Knowing that the amplitude of the total fluctuating force exerted on the foundation is 160 N, determine the radius r.

SOLUTION 2mrω 2f

Pm = 2mrω 2f ,

xm =

2k

1−

With

2kxm = 160 N = ±

( ) ωf ωn

2mrω 2f 1−

M ω 2f

2

,

,

ωn2 =

2k M

ω f = 40π rad/s

2k

Solving for r, 2  ( 300 kg )( 40π s−1 )  160 N 1 − 1200000 N/m      = 0.1493 m r =± 2(0.1 kg)(40π s −1) 2

r = 149.3 mm 

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PROBLEM 19.120 A 360-lb motor is supported by springs of total constant 12.5 kips/ft. The unbalance of the rotor is equivalent to a 0.9-oz weight located 7.5 in. from the axis of rotation. Determine the range of speeds of the motor for which the amplitude of the fluctuating force exerted on the foundation is less than 5 lb.

SOLUTION xm =

From Problem 19.113

r

( Mm ) ( ωω

1−

And

Then

( FT ) m = kxm ,

k = ωn2 , M

(

f n

( )

)

2

ωf 2 ωn

rmω 2f

( FT ) m =

 1 − 

( ) ωf 2  ωn

)

0.9  7.5   16 lb  ft  rm =  = 0.0010918 lb ⋅ s 2 2  12   32.2 ft/s 

ω n2 =

k 12500 lb/ft = = 1118.1 s −2 360 lb M 2 32.2 ft/s

ω 2f

( FT ) m = (0.0010918 lb ⋅ s 2 )

1−

or

ω 2f 1118.1

 ω 2f  ( FT ) m 1 −  = (0.0010918 lb ⋅ s 2 )ω 2f  1118.1   (F )  ( FT ) m =  T m + ( 0.0010918 ) ω 2f 1118.1  

ω 2f =

Then

(a)

1118.1( FT ) m ( FT ) m + 1.2207

( FT ) m = +5: ω 2f =

1118.1( 5 ) = 898.69 s −2 , 5 + 1.2207

ω f ≤ 29.978 rad/s ≤ 286.26 rpm

(b)

( FT ) m = −5: ω 2f =

ω f ≤ 286 rpm 

1118.1 (−5) = 1479.2 s −2 , −5 + 1.2207

ω f > 38.461 rad/s > 367.27 rpm

ω f > 367 rpm 

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PROBLEM 19.121 Figures (1) and (2) show how springs can be used to support a block in two different situations. In Figure (1), they help decrease the amplitude of the fluctuating force transmitted by the block to the foundation. In Figure (2), they help decrease the amplitude of the fluctuating displacement transmitted by the foundation to the block. The ratio of the transmitted force to the impressed force or the ratio of the transmitted displacement to the impressed displacement is called the transmissibility. Derive an equation for the transmissibility for each situation. Give your answer in terms of the ratio ω f /ωn of the frequency ωf of the impressed force or impressed displacement to the natural frequency ωn of the spring-mass system. Show that in order to cause any reduction in transmissibility, the ratio ω f /ωn must be greater than 2.

SOLUTION (1)

xm =

From Equation (19.33):

Force transmitted:

Pm k

1−

( )

ωf 2 ωn

 ( PT )m = kxm = k  1 − 

Pm k

( )

Transmissibility =

Thus,

(2)

ωf ωn

  2  

( PT )m = Pm

1

1−

( )



1



ωf 2 ωn

From Equation (19.33′): Displacement transmitted:

For

( PT ) m Pm

or

xm

δm

xm =

δm 1−

( ) ωf ωn

1

to be less than 1, 1−

( )

ωf 2 ωn

1< 1−

Transmissibility =

2

xm

δm

= 1−

( )

ωf 2 ωn

2  ωn 

ωf > 2 Q.E.D.  ωn

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PROBLEM 19.122 A vibrometer used to measure the amplitude of vibrations consists essentially of a box containing a mass-spring system with a known natural frequency of 120 Hz. The box is rigidly attached to a surface, which is moving according to the equation y = δ m sin ω f t. If the amplitude zm of the motion of the mass relative to the box is used as a measure of the amplitude δ m of the vibration of the surface, determine (a) the percent error when the frequency of the vibration is 600 Hz, (b) the frequency at which the error is zero.

SOLUTION  δm  x= ω 2f  sin ω f t  1 − ω2  n   y = δ m sin ω f t z = relative motion  δm  z = x − y =  ω 2f − δ m  sin ω f t 1 − ω 2  n   ω2

 1  δ m ω 2f 1 − n = zm = δ m  ω 2f 2 ω 1 −  ω2  1− f n   ωn2 ω 2f

(a)

zm

δm

=

ω 2f

1−

600 ( 120 ) = 25 = 1.0417 = 2 600 1 − ( 120 ) 24 2

ωn2 ωn2

 Error = 4.17%

(b)

zm

δm

=1=

1= 2

ff =

ω 2f ωn2

1−

ω 2f ωn2

ω 2f ωn2

2 2 (120) = 84.853 Hz fn = 2 2

f n = 84.9 Hz 

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PROBLEM 19.123 A certain accelerometer consists essentially of a box containing a mass-spring system with a known natural frequency of 2200 Hz. The box is rigidly attached to a surface, which is moving according to the equation y = δ m sin ω f t. If the amplitude zm of the motion of the mass relative to the box times a scale factor ωn2 is used as a measure of the maximum acceleration α m = δ mω 2f of the vibrating surface, determine the percent error when the frequency of the vibration is 600 Hz.

SOLUTION  δm  x= ω 2f  sin ω f t  1 − ω2  n   y = δ m sin ω f t z = relative motion

 δm  z = x − y =  ω 2f − δ m  sin ω f t 1 − ω 2  n   ω2

 1  δ m ω 2f − 1 n = zm = δ m  ω 2f 2 ω 1 −  ω2  1− f n   ωn2

The actual acceleration is The measurement is proportional to Then

am = −ω 2f δ m zmωn2 . zmωn2 zm  ωn  =   am δ m  ω f  1 = ω 2 1 − ωf

2

( ) n

=

1 600 1 − ( 2200 )

2

= 1.0804

Error = 8.04% 

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PROBLEM 19.124 Block A can move without friction in the slot as shown and is acted upon by a vertical periodic force of magnitude P = Pm sin ω f t , where ω f = 2 rad/s and Pm = 20 N. A spring of constant k is attached to the bottom of block A and to a 22-kg block B. Determine (a) the value of the constant k which will prevent a steady-state vibration of block A, (b) the corresponding amplitude of the vibration of block B.

SOLUTION In steady state vibration, block A does not move and therefore, remains in its original equilibrium position. Block A:

ΣF = 0 kx = − Pm sin ω f t

Block B:

(1)

ΣF = mB  x mB  x + kx = 0 x = xm sin ωn t

ωn2 = k/mB From Eq. (1):

kxm sin ωn t = − Pm sin ω f t

ωn = ω f = 2 rad/s kxm = − Pm k mB

ωn =

k = mBωn2 k = (22)(2) 2

(a)

Required spring constant.

(b)

Corresponding amplitude of vibration of B.

k = 88.0 N/m 

kxm = − Pm Pm k 20 N xm = − 88 N/m xm = −

xm = −0.227 m 

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PROBLEM 19.125 A 60-lb disk is attached with an eccentricity e = 0.006 in. to the midpoint of a vertical shaft AB, which revolves at a constant angular velocity ω f . Knowing that the spring constant k for horizontal movement of the disk is 40,000 lb/ft, determine (a) the angular velocity ω f at which resonance will occur, (b) the deflection r of the shaft when ω f = 1200 rpm.

SOLUTION

G describes a circle about the axis AB of radius r + e. Thus,

an = ( r + e)ω 2f

Deflection of the shaft is

F = kr

Thus,

F = man kr = m(r + e)ω 2f

ωn2 =

k m

kr =

k

r=

m=

ωn2 e

k

ωn2

(r + e)ω 2f

ω 2f ωn2

1−

ω 2f ωn2

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PROBLEM 19.125 (Continued)

(a)

Resonance occurs when

ω f = ωn , i.e., r

ωn =



k kg = m W

(40,000)(32.2) 60 = 146.52 rad/s = 1399.1 rpm =

(b)

r=

1200 (0.006 in.) ( 1399.1 ) 1200 1 − ( 1399.1 )

= 0.01670 in.

ωn = ω f = 1399 rpm  2

2

r = 0.01670 in. 

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PROBLEM 19.126 A small trailer and its load have a total mass of 250-kg. The trailer is supported by two springs, each of constant 10 kN/m, and is pulled over a road, the surface of which can be approximated by a sine curve with an amplitude of 40 mm and a wavelength of 5 m (i.e., the distance between successive crests is 5 m and the vertical distance from crest to trough is 80 mm). Determine (a) the speed at which resonance will occur, (b) the amplitude of the vibration of the trailer at a speed of 50 km/h.

SOLUTION Total spring constant

k = 2(10 × 103 N/m) = 20 × 103 N/m k 20 × 103 N/m = = 80 s −2 250 kg m λ =5m

ωn2 =

(a)

δ m = 40 mm = 40 × 10−3 m y = δ m sin

2π x

λ y = δ m sin ω f t ωf =

where

x = vt

2π v

λ

δ m = 40 mm = 40 × 10−3 m From Equation (19.33′):

xm =

Resonance:

ωf =

δm 1 − ω 2f   ω2  n   2π v = ωn = 80 s −1 , 5

v = 7.1176 m/s

(b)

Amplitude at

v = 25.6 km/h 

v = 50 km/h = 13.8889 m/s 2π (13.8889) = 17.4533 rad/s 5 ω 2f = 304.60 s −2

ωf =

xm =

40 × 10−3 = −14.246 × 10−3 m 1 − 304.62 80

xm = −14.25 mm 

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PROBLEM 19.127 Show that in the case of heavy damping (c > cc ), a body never passes through its position of equilibrium O (a) if it is released with no initial velocity from an arbitrary position or (b) if it is started from O with an arbitrary initial velocity.

SOLUTION Since c > cc , we use Equation (19.42), where

λ1 < 0, λ2 < 0 x = C1eλ1t + C2 eλ2t v=

(a)

t = 0, x = x0 ,

dx = C1λ1eλ1t + C2 λ2 eλ2t dt

(1) (2)

v = 0: x0 = C1 + C2

From Eqs. (1) and (2):

0 = C1λ1 + C2 λ2

C1 =

Solving for c1 and c2 ,

λ2

x0

λ2 − λ1 −λ1 C2 = x λ2 − x1 0

Substituting for C1 and C2 in Eq. (1), x =

x2 λ2 eλ1t − λ1eλ2t   λ2 − λ1 

For x = 0: when t ≠ ∞, we must have

λ1eλ2t − λ2 eλ1t = 0

λ2 = e(λ2 −λ1 )t λ1

(3)

Recall that

λ1 < 0, λ2 < 0. Choosing λ1 and λ 2 so that λ1 < λ 2 < 0, we have 0<

λ2 < 1 and λ2 − λ1 > 0 λ1

Thus a positive solution for t > 0 for Equation (3) cannot exist, since it would require that e raised to a positive power be less than 1, which is impossible. Thus, x is never 0. The x − t curve for this case is as shown. 

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PROBLEM 19.127 (Continued)

(b)

t = 0, x = 0, v = v0 :

Equations (1) and (2) yield 0 = C1 + C2

v0 = C1λ1 + C2 λ2

Solving for C1 and C2 ,

C1 = − C2 =

v2 λ2 − λ 1 v0 [eλ2t − eλ1t ] λ2 − λ1

Substituting into Eq. (1),

x=

For x = 0,

t>0

and

v0 λ 2 − λ1

eλ2t = eλ1t

For c > cc , λ1 ≠ λ 2; thus, no solution can exist for t, and x is never 0 when t > 0. The x − t curve for this motion is as shown. 

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PROBLEM 19.128 Show that in the case of heavy damping (c > cc ), a body released from an arbitrary position with an arbitrary initial velocity cannot pass more than once through its equilibrium position.

SOLUTION Substitute the initial conditions, t = 0, x = x0 , v = v0 in Equations (1) and (2) of Problem 19.127. x0 = C1 + C2 C1 = −

Solving for C1 and C2 ,

C2 = x=

And substituting in Eq. (1) For x = 0, t ≠ ∞ :

v0 = C1λ1 + C2 λ2

(v0 − λ2 x0 ) λ2 − λ1

(v0 − λ1 x0 ) λ2 − λ1 1 (v − λ x )eλ2t − (v0 − λ2 x0 )eλ1t  λ 2 − λ1  0 1 0

(v0 − λ1 x0 )eλ2t = (v0 − λ 2 x0 )eλ1t e(λ2 − λ1 )t = t=

(v0 − λ 2 x0 ) (v0 − λ1 x0 ) v0 − λ 2 x0 1 ln (λ 2 − λ1 ) v0 − λ1 x0

This defines one value of t only for x = 0, which will exist if the argument of the natural logarithm is positive, i.e., if

v0 − λ2 x0 v0 − λ1 x0

> 1. Assuming λ1 < λ2 < 0,

this occurs if v0 < λ1 x0 .









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PROBLEM 19.129 In the case of light damping, the displacements x1, x2 , x3, shown in Figure 19.11 may be assumed equal to the maximum displacements. Show that the ratio of any two successive maximum displacements xn and xn +1 is a constant and that the natural logarithm of this ratio, called the logarithmic decrement, is ln

xn 2π (c/cc ) = xn +1 1 − (c/cc ) 2

SOLUTION For light damping, −

( 2cm )t sin(ω t + φ ) 0

Equation (19.46):

x = x0 e

At given maximum displacement,

t = tn , x = xn sin(ω0tn + φ ) = 1 xn = x0 e



( 2cm )tn

t = tn +1 , x = xn +1

At next maximum displacement,

sin(ω0tn +1 + φ ) = 1 xn +1 = x0 e

But



( 2cm )tn + 1

ωD tn +1 − ωD tn = 2π 2π tn +1 − tn = ωD −

c

t

xn x e 2m n = 0−ct xn +1 x0 e 2 m n+1

Ratio of successive displacements:

=e

Thus,

ln

− 2cm ( tn −tn+1 )

ln

c 2π 2 m ωD

(1)

 c  1−    cc 

2

c c ωD = c 1 −   2m c

Thus,

+

xn cπ = xn +1 mωD

ω D = ωn

From Equations (19.45) and (19.41):

=e

xn cπ 2m = xn +1 m cc

1  c  1−    cc 

2

2

ln

2π xn = xn +1 1−

( ) ( ) c cc

c cc

2

Q.E.D. 

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PROBLEM 19.130 In practice, it is often difficult to determine the logarithmic decrement of a system with light damping defined in Problem 19.129 by measuring two successive maximum displacements. Show that the logarithmic decrement can also be expressed as (1/k ) ln ( xn /xn + k ), where k is the number of cycles between readings of the maximum displacement.

SOLUTION As in Problem 19.129, for maximum displacements xn and xn + k at tn and t n + k , sin(ω0 tn + φ ) = 1 and sin(ωn tn + k + φ ) = 1.

xn = x0 e



( 2cm )tn

xn + k = x0 e



( 2cm )(tn+k )

( − c )t

− c t −t xn x e 2m n = 0 −c = e 2 m ( n n+ k ) t xn + k x e( 2 m ) n+k 0

Ratio of maximum displacements: But

ωD tn + k − ωD tn = k (2π ) 2π tn − tn + k = k ωD xn c  2kπ  =+   2m  ω D  xn + k x cπ ln n = k xn + k mωD

Thus,

(2)

But from Problem 19.129, Equation (1): log decrement = ln

xn cπ = xn +1 mωD

log decrement =

Comparing with Equation (2),

x 1 ln n Q.E.D.  k xn + k

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PROBLEM 19.131 In a system with light damping (c < cc ), the period of vibration is commonly defined as the time interval τ d = 2π /ωd corresponding to two successive points where the displacement-time curve touches one of the limiting curves shown in Figure 19.11. Show that the interval of time (a) between a maximum positive displacement and the following maximum negative displacement is 12 τ d , (b) between two successive zero displacements is 12 τ d , (c) between a maximum positive displacement and the following zero displacement is greater than 14 τ d .

SOLUTION

x = x0 e

Equation (19.46): (a)



( 2cm )t sin(ω t + φ ) d

Maxima (positive or negative) when x = 0: − ( c )t  −c  −( 2cm )t sin(ωd t + φ ) + x0ωd e 2 m cos(ωd t + φ ) x = x0  e   2m 

Thus, zero velocities occur at times when x = 0, or tan(ωd t + φ ) =

2mωd c

(1)

The time to the first zero velocity, t1 , is t1 =

 tan −1 

(

2 mωd c

) − φ 

ωd

(2)

The time to the next zero velocity where the displacement is negative is t1′ =

 tan −1 

(

2 mωd c

) − φ + π 

ωd

(3)

Subtracting Eq. (2) from Eq. (3), t1′ − t1 =

π π ⋅τ d τ d Q.E.D. = = 2π 2 ωd

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PROBLEM 19.131 (Continued)  (b)

Zero displacements occur when sin(ωθ t + φ ) = 0 or at intervals of

ωd t + φ = π , 2π nπ Thus,

Time between





(c)

The first maximum occurs at 1:

(t1 )0 = (π − φ )

ωd

and (t1′ )0 =

0′s = (t1′ )0 − (t1 )0 =

2π − π

ωd

(2π − φ )

=

ωd πτ d τ d Q.E.D. = 2π 2

Plot of Equation (1) (ωd t1 + φ )

The first zero occurs at

(ωd (t1 )0 + φ ) = π

From the above plot,

(ωd (t1 )0 + φ ) − (ω D t1 + φ ) >

or

(t1 )0 − t1 >

π 2ωd

π

(t1 )0 − t1 >

2

τd 4

Q.E.D.

Similar proofs can be made for subsequent maximum and minimum.

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PROBLEM 19.132 A loaded railroad car weighing 30,000 lb is rolling at a constant velocity v0 when it couples with a spring and dashpot bumper system (Figure 1). The recorded displacement-time curve of the loaded railroad car after coupling is as shown (Figure 2). Determine (a) the damping constant, (b) the spring constant. (Hint: Use the definition of logarithmic decrement given in Problem 19.129.)

SOLUTION m=

Mass of railroad car:

W 30,000 = g 32.2

= 931.67 lb ⋅ s 2 /ft

The differential equation of motion for the system is mx + cx + kx = 0

For light damping, the solution is given by Eq. (19.44): x=e



( 2cm )t (C sin ω t + C cos ω t ) 1 2 d d

From the displacement versus time curve,

τ d = 0.41 s 2π 2π ωd = = = 15.325 rad/s τ d 0.41 At the first peak, x1 = 0.5 in. and t = t1 . At the second peak, x2 = 0.12 in. and t = t1 + τ d . x2 e ( 2 m ) 1 d − ( c )τ = = e 2m d c −( )t x1 e 2m 1 cτ d x1 = e 2m x2 −

Forming the ratio

x2 x1

,

c

( t +τ )

(1)

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PROBLEM 19.132 (Continued)

(a)

Damping constant. From Eq. (1):

x  cτ d = ln  1  2m  x2 

c=

2m

τd

ln

x1 x2

(2)(931.67) 0.5 ln 0.41 0.12 = 6485.9 lb ⋅ s/ft =

(b)

c = 6.49 kip ⋅ s/ft 

Spring constant. Equation for ωd :

ωd2 =

k  c  − m  2m 

k = mωd2 +

2

c2 4m

= (931.67)(15.325) 2 + = 230 × 103 lb/ft

(6485.9) 2 (4)(931.67) k = 230 kips/ft 

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PROBLEM 19.133 A torsional pendulum has a centroidal mass moment of inertia of 0.3 kg-m2 and when given an initial twist and released is found to have a frequency of oscillation of 200 rpm. Knowing that when this pendulum is immersed in oil and given the same initial condition it is found to have a frequency of oscillation of 180 rpm, determine the damping constant for the oil.

SOLUTION Let the mass be rotated through the small angle θ from the equilibrium position. Couples acting on the mass: Shaft: − Kθ Oil: Equation of motion:

−Cθ

ΣM = I θ : −Kθ − Cθ = I θ I θ + Cθ + Kθ = 0

Solution for light damping: where

θ = e− λt (C1 sin ωd t + C2 sin ωd t ) C 2I K ωn = I

λ=

ωd = ωn2 − λ 2 When there is no oil, assume C ≈ 0.

ωn = 200 rpm = 20.944 rad/s When oil is present,

ωd = 180 rpm = 18.8496 rad/s λ = ωn2 − ωd2 = 9.1293 s −1 Damping constant for oil. C = 2 I λ = (2)(0.3 kg ⋅ m2 )(9.1293 s −1 )

C = 5.48 N ⋅ m ⋅ s 

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PROBLEM 19.134 The barrel of a field gun weighs 1500 lb and is returned into firing position after recoil by a recuperator of constant c = 1100 lb ⋅ s/ft. Determine (a) the constant k which should be used for the recuperator to return the barrel into firing position in the shortest possible time without any oscillation, (b) the time needed for the barrel to move back two-thirds of the way from its maximum-recoil position to its firing position.

SOLUTION (a)

A critically damped system regains its equilibrium position in the shortest time. c = cc = 1100 k m

= 2m

( ) =( ) k=

Then (b)

2

cc 2

m

2 1100 2 1500 lb 32.2 ft/s 2

= 6494 lb/ft

k = 6490 lb/ft 

For a critically damped system, Equation (19.43): x = (C1 + C2 t )e−ωnt

We take t = 0 at maximum deflection x0 . Thus,

x (0) = 0 x(0) = x0

Using the initial conditions,

x(0) = x0 = (C1 + 0)e0 , x = ( x0 + C2 t )e

and

so C1 = x0

−ω n t

x = −ωn ( x0 + C2 t )e−ωnt + C2 e−ωnt x (0) = 0 = −ωn x0 + C2 , so C2 = ωn x0

Thus,

x = x0 (1 + ωn t )e−ωnt

For

x=

Solving by trial for ωn t gives:

ωn t = 2.289

But

ωn =

Then

x0 1 = (1 + ωn t )e −ωnt , 3 3

t=

k 6494 lb/ft = = 11.807 s −1 1500 lb m 2

(

32.2 ft/s

)

ωn t 2.289 = = 0.19387 ωn 11.807

t = 0.1939 s 

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PROBLEM 19.135 A platform of weight 200 lb, supported by two springs each of constant k = 250 lb/in., is subjected to a periodic force of maximum magnitude equal to 125 lb. Knowing that the coefficient of damping is 12 lb ⋅ s/in., determine (a) the natural frequency in rpm of the platform if there were no damping, (b) the frequency in rpm of the periodic force corresponding to the maximum value of the magnification factor, assuming damping, (c) the amplitude of the actual motion of the platform for each of the frequencies found in parts a and b.

SOLUTION (a) No Damping: k = 2(250 lb/in.) = 500 lb/in. = 6000 lb/ft

ωn =

k = m

6000 lb/ft = 31.08 rad/s (2000 lb)/(32.2 ft/s 2 )

f = 4.947 Hz f = 297 rpm 

(b) Damped Motion: c = 12 lb ⋅ s/in. = 144 lb ⋅ s/ft

 200  cc = 2mωn = (2)   (31.08) = 386.09 lb ⋅ s/ft  32.2  c 144 lb ⋅ s/ft = = 0.37297 cc 386.09 lb ⋅ s/ft

From Eq. (19.53): xm

δm

1

=

 1 − 

( )  + 2 ( ) ω ωn

2

c cc

ω ωn

2

ω  For maximum amplitude we set equal to zero the derivative with respect to   of the square of the  ωn  denominator.  ω  2 1 −     ωn  

2

2

        − 2  ω  + 8  c   ω  = 0    ω n    cc   ω n  

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PROBLEM 19.135 (Continued)

Rearranging, we obtain 2  ω   ω   4    −1+  ω n   ω n 

2

ω  c    = 1 − 2   ωn   cc  2

ω    =1−  ωn 

 c 2   cc 

2

=0  

2

2

 c 2   = 1 − 2(0.37297) 2 = 0.72179  cc 

ω = 0.84958 ωn

ω = (0.84958)ωn = (0.84958)(31.08 rad/s)

ω = 26.405 rad/s

f = 4.2025 Hz

f = 252 rpm 

(c) Amplitude: From Eq. (19.53):

xm =

Pm k

 1 − 

( )  ω ωn

2

2

+ 2 

( )( ) c cc

ω ωn

2

For part (a) with Pm = 125 lb and ω = ω n xm =

125 lb 6000 lb/ft

[1 − 1]2 + [2(0.37297) (1)]2

xm = 0.335 in. 

= 0.02793 ft

ω  For part (b) with Pm = 125 lb and   = 0.84958  ωn  xm =

125 lb 6000 lb/ft

[1 − (0.84958) 2 ]2 + [2(0.84958)(0.37297)]2

= 0.0301 ft xm = 0.361 in. 

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PROBLEM 19.136 A 4-kg block A is dropped from a height of 800 mm onto a 9-kg block B which is at rest. Block B is supported by a spring of constant k = 1500 N/m and is attached to a dashpot of damping coefficient c = 230 N ⋅ s/m. Knowing that there is no rebound, determine the maximum distance the blocks will move after the impact.

SOLUTION Velocity of Block A just before impact. v A = 2 gh = 2(9.81)(0.8) = 3.962 m/s

Velocity of Blocks A and B immediately after impact. Conservation of momentum. m Av A + mB vB = (m A + mB )v′ (4)(3.962) + 0 = (4 + 9)v′ v′ = 1.219 m/s x0 = +1.219 m/s = x0

Static deflection (Block A):

mA g k (4)(9.82) =− 1500 = −0.02619 m

x0 = −

x = 0, Equilibrium position for both blocks: cc = 2 km = 2 (1500)(13) = 279.3 N ⋅ s/m

Since c < cc , Equation (19.44):

x=e



( 2cm )t [C sin ω t + C cos ω t ] 1 2 d d

230 c = 2m (2)(13)

= 8.846 s −1 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2437

PROBLEM 19.136 (Continued)

ωd2 =

Expression for ωd :

k  c  −  m  2m 

2

1500  230  −  13  (2)(13) 

ωd =

2

= 6.094 rad/s x = e −8.846t (C1 sin 6.094t + C2 cos 6.094t ) x0 = −0.02619 m (t = 0) x0 = +1.219 m/s

Initial conditions:

'

x0 = −0.02619 = e0 [C1 (0) + C2 (1)] C2 = −0.02619 x (0) = −8.846e( −8.846)0 [C1 (0) + ( −0.02619)(1)] + e( −8.846)(0) [6.094C1 (1) + C2 (0)] = 1.219 1.219 = (−8.846)(−0.02619) + 6.094C1 C1 = 0.16202 x = e−8.846t (0.16202sin 6.094t − 0.02619 cos 6.094t )

Maximum deflection occurs when

x = 0

x = 0 = −8.846e−8.846tm (0.16202sin 6.094tm − 0.02619 cos 6.094tm ) + e −8.846tm [6.094][0.1620cos 6.094tm + 0.02619sin 6.094tm ] 0 = [( −8.846)(0.16202) + (6.094)(0.02619)]sin 6.094tm + [(−8.846)( −0.02619) + (6.094)(0.1620)]cos 6.094tm 0 = −1.274sin 6.094tm + 1.219cos 6.094tm tan 6.094tm =

1.219 = 0.957 1.274

tan −1 0.957 = 0.1253 s 6.094 xm = e− (8.846)(0.1253) [0.1620sin(6.094)(0.1253)

Time at maximum deflection = tm =

− 0.02619 cos(6.094)(0.1253)] xm = (0.3301)(0.1120 − 0.0189) = 0.0307 m

Blocks move, static deflection + xm

Total distance = 0.02619 + 0.0307 = 0.0569 m = 56.9 mm 

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PROBLEM 19.137 A 3-kg slender rod AB is bolted to a 5-kg uniform disk. A dashpot of damping coefficient c = 9 N ⋅ s/m is attached to the disk as shown. Determine (a) the differential equation of motion for small oscillations, (b) the damping factor c/cc .

SOLUTION Data:

r = 100 mm = 0.100 m, l = 400 mm = 0.400 m 1 1 mdisk r 2 = (5 kg)(0.100 m)2 = 0.025 kg ⋅ m 2 2 2 1 1 2 = m AB l = (3 kg)(0.400 m)2 = 0.040 kg ⋅ m 2 2 12 = m AB g = (3 kg)(9.81 m/s 2 ) = 29.43 N

I disk = I AB WAB

c = 9 N ⋅ s/m

Equation of motion: Let the disk and rod assembly be rotated through a small counterclockwise angle θ

ΣM A = Σ(M A )eff :

l l −WAB x − Fd r = I diskθ + I ABθ + mAB  θ  2 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2439

PROBLEM 19.137 (Continued)

where

l sin θ 2 = (29.43 N)(0.2 m) sin θ

WAB x = −WAB

= 5.886 sinθ N ⋅ m ≈ −5.886 θ

Damping force:

Fd = crθ Fd r = cr 2θ = (9 N ⋅ s ⋅ m)(0.100 m) 2θ = 0.09θ = Cθ 2

Inertia:

l I disk + I AB + mAB   = 0.025 + 0.040 + (3)(0.2)2 = 0.185 kg ⋅ m 2 2 − 5.886θ − 0.09θ = 0.185θ

0.185θ + 0.09θ + 5.886θ = 0 M θ + Cθ + Kθ = 0

ωn =

K 5.886 = = 5.6406 rad/s M 0.185

Cc = 2 M ω2 = (2)(0.185)(5.6406) = 2.087 c C 0.09 = = cc Cc 2.087

c = 0.0431  cc

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PROBLEM 19.138 A uniform rod of mass m is supported by a pin at A and a spring of constant k at B and is connected at D to a dashpot of damping coefficient c. Determine in terms of m, k, and c, for small oscillations, (a) the differential equation of motion, (b) the critical damping coefficient cc .

SOLUTION In equilibrium, the force in the spring is mg. For small angles,

sin θ ≈ θ cos θ ≈ 1 l δ yB = θ 2 δ yC = lθ

(a)

ΣM A = (ΣM A )eff

Newton’s Law:

mgl  l l l −  k θ + mg  − clθl = I α + mat 2 2  2 2

Kinematics:

α = θ l l at = α = θ 2 2

2 2   l    l 2  I m θ cl θ k + + +  2  2 θ =0       2

1 l I + m   = ml 2 3 2

(b)

 3c    3k  θ +  4m θ = 0  m  

θ + 

Substituting θ = eλ t into the differential equation obtained in (a), we obtain the characteristic equation, 3k  3c  =0 λ2 +  λ + m 4 m   and obtain the roots

λ=

−3c m



( 3mc ) − ( 3mk ) 2

2

The critical damping coefficient, cc , is the value of c, for which the radicand is zero. 2

Thus,

3k  3cc   m  = m  

cc =

km  3

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PROBLEM 19.139 A machine element weighing 800 lb is supported by two springs, each having a constant of 200 lb/in. A periodic force of maximum value 30 lb is applied to the element with a frequency of 2.5 cycles per second. Knowing that the coefficient of damping is 8 lb ⋅ s/in., determine the amplitude of the steady-state vibration of the element.

SOLUTION Equivalent spring:

k = 2(200) = 400 lb/in. = 4800 lb/ft k 4800 = = 13.90 rad/s m 800/32.2 ω 13.90 fn = n = = 2.212 Hz 2π 2π

Undamped natural frequency:

ωn =

Critical damping coefficient:

 800  cc = 2 mωn = 2   (13.90) = 691 lb ⋅ s/ft  32.2 

Damping coefficient:

c = 8 lb ⋅ s/in. = 96 lb ⋅ s/ft

Damping ratio:

c 96 = = 0.1390 cc 691

Amplitude:

xm =

where

Pm /k  1 − 

( )  ωf ωn

2 2

+  2 cc  c

(1) ωf ωn

 

2

ωf ff 2.5 Hz = = = 1.130 ωn f n 2.212 Hz

Substituting into Eq. (1) with Pm = 30 lb, we have xm =

30 lb/400 lb/in. [1 − (1.130) 2 ]2 + [2(0.1390)(1.130)]2

xm = 0.1791 in. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2442

PROBLEM 19.140 In Problem 19.139, determine the required value of the coefficient of damping if the amplitude of the steady-state vibration of the element is to be 0.15 in. PROBLEM 19.139 A machine element weighing 800 lb is supported by two springs, each having a constant of 200 lb/in. A periodic force of maximum value 30 lb is applied to the element with a frequency of 2.5 cycles per second. Knowing that the coefficient of damping is 8 lb ⋅ s/in., determine the amplitude of the steady-state vibration of the element.

SOLUTION k = 2(200) = 400 lb/in. = 4800 lb/ft

Equivalent spring:

k 4800 = = 13.90 rad/s m 800/32.2 ω 13.90 fn = n = = 2.212 Hz 2π 2π

Undamped natural frequency:

ωn =

Critical damping coefficient:

 800  cc = 2 mωn = 2   (13.90) = 691 lb ⋅ s/ft  32.2 

Amplitude:

xm =

Pm /k  1 − 

( )  ωf ωn

2 2

(1)

+  2 cc  c

ωf ωn

 

2

ωf ff 2.5 Hz = = = 1.130 ωn f n 2.212 Hz

where

Using xm = 0.15 in., Pm = 30 lb, and k = 400 lb/in. 0.15 in. =

30 lb/400 lb/in.  c  [1 − (1.130) ] +  2 (1.130)   cc 

2

2 2

Solving for

c cc

, we find

c cc

= 0.1842.

Since cc = 691 lb ⋅ s/ft, we have c = (0.1842)(691)

c = 1273 lb ⋅ s/ft c = 10.61 lb ⋅ s/in. 

or

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PROBLEM 19.141 In the case of the forced vibration of a system, determine the range of values of the damping factor c/cc for which the magnification factor will always decrease as the frequency ratio ω f /ωn increases.

SOLUTION

From Eq. (19.53)′ : Magnification factor:

Find value of

xm Pm k

1

=

 1 − 

( )

2 ωf 2   ωn

+ 2  

( )( c cc

ωf ωn

)

 

2

ωf x c for which there is no maximum for Pm as increases. m ω cc n k

( ) d( ) d

  −  2 1 −  

xm 2 Pm k

ωf 2 ωn

=

   1 −  

( )  ( −1) + 4 ωf 2 ωn

( )  ωf ωn

2

2

 + 2 

( ) ( ωω c cc

f n

c2 cc2

  

2 

)  

2

=0

ωf c2 −2 + 2  ω  + 4 2 = 0  n cc 2

2  ωf  =1− 2 c  ωn    cc2 2

For

ωf x c2 1 increases. ≥ , there is no maximum for m and the magnification factor will decrease as 2 P m ωn cc 2

( ) k

c 1 ≥ cc 2

c ≥ 0.707  cc

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2444

PROBLEM 19.142 Show that for a small value of the damping factor c/cc , the maximum amplitude of a forced vibration occurs when ω f ≈ ωn and that the corresponding value of the magnification factor is 12 (c/cc ).

SOLUTION From Eq. (19.53′): Magnification factor =

Find value of

xm Pm k

=

1

 1 − 

( ) ωf ωn

2 2

  + 2  

( ) ( ωω c cc

f

n

)

 

2

ωf x for which Pm is a maximum. m ωn k

0=

2

( ) d( ) d

   2 1 −  

xm 2 Pm k

ωf 2 ωn

=−

   1 −  

( )  (−1) + 4 ( ) ωf 2 ωn

( )

2 ωf 2   ωn

+ 2 

c2 cc2

( )( c cc

ωf ωn

)

2

    

2

2

 ωf   c  − 2 + 2  + 4  = 0  cc   ωn 

For small

For

c , cc

ωf ≈1 ωn

ω f ≈ ωn

ωf =1 ωn xm Pm k

=

xm

1 [1 − 1]2 +  2 

( )1 c cc

2

( ) Pm k

=

1 cc  2 c

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2445

PROBLEM 19.143 A counter-rotating eccentric mass exciter consisting of two rotating 14-oz masses describing circles of 6-in. radius at the same speed but in opposite senses is placed on a machine element to induce a steady-state vibration of the element and to determine some of the dynamic characteristics of the element. At a speed of 1200 rpm a stroboscope shows the eccentric masses to be exactly under their respective axes of rotation and the element to be passing through its position of static equilibrium. Knowing that the amplitude of the motion of the element at that speed is 0.6 in. and that the total mass of the system is 300 lb, determine (a) the combined spring constant k, (b) the damping factor c/cc .

SOLUTION Forcing frequency:

ω f = 1200 rpm = 125.664 rad/s

Unbalance of one mass:

w = 14 oz = 0.875 lb r = 6 in. = 0.5 ft

Shaking force:

P = 2 mrω 2f sin ω f t = (2) ( 0.875 (0.5)(125.664) 2 sin ω f t 32.2 ) = 429.11sin ω f t Pm = 429 lb W = 300 lb

Total weight:

By Eqs. (19.48) and (19.52), the vibratory response of the system is x = xm sin(ω f t − ϕ )

xm =

where

tan ϕ =

and Since ϕ = 90° = (a)

π 2

,

Pm

(

k − M ω 2f

)

2

(1) + (cω f )2

cω f

(2)

k − M ω 2f

tan ϕ = ∞ and k − M ω 2f = 0.

Combined spring constant. k = M ω 2f 300 = ( 32.2 ) (125.664)2

= 147.12 × 103 lb/ft

k = 147 kip/ft 

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PROBLEM 19.143 (Continued)

The observed amplitude is From Eq. (1):

Critical damping coefficient:

xm = 0.6 in. = 0.05 ft 2

 Pm  Pm 2   − (k − M ω f ) = ω f  xm  ω f xm 429.11 = (125.664)(0.05) = 68.296 lb ⋅ s/ft

c=

1

cc = 2 kM 300 = 2 (147.12 × 103 ) ( 32.2 )

= 2.3416 × 103 lb ⋅ s/ft

(b)

Damping factor.

c 68.296 = cc 2.3416 × 103

c = 0.0292  cc

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2447

PROBLEM 19.144 A 15-kg motor is supported by four springs, each of constant 40 kN/m. The unbalance of the motor is equivalent to a mass of 20 kg located 125 mm from the axis of rotation. Knowing that the motor is constrained to move vertically and that the damping factor c /cc is equal to 0.4, determine the range of frequencies for which the amplitude of the steady-state vibration of the motor is less than 0.2 mm.

SOLUTION Equivalent spring:

k = (4)(40 × 103 N/m) = 160 × 103 N/m

Mass:

m = 15 kg

Natural frequency:

ωn =

Unbalanced force:

 ωf  Pm = meq rω 2f = meq rωn2    ωn 

k 160 × 103 = = 103.280 rad/s m 15 2

= (0.020 kg)(0.125 m)(130.280 rad/s)2 2

ωf  = 26.667   N  ωn 

At steady state,

xm =

  1 − 

Pm /k   1 − 

1/2 2 c ωf  cc ωn

( )  + ( 2 )  2 ωf 2   ωn

( )  +( 2 ωf 2   ωn 2

1/2 2 c ωf 2c ω  c n

) 

4

=

Pm kxm 2

 ω f   ω f   c ω f   Pm  1− 2  +  +2  =   ω   ωn   cc ωn   kxm 

Amplitude:

(1)

xm = 0.2 mm = 0.2 × 10−3 m Pm 26.667 = kxm (160 × 103 ) (0.2 × 10−3 )

Damping factor:

2

2

 ωf   ωf    = 0.83333    ωn   ωn 

2

c = 0.4 cc

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2448

PROBLEM 19.144 (Continued)

Substituting into Eq. (1),

2 4 2 ωf   ω 2f ωf  ωf    1− 2 + + (2)(0.4) = 0.83333      ωn   ωn2  ωn   ω n   4

   

2

2

ω  ωf  2 2  f  [(1 − (0.83333) ]   − [2 − (2) (0.4) ]   +1 = 0  ωn   ωn  2

4

2

ωf   ωf  0.30556   − 1.36   +1 = 0  ωn   ωn 

Solving the quadratic equation for

( ), ωf 2 ωn

2

ωf    = 3.5216 and 0.92934  ωn 

ωf = 1.8766 and 0.96402 ωn ω f = (1.8766)(103.280 rad/s) = 193.815 rad/s and

ω f = (0.96402)(103.280 rad/s) = 99.564 rad/s

For xm < 0.2 m, the forcing frequency must satisfy

ω f > 193.8 rad/s and ω f < 99.6 rad/s Since

ff =

ωf , 2π f f > 30.8 Hz and

f f < 15.85 Hz 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2449

PROBLEM 19.145 A 220-lb motor is supported by four springs, each of constant 500 lb/in., and is connected to the ground by a dashpot having a coefficient of damping c = 35 lb ⋅ s/in. The motor is constrained to move vertically, and the amplitude of its motion is observed to be 0.08 in. at a speed of 1200 rpm. Knowing that the weight of the rotor is 30 lb, determine the distance between the mass center of the rotor and the axis of the shaft.

SOLUTION Forcing frequency:

ω f = 1200 rpm = 125.664 rad/s

Equivalent spring:

k = (4)(500) = 2000 lb/in. = 24000 lb/ft

Mass:

m=

W 220 lb = = 6.8323 lb ⋅ s 2 /ft g 32.2 ft/s 2

ωn =

k 24000 = = 59.268 rad/s m 6.8323

Natural frequency:

ω f 125.664 rad/s = = 2.12026 ωn 59.268 rad/s Critical damping coefficient:

cc = 2mωn cc = (2)(6.8323)(59.268) = 809.87 lb ⋅ s/ft

Damping coefficient:

c = 35 lb ⋅ s/in. = 420 lb ⋅ s/ft

Damping factor:

c 420 = = 0.51860 cc 809.87

Amplitude:

xm = 0.08 in. = 6.6667 × 10−3 ft xm =

Unbalanced force:

Pm /k

  1 − 

1/2 2 c ωf  cc ωn

( )  + ( 2 )  2 ωf 2   ωn

 Pm = kxm 1 − 

( )  +( 2 ωf 2   ωn

1/2

 ω 2 2 cc ω f  c n

) 

= kxm [(1 − 4.4955)2 + ((2)(0.51860)(2.12026))2 ]1/2 = kxm [12.2185 + 4.8362]1/2 = 4.1297 kxm = (4.1297)(24000)(6.6667 × 10−3 ) = 660.76 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2450

PROBLEM 19.145 (Continued)

But,

Pm = m′eω 2f

where m′ is the mass of the rotor and e is the distance between the mass center of the rotor and the axis of the shaft. 30 lb = 0.93168 lb ⋅ s 2 /ft 32.2 ft/s 2 P 660.76 lb e = m2 = m′ω f (0.93168 lb ⋅ s 2 /ft)(125.664 rad/s) 2

m′ =

= 0.044911 ft e = 0.539 in. 

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PROBLEM 19.146 A 100-lb motor is directly supported by a light horizontal beam which has a static deflection of 0.2 in. due to the weight of the motor. The unbalance of the rotor is equivalent to a mass of 3.5 oz located 3 in. from the axis of rotation. Knowing that the amplitude of the vibration of the motor is 0.03 in. at a speed of 400 rpm, determine (a) the damping factor c/cc , (b) the coefficient of damping c.

SOLUTION k=

Spring constant: Natural undamped circular frequency:

ωn =

W

δ st

=

100 0.2 12

= 6000 lb/ft

k 6000 = 100 = 43.955 rad/s m 32.2

3.5 w = 16 6.7935 × 10−3 slug g 32.2 r = 3 in. = 0.25 ft

m′ =

Unbalance:

Forcing frequency:

ω f = 400 rpm = 41.888 rad/s

Unbalance force:

Pm = m′rω 2f = (6.7935 × 10−3 )(0.25)(41.888) 2 = 2.98 lb

δ st =

Static deflection:

Pm 2.98 = = 0.49666 × 10−3 ft 6000 k

xm = 0.03 in. = 2.5 × 10−3 ft

Amplitude:

ω f 41.888 = = 0.95298 ωn 43.955

Frequency ratio:

xm =

Eq. (19.53):

δ st  1 − 

( )  ωf ωn

2 2

+ 2 

( ) ( ) c cc

ωf ωn

2

2

  ω  2    c   ω   2  δ 2 1 −  f   +  2    f   =  st    ωn     cc   ωn    xm      c [1 − (0.95298) ] +  2    cc 2 2

2

  0.49666 × 10−3   = (0.95298)     −3   2.5 × 10  

2

2

 c 0.0084326 + 3.6327   = 0.039467  cc  2

 c   = 0.0085431  cc  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2452

PROBLEM 19.146 (Continued)

(a)

Damping factor.

c = 0.092429 cc

Critical damping factor.

cc = 2 km

c = 0.0924  cc

100 = 2 (6000) ( 32.2 )

= 273.01 lb ⋅ s/ft

(b)

Coefficient of damping.

 c c=  cc

  cc 

= (0.092429)(273.01)

c = 25.2 lb ⋅ s/ft 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2453

PROBLEM 19.147 A machine element is supported by springs and is connected to a dashpot as shown. Show that if a periodic force of magnitude P = Pm sin ω f t is applied to the element, the amplitude of the fluctuating force transmitted to the foundation is

( ) ( )     1 − ( )  +  2 ( ) ( )      1 + 2 

Fm = Pm

c cc

ωf ωn

2 ωf 2 ωn

c cc

2

2 ωf 2 ωn

SOLUTION x = xm sin(ω f t − φ )

From Equation (19.48), the motion of the machine is The force transmitted to the foundation is Springs:

Fs = kx = kxm sin(ω f t − φ )

Dashpot:

Fd = cx = cxmω f cos(ω f t − φ ) Ft = xm [k sin(ω f t − φ ) + cω f cos(ω f t − φ )]

or recalling the identity, A sin y + B cos y = A2 + B 2 sin( y + ψ ) B sinψ = 2 A + B2 A cosψ = 2 A + B2 Ft =  xm k 2 + (cω f )2  sin(ω f t − φ + ψ )  

Thus, the amplitude of Ft is From Equation (19.53):

Substituting for xm in Equation (1),

Fm = xm k 2 + (cω f )2 xm =

(1)

Pm k

 1 − 

( )

2 ωf 2   ωn

Fm =

ωn2 =

+  2 

( ) c cc

ωf ωn

Pm 1 +  1 − 

( )  ωf ωn

2

2

 

2

( ) cω f k

+ 2 

2

( ) ( ) c cc

(2)

ωf

ωn2

k m

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PROBLEM 19.147 (Continued)

and Equation (19.41),

cc = 2mωn m= cω f k

=

cωn 2 cω f mωn2

 c  ω f  = 2    cc  ωn 

Fm =

Substituting in Eq. (2),

Pm 1 +  2 

 1 − 

( )

2 ωf 2   ωn

( ) ( ) c cc

+ 2 

ωf ωn

2

( ) ( ) c cc

ωf ωn

Q.E.D.  2

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PROBLEM 19.148 A 91-kg machine element supported by four springs, each of constant k = 175 N/m, is subjected to a periodic force of frequency 0.8 Hz and amplitude 89 N. Determine the amplitude of the fluctuating force transmitted to the foundation if (a) a dashpot with a coefficient of damping c = 365 N ⋅ s/m is connected to the machine element and to the ground, (b) the dashpot is removed.

SOLUTION Forcing frequency:

ω f = 2π f f = (2π )(0.8) = 1.6π rad/s

Exciting force amplitude:

Pm = 89 N

Equivalent spring constant: Natural frequency:

Frequency ratio:

k = (4)(175 N/m) = 700 N/m

k 700 = m 91 = 2.7735 rad/s

ωn =

ωf ωn

=

1.6π 2.7735

= 1.8123

Critical damping coefficient:

cc = 2 km = 2 (700)(91) = 504.78 N ⋅ s/m

From the derivation given in Problem 19.147, the amplitude of the force transmitted to the foundation is

Fm =

Pm 1 +  2   1 − 

( )

2 ωf 2   ωn

( ) ( ) c cc

+ 2 

ωf ωn

2

( ) ( ) c cc

ωf ωn

(1) 2

2

 ωf  2 1−   = 1 − (1.8123) = −2.2844 ω  n

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PROBLEM 19.148 (Continued)

(a)

Fm when c = 365 N ⋅ s/m:

c 365 = cc 504.78 = 0.72309

 c  ω f  2   = (2)(0.72309)(1.8123)  cc  ωn  = 2.6209

From Eq. (1):

Fm =

=

(b)

Fm when c = 0:

89 1 + (2.6209) 2 ( −2.2844) 2 + (2.6209) 2 89 7.8692 12.088 Pm

Fm = 1−

( ) ωf ωn

2

=

Fm = 71.8 N  89 2.2844

Fm = 39.0 N 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2457

PROBLEM 19.149 A simplified model of a washing machine is shown. A bundle of wet clothes forms a weight wb of 20 lb in the machine and causes a rotating unbalance. The rotating mass is 40 lb (including mb) and the radius of the washer basket e is 9 in. Knowing the washer has an equivalent spring constant k = 70 lb/ft and damping ratio ζ = c/cc = 0.05 and during the spin cycle the drum rotates at 250 rpm, determine the amplitude of the motion and the magnitude of the force transmitted to the sides of the washing machine.

SOLUTION Forced circular frequency: System mass: Spring constant: Natural circular frequency: Critical damping constant: Damping constant: Unbalance force:

ωf = m=

(2π )(250) = 26.18 rad/s 60 W 40 lb = 32.2 g

k = 70 lb/ft

ωn =

k = m

70 40 32.2

= 7.5067 rad/s

40 cc = 2 km = 2 (70) ( 32.2 ) = 18.650 lb ⋅ s/ft

 c c=  cc mb =

  cc = (0.05)(18.650) = 0.9325 lb ⋅ s/ft 

wb g

Pm = mb eω 2f

 20 lb  9  Pm =  ft  (26.18 rad/s) 2 = 319.28 lb   32.2  12 

The differential equation of motion is mx + cx + kx = Pm sin ω f t The steady state response is x = xm sin(ω f t − ϕ )

x = ω f xm cos(ω f t − ϕ )

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2458

PROBLEM 19.149 (Continued)

xm =

where

Pm

( k − mω ) 2 f

= =

(a)

2

+ (cω f ) 2 319.28

40 [70 − ( 32.2 ) (26.18)2 ]2 + [(0.9325)(26.18)]2

319.28 (−781.42)2 + (24.413) 2

=

319.28 = 0.40839 ft 781.796 xm = 4.90 in. 

Amplitude of vibration. x = 0.40839sin(ω f t − ϕ ) x = (26.18)(0.40839) cos(ω f t − ϕ ) = 10.6917 cos(ω f t − ϕ )

Spring force:

kx = (70)(0.40839)sin(ω f t − ϕ ) = 28.588sin(ω f t − ϕ )

Damping force:

cx = (0.9325)(10.6917) cos(ω f t − ϕ ) = 9.9701cos(ω f t − ϕ )

(b)

Total force:

F = 28.588sin(ω f t − ϕ ) + 9.9701cos(ω f t − ϕ )

Let

F = Fm cosψ sin(ω f t − ϕ ) + Fm sinψ sin(ω f t − ϕ ) = Fm sin(ω f t − ϕ + ψ )

Maximum force.

Fm2 = Fm2 cos 2 ψ + Fm2 sin 2 ψ = (28.588) 2 + (9.9701)2 = 916.65

Fm = 30.3 lb 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2459

PROBLEM 19.150* For a steady-state vibration with damping under a harmonic force, show that the mechanical energy dissipated per cycle by the dashpot is E = π cxm2 ω f , where c is the coefficient of damping, xm is the amplitude of the motion, and ω f is the circular frequency of the harmonic force.

SOLUTION Energy is dissipated by the dashpot. From Equation (19.48), the deflection of the system is x = xm sin(ω f t − ϕ ) Fd = cx

The force on the dashpot.

Fd = cxmω f cos(ω f t − ϕ )

The work done in a complete cycle with

τf =

2π ωf



E = Fd dx (i.e., force × distance) dx = xmω f cos(ω f t − ϕ )dt E= cos 2 (ωD t − ϕ ) =



2π /ω f 0

cxm2 ω 2f cos 2 (ω f t − ϕ )dt

[1 − 2 cos(ω f t − φ )]

E = cxm2 ω 2f

2



2π /ω f

1 − 2 cos(ω f t − φ ) 2

0

dt

2π /ω f

cxm2 ω 2f  2sin(ω f t − φ )  E= t −  2  ωf  0

 cxm2 ω 2f  2π 2 E= (sin(2π − ϕ ) + sin ϕ )  −  2  ω f ω f 

E = π cxm2 ω f Q.E.D. 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2460

PROBLEM 19.151* The suspension of an automobile can be approximated by the simplified spring-and-dashpot system shown. (a) Write the differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross section of amplitude δ m and wave length L. (b) Derive an expression for the amplitude of the vertical displacement of the mass m.

SOLUTION (a)

d x  dx dδ  ΣF = ma : W − k (δ st + x − δ ) − c  − =m 2 dt  dt dt  2

Recalling that W = kδ st , we write m

d2x dx dδ + c + kx = kδ + c 2 dt dt dt

(1)

Motion of wheel is a sine curve, δ = δ m sin ω f t. The interval of time needed to travel a distance L at a speed v is t =

L . v

Thus,

and

ωf =

2π 2π 2π v = L = L τf v

δ = δ m sin ω f t

dδ δ m 2π = L cos ω f t dt v

Thus, Equation (1) is m

d2x dx + c + kx = ( k sin ω f t + cω f cos ω f t )δ m  dt dt

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2461

PROBLEM 19.151* (Continued)

(b)

A sin y + B cos y = A2 + B 2 sin( y + ψ ) B sinψ = A2 + B 2 A cosψ = A2 + B 2

From the identity

We can write the differential equation m

d2x dx + c + kx = δ m k 2 + (cω f ) 2 sin(ω f t + ψ ) 2 dt dt cω f ψ = tan −1 k

The solution to this equation is analogous to Equations 19.47 and 19.48, with Pm = δ m k 2 + (cω f ) 2 x = xm sin(ω f t − ϕ + ψ ) (where analogous to Equations (19.52))  xm =

δ m k 2 + (cω f )2

(k −

mω 2f

)

2

tan ϕ =



+ (cω f )

2

cω f k − mω 2f

tanψ =

cω f k





PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2462

PROBLEM 19.152* Two blocks A and B, each of mass m, are supported as shown by three springs of the same constant k. Blocks A and B are connected by a dashpot, and block B is connected to the ground by two dashpots, each dashpot having the same coefficient of damping c. Block A is subjected to a force of magnitude P = Pm sin ω f t. Write the differential equations defining the displacements x A and xB of the two blocks from their equilibrium positions.

SOLUTION

Since the origins of coordinates are chosen from the equilibrium position, we may omit the initial spring compressions and the effect of gravity For load A, ΣF = ma A : Pm sin ω f t + 2k ( xB − x A ) + c( x B − x A ) = mxA

(1)

ΣF = maB : − 2k ( xB − x A ) − c( x B − x A ) − kxB − 2cxB = mxB

(2)

For load B,

mxA + c( x A − x B ) + 2k ( x A − xB ) = Pm sin ω f t 

Rearranging Equations (1) and (2), we find:

mxB + 3cx B − cx A + 3kxB − 2kx A = 0 

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PROBLEM 19.153 Express in terms of L, C, and E the range of values of the resistance R for which oscillations will take place in the circuit shown when switch S is closed.

SOLUTION For a mechanical system, oscillations take place if c < cc (lightly damped). But from Equation (19.41),

c < 2 km

Therefore, From Table 19.2:

k = 2 km m

cc = 2m

c

R

m

L

k

1 C

(1)

(2)

Substituting in Eq. (1) the analogous electrical values in Eq. (2), we find that oscillations will take place if 1 R < 2   ( L) C 

R
Vector Mechanics for Engineers - Manual Solutions Dynamics 10th edition

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