Solutions Chapter R2

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© 2010 Pearson Education, Education, Inc., Upper Saddle River, River, NJ. All rights reserved. reserved. This material material is protected under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

R2–1. An automobile transmission consists of the planetary gear system shown. If the ring gear R is held fixed so that vR = 0 , and the shaft s and sun gear S , rotates rotates at 20 rad s, determine the angular velocity of each planet gear P and the angular velocity of the connecting rack D , which is free to rotate about the center shaft s .

R

>

P v

R

2 in. S 20 rad/ s

8 in. 4 in.

s

For planet gear P : The velocity of point A is yA = vs rs = 20 vB

a 124 b = 6.667 ft>s.

= vA + vB>A

B a b R 4 12

0 = C 6.667 D +

vP

0 = 6.667 - vP

a b

:

;

a+b :

4 12

vP

>

= 20 rad s

Ans.

For connecting rack D: vC

= vA + vC>A

c d = C 6.667 D + B 20 a 122 b R yC

:

:

;

a+b :

yC

= 6.667 - 20

a 122 b

yC

>

= 3.333 ft s

The rack is rotating about a fixed axis (shaft s). Hence, Hence, yC

3.333 = vD

=

a 126 b

vD rD vD

>

= 6.67 rad s

Ans.

827

D P

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© 2010 Pearson Education, Inc., Upper Saddle River,NJ. All rights reserved. reserved. This material material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.

R2–2. An automobile transmission consists of the planetary gear system shown. If the ring gear R rotates at shaft s and sun gear S , rotate rotatess at vR = 2 rad s, and the shaft 20 rad s, determine the angular velocity of each planet gear P and the angular velocity of the connecting rack D , which is free to rotate about the center shaft s .

R

>

>

P v

R

2 in. S 20 rad/ s

8 in. 4 in.

s

a b

4 For planet gear P : The velocit velocity y of points points A and B are yA = vS rS = 20 12 8 = 6.667 ft s and yB = vB rB = 2 = 1.333 ft s . 12

>

a b

vB

>

= v A + v B>A

c 1.333 d = c 6.667 d + B a 124 b R ;

vP

:

;

a+b :

- 1.333 = 6.667 -

vP

a 124 b

vP

>

= 24 rad s

Ans.

For connecting rack D: vC

= vA + vC>A

c d = c 6.667 d + B 24 a 122 b R yC

:

:

;

a+b :

yC

= 6.667 - 24

a 122 b

yC

>

= 2.667 ft s

The rack is rotating about a fixed axis (shaft s). Hence, Hence, yC

2.667 = vD

=

a 126 b

vD rD vD

>

= 5.33 rad s

Ans.

828

D P

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R2–3. The 6-lb slender rod AB is released from rest when it is in the horizontal position so that it begins to rotate clockwise. A 1-lb ball is thrown at the rod with a velocity v = 50 ft s. The ball strikes the rod at C at the instant the rod is in the vertical vertical position position as shown. Determine Determine the angular velocity of the rod just after the impact. impact. Take e = 0.7 and d = 2 ft.

A

>

d 3 ft C

Datum at A: T1

0 + 0 =

v  50

+ V1 = T2 + V2

c a 32.26 b (3) d = 5.675 rad> s

1 1 2 3

2

v2

B

- 6(1.5)

v

a+

(HA)1 = (HA)2

c a 32.26 b (3) d (5.675) = c 13 a 32.26 b (3) d

1 1 (50)(2) 32.2 3

2

e

2

= 0.7 =

v2

+

1 (v )(2) 32.2 BL

- v BL 50 - [ - 5.675(2)] vC

vC

= 2v2

Solving, v2

vBL vC

ft / s

> = - 35.3 ft> s = 7.61 ft> s

= 3.81 rad s

Ans.

829

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*R2–4. The 6-lb slender rod AB is originally at rest, suspended in the vertical position. A 1-lb ball is thrown at the rod with a velocity v = 50 ft s and strikes the rod at C . Determine the angular velocity of the rod just after the impact. Take e = 0.7 and d = 2 ft .

>

A

d

a + (HA)1 = (HA)2

3 ft C

a 32.21 b (50)(2) = c 13 a 32.26 b (3) d 2

e

= 0.7 = vC

v2

+

1 (v )(2) 32.2 BL

v  50

- vBL 50 - 0

vC

B

= 2v2

Thus, v2

vBL

ft / s

> = - 19.5 ft> s

= 7.73 rad s

Ans.

830

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R2–5. The 6-lb slender rod is originally at rest, suspended in the vertical position. Determine the distance d where the 1-lb ball, traveling traveling at v = 50 ft s , should strike the rod so that it does not create a horizontal impulse at A . What is the rod’s angular velocity just after the impact? Take e = 0.5 .

>

A

d

Rod:

a+

3 ft C

(HG)1 + ©

L

0 +

L

= (HG)2

MG dt

v  50

F dt (d

- 1.5) =

m(vG)1

+ ©

L

L

F dt

= m(1.5v)

0 +

F dt

a 121 (

b

m)(3)2 v

B

= m(vG)2

Thus, m(1.5v)(d

- 1.5) = d

1 (m)(3)2 v 12

= 2 ft

Ans.

This is called the the center of percussion. See Example 19–5.

a+

(HA)1 = (HA)2

c a 32.26 b (3) d

1 1 (50)(2) = 32.2 3 e

= 0.5 =

vC

2

v2

+

1 (v )(2) 32.2 BL

- vBL 50 - 0

vC

= 2v2

Thus, v2

vBL

ft / s

> = - 11.4 ft> s

= 6.82 rad s

Ans.

831

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instant, the wheel rotates rotates with with the R2–6. At a given instant, angular motions shown. Determine the acceleration of the collar at A at this instant. Using instantaneous center method: 8(0.15) yB vAB = = = 4.157 rad s rB IC 0.5 tan 30°

>

>

= 16(0.15)i - 82 (0.15) j = {2.4i - 9.6 j} m s2

aA

= - aA cos 60°i + aA sin 60° j

aB

= a A +

a

* rB>A -

= ak

60

>

aB

a

A

>

rB A

2.4i - 9.6 j = ( - aA cos 60°i + aA sin 60° j) + (ak) * ( - 0.5i) - (4.157)2( - 0.5i) 2.4i - 9.6 j = ( - aA cos 60° + 8.64)i + ( - 0.5a + aA sin 60°) j Equating the i and j components yields:

- 9.6 = - 0.5a + (12.5) sin 60°

a

 16

500 mm 2

rad/ s

150 mm 30

= { - 0.5i} m

>

aA

 8

a

B

v2 rB A

2.4 = - aA cos 60° + 8.64

rad/ s

v

>

= 12.5 m s2

Ans.

;

>

= 40.8 rad s2 d

832

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which has a mass m can be treated as R2–7. The small gear which a uniform disk. If it is released from rest at u = 0° , and rolls along the fixed fixed circular gear rack, determine the angular velocity of the radial line AB at the instant u = 90° .

A u

B R

Potential Energy: Datum is set at point A. When the gear is at its final position (u = 90°), its cent center er of grav graviity is loca locate ted d (R - r ) below the datum. Its gravitational gravitational potential energy at this position is - mg(R - r) . Thus, Thus, the initial and final potential energies are V1

=0

= - mg(R - r)

V2

position (u Kinetic Energy: When gear B is at its final position mass center is yB = vg r or vg =

velocity of its = 90°), the velocity

yB

since the gear rolls without slipping on the r fixed circular gear track. The mass moment of inertia of the gear about its mass 1 center is IB = mr2. Since the gear is at rest initially, initially, the initial kinetic energy is 2 T1 = 0. The final kinetic energy is given by

T2

=

1 1 1 1 my2B + IB v2g = my2B + 2 2 2 2

a 12 b a b mr2

yB

2

r

=

3 my2B 4

Applying Eq. 18–18, we have have Conservation of Energy: Applying

+ V1 = T2 + V2

T1

0 + 0 =

3 my2B + [ - mg(R - r)] 4

yB

=

B

4g(R - r) 3

Thus, the angular velocity of the radical line AB is given by vAB

=

yB

R

-r

=

A

4g

Ans.

3(R - r)

833

r

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*R2–8. The 50-kg cylinder has an angular velocity of 30 rad s when it is brought into contact with the surface at C. If the coefficient of kinetic friction is mk = 0.2 , determine how long it will take for the cylinder to stop spinning.What force is developed in link AB during this time? The axis of the cylinder is connected to two symmetrical symmetrical links. links. (Only For the computation, neglect the weight of AB is shown.) For the links.

500 mm

>

(+ c)

m(yAy)1

+ ©

L

A

v  30 rad / s

B

200 mm C

t2

Fy dt

= m(yAy)2

t1

0 + NC (t) - 50(9.81)(t) = 0

a+b

m(yAx)1

:

+ ©

L

NC

t2

= m(yAx)2

Fx dt

t1

0 + 0.2(490.5)(t)—2FAB (t) = 0 (a + )

IB v1

+ ©

L

= 490.5 N

FAB

= 49.0 N

Ans.

t2

MB dt

= IB v2

t1

c 12

d

- (50)(0.2)2 (30) + 0.2(490.5)(0.2)(t) = 0 t

= 1.53 s

Ans.

R2–9. The gear rack has a mass of 6 kg, and the gears each have a mass of 4 kg and a radius of gyration of k = 30 mm about their their center. center. If the rack rack is originally originally moving moving downward downward at 2 m s , when s = 0 , determine determine the speed of the rack when s = 600 mm . The gears are free to rotate about thei theirr cent center erss, A and and B .

>

A

50 mm

B

50 mm s

>

2 Originally,both gears rotate with an angular velocity of vt = = 40 rad s . After 0.05 the rack has traveled s = 600 mm , both gears rotate with an angular velocity of v2

=

y2

0.05

, where y2 is the speed of the rack at that moment.

Put datum through points A and B.

+ V1 = T2 + V2

T1

b

1 (6)(2)2 + 2

1 C 4(0.03)2 D (40)2 2

r+

y2

0 =

b

a b r - 6(9.81)(0.6)

y2 1 1 (6)y22 + 2 4(0.03)2 D C 2 2 0.05

>

= 3.46 m s

Ans.

834

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R2–10. The gear has a mass of 2 kg and a radius of gyration gyration kA = 0.15 m. The connecting connecting link AB (slender (slender rod) and slider block at B have a mass of 4 kg and 1 kg, respectively. If the gear has an angular velocity v = 8 rad s at the instant u = 45° , determine the gear’s angular velocity when u = 0°.

>

v  8 rad / s

0.2 m A

At position 1: (vAB)1 =

(yA)1

>

rA IC

>

1.6 = 2.6667 rad s 0.6

=

(yB)1 = 0 B

>

(yAB)1 = (vAB)1 rG IC = 2.6667(0.3) = 0.8 m s

>

0.6 m 45

At position 2: (vAB)2 =

(yA)2

>

rA IC

=

v2 (0.2)

0.6 cos 45°

= 0.2357v2

(yB)2 = (vAB)2 rB IC = 0.2357 v2(0.6) = 0.1414v2

>

(yAB)2 = (vAB)2 rG IC = 0.2357 v2(0.6708) = 0.1581v2

>

T1

=

c

d

1 1 1 1 1 C (2)(0.15)2 D (8)2 + 2 (2)(1.6)2 + 2 (4)(0.8)2 + 2 12 (4)(0.6)2 (2.6667)2 2

= 5.7067 J T2

=

1 1 1 (2)(0.15)2 D (v2)2 + (2)(0.2 v2)2 + (4)(0.1581v2)2 C 2 2 2

+

c

d

1 1 1 (4)(0.6)2 (0.2357v2)2 + (1)(0.1414v2)2 2 12 2 T2

= 0.1258 v22

Put datum through bar in position 2. V1

u

= 2(9.81)(0.6 sin 45°) + 4(9.81)(0.3 sin 45°) = 16.6481 J T1

V2

=0

+ V1 = T2 + V2

5.7067 + 16.6481 = 0.1258v22 + 0 v2

>

= 13.3 rad s

Ans.

835

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Page 836


*R2–11. The operation of a doorbell requires the use of an electromagnet, that attracts the iron clapper AB that is pinned at end A and consists of a 0.2-kg slender rod to which is attached a 0.04-kg steel ball having a radius of 6 mm. If the attractive force of the magnet at C is 0.5 N when the switch switch is on, on, determine determine the initial initial angular angular acceleration acceleration of the the clapper. clapper. The spring spring is originall originally y stretched 20 mm.

A

k

 20

N/ m

50 mm

40 mm

C

44 mm

B

Equation of Motion: The spring force is given by Fsp

= kx = 20(0.02) = 0.4 N . The

mass moment of inertia for the clapper AB is (IAB)A =

1 (0.2) A 0.1342 B + 12

#

2 (0.04) A 0.0062 B + 0.04 A 0.142 B = 1.9816 A 10 - 3 B kg m2. 5 Eq. 17–12, 17–12, we have have 0.2 A 0.0672 B +

A p p l y in g

0.4(0.05) - 0.5(0.09) = - 1.9816 A 10 - 3 B a

+ © MA = IA a;

>

= 12.6 rad s2

a

Ans.

*R2–12. The revolving door consists of four doors which are attached to an axle AB . Each door can be assumed to be a 50-lb thin plate. Friction at the axle contributes a moment of 2 lb ft which resists the rotation of the doors. If a woman passes through one door by always pushing with a force P = 15 lb perpendicular to the plane of the door as shown, determine the door’s angular velocity after it has rotated 90°. The doors are originally at rest.

3 ft

#

A

2.5 ft

u

P  15 lb 7 ft

Moment of inertia of the door about axle AB: IAB

=2

100 c 121 a 32.2 b (6) d = 18.6335 slug # ft 2

T1

b

0 +

15(2.5)

B

2

+ © U1-2 = T2

a 2 b - 2 a 2 b r = 12 (18.6335) = 2.45 rad> s p

p

v2

Ans.

v

836

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Page 837


R2–13. The 10-lb cylinder rests on the 20-lb dolly. If the system is released from from rest, determine the angular velocity of the cylinder in 2 s.The cylinder does not slip on the dolly. Neglect the mass of the wheels on the dolly.

0.5 ft

For the cylinder,

L

( +R ) m(yCx¿)1 + ©

t2

Fx¿ dt

= m(yCx¿)2

t1

30

0 + 10 sin 30°(2) - F(2) = (c + ) IC v1 + ©

L

a b

10 y 32.2 C

(1)

t2

MC dt

= IC v2

t1

0 + F(0.5)(2) =

10 c 12 a 32.2 b (0.5) d 2

(2)

v

For the dolly, ( +R ) m(yDx¿)1 + ©

L

t2

Fx¿ dt

= m(yDx¿)2

t1

0 + F(2) + 20 sin 30°(2) =

20 a 32.2 b

(3)

yD

( +R ) vD = vC + v D C

>

vD

= vC - 0.5v

(4)

Solving Eqs. (1) to (4) yields: yields: v

vC

>

= 32.2 ft s

vD

=0

Ans.

>

= 32.2 ft s

F

=0

837

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R2–14. Solve Prob. R2–13 if the coefficients of static and kinetic friction between the cylinder and the dolly are ms = 0.3 and m = 0.2 , respectively.

0.5 ft

For the cylinder, ( +R ) m(yCx¿)1 + ©

L

t2

Fx¿ dt

= m(yCx¿)2

t1

0 + 10 sin 30°(2) - F(2) = (c + ) IC v1 + ©

L

10 a 32.2 b

(1)

yC

30

t2

MC dt

= IC v2

t1

0 + F(0.5)(2) =

10 c 12 a 32.2 b (0.5) d 2

(2)

v

For the dolly, ( +R ) m(yDx¿)1 + ©

L

t2

Fx¿ dt

t1

= m(yDx¿)2

0 + F(2) + 20 sin 30°(2) =

20 a 32.2 b

(3)

yD

( +R ) vD = vC + vD C

>

vD

= vC - 0.5v

(4)

Solving Eqs. (1) to (4) yields: yields: v

vC

>

= 32.2 ft s

vD

=0

Ans.

>

= 32.2 ft s

F

=0

Note: No friction force develops.

838

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R2–15. Gears H and C each have a weight of 0.4 lb and a radius of gyration about their mass center of (kH)B = (kC)A = 2 in. Link AB has a weight of 0.2 lb and a radius of gyration of (kAB)A = 3 in., wh whereas link DE has a weight of 0.15 lb and a radius of gyration of (kDE)B = 4.5 in. If a couple moment of M = 3 lb ft is applied to link AB and the assembly assembly is originally at rest, determine determine the angular velocity of link DE when link AB has rotated 360°. Gear C is prevented prevented from rotating rotating,, and motion motion occurs in the the horizontal plane. Also, gear H and link DE rotate together about the same axle at B .

E

3 in.

#

H

D

For link AB, yB

=

vAB rAB

=

a 126 b = 0.5

vAB

vAB

For gear H , vDE

=

yB

>

rB IC vAB

yB

=

0.5vAB = 2vAB 3 12

=

>

=

1 vDE 2

a 12 b 126 = 0.25 vDE

vDE

Principle of Work and Energy: For the system, T1

0 + 3(2p) =

1 2

+ © U1-2 = T2

2

2

2

0.2 0.4 c a 32.2 b a 123 b d a 12 b + 12 c a 32.2 b a 122 b d 1 0.4 b (0.25 ) + 12 c a 0.15 b a 4.512 b d + 12 a 0.15 b (0.25 + a 2 32.2 32.2 32.2 = 132 rad> s vDE

vDE

2

2

v2DE

v2DE

vDE

839

vDE)2

Ans.

3 in.

B

M

A

C

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Page 840


*R2–16. The inner hub of the roller bearing rotates with an angular velocity velocity of vi = 6 rad s , while the outer hub rotates in the opposite direction at vo = 4 rad s . Determine the angular velocity of each of the rollers if they roll on the hubs without slipping.

>

25 mm

>

50 mm v

o

v

i

>

Since the hub does not slip, yA = vi ri = 6(0.05) = 0.3 m s and yB = vO rO = 4(0.1) = 0.4 m s.

>

vB

= v A + vB>A

c 0. 4 d = c 0. 3 d + B T

(+ T)

c

v(0. 05)

0.4 = - 0.3 + 0.05v

T

v

R >

= 14 rad sb

840

Ans.

 6

rad / s

 4

rad / s

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Page 841


R2–17. The hoop (thin ring) has a mass of 5 kg and is released down the inclined plane such that it has a backspin v = 8 rad s and its center has a velocity vG = 3 m s as shown. If the coefficient coefficient of kinetic friction friction between the hoop and the plane is mk = 0.6 , determine determine how long the hoop rolls before it stops slipping.

>

( +a )

>

mvy1

+ ©

L

Fy dt

v  8 rad / s

G v G  3

m/ s

= mvy2 30

0 + Nh (1) - 5(9.81)t cos 30° = 0

(b+ )

Nh

= 42.479 N

Fh

= 0.6Nh = 0.6(42.479 N) = 25.487 N

mvx1

+ ©

L

Fx dt

0.5 m

= mvx2

5(3) + 5(9.181) sin 30°(t) - 25.487t = 5vG (a + )

(HG)1 + ©

L

MG dt

= (HG)2

- 5(0.5)2(8) + 25.487(0.5)(t) = 5(0.5)2

a 0.5 b vG

Solving, vG t

>

= 2.75 m s

= 1.32 s

Ans.

R2–18. The hoop (thin ring) has a mass of 5 kg and is released down the inclined plane such that it has a backspin v = 8 rad s and its center has a velocity vG = 3 m s as shown. If the coefficient coefficient of kinetic friction friction between the hoop and the plane is mk = 0.6 , determine determine the hoop’s angular velocity 1 s after it is released.

>

>

v  8 rad / s

G v G  3

m/ s

0.5 m

See solution solution to Prob. Prob. R2–17. Since backspin backspin will not stop in t = 1 s 6 1.32 s , then ( +a )

mvy1

+ ©

L

Fy dt

30

= mvy2

0 + Nh (t) - 5(9.81)t cos 30° = 0

a+

Nh

= 42.479 N

Fh

= 0.6Nh = 0.6(42.479 N) = 25.487 N

(HG)1 + ©

L

M dt

= (HG)2

- 5(0.5)2(8) + 25.487(0.5)(1) = - 5(0.5)2 v v

>

= 2.19 rad s d

Ans.

841

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Page 842


R2–19. Determine the angular velocity of rod CD at the instant u = 30°. Rod AB moves to the left at a constant constant speed of vAB = 5 m s .

D

>

v AB v

A

C D

0.3 m u

C

x

#

x

#

=

=

0.3 = 0.3 cot u tan u

yAB

# = - 0.3 csc2 uu

>

Here u = vCD, yAB = - 5 m s and u = 30° .

- 5 = - 0.03 csc2 30°(vCD)

vCD

>

= 4.17 rad s

842

Ans.

B

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Page 843


*R2–20. Determine the angular acceleration of rod CD at the instant u = 30° . R od od AB has zero velocity, i .e .e., vAB = 0 , and an acceleration of aAB = 2 m s2 to the right when u = 30°.

D

>

v AB

A

v C D

0.3 m u

C

x

=

0.3 = 0.3 cot u tan u

# = - 0.3 csc2 uu $ # # $ $ x = aAB = - 0.3 csc2 uu - 2 csc2 u cot uu2 = 0.3 csc2 u 2 cot uu2 - u # $ Here u = vCD, yAB = 0, aAB = 2 m s2 , u = aCD, and u = 30° . #

x

=

yAB

c

d

a

b

>

0 = - 0.3 csc2 30°(vCD) 2 = 0.3 csc2 30° C 2 cot 30°(0)2 - aCD D

vCD aCD

=0

>

= - 1.67 rad s2

843

Ans.

B

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Page 844


R2–21. If the angular velocity of the drum is increased uniformly from 6 rad s when t = 0 to 12 rad s when t = 5 s, determine the magnitudes of the velocity and acceleration of points A and B on the belt when t = 1 s . At this instant the points are located as shown.

>

>

45

B

Angular Motion: The angular acceleration of drum must be determined first. Applying Applying Eq. 16–5, we have have v

=

+

v0

ac t

12 = 6 + ac (5) ac

>

= 1.20 rad s2

The angular velocity of the drum at t = 1 s is given by v

=

v0

+

ac t

>

= 6 + 1.20(1) = 7.20 rad s

Motion of P : The magnitude of the velocity of points A and B can be determined using Eq. Eq. 16–8. yA

=

yB

= vr = 7.20

a 124 b = 2.40 ft>s

Ans.

Also, aA

= (at)A =

ac r

= 1.20

a 124 b = 0.400 ft>s

2

Ans.

The tangential and normal components of the acceleration of points B can be determined using Eqs. 16–11 and 16–12, respectively. (at)B = ac r = 1.20

a 124 b = 0.400 ft>s

(an)B = v2 r = A 7.202 B

2

a 124 b = 17.28 ft>s

2

The magnitude of the acceleration of points B is aB

=

2

(at)2B + (an)2B =

2

>

0.400 0.4002 + 17.282 = 17.3 ft s2

844

Ans.

A

4 in.

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Page 845


R2–22. Pulley A and the attached drum B have a weight of 20 lb and a radius of gyration of kB = 0.6 ft. If pulley P “rolls” “rolls” downward downward on the cord without slipping, slipping, determine determine the speed of the 20-lb crate C at the instant s = 10 ft . Initially, Initially, the crate is released from rest when s = 5 ft. For the calculation, neglect the mass of pulley P and the cord.

0.4 ft B A 0.8 ft s

0.2 ft

Kinematics: Since pulley A is rotating about a fixed point B and pulley P rolls down without without slipping, slipping, the velocity velocity of points D and E on the pulley P are given by Thus,, the yD = 0.4vA and yE = 0.8vA where vA is the angular velocity of pulley A. Thus instantaneous center of zero velocity can be located using similar triangles. x x + 0.4 = 0.4vA 0.8vA

x

= 0.4 ft

Thus, Thus, the velocity of block C is given by yC

0.6

=

0.4vA 0.4

yC

= 0.6vA

block C is at its initial and final Potential Energy: Datumn is set at point B. When block position, its locations locations are 5 ft and 10 ft below the datum. datum. Its initial and final final gravitational potential energies are 20( - 5) = - 100 ft lb and 20( - 10) = - 200 ft lb , respectively. respectively. Thus, the initial and final potential energy are

#

V1

= - 100 ft # lb

V2

#

= - 200 ft # lb

Kinetic Energy: The mass moment of inertia of pulley A about point B is 20 the system is is initially initially at rest, rest, the IB = mk2B = A 0.62 B = 0.2236 slug ft2. Since the 32.2 initial kinetic energy is T1 = 0 . The final kinetic energy is given by

#

T2

=

1 1 m y2 + IB v2A 2 C C 2

=

1 2

20 a 32.2 b (0.6

vA)2

1 2

+ (0.2236) v2A

= 0.2236v2A Applying Eq. 18–19, we have Conservation of Energy: Applying T1

+ V1 = T2 + V2

0 + - 100 = 0.2236v2A + ( - 200) vA

>

= 21.15 rad s

Thus, Thus, the speed of block C at the instant s = 10 ft is yC

>

= 0.6vA = 0.6(21.15) = 12.7 ft s

Ans.

845

P

C

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Page 846


with the finger at B , a thin thin ring R2–23. By pressing down with having a mass m is given an initial velocity v1 and a backspin v1 when the finger is released. If the coefficient of kinetic friction between the table and the ring is m , determine the distance the ring travels forward before the backspin stops.

B v1

v1

r

A

Equations of Motion: The mass moment of inertia of the ring about its mass center is given by IG = mr2. Applying Applying Eq. 17–16, we have

+ c © Fy = m(aG)y ;

- mg = 0

N

= mg

mmg

= maG

aG

= mg

mmgr

= mr2 a

N

+ © Fx = m(aG)x ;

:

a + © MG = IG a;

a

=

mg

r

Kinematics: The time required for the ring to stop back spinning can be determined by applying Eq. Eq. 16–5. v

(c + )

=

v0

+

0 = v1 +

t

=

ac t

a - b mg

r

t

v1 r mg

The distance traveled by the ring just before back spinning stops can be determine by applying Eq. Eq. 12–5.

A

+

;

B

s

= s0 + =0+ =

y0 t

y1

+

1 a t2 2 c

a b + 12 ( - ) a b v1 r mg

mg

v1r

2

mg

v1 r (2y1 - v1r) 2mg

Ans.

846

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Page 847


*R2–24. The pavement roller is traveling down the incline at v1 = 5 ft s when the motor is disengaged. Determine the speed of the roller when it has traveled 20 ft down the plane. plane. The body body of the the roller, roller, excluding excluding the rollers rollers,, has a weight of 8000 lb and a center of gravity at G . Each of the two rear rollers weighs 400 lb and has a radius of gyration o f kA = 3.3 ft. The front roller has a weight of 800 lb and a radius of gyration of kB = 1.8 ft. The rollers do not slip as they turn.

>

3.8 ft A

G

4.5 ft 5 ft 2.2 ft B

10 ft

The wheels roll without slipping, hence v =

T1

=

1 2

800 + 800 a 8000 + 32.2 b (5)

2

+

1 2

yG

r

30

.

800 c a 32.2 b (3.3) d a 3.85 b 2

2

+

1 2

800 c a 32.2 b (1.8) d a 2.25 b 2

2

= 4168.81 ft # lb T2

=

1 2

800 + 800 a 8000 + 32.2 b

y2

+

1 2

800 c a 32.2 b (3.3) d a 3.8 b 2

y

2

+

1 2

800 c a 32.2 b (1.8) d a 2.2 b 2

y

2

= 166.753 y2 Put datum through the mass center of the wheels and body of the roller when it is in the initial position. V1

=0

V2

= - 800(20 sin 30°) - 8000(20 sin 30°) - 800(20 sin 30°) = - 96000 ft # lb T1

+ V1 = T2 + V2

4168.81 + 0 = 166.753y2 - 96000 y

>

= 24.5 ft s

Ans.

R2–25. The cylinder B rolls on the fixed cylinder A without slipping. slipping. If bar CD rotates with an angular velocity vCD = 5 rad s, determine the angular velocity of cylinder B . Point C is a fixed point.

>

D

C

vD

0.3 m

0.1 m

>

A

vC D

 5

rad / s

= 5(0.4) = 2 m s B

>

2 = 6.67 rad s vB = 0.3

Ans.

847

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Page 848


R2–26. The disk has a mass M and a radius R . If a block of mass m is attached attached to the cord, determine determine the angular angular acceleration of the disk when the block is released from rest. Also, Also, what is the distance distance the block falls falls from rest in the time t?

c + © MO = © tMk)0;

a

=

1 MR2 (a) + m(aR)R 2

=

mgR

1 MR2 2

=

I0

2mg R(M + 2m)

h R

-

u0

+

=0+0+ h

=

v0 t

1 2

a

+

1 ac t2 2 2mg

R(M

mg M

Ans.

h R

The displacement displacement h = Ru , hence u =

u

R

+ 2m

+ 2m)

b

t2

t2

Ans.

848

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Page 849


R2–27. The tub of the mixer has a weight of 70 lb and a radius of gyration kG = 1.3 ft about its center of gravity G. If a constant torque M = 60 lb ft is applied to the dumping wheel, determine the angular velocity of the tub when it it has rotated u = 90° . Originally the tub is at rest when u = 0° . Neglect the mass of the wheel.

u

#

+ © U1 - 2 = T2

T1

0 + 60

a2b p

0.8 ft

1 - 70(0.8) = 2

ca b

d

G

c d

70 1 70 (1.3)2 (v)2 + (0.8v)2 32.2 2 32.2

>

= 3.89 rad s

v

M

Ans.

Prob. R2–27 if if the applied applied torque is is *R2–28. Solve Prob. M = (50u) lb ft, where u is in radians.

#

L 0

>

p 2

50u du - 70(0.8) =

0.8 ft

+ © U1 - 2 = T2

T1

0 +

u

1 2 v

70 c a 32.2 b (1.3) d = 1.50 rad> s 2

v2

+

c d

1 70 (0.8v)2 2 32.2 Ans.

G

M

849

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Page 850


R2–29. The spool has a weight of 30 lb and a radius of gyration kO = 0.45 ft. A cord is wrapped around the spool’s inner hub and its end subjected to a horizontal force P = 5 lb. Determine Determine the spool’s spool’s angular velocity velocity in 4 s starting from rest.A ssume the spool rolls without slipping.

0.9 ft O

P  5 lb

0.3 ft

A

( + b)

+ ©

IA v1

L

t2

MA dt

= IAv2

t1

0 + 5(0.6)(4) = v2

30 30 c a 32.2 b (0.45) + a 32.2 b (0.9) d 2

2

v2

>

= 12.7 rad s

Ans.

R2–30. The 75-kg man and 40-kg boy sit on the horizontal seesaw, seesaw, which has negligible negligible mass. mass. At the instant the man lifts his feet from the ground, determine their accelerations if each sits sits upright, upright, i.e., i.e., they do not rotate. rotate. The centers centers of mass of the man and boy are at Gm and Gb , respectively.

a + © MA = © (Mk)A ;

2m

Gb

40(9.81)(2) - 75(9.81)(1.5) [1]

Since the seesaw is rotating about point A, then then

=

ab

2

=

am

or

1.5

am

= 0.75ab

[2]

Solving Eqs. (1) and (2) yields: am

>

= 1.45 m s2

ab

Gm A

= - 40ab (2) - 75am (1.5)

a

1.5 m

>

= 1.94 m s2

Ans.

850

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Page 851


R2–31. A sphere and cylinder are released from rest on the ramp at t = 0 . If each has a mass m and a radius r , determine their angular velocities at time time t . Assume no slipping occurs.

u

Principle of Impulse and Momentum: For the sphere,

( + a)

IA v1

+ ©

L

t2

MA dt

= IAv2

t1

0 + mg sin u(r)(t) = (vS)2 =

c 25

d

mr2

+ mr2 (vS)2

5g sin u t 7r

Ans.

Principle of Impulse and Momentum: For the cyclinder,

( + a)

IA v1

+ ©

L

t2

MA dt

= IAv2

t1

0 + mg sin u(r)(t) = (vC)2 =

c 12

mr2

d

+ mr2 (vC)2

2g sin u t 3r

Ans.

851

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Page 852


instant, link AB has an angular *R2–32. At a given instant, acceleration aAB = 12 rad s2 and an angular velocity vAB = 4 rad s. Determine the angular velocity and angular acceleration of link CD at this instant.

>

>

C S C S B 10

=

30°d

+

2vBC T

45°b

( + ) ;

vC cos 30°

= 10 cos 45° + 0

(+ T)

vC sin 30°

= - 10 sin 45° + 2vBC vBC

vC vCD

>

= 8.16 ft s

>

=

8.16 = 5.44 rad s 1.5

aC

= aB + aC>B

C S C S C S C S c 44.44 c60°

+

(aC)t 30°d

=

R

= 5.58 rad s

>

30

+

45°b

40

Ans.

d B 2 R

+ 2(5.58)2 +

45°d

;

aBC T

( + )

- 44.44 cos 60° + (aC)t cos 30° = 30 cos 45° + 40 cos 45° + 62.21

(+ T)

44.44 sin 60° + (aC)t sin 30° = - 30 sin 45° + 40 sin 45° + 2aBC

;

>

(aC)t = 155 ft s2 aCD

=

aBC

2.5 ft

45 A

>

= 54.4 rad s2

>

155 = 103 rad s2d 1.5

852

Ans.

2 ft

C

1.5 ft

v AB a AB

= vB + vC>B

vC nC

B

60

a C D D

vC D

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Page 853


instant, link CD has an angular R2–33. At a given instant, acceleration aCD = 5 rad s2 and an angular velocity vCD = 2 rad s. Determine the angular velocity and angular acceleration of link AB at this instant.

>

>

rIC - C 2 = sin 45° sin 75° rIC - B

sin 60°

=

vBC

vB

B

2.5 ft

45 A

2 sin 75°

= 1.793 ft

rIC - B

>

3 = 2.0490 rad s 1.464

=

>

= 2.0490(1.793) = 3.6742 ft s

(aB)n =

>

3.6742 = 1.47 rad sd 2.5

=

vAB

v2B rBA

=

(aC)n =

(3.6742)2

rCD

=

Ans.

>

= 5.4000 ft s2

2.5

v2C

(3)2

>

= 6 ft s2

1.5

>

(aC)t = aCD(rCD) = 5(1.5) = 7.5 ft s2 aB

= aC +

a

* rB>C -

v2 rB

>

C

- 5.400 cos 45°i - 5.400 sin 45° j - (aB)t cos 45°i + (aB)t sin 45° j = 6 sin 30°i - 6 cos 30° j - 7.5 cos 30°i - 7.5 sin 30° j + (ak) * ( - 2i) - (2.0490)2( - 2i) - 3.818 - (aB)t(0.7071) = 3 - 6.495 + 8.3971 - 3.818 + (aB)t(0.7071) = - 5.1962 - 3.75 - 2a

>

(aB)t = - 12.332 ft s2 a aAB

=

>

= 1.80 rad s2

>

12.332 = 4.93 rad s2 2.5

b

853

Ans.

C

1.5 ft

v AB a AB

= 1.464 ft

rIC - C

2 ft

60

a C D D

vC D

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Page 854


R2–34. The spool and the wire wrapped around its core have a mass of 50 kg and a centroidal radius of gyration of coefficient of kinetic kinetic friction friction at the kG = 235 mm. If the coefficient surface is mk = 0.15, determine the angular acceleration of the spool after it is released from rest. IG

= mk2G = 500(0.235)2 = 2.76125 kg # m2

+b © Fx¿ = m(aG)x¿ ;

50(9.81) sin 45° - T - 0.15NB = 50aG

+ a© Fy¿ = m(aG)y¿ ; c + © MG = IG a;

G

NB

- 50(9.81) cos 45° = 0

T(0.1)

(1)

45

= 0.1a

(4)

Solving Eqs. (1) to (4) yields: yields:

= 346.8 N

T a

= 281.5 N

>

= 2.66 rad s2b

B

(3)

The spool does not slip at point A, therefore therefore

NB

0.4 m

(2)

- 0.15NB(0.4) = 2.76125a

aG

0.1 m

aG

>

= 0.2659 m s2

Ans.

854

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Page 855


R2–35. The bar is confined to move along the vertical and inclined inclined planes. planes. If the velocity velocity of the roller at A is determine the bar’s angular vA = 6 ft s when u = 45°, determine velocity and the velocity of B at this instant.

>

sB cos 30°

#

sB

$

sB

v A

A

= 5 sin u

u

= 5.774 sin u

5 ft

#

= 5.774 cos uu

(1)

5 cos u = sA + sB sin 30°

30

B

$ # # - 5 sin u u = sA + sB sin 30°

(2) vB

Combine Eqs. Eqs. (1) and (2):

# # - 5 sin u u = - 6 + 5.774 cos u(u)(sin 30°) # # - 3.536u = - 6 + 2.041u v

=

u

>

= 1.08 rad s

Ans.

From From Eq. (1), vB

>

= sB = 5.774 cos 45°(1.076) = 4.39 ft s

Ans.

*R2–36. The bar is confined to move along the vertical and inclined planes. If the roller at A has a constant velocity of vA = 6 ft s, determine the bar’s angular acceleration and the acceleration of B when u = 45° .

>

See solution to Prob. R2–35.

u

5 ft

Taking the time derivatives of Eqs. (1) and (2) yields:

# $ $ = sB = - 5.774 sin u(u)2 + 5.774 cos u(u) # $ $ $ - 5 cos u u2 - 5 sin u(u) = sA + sB sin 30°

aB

30

B

Substitute the data:

vB

$

= - 5.774 sin 45°(1.076)2 + 5.774 cos 45°(u) $ - 5 cos 45°(1.076)2 - 5 sin 45°(u) = 0 + aB sin 30° $ aB = - 4.726 + 4.083 u $ aB = - 8.185 - 7.071 u aB

Solving:

#

v A

A

u

> = - 5.99 ft> s

= - 0.310 rad s2

aB

Ans.

2

Ans.

855

91962_10_R2_p0827-0866

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4:19 PM

Page 856


R2–37. The uniform girder AB has a mass of 8 Mg. Determine Determine the the internal internal axial force, force, shear, and bending bending moment at the center of the girder if a crane gives it an upward acceleration of 3 m s2 .

3 m/ s

>

C

A

Equations of Motion: By considering the entire beam [FBD(a)], [FBD(a)], we have

+ c © Fy = may ;

2T sin 60° - 8000(9.81) = 8000(3) T

= 59166.86 N

From the FBD(b) (beam segment),

a + © MO = © (Mk)O ;

M

+ 4000(9.81)(1) - 59166.86 sin 60°(2) = - 4000(3)(1) M

+ ©Fx = m(aG)x ;

:

= 51240 N # m = 51.2 kN # m

Ans.

59166.86 cos 60° + N = 0 N

+ c © Fy = m(aG)y ;

2

= - 29583.43 N = - 29.6 kN

Ans.

59166.86 sin 60° - 4000(9.81) - V = 4000(3) V

=0

Ans.

856

60

4m

60

B

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Page 857


R2–38. Each gear has a mass of 2 kg and a radius of gyration about its pinned mass centers A and B of kg = 40 mm . Each link has a mass of 2 kg and a radius of gyration about its pinned ends A and B of kl = 50 mm . If originally originally the spring is unstretched when the couple moment M = 20 N m is applied to link AC, determine the angular velocities of the links at the instant link AC rotates u = 45°. Each gear and link is connected together and rotates in the horizontal plane about the fixed pins A and B .

200 mm 50 mm M

A

#

C

k

50 mm

B

 200

N / m

D

Consider the system of both gears and the links. The spring stretches s = 2(0.2 sin 45°) = 0.2828 m . T1

b a b -

0 +

20

p

4

+ © U1 - 2 = T2

1 (200)(0.2828)2 2 v

r= b C 2

1 (2)(0.05)2 + (2)(0.04)2 D v2 2

r

>

= 30.7 rad s

Ans.

Note that work is done by the tangential force between the gears since each move. For the system, system, though, this force is equal but opposite and the work cancels. cancels.

R2–39. The 5-lb rod AB supports the 3-lb disk at its end A . If the disk is given an angular velocity vD = 8 rad s while the rod is held stationary stationary and then released, determine the angular velocity of the rod after the disk has stopped spinning relative to the rod due to frictional resistance at the bearing A. Motion is in the horizontal plane. Neglect friction at the fixed bearing B .

3 ft

>

v

D

A

0.5 ft

Conservation of Momentum:

c + © (HB)1 = © (HB)2

c 12 a 32.23 b (0.5) d (8) + 0 = c 13 a 32.25 b (3) d 2

+

2

v

c 12 a 32.23 b (0.5) d + a 32.23 b (3 )(3) = 0.0708 rad> s 2

v

v

Ans.

v

857

B

91962_10_R2_p0827-0866

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Page 858


*R2–40. A cord is wrapped around the rim of each 10-lb disk. If disk B is released released from rest, determine determine the angular angular velocity of disk A in 2 s. Neglect the mass of the cord.

0.5 ft A

Principle of Impulse and Momentum: The mass moment inertia of disk A about 1 10 point O is IO = Applying Eq. 19–14 to disk A A0.52 B = 0.03882 slug ft2. Applying 2 32.2 [FBD(a)], [FBD(a)], we have

#

a b

IO v1

L

+ ©

t2

= IO v2

MO dt

t1

( + )

0 - [T(2)](0.5) = - 0.03882vA

(1)

1 2 0.03882 slug ft2. Applying Applying Eq. 19–14 to disk disk B [FBD(b)], [FBD(b)], we have

The mass moment inertia of disk B about its mass center is IG =

#

m A yGy B 1

( + c )

+ ©

L

t2

IG v1

(a + )

L

+ ©

2

= m A yGy B 1

Fy dt

t1

0 + T(2) - 10(2) = -

10 a 32.2 b A0.5 B =

10 a 32.2 b

(2)

yG

t2

= IG v2

MG dt

t1

0 - [T(2)](0.5) = - 0.03882vB

(3)

Here,, Kinematics: The speed of point C on disk B is yC = vA rA = 0.5vA. Here directed vertically vertically upward. upward. Applying Applying Eq. 16–15, yC G = vB rC G = 0.5vB which is directed we have

>

>

vC

= v G + vC>G

C S B R C S 0.5vA

=

T

( + c )

yG

+

0.5vB c

T

- 0.5vA = - yG + 0.5vB

(4)

Solving Eqs. Eqs. (1), (2), (3), and (4) yields: yields: vA vB

>

= 51.52 rad s

>

= 51.5 rad s yG

Ans.

>

= 51.52 ft s

T

= 2.00 lb

858

O

B

0.5 ft

91962_10_R2_p0827-0866

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Page 859


R2–41. A cord is wrapped around the rim of each 10-lb disk. If disk B is released released from rest, determine determine how much time t is required before A attains an angular velocity vA = 5 rad s.

0.5 ft

>

A

Principle of Impulse and Momentum: The mass moment inertia of disk A about 1 10 point O is IO = Applying Eq. 19–14 to disk A A0.52 B = 0.03882 slug ft2. Applying 2 32.2 [FBD(a)], [FBD(a)], we have

#

a b

IO v1

+ ©

L

t2

= IO v2

MO dt

t1

( + )

0 - [T(t)](0.5) = - 0.03882(5)

(1)

The mass moment inertia of disk B about its mass center is IG =

A 0.52 B

Applying Eq. Eq. 19–14 19–14 to disk B [FBD(b)], = 0.03882 slug # ft2. Applying

a b

1 10 2 32.2

we have m A yGy B 1

( + c )

+ ©

L

t2

Fy dt

t1

0 + T(t) - 10(t) = -

IG v1

+ ©

L

= m A yGy B 1

10 a 32.2 b

(2)

yG

t2

MG dt

= IG v2

t1

(a + )

0 - [T(t)](0.5) = - 0.03882vB

(3)

>

Kinematics: The speed of point C on disk B is yC = vA rA = 0.5(5) = 2.50 ft s. Here, yC G = vB rC G = 0.5 vB which is directed vertically upward. Ap A pplying Eq. 16–15, we have have

>

>

= v G + vC>G

vC

B R = B R + C 2.50

yG

T

( + c )

T

0.5 vB c

S

- 2.50 = - yG + 0.5 vB

[4]

Solving Eqs. Eqs. (1), (2), (3), and (4) yields: yields: t vB

O

>

= 5.00 rad s

= 0.194 s

yG

Ans.

>

= 5.00 ft s

T

= 2.00 lb

859

B

0.5 ft

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Page 860


disk is pinned at O and is initially at rest. rest. R2–42. The 15-kg disk If a 10-g bullet is fired into the disk with a velocity of 200 m s, as shown, determine the maximum angle u to which the disk swings.The bullet becomes embedded in the disk.

>

O u

30 200 m/ s

a + (HO)1 = (HO)2

c C12 (15)(0.15) = 0.5132 rad> s

2

0.01(200 cos 30°)(0.15) = v

T1

c

+ 15(0.15)2

d

v

+ V1 = T2 + V2

d

1 1 (15)(0.15)2 + 15(0.15)2 (0.5132)2 + 0 = 0 + 15(9.81)(0.15)(1 - cos u) 2 2 u

= 4.45°

Ans.

Note that the calculation neglects the small mass of the bullet after it becomes embedded in the plate, since its position in the plate is is not specified.

860

0.15 m

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Page 861


>

R2–43. The disk rotates at a constant rate of 4 rad s as it falls freely so that its center G has an acceleration of 32.2 ft s2. Determine the accelerations of points A and B on the rim of the disk at the instant shown.

>

= aG + (aA>G)t + (aA>G)n

aA

B R+C ( a A)x

(aA)y

:

a+b

c

S c

d

B

2

= 32.2 + 0 + (4) (1.5) T

T

G

R

v  4 rad / s

(aA)x = 0

:

( + c )

A

>

>

(aA)y = - 32.2 - (4)2 (1.5) = - 56.2 ft s2 = 56.2 ft s2 T

>

aA

= (aA)y = 56.2 ft s2 T

aB

= aG + (aB>G)t + (aB>G)n

Ans.

c ( ) d + B ( ) R = c 32.2 d + 0 + c (4) ( 1.5) d a + b ( ) = - (4) (1.5) = - 24 ft>s = 24 ft>s A + c B ( ) = - 32.2 ft> s = 32.2 ft>s T 24 + 32.2 = 40.2 ft> s = 2 ( ) + ( ) = 2 24 ( ) 32.2 = tan ¢ = tan a b = 53.3° ≤ ( ) 24 aB

aB

x

c

:

aB

:

aB

aB u

2

y

T

;

2

x

2

2

y

aB

-1

2

aB

x

aB

y

aB

x

2

;

2

2

2

y

-1

2

2

Ans.

d

861

Ans.

1.5 ft

B

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operation of “reverse” “reverse” for a three-speed three-speed *R2–44. The operation automotive transmission is illustrated schematically in the figure. If the shaft G is turning with an angular velocity of vG = 60 rad s, determine the angular velocity of the drive shaft H. Each of the gears rotates about a fixed fixed axis. axis. Note th at at gears A an d B, C and D , E a nd nd F are in mesh. The radius of each of these gears is reported in the figure.

A

vG

>

v

H

G

H F

C

B

E

=

vC

vB

=

vD

vH

=

vE

= = rE rF

rA rB rC rD vE

vG

vC

=

= =

>

D

90 (60) = 180 rad s 30

r

 90

r



r D

 50

>

r

 70

mm

r

 60

mm

A B

30 (180) = 108 rad s 50

E

F

>

70 (108) = 126 rad s 60

r

mm

C

 30

mm

mm

Ans.

internal gearing of a “spinner” used R2–45. Shown is the internal for drilling drilling wells. wells. With With constant angular accelerati acceleration, on, the motor M rotates the shaft S to 100 rev min in t = 2 s starting from rest. Determine the angular acceleration of the drill-pipe connection D and the number of revolutions it makes during the 2-s startup.

>

D

150 mm

60 mm

For shaft S,

=

v

v0

+

100(2p) 60 u

S

ac t

=0+ =

u0

+

aS (2)

aS

v0 t +

1 a t2 2 c

>

M

= 5.236 rad s2

1 2

uS

= 0 + 0 + (5.236)(2)2 = 10.472 rad

aS

=

60 (5.236) = 2.09 rad s2 150

Ans.

uS

=

60 (10.472) = 4.19 rad = 0.667 rev 150

Ans.

For connection D, aD

uD

= =

rS rD rS

rD

 60

>

862

rad / s

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Page 863


R2–46. Gear A has a mass of 0.5 kg and a radius of gyration gyration of kA = 40 mm , and gear B has a mass of 0.8 kg and a radius of gyration of kB = 55 mm . The link is pinned at C and has a mass of 0.35 kg. If the the link can be treated as a slender rod, determine the angular velocity of the link after the assembly is released from rest when u = 0° and falls to u = 90°.

A

125 mm

50 mm

B

u

125 mm

Kinematics: The velocity of the mass center of gear A is yD = 0.25 vCD, and since is rolls without without slipping slipping on the fixed circular circular gear track, track, the location location of the instantaneous center of zero velocity is as shown.Thus, vA

=

yD

>

rD IC

=

0.25 vCD = 5vCD 0.05

yE

=

>

vA rE

IC

= 5vCD (0.1) = 0.5 vCD

The velocity of the mass center of gear B is yF = 0.125vCD. The location location of the instantaneous center of zero velocity is as shown.Thus, vB

yE

=

>

rE (IC)1

=

0.5 vCD = 5vCD 0.1

gears A, B and link AC are at their Potential Energy: Datum is set at point C . When gears initial initial position position (u = 0°) , their centers centers of gravity gravity are located located 0.25 m, 0.125 m, and 0.125 m above the datum, respectively.The total gravitational potential energy when they are at these positions is 0.5(9.81)(0.25) + 0.8(9.81)(0.125) + 0.35(9.81)(0.125) = 2.636 N m. Thus, the initial and final potential energy is

#

V1

= 2.636 N # m

V2

=0

Kinetic Energy: The mass moment of inertia of gears A and B about their mass center

#

#

is ID = 0.5 A 0.042 B = 0.8 A 10- 3 B kg m2 and IF = 0.8 A 0.0552 B = 2.42 A 10-3 B kg m2. 1 The mass moment of inertia of link CD about point C is (ICD)C = (0.35) A 0.252 B + 12 0.35 A 0.1252 B = 7.292 A 10-3 B kg m2. Since the system is at rest initially initially,, the initial kinetic kinetic

#

energy is T1 = 0. The final kinetic energy is given by

T2

=

1 1 1 1 1 mA y2D + ID v2A + mB y2F + IFv2B + (ICD)C v2CD 2 2 2 2 2 1 2

= (0.5)(0.25 vCD)2 + +

1 1 C 0.8 A 10-3 B D(5vCD)2 + 2(0.8)(0.125 vCD)2 2

1 1 2.42 A 10-3 B D(5 vCD)2 + C 7.292 A 10-3 B D A v2CD B C 2 2

= 0.06577 v2CD Applying Eq. 18–19, we have have Conservation of Energy: Applying T1

+ V1 = T2 + V2

0 + 2.636 = 0.06577 v2CD vCD

>

= 6.33 rad s

Ans.

863

75 mm C

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Page 864


R2–47. The 15-kg cylinder rotates with an angular velocity velocity of v = 40 rad s. If a force F = 6 N is applied applied to bar AB, as shown, determine determine the time needed to stop the rotation. The coefficient of kinetic friction between AB and the cylinder is mk = 0.4 . Neglect the thickness of the bar.

F

>

 6

N

400 mm

A

For link AB,

500 mm

B

150 mm C v

a + © MB = 0;

6(0.9) - NE(0.5) = 0

=

IC

NE

= 10.8 N

#

1 1 mr2 = (15)(0.15)2 = 0.16875 kg m2 2 2

a + © MC = IC a;

- 0.4(10.8)(0.15) = 0.16875(a)

a

>

= - 3.84 rad s2

a + v = v0 + at 0 = 40 + ( - 3.84) t t

= 10.4 s

Ans.

>

determine *R2–48. If link AB rotates at vAB = 6 rad s , determine the angular velocities of links BC and CD at the instant shown.

A

250 mm 30 B

Link AB rotates about the fixed point A. Hence, Hence, yB

=

>

 6

rad / s C

300 mm

>

= 6(0.25) = 1.5 m s

vAB rAB

400 mm

For link BC, rB IC

v AB

= 0.3 cos 30° = 0.2598 m vBC yC

=

=

vBC

yB

> r >

=

rB IC C IC

>

rC IC

= 0.3 cos 60° = 0.15 m 60

>

1.5 = 5.77 rad s 0.2598

Ans.

>

= 5.77(0.15) = 0.8660 m s

Link CD rotates about the fixed point D. Hence Hence,, yC

=

0.8660 = vCD (0.4)

vCD rCD vCD

>

= 2.17 rad s

Ans.

864

D

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Page 865


R2–49. If the thin hoop has a weight W and radius r and is thrown onto a rough surface with a velocity vG parallel to the surf surface ace,, determ determine ine the the backs backspin, pin, V , it must must be be given given so that it stops spinning at the same instant that its forward velocity is zero. It is not necessary to know the coefficient of kinetic friction at A for the calculation.

v

vG

G

Equations of Motion: The mass moment of inertia of the hoop about its mass center W 2 is given by IG = mr2 = Applying Eq. 17–16, we have r . Applying g

+ c ©Fy = m(aG)y ;

N

+ ©Fx = m (aG)x ;

mW

:

c + © MG = IG a;

-W=0

N

=W

W a g G

aG

= mg

=

mWr

W 2 r a g

=

a

=

A

mg

r

Kinematics: The time required for the hoop to stop back spinning can be determined by applying Eq. 16–5. v

(c + )

=

v0

+

0 = v +

t1

=

a t1

a - b mg

r

t1

vr mg

The time required for the hoop to stop can be determined by applying Eq. 12–4.

A

+

;

B

y

=

y0

+ a t2

0 = yG + ( - mg) t2 t2

=

yG mg

It is required that t1 = t2. Thus, vr mg v

= =

r

yG mg yG

Ans.

r

865

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Page 866


R2–50. The wheel has a mass of 50 kg and a radius of gyration gyration kG = 0.4 m. If it rolls without without slipping slipping down the inclined inclined plank, determine determine the horizonta horizontall and vertical vertical components of reaction at A , and the normal reaction at the smooth support B at the the instant the wheel is located at the midpoint of the plank. The plank has negligible thickness and has a mass of 20 kg.

B G 0.6 m

2m A

rolls down the plane without without slipping, then Equations of Motion: Since the tire rolls aG = ar = 0.6a. The mass moment of inertia of the tire about its mass center is given

#

by IG = mk2G = 50 A 0.42 B = 8.00 kg m2. Applying Applying Eq. 17–16 to [FBD(a)],we [FBD(a)],we have

a+ © Fy¿ = m(aG)y¿ ;

N

- 50(9.81) cos 30° = 0

b+ © Fx¿ = m(aG)x¿ ;

N = 424.79 N

50(9.81) sin 30° - Ff = 50(0.6a)

a + © MG = IG a;

F f(0.6)

= 8.00a

(1) (2)

Solving Eqs. [1] and [2] yields yields F f

= 75.46 N

a

>

= 5.660 rad s2

Equations of Equilibrium: From FBD(b).

a + © MA = 0;

NB (4)

- 20(9.81) cos 30°(2) - 424.79(2) = 0 NB

+ c © Fy = 0;

Ay

= 297.35 N = 297 N

Ans.

+ 297.35 cos 30° - 20(9.81) - 424.79 cos 30° - 75.46 sin 30° = 0 Ay

+ ©Fx = 0;

:

Ax

= 344 N

Ans.

+ 424.79 sin 30° - 75.46 cos 30° - 297.35 sin 30° = 0 Ax

= 1.63 N

Ans.

866

30

2m

Solutions Chapter R2

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