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15–149. The chain has a total length L 6 d and a mass per unit length of m¿ . If a portion h of the chain is suspended over the table and released, determine the velocity of its end A as a function of its position y. Neglect friction.

h y d A

SOLUTION ©Fs = m

dm e dv + vD>e dt dt

m¿gy = m¿y

dv + v(m¿v) dt

m¿gy = m¿ ay

Since dt =

dy , we have v or

dv + v2 dy in

teaching

Web)

laws

gy = vy

dv + v2 b dt

Dissemination

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copyright

Multiply by 2y and integrate:

the

not

instructors

States

World

permitted.

dv + 2yv2 b dy dy

is use

United

on

learning.

L

a 2vy2

of

L

2gy2 dy =

the

(including

for

work

student

the

by

and

2 3 3 g y + C = v2y2 3

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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is

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and

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will

y3 - h3 2 ga b B3 y2

their

v =

2 y3 - h3 ga b 3 y2

sale

Thus, v2 =

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2 when v = 0, y = h, so that C = - gh3 3

Ans.

16–1. The angular velocity of the disk is defined by v = 15t2 + 22 rad>s, where t is in seconds. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = 0.5 s.

A

0.8 m

SOLUTION v = (5 t2 + 2) rad>s a =

dv = 10 t dt

t = 0.5 s v = 3.25 rad>s a = 5 rad>s2 vA = vr = 3.25(0.8) = 2.60 m>s

Ans. laws

or

a z = ar = 5(0.8) = 4 m>s2 in

teaching

Web)

a n = v2r = (3.25)2(0.8) = 8.45 m>s2

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Dissemination

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Ans.

copyright

a A = 2(4)2 + (8.45)2 = 9.35 m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

16–2. A flywheel has its angular speed increased uniformly from 15 rad>s to 60 rad>s in 80 s. If the diameter of the wheel is 2 ft, determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel when t = 80 s, and the total distance the point travels during the time period.

SOLUTION v = v0 + ac t 60 = 15 + ac(80) ac = 0.5625 rad/s2 a t = ar = (0.5625)(1) = 0.562 ft/s2

Ans.

a n = v2r = (60)2(1) = 3600 ft/s2

Ans.

v2 = v20 + 2ac (u - u0) (60)2 = (15)2 + 2(0.5625)(u-0) laws

or

u = 3000 rad

teaching

Web)

s = ur = 3000(1) = 3000 ft

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of

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permitted.

Dissemination

Wide

copyright

in

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

16–3. The disk is originally rotating at v0 = 8 rad>s. If it is subjected to a constant angular acceleration of a = 6 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point A at the instant t = 0.5 s.

V0 B 1.5 ft 2 ft

SOLUTION v = v0 + ac t v = 8 + 6(0.5) = 11 rad>s v = rv;

vA = 2(11) = 22 ft>s

Ans.

at = ra;

(aA)t = 2(6) = 12.0 ft>s2

Ans.

(aA)n = (11)2(2) = 242 ft>s2

Ans.

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Dissemination

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in

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laws

or

an = v2r;

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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8 rad/s

A

*16–4. The disk is originally rotating at v0 = 8 rad>s. If it is subjected to a constant angular acceleration of a = 6 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point B just after the wheel undergoes 2 revolutions.

V0 B 1.5 ft 2 ft

SOLUTION v2 = v20 + 2ac (u - u0) v2 = (8)2 + 2(6)[2(2p) - 0] v = 14.66 rad>s Ans.

(aB)t = ar = 6(1.5) = 9.00 ft>s2

Ans.

(aB)n = v2r = (14.66)2(1.5) = 322 ft>s2

Ans.

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Dissemination

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laws

or

vB = vr = 14.66(1.5) = 22.0 ft>s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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8 rad/s

A

16–5. Initially the motor on the circular saw turns its drive shaft at v = 120t2>32 rad>s, where t is in seconds. If the radii of gears A and B are 0.25 in. and 1 in., respectively, determine the magnitudes of the velocity and acceleration of a tooth C on the saw blade after the drive shaft rotates u = 5 rad starting from rest.

SOLUTION v = 20 t2>3 dv 40 -1>3 = t dt 3

a =

du = v dt t

u

L0

du =

L0

20 t2>3 dt

3 u = 20a b t5>3 5

teaching

Web)

laws

or

When u = 5 rad

Wide

copyright

in

t = 0.59139 s

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permitted.

Dissemination

a = 15.885 rad>s2

on

learning.

is

of

the

not

instructors

v = 14.091 rad>s

the

by

and

use

United

v A r A = vB r B

the

(including

for

work

student

14.091(0.25) = vB(1) assessing

is

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solely

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vB = 3.523 rad>s

Ans.

part

the

is and

any

courses

aA rA = aB rB

This

provided

integrity

of

and

work

this

vC = vB r = 3.523(2.5) = 8.81 in.>s

aB = 3.9712 rad>s2

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destroy

of

15.885(0.25) = aB(1)

(aC)t = aB r = 3.9712(2.5) = 9.928 in.>s2 (aC)n = v2B r = (3.523)2 (2.5) = 31.025 in.>s2 aC = 2(9.928)2 + (31.025)2 = 32.6 in.>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

16–6. A wheel has an initial clockwise angular velocity of 10 rad>s and a constant angular acceleration of 3 rad>s2. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity of 15 rad>s. What time is required?

SOLUTION v2 = v20 + 2ac(u - u0) (15)2 = (10)2 + 2(3)(u- 0) 1 ≤ = 3.32 rev. 2p

u = 20.83 rad = 20.83 ¢

Ans.

v = v0 + ac t 15 = 10 + 3t t = 1.67 s

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in

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laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

16–7. If gear A rotates with a constant angular acceleration of aA = 90 rad>s2, starting from rest, determine the time required for gear D to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear D to attain this angular velocity. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

F

SOLUTION aB rB = aA rA rA 15 aB = a baA = a b (90) = 27 rad>s2 rB 50 Since gears C and B share the same shaft, aC = aB = 27 rad>s2. Also, gear D is in mesh with gear C. Thus,

or

aD rD = aC rC rC 25 aD = a b aC = a b (27) = 9 rad>s2 rD 75 600 rev 2p rad 1 min ba ba b = min 1 rev 60 s 20p rad>s. Applying the constant acceleration equation, Dissemination

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copyright

in

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laws

The final angular velocity of gear D is vD = a

of

the

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permitted.

vD = (vD)0 + aD t

and

use

United

on

learning.

is

20p = 0 + 9t

Ans. (including

for

work

student

the

by

t = 6.98 s

is

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protected

the

and

any

destroy

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and

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1 rev b 2p rad

courses

part

the

is

This

(20p)2 = 02 + 2(9)(uD - 0)

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integrity

of

and

work

this

assessing

vD2 = (vD)0 2 + 2aD [uD - (uD)0]

uD = (219.32 rad) a

sale

will

= 34.9 rev

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

B C

Gear B is in mesh with gear A. Thus,

Ans.

A

*16–8. If gear A rotates with an angular velocity of vA = (uA + 1) rad>s, where uA is the angular displacement of gear A, measured in radians, determine the angular acceleration of gear D when uA = 3 rad, starting from rest. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

F

SOLUTION aA duA = vA dvA aA duA = (uA + 1) d(uA + 1) aA duA = (uA + 1) duA aA = (uA + 1) At uA = 3 rad,

laws

or

aA = 3 + 1 = 4 rad>s2

in

teaching

Web)

Motion of Gear D: Gear A is in mesh with gear B. Thus,

of

the

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permitted.

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aB rB = aA rA rA 15 aB = a baA = a b (4) = 1.20 rad>s2 rB 50

the

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aD rD = aC rC rC 25 aD = a b aC = a b (1.20) = 0.4 rad>s2 rD 75

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Since gears C and B share the same shaft aC = aB = 1.20 rad>s2. Also, gear D is in mesh with gear C. Thus,

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

B C

Motion of Gear A:

Ans.

A

16–9. The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of ac = 0.5 rad>s2, determine the magnitude of the velocity and acceleration of points A and B on the blade after the blade has rotated through two revolutions.

ac

0.5 rad/s2

B

SOLUTION

10 ft

Angular Motion: The angular velocity of the blade after the blade has rotated 2(2p) = 4p rad can be obtained by applying Eq. 16–7.

20 ft

v2 = v20 + 2ac (u - u0) v2 = 02 + 2(0.5)(4p - 0) v = 3.545 rad>s Motion of A and B: The magnitude of the velocity of point A and B on the blade can be determined using Eq. 16–8. Ans.

vB = vrB = 3.545(10) = 35.4 ft>s

Ans. in

teaching

Web)

laws

or

vA = vrA = 3.545(20) = 70.9 ft>s

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permitted.

Dissemination

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The tangential and normal components of the acceleration of point A and B can be determined using Eqs. 16–11 and 16–12 respectively.

on

learning.

is

of

the

not

instructors

(at)A = arA = 0.5(20) = 10.0 ft>s2

student

the

by

and

use

United

(an)A = v2 rA = A 3.5452 B (20) = 251.33 ft>s2

the

(including

for

work

(at)B = arB = 0.5(10) = 5.00 ft>s2

work

this

assessing

is

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solely

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protected

(an)B = v2 rB = A 3.5452 B (10) = 125.66 ft>s2

part

the

is

This

provided

integrity

of

and

The magnitude of the acceleration of points A and B are

Ans.

and

any

courses

(a)A = 2(at)2A + (an)2A = 210.02 + 251.332 = 252 ft>s2 5.002 + 125.662 = 126 ft s2 destroy

of

sale

will

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

their

(a)B = =

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

A

16–10. The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of ac = 0.5 rad>s2, determine the magnitude of the velocity and acceleration of points A and B on the blade when t=4 s.

ac

0.5 rad/s2

B

SOLUTION

10 ft

Angular Motion: The angular velocity of the blade at t = 4 s can be obtained by applying Eq. 16–5.

20 ft

v = v0 + ac t = 0 + 0.5(4) = 2.00 rad>s Motion of A and B: The magnitude of the velocity of points A and B on the blade can be determined using Eq. 16–8. vA = vrA = 2.00(20) = 40.0 ft>s

Ans.

vB = vrB = 2.00(10) = 20.0 ft>s

Ans.

teaching

Web)

laws

or

The tangential and normal components of the acceleration of points A and B can be determined using Eqs. 16–11 and 16–12 respectively.

Wide

copyright

in

(at)A = arA = 0.5(20) = 10.0 ft>s2

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permitted.

Dissemination

(an)A = v2 rA = A 2.002 B (20) = 80.0 ft>s2

on

learning.

is

of

the

not

(at)B = arB = 0.5(10) = 5.00 ft>s2

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(an)B = v2 rB = A 2.002 B (10) = 40.0 ft>s2

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protected

the

(including

The magnitude of the acceleration of points A and B are

Ans.

this

assessing

is

work

solely

(a)A = 2(at)2A + (an)2A = 210.02 + 80.02 = 80.6 ft>s2

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(a)B = 2(at)2B + (an)2B = 25.002 + 40.02 = 40.3 ft>s2

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

A

16–11. 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

SOLUTION v = v0 + act

A

a = 1.2 rad>s2

12 = 6 + a(5) At t = 1 s,

v = 6 + 1.2(1) = 7.2 rad>s vA = vB = v r = 7.2 ¢

Ans.

4 ≤ = 0.4 ft>s2 12

Ans. laws

or

aA = ar = 1.2 ¢

4 ≤ = 2.4 ft>s 12

in

teaching

Web)

4 ≤ = 0.4 ft>s2 12

Wide

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(aB)t = ar = 1.2 ¢

World

permitted.

Dissemination

= 17.28 ft>s2

is

of

the

not

instructors

4 12

States

(aB)n = v2r = (7.2)2

Ans.

on

learning.

0.42 + 17.282 = 17.3 ft s2

United

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student

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by

and

(aB)2t + (aB)2n =

use

aB =

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4 in.

*16–12. A motor gives gear A an angular acceleration of aA = 10.25u3 + 0.52 rad>s2, where u is in radians. If this gear is initially turning at 1vA20 = 20 rad>s, determine the angular velocity of gear B after A undergoes an angular displacement of 10 rev.

(vA)0 ⫽ 20 rad/s B 0.05 m

A

AA

SOLUTION aA = 0.25 u3 + 0.5 a du = v dv vA

20p

L0

(0.25 u3 + 0.5)duA = 20p

(0.0625 u4 + 0.5 u)冷 0

=

vA dvA

L20

vA 1 (vA)2冷 20 2

vA = 1395.94 rad>s vA rA = vB rB laws

or

1395.94(0.05) = vB (0.15) vB = 465 rad>s

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the

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work

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of

the

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permitted.

Dissemination

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teaching

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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0.15 m

16–13. A motor gives gear A an angular acceleration of aA = 14t32 rad>s2, where t is in seconds. If this gear is initially turning at 1vA20 = 20 rad>s, determine the angular velocity of gear B when t = 2 s.

(vA)0 ⫽ 20 rad/s B 0.05 m

A 0.15 m AA

SOLUTION a A = 4 t3 dv = a dt t

vA

L20

dvA =

L0

t

aA dt =

4 t3dt

L0

vA = t4 + 20 When t = 2 s, vA = 36 rad>s laws

or

vA rA = vB rB in

teaching

Web)

36(0.05) = vB (0.15)

Wide

copyright

vB = 12 rad>s

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the

(including

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the

by

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learning.

is

of

the

not

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permitted.

Dissemination

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

16–14. The disk starts from rest and is given an angular acceleration a = (2t 2) rad>s2, where t is in seconds. Determine the angular velocity of the disk and its angular displacement when t = 4 s.

P 0.4 m

SOLUTION dv = 2 t2 dt

a =

t

v

2

dv =

L0

L0

v =

2 32t t 3 0

v =

23 t 3

2 t dt

When t = 4 s, 2 3 (4) = 42.7 rad>s 3

in

teaching

Web)

2 3 t dt L0 3

permitted.

Dissemination

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1 4 t 6 States

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u =

du =

laws

t

u

L0

Ans. or

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learning.

is

of

the

not

instructors

When t = 4 s, by

and

use

United

on

1 4 (4) = 42.7 rad 6

sale

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of

and

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the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

u =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

16–15. The disk starts from rest and is given an angular acceleration a = (5t1>2) rad>s2, where t is in seconds. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when t = 2 s.

P 0.4 m

SOLUTION Motion of the Disk: Here, when t = 0, v = 0. dv = adt t

v

L0 v2

dv =

v

= 0

v = e

L0

1

5t2dt

10 3 2 t t2 3 0 10 3 t2 f rad>s 3

teaching

Web)

laws

10 3 A 2 2 B = 9.428 rad>s 3

in

v =

or

When t = 2 s,

Dissemination

Wide

copyright

When t = 2 s,

permitted.

a = 5 A 2 2 B = 7.071 rad>s2 United

on

learning.

is

of

the

not

instructors

States

World

1

for

work

student

the

by

and

use

Motion of point P: The tangential and normal components of the acceleration of point P when t = 2 s are Ans.

is

work

solely

of

protected

the

(including

at = ar = 7.071(0.4) = 2.83 m>s2

Ans.

sale

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and

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the

is

This

provided

integrity

of

and

work

this

assessing

a n = v2r = 9.4282(0.4) = 35.6 m>s2

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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*16–16. The disk starts at v0 = 1 rad>s when u = 0, and is given an angular acceleration a = (0.3u) rad>s2, where u is in radians. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when u = 1 rev.

P 0.4 m

SOLUTION a = 0.3u v

L1

vdv =

u

L0

0.3udu

u 1 22v v = 0.15u2 2 2 1 0

v2 - 0.5 = 0.15u2 2

laws

or

v = 20.3u2 + 1

in

teaching

Web)

At u = 1 rev = 2p rad Dissemination

Wide

copyright

v = 20.3(2p)2 + 1

learning.

is

of

the

not

instructors

States

World

permitted.

v = 3.584 rad>s

Ans.

the

by

and

use

United

on

a t = ar = 0.3(2p) rad>s 2 (0.4 m) = 0.7540 m>s2

protected

the

(including

for

work

student

a n = v2r = (3.584 rad>s)2(0.4 m) = 5.137 m>s2

sale

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and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

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solely

of

a p = 2(0.7540)2 + (5.137)2 = 5.19 m>s2

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

16–17. Starting at (vA)0 = 3 rad>s, when u = 0, s = 0, pulley A is given an angular acceleration a = (0.6u) rad>s2, where u is in radians. Determine the speed of block B when it has risen s = 0.5 m. The pulley has an inner hub D which is fixed to C and turns with it.

150 mm C

D A 50 mm

75 mm

SOLUTION aa = 0.6uA uC =

B

0.5 = 6.667 rad 0.075 s

uA(0.05) = (6.667)(0.15) uA = 20 rad adu = vdv vA

0.6uAduA =

L3

vAdvA laws

L0

or

20

teaching

Web)

1 2 vA v 2 2 A 3

Wide States

World

permitted.

Dissemination

0

=

in

20

copyright

0.3u2A 2

United

on

learning.

is

of

the

not

instructors

1 2 v - 4.5 2 A by

and

use

120 =

the

(including

for

work

student

the

vA = 15.780 rad>s

assessing

is

work

solely

of

protected

15.780(0.05) = vC (0.15)

the

is

This

provided

integrity

of

and

work

this

vC = 5.260 rad>s

part

vB = 5.260(0.075) = 0.394 m>s

sale

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of

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courses

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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16–18. Starting from rest when s = 0, pulley A is given a constant angular acceleration ac = 6 rad>s2. Determine the speed of block B when it has risen s = 6 m. The pulley has an inner hub D which is fixed to C and turns with it.

150 mm C

D A 50 mm

75 mm

SOLUTION aA r A = aC r C 6(50) = aC(150)

B

aC = 2 rad>s2

s

aB = aC rB = 2(0.075) = 0.15 m>s2 (+ c)

v2 = v20 + 2 ac (s - s0) v2 = 0 + 2(0.15)(6 - 0) v = 1.34 m>s

sale

will

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destroy

of

and

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courses

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the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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16–19. The vacuum cleaner’s armature shaft S rotates with an angular acceleration of a = 4v3>4 rad>s2, where v is in rad>s. Determine the brush’s angular velocity when t = 4 s, starting from v0 = 1 rad>s, at u = 0. The radii of the shaft and the brush are 0.25 in. and 1 in., respectively. Neglect the thickness of the drive belt.

SOLUTION Motion of the Shaft: The angular velocity of the shaft can be determined from dt =

L t

L0 2

t

dt =

A

dvS L aS vs

dvS

L1 4vS 3>4 2

vs

t 0 = vS 1>4 1 t = vS 1>4 – 1

laws

or

vS = (t+1) 4

in

teaching

Web)

When t = 4 s

States

World

permitted.

Dissemination

Wide

copyright

vs = 54 = 625 rad>s

and

use

United

on

learning.

is

of

the

for

work

student

the

by

vB rB = vs rs

work

solely

of

protected

the

(including

rs 0.25 b(625) = 156 rad>s ≤v = a rB s 1

sale

will

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the

is

This

provided

integrity

of

and

work

this

assessing

is

vB = ¢

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Ans.

not

instructors

Motion of the Beater Brush: Since the brush is connected to the shaft by a non-slip belt, then

S

A

S

*16–20. The operation of “reverse” for a three-speed automotive transmission is illustrated schematically in the figure. If the crank shaft G is turning with an angular speed of 60 rad>s, determine the angular speed of the drive shaft H. Each of the gears rotates about a fixed axis. Note that gears A and B, C and D, E and F are in mesh. The radii of each of these gears are reported in the figure.

A

vH H

G

F

C B D

E

rA ⫽ 90 mm rB ⫽ rC ⫽ 30 mm rD ⫽ 50 mm rE ⫽ 70 mm rF ⫽ 60 mm

SOLUTION 60(90) = vBC (30) vBC = 180 rad>s 180(30) = 50(vDE) vDE = 108 rad>s 108(70) = (60) (vH) vH = 126 rad>s

will

their

destroy

of

and

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courses

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the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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vG ⫽ 60 rad/s

16–21. V0 ⫽ 6 rad/s

A motor gives disk A an angular acceleration of aA = 10.6t2 + 0.752 rad>s2, where t is in seconds. If the initial angular velocity of the disk is v0 = 6 rad>s, determine the magnitudes of the velocity and acceleration of block B when t = 2 s.

A

0.15 m

SOLUTION dv = a dt v

L6

2

dv =

L0

(0.6 t2 + 0.75) dt

B

v - 6 = (0.2 t3 + 0.75 t)|20 v = 9.10 rad>s Ans.

aB = at = ar = [0.6(2)2 + 0.75](0.15) = 0.472 m>s2

Ans.

sale

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and

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courses

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the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

vB = vr = 9.10(0.15) = 1.37 m>s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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16–22. For a short time the motor turns gear A with an angular acceleration of aA = (30t1>2) rad>s2, where t is in seconds. Determine the angular velocity of gear D when t = 5 s, starting from rest. Gear A is initially at rest. The radii of gears A, B, C, and D are rA = 25 mm, rB = 100 mm, rC = 40 mm, and rD = 100 mm, respectively.

A C

SOLUTION Motion of the Gear A: The angular velocity of gear A can be determined from dvA =

L

adt

L

t

vA

dvA =

L0 vA

vA 0

30t1>2dt

L0

t

= 20t3>2 2

0

vA = A 20t

3>2

B rad>s laws

or

When t = 5 s

Wide

copyright

in

teaching

Web)

vA = 20 A 53>2 B = 223.61 rad>s

of

the

not

instructors

States

World

and

use

United

on

learning.

is

vB rB = vA rA

the

(including

for

work

student

the

by

rA 25 b(223.61) = 55.90 rad>s ≤v = a rB A 100 work

solely

of

protected

vC = v B = ¢

work

this

assessing

is

Also, gear D is in mesh with gear C. Then

part

the

is

This

provided

integrity

of

and

vD rD = vC rC

any

courses

rC 40 b (55.90) = 22.4 rad>s ≤v = a rD C 100 sale

will

their

destroy

of

and

vD = ¢

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

permitted.

Dissemination

Motion of Gears B, C, and D: Gears B and C which are mounted on the same axle will have the same angular velocity. Since gear B is in mesh with gear A, then

B D

16–23. The motor turns gear A so that its angular velocity increases uniformly from zero to 3000 rev>min after the shaft turns 200 rev. Determine the angular velocity of gear D when t = 3 s. The radii of gears A, B, C, and D are rA = 25 mm, rB = 100 mm, rC = 40 mm, and rD = 100 mm, respectively.

A C

SOLUTION Motion of Wheel A: Here, vA = a 3000

1 min 2p rad rev ba ba b = 100p rad>s min 60 s 1rev

2p rad b = 400p rad. Since the angular acceleration of gear 1 rev A is constant, it can be determined from when uA = (200 rev) a

vA 2 = (vA)0 2 + 2aA C uA - (uA)0 D (100p)2 = 02 + 2aA (400p - 0)

laws

or

aA = 39.27 rad>s2

teaching

Web)

Thus, the angular velocity of gear A when t = 3 s is

Dissemination

Wide

copyright

in

vA = A v A B 0 + a A t

not

instructors

States

World

permitted.

= 0 + 39.27(3)

by

and

use

United

on

learning.

is

of

the

= 117.81 rad>s

the

(including

for

work

student

the

Motion of Gears B, C, and D: Gears B and C which are mounted on the same axle will have the same angular velocity. Since gear B is in mesh with gear A, then assessing

is

work

solely

of

protected

vB rB = vB rA provided

integrity

of

and

work

this

rA 25 b (117.81) = 29.45 rad>s ≤v = a rB A 100 and

any

courses

part

the

is

This

v C = vB = ¢

vD = ¢

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sale

vD rD = vC rC

will

their

destroy

of

Also, gear D is in mesh with gear C. Then

rC 40 b (29.45) = 11.8 rad>s ≤v = a rD C 100

Ans.

B D

*16–24. The gear A on the drive shaft of the outboard motor has a radius rA = 0.5 in. and the meshed pinion gear B on the propeller shaft has a radius rB = 1.2 in. Determine the angular velocity of the propeller in t = 1.5 s, if the drive shaft rotates with an angular acceleration a = (400t3) rad>s2, where t is in seconds. The propeller is originally at rest and the motor frame does not move.

A

2.20 in. P

SOLUTION Angular Motion: The angular velocity of gear A at t = 1.5 s must be determined first. Applying Eq. 16–2, we have dv = adt 1.5 s

L0

dv =

400t3 dt

L0

laws

or

s vA = 100t4 |1.5 = 506.25 rad>s 0

teaching in

Ans. World

Dissemination

Wide

copyright

rA 0.5 b(506.25) = 211 rad>s v = a rB A 1.2

sale

will

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of

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the

is

This

provided

integrity

of

and

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this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

vB =

Web)

However, vA rA = vB rB where vB is the angular velocity of propeller. Then,

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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permitted.

vA

B

16–25. For the outboard motor in Prob. 16–24, determine the magnitude of the velocity and acceleration of point P located on the tip of the propeller at the instant t = 0.75 s.

A

2.20 in. P

SOLUTION Angular Motion: The angular velocity of gear A at t = 0.75 s must be determined first. Applying Eq. 16–2, we have dv = adt 0.75 s

vA

L0

dv =

400t3 dt

L0

Web)

laws

or

s vA = 100t4 |0.75 = 31.64 rad>s 0

Wide

copyright

in

teaching

The angular acceleration of gear A at t = 0.75 s is given by

by

and

use

United

on

learning.

is

of

the

However, vA rA = vB rB and aA rA = aB rB where vB and aB are the angular velocity and acceleration of propeller. Then,

the

(including

for

work

student

the

rA 0.5 b (31.64) = 13.18 rad>s v = a rB A 1.2

this

part

the

is

This

provided

integrity

of

and

work

aB =

assessing

is

rA 0.5 a = a b(168.75) = 70.31 rad>s2 rB A 1.2

work

solely

of

protected

vB =

their

destroy

of

and

any

courses

Motion of P: The magnitude of the velocity of point P can be determined using Eq. 16–8. will

2.20 b = 2.42 ft>s 12 sale

vP = vB rP = 13.18 a

Ans.

The tangential and normal components of the acceleration of point P can be determined using Eqs. 16–11 and 16–12, respectively. ar = aB rP = 70.31 a

2.20 b = 12.89 ft>s2 12

an = v2B rP = A 13.182 B a

2.20 b = 31.86 ft>s2 12

The magnitude of the acceleration of point P is aP = 2a2r + a2n = 212.892 + 31.862 = 34.4 ft>s2

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Ans.

not

instructors

States

World

permitted.

Dissemination

aA = 400 A 0.753 B = 168.75 rad>s2

B

16–26. The pinion gear A on the motor shaft is given a constant angular acceleration a = 3 rad>s2. If the gears A and B have the dimensions shown, determine the angular velocity and angular displacement of the output shaft C, when t = 2 s starting from rest. The shaft is fixed to B and turns with it.

B C

125 mm A

SOLUTION v = v0 + a c t vA = 0 + 3(2) = 6 rad>s 1 ac t2 2

u = u0 + v 0 t + uA = 0 + 0 +

1 (3)(2)2 2

uA = 6 rad

or

vA rA = vB rB teaching

Web)

laws

6(35) = vB(125) in

vC = vB = 1.68 rad>s

Wide

copyright

Ans.

not

instructors

States

World

permitted.

Dissemination

uA rA = uB rB

use

United

on

learning.

is

of

the

6(35) = uB (125)

and

uC = uB = 1.68 rad

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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35 mm

16–27. For a short time, gear A of the automobile starter rotates with an angular acceleration of aA = (450t2 + 60) rad>s2, where t is in seconds. Determine the angular velocity and angular displacement of gear B when t = 2 s, starting from rest. The radii of gears A and B are 10 mm and 25 mm, respectively.

A

B

SOLUTION Motion of Gear A: Applying the kinematic equation of variable angular acceleration, dvA =

L

t

vA

dvA =

L0

L0

aAdt

L

A 450t2 + 60 B dt

vA

vA 0 = 150t3 + 60t 2

t 0

vA = A 150t + 60t B rad>s or

3

teaching

Web)

laws

When t = 2 s,

Dissemination

Wide

copyright

in

vA = 150(2)3 + 60(2) = 1320 rad>s

World

permitted.

vA dt

on United by

and

use

A 150t3 + 60t B dt

student

L0

the

(including

for

work

t

of

0 is

work

solely

uA

uA 0 = 37.5t4 + 30t2 2

protected

L0

duA =

the

t

uA

learning.

is

of

the

not

instructors

L

States

duA =

L

B rad

this

provided

integrity

of

and

work

2

assessing

uA = A 37.5t + 30t 4

courses

part

the

is

This

When t = 2 s

will

their

destroy

of

and

any

uA = 37.5(2)4 + 30(2)2 = 720 rad sale

Motion of Gear B: Since gear B is meshed with gear A, Fig. a, then vp = vA rA = vB rB vB = vA ¢

rA ≤ rB

= (1320) ¢

0.01 ≤ 0.025

= 528 rad>s uB = uA ¢ = 720

rA ≤ rB 0.01 0.025

= 288 rad

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Ans.

Ans.

*16–28. For a short time, gear A of the automobile starter rotates with an angular acceleration of aA = (50v1>2) rad>s2, where v is in rad>s. Determine the angular velocity of gear B when t = 1 s. Orginally (vA)0 = 1 rad>s when t = 0. The radii of gears A and B are 10 mm and 25 mm, respectively.

A

B

SOLUTION Motion of Gear A: We have

L t

L0

dvA L aA

dt =

vA

dt = t

t0 =

L1

50vA 1>2

vA 1 vA 1>2 2 25 1

1 1 v 1>2 25 25 A or

t=

dvA

in

teaching

Web)

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vA = (25t + 1)2

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When t = 1 s, vA = 676 rad>s

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vp = vA rA = vB rB

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= 270 rad>s

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Ans.

16–29. A mill in a textile plant uses the belt-and-pulley arrangement shown to transmit power. When t = 0 an electric motor is turning pulley A with an angular velocity of vA = 5 rad>s. If this pulley is subjected to a constant angular acceleration 2 rad>s2, determine the angular velocity of pulley B after B turns 6 revolutions. The hub at D is rigidly connected to pulley C and turns with it.

B 4 in.

3 in. D

SOLUTION

5 in.

When uB = 6 rev; 4(6) = 3 uC

4.5 in.

uC = 8 rev

A

8(5) = 4.5(uA)

ω A = 5 rad/s

uA = 8.889 rev (vA)22 = (vA)21 + 2aC[(uA)2 - (uA)1]

or

(vA)22 = (5)2 + 2(2)[(8.889)(2p) - 0]

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(vA)2 = 15.76 rad>s

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15.76(4.5) = 5vC

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vC = 14.18 rad>s

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14.18(3) = 4(vB)2

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by

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(vB)2 = 10.6 rad s

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Ans.

C

16–30. A tape having a thickness s wraps around the wheel which is turning at a constant rate V. Assuming the unwrapped portion of tape remains horizontal, determine the acceleration of point P of the unwrapped tape when the radius of the wrapped tape is r. Hint: Since vP = vr, take the time derivative and note that dr>dt = v(s>2p).

s P aP r

SOLUTION vP = vr dvP dv dr = r + v a = dt dt dt Since

v

dv = 0, dt

a = va

dr b dt

In one revolution r is increased by s, so that

laws

or

2p s = u ¢r teaching

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Hence, in

s u 2p

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dr s = v dt 2p

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s 2 v 2p

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a =

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16–31. Due to the screw at E, the actuator provides linear motion to the arm at F when the motor turns the gear at A. If the gears have the radii listed in the figure, and the screw at E has a pitch p = 2 mm, determine the speed at F when the motor turns A at vA = 20 rad>s. Hint: The screw pitch indicates the amount of advance of the screw for each full revolution.

F

D

SOLUTION

E

B

C A

vA r A = v B r B vC rC = vD rD Thus, rA r C 10 15 20 = 1 rad>s v = rB r D A 50 60

vF =

1 rad > s 1 rev (2 mm) = 0.318 mm >s 2p rad

Ans.

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or

vD =

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rA ⫽ 10 mm rB ⫽ 50 mm rC ⫽ 15 mm rD ⫽ 60 mm

*16–32. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If A has a constant angular acceleration of aA = 30 rad>s2, determine the tangential and normal components of acceleration of a point located at the rim of B when t = 3 s, starting from rest.

200 mm vA

Motion of Wheel A: Since the angular acceleration of wheel A is constant, its angular velocity can be determined from vA = (vA)0 + aCt = 0 + 30(3) = 90 rad>s Motion of Wheel B: Since wheels A and B are connected by a nonslip belt, then vBrB = vArA

laws

or

rA 200 bv = a b (90) = 144 rad>s rB A 125

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aBrB = aArA

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rA 200 ba = a b (30) = 48 rad>s2 rB A 125 by

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aB = a

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Thus, the tangential and normal components of the acceleration of point P located at the rim of wheel B are (ap)t = aBrB = 48(0.125) = 6 m>s2

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(ap)n = vB 2rB = (144 2)(0.125) = 2592 m>s2

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125 mm

vB A

SOLUTION

vB = a

B

Ans.

16–33. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If the angular displacement of A is uA = (5t3 + 10t2) rad, where t is in seconds, determine the angular velocity and angular acceleration of B when t = 3 s.

200 mm vA

Motion of Wheel A: The angular velocity and angular acceleration of wheel A can be determined from duA = A 15t2 + 20t B rad>s dt

aA =

dvA = A 30t + 20 B rad>s dt laws

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When t = 3 s,

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vA = 15 A 32 B + 20(3) = 195 rad>s

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aA = 30(3) + 20 = 110 rad>s

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Motion of Wheel B: Since wheels A and B are connected by a nonslip belt, then

Ans.

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rA 200 bvA = a b (195) = 312 rad>s rB 125

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vBrB = vArA vB = a

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aBrB = aArA

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rA 200 ba = a b (110) = 176 rad>s2 rB A 125 sale

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aB = a

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125 mm

vB A

SOLUTION

vA =

B

Ans.

16–34. The rope of diameter d is wrapped around the tapered drum which has the dimensions shown. If the drum is rotating at a constant rate of v, determine the upward acceleration of the the block. Neglect the small horizontal displacement of the block.

L r1 r2

SOLUTION

ω d

v = vr

dr dv r + v dt dt

= v(

dr ) dt

r = r1 + (

r2 - r1 ) dx L in

teaching

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dr = (

r2 - r1 )x L or

=

d(vr) dt

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du –d 2p

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1 r2 - r1 du dr = ( ) d( ) dt 2p L dt

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1 r2 - r1 ( ) dv 2p L is

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v2 r2 - r1 ( )d 2p L

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Thus, a =

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Ans.

16–35. z

If the shaft and plate rotates with a constant angular velocity of v = 14 rad>s, determine the velocity and acceleration of point C located on the corner of the plate at the instant shown. Express the result in Cartesian vector form.

A v

0.6 m

a 0.2 m C

D

SOLUTION We will first express the angular velocity v of the plate in Cartesian vector form. The unit vector that defines the direction of v is uOA =

- 0.3i + 0.2j + 0.6k 2( - 0.3)2 + 0.22 + 0.62

x

0.4 m

3 2 6 = - i + j + k 7 7 7

B

2 6 3 v = vuOA = 14 a - i + j + k b = [- 6i + 4j + 12k] rad>s 7 7 7

or

Since v is constant

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a = 0

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For convenience, rC = [ -0.3i + 0.4j] m is chosen. The velocity and acceleration of point C can be determined from

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vC = v * rC

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on

= ( -6i + 4j + 12k) * ( -0.3i + 0.4j)

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= [-4.8i - 3.6j - 1.2k] m>s

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aC = a * rC + V * (V * rc)

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= 0 + ( -6i + 4j + 12k) * [( -6i + 4j + 12k) * ( -0.3i + 0.4j)] = [38.4i - 64.8j + 40.8k]m s2

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0.3 m 0.3 m

Thus,

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O

0.4 m

Ans.

y

*16–36. z

At the instant shown, the shaft and plate rotates with an angular velocity of v = 14 rad>s and angular acceleration of a = 7 rad>s2. Determine the velocity and acceleration of point D located on the corner of the plate at this instant. Express the result in Cartesian vector form.

A v

0.6 m

a 0.2 m C

D

SOLUTION We will first express the angular velocity v of the plate in Cartesian vector form. The unit vector that defines the direction of v and a is uOA =

O

0.4 m

- 0.3i + 0.2j + 0.6k

x

3 2 6 = - i + j + k 7 7 7

2( - 0.3)2 + 0.22 + 0.62

0.3 m 0.3 m 0.4 m B

Thus, 2 6 3 v = vuOA = 14 a - i + j + k b = [- 6i + 4j + 12k] rad>s 7 7 7

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For convenience, rD = [- 0.3i + 0.4j] m is chosen. The velocity and acceleration of point D can be determined from

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3 2 6 a = auOA = 7 a - i + j + k b = [-3i + 2j + 6k] rad>s 7 7 7

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vD = v * rD

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= (- 6i + 4j + 12k) * (0.3i - 0.4j) by

= [4.8i + 3.6j + 1.2k]m>s

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aD = a * rD - v2 rD part

the

is

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= ( - 3i + 2j + 6k) * ( -0.3i + 0.4j) + (-6i + 4j + 12k) * [(-6i + 4j + 12k) * (-0.3i + 0.4j)] any

courses

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= [ - 36.0i + 66.6j - 40.2k]m s2

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y

16–37. The rod assembly is supported by ball-and-socket joints at A and B. At the instant shown it is rotating about the y axis with an angular velocity v = 5 rad>s and has an angular acceleration a = 8 rad>s2. Determine the magnitudes of the velocity and acceleration of point C at this instant. Solve the problem using Cartesian vectors and Eqs. 16–9 and 16–13.

z

C 0.3 m

0.4 m

B A A 0.4 m

SOLUTION

x

vC = v * r vC = 5j * ( -0.4i + 0.3k) = {1.5i + 2k} m>s vC = 21.52 + 2 2 = 2.50 m>s

Ans.

a C = a * r - v2r = 8j * ( -0.4i + 0.3k) -52 ( -0.4i + 0.3k) = {12.4i - 4.3k} m>s2 a C = 212.4 2 + (-4.3)2 = 13.1 m>s2

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Ans.

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V

y

16–38. Rotation of the robotic arm occurs due to linear movement of the hydraulic cylinders A and B. If this motion causes the gear at D to rotate clockwise at 5 rad>s, determine the magnitude of velocity and acceleration of the part C held by the grips of the arm.

4 ft 2 ft

45

C

SOLUTION D

Motion of Part C: Since the shaft that turns the robot’s arm is attached to gear D, then the angular velocity of the robot’s arm vR = vD = 5.00 rad>s. The distance of part C from the rotating shaft is rC = 4 cos 45° + 2 sin 45° = 4.243 ft. The magnitude of the velocity of part C can be determined using Eq. 16–8. vC = vR rC = 5.00(4.243) = 21.2 ft>s

B

A

Ans.

The tangential and normal components of the acceleration of part C can be determined using Eqs. 16–11 and 16–12 respectively. at = arC = 0

laws

or

an = v2R rC = A 5.002 B (4.243) = 106.07 ft>s2

Ans.

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Dissemination

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aC = 2a2t + a2n = 202 + 106.072 = 106 ft>s2

Wide

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The magnitude of the acceleration of point C is

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3 ft

16–39. The bar DC rotates uniformly about the shaft at D with a constant angular velocity V. Determine the velocity and acceleration of the bar AB, which is confined by the guides to move vertically.

C A

V D

B U l

SOLUTION y = l sin u # # y = vy = l cos uu $ # $ y = a y = l(cos uu - sin uu) $ $ Here vy = vAB , ay = aAB, and u = v, u = a = 0. Ans.

aAB = l C cos u(0) - sin u(v)2 D = - v2 l sin u

Ans.

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vAB = l cos u (v) = v l cos u

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*16–40. The mechanism is used to convert the constant circular motion v of rod AB into translating motion of rod CD and the attached vertical slot. Determine the velocity and acceleration of CD for any angle u of AB .

B D

l V

u x

SOLUTION x = l cos u # # x = vx = - l sin uu $ # # x = a x = - l(sin uu + cos u u2) # $ Here vx = vCD, ax = aCD, and u = v, u = a = 0. vCD = - l sin u (v) = - vl sin u

Ans.

a CD = -l C sin u(0) + cos u(v)2 D = - v2 l cos u

Ans.

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Negative signs indicate that both vCD and aCD are directed opposite to positive x.

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A C

16–41. At the instant u = 50°, the slotted guide is moving upward with an acceleration of 3 m>s2 and a velocity of 2 m>s. Determine the angular acceleration and angular velocity of link AB at this instant. Note: The upward motion of the guide is in the negative y direction.

B 300 mm V, A

y U

A

SOLUTION y = 0.3 cos u # # y = vy = - 0.3 sin uu $ # ## y = a y = -0.3 A sin uu + cos uu2 B

v a

# # $ Here vy = - 2 m>s, ay = - 3 m>s2, and u = v, u = v, u = a, u = 50°. v = 8.70 rad>s

Ans.

- 3 = -0.3[sin 50°(a) + cos 50°(8.70)2]

a = - 50.5 rad>s 2

Ans.

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-2 = -0.3 sin 50°(v)

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2 m/s 3 m/s2

16–42. The mechanism shown is known as a Nuremberg scissors. If the hook at C moves with a constant velocity of v, determine the velocity and acceleration of collar A as a function of u. The collar slides freely along the vertical guide.

A

L/2

u L/2

u

B

SOLUTION

C

x = 3L sin u # # v = x = 3L cos u u y = L cos u # # y = - L sin u u # # y L sin u u # = v 3L cos u u # y = (v tan u)>3 T

Ans.

in

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laws

or

# v 1 v v $ b ba y = (sec2u u) = a 2 3 3 cos u 3L cos u v2 T 9L cos3 u

Wide

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Ans.

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Dissemination

$ y =

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v

16–43. The crankshaft AB is rotating at a constant angular velocity of v = 150 rad>s. Determine the velocity of the piston P at the instant u = 30°.

B 0.75 ft

0.2 ft U

A

V x

SOLUTION x = 0.2 cos u + 2(0.75)2 - (0- 2 sin u)2 $ $ 1 1 # x = -0.2 sin uu + C (0.75)2 -(0.2 sin u)2 D - 2( -2)(0.2 sin u)(0.2 cos u)u 2 (0.2)2v sin 2 u 1 vP = -0.2v sin u - a b 2 2(0.75)2 - (0.2 sin u)2 At u = 30°, v = 150 rad>s (0.2)2(150) sin 60° 1 vP = -0.2(150) sin 30° - a b 2 2(0.75)2 - (0.2 sin 30°)2 or

vP = -18.5 ft>s = 18.5 ft>s ;

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P

*16–44. v

Determine the velocity and acceleration of the follower rod CD as a function of u when the contact between the cam and follower is along the straight region AB on the face of the cam. The cam rotates with a constant counterclockwise angular velocity V.

A r O

u

D

C B

SOLUTION Position Coordinate: From the geometry shown in Fig. a, xC =

r = r sec u cos u

Time Derivative: Taking the time derivative, # # vCD = xC = r sec u tan uu

(1)

laws

or

# Here, u = + v since v acts in the positive rotational sense of u. Thus, Eq. (1) gives vCD = rv sec u tan u :

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The time derivative of Eq. (1) gives $ # # # $ aCD = xC = r{sec u tan uu + u[sec u(sec2uu) + tan u(sec u tan uu)]} $ # aCD = r[sec u tan u u + (sec3 u + sec u tan2 u)u2]

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# $ Since u = v is constant, u = a = 0. Then, solely

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a CD = r[sec u tan u(0) + (sec3 u + sec u tan2 u)v2] assessing

is

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= rv2 A sec3 u + sec u tan2 u B :

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Ans.

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16–45. Determine the velocity of rod R for any angle u of the cam C if the cam rotates with a constant angular velocity V. The pin connection at O does not cause an interference with the motion of A on C.

V

r1

C

r2

SOLUTION

O R

Position Coordinate Equation: Using law of cosine. (1)

Time Derivatives: Taking the time derivative of Eq. (1).we have 0 = 2x

dx du dx - 2r1 ¢ -x sin u + cos u ≤ dt dt dt

(2)

du dx and v = . From Eq.(2), dt dt laws

or

0 = xv - r1(v cos u - xv sin u) in

teaching

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r1xv sin u r1 cos u - x

(3)

Dissemination

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copyright

v =

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x = r1 cos u + 2r21 cos 2u + r22 + 2r1r2

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Substitute into Eq.(3),we have

the

r21v sin 2u

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+ r1v sin u

2 2r21 cos2u + r22 + 2r# 1r2

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v = -

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Note: Negative sign indicates that v is directed in the opposite direction to that of positive x.

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A x

(r1 + r2)2 = x2 + r12 - 2r1x cos u

However v =

u

16–46. The bridge girder G of a bascule bridge is raised and lowered using the drive mechanism shown. If the hydraulic cylinder AB shortens at a constant rate of 0.15 m>s, determine the angular velocity of the bridge girder at the instant u = 60°.

G

B C

A

u 3m

SOLUTION

5m

Position Coordinates: Applying the law of cosines to the geometry shown in Fig. a, s2 = 32 + 52 - 2(3)(5) cos A 180°- u B s2 = 34 - 30 cos A 180° - u B However, cos A 180°-u B = - cos u. Thus, s2 = 34 + 30 cos u Time Derivatives: Taking the time derivative,

laws

or

# # 2ss = 0 + 30 A - sin uu B # # ss = - 15 sin uu

teaching

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(1)

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permitted.

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# # When u = 60°, s = 234 + 30 cos 60° = 7 m. Also, s = - 0.15 m>s since s is directed towards the negative sense of s. Thus, Eq. (1) gives # 7 A - 0.15 B = - 15 sin 60°u # v = u = 0.0808 rad>s Ans.

16–47. v

The circular cam of radius r is rotating clockwise with a constant angular velocity V about the pin at O, which is at an eccentric distance e from the center of the cam. Determine the velocity and acceleration of the follower rod A as a function of u.

e

O

u

SOLUTION Position Coordinates: From the geometry shown in Fig. a, xA = e cos u + r

(1)

Time Derivatives: Taking the time derivative, # # vA = xA = - e sin uu

(2)

# Since v acts in the negative rotational sense of u, then u = - v. Thus, Eq. (2) gives vA = - e sin u( -v) = ev sin u :

or

Ans.

teaching

Web)

laws

Taking the time derivative of Eq. (2) gives $ # $ aA = xA = - e A sin uu + cos uu2 B

Wide

copyright

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(3)

not

instructors

States

World

permitted.

Dissemination

$ Since v is constant, u = a = 0. Then Eq. (3) gives

United

on

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is

of

the

aA = - e[sin u(0) + cos u(v2)]

the

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= - ev2 cos u = ev2 cos u ;

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The negative sign indicates that aA acts towards the negative sense of xA.

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Ans.

r

A

*16–48. Peg B mounted on hydraulic cylinder BD slides freely along the slot in link AC. If the hydraulic cylinder extends at a constant rate of 0.5 m>s, determine the angular velocity and angular acceleration of the link at the instant u = 45°.

C

B

0.5 m/s u D

SOLUTION 0.6 m

Position Coordinate: From the geometry shown in Fig. a, yC = 0.6 tan u m Time Derivatives: Taking the time derivative, # # vC = yC = (0.6 sec2 uu) m>s

(1)

laws

or

Here, vC = 0.5 m>s since vC acts in the positive sense of yC. When u = 45°, Eq. (1) gives # 0.5 = A 0.6 sec2 45° B u # Ans. vAB = u = 0.4167 rad>s = 0.417 rad>s

permitted.

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copyright

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The time derivative of Eq. (1) gives $ # $ aC = yC = 0.6 A sec2 uu + 2 sec u sec u tan uu2 B $ # aC = 0.6 sec2 u A u + 2 tan uu2 B

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Since vC is constant, aC = 0. Thus, Eq. (2) gives $ 0 = u + 2 tan 45° A 0.41672 B $ aAB = u = - 0.3472 rad>s2 = 0.347 rad>s2d

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integrity

of

and

work

The negative sign indicates that aAB acts counterclockwise.

Ans.

A

16–49. Bar AB rotates uniformly about the fixed pin A with a constant angular velocity V. Determine the velocity and acceleration of block C, at the instant u = 60°.

B

L

V

L A

u

SOLUTION L cos u + L cos f = L

C

cos u + cos f = 1 # # sin u u] + sin f f = 0 $ $ # # cos u(u)2 + sin uu + sinf f + cos f ( f)2 = 0

L

(1) (2)

When u = 60°, f = 60°, # # thus, u = -f = v (from Eq. (1)) $ u = 0

in

teaching

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laws

or

$ f = -1.155v2 (from Eq.(2))

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Also, sC = L sin f - L sin u # # vC = L cos f f - L cos u u # $ $ # aC = -L sin f (f)2 + L cos f (f) - L cos u(u) + L sin u(u)2

(including

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the

At u = 60°, f = 60°

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the

sC = 0

assessing

vC = L(cos 60°)( - v) - L cos 60°(v) = - Lv = Lv c

integrity

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and

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this

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aC = - L sin 60°( -v)2 + L cos 60°(-1.155v2) + 0 + L sin 60°(v)2 any

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aC = - 0.577 Lv2 = 0.577 Lv2 c

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16–50. The block moves to the left with a constant velocity v0. Determine the angular velocity and angular acceleration of the bar as a function of u.

a u x

SOLUTION Position Coordinate Equation: From the geometry, x =

a = a cot u tan u

(1)

Time Derivatives: Taking the time derivative of Eq. (1),we have du dx = - a csc2 u dt dt dx du = -v0. Also, = v. dt dt laws

or

Since y0 is directed toward negative x, then

(2)

Wide

copyright

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teaching

Web)

From Eq.(2),

instructors

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permitted.

Dissemination

- v0 = - a csc2 u(v) of

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v0 sin2 u a

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=

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v0 a csc2 u

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dv . Then from the above expression dt is

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Here, a =

assessing

v0 du (2 sin u cos u) a dt

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a =

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v0 du sin2 u. Substitute these values into = a dt destroy

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and

However, 2 sin u cos u = sin 2u and v =

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Eq.(3) yields a =

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v0 v0 2 v0 sin 2u a sin2u b = a b sin 2u sin2 u a a a

Ans.

v0

16–51. The bar is confined to move along the vertical and inclined planes. If the velocity of the roller at A is vA = 6 ft>s when u = 45°, determine the bar’s angular velocity and the velocity of roller B at this instant.

vA

A u 5 ft

SOLUTION 30⬚

sB cos 30° = 5 sin u B

sB = 5.774 sin u # s

B

# = 5.774 cos u u

vB

(1)

5 cos u = sA + sB sin 30° # # # -5 sin u u = sA + sB sin 30°

(2)

teaching

Web)

laws

or

Combine 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

Dissemination

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Ans.

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From Eq.(1): not

instructors

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

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Ans.

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*16–52. Arm AB has an angular velocity of V and an angular acceleration of A. If no slipping occurs between the disk and the fixed curved surface, determine the angular velocity and angular acceleration of the disk.

C

A r

ω ', α '

ω ,α

SOLUTION

B

ds = (R - r) du = - r df (R - r) ¢

R

df du ≤ = -r ¢ ≤ dt dt

(R - r) v r

Ans.

a¿ = -

(R - r) a r

Ans.

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laws

or

v¿ = -

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16–53. If the wedge moves to the left with a constant velocity v, determine the angular velocity of the rod as a function of u.

L v

u

SOLUTION Position Coordinates:Applying the law of sines to the geometry shown in Fig. a, xA L = sin(f - u) sin A 180° - f B xA =

L sin(f - u) sin A 180° - f B

However, sin A 180° - f B = sinf. Therefore, L sin (f - u) sin f laws

or

xA =

Time Derivative: Taking the time derivative,

Dissemination

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# L cos (f - u)(- u) # xA = sin f

not

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# L cos (f - u)u # vA = xA = sin f by

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(1)

is

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v sin f L cos (f - u)

this

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# u =

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Since point A is on the wedge, its velocity is vA = - v. The negative sign indicates that vA is directed towards the negative sense of xA. Thus, Eq. (1) gives

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f

16–54. The slotted yoke is pinned at A while end B is used to move the ram R horizontally. If the disk rotates with a constant angular velocity V, determine the velocity and acceleration of the ram. The crank pin C is fixed to the disk and turns with it.

R

B

V

u

SOLUTION (1)

x = l tan f However

r s s = = sin f sin (180° - u) sin u d = s cos f - r cos u

A

r sin u s

sin f =

d + r cos u s

cos f =

not and

by

(d + r cos u)4

d

the

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(d + r cos u)3

protected

lr sin u(2r2 - d2 + rd cos u)

is

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x= a = lrv c ¶

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# # (d + r cos u)2( -d sin uu) - (r + d cos u)(2)(d + r cos u)( -r sin uu)

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Ans.

permitted.

Ans. World

v

instructors

(d + r cos u)2

States

lr(r + d cos u)

Dissemination

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copyright

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teaching

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laws

or

r sin u sin f s lr sin u b = l§ ¥ = From Eq. (1) x = l a cos f d + r cos u d + r cos u s # # # (d + r cos u)(lr cos uu) - (lr sin u)(-r sin uu) xª = v = Where u = v (d + r cos u)2 =

f

d

r

C l

16–55. The Geneva wheel A provides intermittent rotary motion vA for continuous motion vD = 2 rad>s of disk D. By choosing d = 100 22 mm, the wheel has zero angular velocity at the instant pin B enters or leaves one of the four slots. Determine the magnitude of the angular velocity V A of the Geneva wheel at any angle u for which pin B is in contact with the slot.

d ⫽ 100 2 mm

A 100 mm · vD ⫽ u

f

u

100 mm

D

SOLUTION tan f =

B

0.1 sin u 0.1 A 22 - cos u B

# sec2 f f =

sin u

=

22 - cos u # ª) A 22 - cos u B (cos uu) - sin u(sin uu

A 22 - cos u B

22 cos u - 1 # u A 22 - cos u B 2

=

2

(1)

From the geometry: r 2 = (0.1 sin u)2 + [0.1 A 22 - cos u B ]2 = 0.01 A 3 - 2 22 cos u B 0.01 A 3 - 2 22 cos u B

(3 - 222 cos u)

=

[0.1 A 22 - cosu B ]

2

A 22 - cos u B 2

teaching

Web)

[0.1 A 22 - cos u B ]

2

=

or

r2

laws

sec2 f =

Wide

copyright

in

From Eq. (1)

World

permitted.

Dissemination

22 cos u - 1 # u A 22 - cos u B 2

the

not

instructors

is

fª =

States

A 22 - cos u B

2

of

A 3 - 2 22 cos u B

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# Here f = vA and u = vD = 2 rad>s the

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3 - 2 22 cos u

protected

22 cos u - 1

is

vA = 2a

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student

the

# 22 cos u - 1 # u f = 3 - 2 22 cos u

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Ans.

· vA ⫽ f

*16–56. At the instant shown, the disk is rotating with an angular velocity of V and has an angular acceleration of A. Determine the velocity and acceleration of cylinder B at this instant. Neglect the size of the pulley at C.

A 3 ft V, A

u

C

5 ft

SOLUTION

B

s = 232 + 52 - 2(3)(5) cos u # 1 1 # vB = s = (34 - 30 cos u)- 2(30 sin u)u 2 15 v sin u

# # 15 v cos uu + 15v sin u

-

225 v2 sin2 u

Ans.

3

(34 - 30 cos u) 2

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(34 - 30 cos u)

1 2

3

(34 - 30 cos u) 2 or

15 (v2 cos u + a sin u)

+

Web)

234 - 30 cos u

# 1 a - b(15v sin u) a 30 sin uu b 2

laws

# aB = s =

=

Ans.

1

(34 - 30 cos u) 2

teaching

vB =

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16–57. If h and u are known, and the speed of A and B is vA = vB = v, determine the angular velocity V of the body and the direction f of vB.

vA u A

V

SOLUTION

vB f

vB = vA + v * rB>A -v cos fi + v sin fj = v cos ui + v sin uj + ( -vk) * ( -hj) + ) (:

- v cos f = v cos u - vh

(1)

(+ c )

v sin f = v sin u

(2)

From Eq. (2),

f = u

From Eq. (1),

v =

Ans.

2v cos u h

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Ans.

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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h

B

16–58. If the block at C is moving downward at 4 ft/s, determine the angular velocity of bar AB at the instant shown.

C 3 ft

A

30°

ω AB

SOLUTION

2 ft

Kinematic Diagram: Since link AB is rotating about fixed point A, then vB is always directed perpendicular to link AB and its magnitude is vB = vAB rAB = 2vAB. At the instant shown, vB is directed towards the negative y axis. Also, block C is moving downward vertically due to the constraint of the guide. Then vc is directed toward negative y axis. Velocity Equation: Here, rC>A = {3 cos 30°i + 3 sin 30°j} ft = {2.598i + 1.50j} ft. Applying Eq. 16–16, we have vC = vB + vBC * rC>B

or

- 4j = - 2vAB j + (vBCk) * (2.598i + 1.50j)

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-4j = - 1.50vBCi + (2.598vBC - 2vAB)j

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in

Equating i and j components gives

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Dissemination

vBC = 0

0 = - 1.50vBC -4 = 2.598(0) - 2vAB

not

vAB = 2.00 rad s

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Ans.

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B

vC = 4 ft/s

16–59. The velocity of the slider block C is 4 ft>s up the inclined groove. Determine the angular velocity of links AB and BC and the velocity of point B at the instant shown.

B

C

1 ft

SOLUTION For link BC

45⬚

A

vC = {- 4 cos 45°i + 4 sin 45°j} ft>s

vB = -vB i

v = vBC k

rB>C = {1i} ft vC = vB + v * rC>B - 4 cos 45°i + 4 sin 45°j = -vBi + (vBC k) * (1i) - 4 cos 45°i + 4 sin 45°j = -vB i + vBCj

or

Equating the i and j components yields: Ans.

vBC = 2.83 rad>s

Ans. Wide

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-4 cos 45° = vBC

in

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laws

vB = 2.83 ft>s

-4 cos 45° = - vB

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Dissemination

For link AB: Link AB rotates about the fixed point A. Hence

use

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vB = vAB rAB

and

vAB = 2.83 rad>s

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Ans.

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2.83 = vAB(1)

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vC

1 ft

4 ft/s

*16–60. The epicyclic gear train consists of the sun gear A which is in mesh with the planet gear B. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the connecting link DE pinned to B and C is rotating at vDE = 18 rad>s about the pin at E, determine the angular velocities of the planet and sun gears.

100 mm 600 mm

A E

B C D

SOLUTION

200 mm vD E

vD = rDE vDE = (0.5)(18) = 9 m>s c The velocity of the contact point P with the ring is zero.

R

vD = vP + v * rD>P 9j = 0 + ( -vB k) * ( - 0.1i) vB = 90 rad>s

b

Ans.

Let P¿ be the contact point between A and B.

or

vP¿ = vP + v * rP¿>P teaching

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laws

vP¿ j = 0 + (- 90k) * ( -0.4i)

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vP¿ = 36 m>s c

permitted.

d

World

Ans. the

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vP¿ 36 = 180 rad>s = rA 0.2

of

vA =

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18 rad/s 300 mm

16–61. The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the velocity of the slider block C at the instant u = 60°, if link AB is rotating at 4 rad>s.

125 mm C 45

B

SOLUTION vAB

vC = vB + v * rC>B

4 rad/s 300 mm

-vCi = -4(0.3) sin 30°i + 4(0.3) cos 30°j + vk * ( -0.125 cos 45°i + 0.125 sin 45°j)

u

-vC = - 1.0392 - 0.008839v A

0 = 0.6 - 0.08839v Solving, v = 6.79 rad > s vC = 1.64 m>s

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laws

or

Ans.

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16–62. If the flywheel is rotating with an angular velocity of vA = 6 rad>s, determine the angular velocity of rod BC at the instant shown.

1.5 m A 0.3 m

vA ⫽ 6 rad/s

O C B

SOLUTION

60⬚ 0.6 m D

Rotation About a Fixed Axis: Flywheel A and rod CD rotate about fixed axes, Figs. a and b. Thus, the velocity of points B and C can be determined from vB = vA * rB = (-6k) * ( -0.3j) = [-1.8i] m>s vC = vCD * rC = (vCDk) * (0.6 cos 60° i + 0.6 sin 60° j) = - 0.5196vCD i + 0.3vCD j

or

General Plane Motion: By referring to the kinematic diagram of link BC shown in Fig. c and applying the relative velocity equation, we have

teaching

Web)

laws

vB = vC + vBC * rB>C

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copyright

in

- 1.8i = - 0.5196vCD i + 0.3vCD j + (vBC k) * ( -1.5i)

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Dissemination

- 1.8i = - 0.5196vCD i + (0.3vCD - 1.5vBC)j

use

United

on

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not

Equating the i and j components

work

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- 1.8 = - 0.5196vCD

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0 = 0.3vCD - 1.5vBC

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vCD = 3.46 rad>s

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and

vBC = 0.693 rad>s

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16–63. If the angular velocity of link AB is vAB = 3 rad>s, determine the velocity of the block at C and the angular velocity of the connecting link CB at the instant u = 45° and f = 30°.

C θ = 45°

3 ft

A

ω A B = 3 rad/s

SOLUTION

2 ft φ = 30° B

vC = vB + vC>B

B vC R = C 6 ;

30°c

S + D vCB (3) T 45°b

+ ) (:

-vC = 6 sin 30° - vCB (3) cos 45°

(+ c )

0 = - 6 cos 30° + vCB (3) sin 45° d

Ans. or

vCB = 2.45 rad>s vC = 2.20 ft>s ;

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Ans.

Wide

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Also,

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vC = vB + v * rC>B

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the

not

-vC i = (6 sin 30°i - 6 cos 30°j) + (vCB k) * (3 cos 45°i + 3 sin 45°j) -vC = 3 - 2.12vCB

(+ c )

0 = - 5.196 + 2.12vCB

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+b a:

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d

Ans.

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is

vCB = 2.45 rad s

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vC = 2.20 ft s ;

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Ans.

*16–64. Pinion gear A rolls on the fixed gear rack B with an angular velocity v = 4 rad>s. Determine the velocity of the gear rack C.

C

A 0.3 ft

v

B

SOLUTION vC = vB + vC>B + ) (;

vC = 0 + 4(0.6) vC = 2.40 ft>s

Ans.

Also: vC = vB + v * rC>B -vC i = 0 + (4k) * (0.6j) vC = 2.40 ft>s

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Ans.

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16–65. The pinion gear rolls on the gear racks. If B is moving to the right at 8 ft>s and C is moving to the left at 4 ft>s, determine the angular velocity of the pinion gear and the velocity of its center A.

C

A V

0.3 ft

B

SOLUTION vC = vB + vC>B + ) (:

- 4 = 8 - 0.6(v) v = 20 rad>s

Ans.

vA = vB + vA>B + ) (:

vA = 8 - 20(0.3) vA = 2 ft>s :

Ans.

or

Also:

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laws

vC = vB + v * rC>B

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- 4i = 8i + (vk) * (0.6j)

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-4 = 8 - 0.6v

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the

Ans.

student

the

by

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vA = vB + v * rA>B

the

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work

vAi = 8i + 20k * (0.3j)

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protected

vA = 2 ft>s :

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Ans.

not

instructors

v = 20 rad>s

16–66. Determine the angular velocity of the gear and the velocity of its center O at the instant shown.

A 45 4 ft/s

SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, vB = vC + v * rB>C 3i = -4i +

A - vk B * A 2.25j B

3i = A 2.25v - 4 B i Equating the i components yields 3 = 2.25v - 4

(1)

v = 3.111 rad>s

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or

Ans. (2)

Wide

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For points O and C,

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Dissemination

vO = vC + v * rO>C the

not

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States

A -3.111k B * A 1.5j B

on

learning.

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= - 4i +

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= [0.6667i] ft>s

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vO = 0.667 ft>s :

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Ans.

0.75 ft O 1.50 ft

3 ft/s

16–67. Determine the velocity of point A on the rim of the gear at the instant shown.

A 45 4 ft/s

SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, vB = vC + v * rB>C 3i = - 4i +

A -vk B * A 2.25j B

3i = A 2.25v - 4 B i

(1)

v = 3.111 rad>s

(2) teaching

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laws

3 = 2.25v - 4

or

Equating the i components yields

Wide

copyright

in

For points A and C,

World

permitted.

Dissemination

vA = vC + v * rA>C

learning.

is

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States

A vA B x i + A vA B y j = - 4i + A - 3.111k B * A - 1.061i + 2.561j B

work

student

the

by

and

use

United

on

A vA B x i + A vA B y j = 3.9665i + 3.2998j

the

(including

for

Equating the i and j components yields

A vA B y = 3.2998 ft>s work

this

assessing

is

work

solely

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protected

A vA B x = 3.9665 ft>s

part

the

is

This

provided

integrity

of

and

Thus, the magnitude of vA is

Ans.

destroy

of

and

any

courses

vA = 2 A vA B x 2 + A vA B y 2 = 23.96652 + 3.29982 = 5.16 ft>s

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will

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and its direction is u = tan - 1 C

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A vA B y A vA B x

S = tan - 1 ¢

3.2998 ≤ = 39.8° 3.9665

Ans.

0.75 ft O 1.50 ft

3 ft/s

*16–68. Part of an automatic transmission consists of a fixed ring gear R, three equal planet gears P, the sun gear S, and the planet carrier C, which is shaded. If the sun gear is rotating at vS = 6 rad>s, determine the angular velocity vC of the planet carrier. Note that C is pin connected to the center of each of the planet gears.

C

P

S

4 in.

SOLUTION

vC

vD = vA + vD>A

2 in.

24 = 0 + 4(vP) b b vP = 6 rad>s

R P

vE = vA + vE>A vE = 0 + 6(2) b b vE = 12 in.>s 12 = 2 rad>s 6

Ans.

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vC =

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P

vS vP

16–69. If the gear rotates with an angular velocity of v = 10 rad>s and the gear rack moves at vC = 5 m>s, determine the velocity of the slider block A at the instant shown.

A

0.5 m

SOLUTION General Plane Motion: Referring to the diagram shown in Fig. a and applying the relative velocity equation, vC

= - 5i + ( -10k) * (0.075j) = [ -4.25i] m>s Then, applying the relative velocity equation to link AB shown in Fig. b, vA = vB + vAB * rA>B laws

or

vA j = - 4.25i + ( - vAB k) * ( -0.5 cos 60° i + 0.5 sin 60° j)

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vA j = (0.4330vAB - 4.25)i + 0.25vAB j

States

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permitted.

Dissemination

Equating the i and j components, yields

(1)

learning.

is

of

the

not

instructors

0 = 0.4330vAB - 4.25

(2)

work

student

the

by

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use

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on

vA = 0.25vAB

of

protected

the

(including

for

Solving Eqs. (1) and (2) yields

this

assessing

is

work

solely

vAB = 9.815 rad>s

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vA = 2.45 m>s c

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10 rad/s

75 mm

5 m/s C

vB = vC + v * rB>C

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

v 60

Ans.

16–70. If the slider block C is moving at vC = 3 m>s, determine the angular velocity of BC and the crank AB at the instant shown.

B 0.5 m 60⬚

1m

A 45⬚

vC ⫽ 3 m/s

SOLUTION Rotation About a Fixed Axis: Referring to Fig. a, vB = vAB * rB = ( -vAB k) * (0.5 cos 60° i + 0.5 sin 60° j) = 0.4330vAB i - 0.25vAB j General Plane Motion: Applying the relative velocity equation and referring to the kinematic diagram of link BC shown in Fig. b, vB = vC + vBC * rB>C laws

or

0.4330vAB i - 0.25vAB j = - 3j + ( - vBC k) * ( -1 cos 45° i + 1 sin 45° j)

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0.4330vAB i - 0.25vAB j = 0.7071vBC i + (0.7071vBC - 3)j

States

World

permitted.

Dissemination

Equating the i and j components yields,

on

learning.

is

of

the

not

instructors

0.4330vAB = 0.7071vBC

work

student

the

by

and

use

United

-0.25vAB = 0.7071vBC - 3

of

protected

the

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for

Solving,

Ans.

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is

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solely

vBC = 2.69 rad>s

sale

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vAB = 4.39 rad>s

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C

Ans.

16–71. The two-cylinder engine is designed so that the pistons are connected to the crankshaft BE using a master rod ABC and articulated rod AD. If the crankshaft is rotating at v = 30 rad>s, determine the velocities of the pistons C and D at the instant shown.

D

C 250 mm 250 mm

45°

50 mm

SOLUTION

A 45°

vC = vB + vC>B

50 mm

+ (0.25) v vC = 1.5 ; c30° 45°d + (;) vC cos 45° = 1.5 - vC(0.25)(cos 30°) (+ T)

vC sin 45° = 0 + vC(0.25) (sin 30°)

vC = 0.776 m>s

Ans.

v C = 4.39 rad>s vA = vB + vA>B or Web) Wide

copyright

vD = vA + vD>A

[0.05(4.39) = 0.2195] 45°a

laws

+

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:

in

= 1.5 ;

vA

instructors

States

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permitted.

Dissemination

vD = 1.5 + 0.21 95 + v¿(0 .25) Q Q 45°d 45°d the

not

vD = - 1.5 sin 45°

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(R+)

Ans.

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vD = 1.06 m>s a

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vB = v * rB>E

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vC = vB + vBC * rC>B

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- vC cos 45° = - 1.5 - vBC (0.2165)

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- vC cos 45°i - vC sin 45°j = (30k) * (0.05j) + (vBC k) * (0.25 cos 60°i + 0.25 sin 60°j)

- vC sin 45° = 0.125vBC vC = 0.776 m>s

Ans.

v BC = 4.39 rad>s vA = vB + vBC * rA>B vD = vA + vAD * rD>A vD cos 45°i - vD sin 45°j = (30k) * (0.05j) + ( -4.39k) * (0.05 cos 45°i + 0.05 sin 45°j) + (vAD k) * (- 0.25 cos 45°i + 0.25 sin 45°j) vD cos 45° = - 1.5 + 0.1552 - vAD (0.1768) - vD sin 45° = 0.1552 - 0.1768vAD vAD = 3.36 rad s vD = 1.06 m s

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Ans.

60° B ω = 30 rad/s E

*16–72. Determine the velocity of the center O of the spool when the cable is pulled to the right with a velocity of v. The spool rolls without slipping.

A

O R

SOLUTION Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point P is zero. The kinematic diagram of the spool is shown in Fig. a. General Plane Motion:Applying the relative velocity equation and referring to Fig. a, vB = vP + v * rB>D vi = 0 + ( -vk) * C (R - r)j D vi = v(R - r)i

or

Equating the i components, yields

teaching

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laws

v R - r

v =

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v = v(R - r)

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vO = vP + v * rO>P

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vO = ¢

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

r

v

16–73. Determine the velocity of point A on the outer rim of the spool at the instant shown when the cable is pulled to the right with a velocity of v. The spool rolls without slipping.

A

O R

SOLUTION Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point P is zero. The kinematic diagram of the spool is shown in Fig. a. General Plane Motion:Applying the relative velocity equation and referring to Fig. a, vB = vP + v * rB>D vi = 0 + ( -vk) * C (R - r)j D vi = v(R - r)i

or

Equating the i components, yields

teaching

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laws

v R - r

v =

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v = v(R - r)

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vA = vP + v * rA>P

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v k ≤ * 2Rj R - r

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2R ≤ vR i R - r

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= B¢

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will sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

and

2R ≤v : R - r

their

vA = ¢

r

v

16–74. If crank AB rotates with a constant angular velocity of vAB = 6 rad>s, determine the angular velocity of rod BC and the velocity of the slider block at the instant shown. The rod is in a horizontal position.

0.5 m 0.3 m

B

C 60⬚

30⬚ vAB ⫽ 6 rad/s

A

SOLUTION Rotation About a Fixed Axis: Referring to Fig. a, vB = vAB * rB = (6k) * (0.3 cos 30° i + 0.3 sin 30° j) = [ -0.9i + 1.559j] General Plane Motion: Applying the relative velocity equation to the kinematic diagram of link BC shown in Fig. b, vB = vC + vBC * rB>C laws

or

(- 0.9i + 1.559j) = (- vC cos 60° i - vC sin 60° j) + ( -vBC k) * ( -0.5i)

Wide

copyright

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- 0.9i + 1.559j = - 0.5vC i + (0.5vBC - 0.8660vC)j

States

World

permitted.

Dissemination

Equating the i and j components yields

(1)

United

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-0.9 = - 0.5vC

(2)

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1.559 = 0.5vBC - 0.8660vC

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Solving Eqs. (1) and (2) yields

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vC = 1.80 m>s

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vBC = 6.24 rad>s

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or transmission in any form or by any means, electronic, mechanical,

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Ans.

16–75. If the slider block A is moving downward at vA = 4 m>s, determine the velocity of block B at the instant shown.

E

B

300 mm 4

250 mm

D

400 mm

SOLUTION

30° A

C

vB = 4 T + v (0.55) AB : 3 + ) (: vB = 0 + vAB (0.55)( ) 5 4 0 = - 4 + vAB (0.55)( ) 5

Solving, vAB = 9.091 rad>s vB = 3.00 m>s

Ans.

Also: laws

or

vB = vA + vAB * rB>A in

teaching

Web)

-4 3 (0.55)i + (0.55) j } 5 5

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copyright

vB i = - 4j + ( -vAB k) * {

permitted.

Dissemination

vB = vAB (0.33)

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0 = - 4 + 0.44vAB by

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vAB = 9.091 rad>s

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vB = 3.00 m>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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vA = 4 m/s

300 mm

vB = vA + vB>A

(+ c )

3 5

Ans.

*16–76. If the slider block A is moving downward at vA = 4 m>s, determine the velocity of block C at the instant shown.

E

B

300 mm 4

250 mm

3 5

D

400 mm

vA ⫽ 4 m/s

300 mm 30⬚ A

C

SOLUTION

General Plane Motion: Applying the relative velocity equation by referring to the kinematic diagram of link AB shown in Fig. a, vB = vA + V AB * rB>A 4 3 vB i = - 4j + ( -vAB k) * c - 0.55 a b i + 0.55a bj d 5 5 vB i = 0.33vAB i + (0.44vAB - 4) j Equating j component, 0 = 0.44vAB - 4

vAB = 9.091 rad>sb

teaching

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laws

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Using the result of vAB, in

vD = vA + V AB * rD>A

States

World

permitted.

Dissemination

Wide

copyright

3 4 = - 4j + ( - 9.091k) * c -0.3 a b i + 0.3 a bj d 5 5

use

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= {1.636i - 1.818j} m>s

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vC = vD + V CD * rC>D assessing

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vC i = (1.636i - 1.818j) + ( -vCD k) * (-0.4 cos 30° i - 0.4 sin 30° j)

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vC i = (1.636 - 0.2vCD)i + (0.3464vCD - 1.818)j

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Equating j and i components,

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0 = 0.3464vCD - 1.818

vC = 1.636 - 0.2(5.249) = 0.587 m>s :

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

16–77. The planetary gear system is used in an automatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, vR = 0, and the sun gear S is rotating at vS = 5 rad>s. Determine the angular velocity of each of the planet gears P and shaft A.

40 mm vR P

vS

R

S

SOLUTION

A

80 mm

vA = 5(80) = 400 mm>s ; vB = 0 vB = vA + v * rB>A 0 = -400i + (vp k) * (80j)

40 mm

0 = -400i - 80vp i vP = -5 rad>s = 5 rad>s

Ans.

vC = vB + v * rC>B

teaching

Web)

laws

or

vC = 0 + ( -5k) * (-40j) = - 200i in

200 = 1.67 rad>s 120

Ans.

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copyright

vA =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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16–78. If the ring gear D rotates counterclockwise with an angular velocity of vD = 5 rad>s while link AB rotates clockwise with an angular velocity of vAB = 10 rad>s, determine the angular velocity of gear C.

D 0.5 m A 0.125 m vAB

SOLUTION

10 rad/s C

Rotation About a Fixed Axis: Since link AB and gear D rotate about a fixed axis, Fig. a, the velocity of the center B and the contact point of gears D and C is 0.375 m

vB = vAB rB = 10(0.375) = 3.75 m>s vP = vD rP = 5(0.5) = 2.5 m>s General Plane Motion: Applying the relative velocity equation and referring to the kinematic diagram of gear C shown in Fig. b, vB = vP + vC * rB>P

laws

or

-3.75i = 2.5i + (vC k) * (0.125j)

in

teaching

Web)

-3.75i = (2.5 - 0.125vC)i

Dissemination

Wide

copyright

Thus,

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-3.75 = 2.5 - 0.125vC

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vC = 50 rad>s

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Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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B

16–79. The differential drum operates in such a manner that the rope is unwound from the small drum B and wound up on the large drum A. If the radii of the large and small drums are R and r, respectively, and for the pulley it is (R + r)>2, determine the speed at which the bucket C rises if the man rotates the handle with a constant angular velocity of v. Neglect the thickness of the rope.

A B

SOLUTION Rotation About a Fixed Axis: Since the datum rotates about a fixed axis, Fig. a, we obtain vB = vr and vA = vR. General Plane Motion: Applying the relative velocity equation and referring to the kinematic diagram of pulley shown in Fig. b,

C

vA = vB + V P * rA>B vRj = - vrj + ( - vPk) * ( - 2rPi) vRj = (-vr + 2vPrP)j Equating the j components, yields or

vR = - vr + 2vPrP teaching

Web)

laws

v(R + r) 2rP

Wide

copyright

in

vP =

permitted.

Dissemination

Since rP = (R + r)>2, then

on

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v(R + r) = v (R + r)

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vP =

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Using this result, we have

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vC = vB + V P * rC>B

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v = c (R - r) d j 2

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v(R + r) dj 2

This

= c -vr +

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= - vrj + ( -v k) * ( - rP i)

Thus, vC =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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v (R - r) c 2

Ans.

*16–80. Mechanical toy animals often use a walking mechanism as shown idealized in the figure. If the driving crank AB is propelled by a spring motor such that vAB = 5 rad>s, determine the velocity of the rear foot E at the instant shown. Although not part of this problem, the upper end of the foreleg has a slotted guide which is constrained by the fixed pin at G.

G

3 in. C

B

1 in.

2.5 in. 0.5 in. 50° A ω AB = 5 rad/s

D

SOLUTION

2 in.

vC = vB + vC>B

60°

vC = 2.5 + 3v T c30° c 50° + ) (:

E

F

yC cos 30° = 2.5 sin 50° + 0

vC = 2.21 in.>s 2.21 = 2.21 rad>s 1

vE = (2.21) (2) = 4.42 in.>s

Ans. laws

or

vEC =

in

teaching

Web)

Also:

permitted.

Dissemination

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copyright

vB = vAB * rB>A

is

of

the

not

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vC = vEC * rC>D

student

the

by

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learning.

vC = vB + v * rC>B

the

(including

for

work

(vEC k) * (cos 60°i + sin 60°j) = (- 5k) * (0.5 cos 50°i + 0.5 sin 50°j) + (vk) * ( -3i)

this

assessing

is

work

solely

of

protected

-0.866vEC = 1.915

the

is

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provided

integrity

of

and

work

0.5vEC = - 1.607 - 3v

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vEC = - 2.21 rad s v = - 0.167 rad s vE = 2(2.21) = 4.42 in. s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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45°

Ans.

16–81. In each case show graphically how to locate the instantaneous center of zero velocity of link AB. Assume the geometry is known.

B

V

A

A

(a) V

V A

SOLUTION

B

a) (c)

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

b)

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of

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not

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permitted.

c)

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(b)

C B

16–82. Determine the angular velocity of link AB at the instant shown if block C is moving upward at 12 in.>s.

A

VAB C 5 in.

45⬚

4 in. 30⬚

B

SOLUTION rIC-B rIC - C 4 = = sin 45° sin 30° sin 105° rIC-C = 5.464 in. rIC-B = 2.828 in. vC = vB C(rIC - C) 12 = vB C(5.464) vB C = 2.1962 rad>s

laws

or

vB = vB C(rIC-B)

in

teaching

Web)

= 2.1962(2.828) = 6.211 in.>s

Dissemination

Wide

copyright

vB = vAB rAB

of

the

not

instructors

States

World

permitted.

6.211 = vAB(5)

Ans.

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is

vAB = 1.24 rad>s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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16–83. At the instant shown, the disk is rotating at v = 4 rad>s. Determine the velocities of points A, B, and C.

D

C

A

SOLUTION

0.15 m

The instantaneous center is located at point A. Hence, vA = 0 rC>IC = 20.152 + 0.152 = 0.2121 m

Ans.

E

rB>IC = 0.3 m

vB = v rB>IC = 4(0.3) = 1.2 m>s

Ans.

vC = v rC>IC = 4(0.2121) = 0.849 m>s

Ans.

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c 45°

will

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sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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O

v ⫽ 4 rad/s B

*16–84. If link CD has an angular velocity of vCD = 6 rad>s, determine the velocity of point B on link BC and the angular velocity of link AB at the instant shown.

0.3 m 0.3 m C

B E

0.6 m 30

A

SOLUTION Rotation About Fixed Axis: Referring to Fig. a and b, vC = vCD rC = 6(0.6) = 3.60 m>s ; vB = vAB rB = vAB(1.2) 60° b

(1)

General Plane Motion: The location of IC for link BC is indicated in Fig. c. From the geometry of this figure, rC>IC = 0.6 tan 30° = 0.3464 m rB>IC =

0.6 = 0.6928 m cos 30°

Thus, the angular velocity of link BC can be determined from laws

or

3.60 = 10.39 rad>s 0.3464

Web)

rC>IC

=

teaching

vC

in

vBC =

vB = vBC rB>IC = 10.39 (0.6928) = 7.20 m>s

Ans. is

of

the

not

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States

World

60° b

the

by

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United

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learning.

Substitute this result into Eq. (1),

(including

for

work

student

7.20 = vAB (1.2)

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the

vAB = 6 rad>s d

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

permitted.

Dissemination

Wide

copyright

Then

vCD

6 rad/s D

16–85. If link CD has an angular velocity of vCD = 6 rad>s, determine the velocity of point E on link BC and the angular velocity of link AB at the instant shown.

0.3 m 0.3 m C

B E

0.6 m 30

A

SOLUTION vC = vCD (rCD) = (6)(0.6) = 3.60 m>s vBC =

vC 3.60 = 10.39 rad>s = rC>IC 0.6 tan 30°

vB = vBC rB>IC = (10.39) a vAB =

vB = rAB

0.6 b = 7.20 m>s cos 30°

7.20 = 6 rad>s 0.6 b a sin 30°

d

Ans.

vE = vBC rE>IC = 10.39 2(0.6 tan 30°)2 + (0.3)2 = 4.76 m>s

laws

or

Ans.

0.3 b = 40.9° b 0.6 tan 30°

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u = tan - 1 a

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vCD

6 rad/s D

16–86. At the instant shown, the truck travels to the right at 3 m>s, while the pipe rolls counterclockwise at v = 6 rad>s without slipping at B. Determine the velocity of the pipe’s center G.

G 1.5 m B

SOLUTION Kinematic Diagram: Since the pipe rolls without slipping, then the velocity of point B must be the same as that of the truck, i.e; yB = 3 m>s. Instantaneous Center: rB>IC must be determined first in order to locate the the instantaneous center of zero velocity of the pipe. yB = vrB>IC 3 = 6(rB>IC) rB>IC = 0.5 m

laws

or

Thus, rG>IC = 1.5 - rB>IC = 1.5 - 0.5 = 1.00 m. Then yG = vrG>IC = 6(1.00) = 6.00 m>s ;

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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3 m/s

16–87. If crank AB is rotating with an angular velocity of vAB = 6 rad>s, determine the velocity of the center O of the gear at the instant shown.

0.6 m

B

C O

0.4 m vAB 60

A

SOLUTION Rotation About a Fixed Axis: Referring to Fig. a, vB = vAB rB = 6(0.4) = 2.4 m>s General Plane Motion: Since the gear rack is stationary, the IC of the gear is located at the contact point between the gear and the rack, Fig. b. Thus, vO and vC can be related using the similar triangles shown in Fig. b, vg =

vC vO = rC>IC rO>IC

laws

or

vO vC = 0.2 0.1

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copyright

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vC = 2vO

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permitted.

Dissemination

The location of the IC for rod BC is indicated in Fig. c. From the geometry shown,

the

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0.6 = 1.2 m cos 60°

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rC>IC = 0.6 tan 60° = 1.039 m

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Thus, the angular velocity of rod BC can be determined from

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vB 2.4 = 2 rad>s = rB>IC 1.2 any

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vC = vBC rC>IC 2vO = 2(1.039) vO = 1.04 m>s :

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Ans.

6 rad/s

0.1 m 0.1 m

*16–88. If link AB is rotating at vAB = 6 rad>s, determine the angular velocities of links BC and CD at the instant u = 60°.

A 250 mm

ω AB = 6 rad/s

30° C

B 300 mm

400 mm

SOLUTION rIC - B = 0.3 cos 30° = 0.2598 m

θ

rIC - C = 0.3 cos 60° = 0.1500 m vBC =

1.5 = 5.774 = 5.77 rad>s 0.2598

Ans.

vC = 5.774(0.15) = 0.8661 m>s 0.8661 = 2.17 rad s 0.4

Ans.

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vCD =

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D

16–89. The oil pumping unit consists of a walking beam AB, connecting rod BC, and crank CD. If the crank rotates at a constant rate of 6 rad>s, determine the speed of the rod hanger H at the instant shown. Hint: Point B follows a circular path about point E and therefore the velocity of B is not vertical.

9 ft 1.5 ft

9 ft E A

B

9 ft

10 ft 3 ft 6 rad/s H

D C

SOLUTION 1.5 b = 9.462° and 9 rBE = 292 + 1.52 = 9.124 ft. Since crank CD and beam BE are rotating about fixed points D and E, then vC and vB are always directed perpendicular to crank CD and beam BE, respectively. The magnitude of vC and vB are yC = vCD rCD = 6(3) = 18.0 ft>s and yB = vBE rBE = 9.124vBE. At the instant shown, vC is directed vertically while vB is directed with an angle 9.462° with the vertical. u = tan - 1 a

geometry,

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Kinematic Diagram: From

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Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. From the geometry 10 = 60.83 ft sin 9.462°

rC>IC =

10 = 60.0 ft tan 9.462° destroy

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rB>IC =

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The angular velocity of link BC is given by vBC =

yC rC>IC

=

18.0 = 0.300 rad>s 60.0

Thus, the angular velocity of beam BE is given by yB = vBC rB>IC 9.124vBE = 0.300(60.83) vBE = 2.00 rad>s The speed of rod hanger H is given by yH = vBErEA = 2.00(9) = 18.0 ft>s

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Ans.

16–90. vB ⫽ 10 ft/s B

Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point D at this instant.

D E

45 C A

SOLUTION 1.6 - x x = 5 10 5x = 16 - 10x x = 1.06667 ft v =

10 = 9.375 rad>s 1.06667

rIC-D = 2(0.2667)2 + (0.8)2 - 2(0.2667)(0.8) cos 135° = 1.006 ft

laws

or

sin f sin 135° = 0.2667 1.006 in

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f = 10.80°

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copyright

vC = 0.2667(9.375) = 2.50 ft>s

permitted.

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vD = 1.006(9.375) = 9.43 ft>s

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u = 45° + 10.80° = 55.8°

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0.8 ft 30 F vA

5 ft/s

16–91. vB ⫽ 10 ft/s B

Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point E at this instant.

D E

45 C A

SOLUTION x 1.6 - x = 5 10 5x = 16 - 10x x = 1.06667 ft v =

10 = 9.375 rad>s 1.06667

vC = v(rIC - C)

or

= 9.375(1.06667 - 0.8) laws

= 2.50 ft>s

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vE = v(rIC - E)

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Dissemination

= 9.3752(0.8)2 + (0.26667)2

not

= 7.91 ft>s

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the

Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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0.8 ft 30 F vA ⫽ 5 ft/s

*16–92. Knowing that the angular velocity of link AB is vAB = 4 rad>s, determine the velocity of the collar at C and the angular velocity of link CB at the instant shown. Link CB is horizontal at this instant.

350 mm C

SOLUTION

B

45⬚

rIC-B rIC-C 0.350 = = sin 75° sin 45° sin 60°

60⬚

rIC-B = 0.2562 m rIC-C = 0.3138 m vCB =

A

2 = 7.8059 = 7.81 rad>s 0.2562

Ans.

vC = 7.8059(0.3138) = 2.45 m>s

any

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or

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destroy

of

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sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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500 mm

VAB

16–93. If the collar at C is moving downward to the left at vC = 8 m>s, determine the angular velocity of link AB at the instant shown. 350 mm C

SOLUTION

B

45⬚

rIC-B rIC-C 0.350 = = sin 75° sin 45° sin 60°

60⬚

rIC-B = 0.2562 m rIC-C = 0.3138 m vCB =

A

8 = 25.494 rad>s 0.3138

vB = 25.494(0.2562) = 6.5315 m>s 6.5315 = 13.1 rad>s 0.5

or

Ans.

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vAB =

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500 mm

VAB

16–94. If the roller is given a velocity of vA = 6 ft>s to the right, determine the angular velocity of the rod and the velocity of C at the instant shown.

C

B 6 ft

SOLUTION v =

4 ft

6 ft>s = 1.125 rad>s = 1.12 rad>s 5.334 ft

Ans.

vC = (1.125 rad>s)(3.003 ft) = 3.38 ft>s

Ans. 60⬚

vA ⫽ 6 ft/s

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laws

or

A

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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16–95. As the car travels forward at 80 ft/s on a wet road, due to slipping, the rear wheels have an angular velocity v = 100 rad>s. Determine the speeds of points A, B, and C caused by the motion.

80 ft/s C B 1.4 ft

SOLUTION 80 = 0.8 ft 100

r =

vA = 0.6(100) = 60.0 ft s :

Ans.

vC = 2.2(100) = 220 ft s ;

Ans. 60.3°

Ans.

b

sale

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Dissemination

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laws

or

vB = 1.612(100) = 161 ft s

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A 100 rad/s

*16–96. Determine the angular velocity of the double-tooth gear and the velocity of point C on the gear.

0.3 m C B

SOLUTION General Plane Motion: The location of the IC can be found using the similar triangles shown in Fig. a. rA>IC 4

0.45 - rA>IC =

rA>IC = 0.18 m

6

Then, y = 0.3 - rA>IC = 0.3 - 0.18 = 0.12 m

laws

or

and

in

teaching

Web)

rC>IC = 20.32 + 0.122 = 0.3231 m 0.12 b = 21.80° 0.3 learning.

is

of

the

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instructors

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permitted.

Dissemination

Wide

copyright

f = tan - 1 a

the

by

and

use

United

on

Thus, the angular velocity of the gear can be determined from

work

student

vA 4 = = 22.22 rad>s = 22.2 rad>s rA>IC 0.18

the

(including

for

Ans.

is

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v =

and

work

this

assessing

Then

integrity

of

vC = vrC>IC = 22.2(0.3231) = 7.18 m>s any

courses

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Ans.

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And its direction is

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f = 90° - f = 90° - 21.80° = 68.2°

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A

vA ⫽ 4 m/s

Ans.

0.15 m vB ⫽ 6 m/s

16–97. The wheel is rigidly attached to gear A, which is in mesh with gear racks D and E. If D has a velocity of vD = 6 f t>s to the right and the wheel rolls on track C without slipping, determine the velocity of gear rack E.

D vD

6 ft/s

1.5 ft A 0.75 ft

O C C

vE E

SOLUTION General Plane Motion: Since the wheel rolls without slipping on track C, the IC is located there, Fig. a. Here, rD>IC = 2.25 ft

rE>IC = 0.75 ft

Thus, the angular velocity of the gear can be determined from v =

vD 6 = 2.667 rad>s = rD>IC 2.25

Then, vE = vrE>IC = 2.667(0.75) = 2 ft>s ;

sale

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the

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Dissemination

Wide

copyright

in

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Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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16–98. The wheel is rigidly attached to gear A, which is in mesh with gear racks D and E. If the racks have a velocity of vD = 6 ft>s and vE = 10 ft>s, show that it is necessary for the wheel to slip on the fixed track C. Also find the angular velocity of the gear and the velocity of its center O.

D vD

6 ft/s

1.5 ft A 0.75 ft

O C C

vE E

SOLUTION General Plane Motion: The location of the IC can be found using the similar triangles shown in Fig. a, rD>IC 6

=

3 - rD>IC

rD>IC = 1.125 ft

10

Thus, rO>IC = 1.5 - rD>IC = 1.5 - 1.125 = 0.375ft rF>IC = 2.25 - rD>IC = 2.25 - 1.125 = 1.125ft

laws

or

Thus, the angular velocity of the gear is Web)

vD 6 = = 5.333 rad>s = 5.33 rad>s rD>IC 1.125

in

teaching

Ans. Dissemination

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v =

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instructors

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The velocity of the contact point F between the wheel and the track is

and

use

United

on

learning.

is

of

the

vF = vrF>IC = 5.333(1.125) = 6 ft>s ; by

Since vF Z 0, the wheel slips on the track

(including

for

work

student

the

(Q.E.D.)

is

work

solely

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protected

the

The velocity of center O of the gear is

assessing

vO = vrO>IC = 5.333(0.375) = 2ft>s ;

sale

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Ans.

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16–99. The epicyclic gear train is driven by the rotating link DE, which has an angular velocity vDE = 5 rad>s. If the ring gear F is fixed, determine the angular velocities of gears A, B, and C.

C E

30 mm 40 mm

B vDE ⫽ 5 rad/s A

SOLUTION vE = 0.

= 0.8 m>s

0 vC = 0.

26.7 rad>s

vP = (0.

6.7) = 1.6 m>s

F

Ans.

1.6 = x x

or

x = 0.05 ; = 28.75 rad>s

Ans. in

teaching

Web)

0

laws

vB =

Dissemination

Wide

copyright

vP¿ = 28.75(0.08 - 0.05565) = 0.700 m>s ;

permitted.

0.700 = 14.0 rad>s 0.05

the

not

instructors

States

World

Ans.

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vA =

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D

50 mm

*16–100. The similar links AB and CD rotate about the fixed pins at A and C. If AB has an angular velocity vAB = 8 rad>s, determine the angular velocity of BDP and the velocity of point P.

B

300 mm

300 mm

300 mm

300 mm 60°

60° A

C

ωAB = 8 rad/s 700 mm

SOLUTION Kinematic Diagram: Since link AB and CD is rotating about fixed points A and C. then vB and vD are always directed perpendicular to link AB and CD respectively. The magnitude of vB and vD are vB = vAB rAB = 8(0.3) = 2.40 m>s and vD = vCD rCD = 0.3 vCD. At the instant shown. vB and vD are directed at 30° with the horizontal.

P

Instantaneous Center: The instantaneous center of zero velocity of link BDP at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vD. From the geometry. 0.3 = 0.600 m cos 60° or

rB>IC =

in

teaching

Web)

laws

rP>IC = 0.3 tan 60° + 0.7 = 1.220 m

Dissemination

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copyright

The angular velocity of link BDP is given by

permitted.

vB 2.40 = 4.00 rad s = rB>IC 0.600

the

not

instructors

States

World

Ans.

use

United

on

learning.

is

of

vBDP =

student

the

by

and

Thus, the velocity of point P is given by

work

vP = vBDP rP IC = 4.00(1.220) = 4.88 m s ;

the

(including

for

Ans.

work

solely

of

protected

this

assessing

is

integrity

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and

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provided

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This

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D

16–101. If rod AB is rotating with an angular velocity vAB = 3 rad>s, determine the angular velocity of rod BC at the instant shown.

C 3 ft 4 ft

B 2 ft vAB

3 rad/s A

45

SOLUTION 4 sin 60° - 2 sin 45° b = 43.10°. 3 Since links AB and CD is rotating about fixed points A and D, then vB and vC are always Kinematic Diagram: From the geometry, u = sin - 1 a

directed perpendicular to links AB and CD, respectively. The magnitude of vB and vC are yB = vAB rAB = 3(2) = 6.00 ft>s and yC = vCDrCD = 4vCD. At the instant

laws

or

shown, vB is directed at an angle of 45° while vC is directed at 30°

Dissemination

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in

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Web)

Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. Using law of sines, we have 3 sin 75°

rB>IC = 3.025 ft

=

3 sin 75°

rC>IC = 0.1029 ft

the

not

instructors

States

World

permitted.

=

is and by

the

work

student

the

(including

sin 1.898°

for

rC>IC

use

United

on

learning.

sin 103.1°

of

rB>IC

assessing

is

work

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protected

The angular velocity of link BC is given by

provided

part

the

is

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will

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integrity

of

and

work

this

yB 6.00 = = 1.983 rad>s = 1.98 rad>s rB>IC 3.025 This

vBC =

Ans.

60

D

16–102. If rod AB is rotating with an angular velocity vAB = 3 rad>s, determine the angular velocity of rod CD at the instant shown.

C 3 ft 4 ft

B 2 ft vAB

3 rad/s A

45

SOLUTION 4sin 60° - 2 sin 45° b = 43.10°. 3 Since links AB and CD is rotating about fixed points A and D, then vB and vC are Kinematic Diagram: From the geometry. u = sin - 1 a

always directed perpendicular to links AB and CD, respectively. The magnitude of vB and vC are yB = vAB rAB = 3(2) = 6.00 ft>s and yC = vCD rCD = 4vCD. At the

Dissemination

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in

teaching

Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. Using law of sines, we have

Web)

laws

or

instant shown, vB is directed at an angle of 45° while vC is directed at 30°.

3 sin 75°

rB>IC = 3.025 ft

=

3 sin 75°

rC>IC = 0.1029 ft

the

not

instructors

States

World

permitted.

=

is United

on

learning.

sin 103.1°

of

rB>IC

by

work

student

the

the

(including

for

sin 1.898°

and

use

rC>IC

assessing

is

work

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The angular velocity of link BC is given by

provided

integrity

of

and

work

this

yB 6.00 = 1.983 rad>s = rB>IC 3.025

sale

yC = vBCrC>IC

will

their

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of

and

Thus, the angular velocity of link CD is given by

any

courses

part

the

is

This

vBC =

4vCD = 1.983(0.1029) vCD = 0.0510 rad>s

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Ans.

60

D

16–103. At a given instant the top end A of the bar has the velocity and acceleration shown. Determine the acceleration of the bottom B and the bar’s angular acceleration at this instant.

vA ⫽ 5 ft/s aA ⫽ 7 ft/s2

A

10 ft

SOLUTION

60 B

v =

5 = 1.00 rad>s 5

aB = aA + aB>A aB = 7 + 10 + a(10) : a 30° T h 30° + ) (:

aB = 0 - 10 sin 30° + a(10) cos 30°

(+ c)

0 = - 7 + 10 cos 30° + a(10) sin 30° Ans.

aB = - 7.875 ft>s2 = 7.88 ft>s2 ;

Ans. teaching

Web)

laws

or

a = - 0.3321 rad>s2 = 0.332 rad>s2 b

Wide

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in

Also:

on

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is

of

the

aBi = - 7j - (1)2(10 cos 60°i - sin 60°j) + (ak) * (10 cos 60°i - 10 sin 60°j)

not

instructors

States

World

permitted.

Dissemination

aB = aA - v2rB>A + a * rB>A

aB = - 10 cos 60° + a(10 sin 60°)

(+ c)

0 = - 7 + 10 sin 60° + a(10 cos 60°)

of

protected

the

(including

for

work

student

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+ :

Ans.

this

assessing

is

work

solely

a = -0.3321 rad>s2 = 0.332 rad>s2 b

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provided

integrity

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work

aB = - 7.875 ft>s2 = 7.88 ft>s2 ;

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Ans.

*16–104. At a given instant the bottom A of the ladder has an acceleration aA = 4 ft>s2 and velocity vA = 6 ft>s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant. 16 ft

SOLUTION v =

6 = 0.75 rad>s 8

30⬚

A

aB = aA + (aB>A)n + (aB>A)t aB = 4 + (0.75)2 (16) + a(16) ; 30° d 30° f T + ) (;

0 = 4 + (0.75)2(16) cos 30° - a(16) sin 30°

(+ T )

aB = 0 + (0.75)2(16) sin 30° + a(16) cos 30°

or

Solving, Ans.

aB = 24.9 ft>s2 T

Ans. Dissemination

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a = 1.47 rad>s2

instructors

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Also:

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on

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the

not

aB = aA + a * rB>A - v2rB>A

for

work

student

the

by

and

- aBj = - 4i + (ak) * (16 cos 30°i + 16 sin 30°j) - (0.75)2(16 cos 30°i + 16 sin 30°j)

solely

of

protected

the

(including

0 = - 4 - 8a - 7.794

this

assessing

is

work

- aB = 13.856a - 4.5

Ans.

part

the

is

This

provided

integrity

of

and

work

a = 1.47 rad>s2

any

courses

Ans.

sale

will

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destroy

of

and

aB = 24.9 ft>s2 T

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B

16–105. At a given instant the top B of the ladder has an acceleration aB = 2 ft>s2 and a velocity of vB = 4 ft>s, both acting downward. Determine the acceleration of the bottom A of the ladder, and the ladder’s angular acceleration at this instant. 16 ft

SOLUTION 4 = 0.288675 rad>s 16 cos 30°

v =

30⬚

A

aA = aB + a * rA>B - v2rA>B - aAi = - 2j + (ak) * ( - 16 cos 30°i - 16 sin 30°j) - (0.288675)2( -16 cos 30°i - 16 sin 30°j) - aA = 8a + 1.1547

Ans.

aA = -0.385 ft>s2 = 0.385 ft>s2 :

Ans.

sale

will

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of

and

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the

is

This

provided

integrity

of

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this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

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World

permitted.

Dissemination

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laws

a = -0.0962 rad>s2 = 0.0962 rad>s2b

or

0 = -2 - 13.856 a + 0.6667

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B

16–106. Crank AB is rotating with an angular velocity of vAB = 5 rad>s and an angular acceleration of aAB = 6 rad>s2. Determine the angular acceleration of BC and the acceleration of the slider block C at the instant shown.

0.5 m

C 0.3 m

SOLUTION Angular Velocity: Since crank AB rotates about a fixed axis, Fig. a, vB = vAB rB = 5(0.3) = 1.5 m>s : The location of the IC for link BC is indicated in Fig. b. From the geometry of this figure, rB>IC = 0.5 tan 45° = 0.5 m Then, or

vB 1.5 = = 3 rad>s rB>IC 0.5 in

teaching

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laws

vBC =

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World

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Dissemination

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Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis, Fig. c,

learning.

is

of

the

not

instructors

a B = aAB * rB - vAB2rB by

and

use

United

on

= ( -6k) * (0.3j) - 52(0.3j)

the

(including

for

work

student

the

= [1.8i - 7.5j] m>s2

is

work

solely

of

protected

Using this result and applying the relative acceleration equation by referring to Fig. d, integrity

of

and

work

this

assessing

aC = aB + aBC * rC>B - vBC2 rC>B part

the

is

This

provided

aC cos 45°i + aC sin 45°j = (1.8i - 7.5j) + (aBCk) * (0.5i) - 32(0.5i)

their

destroy

of

and

any

courses

0.7071aCi + 0.7071aCj = - 2.7i + (0.5aBC - 7.5)j sale

will

Equating the i and j components, 0.7071aC = - 2.7

(1)

0.7071aC = 0.5aBC -7.5

(2)

Solving Eqs. (1) and (2), aC = - 3.818 m>s2 = 3.82 m>s2 b

Ans.

aBC = 9.60 rad>s2d

Ans.

The negative sign indicates that aC acts in the opposite sense to that shown in Fig. c.

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45°

B

vAB ⫽ 5 rad/s aAB ⫽ 6 rad/s2 A

16–107. At a given instant, the slider block A has the velocity and deceleration shown. Determine the acceleration of block B and the angular acceleration of the link at this instant.

B 300 mm

45°

SOLUTION vAB =

vB rA>IC

A v A = 1.5 m/s

1.5 = 7.07 rad>s = 0.3 cos 45°

aB = aA + a * rB>A - v2 rB>A -aB j = 16i + (ak) * (0.3 cos 45°i + 0.3 sin 45° j) - (7.07)2 (0.3 cos 45°i + 0.3 sin 45°j) + b a: (+ T)

0 = 16 - a (0.3) sin 45° - (7.07)2 (0.3) cos 45° aB = 0 - a(0.3) cos 45° + (7.07)2 (0.3) sin 45°

Ans.

aB = 5.21 m s2

Ans.

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This

provided

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this

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is

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the

(including

for

work

student

the

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on

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of

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T

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aA>B = 25 4 rad>s2 d

or

Solving:

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aA = 16 m/s2

*16–108. As the cord unravels from the cylinder, the cylinder has an angular acceleration of a = 4 rad>s2 and an angular velocity of v = 2 rad>s at the instant shown. Determine the accelerations of points A and B at this instant. A

α = 4 rad/s2 ω = 2 rad/s

aC = 4(0.75) = 3 ft>s2

0.75 ft

T

aA = aC + a * rA>C - v2rA>C aA = - 3j + 4k * (0.75j) - (2)2 (0.75j) aA = { - 3i - 6j} ft>s2 aA = 2(-3)2 + ( - 6)2 = 6.71 ft>s2

Ans.

6 u = tan - 1 a b = 63.4°d 3

Ans.

laws

or

aB = aC + a * rB>C - v2 rB>C in

teaching

Web)

aB = - 3j + 4k * (- 0.75i) - (2)2 ( - 0.75i) Dissemination

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copyright

aB = {3i - 6j} ft>s2

permitted.

aB = 2(3)2 + ( -6)2 = 6.71 ft>s2

the

not

instructors

States

World

Ans. is

of sale

will

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of

and

any

courses

part

the

is

This

provided

integrity

of

and

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this

assessing

is

work

solely

of

protected

the

(including

for

work

student

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and

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on

learning.

6 f = tan - 1 a b = 63.4°c 3

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C

B

SOLUTION

Ans.

16–109. The hydraulic cylinder is extending with a velocity of vC = 3 ft>s and an acceleration of aC = 1.5 ft>s2. Determine the angular acceleration of links BC and AB at the instant shown.

B C 1.5 ft 45⬚

SOLUTION Angular Velocity: Since link AB rotates about a fixed axis, Fig. a, then vB = vAB rB = vAB (1.5) The location of the IC for link BC is indicated in Fig. b. From the geometry of this figure, 3 = 4.243 ft cos 45°

rB>IC =

rC>IC = 3 tan 45° = 3 ft Then

or

vC 3 = = 1 rad>s rC>IC 3 teaching

Web)

laws

vBC =

Wide

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Dissemination

vB = vBC rB>IC

is

of

the

not

instructors

States

vAB(1.5) = (1)(4.243)

the

by

and

use

United

on

learning.

vAB = 2.828 rad>s

of

protected

the

(including

for

work

student

Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis, Fig. c, then assessing

is

work

solely

a B = aAB * rB - vAB2rB provided

integrity

of

and

work

this

= (aAB k) * (- 1.5 cos 45°i + 1.5 sin 45°j) - 2.8282( -1.5 cos 45°i + 1.5 sin 45°j)

and

any

courses

part

the

is

This

= (8.485 - 1.061aAB)i - (8.485 + 1.061aAB)j

sale

will

their

destroy

of

Using this result and applying the relative acceleration equation by referring to Fig. d, aB = aC + aBC * rB>C - vBC2 rB>C

(8.485 - 1.061aAB)i - (8.485 + 1.061aAB)j = - 1.5i + (aBCk) * (-3i) - 12( -3i) (8.485 - 1.061aAB)i - (8.485 + 1.061aAB)j = 1.5i - 3aBC j Equating the i and j components, yields 8.485 - 1.061aAB = 1.5

(1)

-(8.485 + 1.061aAB) = - 3aBC

(2)

Solving Eqs. (1) and (2),

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vC ⫽ 3 ft/s aC ⫽ 1.5 ft/s2 D

3 ft

aAB = 6.59 rad>s2

Ans.

aBC = 5.16 rad>s2

Ans.

A

16–110. At a given instant the wheel is rotating with the angular motions shown. Determine the acceleration of the collar at A at this instant.

A 60° 500 mm

ω = 8 rad/s α = 16 rad/s2

150 mm

B

SOLUTION

30°

Using instantaneous center method: 8(0.15) vB vAB = = = 4.157 rad>s rB>IC 0.5 tan 30° aA = aB + aA>B = 2.4 + 9.6 + (4.157 )2(0.5) 60°d a60° c30°

aA ;

+ ) (:

+ a(0.5) a60°

-aA = 2.4 cos 60° + 9.6 cos 30° - 8.65 cos 60° - a(0.5) sin 60°

( + c ) 0 = 2.4 sin 60° - 9.6 sin 30° - 8.65 sin 60° + a(0.5) cos 60° a = 40.8 rad>s2 d or

aA = 12.5 m>s2 ;

teaching

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laws

Ans.

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Dissemination

aA = aB + a * rA>B - v2 rA>B

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of

the

not

aA i = (8)2(0.15)(cos 30°)i - (8)2(0.15) sin 30°j + (16)(0.15) sin 30°i + (16)(0.15) cos 30° j

student

the

by

and

+ (ak) * (0.5 cos 60°i + 0.5 sin 60°j) - (4.157 2) (0.5 cos 60°i + 0.5 sin 60°j)

of

protected

the

(including

for

work

aA = 8.314 + 1.200 - 0.433 a - 4.326

this

assessing

is

work

solely

0 = - 4.800 + 2.0785 + 0.25a - 7.4935

the

is

This

provided

integrity

of

and

work

a = 40.8 rad s2 d

part

aA = 12.5 m s2 ;

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courses

Ans.

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16–111. Crank AB rotates with the angular velocity and angular acceleration shown. Determine the acceleration of the slider block C at the instant shown.

0.4 m B

vAB ⫽ 4 rad/s aAB ⫽ 2 rad/s2 30⬚ 0.4 m

A

SOLUTION Angular Velocity: Since crank AB rotates about a fixed axis, Fig. a, vB = vAB rB = 4(0.4) = 1.6 m>s The location of the IC for link BC is indicated in Fig. b. From the geometry of this figure, rB>IC = 0.4 m Then

or

vB 1.6 = = 4 rad>s rB>IC 0.4 laws

vBC =

Wide

copyright

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Web)

Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis, Fig. a States

World

permitted.

Dissemination

aB = aAB * rB - vAB2rB learning.

is

of

the

not

instructors

= (- 2k) * (0.4 cos 30°i + 0.4 sin 30°j) - 4 2(0.4 cos 30°i + 0.4 sin 30°j)

student

the

by

and

use

United

on

= [- 5.143i - 3.893j] m>s2

the

(including

for

work

Using this result and applying the relative acceleration equation by referring to Fig. c,

assessing

is

work

solely

of

protected

aC = aB + aBC * rC>B - vBC2 rC>B

provided

integrity

of

and

work

this

aCi = (- 5.143i - 3.893j) + (aBCk) * (0.4 cos 30°i - 0.4 sin 30°j) - 42(0.4 cos 30°i - 0.4 sin 30°j)

and

any

courses

part

the

is

This

aCi = (0.2aBC - 10.69)i + (0.3464aBC - 0.6928)j

0 = 0.3464aBC - 0.6928

sale

aC = 0.2aBC - 10.69

will

their

destroy

of

Equating the i and j components, yields

(1) (2)

Solving Eqs. (1) and (2), aBC = 2 rad>s2 aC = - 10.29 m>s2 = 10.3 m>s2 ;

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Ans.

30⬚ C

*16–112. D

The wheel is moving to the right such that it has an angular velocity v = 2 rad>s and angular acceleration a = 4 rad>s2 at the instant shown. If it does not slip at A, determine the acceleration of point B.

B 4 rad/s2 2 rad/s

a v

1.45 ft

60

30

C

SOLUTION Since no slipping aC = ar = 4(1.45) = 5.80 ft>s2 aB = aC + aB>C

A

aB = 5.80 + (2)2(1.45) + 4(1.45) : c 30° g 30° + ) (:

(aB)x = 5.80 + 5.02 + 2.9 = 13.72

(+ c )

(aB)y = 0 - 2.9 + 5.02 = 2.12

laws

or

aB = 2(13.72)2 + (2.12)2 = 13.9 ft>s2 2.123 b = 8.80° a u 13.72

in

teaching

Web)

Ans. Wide

copyright

u = tan - 1 a

instructors

States

World

permitted.

Dissemination

Also:

United

on

learning.

is

of

the

not

aB = aC + a * rB>C - v2rB>C

student

the

by

and

use

aB = 5.80i + (-4k) * ( - 1.45 cos 30°i + 1.45 sin 30°j) - (2)2( - 1.45 cos 30°i + 1.45 sin 30°j)

protected

the

(including

for

work

aB = {13.72i + 2.123j} ft>s2

integrity

of

and provided

Ans.

part

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and

any

courses

2.123 b = 8.80° a u 13.72

This

u = tan - 1 a

work

this

assessing

is

work

solely

of

aB = 2(13.72)2 + (2.123)2 = 13.9 ft>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

16–113. v ⫽ 3 rad/s a ⫽ 8 rad/s2

The disk is moving to the left such that it has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point B.

D 30⬚

45⬚

0.5 m B

A

SOLUTION aC = 0.5(8) = 4 m>s2 aB = aC + aB>C

+ B A:

(aB)x = -4 + 4.5 cos 30° + 4 sin 30° = 1.897 m>s2

A+cB

(aB)y = 0 + 4.5 sin 30° - 4 cos 30° = -1.214 m>s2 Ans.

1.214 b = 32.6° c 1.897

Ans.

teaching

in

Dissemination

Wide

copyright

u = tan - 1 a

Web)

laws

aB = 2(1.897)2 + ( - 1.214)2 = 2.25 m>s2

or

aB = c 4 d + D (3)2 (0.5) T + D (0.5) (8 ) T ; f 30° a 30°

instructors

States

World

permitted.

Also,

use

United

on

learning.

is

of

the

not

aB = aC + a * rB>C - v2 rB>C

student

the

by

and

(aB)x i + (aB)y j = - 4i + (8k) * ( - 0.5 cos 30°i - 0.5 sin 30°j) - (3)2 (-0.5 cos 30°i - 0.5 sin 30°j) (aB)x = -4 + 8(0.5 sin 30°) + (3)2(0.5 cos 30°) = 1.897 m>s2

A+cB

(aB)y = 0 - 8(0.5 cos 30°) + (3)2 (0.5 sin 30°) = - 1.214 m>s2 integrity

of provided

part

the

is

Ans.

and

any

courses

1.214 b = 32.6° c 1.897

This

u = tan - 1 a

and

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this

assessing

is

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protected

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+ B A:

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aB = 2(1.897)2 + ( - 1.214)2 = 2.25 m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

C

16–114. v ⫽ 3 rad/s a ⫽ 8 rad/s2

The disk is moving to the left such that it has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point D.

D 30⬚

45⬚

0.5 m B

SOLUTION aC = 0.5(8) = 4 m>s2 aD = aC + aD>C aD = c 4 d + D (3)2 (0.5 )T + D 8 (0.5) T ; 45° b e 45° + B A:

(aD)x = - 4 - 4.5 sin 45° - 4 cos 45° = -10.01 m>s2

A+cB

(aD)y = 0 - 4.5 cos 45° + 4 sin 45° = - 0.3536 m>s2 0.3536 b = 2.02° d 10.01

or

Ans. laws

u = tan - 1 a

teaching

Web)

aD = 2( -10.01)2 + (- 0.3536)2 = 10.0 m>s2

Wide

copyright

in

Ans.

States

World

permitted.

Dissemination

Also,

on

learning.

is

of

the

not

instructors

aD = aC + a * rD>C - v2 rD>C

the

by

and

use

United

(aD)x i + (aD)y j = - 4i + (8k) * (0.5 cos 45°i + 0.5 sin 45°j) - (3)2 (0.5 cos 45°i + 0.5 sin 45°j) (aD)x = - 4 - 8(0.5 sin 45°) - (3)2(0.5 cos 45°) = - 10.01 m>s2

A+cB

(aD)y = +8(0.5 cos 45°) - (3)2 (0.5 sin 45°) = -0.3536 m>s2 this

assessing

is

work

solely

of

protected

the

(including

for

work

student

+ B A:

provided

integrity

of

and

work

0.3536 b = 2.02° d 10.01

Ans.

part

the

is

This

u = tan - 1 a

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destroy

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and

any

courses

aD = 2( -10.01)2 + ( -0.3536)2 = 10.0 m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

A

C

16–115. A cord is wrapped around the inner spool of the gear. If it is pulled with a constant velocity v, determine the velocities and accelerations of points A and B. The gear rolls on the fixed gear rack.

B 2r A

r

G v

SOLUTION Velocity analysis: v r

v =

vB = vrB>IC =

v (4r) = 4v : r

vA = v rA>IC =

v A 2(2r)2 + (2r)2 B = 2 22v r

Ans.

Ans.

a45°

or

Acceleration equation: From Example 16–3, Since aG = 0, a = 0 rA>G = - 2r i

teaching

Web)

laws

rB>G = 2r j

Wide

copyright

in

aB = aG + a * rB>G - v2rB>G

is

of

the

not

instructors

States

World

permitted.

Dissemination

2v2 v 2 j = 0 + 0 - a b (2rj) = r r

Ans.

by

and

use

United

on

learning.

2v2 T r

for

work

student

the

aB =

solely

of

protected

the

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a A = aG + a * rA>G - v2rA>G

part

the

is

any

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Ans.

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destroy

of

2v2 : r

and

aA =

This

provided

integrity

of

and

work

this

assessing

is

work

2v2 v 2 i = 0 + 0 - a b ( - 2ri) = r r

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*16–116. At a given instant, the gear racks have the velocities and accelerations shown. Determine the acceleration of point A.

a v

2 ft/s2 6 ft/s

SOLUTION Velocity Analysis: The angular velocity of the gear can be obtained by using the method of instantaneous center of zero velocity. From similar triangles, yD yC v = = rD>IC rC>IC 6 rD>IC

=

2

A 0.25 ft

(1)

rC>IC

B

Where rD>IC + rC>IC = 0.5

(2)

a v

Solving Eqs.(1) and (2) yields rD>IC = 0.375 ft Thus,

yD 6 = 16.0 rad>s = rD>IC 0.375

Dissemination

Wide

copyright

in

teaching

Acceleration Equation: The angular acceleration of the gear can be obtained by analyzing the angular motion of points C and D. Applying Eq. 16–18 with rD>C = { -0.5i} ft, we have

Web)

laws

or

v =

rC>IC = 0.125 ft

not

instructors

States

World

permitted.

aD = aC + a * rD>C - v2 rD>C

use

United

on

learning.

is

of

the

(aD)ni + 2j = -(aC)ni - 3j + (- ak) * ( -0.5i) - 16.02 (-0.5i)

for

work

student

the

by

and

(aD)ni + 2j = -(aC)ni + (0.5a - 3)j + 128i

of

protected

the

(including

Equating the j components, we have a = 10.0 rad>s2 this

assessing

is

work

solely

2 = 0.5 a - 3

any

destroy

of

and

aA = aC + a * rA>C - v2 rA>C

courses

part

the

is

This

provided

integrity

of

and

work

The acceleration of point A can be obtained by analyzing the angular motion of points A and C. Applying Eq. 16–18 with rA>C = {- 0.25i} ft, we have

sale

will

their

aAj = - (aC)ni - 3j + (-10.0k) * ( - 0.25i) - 16.02 (-0.25i) Equating the i and j components, we have aA = 0.500 ft>s2 T (aC)n = 64 m>s2 d

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

3 ft/s2 2 ft/s

16–117. At a given instant, the gear racks have the velocities and accelerations shown. Determine the acceleration of point B.

a ⫽ 2 ft/s2 v ⫽ 6 ft/s

A

SOLUTION 0.25 ft

Angular Velocity: The method of IC will be used. The location of IC for the gear is indicated in Fig. a. using the similar triangle, 2 6 = rD>IC 0.5 - rD>IC

B

rD>IC = 0.125 ft a ⫽ 3 ft/s2 v ⫽ 2 ft/s

Thus, v =

vD rD>IC

=

2 = 16 rad>s b 0.125 ft

laws

or

Acceleration and Angular Acceleration: Applying the relative acceleration equation for points C and D by referring to Fig. b, in

teaching

Web)

aC = aD + A * rC>D - v2rC>D Dissemination

Wide

copyright

(aC)ni + 2j = - (aD)ni - 3j + ( - ak) * ( -0.5i) - 162(-0.5i)

of

the

not

instructors

States

World

permitted.

(aC)ni + 2j = [128 - (aD)n]i + (0.5a - 3)j

and

use

United

on

learning.

is

Equating j component,

(including

for

work

student

the

by

2 = 0.5a - 3

assessing

is

work

solely

of

protected

the

Using this result, the relative acceleration equation applied to points A and C, Fig. b, gives

provided

integrity

of

and

work

this

aC = aA + A * rC>A - v2rC>A

any their

destroy

of

and

(aC)ni + 2j = 64i + (aA + 2.5)j

courses

part

the

is

This

(aC)ni + 2j = aAj + (- 10k) * ( -0.25i) - 162( -0.25i)

2 = aA + 2.5

sale

will

Equating j component,

aA = - 0.5 ft>s2 = 0.5 ft>s2 T

Using this result to apply the relative acceleration equation to points A and B, aB = aA + A * rB>A - v2rB>A - (aB)ti + (aB)nj = - 0.5j + ( -10k) * ( - 0.25j) - 162(- 0.25j) -(aB)ti + (aB)n j = - 2.5i + 63.5j Equating i and j components, (aB)t = 2.50 ft>s2

(aB)n = 63.5 ft>s2

Thus, the magnitude of aB is aB = 2(aB)2t + (aB)2n = 22.502 + 63.52 = 63.55 ft>s2

Ans.

and its direction is u = tan - 1 c © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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(aB)n 63.5 d = tan - 1 a b = 87.7° ub (aB)t 2.50

Ans.

16–118. At a given instant gears A and B have the angular motions shown. Determine the angular acceleration of gear C and the acceleration of its center point D at this instant. Note that the inner hub of gear C is in mesh with gear A and its outer rim is in mesh with gear B.

5 in. A

vA ⫽ 4 rad/s aA ⫽ 8 rad/s2

20 in. D 5 in.

SOLUTION

B

aP = aP¿ + aP>P¿ + ) (:

vB ⫽ 1 rad/s aB ⫽ 6 rad/s2

120 = -40 + a(15)

a = 10.67 rad>s2 d

Ans.

aP = aD + aP>D + ) (:

120 = (aD)t + (10.67)(10)

(aD)t = 13.3 in. >s2 : vP = vP¿ + vP>P¿ or

20 = -20 + v(15)

Web)

laws

+ ) (:

Wide

copyright

in

teaching

v = 2.667 rad>s

States

the

not

instructors

vD = -20 + 10(2.667)

on

learning.

is

of

+ ) (;

World

permitted.

Dissemination

vD = vP + vD>P

the

by

and

use

United

vD = 6.67 in. >s

the

(including

for

work

student

(6.67)2 = 4.44 in. >s2 c 10 is

work

solely

of

protected

(aD)n =

provided

integrity

of

and

work

this

assessing

4.44 ) = 18.4° 13.3

part

the

is

This

u = tan - 1 (

Ans. any

courses

(4.44)2 + (13.3)2 = 14.1 in. s2

sale

will

their

destroy

of

and

aD =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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C

10 in.

16–119. The wheel rolls without slipping such that at the instant shown it has an angular velocity V and angular acceleration A. Determine the velocity and acceleration of point B on the rod at this instant.

A 2a B

SOLUTION vB = vA + vB/A (Pin) + v = ; B

1 Qv 22aR + 2av¿ a b 2 22

+c O = v¿ =

1

1 22

Qv22aR + 2av¿ a

23 b 2

v 23

vB = 1.58 va

Ans.

Web)

laws

or

a A = aO + aA/O (Pin)

T

in

:

Wide

;

Dissemination

T

copyright

;

teaching

(a A)x + (aA)y = aa + a(a) + v2a

is

of

the

not

instructors

States

World

permitted.

(a A)x = aa - v2a

by

and

use

United

on

learning.

(a A)y = aa

(including

for

work

student

the

a B = aA + aB/A (Pin)

work

this

assessing

is

work

solely

of

protected

the

v 2 23 1 a B = aa - v2a + 2a(a¿)a b - 2a a b 2 2 23 2 1 b a b 2 23

provided

integrity

of

and

v

the part

and

any

courses

23

b + 2a a

is

2

This

O = -aa + 2aa¿ a

a B = 1.58aa - 1.77v2a

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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sale

will

their

destroy

of

a¿ = 0.577a - 0.1925v2

Ans.

O a

V, A

*16–120. The center O of the gear and the gear rack P move with the velocities and accelerations shown. Determine the angular acceleration of the gear and the acceleration of point B located at the rim of the gear at the instant shown.

B

vO ⫽ 3 m/s aO ⫽ 6 m/s2 vP ⫽ 2 m/s aP ⫽ 3 m/s2

O

A

SOLUTION Angular Velocity: The location of the IC is indicated in Fig. a. Using similar triangles, 3 2 = rO>IC 0.15 - rO>IC

rO>IC = 0.09 m

Thus, v =

vO 3 = = 33.33 rad>s rO>IC 0.09

Acceleration and Angular Acceleration: Applying the relative acceleration equation to points O and A and referring to Fig. b,

laws

or

aA = aO + a * rA>O - v2rA>O

in

teaching

Web)

-3i + (aA)n j = 6i + ( -ak) * (- 0.15j) - 33.332( -0.15j)

permitted.

Dissemination

Wide

copyright

- 3i + (aA)n j = (6 - 0.15a)i + 166.67j

learning.

is

of

the

not

instructors

States

World

Equating the i components,

the

by

and

use

United

on

- 3 = 6 - 0.15a

Ans.

the

(including

for

work

student

a = 60 rad>s2

and

work

this

assessing

is

work

solely

of

protected

Using this result, the relative acceleration equation is applied to points O and B, Fig. b, which gives

part

the

is

This

provided

integrity

of

aB = aO + a * rB>O - v2rB>O

destroy

of

and

any

courses

(aB)ti - (aB)n j = 6i + ( - 60k) * (0.15j) - 33.332(0.15j) sale

will

their

(aB)ti - (aB)n j = 15i - 166.67j Equating the i and j components, (aB)t = 15 m>s2

(aB)n = 166.67 m>s2

Thus, the magnitude of aB is aB = 3(aB)t 2 + (aB)n 2 = 3152 + 166.672 = 167 m>s2 and its direction is u = tan-1 c

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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(aB)n 166.67 d = tan-1 a b = 84.9° c (aB)t 15

150 mm

Ans.

P

16–121. The tied crank and gear mechanism gives rocking motion to crank AC, necessary for the operation of a printing press. If link DE has the angular motion shown, determine the respective angular velocities of gear F and crank AC at this instant, and the angular acceleration of crank AC.

C F

100 mm

50 mm

v DE ⫽ 4 rad/s

75 mm

SOLUTION

B

100 mm

Velocity analysis:

G 30

yD = vDErD>E = 4(0.1) = 0.4 m>s c 150 mm

vB = vD + vB>D yB = 0.4 + (vG)(0.075) c

a 30°

+ ) (:

T

yB cos 30° = 0,

yB = 0

(+ c ) vG = 5.33 rad>s vAC = 0

Ans.

or

yC = 0,

laws

Since yB = 0,

in

teaching

Web)

vFrF = vGrG

Wide

copyright

100 b = 10.7 rad>s 50

permitted.

Dissemination

Ans. of

the

not

instructors

States

World

vF = 5.33 a

and

use

United

on

learning.

is

Acceleration analysis:

for

work

student

the

by

(aD)n = (4)2(0.1) = 1.6 m>s2 :

is

work

solely

of

protected

the

(including

(aD)t = (20)(0.1) = 2 m>s2 c

and

work

this

assessing

(aB)n + (aB)t = (aD)n + (aD)t + (aB>D)n + (aB>D)t provided

integrity

of

= 1.6 + 2 + (5.33)2(0.075) + aG (0.075) :

part

c

the

is

:

c and

any

courses

a 30°

This

0 + (a B)t

(aB)t sin 30° = 0 + 2 + 0 + aG (0.075)

+ ) (:

(aB)t cos 30° = 1.6 + 0 + (5.33)2(0.075) + 0 sale

will

their

destroy

of

(+ c )

Solving, (aB)t = 4.31 m>s2,

aG = 2.052 rad>s2b

Hence, aAC =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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E D

(a B)t 4.31 = 28.7 rad>s2b = rB>A 0.15

Ans.

A

aDE ⫽ 20 rad/s2

16–122. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the angular acceleration of pulley B at the instant shown.

50 mm

vA aA

40 rad/s 5 rad/s2

SOLUTION

50 mm

Angular Velocity: Since pulley A rotates about a fixed axis,

B

vC = vA rA = 40(0.05) = 2 m>s c

125 mm

The location of the IC is indicated in Fig. a. Thus, vB =

E

vC 2 = 11.43 rad>s = rC>IC 0.175

Acceleration and Angular Acceleration: For pulley A,

laws

or

(aC)t = aArA = 5(0.05) = 0.25 m>s2 c

in

teaching

Web)

Using this result and applying the relative acceleration equation to points C and D by referring to Fig. b,

permitted.

Dissemination

Wide

copyright

aD = aC + aB * rD>C - vB 2rD>C

is

of

the

not

instructors

States

World

(aD)n i = (aC)n i + 0.25j + ( -aB k) * (0.175i)-11.432(0.175i)

the

by

and

use

United

on

learning.

(aD)n i = [(aC)n - 22.86]i + (0.25 - 0.175aB)j

the

(including

for

work

student

Equating the j components,

assessing

is

work

solely

of

protected

0 = 0.25 - 0.175aB

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

aB = 1.43 rad>s2

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16–123. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the acceleration of block E at the instant shown.

50 mm

vA aA

40 rad/s 5 rad/s2

SOLUTION

50 mm

Angular Velocity: Since pulley A rotates about a fixed axis,

B

vC = vArA = 40(0.05) = 2 m>s c

125 mm

The location of the IC is indicated in Fig. a. Thus, vB =

E

vC 2 = 11.43 rad>s = rC>IC 0.175

Acceleration and Angular Acceleration: For pulley A,

laws

or

(aC)t = aA rA = 5(0.05) = 0.25 m>s2 c

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permitted.

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aD = aC + aB * rD>C - vB 2rD>C

is

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(aD)n i = (aC)n i + 0.25j + ( - aBk) * (0.175i) - 11.432(0.175i)

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(aD)n i = [(aC)n - 22.86]i + (0.25 - 0.175aB)j

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Equating the j components,

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aB = 1.429 rad>s = 1.43 rad>s2

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0 = 0.25 - 0.175a B

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aE = aC + aB * rE>C - vB 2rE>C

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Using this result, the relative acceleration equation applied to points C and E, Fig. b, gives

aE j = [(aC)n i + 0.25j] + (- 1.429k) * (0.125i) - 11.432(0.125i) aE j = [(aC)n - 16.33]i + 0.0714j Equating the j components, aE = 0.0714 m>s2 c

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or transmission in any form or by any means, electronic, mechanical,

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Ans.

*16–124. At a given instant, the gear has the angular motion shown. Determine the accelerations of points A and B on the link and the link’s angular acceleration at this instant.

B 60⬚ 3 in.

v ⫽ 6 rad/s a ⫽ 12 rad/s2

2 in.

SOLUTION

yA = vrA>IC = 6(1) = 6 in.>s aO = -12(3)i = { - 36i} in.>s2

rA>O = {- 2j} in.

a = {12k} rad>s2

aA = a0 + a * rA>O - v2rA>O = -36i + (12k) * ( -2j) - (6)2( - 2j) = { -12i + 72j} in.>s2 aA = 2(- 12)2 + 722 = 73.0 in.>s2

or

Ans. laws

72 b = 80.5° b 12

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Ans.

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For link AB

use by

and

6 = 0 q

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rB>A = {8 cos 60°i + 8 sin 60°j} in.

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aAB = - aAB k

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aB = aB i

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The IC is at q , so vAB = 0, i.e., vAB =

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aB = aA + aAB * rB>A - v2 rB>A

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aB i = (- 12i + 72j) + ( - aAB k) * (8 cos 60°i + 8 sin 60°j) - 0 aB = - 12 + 8 sin 60°(18) = 113 in.>s2 :

(+ c)

0 = 72 - 8 cos 60°aAB

Ans.

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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their

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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

8 in. A

For the gear

u = tan-1 a

O

sale

will

aAB = 18 rad>s2 b

Ans.

16–125. The ends of the bar AB are confined to move along the paths shown. At a given instant, A has a velocity of vA = 4 ft>s and an acceleration of a A = 7 ft>s2. Determine the angular velocity and angular acceleration of AB at this instant.

B

2 ft 60

2 ft

SOLUTION vA aA

vB = vA + vB>A 30°

4 ft/s 7 ft/s2

= 4+ v(4.788) T h 51.21°

vB b

+ ) (:

-vB cos 30° = 0 - v(4.788) sin 51.21°

(+ c )

vB sin 30° = - 4 + v(4.788) cos 51.21°

vB = 20.39 ft>s

30° b

v = 4.73 rad>sd

Ans.

laws

or

a B = a A + a B>A = 7 + 107.2 + 4.788(a) d 51.21° T h 51.21°

Web)

2 07.9 60° d

teaching

+

in

at 30° b

at cos 30° + 207.9 cos 60° = 0 + 107.2 cos 51.21° + 4.788a(sin 51.21°)

(+ c)

at sin 30° - 207.9 sin 60° = - 7 - 107.2 sin 51.21° + 4.788a(cos 51.21°) on

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at(0.866) - 3.732a = -36.78

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at = - 607 ft>s2

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a = -131 rad>s2 = 131 rad>s2b

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vB = vA = v * rB>A

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Also:

-vB cos 30°i + vB sin 30°j = - 4j + (vk) * (3i + 3.732j) -vB cos 30° = - v(3.732) vB sin 30° = -4 + v(3) v = 4.73 rad>sd

Ans.

vB = 20.39 ft>s a B = a A - v2rB>A + a * rB>A (-at cos 30°i + at sin 30°j) + ( -207.9 cos 60°i - 207.9 sin 60°j) = - 7j - (4.732)2(3i + 3.732j) + (ak) * (3i + 3.732j) -at cos 30° - 207.9 cos 60° = - (4.732)2(3) - a(3.732) at sin 30° - 207.9 sin 60° = - 7 -(4.732)2(3.732) + a(3) at = - 607 ft>s2 a = -131 rad>s2 = 131 rad>s2b © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

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Ans.

A

16–126. At a given instant, the cables supporting the pipe have the motions shown. Determine the angular velocity and angular acceleration of the pipe and the velocity and acceleration of point B located on the pipe.

v

a

5 ft/s

2 ft

v = 0.25 rad>s b

Ans.

vB = 5.00 ft>s T

Ans.

aB = aA + aB>A 1.5 + (aB)x = 2 + (aA)x + (a) (4) + (0.25) 2 (4) :

T

:

T

1.5 = - 2 + a(4) laws

or

(+ T )

;

c

a = 0.875 rad>s2 d

in

teaching

Web)

Ans.

Dissemination

Wide

copyright

aB = aO + aB>O

World

:

of

the

not

T

T

permitted.

= aO + 0.875(2) + (0.25)2 (2)

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T

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1.5 + (aB)x

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+ ) (a ) = (0.25)2(2) = 0.125 ft>s2 (: B x

Ans.

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Ans.

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1.5 b = 85.2° 0.125

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aB = 2(1.5)2 + (0.125)2 = 1.51 ft>s2 u = tan - 1 a

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5 = 0.25 rad>s 20

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v =

vB = 5.00 ft>s T

Ans.

aB = aA + a * rB>A - v2rB>A - 1.5j + (aB)x i = 2j - (aA)x i + (ak) * ( - 4i) - (0.25)2(- 4i) - 1.5 = 2 - 4a a = 0.875 rad>s2 d

Ans.

aB = aO + a * rB>O - v2rB>O - 1.5j + (aB)x i = -aO j + (ak) * ( - 2i) - (0.25)2 ( - 2i) (aB)x = (0.25)2(2) = 0.125 aB = 2(1.5)2 + (0.125)2 = 1.51 ft>s2 u = tan - 1

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.5 0.125

O

B

( + T ) 5 = 6 - v (4)

= 85.2°

cu

Ans. Ans.

6 ft/s

a

2 ft/s2

1.5 ft/s2

SOLUTION vB = vA + vB>A

v

A

16–127. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the angular acceleration of rod AB at the instant shown. 1.5 ft A vB aB

30 2 ft B

SOLUTION Angular Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vAB =

vB rB>IC

=

5 = 5 rad>s 1 laws

or

Then

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vA = vAB rA>IC = 5(1.732) = 8.660 ft>s

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permitted.

Acceleration and Angular Acceleration: Since point A travels along the circular

the

(including

for

work

student

the

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learning.

is

of

the

slot, the normal component of its acceleration has a magnitude of vA 2 8.6602 = 50 ft>s2 and is directed towards the center of the circular (aA)n = = r 1.5 slot. The tangential component is directed along the tangent of the slot. Applying

assessing

is

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the relative acceleration equation and referring to Fig. b,

provided

integrity

of

and

work

this

aA = aB + aAB * rA>B - vAB 2 rA>B

courses

part

the

is

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50i - (aA)t j = 3i + (aAB k) * ( - 2 cos 30°i + 2 sin 30°j) - 52(- 2 cos 30° i + 2 sin 30°j)

Equating the i components,

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50i - (aA)t j = (46.30 - aAB)i + (1.732aAB + 25)j

50 = 46.30 - aAB aAB = -3.70 rad>s2 = 3.70 rad>s2 b

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

5 ft/s 3 ft/s2

*16–128. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the acceleration of A at the instant shown. 1.5 ft A vB aB

30 2 ft B

SOLUTION Angualr Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vB 5 = 5 rad>s = rB>IC 1 or

vAB =

teaching

Web)

laws

Then

permitted.

Dissemination

Wide

copyright

in

vA = vAB rA>IC = 5 A 1.732 B = 8.660 ft>s

of

the

not

instructors

States

World

Acceleration and Angular Acceleration: Since point A travels along the circular

of

protected

the

(including

for

work

student

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on

learning.

is

slot, the normal component of its acceleration has a magnitude of vA 2 8.6602 = = 50 ft>s2 and is directed towards the center of the circular A aA B n = r 1.5 slot. The tangential component is directed along the tangent of the slot. Applying assessing

is

work

solely

the relative acceleration equation and referring to Fig. b, provided

integrity

of

and

work

this

aA = aB + aAB * rA>B - vAB 2 rA>B

A -2cos 30°i + 2 sin 30°j B -52 A -2 cos 30°i + 2 sin 30°j B and

any

courses

part

the

is

This

50i - A aA B t j = 3i + A aAB k B *

sale

will

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destroy

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50i - A aA B t j = A 46.30 - aAB B i - A 1.732aAB + 25 B j Equating the i and j components, 50 = 46.30-aAB - A aA B t = - A 1.732aAB + 25 B Solving, aAB = - 3.70 rad>s2

A aA B t = 18.59 ft>s2 T Thus, the magnitude of aA is aA = 4A aA B t 2 + A aA B n 2 = 218.592 + 502 = 53.3ft>s2

Ans.

and its direction is u = tan-1 C

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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A aA B t A aA B n

S = tan-1 a

18.59 b = 20.4° c 50

Ans.

5 ft/s 3 ft/s2

16–129. Ball C moves along the slot from A to B with a speed of 3 ft>s, which is increasing at 1.5 ft>s2, both measured relative to the circular plate. At this same instant the plate rotates with the angular velocity and angular deceleration shown. Determine the velocity and acceleration of the ball at this instant.

z v a

6 rad/s 1.5 rad/s2

B

SOLUTION C

Reference Frames: The xyz rotating reference frame is attached to the plate and coincides with the fixed reference frame XYZ at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vO = aO = 0

A

# v = a = [-1.5k] rad>s2

v = [6k] rad>s

2 ft

x

y

For the motion of ball C with respect to the xyz frame, (vrel)xyz = (-3 sin 45°i - 3 cos 45°j) ft>s = [-2.121i - 2.121j] ft>s (arel)xyz = ( -1.5 sin 45°i - 1.5 cos 45°j) ft>s2 = [- 1.061i - 1.061j] ft>s2

laws

or

From the geometry shown in Fig. b, rC>O = 2 cos 45° = 1.414 ft. Thus,

copyright

in

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Web)

rC>O = (- 1.414 sin 45°i + 1.414 cos 45°j)ft = [-1i + 1j] ft

permitted.

Dissemination

Wide

Velocity: Applying the relative velocity equation,

is

of

the

not

instructors

States

World

vC = vO + v * rC>O + (vrel)xyz

by

and

use

United

on

learning.

= 0 + (6k) * (- 1i + 1j) + ( -2.121i - 2.121j)

Ans.

(including

for

work

student

the

= [-8.12i - 8.12j] ft>s

of

protected

the

Acceleration: Applying the relative acceleration equation, we have

integrity

of

and

work

this

assessing

is

work

solely

# aC = aO + v * rC>O + v * (v * rC>O) + 2v * (vrel)xyz + (a rel)xyz part

the

is

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provided

= 0 + (1.5k) * ( -1i + 1j) + (6k) * [(6k) * ( - 1i + 1j)] + 2(6k) * ( - 2.121i - 2.121j) + ( -1.061i - 1.061j) any

courses

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= [61.9i - 61.0j]ft>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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2 ft

45

16–130. The crane’s telescopic boom rotates with the angular velocity and angular acceleration shown. At the same instant, the boom is extending with a constant speed of 0.5 ft>s, measured relative to the boom. Determine the magnitudes of the velocity and acceleration of point B at this instant.

60 ft B vAB aAB

SOLUTION

30

Reference Frames: The xyz rotating reference frame is attached to boom AB and coincides with the XY fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xy frame with respect to the XY frame is vA = aA = 0

# vAB = a = [-0.01k] rad>s2

vAB = [-0.02k] rad>s

For the motion of point B with respect to the xyz frame, we have rB>A = [60j] ft

(vrel)xyz = [0.5j] ft>s

(arel)xyz = 0

Velocity: Applying the relative velocity equation, laws

or

vB = vA + vAB * rB>A + (v rel)xyz in

teaching

Web)

= 0 + ( -0.02k) * (60j) + 0.5j

States

World

permitted.

Dissemination

Wide

copyright

= [1.2i + 0.5j] ft > s

learning.

is

of

the

not

instructors

Thus, the magnitude of vB, Fig. b, is

Ans.

(including

for

Acceleration: Applying the relative acceleration equation,

work

student

the

by

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use

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on

vB = 21.22 + 0.52 = 1.30 ft>s

assessing

is

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solely

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protected

the

# aB = aA + vAB * rB>A + vAB * (vAB * rB>A) + 2vAB * (vrel)xyz + (a rel)xyz

part

the

is sale

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Thus, the magnitude of aB, Fig. c, is

destroy

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and

any

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= [0.62i - 0.024 j] ft>s2

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provided

integrity

of

and

work

this

= 0 + ( -0.01k) * (60j) + (- 0.02k) * [(-0.02k) * (60j)] + 2( -0.02k) * (0.5j) + 0

aB = 20.622 + ( - 0.024)2 = 0.6204 ft>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

0.02 rad/s 0.01 rad/s2 A

16–131. While the swing bridge is closing with a constant rotation of 0.5 rad>s, a man runs along the roadway at a constant speed of 5 ft>s relative to the roadway. Determine his velocity and acceleration at the instant d = 15 ft.

d

z O

y

x v

SOLUTION Æ = {0.5k} rad>s Æ = 0 rm>o = {-15 j} ft (vm>o)xyz = {- 5j} ft>s (am>o)xyz = 0 vm = vo + Æ * rm>o + (vm>o)xyz vm = 0 + (0.5k) * ( -15j) - 5j vm = {7.5i - 5j} ft>s

laws

or

Ans.

in

teaching

Web)

am = aO + Æ * rm>O + Æ * (Æ * rm>O) + 2Æ * (vm>O)xyz + (am>O)xyz Dissemination

Wide

copyright

am = 0 + 0 + (0.5k) * [(0.5k) * (-15j)] + 2(0.5k) * ( -5j) + 0

permitted.

am = {5i + 3.75j} ft>s2

sale

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provided

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the

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on

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is

of

the

not

instructors

States

World

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.5 rad/s

*16–132. While the swing bridge is closing with a constant rotation of 0.5 rad>s, a man runs along the roadway such that when d = 10 ft he is running outward from the center at 5 ft>s with an acceleration of 2 ft>s2, both measured relative to the roadway. Determine his velocity and acceleration at this instant.

d

z O

y

x v

SOLUTION Æ = {0.5k} rad>s Æ = 0 rm>o = {- 10 j} ft (vm>O)xyz = {- 5j} ft>s (am>O)xyz = {- 2j} ft>s2 vm = vo + Æ * rm>o + (vm>o)xyz vm = 0 + (0.5k) * (- 10j) - 5j or

vm = {5i - 5j} ft>s # am = aO + Æ * rm>O + Æ * (Æ * rm>O) + 2Æ * (vm>O)xyz + (am>O)xyz

Wide

copyright

in

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laws

Ans.

States

World

permitted.

Dissemination

am = 0 + 0 + (0.5k) * [(0.5k) * ( -10j)] + 2(0.5k) * ( -5j) - 2j instructors

am = {5i + 0.5j} ft>s2

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and

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is

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the

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

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0.5 rad/s

16–133. Collar C moves along rod BA with a velocity of 3 m>s and an acceleration of 0.5 m>s2, both directed from B towards A and measured relative to the rod. At the same instant, rod AB rotates with the angular velocity and angular acceleration shown. Determine the collar’s velocity and acceleration at this instant.

z

0.5 m B

x v ⫽ 6 rad/s a ⫽ 1.5 rad/s2

SOLUTION

y

# vAB = a = [1.5k] rad>s2

vB = v = [6k] rad>s

For the motion of collar C with respect to the xyz frame, we have (arel)xyz = [0.5j] m>s2

(vrel)xyz = [3j] m>s

or

rC>B = [0.5j] m

teaching

Web)

laws

Velocity: Applying the relative velocity equation,

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copyright

in

vC = vB + vAB * rC>B + (vrel)xyz

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use

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learning.

is

of

the

instructors

= [ -3i + 3j] m>s

student

the

by

and

Acceleration: Applying the relative acceleration equation,

of

protected

the

(including

for

work

# aC = aB + vAB * rC>B + vAB * (vAB * rC>B) + 2vAB * (vrel)xyz + (arel)xyz

this

assessing

is

work

solely

= 0 + (1.5k) * (0.5j) + (6k) * (6k * 0.5j) + 2(6k) * (3j) + (0.5j)

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= [- 36.75i - 17.5j] m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

not

States

World

permitted.

Dissemination

= 0 + (6k) * (0.5j) + 3j

C A

Reference Frames: The xyz rotating reference frame is attached to rod AB and coincides with the XYZ reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vB = aB = 0

3 m/s 0.5 m/s2

16–134. y

Block A, which is attached to a cord, moves along the slot of a horizontal forked rod. At the instant shown, the cord is pulled down through the hole at O with an acceleration of 4 m>s2 and its velocity is 2 m>s. Determine the acceleration of the block at this instant. The rod rotates about O with a constant angular velocity v = 4 rad>s.

A V

Motion of moving reference. vO = 0 aO = 0 Æ = 4k # Æ = 0 Motion of A with respect to moving reference.

laws

or

rA>O = 0.1i teaching

Web)

vA>O = -2i

Dissemination

Wide

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in

aA>O = -4i States

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permitted.

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learning.

is

of

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instructors

# aA = aO + Æ * rA>O + Æ * (Æ * rA>O) + 2Æ * (vA>O)xyz + (aA>O)xyz by

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a A = {- 5.60i - 16j} m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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O 100 mm

SOLUTION

= 0 + 0 + (4k) * (4k * 0.1i) + 2(4k * ( - 2i)) - 4i

x

Ans.

16–135. A girl stands at A on a platform which is rotating with a constant angular velocity v = 0.5 rad>s. If she walks at a constant speed of v = 0.75 m>s measured relative to the platform, determine her acceleration (a) when she reaches point D in going along the path ADC, d = 1 m; and (b) when she reaches point B if she follows the path ABC, r = 3 m.

O A r

SOLUTION

D B

(a) # aD = aO + Æ * rD>O + Æ * (Æ * rD>O) + 2Æ * (vD>O)xyz + (aD>O)xyz Motion of moving reference

x

(1)

Motion of D with respect to moving reference

aO = 0

rD>O = {1i} m

Æ = {0.5k} rad>s # Æ = 0

(vD>O)xyz = {0.75j} m>s (aD>O)xyz = 0 laws

or

Substitute the data into Eq.(1): in

teaching

Web)

aB = 0 + (0) * (1i) + (0.5k) * [(0.5k) * (1i)] + 2(0.5k) * (0.75j) + 0

permitted.

Dissemination

Wide

Ans.

copyright

= { -1i} m>s2

learning.

is

of

# aB = aO + Æ * rB>O + Æ * (Æ * rB>O) + 2Æ * (vB>O)xyz + (aB>O)xyz

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(2)

Motion of B with respect to moving reference solely

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Motion of moving reference

rB>O = {3i} m this

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ao = 0

(vB>O)xyz = {0.75j} m>s part

the

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and

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Æ = {0.5k} rad>s

any

courses

(aB>O)xyz = -(aB>O)n i + (aB>O)t j 2

= -(0.75 3 )i

sale

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destroy

of

and

Æ = 0

= {-0.1875i} m>s Substitute the data into Eq.(2): aB = 0 + (0) * (3i) + (0.5k) * [(0.5k) * (3i)] + 2(0.5k) * (0.75j) + (- 0.1875i) = { -1.69i} m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

y d

C

*16–136. A girl stands at A on a platform which is rotating with an angular acceleration a = 0.2 rad>s2 and at the instant shown has an angular velocity v = 0.5 rad>s. If she walks at a constant speed v = 0.75 m>s measured relative to the platform, determine her acceleration (a) when she reaches point D in going along the path ADC, d = 1 m; and (b) when she reaches point B if she follows the path ABC, r = 3 m.

O A r

SOLUTION # aD = aO + Æ * rD>O + Æ * (Æ * rD>O) + 2Æ * (vD>O)xyz + (aD>O)xyz

x

(1)

Motion of moving reference

Motion of D with respect to moving reference

aO = 0

rD>O = {1i} m

Æ = {0.5k} rad>s # Æ = {0.2k} rad>s2

(vD>O)xyz = {0.75j} m>s (aD>O)xyz = 0 laws

or

Substitute the data into Eq.(1):

Ans.

permitted.

Dissemination

copyright

= {-1i + 0.2j} m>s2

Wide

in

teaching

Web)

aB = 0 + (0.2k) * (1i) + (0.5k) * [(0.5k) * (1i)] + 2(0.5k) * (0.75j) + 0

not

instructors

States

World

(b)

learning.

is

of

the

# aB = aO + Æ * rB>O + Æ * (Æ * rB>O) + 2Æ * (vB>O)xyz + (aB>O)xyz by

and

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on

(2)

the

Motion of B with respect to moving reference

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Motion of moving reference

rB>O = {3i} m

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aO = 0

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(vB>O)xyz = {0.75j} m>s

any

courses

(aB>O)xyz = -(aB>O)n i + (a B>O)t j = - A 0.75 3 B i 2

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and

Æ = {0.5k} rad>s # Æ = {0.2k} rad>s2

= {-0.1875i} m>s Substitute the data into Eq.(2): aB = 0 + (0.2k) * (3i) + (0.5k) * [(0.5k) * (3i)] + 2(0.5k) * (0.75j) + ( -0.1875i) = {-1.69i + 0.6j} m s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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D B

(a)

Ans.

y d

C

16–137. At the instant shown, rod AB has an angular velocity vAB = 3 rad>s and an angular acceleration aAB = 5 rad>s2. Determine the angular velocity and angular acceleration of rod CD at this instant. The collar at C is pin-connected to CD and slides over AB.

vAB aAB

A

3 rad/s 5 rad/s2

60 C 0.75 m B 0.5 m

SOLUTION rC>A = (0.75 sin 60°)i - (0.75 cos 60°)j

D

rC>A = {0.6495i - 0.375j} m vC = vCD * rC>D = (vCDk) * (0.5j) = { -0.5vCDi} m>s a C = aCD * rCD - v2CD rCD = (aCDk) * (0.5j) - v2CD(0.5j) laws

or

aC = { -0.5 aCDi - v2CD(0.5)j} m>s2

Wide

copyright

in

teaching

Web)

vC = vA + Æ * rC>A + (vC>A)xyz

World

permitted.

Dissemination

-0.5vCDi = 0 + (3k) * (0.6495i - 0.375j) + vC>A sin 60°i - vC>A cos 60°j

learning.

is

of

the

not

instructors

States

-0.5vCD = 1.125 + 0.866vC>A

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0 = 1.9485 - 0.5vC>A

the

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student

vC>A = 3.897 m>s

Ans.

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and

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this

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is

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protected

vCD = - 9.00 rad>s = 9.00 rad>sb # a C = a A + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (a C>A)xyz

and

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courses

part

the

is

This

a C = 0 + (5k) * (0.6495i - 0.375j) + (3k) * [(3k) * (0.6495i - 0.375j)]

will

their

destroy

of

+ 2(3k) * [3.897(0.866)i - 0.5(3.897)j] + 0.866a C>A i - 0.5a C>A j sale

0.5 aCD i - ( -9.00)2(0.5)j = 0 + 1.875i + 3.2475j - 5.8455i + 3.375j + 11.6910i + 20.2488j + 0.866a C>A i - 0.5a C>A j 0.5 aCD = 7.7205 + 0.866a C>A -40.5 = 26.8713 - 0.5aC>A aC>A = 134.7 m>s2 aCD = 249 rad>s2b

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

16–138. Collar B moves to the left with a speed of 5 m>s, which is increasing at a constant rate of 1.5 m>s2, relative to the hoop, while the hoop rotates with the angular velocity and angular acceleration shown. Determine the magnitudes of the velocity and acceleration of the collar at this instant.

A

v ⫽ 6 rad/s a ⫽ 3 rad/s2

450 mm

SOLUTION 200 mm

Reference Frames: The xyz rotating reference frame is attached to the hoop and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vA = aA = 0

B

# v = a = [- 3k] rad>s2

v = [ -6k] rad>s

For the motion of collar B with respect to the xyz frame, rB>A = [- 0.45j] m (vrel)xyz = [- 5i] m>s (vrel)xyz2

52 = 125 m>s2. Thus, 0.2

=

laws

r

or

The normal components of (arel)xyz is [(arel)xyz]n =

in

teaching

Web)

(arel)xyz = [- 1.5i + 125j] m>s

permitted.

Dissemination

Wide

copyright

Velocity: Applying the relative velocity equation,

is

of

the

not

instructors

States

World

vB = vA + v * rB>A + (vrel)xyz

the

by

and

use

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on

learning.

= 0 + ( -6k) * ( -0.45j) + ( - 5i)

the

(including

for

work

student

= [- 7.7i] m>s

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is

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Thus,

Ans.

is

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of

and

work

this

vB = 7.7 m>s ;

and

any

courses

part

the

Acceleration: Applying the relative acceleration equation,

#

will

their

destroy

of

aB = aA + v * rB>A + v * (v * rB>A) + 2v * (vrel)xyz + (arel)xyz sale

= 0 + (- 3k) * (- 0.45j) + ( - 6k) * [( -6k) * ( -0.45j)] + 2( - 6k) * (- 5i) + ( - 1.5i + 125j) = [ - 2.85i + 201.2j] m>s2 Thus, the magnitude of aB is therefore aB = 32.852 + 201.22 = 201 m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

16–139. D

Block B of the mechanism is confined to move within the slot member CD. If AB is rotating at a constant rate of vAB = 3 rad>s, determine the angular velocity and angular acceleration of member CD at the instant shown.

100 mm

A

SOLUTION vAB

Coordinate Axes: The origin of both the fixed and moving frames of reference are located at point C. The x, y, z moving frame is attached to and rotates with rod CD since peg B slides along the slot in member CD.

B 3 rad/s

vCD, aCD

Kinematic Equation: Applying Eqs. 16–24 and 16–27, we have 30

vB = vC + Æ * rB>C + (vB>C)xyz # aB = aC + Æ * rB>C + Æ * (Æ * rB>C) + 2Æ * (vB>C)xyz + (aB>C)xyz

(1) (2)

vC = 0

rB>C = {0.2i} m

aC = 0

(vB>C)xyz = (vB>C)xyz i

Æ = -vCD k # Æ = -aCD k

(a B>C)xyz = (a B>C)xyz i

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of

the work the

(including

for

= {- 0.2598i - 0.150j} m>s

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vB = vAB * rB>A = -3k * (0.05i - 0.08660j)

assessing

is

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protected

aB = aAB * rB>A - v2ABrB>A = 0 - 32 (0.05i - 0.08660j)

is

This

provided

integrity

of

and

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this

= {- 0.450i + 0.7794j} m>s

and

any

courses

part

the

Substitute the above data into Eq.(1) yields

will

their

destroy

of

vB = vC + Æ * rB>C + (vB>C)xyz sale

-0.2598i - 0.150j = 0 + (-vCDk) * 0.2i + (vB>C)xyz i -0.2598i - 0.150j = (vB>C)xyzi - 0.2vCD j Equating i and j components, we have (vB>C)xyz = - 0.2598 m>s vCD = 0.750 rad>s

Ans.

Substitute the above data into Eq.(2) yields # aB = aC + Æ * rB>C + Æ * (Æ * rB>C) + 2Æ * (vB>C)xyz + (aB>C)xyz -0.450i + 0.7794j = 0 + ( -aCDk) * 0.2i + ( -0.750k) * [( -0.750k) * 0.2i] + 2(- 0.750k) * (-0.2598i) + (aB>C)xyz i -0.450i + 0.7794j = C (aB>C)xyz - 0.1125 D i + (0.3897 - 0.2aCD)j Equating i and j components, we have (a B>C)xyz = - 0.3375 m>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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aCD = -1.95 rad>s2 = 1.95 rad>s2

Ans.

not

instructors

States

World

The velocity and acceleration of peg B can be determined using Eqs. 16–9 and 16–14 with rB>A = {0.1 cos 60°i - 0.1 sin 60°j} m = {0.05i - 0.08660j} m.

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Motion of C with respect to moving reference

laws

Motion of moving reference

or

C

200 mm

*16–140. At the instant shown rod AB has an angular velocity vAB = 4 rad>s and an angular acceleration aAB = 2 rad>s2. Determine the angular velocity and angular acceleration of rod CD at this instant.The collar at C is pin connected to CD and slides freely along AB.

vAB aAB

A 60

0.75 m

Coordinate Axes: The origin of both the fixed and moving frames of reference are located at point A. The x, y, z moving frame is attached to and rotate with rod AB since collar C slides along rod AB. Kinematic Equation: Applying Eqs. 16–24 and 16–27, we have vC = vA + Æ * rC>A + (vC>A)xyz # + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (v C>A)xyz + (a C>A)xyz

(1) (2)

Motion of C with respect to moving reference rC>A = 50.75i6m or

Motion of moving reference vA = 0 aA = 0 Æ = 4k rad>s # Æ = 2k rad>s2

laws

(vC>A)xyz = (yC>A)xyz i

permitted.

Dissemination

Wide

copyright

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(a C>A)xyz = (aC>A)xyz i

United

on

learning.

is

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the

not

instructors

States

World

The velocity and acceleration of collar C can be determined using Eqs. 16–9 and 16–14 with rC>D = { -0.5 cos 30°i - 0.5 sin 30°j }m = { -0.4330i - 0.250j} m.

protected

the

(including

for

= -0.250vCDi + 0.4330vCDj

work

student

the

by

and

use

vC = vCD * rC>D = -vCDk * ( -0.4330i - 0.250j)

this

assessing

is

work

solely

of

aC = a CD * rC>D - v2CD rC>D

the

is

This

provided

integrity

of

and

work

= -aCD k * ( -0.4330i - 0.250j) - v2CD( -0.4330i - 0.250j)

destroy

of

and

any

courses

part

= A 0.4330v2CD - 0.250 aCD B i + A 0.4330aCD + 0.250v2CD B j

sale

will

their

Substitute the above data into Eq.(1) yields v C = vA + Æ * rC>A + (vC>A)xyz -0.250 vCD i + 0.4330vCDj = 0 + 4k * 0.75i + (yC>A)xyz i -0.250vCD i + 0.4330vCD j = (yC>A)xyz i + 3.00j Equating i and j components and solve, we have (yC>A)xyz = - 1.732 m>s vCD = 6.928 rad>s = 6.93 rad>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

0.5 m C

D B

SOLUTION

aC = aA

4 rad/s 2 rad/s2

Ans.

16–140. continued Substitute the above data into Eq.(2) yields # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz

C 0.4330 A 6.9282 B - 0.250 aCD D i + C 0.4330aCD + 0.250 A 6.9282 B D j = 0 + 2k * 0.75i + 4k * (4k * 0.75i) + 2 (4k) * ( -1.732i) + (aC>A)xyz i (20.78 - 0.250aCD)i + (0.4330 aCD + 12)j = C (aC>A)xyz - 12.0 D i - 12.36j Equating i and j components, we have (aC>A)xyz = 46.85 m>s2 d

Ans.

sale

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destroy

of

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is

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the

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permitted.

Dissemination

Wide

copyright

in

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laws

or

aCD = - 56.2 rad>s2 = 56.2 rad>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

16–141. The “quick-return” mechanism consists of a crank AB, slider block B, and slotted link CD. If the crank has the angular motion shown, determine the angular motion of the slotted link at this instant.

D

100 mm vAB aAB

3 rad/s 9 rad/s2

30 A

SOLUTION vB = 3(0.1) = 0.3 m>s (aB)t = 9(0.1) = 0.9 m>s

30 2

(aB )n = (3)2 (0.1) = 0.9 m>s2

vCD, aCD

vB = vC + Æ * rB>C + (vB>C)xyz 0.3 cos 60°i + 0.3 sin 60°j = 0 + (vCDk) * (0.3i) + vB>C i vB>C = 0.15 m>s vCD = 0.866 rad>s d # aB = aC + Æ * rB>C + Æ * (Æ * rB>C) + 2Æ * (vB>C)xyz + (aB>C)xyz

Web)

laws

or

Ans.

Wide

copyright

in

teaching

0.9 cos 60°i - 0.9 cos 30°i + 0.9 sin 60°j + 0.9 sin 30°j = 0 + (aCD k) * (0.3i)

States

World

permitted.

Dissemination

+(0.866k) * (0.866k * 0.3i) + 2(0.866k * 0.15i) + aB>C i

on

learning.

is

of

the

not

instructors

-0.3294i + 1.2294j = 0.3aCD j - 0.225i + 0.2598j + aB>C i

the

by

and

use

United

aB>C = - 0.104 m>s2

sale

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the

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student

aCD = 3.23 rad>s2 d

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

C

300 mm

B

16–142. At the instant shown, the robotic arm AB is rotating counter clockwise at v = 5 rad>s and has an angular acceleration a = 2 rad>s2. Simultaneously, the grip BC is rotating counterclockwise at v¿ = 6 rad>s and a¿ = 2 rad>s2, both measured relative to a fixed reference. Determine the velocity and acceleration of the object heldat the grip C.

125 mm

y

ω ,α

SOLUTION

A

vC = vB + Æ * rC>B + (vC>B)xyz # aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz

(1) (2)

Motion of C with respect to moving reference rC>B = {0.125 cos 15°i + 0.125 sin 15°j} m

Æ = {6k} rad>s # Æ = {2k} rad>s2

(vC>B)xyz = 0 (aC>B)xyz = 0

laws

or

Motion of B:

in

teaching

Web)

vB = v * rB>A Dissemination

Wide

copyright

= (5k) * (0.3 cos 30°i + 0.3 sin 30°j)

is

of

the

not

instructors

States

World

permitted.

= {- 0.75i + 1.2990j} m>s

by

and

use

United

on

learning.

aB = a * rB>A - v2rB>A

(including

for

work

student

the

= (2k) * (0.3 cos 30°i + 0.3 sin 30°j) - (5)2(0.3 cos 30°i + 0.3 sin 30°j)

assessing

is

work

solely

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protected

the

= { -6.7952i - 3.2304j} m>s2

provided

integrity

of

and

work

this

Substitute the data into Eqs. (1) and (2) yields:

Ans.

their

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and

any

courses

part

the

is

This

vC = (- 0.75i + 1.2990j) + (6k) * (0.125 cos 15°i + 0.125 sin 15°j) + 0 = { -0.944i + 2.02j} m>s

sale

will

aC = (- 6.79527i - 3.2304j) + (2k) * (0.125 cos 15°i + 0.125 sin 15°j) + (6k) * [(6k) * (0.125 cos 15°i + 0.125 sin 15°j)] + 0 + 0 = { -11.2i - 4.15j} m s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

15°

300 mm

30°

Motion of moving reference

C

B

Ans.

ω, α

x

16–143. Peg B on the gear slides freely along the slot in link AB. If the gear’s center O moves with the velocity and acceleration shown, determine the angular velocity and angular acceleration of the link at this instant.

150 mm vO aO

B

3 m/s 1.5 m/s2

600 mm O

150 mm

SOLUTION Gear Motion: The IC of the gear is located at the point where the gear and the gear rack mesh, Fig. a. Thus, vO 3 = 20 rad>s = v = rO>IC 0.15 Then, vB = vrB>IC = 20(0.3) = 6 m>s : aO 1.5 = = 10 rad>s. By referring to Fig. b, r 0.15

Since the gear rolls on the gear rack, a =

aB = aO + a * rB>O - v2 rB>O (aB)t i - (aB)n j = 1.5i + ( - 10k) * 0.15j - 202(0.15j) (aB)t i - (aB)n j = 3i - 60j Thus, or

(aB)n = 60 m>s2

Dissemination

Wide

copyright

in

teaching

Reference Frame: The x¿y¿z¿ rotating reference frame is attached to link AB and coincides with the XYZ fixed reference frame, Figs. c and d. Thus, vB and aB with respect to the XYZ frame is

Web)

laws

(aB)t = 3 m>s2

not

instructors

States

World

permitted.

vB = [6 sin 30°i - 6 cos 30° j] = [3i - 5.196j] m>s

and

use

United

on

learning.

is

of

the

aB = (3 sin 30° - 60 cos 30°)i + (-3 cos 30° - 60 sin 30°)j

work

student

the

by

= [ -50.46i - 32.60j] m>s2

assessing

is

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solely

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protected

the

(including

for

For motion of the x¿y¿z¿ frame with reference to the XYZ reference frame, # vA = aA = 0 vAB = -vABk vAB = - aAB k provided

integrity

of

and

work

this

For the motion of point B with respect to the x¿y¿z¿ frame is (vrel)x¿y¿z¿ = (vrel)x¿y¿z¿ j

part

the

is

This

rB>A = [0.6j]m

(arel)x¿y¿z¿ = (arel)x¿y¿z¿ j

destroy

of

and

any

courses

Velocity: Applying the relative velocity equation, sale

will

their

vB = vA + vAB * rB>A + (vrel)x¿y¿z¿ 3i - 5.196j = 0 + ( -vABk) * (0.6j) + (vrel)x¿y¿z¿ j 3i - 5.196j = 0.6vAB i + (vrel)x¿y¿z¿j Equating the i and j components yields vAB = 5 rad>s

3 = 0.6vAB

Ans.

(vrel)x¿y¿z¿ = - 5.196 m>s Acceleration: Applying the relative acceleration equation. # aB = aA + vAB * rB>A + vAB * (vAB * rB>A) + 2vAB * (vrel)x¿y¿z¿ + (a rel)x¿y¿z¿ -50.46i - 32.60j = 0 + ( - aABk) * (0.6j) + ( -5k) * [(- 5k) * (0.6j)] + 2( -5k) * ( -5.196j) + (arel)x¿y¿z¿j -50.46i - 32.60j = (0.6aAB - 51.96)i + C (arel)x¿y¿z¿ - 15 D j Equating the i components, - 50.46 = 0.6a AB - 51.96 aAB = 2.5 rad>s2 © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

A

*16–144. The cars on the amusement-park ride rotate around the axle at A with a constant angular velocity vA>f = 2 rad>s, measured relative to the frame AB. At the same time the frame rotates around the main axle support at B with a constant angular velocity vf = 1 rad>s. Determine the velocity and acceleration of the passenger at C at the instant shown.

y

D 8 ft 8 ft A

x

SOLUTION vC = vA + Æ * rC>A + (vC>A)xyz # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz

C vA/f ⫽ 2 rad/s

(1)

15 ft 30

(2)

vf ⫽ 1 rad/s

Motion of moving refernce

Motion of C with respect to moving reference

Æ = {3k} rad>s # Æ = 0

rC>A = {- 8i} ft (vC>A)xyz = 0

or

(a C>A)xyz = 0

teaching

Web)

laws

Motion of A:

Wide

copyright

in

vA = v * rA>B

States

World

permitted.

Dissemination

= (1k) * ( - 15 cos 30°i + 15 sin 30°j)

on

learning.

is

of

the

not

instructors

= {- 7.5i - 12.99j} ft>s the

by

and

use

United

aA = a * rA>B - v2 rA>B

the

(including

for

work

student

= 0 - (1)2(- 15 cos 30°i + 15 sin 30°j)

this

assessing

is

work

solely

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protected

= {12.99i - 7.5j} ft>s2

part

the

is

This

provided

integrity

of

and

work

Substitute the data into Eqs.(1) and (2) yields:

destroy

Ans.

sale

will

their

= {-7.5i - 37.0j }ft>s

of

and

any

courses

vC = ( - 7.5i - 12.99j) + (3k) * ( -8i) + 0

aC = (12.99i - 7.5j) + 0 + (3k) * [(3k) * (-8i) + 0 + 0] = {85.0i - 7.5j}ft>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

B

16–145. The cars on the amusement-park ride rotate around the axle at A with a constant angular velocity vA>f = 2 rad>s, measured relative to the frame AB. At the same time the frame rotates around the main axle support at B with aconstant angular velocity vf = 1 rad>s. Determine the velocity and acceleration of the passenger at D at the instant shown.

y

D 8 ft 8 ft A

SOLUTION

C ωA = 2 rad/s /f

vD = vA + Æ * rD>A + (vD>A)xyz aD = aA

(1)

# + v * rD>A + Æ * (Æ * rD>A) + 2Æ * (vD>A)xyz + (aD>A)xyz

Motion of moving reference

30°

(2)

rD>A = {8j} ft (vD>A)xyz = 0 (aD>A)xyz = 0

laws

or

Motion of A:

in

teaching

Web)

vA = v * rA>B Dissemination

Wide

copyright

= (1k) * (-15 cos 30°i + 15 sin 30°j)

is

of

the

not

instructors

States

World

permitted.

= {- 7.5i - 12.99j} ft>s

by

and

use

United

on

learning.

aA = a * rA>B - v2rA>B

(including

for

work

student

the

= 0 - (1)2(-15 cos 30°i + 15 sin 30°j)

assessing

is

work

solely

of

protected

the

= {12.99i - 7.5j} ft>s2

provided

integrity

of

and

work

this

Substitute the data into Eqs. (1) and (2) yields:

destroy

of

and

Ans.

their

= { -31.5i - 13.0j} ft>s

any

courses

part

the

is

This

vD = ( -7.5i - 12.99j) + (3k) * (8j) + 0

sale

will

aD = (12.99i - 7.5j) + 0 + (3k) * [(3k) * (8j)] + 0 + 0 = {13.0i - 79.5j} ft s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

15 ft

ω f = 1 rad/s

Motion of D with respect to moving reference

Æ = {3k} rad>s # Æ = 0

x

B

16–146. If the slotted arm AB rotates about the pin A with a constant angular velocity of vAB = 10 rad>s, determine the angular velocity of link CD at the instant shown.

B D

600 mm

vAB ⫽ 10 rad/s

SOLUTION

30⬚ A

Reference Frame: The xyz rotating reference frame is attached to link AB and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vA = 0

#

vAB = 0

vAB = [10k] rad>s

For the motion of point D relative to the xyz frame, we have rD>A = [0.6i] m

(vrel)xyz = (vrel)xyz i

Since link CD rotates about a fixed axis, vD can be determined from

laws

or

vD = vCD * rD

in

teaching

Web)

= (vCD k) * (0.45 cos 15° i + 0.45 sin 15° j)

permitted.

Dissemination

Wide

copyright

= - 0.1165vCD i + 0.4347vCD j

is

of

the

not

instructors

States

World

Velocity: Applying the relative velocity equation, we have

the

by

and

use

United

on

learning.

vD = vA + vAB * rD>A + (vrel)xyz

the

(including

for

work

student

- 0.1165vCD i + 0.4347vCD j = 0 + (10k) * (0.6i) + (vrel)xyz i

assessing

is

work

solely

of

protected

- 0.1165vCD i + 0.4347vCD j = (vrel)xyzi + 6j

part

the

is and

any

courses

- 0.1165vCD = (vrel)xyz

This

provided

integrity

of

and

work

this

Equating the i and j components

will

their

destroy

of

0.4347vCD = 6 sale

Solving, vCD = 13.80 rad>s = 13.8 rad>s (vrel)xyz = - 1.608 m>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

C

450 mm 45⬚

16–147. At the instant shown, boat A travels with a speed of 15 m>s, which is decreasing at 3 m>s2, while boat B travels with a speed of 10 m>s, which is increasing at 2 m>s2. Determine the velocity and acceleration of boat A with respect to boat B at this instant.

30 m 15 m/s 50 m

SOLUTION

A

B 3 m/s2

50 m 10 m/s 2 m/s2

Reference Frame: The xyz rotating reference frame is attached to boat B and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since

vB = [-10j] m>s

aA = [ -4.5i - 3j] m>s2

aB = [2i - 2j] m>s2 laws

vA = [15j] m>s

or

boats A and B move along the circular paths, their normal components of vA 2 vB 2 152 102 = 4.5 m>s2 and (aB)n = = 2 m>s2. = = acceleration are (aA)n = r r 50 50 Thus, the motion of boats A and B with respect to the XYZ frame are

v = [0.2k] rad>s

World

vB 10 = 0.2 rad>s = r 50

learning.

is

of

the

not

instructors

States

v =

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are

the

(including

for

work

student

the

by

and

use

United

on

# v = [0.04k] rad>s2

(aB)t 2 # = 0.04 rad>s2 = v = r 50

is

work

solely

of

protected

And the position of boat A with respect to B is

provided

integrity

of

and

work

this

assessing

rA>B = [-20i] m

any

courses

part

the

is

This

Velocity: Applying the relative velocity equation,

their

destroy

of

and

vA = vB + v * rA>B + (vrel)xyz sale

will

15j = - 10j + (0.2k) * ( - 20i) + (vrel)xyz 15j = - 14j + (vrel)xyz (vrel)xyz = [29j] m>s

Ans.

Acceleration: Applying the relative acceleration equation, # aA = aB + v * rA>B + v * (v * rA>B) + 2v * (vrel)xyz + (arel)xyz (- 4.5i - 3j) = (2i - 2j) + (0.04k) * (-20i) + (0.2k) * C (0.2k) * ( -20i) D + 2(0.2k) * (29j) + (arel)xyz -4.5i - 3j = - 8.8i - 2.8j + (arel)xyz (arel)xyz = [4.3i - 0.2j] m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

*16–148. At the instant shown, boat A travels with a speed of 15 m>s, which is decreasing at 3 m>s2, while boat B travels with a speed of 10 m>s, which is increasing at 2 m>s2. Determine the velocity and acceleration of boat B with respect to boat A at this instant.

30 m 15 m/s 50 m

SOLUTION

A

B 3 m/s2

50 m 10 m/s 2 m/s2

Reference Frame: The xyz rotating reference frame is attached to boat A and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since

vB = [-10j] m>s

aA = [-4.5i - 3j] m>s2

aB = [2i - 2j] m>s2 laws

vA = [15j] m>s

or

boats A and B move along the circular paths, their normal components of vA 2 vB 2 152 102 = 4.5 m>s2 and (aB)n = = 2 m>s2. = = acceleration are (aA)n = r r 50 50 Thus, the motion of boats A and B with respect to the XYZ frame are

Dissemination

Wide

copyright

in

teaching

Web)

Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are vA 15 = 0.3 rad>s = r 50

World

permitted.

v = [0.3k] rad>s

# v = [-0.06k] rad>s2 the

(including

for

work

student

the

by

and

use

United

(aA)t 3 # = v = = 0.06 rad>s2 r 50

on

learning.

is

of

the

not

instructors

States

v =

is

work

solely

of

protected

And the position of boat B with respect to boat A is

provided

integrity

of

and

work

this

assessing

rB>A = [20i] m

any

courses

part

the

is

This

Velocity: Applying the relative velocity equation,

their

destroy

of

and

vB = vA + v * rB>A + (vrel)xyz sale

will

-10j = 15j + (0.3k) * (20i) + (vrel)xyz - 10j = 21j + (vrel)xyz (vrel)xyz = [-31j] m>s

Ans.

Acceleration: Applying the relative acceleration equation, # aB = aA + v * rB>A + v(v * rB>A) + 2v * (vrel)xyz + (arel)xyz (2i - 2j) = ( -4.5i - 3j) + ( - 0.06k) * (20i) + (0.3k) * C (0.3k) * (20i) D + 2(0.3k) * ( -31j) + (arel)xyz 2i - 2j = 12.3i - 4.2j + (arel)xyz (arel)xyz = [-10.3i + 2.2j] m>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

16–149. vA ⫽ 3 m/s aA ⫽ 1.5 m/s2

If the piston is moving with a velocity of vA = 3 m>s and acceleration of aA = 1.5 m>s2, determine the angular velocity and angular acceleration of the slotted link at the instant shown. Link AB slides freely along its slot on the fixed peg C.

A

30⬚ C 0.5 m

SOLUTION

B

Reference Frame: The xyz reference frame centered at C rotates with link AB and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is aAB = - aABk

vAB = - vABk

vC = aC = 0

The motion of point A with respect to the xyz frame is rA>C = [ -0.5i] m

(arel)xyz = (arel)xyzi

(vrel)xyz = (vrel)xyz i

The motion of point A with respect to the XYZ frame is

or

vA = 3 cos 30°i + 3 sin 30°j = [2.598i + 1.5j] m>s

teaching

Web)

laws

aA = 1.5 cos 30°i + 1.5 sin 30°j = [1.299i + 0.75j] m>s

Wide

copyright

in

Velocity: Applying the relative velocity equation,

not

instructors

States

World

permitted.

Dissemination

vA = vC + vAB * rA>C + (vrel)xyz

and

use

United

on

learning.

is

of

the

2.598i + 1.5j = 0 + (- vAB k) * ( -0.5i) + (vrel)xyz i

for

work

student

the

by

2.598i + 1.5j = (vrel)xyzi + 0.5vAB j

is

work

solely

of

protected

the

(including

Equating the i and j components,

of

and

work

this

assessing

(vrel)xyz = 2.598 m>s provided

integrity

vAB = 3 rad>s

Ans.

part

the

is

This

0.5vAB = 1.5

destroy

of

and

any

courses

Acceleration: Applying the relative acceleration equation, sale

will

their

# aA = aC + vAB * rA>C + vAB * (vAB * rA>C) + 2vAB * (vrel)xyz + (arel)xyz 1.299i + 0.75j = 0 + (- aABk) * (- 0.5i) + (- 3k) * [( -3k) * ( - 0.5i)] + 2(- 3k) * (2.598i) + (arel)xyzi 1.299i + 0.75j = C 4.5 + (arel)xyz D i + (0.5aAB - 15.59)j Equating the j components, 0.75 = 0.5aAB - 15.59 aAB = 32.68 rad>s2 = 32.7 rad>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Ans.

16–150. B

The two-link mechanism serves to amplify angular motion. Link AB has a pin at B which is confined to move within the slot of link CD. If at the instant shown, AB (input) has an angular velocity of vAB = 2.5 rad>s, determine the angular velocity of CD (output) at this instant.

D

150 mm

C 30 45

SOLUTION

vAB

rBA 0.15 m = sin 120° sin 45° rBA = 0.1837 m vC = 0 aC = 0 Æ = -vDCk # Æ = -aDCk laws

or

rB>C = {- 0.15 i} m in

teaching

Web)

(vB>C)xyz = (yB>C)xyzi Dissemination

Wide

copyright

(aB>C)xyz = (aB>C)xyzi

is

of

the

not

instructors

States

World

permitted.

vB = vAB * rB>A = ( -2.5k) * ( -0.1837 cos 15°i + 0.1837 sin 15°j)

by

and

use

United

on

learning.

= {0.1189i + 0.4436j} m>s

(including

for

work

student

the

vB = vC + Æ * rB>C + (vB>C)xyz

is

work

solely

of

protected

the

0.1189i + 0.4436j = 0 + (- vDCk) * ( - 0.15i) + (vB>C)xyz i

sale

vDC = 2.96 rad>s b

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

will

their

destroy

of

and

(vB>C)xyz = 0.1189 m>s

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

0.1189i + 0.4436j = (vB>C)xyz i + 0.15vDC j Solving:

A

Ans.

2.5 rad/s

16–151. The gear has the angular motion shown. Determine the angular velocity and angular acceleration of the slotted link BC at this instant. The peg at A is fixed to the gear.

A

2 ft

0.5 ft 0.7 ft

SOLUTION

B

yA = (1.2)(2) = 2.4 ft>s ; aO = 4(0.7) = 2.8 ft>s2 a A = a O + a A>O aA = 2.8 + 4(0.5) + (2)2(0.5) ; ; T a A = 4.8 + 2 ; T vA = vB + Æ * rA>B + (vA>B)xyz

in

teaching

Web)

laws

or

4 3 -2.4i = 0 + (Æk) * (1.6i + 1.2j) + yA>B a b i + yA>B a b j 5 5

Dissemination

Wide

copyright

-2.4i = 1.6Æj - 1.2Æi + 0.8yA>B i + 0.6yA>B j

of

the

not

instructors

States

World

permitted.

-2.4 = -1.2Æ + 0.8yA>B

the

by

and

use

United

on

learning.

is

0 = 1.6Æ + 0.6yA>B

(including

for

work

student

Solving,

Ans.

is

work

solely

of

protected

the

vBC = Æ = 0.720 rad>sd

integrity

of

and

work

this

assessing

yA>B = -1.92 ft>s

their

destroy

of

and

any

courses

part

the

is

This

provided

a A = a B + Æ * rA>B + Æ * (Æ * rA>B) + 2Æ * (vA>B)xyz + (a A>B)xyz # -4.8i - 2j = 0 + (Æ k) * (1.6i + 1.2j) + (0.72k) * (0.72k) * (1.6i + 1.2j)) sale

will

+2(0.72k) * [- (0.8)(1.92)i - 0.6(1.92)j] + 0.8aB>A i + 0.6aB>A j # # -4.8i - 2j = 1.6Æ j - 1.2Æ i - 0.8294i - 0.6221j - 2.2118j + 1.6589i + 0.8aB>A i + 0.6aB>A. # -4.8 = -1.2Æ - 0.8294 + 1.6589 + 0.8aB/A # -2 = 1.6Æ - 0.6221 - 2.2118 + 0.6aB/A # -4.6913 = - Æ + 0.667aB/A # 0.5212 = Æ + 0.357a B/A # Ans. aBC = Æ = 2.02 rad/s2 d aB/A = -4.00 ft>s2

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C

O

v ⫽ 2 rad/s a ⫽ 4 rad/s2

*16–152. vB

The Geneva mechanism is used in a packaging system to convert constant angular motion into intermittent angular motion. The star wheel A makes one sixth of a revolution for each full revolution of the driving wheel B and the attached guide C. To do this, pin P, which is attached to B, slides into one of the radial slots of A, thereby turning wheel A, and then exits the slot. If B has a constant angular velocity of vB = 4 rad>s, determine V A and AA of wheel A at the instant shown.

B

The circular path of motion of P has a radius of rP = 4 tan 30° = 2.309 in. Thus, vP = - 4(2.309)j = -9.238j aP = -(4)2(2.309)i = - 36.95i

laws

or

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Web)

vP = vA + Æ * rP>A + (vP>A)xyz

permitted.

Dissemination

Wide

copyright

- 9.238j = 0 + (vA k) * (4j) - vP>A j

is

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# aP = aA + Æ * rP>A + Æ * (Æ * rP>A) + 2Æ * (vP>A)xyz + (aP>A)xyz

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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will

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aA = 9.24 rad>s2 d

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

4 in.

A u

-36.95 = -4aA

C P

SOLUTION

Solving,

4 rad/s

Ans.

30

17–1. z

Determine the moment of inertia Iy for the slender rod. The rod’s density r and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m. l

SOLUTION

A x

Iy =

LM

x 2 dm

l

= =

L0

x 2 (r A dx)

1 r A l3 3

m = rAl

laws

or

Thus, 1 m l2 3

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Iy =

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y

17–2. z

The solid cylinder has an outer radius R, height h, and is made from a material having a density that varies from its center as r = k + ar2, where k and a are constants. Determine the mass of the cylinder and its moment of inertia about the z axis.

R

h

SOLUTION Consider a shell element of radius r and mass dm = r dV = r(2p r dr)h R

m =

(k + ar2)(2p r dr)h

L0

m = 2p h(

aR4 kR2 + ) 2 4

m = p h R2(k +

aR2 ) 2

Ans.

laws

or

dI = r2 dm = r2(r)(2p r dr)h

teaching in

r2(k + ar2)(2p r dr) h Dissemination

Wide

copyright

L0

Web)

R

Iz =

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not

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States

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(k r3 + a r5) dr United

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of

L0

permitted.

R

Iz = 2ph

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k R4 aR6 + ] 4 6

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2 aR2 p h R4 [k + ] 2 3

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Iz =

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Ans.

17–3. y

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

R x

SOLUTION 2p

Iz =

L0

r A(R du)R2 = 2p r A R3 2p

m =

L0

r A R du = 2p r A R

Thus, I z = m R2

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Ans.

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*17–4. Determine the moment of inertia of the semiellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density r.

y

y2 x2 ⫹ 2⫽1 2 a b b x

SOLUTION dIx = m =

y2dm 2

Lv

a

rdV

a

2

L0

2

x

2

a

b dx

2 rpab2 3 a

2

1 x 2 4 rp b a1 - 2 b dx 2 L0 a

teaching

Web)

Ix =

or

=

rpb a 1 -

laws

=

in

4 rpab4 15

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=

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2 mb2 5

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Ix =

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Ans.

17–5. y

The sphere is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the sphere. The material has a constant density r.

x2 + y2 = r2

x

SOLUTION dIx =

y2 dm 2

dm = r dV = r(py2 dx) = r p(r2 - x2) dx dIx =

1 r p(r2 - x2)2 dx 2 r

8 pr r5 15 laws

=

1 2 2 2 r p(r - x ) dx L- r 2

or

Ix =

r

2

Ix =

2 m r2 5

Web) teaching in

Wide

L- r

2

r p(r - x ) dx copyright

m =

World

permitted.

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4 r p r3 3

on

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Thus,

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Ans.

17–6. Determine the mass moment of inertia Iz of the cone formed by revolving the shaded area around the z axis. The density of the material is r. Express the result in terms of the mass m of the cone.

z z ⫽ hr (r ⫺ y)

SOLUTION

h

Differential Element: The mass of the disk element shown shaded in Fig. a is ro ro 2 dm = r dV = rpr2dz. Here, r = y = ro z. Thus, dm = rp aro - zb dz. The h h mass moment of inertia of this element about the z axis is ro 4 1 1 1 1 dIz = dmr2 = (rpr2dz)r2 = rpr4dz = rparo - zb dz 2 2 2 2 h

y

x

Mass: The mass of the cone can be determined by integrating dm. Thus, h ro 2 rparo - z b dz m = dm = h L L0

laws

or

h ro 3 1 h 1 = rpc aro - z b a - b d 2 = rpro2h ro 3 h 3 0

Dissemination

Wide

copyright

in

teaching

Web)

Mass Moment of Inertia: Integrating dIz, we obtain h ro 4 1 dIz = rp aro - z b dz Iz = h L L0 2

the

not

instructors

States

World

permitted.

h ro 3 1 1 h 1 rpc aro - z b a - b d 2 = rpro4h ro 2 5 h 10 0 United

on

learning.

is

of

=

by

and

use

From the result of the mass, we obtain rpro2h = 3m. Thus, Iz can be written as the

(including

for

work

student

the

1 1 3 A rpro2h B ro2 = (3m)ro2 = mro2 10 10 10

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Iz =

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Ans.

r

17–7. y

The solid is formed by revolving the shaded area around the y axis. Determine the radius of gyration ky . The specific weight of the material is g = 380 lb>ft3.

3 in. y3

x

SOLUTION

3 in.

The moment of inertia of the solid : The mass of the disk element 1 dm = rpx2 dy = 81 rpy6 dy. dIy =

1 dmx2 2

=

1 A rpx2 dy B x2 2

=

1 1 rpx4 dy = rpy12 dy 2 2(94) 3

or

1 y12 dy rp 2(94) L0

teaching

Web)

L

dIy =

laws

Iy =

Dissemination

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copyright

in

= 29.632r

not

instructors

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permitted.

The mass of the solid:

of

is on

learning. United

and

use student

the

Lm

1 rp y6 dy = 12.117r 81 L0

by

dm =

the

3

m =

the

(including

for

work

29.632r Iy = = 1.56 in. Am A 12.117r

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ky =

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9x

Ans.

*17–8. y

The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of inertia Iy . The specific weight of concrete is g = 150 lb>ft3.

6 in.

y

SOLUTION d Iy = =

1 1 (dm)(10)2 - (dm)x2 2 2 1 1 [pr(10)2 dy](10)2 - prx2 dyx2 2 2 8

Iy = =

8

1 9 2 (10)4 dy a b y2dy R pr B 2 L0 L0 2 1 2 p(150) 3

32.2(12)

2

9 2

1 3

B (10)4(8) - a b a b (8)3 R

or

= 324.1 slug # in2 laws

Iy = 2.25 slug # ft 2

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in

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Web)

Ans.

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4 in.

2 2 8 in. 9x

x

17–9. z

Determine the moment of inertia Iz of the torus. The mass of the torus is m and the density r is constant. Suggestion: Use a shell element.

R

SOLUTION

a

dm = 2p(R - x)(2z¿r dx) dlz = (R - x)2 dm = 4pr[(R3 - 3R2 x + 3Rx2 - x3) 2a2 - x2 dx] a

Iz = 4pr[R3

L-a

a

2a2 - x2 dx - 3R 2

L-a

a

x3 2a 2 - x 2 dx + 3R

a

2a 2 - x 2 dx -

L-a

L-a

3

x 2a 2 - x 2

3 2 a) 4

= 2p2 rRa2 (R2 +

laws

or

Since m = rV = 2pRrpa2 3 2 a ) 4

in

teaching

Web)

Ans.

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Dissemination

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Iz = m(R2 +

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17–10. O

Determine the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through point O. The slender rod has a mass of 10 kg and the sphere has a mass of 15 kg.

450 mm

SOLUTION

A

Composite Parts: The pendulum can be subdivided into two segments as shown in Fig. a. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated.

100 mm B

Moment of Inertia: The moment of inertia of the slender rod segment (1) and the sphere segment (2) about the axis passing through their center of mass can be 1 2 computed from (IG)1 = ml2 and (IG)2 = mr2. The mass moment of inertia of 12 5 each segment about an axis passing through point O can be determined using the parallel-axis theorem. IO = ©IG + md2 1 2 (10)(0.452) + 10(0.2252) d + c (15)(0.12) + 15(0.552) d 12 5

or

= c

laws

= 5.27 kg # m2

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in

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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17–11. The slender rods have a weight of 3 lb>ft. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A.

A 2 ft

1 ft

SOLUTION Ans.

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laws

or

IA

1.5 ft

3(3) 1 3(3) 1 3(3) d (3)2 + c d(3)2 + c d(2)2 = 2.17 slug # ft2 = c 3 32.2 12 32.2 32.2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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1.5 ft

*17–12. Determine the moment of inertia of the solid steel assembly about the x axis. Steel has a specific weight of gst = 490 lb>ft3.

0.25 ft 0.5 ft

2 ft

SOLUTION Ix =

1 3 3 m (0.5)2 + m (0.5)2 m (0.25)2 2 1 10 2 10 3

1 3 1 3 1 490 = c p(0.5)2(3)(0.5)2 + a b p(0.5)2 (4)(0.5)2 a bp(0.25)2(2)(0.25)2 d a b 2 10 3 10 2 32.2 = 5.64 slug # ft2

sale

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permitted.

Dissemination

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copyright

in

teaching

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laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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3 ft

x

17–13. The wheel consists of a thin ring having a mass of 10 kg and four spokes made from slender rods, each having a mass of 2 kg. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A.

500 mm

SOLUTION A

IA = Io + md3 = c2 c

1 (4)(1)2 d + 10(0.5)2 d + 18(0.5)2 12

= 7.67 kg # m2

sale

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destroy

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Dissemination

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copyright

in

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laws

or

Ans.

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17–14. If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

4 ft

1 ft O

SOLUTION A

Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a 3 distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2). 2 Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = a

100 1 20 20 15 b (4 2) + 8c a b(32) + a b(2.52) d + a b(12) 32.2 12 32.2 32.2 32.2

laws

or

= 84.94 slug # ft2

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem 100 20 15 IA = IO + md 2, where m = + 8a b + = 8.5404 slug and d = 4 ft. 32.2 32.2 32.2 Thus, is

of

the

not

IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2

sale

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solely

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the

(including

for

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student

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Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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17–15. Determine the moment of inertia about an axis perpendicular to the page and passing through the pin at O. The thin plate has a hole in its center. Its thickness is 50 mm, and the material has a density r = 50 kg>m3.

O

150 mm

SOLUTION IG =

1.40 m

1 1 C 50(1.4)(1.4)(0.05) D C (1.4)2 + (1.4)2 D - C 50(p)(0.15)2(0.05) D (0.15)2 12 2

= 1.5987 kg # m2 IO = IG + md2 m = 50(1.4)(1.4)(0.05) - 50(p)(0.15)2(0.05) = 4.7233 kg IO = 1.5987 + 4.7233(1.4 sin 45°)2 = 6.23 kg # m2

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1.40 m

*17–16. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

200 mm

O

200 mm

SOLUTION Composite Parts: The plate can be subdivided into two segments as shown in Fig. a. Since segment (2) is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated.

200 mm

Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed as m1 = p(0.2 2)(20) = 0.8p kg and m2 = (0.2)(0.2)(20) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem. IO = ©IG + md2 laws

or

1 1 = c (0.8p)(0.22) + 0.8p(0.22) d - c (0.8)(0.22 + 0.22) + 0.8(0.22) d 2 12

Web)

= 0.113 kg # m2

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17–17. The assembly consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass of 2 kg/m. Determine the length L of DC so that the center of mass is at the bearing O. What is the moment of inertia of the assembly about an axis perpendicular to the page and passing through O?

0.8 m

0.5 m

D

0.2 m

L A

O

B C

SOLUTION Measured from the right side, y =

6(1.5) + 2(1.3)(0.65) = 0.5 6 + 1.3(2) + L(2)

L = 6.39 m IO =

Ans.

1 1 1 (6)(0.2)2 + 6(1)2 + (2)(1.3)(1.3)2 + 2(1.3)(0.15)2 + (2)(6.39)(6.39)2 + 2(6.39)(0.5)2 2 12 12

IO = 53.2 kg # m2

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17–18. The assembly consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass of 2 kg/m. If L = 0.75 m, determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through O.

0.8 m

0.5 m

D

0.2 m

L A

O

B C

SOLUTION IO =

1 1 1 (6)(0.2)2 + 6(1)2 + (2)(1.3)(1.3)2 + 2(1.3)(0.15)2 + (2)(0.75)(0.75)2 + 2(0.75)(0.5)2 2 12 12

IO = 6.99 kg # m2

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17–19. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg>m. The thin plate has a mass of 12 kg>m2. Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

0.4 m

0.4 m

A

B O

_ y

G

SOLUTION y =

1.5(3)(0.75) + p(0.3)2(12)(1.8) - p(0.1)2(12)(1.8)

C

1.5(3) + p(0.3)2(12) - p(0.1)2(12) + 0.8(3)

0.1 m

= 0.8878 m = 0.888 m IG =

Ans. 0.3 m

1 (0.8)(3)(0.8)2 + 0.8(3)(0.8878)2 12 +

1 (1.5)(3)(1.5)2 + 1.5(3)(0.75 - 0.8878)2 12

laws

or

1 + [p(0.3)2(12)(0.3)2 + [p(0.3)2(12)](1.8 - 0.8878)2 2 teaching

Web)

1 [p(0.1)2(12)(0.1)2 - [p(0.1)2(12)](1.8 - 0.8878)2 2

in

-

Ans.

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IG = 5.61 kg # m2

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1.5 m

*17–20. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg>m2. Determine the moment of inertia of the pendulum about an axis perpendicular to the page and passing through the pin at O.

0.4 m

0.4 m

A

B O

–y

G

SOLUTION Io =

1 1 1 [3(0.8)](0.8)2 + [3(1.5)](1.5)2 + [12(p)(0.3)2](0.3)2 12 3 2

+ [12(p)(0.3)2](1.8)2 -

C 0.1 m

1 [12(p)(0.1)2](0.1)2 - [12(p)(0.1)2](1.8)2 2

0.3 m

= 13.43 = 13.4 kg # m2

Ans.

Also, from the solution to Prob. 17–16, m = 3(0.8 + 1.5) + 12[p(0.3)2 - p(0.1)2] = 9.916 kg

laws

or

Io = IG + m d2 teaching

Web)

= 5.61 + 9.916(0.8878)2 in

= 13.4 kg # m2

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Ans.

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1.5 m

17–21. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

O

y 2m

G

SOLUTION y =

0.5 m

© ym 1(3) + 2.25(5) = = 1.781 m = 1.78 m ©m 3 + 5

Ans. 1m

IG = ©IG + md 2 =

1 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12

= 4.45 kg # m2

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17–22. 20 mm

Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION mc = 7.85(10 ) A (0.05)p(0.01) 3

2

x¿

B = 0.1233 kg

30 mm 20 mm

mr = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg

50 mm

1 Ix = 2 c (0.1233)(0.01)2 + (0.1233)(0.06)2 d 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B d 12

= 0.00325 kg # m2 = 3.25 g # m2

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photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

30 mm

17–23. 20 mm

Determine the moment of inertia of the overhung crank about the x¿ axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION mc = 7.85 A 10

3

x¿

B A (0.05)p(0.01) B = 0.1233 kg

30 mm

2

mp = 7.85 A 103 B A (0.03)(0.180)(0.02) B = 0.8478 kg

20 mm

50 mm

1 1 Ix = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d 2 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B + (0.8478)(0.06)2 d 12

= 0.00719 kg # m2 = 7.19 g # m2

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30 mm

*17–24. The door has a weight of 200 lb and a center of gravity at G. Determine how far the door moves in 2 s, starting from rest, if a man pushes on it at C with a horizontal force F = 30 lb. Also, find the vertical reactions at the rollers A and B.

6 ft

3 ft

200 )a 30 = ( 32.2 G

+ ©F = m(a ) ; : x G x

aG = 4.83 ft>s2 a+©MA = ©(Mk)A;

NB(12) - 200(6) + 30(9) = (

200 )(4.83)(7) 32.2

NB = 95.0 lb + c ©Fy = m(aG)y ;

Ans.

NA + 95.0 - 200 = 0 NA = 105 lb

Web)

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Ans. in

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1 2 a t 2 G

Dissemination

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s = s0 + v0t +

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not

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permitted.

1 (4.83)(2)2 = 9.66 ft 2

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

12 ft

F

SOLUTION

s = 0 + 0 +

B

G

C

+ ) (:

6 ft

A

5 ft

17–25. The door has a weight of 200 lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12 ft to the right in 5 s, starting from rest. Also, find the vertical reactions at the rollers A and B.

6 ft

3 ft

+ )s = s + v t + 1 a t2 (: 0 0 G 2 1 aG(5)2 2

12 = 0 + 0 + ac = 0.960 ft>s2 F =

200 (0.960) 32.2

F = 5.9627 lb = 5.96 lb

Ans.

teaching

Web)

laws

or

200 (0.960)(7) 32.2 in

NB(12) - 200(6) + 5.9627(9) =

Ans. not

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NB = 99.0 lb

United

on

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is

of

the

NA + 99.0 - 200 = 0

the

by

and

use

+ c ©Fy = m(aG)y ;

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NA = 101 lb

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or transmission in any form or by any means, electronic, mechanical,

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12 ft

F

SOLUTION

a + ©MA = ©(Mk)A ;

B

G

C

+ ©F = m(a ) ; : x G x

6 ft

A

Ans.

5 ft

17–26. The uniform pipe has a weight of 500 lb/ft and diameter of 2 ft. If it is hoisted as shown with an acceleration of 0.5 ft>s2, determine the internal moment at the center A of the pipe due to the lift.

0.5 ft/s2

5 ft A 1 ft 5 ft

SOLUTION Pipe: + c ©Fy = m ay ;

T-10 000 =

10 000 (0.5) 32.2

T = 10 155.27 lb Cables: + c ©Fy = 0;

10 155.27 - 2P cos 45° = 0 P = 7 180.867 lb

Web)

laws

or

) = 5 077.64 lb

teaching

1 22

-5000 (0.5)(5) 32.2 in

Px = Py = 7 180.867(

MA + 5000(5) - 5077.64(5) - 5077.64(1) =

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a + ©MA = ©(Mk)A ;

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not

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permitted.

MA = 5077.6 lb # ft = 5.08(103) lb # ft

5 ft

5 ft

5 ft

17–27. The drum truck supports the 600-lb drum that has a center of gravity at G. If the operator pushes it forward with a horizontal force of 20 lb, determine the acceleration of the truck and the normal reactions at each of the four wheels. Neglect the mass of the wheels.

20 lb G 2 ft

SOLUTION + ©F = m(a ) ; ; x G x

20 = (

A

600 )a 32.2 G Ans. 600 (1.0733)(2) 32.2

20(4) - 600(0.5) + 2NB (1.5) = NB = 86.7 lb

+ c ©Fy = m(aG)y ;

Ans.

2NA + 2(86.7) - 600 = 0 NA = 213 lb

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B

0.5 ft 1 ft

aG = 1.0733 ft>s2 = 1.07 ft>s2 a + ©MA = ©(Mk)A ;

4 ft

*17–28. 1 ft

If the cart is given a constant acceleration of a = 6 ft>s2 up the inclined plane, determine the force developed in rod AC and the horizontal and vertical components of force at pin B. The crate has a weight of 150 lb with center of gravity at G, and it is secured on the platform, so that it does not slide. Neglect the platform’s weight.

G

1 ft C 30⬚

SOLUTION A

Equations of Motion: FAC can be obtained directly by writing the moment equation of motion about B,

30⬚

+ ©MB = ©(Mk)B; 150(2) - FAC sin 60°(3) = - a

150 150 b (6) cos 30°(1) - a b(6) sin 30°(2) 32.2 32.2 FAC = 135.54 lb = 136 lb

Ans.

Using this result and writing the force equations of motion along the x and y axes, 150 b (6) cos 30° 32.2 Bx = 43.57 lb = 43.6 lb

Ans.

150 (6) sin 30° 32.2 By = 46.59 lb = 46.6 lb

Ans.

or

135.54 cos 60° - Bx = a

laws

+ ©F = m(a ) ; : x G x

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By + 135.54 sin 60° - 150 =

copyright

+ c ©Fy = m(aG)y;

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2 ft

a B

P

17–29. 1 ft

If the strut AC can withstand a maximum compression force of 150 lb before it fails, determine the cart’s maximum permissible acceleration. The crate has a weight of 150 lb with center of gravity at G, and it is secured on the platform, so that it does not slide. Neglect the platform’s weight.

G

1 ft C 30⬚

SOLUTION A

Equations of Motion: FAC in terms of a can be obtained directly by writing the moment equation of motion about B.

30⬚

+ ©MB = ©(Mk)B; 150(2) - FAC sin 60°(3) = - a

150 150 b a cos 30°(1) - a ba sin 30°(2) 32.2 32.2

FAC = (3.346a + 115.47) lb Assuming AC is about to fail, FAC = 150 = 3.346a + 115.47 a = 10.3 ft>s2

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Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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2 ft

a B

P

17–30. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the tension in the cable when the truck begins to accelerate at 5 m>s2. Also, what are the horizontal and vertical components of reaction at the hinge C?

B

A

30⬚

G C

1.5 m 45⬚

SOLUTION a + ©MC = ©(Mk)C ;

T sin 30°(2.5) - 12 262.5(1.5 cos 45°) = 1250(5)(1.5 sin 45°) T = 15 708.4 N = 15.7 kN

+ ©F = m(a ) ; ; x G x

Ans.

- Cx + 15 708.4 cos 15° = 1250(5) Cx = 8.92 kN

+ c ©Fy = m(aG)y ;

Ans.

Cy - 12 262.5 - 15 708.4 sin 15° = 0 Cy = 16.3 kN

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1m

17–31. The pipe has a length of 3 m and a mass of 500 kg. It is attached to the back of the truck using a 0.6-m-long chain AB. If the coefficient of kinetic friction at C is mk = 0.4, determine the acceleration of the truck if the angle u = 10° with the road as shown.

B 3m

θ = 10°

C

SOLUTION f = sin - 1 ¢

0.4791 ≤ = 52.98° 0.6

+ ©F = m(a ) ; : x G x

T cos 52.98° - 0.4NC = 500aG

+ c ©Fy = m(aG)y ;

NC - 500(9.81) + T sin 52.98° = 0

a + ©MC = ©(Mk)C;

- 500(9.81)(1.5 cos 10°) + T sin (52.98° - 10°)(3) = -500aG(0.2605) T = 3.39 kN NC = 2.19 kN aG = 2.33 m>s2

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sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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A 1m

*17–32. The mountain bike has a mass of 40 kg with center of mass at point G1, while the rider has a mass of 60 kg with center of mass at point G2. Determine the maximum deceleration when the brake is applied to the front wheel, without causing the rear wheel B to leave the road. Assume that the front wheel does not slip. Neglect the mass of all the wheels.

G2

1.25 m

G1 0.4 m A

SOLUTION

B 0.4 m

Equations of Motion: Since the rear wheel B is required to just leave the road, NB = 0. Thus, the acceleration a of the bike can be obtained directly by writing the moment equation of motion about point A. + ©MA = (Mk)A; - 40(9.81)(0.4) - 60(9.81)(0.6) = - 40a(0.4) - 60a(1.25) a = 5.606 m>s2 = 5.61 m>s2

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0.4 m 0.2 m

17–33. The mountain bike has a mass of 40 kg with center of mass at point G1, while the rider has a mass of 60 kg with center of mass at point G2. When the brake is applied to the front wheel, it causes the bike to decelerate at a constant rate of 3 m>s2. Determine the normal reaction the road exerts on the front and rear wheels. Assume that the rear wheel is free to roll. Neglect the mass of all the wheels.

G2

1.25 m

G1 0.4 m A

SOLUTION

B 0.4 m

Equations of Motion: NB can be obtained directly by writing the moment equation of motion about point A. + ©MA = (Mk)A; NB(1) - 40(9.81)(0.4) - 60(9.81)(0.6) = - 60(3)(1.25) - 40(3)(0.4) NB = 237.12 N = 237 N

Ans.

Using this result and writing the force equations of motion along the y axis, NA + 237.12 - 40(9.81) - 60(9.81) = 0 NA = 743.88 N = 744 N

Ans.

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+ c ©Fy = m(aG)y;

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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0.4 m 0.2 m

17–34. The trailer with its load has a mass of 150 kg and a center of mass at G. If it is subjected to a horizontal force of P = 600 N, determine the trailer’s acceleration and the normal force on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass.

G P

1.25 m 0.25 m

SOLUTION

0.75 m

Equations of Motion: Writing the force equation of motion along the x axis, + ©F = m(a ) ; : x G x

600 = 150a

a = 4 m>s2 :

Ans.

Using this result to write the moment equation about point A, a + ©MA = (Mk)A ;

150(9.81)(1.25) - 600(0.5) - NB(2) = -150(4)(1.25) NB = 1144.69 N = 1.14 kN

Ans.

Using this result to write the force equation of motion along the y axis, or

NA + 1144.69 - 150(9.81) = 150(0) laws

+ c ©Fy = m(a G)y ;

NA = 326.81 N = 327 N

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sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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B

0.25 m

1.25 m

A

600 N

0.5 m

17–35. At the start of a race, the rear drive wheels B of the 1550-lb car slip on the track. Determine the car’s acceleration and the normal reaction the track exerts on the front pair of wheels A and rear pair of wheels B. The coefficient of kinetic friction is mk=0.7, and the mass center of the car is at G. The front wheels are free to roll. Neglect the mass of all the wheels.

0.75 ft G A

SOLUTION Equations of Motion: Since the rear wheels B are required to slip, the frictional force developed is FB = msNB = 0.7NB. 1550 a 32.2

+ ©F = m(a ) ; ; x G x

0.7NB =

+ c ©Fy = m(aG)y;

NA + NB - 1550 = 0

(2)

a + ©MG = 0;

NB(4.75) - 0.7NB(0.75) - NA(6) = 0

(3)

(1)

Solving Eqs. (1), (2), and (3) yields a = 13.2 ft>s2

or

Ans.

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NB = 909.54 lb = 910 lb

laws

NA = 640.46 lb = 640 lb

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B 6 ft

4.75 ft

*17–36. Determine the maximum acceleration that can be achieved by the car without having the front wheels A leave the track or the rear drive wheels B slip on the track. The coefficient of static friction is ms = 0.9. The car’s mass center is at G, and the front wheels are free to roll. Neglect the mass of all the wheels.

0.75 ft G A

B 6 ft

4.75 ft

SOLUTION Equations of Motion: + ©F = m(a ) ; ; x G x + c ©Fy = m(a G)y;

FB =

1550 a 32.2

(1)

NA + NB - 1550 = 0

a + ©MG = 0;

(2)

NB(4.75) - FB(0.75) - NA(6) = 0

(3)

If we assume that the front wheels are about to leave the track, NA = 0. Substituting this value into Eqs. (2) and (3) and solving Eqs. (1), (2), (3), a = 203.93 ft>s2

or

FB = 9816.67 lb

laws

NB = 1550 lb

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Since FB 7 (FB)max = msNB = 0.9(1550) lb = 1395 lb, the rear wheels will slip. Thus, the solution must be reworked so that the rear wheels are about to slip. permitted.

(4) not

instructors

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Dissemination

FB = msNB = 0.9NB

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

and

NB = 923.08 lb for

work

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NA = 626.92 lb

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a = 17.26 ft>s2 = 17.3 ft>s2

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

,

17–37. If the 4500-lb van has front-wheel drive, and the coefficient of static friction between the front wheels A and the road is ms = 0.8, determine the normal reactions on the pairs of front and rear wheels when the van has maximum acceleration. Also, find this maximum acceleration. The rear wheels are free to roll. Neglect the mass of the wheels.

G 2.5 ft B

SOLUTION Equations of Motion: The maximum acceleration occurs when the front wheels are about to slip. Thus, FA = msNA = 0.8 NA. Referring to the free-body diagram of the van shown in Fig. a, we have + ©F = m(a ) ; : x G x

0.8NA = a

4500 ba 32.2 max

(1)

+ c ©Fy = m(aG)y;

NA + NB - 4500 = 0

(2)

+ ©MG = 0;

NA(3.5) + 0.8NA(2.5) - NB(6) = 0

(3)

Solving Eqs. (1), (2), and (3) yields NB = 2152.171 lb = 2.15 kip

Ans. laws

or

NA = 2347.82 lb = 2.35 kip amax = 13.44 ft>s2 = 13.4 ft>s2

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6 ft

A 3.5 ft

17–38. If the 4500-lb van has rear-wheel drive, and the coefficient of static friction between the front wheels B and the road is ms = 0.8, determine the normal reactions on the pairs of front and rear wheels when the van has maximum acceleration. The front wheels are free to roll. Neglect the mass of the wheels.

G 2.5 ft B

SOLUTION Equations of Motion: The maximum acceleration occurs when the rear wheels are about to slip. Thus, FB = msNB = 0.8 NB. Referring to Fig. a, + ©F = m(a ) ; : x G x

0.8NB = a

4500 ba 32.2 max

+ c ©Fy = m(aG)y;

NA + NB - 4500 = a

+ ©MG = 0;

NA(3.5) + 0.8NB(2.5) - NB(6) = 0

(1) 4500 b(0) 32.2

(2) (3)

Solving Eqs. (1), (2), and (3) yields amax = 12.02 ft>s2 = 12.0 ft>s2

Ans. or

NB = 2.10 kip

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NA = 2.40 kip

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6 ft

A 3.5 ft

17–39. A

The uniform bar of mass m is pin connected to the collar, which slides along the smooth horizontal rod. If the collar is given a constant acceleration of a, determine the bar’s inclination angle u. Neglect the collar’s mass. L

SOLUTION Equations of Motion: Writing the moment equation of motion about point A, + ©MA = (Mk)A;

mg sin ua

L L b = ma cos u a b 2 2

a u = tan-1 a b g

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u

a

*17–40. The lift truck has a mass of 70 kg and mass center at G. If it lifts the 120-kg spool with an acceleration of 3 m>s2 , determine the reactions on each of the four wheels. The loading is symmetric. Neglect the mass of the movable arm CD.

0.7 m

SOLUTION a + ©MB = ©(Mk)B ;

D

C

70(9.81)(0.5) + 120(9.81)(0.7) - 2NA(1.25)

G

= -120(3)(0.7)

0.4 m

NA = 567.76 N = 568 N

A

Ans.

B 0.75 m

+ c ©Fy = m(a G)y ;

2(567.76) + 2NB - 120(9.81) - 70(9.81) = 120(3) NB = 544 N

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0.5 m

17–41. The lift truck has a mass of 70 kg and mass center at G. Determine the largest upward acceleration of the 120-kg spool so that no reaction on the wheels exceeds 600 N. 0.7 m

SOLUTION Assume NA = 600 N. a + ©MB = ©(Mk)B ;

G

70(9.81)(0.5) + 120(9.81)(0.7) - 2(600)(1.25) = -120a(0.7)

0.4 m A

a = 3.960 m>s2 + c ©Fy = m(a G)y ;

2(600) + 2NB - 120(9.81) - 70(9.81) = 120(3.960) OK

Thus a = 3.96 m>s2

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sale Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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B 0.75 m

NB = 570 N 6 600 N

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

D

C

0.5 m

17–42. The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and the cart is ms = 0.5.

0.6 m

F

1m

SOLUTION 15

Equations of Motion: Assume that the crate slips, then Ff = ms N = 0.5N. a + ©MA = ©(Mk)A ;

50(9.81) cos 15°(x) - 50(9.81) sin 15°(0.5) = 50a cos 15°(0.5) + 50a sin 15°(x)

(1)

+Q©Fy¿ = m(a G)y¿ ;

N - 50(9.81) cos 15° = -50a sin 15°

(2)

R+ ©Fx¿ = m(aG)x¿ ;

50(9.81) sin 15° - 0.5N = -50a cos 15°

(3)

Solving Eqs. (1), (2), and (3) yields x = 0.250 m or

N = 447.81 N

a = 2.01 m>s2

teaching

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laws

Ans. in

Since x 6 0.3 m , then crate will not tip. Thus, the crate slips.

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Ans.

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17–43. 1 ft

Determine the acceleration of the 150-lb cabinet and the normal reaction under the legs A and B if P = 35 lb. The coefficients of static and kinetic friction between the cabinet and the plane are ms = 0.2 and mk = 0.15, respectively. The cabinet’s center of gravity is located at G. P

Equations of Equilibrium: The free-body diagram of the cabinet under the static condition is shown in Fig. a, where P is the unknown minimum force needed to move the cabinet. We will assume that the cabinet slides before it tips. Then, FA = msNA = 0.2NA and FB = msNB = 0.2NB. + ©F = 0; : x

P - 0.2NA - 0.2NB = 0

(1)

+ c ©Fy = 0;

NA + NB - 150 = 0

(2)

+ ©MA = 0;

NB(2) - 150(1) - P(4) = 0

(3)

NB = 135 lb laws

or

NA = 15 lb

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Since P 6 35 lb and NA is positive, the cabinet will slide.

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NA + NB - 150 = 0

(4) (5)

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+ ©F = m(a ) ; : x G x

150 ba 32.2 by

35 - 0.15NA - 0.15NB = a

the

+ ©F = m(a ) ; : x G x

work

solely

NB(1) - 0.15NB(3.5) - 0.15NA(3.5) - NA(1) - 35(0.5) = 0

(6)

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assessing

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+ ©MG = 0;

the

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Solving Eqs. (4), (5), and (6) yields

part

a = 2.68 ft>s2 and

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NB = 123 lb sale

will

their

NA = 26.9 lb

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Ans.

permitted.

Dissemination

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copyright

Equations of Motion: Since the cabinet is in motion, FA = mkNA = 0.15NA and FB = mkNB = 0.15NB. Referring to the free-body diagram of the cabinet shown in Fig. b,

3.5 ft

A

Solving Eqs. (1), (2), and (3) yields P = 30 lb

G 4 ft

SOLUTION

1 ft

B

*17–44. The assembly has a mass of 8 Mg and is hoisted using the boom and pulley system. If the winch at B draws in the cable with an acceleration of 2 m>s2, determine the compressive force in the hydraulic cylinder needed to support the boom. The boom has a mass of 2 Mg and mass center at G.

6m

2m

SOLUTION

G

sB + 2sL = l aB = - 2aL

4m

2 = - 2aL

B

C

60

aL = - 1 m>s

2

A

1m

D

Assembly: + c ©Fy = m ay ;

2m

2T - 8(103)(9.81) = 8(103)(1)

laws

or

T = 43.24 kN

teaching

Web)

Boom: in

FCD(2) - 2(103)(9.81)(6 cos 60°) - 2(43.24)(103)(12 cos 60°) = 0

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FCD = 289 kN

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a + ©MA = 0;

17–45. The 2-Mg truck achieves a speed of 15 m>s with a constant acceleration after it has traveled a distance of 100 m, starting from rest. Determine the normal force exerted on each pair of front wheels B and rear driving wheels A. Also, find the traction force on the pair of wheels at A. The front wheels are free to roll. Neglect the mass of the wheels.

G 0.75 m A 2m

SOLUTION Kinematics: The acceleration of the truck can be determined from v2 = v02 + 2ac(s - s0) 152 = 0 + 2a(100 - 0) a = 1.125 m>s2 Equations of Motion: NB can be obtained directly by writing the moment equation of motion about point A. + ©MA = (Mk)A;

NB(3.5) - 2000(9.81)(2) = - 2000(1.125)(0.75) NB = 10 729.29 N = 10.7 kN

Ans. laws

or

Using this result and writing the force equations of motion along the x and y axes, Web)

NA + 10 729.29 - 2000(9.81) = 0

teaching

+ c ©Fy = m(aG)y;

Ans. in

FA = 2000(1.125) = 2250 N = 2.25 kN NA = 8890.71 N = 8.89 kN

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+ ©F = m(a ) ; : x G x

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B 1.5 m

17–46. Determine the shortest time possible for the rear-wheel drive, 2-Mg truck to achieve a speed of 16 m>s with a constant acceleration starting from rest. The coefficient of static friction between the wheels and the road surface is ms = 0.8. The front wheels are free to roll. Neglect the mass of the wheels.

G 0.75 m A 2m

SOLUTION Equations of Motion: The maximum acceleration of the truck occurs when its rear wheels are on the verge of slipping. Thus, FA = msNA = 0.8NA. Referring to the free-body diagram of the truck shown in Fig. a, we can write + ©F = m(a ) ; : x G x

0.8NA = 2000a

(1)

+ c ©Fy = m(aG)y;

NA + NB - 2000(9.81) = 0

(2)

+ ©MG = 0;

NB(1.5) + 0.8NA(0.75) - NA(2) = 0

(3)

Solving Eqs. (1), (2), and (3) yields a = 4.059 m>s2

NB = 9471.72 N

or

NA = 10 148.28 N

teaching

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laws

Kinematics: Since the acceleration of the truck is constant, we can apply in

v = v0 + at

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+ B A:

Ans.

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t = 3.94 s

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16 = 0 + 4.059t

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B 1.5 m

17–47. 0.5 ft

The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. If the acceleration is a = 20 ft>s2, determine the maximum height h of G2 of the rider so that the snowmobile’s front skid does not lift off the ground. Also, what are the traction (horizontal) force and normal reaction under the rear tracks at A?

a G2 G1 1 ft

SOLUTION

1.5 ft

Equations of Motion: Since the front skid is required to be on the verge of lift off, NB = 0. Writing the moment equation about point A and referring to Fig. a, a + ©MA = (Mk)A ;

150 250 (20)(hmax) + (20)(1) 32.2 32.2

250(1.5) + 150(0.5) =

hmax = 3.163 ft = 3.16 ft

Ans.

Writing the force equations of motion along the x and y axes, + ©F = m(a ) ; ; x G x

FA =

250 150 (20) + (20) 32.2 32.2 FA = 248.45 lb = 248 lb

laws

or

Ans. Web)

NA - 250 - 150 = 0 teaching

+ c ©Fy = m(a G)y ;

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Ans.

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NA = 400 lb

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h A

*17–48. The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. If h = 3 ft, determine the snowmobile’s maximum permissible acceleration a so that its front skid does not lift off the ground. Also, find the traction (horizontal) force and the normal reaction under the rear tracks at A.

0.5 ft

a G2 G1 1 ft

SOLUTION

1.5 ft

Equations of Motion: Since the front skid is required to be on the verge of lift off, NB = 0. Writing the moment equation about point A and referring to Fig. a, a + ©MA = (Mk)A ;

150 250 a b (3) + a a b(1) 32.2 max 32.2 max

250(1.5) + 150(0.5) = a a max = 20.7 ft>s2

Ans.

Writing the force equations of motion along the x and y axes and using this result, we have FA =

150 250 (20.7) + (20.7) 32.2 32.2 or

+ ©F = m(a ) ; ; x G x

laws

FA = 257.14 lb = 257 lb

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+ c ©Fy = m(a G)y ; NA - 150 - 250 = 0 NA = 400 lb

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Ans.

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h A

17–49. A

If the cart’s mass is 30 kg and it is subjected to a horizontal force of P = 90 N, determine the tension in cord AB and the horizontal and vertical components of reaction on end C of the uniform 15-kg rod BC.

30⬚ 1m

C

SOLUTION Equations of Motion: The acceleration a of the cart and the rod can be determined by considering the free-body diagram of the cart and rod system shown in Fig. a. + ©F = m(a ) ; : x G x

a = 2 m>s2

90 = (15 + 30)a

The force in the cord can be obtained directly by writing the moment equation of motion about point C by referring to Fig. b. + ©MC = (Mk)C;

FAB sin 30°(1) - 15(9.81) cos 30°(0.5) = - 15(2) sin 30°(0.5) FAB = 112.44 N = 112 N

or

Ans.

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Using this result and applying the force equations of motion along the x and y axes, in

-Cx + 112.44 sin 30° = 15(2)

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+ ©F = m(a ) ; : x G x

Ans.

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Cx = 26.22 N = 26.2 N the

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Cy + 112.44 cos 30° - 15(9.81) = 0

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of

+ c ©Fy = m(aG)y;

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Cy = 49.78 N = 49.8 N

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Ans.

30⬚

B

P

17–50. A

If the cart’s mass is 30 kg, determine the horizontal force P that should be applied to the cart so that the cord AB just becomes slack. The uniform rod BC has a mass of 15 kg.

30⬚ 1m

30⬚

C

SOLUTION Equations of Motion: Since cord AB is required to be on the verge of becoming slack, FAB = 0. The corresponding acceleration a of the rod can be obtained directly by writing the moment equation of motion about point C. By referring to Fig. a. + ©MC = ©(MC)A;

-15(9.81) cos 30°(0.5) = - 15a sin 30°(0.5) a = 16.99 m>s2

Using this result and writing the force equation of motion along the x axis and referring to the free-body diagram of the cart and rod system shown in Fig. b, P = (30 + 15)(16.99) or

+ B ©F = m(a ) ; A: x G x

Ans.

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= 764.61 N = 765 N

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B

P

17–51. at

The pipe has a mass of 800 kg and is being towed behind the truck. If the acceleration of the truck is a t = 0.5 m>s2, determine the angle u and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A G 0.4 m

SOLUTION + ©F = ma ; : x x

- 0.1NC + T cos 45° = 800(0.5)

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

- 0.1NC(0.4) + T sin f(0.4) = 0

C

NC = 6770.9 N T = 1523.24 N = 1.52 kN 0.1(6770.9) 1523.24

f = 26.39°

u = 45° - f = 18.6°

Ans.

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or

sin f =

Ans.

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u

45

*17–52. at

The pipe has a mass of 800 kg and is being towed behind a truck. If the angle u = 30°, determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A

SOLUTION + ©F = ma ; : x x

T cos 45° - 0.1NC = 800a

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

u

G 0.4 m C

T sin 15°(0.4) - 0.1NC(0.4) = 0

NC = 6161 N Ans.

a = 1.33 m>s2

Ans.

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T = 2382 N = 2.38 kN

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45

17–53. The arched pipe has a mass of 80 kg and rests on the surface of the platform. As it is hoisted from one level to the next, a = 0.25 rad>s2 and v = 0.5 rad>s at the instant u = 30°. If it does not slip, determine the normal reactions of the arch on the platform at this instant.

500 mm G A 1m

θ

B

200 mm

ω, α 1m

SOLUTION + c ©Fy = m(aG)y ;

NA + NB - 80(9.81) = 20 sin 60° - 20 cos 60° NA + NB = 792.12

a + ©MA = ©(Mk)A ;

NB(1) - 80(9.81)(0.5) = 20 cos 60°(0.2) + 20 sin 60°(0.5) - 20 cos 60°(0.5) + 20 sin 60°(0.2) Ans.

NA = 391 N

Ans.

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NB = 402 N

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17–54. The arched pipe has a mass of 80 kg and rests on the surface of the platform for which the coefficient of static friction is ms = 0.3. Determine the greatest angular acceleration a of the platform, starting from rest when u = 45°, without causing the pipe to slip on the platform.

500 mm G A 1m

θ

B

200 mm

ω, α 1m

SOLUTION aG = (aG)t = (1)(a) a + ©MA = ©(Mk)A ;

NB(1) - 80(9.81)(0.5) = 80(1a)(sin 45°)(0.2) + 80(1a)(cos 45°)(0.5)

+ ©F = m(a ) ; ; x G x

0.3NA + 0.3NB = 80(1a) sin 45°

+ c ©Fy = m(aG)y ;

NA + NB - 80(9.81) = 80(1a) cos 45°

Solving, a = 5.95 rad>s2

Ans.

or

NA = 494 N

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NB = 628 N

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17–55. 1 ft

At the instant shown, link CD rotates with an angular velocity of vCD = 8 rad>s. If link CD is subjected to a couple moment of M = 650 lb # ft, determine the force developed in link AB and the angular acceleration of the links at this instant. Neglect the weight of the links and the platform. The crate weighs 150 lb and is fully secured on the platform.

D

M ⫽ 650 lb⭈ft

Equilibrium: Since the mass of link CD can be neglected, Dt can be obtained directly by writing the moment equation of equilibrium about point C using the free-body diagram of link CD, Fig. a, Dt(4) - 650 = 0

Dt = 162.5 lb

150 C a(4) D 32.2

a = 8.72 rad>s2

Ans.

150 (256) 32.2

©Fn = m(aG)n;

Dn + FAB + 150 =

a ©MG = 0;

Dn(1) - FAB(2) + 162.5(1) = 0

(1) teaching

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162.5 =

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Dissemination

Solving Eqs. (1) and (2), we obtain

Ans.

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FAB = 402 lb

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Dn = 641 lb

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4 ft A

C 3 ft

Equations of Motion: Since the crate undergoes curvilinear translation, (aG)n = v2rG = 82(4) = 256 ft>s2 and (aG)t = arG = a(4). Referring to the freebody diagram of the crate, Fig. b, we have ©Ft = m(aG)t;

B

vCD ⫽ 8 rad/s

SOLUTION

a + ©MC = 0;

G

1 ft

*17–56. A

Determine the force developed in the links and the acceleration of the bar’s mass center immediately after the cord fails. Neglect the mass of links AB and CD. The uniform bar has a mass of 20 kg.

45⬚ 0.4 m D

B 0.6 m

SOLUTION Equations of Motion: Since the bar is still at rest at the instant the cord fails, vG = 0. v2G = 0. Referring to the free-body diagram of the bar, Fig. a, Thus, (aG)n = r ©Fn = m(aG)n;

TAB + TCD - 20(9.81) cos 45° + 50 cos 45° = 0

©Ft = m(aG)t;

20(9.81) sin 45° + 50 sin 45° = 20(aG)t

+ ©MG = 0;

TCD cos 45°(0.3) - TAB cos 45°(0.3) = 0

Solving, TAB = TCD = 51.68 N = 51.7 N

or

Ans.

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(aG)t = 8.704 m>s2

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Since (aG)n = 0, then

Dissemination

aG = (aG)t = 8.70 m>s2 R

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Ans.

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C

P ⫽ 50 N

17–57. The 10-kg wheel has a radius of gyration kA = 200 mm. If the wheel is subjected to a moment M = 15t2 N # m, where t is in seconds, determine its angular velocity when t = 3 s starting from rest. Also, compute the reactions which the fixed pin A exerts on the wheel during the motion.

M A

SOLUTION + ©F = m(a ) ; : x G x

Ax = 0

+ c ©Fy = m(aG)y;

Ay - 10(9.81) = 0

c + ©MA = Ia a;

5t = 10(0.2)2a a =

dv = 12.5t dt 3

v =

L0

12.5 2 (3) 2

12.5t dt =

v = 56.2 rad>s

or

Ans. Ans.

Ay = 98.1 N

Ans.

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Ax = 0

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17–58. The 80-kg disk is supported by a pin at A. If it is released from rest from the position shown, determine the initial horizontal and vertical components of reaction at the pin. A 1.5 m

SOLUTION + ©F = m(a ) ; ; x G x

Ax = 0

+ c ©Fy = m(aG)y ;

Ay - 80(9.81) = - 80(1.5)(a)

c + ©MA = IAa;

3 80(9.81)(1.5) = [ (80)(1.5)2]a 2

Ans.

a = 4.36 rad>s2 Ay = 262 N

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17–59. The uniform slender rod has a mass m. If it is released from rest when u = 0°, determine the magnitude of the reactive force exerted on it by pin B when u = 90°.

L 3 A B

u

2 L 3

SOLUTION C

Equations of Motion: Since the rod rotates about a fixed axis passing through point L L B, (aG)t = a rG = a a b and (aG)n = v2rG = v2 a b . The mass moment of inertia 6 6 1 2 mL . Writing the moment equation of motion about of the rod about its G is IG = 12 point B, - mg cos u a

+ ©MB = ©(Mk)B;

a =

L L L 1 b = - m c a a b d a b - a mL2 ba 6 6 6 12

3g cos u 2L

laws

or

1 This equation can also be obtained by applying ©MB = IBa, whereIB = mL2 + 12 2 1 L ma b = mL2. Thus, 6 9 in

teaching

Web)

L 1 b = - a mL2 b a 6 9

Wide

copyright

-mg cos u a

World

permitted.

Dissemination

3g cos u 2L

and

use

United

on

learning.

Using this result and writing the force equation of motion along the n and t axes,

is

of

the

not

instructors

a =

States

+ ©MB = IBa;

work

student

the

by

3g L cos u b a b d 2L 6

(including

for

mg cos u - Bt = mc a

©Ft = m(aG)t;

assessing

1 mv2L + mg sin u 6

this

integrity

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the part

(2)

and

any

courses

Bn =

and

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L bd 6 provided

Bn - mg sin u = m c v2 a

will

their

destroy

©Fn = m(aG)n;

(1)

is

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solely

of

protected

the

3 mg cos u 4

of

Bt =

sale

Kinematics: The angular velocity of the rod can be determined by integrating L

vdv = v

L0

adu u

vdv =

v =

L

3g cos u du 2L L0

3g sin u BL

When u = 90°, v =

3g . Substituting this result and u = 90° into Eqs. (1) and (2), AL

3 mg cos 90° = 0 4 3g 1 3 Bn = ma b(L) + mg sin 90° = mg 6 L 2

Bt =

FA = 3At2 + An2 =

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2 3 3 02 + a mg b = mg C 2 2

Ans.

*17–60. ω

The drum has a weight of 80 lb and a radius of gyration kO = 0.4 ft. If the cable, which is wrapped around the drum, is subjected to a vertical force P = 15 lb, determine the time needed to increase the drum’s angular velocity from v1 = 5 rad>s to v2 = 25 rad>s. Neglect the mass of the cable.

0.5 ft O

SOLUTION c + ©MO = IO a;

15(0.5) = [

80 (0.4)2] a 32.2

P

(1)

a = 18.87 rad>s2 (c + )v = v0 + a t 25 = 5 + 18.87 t t = 1.06 s

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or

Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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17–61. Cable is unwound from a spool supported on small rollers at A and B by exerting a force of T = 300 N on the cable in the direction shown. Compute the time needed to unravel 5 m of cable from the spool if the spool and cable have a total mass of 600 kg and a centroidal radius of gyration of kO = 1.2 m. For the calculation, neglect the mass of the cable being unwound and the mass of the rollers at A and B. The rollers turn with no friction.

T

300 N 1.5 m 30

0.8 m O

A

B

SOLUTION 1m

Equations of Motion: The mass moment of inertia of the spool about point O is given by IO = mk2O = 600 A 1.2 2 B = 864 kg # m2. Applying Eq. 17–16, we have a + ©MO = IO a;

- 300(0.8) = - 864a

a = 0.2778 rad>s2

5 s Kinematics: Here, the angular displacement u = = = 6.25 rad. Applying r 0.8 1 2 equation u = u0 + v0 t + at , we have 2 (c +)

6.25 = 0 + 0 +

1 (0.2778)t2 2 or

t = 6.71 s

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Ans.

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17–62. The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so that the torque developed is M = 15u2 lb # ft, where u is in radians. If the bar is released from rest when it is vertical at u = 90°, determine its angular velocity at the instant u = 0°.

1 ft u 1 ft O

SOLUTION c + ©MO = IOa; - 5u = [

1 10 ( )(2)2]a 12 32.2

- 48.3 u = a a du = v dv o

-

Lp2

48.3 u du =

v

L0

v dv

1 48.3 p 2 ( ) = v2 2 2 2 v = 10.9 rad/s

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Dissemination

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laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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17–63. The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so that the torque developed is M = 15u2 lb # ft, where u is in radians. If the bar is released from rest when it is vertical at u = 90°, determine its angular velocity at the instant u = 45°.

1 ft u 1 ft O

SOLUTION c + ©MO = IOa;

5u = [

1 10 ( )(2)2]a 12 32.2

a = - 48.3u a du = v dv -

p 4

Lp2

v

48.3u du =

L0

v dv

p p 1 - 24.15 a( )2 - ( )2 b = v2 4 2 2 v = 9.45 rad s

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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*17–64. If shaft BC is subjected to a torque of M = (0.45t1>2) N # m, where t is in seconds, determine the angular velocity of the 3-kg rod AB when t = 4 s, starting from rest. Neglect the mass of shaft BC.

B 300 mm

C

A

SOLUTION Equations of Motion: The mass moment of inertia of the rod about the z axis is 1 Iz = IG + md2 = (3) A 0.32 B + 3 A 0.152 B = 0.09 kg # m2. Writing the moment 12 equation of motion about the z axis, 0.45t1>2 = 0.09a

©Mz = Iza;

a = 5t1>2 rad>s2

Kinematics: The angular velocity of the rod can be obtained by integration. L

a dt

v

L0

dv =

t 1>2

L0

5t dt or

dv =

laws

L

Wide

copyright

in

teaching

Web)

v = A 3.333t3>2 B rad>s

permitted.

Dissemination

When t = 4 s,

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is

This

provided

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assessing

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the

(including

for

work

student

the

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on

learning.

is

of

the

Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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not

instructors

States

World

v = 3.333 A 43>2 B = 26.7 rad>s

M ⫽ (0.45 t1/2) N⭈m

17–65. 0.6 m

Determine the vertical and horizontal components of reaction at the pin support A and the angular acceleration of the 12-kg rod at the instant shown, when the rod has an angular velocity of v = 5 rad>s.

A v ⫽ 5 rad/s

SOLUTION Equations of Motion: Since the rod rotates about a fixed axis passing through

point A, (aG)t = arG = a(0.3) and (aG)n = v2rG = A 52 B (0.3) = 7.5 m>s2. The mass 1 mL2 = moment of inertia of the rod about its mass center is IG = 12 1 (12) A 0.62 B = 0.36 kg # m2. Writing the moment equation of motion about point A 12 and referring to Fig. a, - 12(9.81)(0.3) = - 0.36a - 12[a(0.3)](0.3) a = 24.525 rad>s2 = 24.5 rad>s2

Ans. 1 2 ml = 3 laws

This result can also be obtained by applying ©MA = IAa, where IA =

or

+ ©MA = ©(Mk)A;

teaching in

-12(9.81)(0.3) = - 1.44a

Ans. is

of

the

not

instructors

States

World

a = 24.525 rad>s2 = 24.5 rad>s2

for

work

student

the

by

and

use

United

on

learning.

Using this result to write the force equations of motion along the x and y axes, we have Ax = 12(7.5) = 90 N

+ c ©Fy = m(aG)y;

Ay = 12(9.81) = - 12[24.525(0.3)]

Ans.

integrity

of provided

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is

This

Ay = 29.43 N = 29.4 N

and

work

this

assessing

is

work

solely

of

protected

the

(including

+ ©F = m(a ) ; ; x G x

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Ans.

permitted.

Dissemination

Wide

copyright

+ ©MA = IAa;

Web)

1 (12)(0.62) = 1.44 kg # m2. Thus, 3

B

17–66. a

The kinetic diagram representing the general rotational motion of a rigid body about a fixed axis passing through O is shown in the figure. Show that IGA may be eliminated by moving the vectors m(aG)t and m(aG)n to point P, located a distance rGP = k2G>rOG from the center of mass G of the body. Here kG represents the radius of gyration of the body about an axis passing through G. The point P is called the center of percussion of the body.

P m(aG)t G m(aG)n

rOG

m(aG)t rOG + IG a = m(aG)t rOG + A mk2G B a However, (aG)t rOG

k2G = rOG rGP and a =

m(aG)t rOG + IG a = m(aG)t rOG + (mrOG rGP) c = m(aG)t(rOG + rGP)

(aG)t d rOG

any

courses

part

the

is

This

provided

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is

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the

(including

for

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the

by

and

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on

learning.

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of

the

not

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States

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Dissemination

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copyright

in

teaching

Web)

laws

or

Q.E.D.

destroy

of

and

will

their

sale Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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rGP

O

SOLUTION

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

IG a

17–67. Determine the position rP of the center of percussion P of the 10-lb slender bar. (See Prob. 17–66.) What is the horizontal component of force that the pin at A exerts on the bar when it is struck at P with a force of F=20 lb?

A

rP 4 ft

P

SOLUTION Using the result of Prob 17–66,

rGP =

k2G rAG

B =

1 ml2 2 a bR B 12 m

1 l 6

=

l 2

Thus, 1 2 2 1 l + l = l = (4) = 2.67 ft 6 2 3 3

Ans. laws

or

rP =

Web)

1 10 a b (4)2 da 3 32.2

teaching

20(2.667) = c

in

c + ©MA = IA a;

permitted.

Dissemination

Wide

copyright

a = 32.2 rad>s2

is

of

the

not

instructors

States

World

(a G)t = 2(32.2) = 64.4 ft>s2

and

use

United

on

learning.

10 b (64.4) 32.2

by

student

-Ax + 20 = a

the

+ ©F = m(a ) ; ; x G x

work

Ax = 0

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This

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the

(including

for

Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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F

*17–68. The disk has a mass M and a radius R. If a block of mass m is attached to the cord, determine the angular acceleration of the disk when the block is released from rest. Also,what is the velocity of the block after it falls a distance 2R starting from rest?

R

SOLUTION c + ©MO = ©(Mk) O;

mgR = 12 MR2 (a) + m (aR) R 2mg R(M + 2m)

a =

Ans.

a = aR v2 = v20 + 2 a(s - s0)

8mgR (M + 2m)

Ans.

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Dissemination

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laws

v =

2mgR ) (2R - 0) R(M + 2m

or

v2 = 0 + 2(

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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17–69. The door will close automatically using torsional springs mounted on the hinges. Each spring has a stiffness k = 50 N # m>rad so that the torque on each hinge is M = 150u2 N # m, where u is measured in radians. If the door is released from rest when it is open at u = 90°, determine its angular velocity at the instant u = 0°. For the calculation, treat the door as a thin plate having a mass of 70 kg.

M 1.5 m

A

SOLUTION IAB

0.4 m

1 1 ml2 + md2 = (70)(1.2)2 + 70(0.6)2 = 33.6 kg # m2 = 12 12

©MAB = IAB a ;

2(50u) = - 33.6(a)

v

L0

0

Lp2

2.9762udu

v = 2.71 rad>s

will

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Dissemination

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Ans.

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M

1.2 m

a = - 2.9762u

vdv = adu vdv = -

θ

B

0.4 m

17–70. The door will close automatically using torsional springs mounted on the hinges. If the torque on each hinge is M = ku, where u is measured in radians, determine the required torsional stiffness k so that the door will close 1u = 0°2 with an angular velocity v = 2 rad>s when it is released from rest at u = 90°. For the calculation, treat the door as a thin plate having a mass of 70 kg.

M 1.5 m

A

SOLUTION ©MA = IAa;

2M = - B

0.4 m

1 (70)(1.2)2 + 70(0.6)2 R (a) 12

ku = - 16.8a a du = v dv 0

Lp2

udu = 16.8

2

L0

v dv

laws

or

16.8 k p 2 ( ) = (2)2 2 2 2

teaching

Web)

k = 27.2 N # m>rad

will

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This

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the

(including

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is

of

the

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permitted.

Dissemination

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copyright

in

Ans.

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

M

1.2 m

M = - 16.8a

-k

θ

B

0.4 m

17–71. The pendulum consists of a 10-kg uniform slender rod and a 15-kg sphere. If the pendulum is subjected to a torque of M = 50 N # m, and has an angular velocity of 3 rad>s when u = 45°, determine the magnitude of the reactive force pin O exerts on the pendulum at this instant.

O u

M ⫽ 50 N⭈m

600 mm A

100 mm

SOLUTION Equations of Motion: Since the pendulum rotates about a fixed axis passing through point O, [(aG)OA]t = a(rG)OA = a(0.3), [(aG)B]t = a(rG)B = a(0.7), [(aG)OA]n = v2(rG)OA = (32)(0.3) = 2.7 m>s2, and [(aG)B]n = v2(rG)B = (32)(0.7) = 6.3 m>s2. The mass moment of inertia of the rod and sphere about their respective 1 1 (IG)OA = ml2 = (10)(0.62) = 0.3 kg # m2 mass centers are and 12 12 2 2 (IG)B = mr2 = (15)(0.12) = 0.06 kg # m2. Writing the moment equation of 5 5 motion about point O, we have

B

+ ©MO = ©(Mk)O; -10(9.81) cos 45°(0.3) - 15(9.81) cos 45°(0.7) - 50 =

laws

or

-10[a(0.3)](0.3) - 0.3a - 15[a(0.7)](0.7) - 0.06a

in

teaching

Web)

a = 16.68 rad>s2

of

the

not

instructors

States

World

on

learning.

is

-10(9.81) cos 45°(0.3) - 15(9.81) cos 45°(0.7) - 50 = - 8.61a and

use

United

+ ©MO = IOa;

(including

for

work

student

the

by

a = 16.68 rad>s2

is

work

solely

of

protected

the

Using this result to write the force equations of motion along the n and t axes, this

assessing

10(9.81) cos 45° + 15(9.81) cos 45° + Ot integrity

of

and

work

©Ft = m(aG)t;

any their

destroy

of

and

Ot = 51.81 N

courses

part

the

is

This

provided

= 10[16.68(0.3)] + 15[16.68(0.7)]

will

On - 10(9.81) sin 45° - 15(9.81) sin 45° = 10(2.7) + 15(6.3) sale

©Fn = m(aG)n;

On = 294.92 N Thus, FO = 4Ot 2 + On2 = 451.812 + 294.922 = 299.43 N = 299 N

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Ans.

permitted.

Dissemination

Wide

copyright

This result can also be obtained by applying ©MO = IOa, where IO = ©IG + md2 = 2 1 (10)(0.62) + 10(0.32) + (15)(0.12) + 15(0.72) = 8.61 kg # m2. Thus, 12 5

*17–72. ␻

The disk has a mass of 20 kg and is originally spinning at the end of the strut with an angular velocity of v = 60 rad>s. If it is then placed against the wall, where the coefficient of kinetic friction is mk = 0.3, determine the time required for the motion to stop. What is the force in strut BC during this time?

B

150 mm

SOLUTION + ©F = m(a ) ; : x G x

FCB sin 30° - NA = 0

+ c ©Fy = m(aG)y ;

FCB cos 30° - 20(9.81) + 0.3NA = 0

a + ©MB = IB a;

60⬚ C

1 0.3NA (0.15) = c (20)(0.15)2 da 2 NA = 96.6 N FCB = 193 N

Ans.

a = 19.3 rad>s2 v = v0 + ac t laws

or

c+

in

teaching

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0 = 60 + ( -19.3) t

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Ans.

copyright

t = 3.11 s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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A

17–73. The slender rod of length L and mass m is released from rest when u = 0°. Determine as a function of u the normal and the frictional forces which are exerted by the ledge on the rod at A as it falls downward. At what angle u does the rod begin to slip if the coefficient of static friction at A is m?

A u L

SOLUTION Equations of Motion: The mass moment inertia of the rod about its mass center is 1 given by IG = mL2. At the instant shown, the normal component of acceleration 12 L of the mass center for the rod is (a G)n = v2rG = v2 a b . The tangential component 2 L of acceleration of the mass center for the rod is (aG)t = ars = aa b . 2 L L 1 L b = - a mL2 b a - m caa b d a b 2 12 2 2 3g cos u 2L

teaching

Web)

a =

or

-mg cos u a

laws

a + ©MA = ©(Mk)O ;

in

3g L cos ua b d 2L 2

Wide Dissemination

mg cos u - NA = m c

copyright

+b©Ft = m(aG)t ;

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permitted.

mg cos u 4

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is

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NA =

and

use

United

L bd 2

(1) (including

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by

Ff - mg sin u = mc v2 a

the

a + ©Fn = m(aG)n ;

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the

Kinematics:Applying equation v dv = a du, we have this

integrity

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3g cos u du L0° 2L any

destroy

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and

3g sin u L

3g sin u into Eq. (1) gives L Ff =

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L0

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v dv =

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v

5mg sin u 2

Ans.

If the rod is on the verge of slipping at A, Ff = mNA. Substitute the data obtained above, we have 5mg mg sin u = ma cos u b 2 4 u = tan-1 a

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m b 10

Ans.

17–74. The 5-kg cylinder is initially at rest when it is placed in contact with the wall B and the rotor at A. If the rotor always maintains a constant clockwise angular velocity v = 6 rad>s, determine the initial angular acceleration of the cylinder. The coefficient of kinetic friction at the contacting surfaces B and C is mk = 0.2. B 125 mm

SOLUTION

v 45

Equations of Motion: The mass moment of inertia of the cylinder about point O is 1 1 given by IO = mr 2 = (5)(0.1252) = 0.0390625 kg # m2. Applying Eq. 17–16, 2 2 we have + ©F = m(a ) ; : x G x

NB + 0.2NA cos 45° - NA sin 45° = 0

(1)

+ c ©Fy = m(a G)y ;

0.2NB + 0.2NA sin 45° + NA cos 45° - 5(9.81) = 0

(2)

a + ©MO = IO a;

0.2NA (0.125) - 0.2NB (0.125) = 0.0390625a

C

(3)

Solving Eqs. (1), (2), and (3) yields; or

NB = 28.85 N

laws

NA = 51.01 N

Web)

a = 14.2 rad>s2

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Ans.

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A

17–75. The wheel has a mass of 25 kg and a radius of gyration kB = 0.15 m. It is originally spinning at v1 = 40 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is mC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of reaction which the pin at A exerts on AB during this time? Neglect the mass of AB.

0.4 m A 0.3 m B

0.2 m

V

SOLUTION

C

IB = mk2B = 25(0.15)2 = 0.5625 kg # m2 + c ©Fy = m(aG)y ; + ©F = m(a ) ; : x G x a + ©MB = IBa;

A 35 B FAB + NC - 25(9.81) = 0

(1)

A 45 B FAB = 0

(2)

0.5NC -

0.5NC(0.2) = 0.5625(- a)

(3)

Solvings Eqs. (1),(2) and (3) yields: FAB = 111.48 N

NC = 178.4 N

or

a = - 31.71 rad>s2 Ans.

Ay = 35 FAB = 0.6(111.48) = 66.9 N

Ans. Wide

copyright

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Web)

laws

Ax = 45FAB = 0.8(111.48) = 89.2 N

instructors

States

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permitted.

Dissemination

v = v0 + ac t

United

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of

the

not

0 = 40 + ( -31.71) t

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t = 1.26 s

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Ans.

*■17–76. A 40-kg boy sits on top of the large wheel which has a mass of 400 kg and a radius of gyration kG = 5.5 m. If the boy essentially starts from rest at u = 0°, and the wheel begins to rotate freely, determine the angle at which the boy begins to slip. The coefficient of static friction between the wheel and the boy is ms = 0.5. Neglect the size of the boy in the calculation.

u

SOLUTION c + ©MO = ©(Mk)O ;

392.4(8 sin u) = 400(5.5)2 a + 40(8)(a)(8)

0.2141 sin u = a a du = v dv u

L0

v

0.2141 sin u du =

vdv

L0

u

1 2 v 2

-0.2141 cos u 2 = 0

v = 0.4283(1 - cos u) laws

or

2

Web)

392.4 cos u - N = 40(v2)(8)

teaching

b ©Fy¿ = m(aG)y¿;

in

+

392.4 sin u - 0.5 N = 40(8)(a)

Wide Dissemination

= m(aG)x¿ ;

copyright

+R©Fx¿

not

instructors

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World

permitted.

N = 392.4 cos u - 137.05(1 - cos u) = 529.45 cos u - 137.05

and

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is

of

the

392.4 sin u - 0.5(529.45 cos u - 137.05) = 320(0.2141 sin u)

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by

323.89 sin u - 264.73 cos u + 68.52 = 0

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- sin u + 0.8173 cos u = 0.2116

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Solve by trial and error part

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u = 29.8°

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Note: The boy will loose contact with the wheel when N = 0, i.e. sale

will

N = 529.45 cos u - 137.05 = 0 u = 75.0° 7 29.8° Hence slipping occurs first.

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Ans.

8m

17–77. M ⫽ 200(1 ⫺ e⫺0.2 t ) N⭈m

Gears A and B have a mass of 50 kg and 15 kg, respectively. Their radii of gyration about their respective centers of mass are kC = 250 mm and kD = 150 mm. If a torque of M = 200(1 - e-0.2t) N # m, where t is in seconds, is applied to gear A, determine the angular velocity of both gears when t = 3 s, starting from rest.

300 mm C

SOLUTION

B

rA Equations of Motion: Since gear B is in mesh with gear A, aB = a baA = rB 0.3 a b a = 1.5aA.The mass moment of inertia of gears A and B about their respective 0.2 A

A

centers are IC = mA kC 2 = 50(0.252) = 3.125 kg # m2 and ID = mB kD2 = 15(0.152) = 0.3375 kg # m2. Writing the moment equation of motion about the gears’ center using the free-body diagrams of gears A and B, Figs. a and b, a + ©MC = ICaA;

F(0.3) - 200(1 - e-0.2t) = - 3.125aA

(1)

F(0.2) = 0.3375(1.5aA)

(2)

and

laws

or

a + ©MD = IDaB;

teaching

Web)

Eliminating F from Eqs. (1) and (2) yields

World

is

of

the

not

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States

Kinematics: The angular velocity of gear A can be determined by integration.

permitted.

Dissemination

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copyright

in

aA = 51.49(1 - e-0.2t) rad>s2

on

learning.

aA dt

United

and

L

use

dvA =

the

by

L

work

student

t

for

vA

51.49(1 - e-0.2t)dt L0 L0 vA = 51.49(t + 5e-0.2t - 5) rad>s

integrity

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dvA =

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provided

When t = 3 s,

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courses

vA = 51.49(3 + 5e-0.2(3) - 5) = 38.31 rad>s = 38.3 rad>s sale

will

Then vB = a

rA 0.3 bv = a b (38.31) rB A 0.2

= 57.47 rad>s = 57.5 rad>s

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200 mm D

Ans.

17–78. Block A has a mass m and rests on a surface having a coefficient of kinetic friction mk. The cord attached to A passes over a pulley at C and is attached to a block B having a mass 2m. If B is released, determine the acceleration of A. Assume that the cord does not slip over the pulley. The pulley can be approximated as a thin disk of radius r and mass 14m. Neglect the mass of the cord.

r A

SOLUTION

B

Block A: + ©F = ma ; : x x

T1 - mk mg = ma

(1)

2mg - T2 = 2ma

(2)

Block B: + T ©Fy = may; Pulley C: 1 1 a T2r - T1r = c a m br2 d a b r 2 4 1 ma 8

(3) laws

T2 - T1 =

or

c + ©MC = IG a;

25 a 8

permitted.

(2 - mk)g =

Wide

1 ma 8

Dissemination

2mg - 2ma - (ma + mk mg) =

copyright

in

teaching

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Substituting Eqs. (1) and (2) into (3),

World

8 (2 - mk)g 25

Ans.

the

a =

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sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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not

instructors

States

1 ma + 3ma 8

of

2mg - mk mg =

C

17–79. The two blocks A and B have a mass of 5 kg and 10 kg, respectively. If the pulley can be treated as a disk of mass 3 kg and radius 0.15 m, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. r O

SOLUTION

A

Kinematics: Since the pulley rotates about a fixed axis passes through point O, its angular acceleration is a =

B

a a = = 6.6667a r 0.15

The mass moment of inertia of the pulley about point O is 1 1 Io = Mr2 = (3)(0.152) = 0.03375 kg # m2 2 2 Equation of Motion: Write the moment equation of motion about point O by referring to the free-body and kinetic diagram of the system shown in Fig. a, 5(9.81)(0.15) - 10(9.81)(0.15) laws

or

a + ©Mo = ©(Mk)o;

teaching

Web)

= - 0.03375(6.6667a) - 5a(0.15) - 10a(0.15) in

a = 2.973 m>s2 = 2.97 m>s2

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Dissemination

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copyright

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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*17–80. The two blocks A and B have a mass mA and mB , respectively, where mB 7 mA . If the pulley can be treated as a disk of mass M, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. r O

SOLUTION A

a = ar B

c + ©MC = © (Mk)C ;

1 mB g(r) - mA g(r) = a Mr2 b a + mB r2 a + mA r2 a 2 g(mB - mA)

a =

1 r a M + mB + mA b 2 g(mB - mA)

a =

Ans.

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copyright

in

teaching

Web)

laws

or

1 a M + mB + mA b 2

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17–81. Determine the angular acceleration of the 25-kg diving board and the horizontal and vertical components of reaction at the pin A the instant the man jumps off. Assume that the board is uniform and rigid, and that at the instant he jumps off the spring is compressed a maximum amount of 200 mm, v = 0, and the board is horizontal. Take k = 7 kN>m.

1.5 m

1.5 m

A k

SOLUTION a + a MA = IAa; + c a Ft = m(aG)t ; + ; a Fn = m(aG)n ;

1 1.5(1400 - 245.25) = c (25)(3)2 da 3 1400 - 245.25 - Ay = 25(1.5a) Ax = 0

Ans.

A y = 289 N

Ans.

a = 23.1 rad>s2

Ans.

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laws

Ax = 0

or

Solving,

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or transmission in any form or by any means, electronic, mechanical,

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17–82. The lightweight turbine consists of a rotor which is powered from a torque applied at its center. At the instant the rotor is horizontal it has an angular velocity of 15 rad/s and a clockwise angular acceleration of 8 rad>s2. Determine the internal normal force, shear force, and moment at a section through A. Assume the rotor is a 50-m-long slender rod, having a mass of 3 kg/m.

10 m A 25 m

SOLUTION + ©F = m(a ) ; ; n G n + T ©Ft = m(aG)t ;

NA = 45(15)2 (17.5) = 177kN

Ans.

VA + 45(9.81) = 45(8)(17.5) VA = 5.86 kN

c + ©MA = ©(Mk)A ;

Ans.

MA + 45(9.81)(7.5) = c

1 (45)(15)2 d(8) + [45(8)(17.5)](7.5) 12

MA = 50.7 kN # m

sale

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Dissemination

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in

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laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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17–83. The two-bar assembly is released from rest in the position shown. Determine the initial bending moment at the fixed joint B. Each bar has a mass m and length l.

A

l

B

l

SOLUTION

C

Assembly: IA =

1 2 1 l (m)(l)2 + m(l2 + ( )2) ml + 3 12 2

= 1.667 ml2 c + ©MA = IA a;

l mg( ) + mg(l) = (1.667ml2)a 2 0.9 g l or

a =

teaching

Web)

laws

Segment BC: in

l>2 1 l l ml2 d a + m(l 2 + ( )2)1>2 a( )( ) l 2 2 2 12 2 l + (2)

Wide

permitted.

Dissemination

M = c

copyright

c + ©MB = ©(Mk)B;

on

learning.

is

of

the

not

instructors

States

World

0.9g 1 2 1 ml a = ml2 ( ) 3 3 l

and

use

United

M =

by

M = 0.3gml

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the

(including

for

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student

the

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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*17–84. The armature (slender rod) AB has a mass of 0.2 kg and can pivot about the pin at A. Movement is controlled by the electromagnet E, which exerts a horizontal attractive force on the armature at B of FB = 10.2110-32l-22 N, where l in meters is the gap between the armature and the magnet at any instant. If the armature lies in the horizontal plane, and is originally at rest, determine the speed of the contact at B the instant l = 0.01 m. Originally l = 0.02 m.

l B

E

150 mm

SOLUTION Equation of Motion: The mass moment of inertia of the armature about point A is 1 given by IA = IG + m r2G = (0.2) A 0.152 B + 0.2 A 0.0752 B = 1.50 A 10-3 B kg # m2 12 Applying Eq. 17–16, we have 0.2(10-3)

a + ©MA = IAa;

l2

A

(0.15) = 1.50 A 10-3 B a 0.02 l2 or

a =

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

Kinematic: From the geometry, l = 0.02 - 0.15 u. Then dl = - 0.15 du or dv dl v . Also, v = hence dv = . Substitute into equation vdv = adu, du = 0.15 0.15 0.15 we have

the

by

and

use

United

on

learning.

is

of

the

dv dl v ¢ ≤ = a¢ ≤ 0.15 0.15 0.15

the

(including

for

work

student

vdv = - 0.15 adl is

work

solely

of

protected

0.02 dl l2

- 0.15

this

L0.02 m

integrity

of

and

work

L0

0.01 m

vdv =

assessing

v

Ans.

sale

will

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part

the

is

This

provided

v = 0.548 m s

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17–85. The bar has a weight per length of w. If it is rotating in the vertical plane at a constant rate V about point O, determine the internal normal force, shear force, and moment as a function of x and u.

O V U L

SOLUTION

x

a = v2 a L -

x u bh z

Forces: wx 2 x v aL - b u = N u + S au + wx T h g z h

(1)

Moments:

or

x Ia = M - S a b 2 laws

1 Sx 2

teaching

Web)

(2) Wide

copyright

in

O = M -

States

World

permitted.

Dissemination

Solving (1) and (2),

not

instructors

v2 x A L - B + cos u R g 2

United

on

learning.

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the

Ans.

by

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N = wx B

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the

S = wx sin u

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the

1 wx2 sin u 2

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M =

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Ans.

17–86. A force F = 2 lb is applied perpendicular to the axis of the 5-lb rod and moves from O to A at a constant rate of 4 ft>s. If the rod is at rest when u = 0° and F is at O when t = 0, determine the rod’s angular velocity at the instant the force is at A. Through what angle has the rod rotated when this occurs? The rod rotates in the horizontal plane.

O u 4 ft/s

4 ft

F ⫽ 2 lb

SOLUTION IO = a + a MO = IOa;

A

1 1 5 mR2 = a b (4)2 = 0.8282 slug # ft2 3 3 32.2 2(4t) = 0.8282(a) a = 9.66t dv = adt v

L0

t

#

dv =

L0

9.66t dt

teaching

Web)

laws

or

v = 4.83t2

Wide

copyright

in

When t = 1 s, v = 4.83(1)2 = 4.83 rad>s States

World

permitted.

Dissemination

Ans.

learning.

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du = v dt u

on United

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use by

4.83t2 dt

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du =

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the

u = 1.61 rad = 92.2°

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Ans.

17–87. A

The 15-kg block A and 20-kg cylinder B are connected by a light cord that passes over a 5-kg pulley (disk). If the system is released from rest, determine the cylinder’s velocity after its has traveled downwards 2 m. Neglect friction between the plane and the block, and assume the cord does not slip over the pulley.

100 mm O

SOLUTION Equations of Motion: Since the pulley rotates about a fixed axis passing through a a = = 10a. The mass moment of inertia point O, its angular acceleration is a = r 0.1 1 2 1 of the pulley about point O is IO = mr = (5)(0.12) = 0.025 kg # m2. Writing the 2 2 moment equation of equilibrium about point O and realizing that the moment of 15(9.81) N and NA cancel out, we have + ©MO = ©(Mk)O;

B

- 20(9.81)(0.1) = - 15a(0.1) - 20a(0.1) - 0.025(10a) a = 5.232 m>s2

or

Kinematics: Since the angular acceleration is constant, v2 = v0 2 + 2ac(s - s0)

teaching

Web)

laws

(+ T)

Wide

copyright

in

v2 = 02 + 2(5.232)(2 - 0)

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Dissemination

v = 4.57 m>s T

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*17–88. A

The 15-kg block A and 20-kg cylinder B are connected by a light cord that passes over a 5-kg pulley (disk). If the system is released from rest, determine the cylinder’s velocity after its has traveled downwards 2 m. The coefficient of kinetic friction between the block and the horizontal plane is mk = 0.3. Assume the cord does not slip over the pulley.

100 mm O

SOLUTION Equations of Motion: Since the block is in motion, FA = mk NA = 0.3NA. Referring to the free-body diagram of block A shown in Fig. a, + c ©Fy = m(aG)y;

NA - 15(9.81) = 0

+ ©F = m(a ) ; : x G x

T1 - 0.3(147.15) = 15a

B

NA = 147.15 N (1)

Referring to the free-body diagram of the cylinder, Fig. b, + c ©Fy = m(aG)y;

T2 - 20(9.81) = - 20a

(2)

a a = = 10a. r 0.1 1 2 The mass moment of inertia of the pulley about O is IO = mr = 2 1 (5)(0.12) = 0.025 kg # m2. Writing the moment equation of motion about point O 2 using the free-body diagram of the pulley shown in Fig. c, Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Since the pulley rotates about a fixed axis passing through point O, a =

permitted.

(3) World

T1(0.1) - T2(0.1) = - 0.025(10a)

is

of

the

not

instructors

States

+ ©MO = IOa;

T1 = 104.967 N work

student

T2 = 115.104 N

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(including

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a = 4.0548 m>s2

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Kinematics: Since the acceleration is constant,

this

v2 = v02 + 2ac(s - s0)

part

the

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v2 = 02 + 2(4.0548)(2 - 0)

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v = 4.027 m>s = 4.03 m>s T

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Ans.

17–89. The “Catherine wheel” is a firework that consists of a coiled tube of powder which is pinned at its center. If the powder burns at a constant rate of 20 g>s such as that the exhaust gases always exert a force having a constant magnitude of 0.3 N, directed tangent to the wheel, determine the angular velocity of the wheel when 75% of the mass is burned off. Initially, the wheel is at rest and has a mass of 100 g and a radius of r = 75 mm. For the calculation, consider the wheel to always be a thin disk.

r C 0.3 N

SOLUTION Mass of wheel when 75% of the powder is burned = 0.025 kg Time to burn off 75 % =

0.075 kg = 3.75 s 0.02 kg>s

m(t) = 0.1 - 0.02 t Mass of disk per unit area is

or

0.1 kg m = = 5.6588 kg>m2 A p(0.075 m)2 laws

r0 =

in

teaching

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At any time t,

permitted.

Dissemination

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copyright

0.1 - 0.02t pr2

not

instructors

States

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5.6588 =

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on

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0.1 - 0.02t A p(5.6588)

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by

r(t) =

work

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1 2 mr a 2

the

(including

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0.6 0.1 - 0.02t A p(5.6588) integrity

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0.6 = mr This

a =

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0.3r =

protected

+ ©MC = ICa;

3

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a = 0.6 A 2p(5.6588) B [0.1 - 0.02t]- 2 3

sale

will

a = 2.530[0.1 - 0.02t]- 2 dv = a dt v

L0

t

dv = 2.530

L0

3

[0.1 - 0.02t]- 2 dt 1

v = 253 C (0.1 - 0.02t)- 2 - 3.162 D For t = 3.75 s, v = 800 rad s

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Ans.

17–90. If the disk in Fig. 17–21a rolls without slipping, show that when moments are summed about the instantaneous center of zero velocity, IC, it is possible to use the moment equation ©MIC = IICa, where IIC represents the moment of inertia of the disk calculated about the instantaneous axis of zero velocity.

SOLUTION c + ©MlC = ©(MK)lC;

©MlC = IGa + (maG)r

Since there is no slipping, aG = ar Thus, ©MIC = A IG + mr2 B a By the parallel–axis thoerem, the term in parenthesis represents IIC. Thus, ©MIC = IICa

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Dissemination

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Q.E.D.

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17–91. The 20-kg punching bag has a radius of gyration about its center of mass G of kG = 0.4 m. If it is initially at rest and is subjected to a horizontal force F = 30 N, determine the initial angular acceleration of the bag and the tension in the supporting cable AB.

A 1m B 0.3 m G

SOLUTION

0.6 m

+ ©F = m(a ) ; : x G x

30 = 20(aG)x

+ c ©Fy = m(aG)y;

T - 196.2 = 20(aG)y

a + ©MG = IGa;

F

30(0.6) = 20(0.4)2a a = 5.62 rad>s2

Ans.

(aG)x = 1.5 m>s2 aB = aG + aB>G aB i = (aG)y j + (aG)xi - a(0.3)i or

(aG)y = 0

teaching

Web)

laws

(+ c )

Wide

copyright

in

Thus,

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Dissemination

T = 196 N

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*17–92. FA ⫽ 100 lb

The uniform 150-lb beam is initially at rest when the forces are applied to the cables. Determine the magnitude of the acceleration of the mass center and the angular acceleration of the beam at this instant.

FB ⫽ 200 lb

A

B

12 ft

SOLUTION Equations of Motion: The mass moment of inertia of the beam about its mass center 1 1 150 ml2 = a b A 122 B = 55.90 slug # ft2. is IG = 12 12 32.2 + ©F = m(a ) ; : x G x

200 cos 60° =

150 (a ) 32.2 G x

(aG)x = 21.47 ft>s2 + c ©Fy = m(aG)y;

150 (a ) 32.2 G y

100 + 200 sin 60° - 150 = (aG)y = 26.45 ft>s2

200 sin 60°(6) - 100(6) = 55.90a or

+ ©MG = IGa;

laws

a = 7.857 rad>s2 = 7.86 rad>s2

in

teaching

Web)

Ans.

Dissemination

Wide

copyright

Thus, the magnitude of aG is aG = 3(aG)x2 + (aG)y2 = 321.472 + 26.452 = 34.1 ft>s2

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Ans.

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60⬚

17–93. B

The rocket has a weight of 20 000 lb, mass center at G, and radius of gyration about the mass center of kG = 21 ft when it is fired. Each of its two engines provides a thrust T = 50 000 lb. At a given instant, engine A suddenly fails to operate. Determine the angular acceleration of the rocket and the acceleration of its nose B.

30 ft

G

SOLUTION a + ©MG = IGa;

50 000(1.5) =

20 000 (21)2a 32.2

a = 0.2738 rad>s2 = 0.274 rad>s2 + c ©Fy = m(aG)y;

Ans.

20 000 a 32.2 G

50 000 - 20 000 =

A

aG = 48.3 ft>s2

1.5 ft

aB = aG + aB>G

1.5 ft

or

Since v = 0 laws

T

in

teaching

Web)

aB = 48.3j - 0.2738(30)i

Ans. instructors

States

World

aB = 2(48.3)2 + (8.214)2 = 49.0 ft>s2

of

the

not

48.3 = 80.3° ub 8.214

United

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by

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u = tan - 1

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= 48.34j - 8.214i

T

17–94. The wheel has a weight of 30 lb and a radius of gyration of kG = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are ms = 0.2 and mk = 0.15, determine the wheel’s angular acceleration as it rolls down the incline. Set u = 12°.

G 1.25 ft

SOLUTION

u

+ b©Fx = m(aG)x ; +a©Fy

= m(aG)y ;

a + ©MG = IG a;

30 ba 30 sin 12° - F = a 32.2 G N - 30 cos 12° = 0 F(1.25) = c a

30 b (0.6)2 da 32.2

Assume the wheel does not slip. aG = (1.25)a

laws

or

Solving:

in

teaching

Web)

F = 1.17 lb

Dissemination

Wide

copyright

N = 29.34 lb

of

the

not

instructors

States

World

permitted.

aG = 5.44 ft>s2

is

a = 4.35 rad>s2 by

and

use

United

on

learning.

Ans. OK

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student

the

Fmax = 0.2(29.34) = 5.87 lb 7 1.17 lb

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17–95. The wheel has a weight of 30 lb and a radius of gyration of kG = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are ms = 0.2 and mk = 0.15, determine the maximum angle u of the inclined plane so that the wheel rolls without slipping.

G 1.25 ft

SOLUTION

u

Since wheel is on the verge of slipping: + b©Fx = m(aG)x ;

30 sin u - 0.2N = a

+ a©Fy = m(aG)y ;

N - 30 cos u = 0

a + ©MC = IG a;

0.2N(1.25) = c a

30 b (1.25a) 32.2

(1) (2)

30 b (0.6)2 da 32.2

(3)

Substituting Eqs.(2) and (3) into Eq. (1),

laws

or

30 sin u - 6 cos u = 26.042 cos u

in

teaching

Web)

30 sin u = 32.042 cos u

Ans.

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u = 46.9°

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tan u = 1.068

*17–96. The spool has a mass of 100 kg and a radius of gyration of kG = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 50 N.

P 250 mm

G

A

SOLUTION + ©F = m(a ) ; : x G x

50 + FA = 100aG

+ c ©Fy = m(aG)y ;

NA - 100(9.81) = 0

c + ©MG = IG a;

50(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4a a = 1.30 rad>s2 aG = 0.520 m>s2

Ans. NA = 981 N

FA = 2.00 N

Since (FA)max = 0.2(981) = 196.2 N 7 2.00 N

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in

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laws

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OK

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400 mm

17–97. Solve Prob. 17–96 if the cord and force P = 50 N are directed vertically upwards.

P 250 mm

G

A

SOLUTION + ©F = m(a )x; F = 100a : x G A G + c ©Fy = m(aG)y;

NA + 50 - 100(9.81) = 0

c + ©MG = IG a;

50(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4 a a = 0.500 rad>s2 aG = 0.2 m>s2

Ans. FA = 20 N

NA = 931 N

OK

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the

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Since (FA)max = 0.2(931) = 186.2 N 7 20 N

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400 mm

17–98. The spool has a mass of 100 kg and a radius of gyration kG = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 600 N.

P 250 mm

G

A

SOLUTION + ©F = m(a ) ; : x G x

600 + FA = 100aG

+ c ©Fy = m(aG)y;

NA - 100(9.81) = 0

c + ©MG = IG a;

600(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4a a = 15.6 rad>s2 aG = 6.24 m>s2

Ans. NA = 981 N

FA = 24.0 N

Since (FA)max = 0.2(981) = 196.2 N 7 24.0 N

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Dissemination

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OK

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400 mm

17–99. The upper body of the crash dummy has a mass of 75 lb, a center of gravity at G, and a radius of gyration about G of kG = 0.7 ft. By means of the seat belt this body segment is assumed to be pin-connected to the seat of the car at A. If a crash causes the car to decelerate at 50 ft>s2, determine the angular velocity of the body when it has rotated to u = 30°.

θ

G

SOLUTION

1.9 ft A

75 75 (a ) R (1.9) ≤ (0.7)2 R a + B 32.2 32.2 G t

75(1.9 sin u) = B ¢

a + ©MA = ©(Mk)A;

50 ft/s2

+b aG = aA + (aG>A)n + (aG>A)t (aG)t = - 50 cos u + 0 + (a )(1.9) 142.5 sin u = 1.1413a - 221.273 cos u + 8.4084a 142.5 sin u + 221.273 cos u = 9.5497a v dv = a du v

30°

L0

laws

or

(14.922 sin u + 23.17 cos u)du

teaching

Web)

L0

v dv =

permitted.

Dissemination

Wide

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1 2 v = - 14.922(cos 30° - cos 0°) + 23.17(sin 30° - sin 0°) 2 States

World

v = 5.21 rad s

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This

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is

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protected

the

(including

for

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student

the

by

and

use

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on

learning.

is

of

the

not

instructors

Ans.

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*17–100. A uniform rod having a weight of 10 lb is pin supported at A from a roller which rides on a horizontal track. If the rod is originally at rest, and a horizontal force of F = 15 lb is applied to the roller, determine the acceleration of the roller. Neglect the mass of the roller and its size d in the computations.

d

A

F

2 ft

SOLUTION Equations of Motion: The mass moment of inertia of the rod about its mass center is 10 1 1 ml2 = a b(2 2) = 0.1035 slug # ft2. At the instant force F is given by IG = 12 12 32.2 applied, the angular velocity of the rod v = 0. Thus, the normal component of acceleration of the mass center for the rod (aG)n = 0. Applying Eq. 17–16, we have ©Ft = m(a G)t;

15 = a

a + ©MA = ©(Mk)A ;

10 ba a G = 48.3 ft>s2 32.2 G 0 = a

10 b(48.3)(1) - 0.1035 a 32.2 laws

or

a = 144.9 rad >s2

Wide

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teaching

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Kinematics: Since v = 0, (aG>A)n = 0. The acceleration of roller A can be obtain by analyzing the motion of points A and G. Applying Eq. 16–17, we have

instructors

States

World

permitted.

Dissemination

aG = aA + (aG>A)t + (aG>A)n the

is

of United

on

learning.

;

and by

48.3 = aA - 144.9

for

work

student

the

+ ) (:

:

use

:

not

c 48.3 d = c aA d + c 144.9(1) d + c0 d

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This

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protected

the

(including

aA = 193 ft>s2

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Ans.

17–101. Solve Prob. 17–100 assuming that the roller at A is replaced by a slider block having a negligible mass. The coefficient of kinetic friction between the block and the track is mk=0.2. Neglect the dimension d and the size of the block in the computations.

d

A

F

2 ft

SOLUTION Equations of Motion: The mass moment of inertia of the rod about its mass center is 1 1 10 ml2 = a b (2 2) = 0.1035 slug # ft2. At the instant force F is given by IG = 12 12 32.2 applied, the angular velocity of the rod v = 0. Thus, the normal component of acceleration of the mass center for the rod (aG)n = 0. Applying Eq. 17–16, we have ©Fn = m(aG)n ;

10 - N = 0

N = 10.0 lb

©Ft = m(aG)t ;

15 - 0.2(10.0) = a

a + ©MA = ©(Mk)A ;

0 = a

10 ba 32.2 G

aG = 41.86 ft>s2

laws

or

10 b (41.86)(1) - 0.1035a 32.2

not

instructors

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on

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is

of

the

aG = aA + (aG>A)t + (aG>A)n ;

work

student

(including

:

for

:

the

c 41. 8 6 d = c a A d + c125.58(1) d + C 0 D

the

41.86 = aA - 125.58 is

work

solely

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protected

+ ) (:

Ans.

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this

assessing

aA = 167 ft>s2

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permitted.

Dissemination

Kinematics: Since v = 0, (aG>A)n = 0. The acceleration of block A can be obtain by analyzing the motion of points A and G. Applying Eq. 16–17, we have

Wide

copyright

in

teaching

Web)

a = 125.58 rad >s2

17–102. The 2-kg slender bar is supported by cord BC and then released from rest at A. Determine the initial angular acceleration of the bar and the tension in the cord.

C

30° B

A 300 mm

SOLUTION + ©F = m(a ) ; : x G x

T cos 30° = 2(aG)x

+ c ©Fy = m(aG)y ;

T sin 30° - 19.62 = 2(aG)y

a + ©MG = IGa ;

T sin 30°(0.15) = [

1 (2)(0.3)2]a 12

aB = aG + aB>G

(aB) sin 30° = (aG)x

(+ c)

(aB) cos 30° = - (aG)y - a (0.15)

laws

+ ) (:

or

aB sin 30°i - aB cos 30°j = (aG)xi + (aG)y j + a (0.15)j

copyright

in

teaching

Web)

Thus,

permitted.

Dissemination

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1.7321(aG)x = - (aG)y - 0.15a

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the

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on

(aG)x = 2.43 m s2

the

(including

for

work

student

(aG)y = - 8.41 m s2

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This

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a = 28.0 rad s2

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Ans.

not

States

World

T = 5.61 N

17–103. If the truck accelerates at a constant rate of 6 m>s2, starting from rest, determine the initial angular acceleration of the 20-kg ladder. The ladder can be considered as a uniform slender rod. The support at B is smooth.

C 1.5 m

B 2.5 m

60

SOLUTION Equations of Motion: We must first show that the ladder will rotate when the acceleration of the truck is 6 m>s2. This can be done by determining the minimum acceleration of the truck that will cause the ladder to lose contact at B, NB = 0. Writing the moment equation of motion about point A using Fig. a, a + ©MA = ©(Mk)A;

20(9.81) cos 60°(2) = 20amin (2 sin 60°) amin = 5.664 m>s2

teaching

Web)

laws

or

Since a min 6 6 m>s2, the ladder will in the fact rotate.The mass moment of inertia about 1 1 its mass center is IG = ml2 = (20) A 4 2 B = 26.67 kg # m2. Referring to Fig. b, 12 12 Wide of

the

not

instructors

States

World

(1)

the

by

and

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United

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learning.

is

Kinematics: The acceleration of A is equal to that of the truck. Thus, a A = 6 m>s2 ; . Applying the relative acceleration equation and referring to Fig. c, the

(including

for

work

student

aG = aA + a * rG>A - v2 rG>A

part

the

is

This

provided

integrity

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and

work

(aG)x i + (aG)y j = (2 sin 60° a - 6)i + aj

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is

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(aG)x i + (aG)y j = - 6i + ( -ak) * ( -2 cos 60° i + 2 sin 60° j) - 0

and

any

courses

Equating the i and j components, (2)

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of

(a G)x = 2 sin 60° a - 6 (a G)y = a

(3)

Substituting Eqs. (2) and (3) into Eq. (1), a = 0.1092 rad>s2 = 0.109 rad>s2

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Ans.

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Dissemination

- 20(a G)y (2 cos 60°) - 26.67a

in

20(9.81) cos 60°(2) = -20(a G)x (2 sin 60°)

copyright

a + ©MA = ©(Mk)A;

A

*17–104. P

If P = 30 lb, determine the angular acceleration of the 50-lb roller. Assume the roller to be a uniform cylinder and that no slipping occurs.

1.5 ft 30⬚

SOLUTION Equations of Motion: The mass moment of inertia of the roller about its mass center 1 1 50 is IG = mr2 = a b A 1.52 B = 1.7469 slug # ft2. We have 2 2 32.2 50 a 32.2 G

+ ©F = m(a ) ; : x G x

30 cos 30° - Ff =

+ c ©Fy = m(aG)y;

N - 50 - 30 sin 30° = 0

+ ©MG = IGa;

Ff(1.5) = 1.7469a

(1) N = 65 lb (2)

Since the roller rolls without slipping, aG = ar = a(1.5)

or

(3)

teaching

Web)

laws

Solving Eqs. (1) through (3) yields a = 7.436 rad>s2 = 7.44 rad>s2

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Ans.

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the

(including

for

work

student

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by

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use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

aG = 11.15 ft>s2

Ff = 8.660 lb

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17–105. P

If the coefficient of static friction between the 50-lb roller and the ground is ms = 0.25, determine the maximum force P that can be applied to the handle, so that roller rolls on the ground without slipping. Also, find the angular acceleration of the roller. Assume the roller to be a uniform cylinder.

1.5 ft 30⬚

SOLUTION Equations of Motion: The mass moment of inertia of the roller about its mass center 1 1 50 b A 1.52 B = 1.7469 slug # ft2. We have is IG = mr2 = a 2 2 32.2 50 a 32.2 G

+ ©F = m(a ) ; : x G x

P cos 30° - Ff =

+ c ©Fy = m(aG)y;

N - P sin 30° - 50 = 0

(2)

+ ©MG = IGa;

Ff(1.5) = 1.7469a

(3)

(1)

Since the roller is required to be on the verge of slipping, (4)

Ff = msN = 0.25N

(5) teaching

Web)

laws

or

aG = ar = a(1.5)

a = 18.93 rad>s2 = 18.9 rad>s2

Ans.

instructors

States

World

Dissemination

P = 76.37 lb = 76.4 lb

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learning.

Ff = 22.05 lb

the

not

aG = 28.39 ft>s2

of

N = 88.18 lb

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Solving Eqs. (1) through (5) yields

17–106. The spool has a mass of 500 kg and a radius of gyration kG = 1.30 m. It rests on the surface of a conveyor belt for which the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction is mk = 0.4. If the conveyor accelerates at a C = 1 m>s2, determine the initial tension in the wire and the angular acceleration of the spool. The spool is originally at rest.

0.8 m G aC

SOLUTION + : a Fx = m(aG)x; + c a Fy = m(aG)y; c + a MG = IGa;

-Fs + T = 500aG Ns - 500(9.81) = 0 Fs(1.6) - T(0.8) = 500(1.30)2a ap = aG + ap>G (ap)yj = aG i - 0.8ai aG = 0.8a

or

Ns = 4905 N

teaching

Web)

laws

Assume no slipping in

ac 1 = = 1.25 rad>s 0.8 0.8

permitted.

Dissemination

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Ans.

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a =

is

of

the

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aG = 0.8(1.25) = 1 m>s2

learning.

T = 2.32 kN

on United

and work

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(including

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this

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is

the

is

This

provided

integrity

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and

work

(Fs)max = 0.5(4.905) = 2.45 7 1.82

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of

and

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part

(No slipping occurs)

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Ans.

use by

the

Fs = 1.82 kN Since

1.6 m

17–107. The spool has a mass of 500 kg and a radius of gyration kG = 1.30 m. It rests on the surface of a conveyor belt for which the coefficient of static friction is ms = 0.5. Determine the greatest acceleration a C of the conveyor so that the spool will not slip. Also, what are the initial tension in the wire and the angular acceleration of the spool? The spool is originally at rest.

0.8 m G aC

SOLUTION + : a Fx = m(aG)x; + c a Fy = m(aG)y; c + a MG = IGa,

T - 0.5NS = 500aG Ns - 500(9.81) = 0 0.5Ns(1.6) - T(0.8) = 500(1.30)2a

ap = aC + ap>G (ap)yj = aGi - 0.8ai aG = 0.8a

or

Solving;

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laws

Ns = 4905 N T = 3.13 kN

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Ans. Ans.

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Dissemination

a = 1.684 rad>s

and

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learning.

is

of

the

not

aG = 1.347 m>s2

for

work

student

the

by

Since no slipping

solely

of

protected

the

(including

aC = aG + aC>G

this

assessing

is

work

ac = 1.347i - (1.684) (1.6)i

any

destroy

of

and

Also,

courses

part

the

is

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provided

integrity

of

and

work

aC = 1.35 m>s2

will

their

0.5Ns(0.8) = [500(1.30)2 + 500(0.8)2]a sale

c + a MIC = IICa; Since NS = 4905 N

a = 1.684 rad>s

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1.6 m

Ans.

*17–108. The semicircular disk having a mass of 10 kg is rotating at v = 4 rad>s at the instant u = 60°. If the coefficient of static friction at A is ms = 0.5, determine if the disk slips at this instant.

v 0.4 m

O

G

u A

SOLUTION Equations of Motion:The mass moment of inertia of the semicircular disk about its center 1 of mass is given by IG = (10) A 0.4 2 B - 10 (0.16982) = 0.5118 kg # m2. From the 2 geometry, rG>A = 20.16982 + 0.4 2 - 2(0.1698) (0.4) cos 60° = 0.3477 m Also, using sin 60° sin u = law of sines, , u = 25.01°. Applying Eq. 17–16, we have 0.1698 0.3477 a + ©MA = ©(Mk)A ;

10(9.81)(0.1698 sin 60°) = 0.5118a + 10(a G)x cos 25.01°(0.3477) + 10(a G)y sin 25.01°(0.3477)

or

(1) laws

+ ©F = m(a ) ; ; x G x

Ff = 10(a G)x

teaching

Web)

(2) in

N - 10(9.81) = - 10(aG)y

the

by

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States

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Kinematics:Assume that the semicircular disk does not slip at A, then (a A)x = 0. Here, rG>A = {-0.3477 sin 25.01°i + 0.3477 cos 25.01°j} m = { -0.1470i + 0.3151j} m. Applying Eq. 16–18, we have

(including

for

work

student

a G = a A + a * rG>A - v2rG>A

assessing

is

work

solely

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protected

the

-(aG)x i - (aG)y j = 6.40j + ak * ( -0.1470i + 0.3151j) - 42( -0.1470i + 0.3151j)

part

the

is and

any

courses

Equating i and j components, we have

This

provided

integrity

of

and

work

this

-(a G)x i - (aG)y j = (2.3523 - 0.3151 a) i + (1.3581 - 0.1470a)j

(4)

their

destroy

of

(a G)x = 0.3151a - 2.3523 sale

will

(a G)y = 0.1470a - 1.3581

(5)

Solving Eqs. (1), (2), (3), (4), and (5) yields: a = 13.85 rad>s2

(aG)x = 2.012 m>s2 Ff = 20.12 N

(aG)y = 0.6779 m>s2

N = 91.32 N

Since Ff 6 (Ff)max = msN = 0.5(91.32) = 45.66 N, then the semicircular disk does not slip.

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Ans.

permitted.

Dissemination

Wide

(3)

copyright

+ c Fy = m(a G)y ;

4 (0.4) ——— m 3p

17–109. The 500-kg concrete culvert has a mean radius of 0.5 m. If the truck has an acceleration of 3 m>s2, determine the culvert’s angular acceleration. Assume that the culvert does not slip on the truck bed, and neglect its thickness.

0.5m

SOLUTION Equations of Motion: The mass moment of inertia of the culvert about its mass center is IG = mr 2 = 500 A 0.52 B = 125 kg # m2. Writing the moment equation of motion about point A using Fig. a, a + ©MA = ©(Mk)A ;

0 = 125a - 500a G(0.5)

(1)

Kinematics: Since the culvert does not slip at A, (aA)t = 3 m>s2. Applying the relative acceleration equation and referring to Fig. b, aG = a A + a * rG>A - v2rG>A a Gi - 3i + (a A)n j + (ak * 0.5j) - v2(0.5j)

teaching

Web)

laws

or

aGi = (3 - 0.5a)i + C (aA)n - 0.5v2 D j

Wide

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in

Equating the i components,

(2) not

instructors

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permitted.

Dissemination

a G = 3 - 0.5a

use

United

on

learning.

is

of

the

Solving Eqs. (1) and (2) yields

work

student

the

by

and

aG = 1.5 m>s2 :

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destroy

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and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

a = 3 rad>s2

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3 m/s2

4m

Ans.

17–110. v0

The 10-lb hoop or thin ring is given an initial angular velocity of 6 rad>s when it is placed on the surface. If the coefficient of kinetic friction between the hoop and the surface is mk = 0.3, determine the distance the hoop moves before it stops slipping.

6 in. O

SOLUTION + c ©Fy = m(aG)y ; + ©F = m(a ) ; ; x G x c + ©MG = IGa;

N -10 = 0 0.3(10) =

N = 10 lb

A

6 0.3(10) A 12 B =

10 32.2

B aG

aG = 9.66 ft>s2

10 A 32.2 B A 126 B 2 a

a = 19.32 rad>s2

When slipping ceases, vG = v r = 0.5v

(1)

v = v0 + at

(a + )

v = 6 + (- 19.32)t + B A;

(2)

or

vG = (vG)0 + aGt laws

vG = 0 + 9.66t in

teaching

Web)

(3)

Dissemination

Wide

copyright

Solving Eqs. (1) to (3) yields:

World

permitted.

v = 3 rad>s

on

learning.

is

of

the

s = s0 + (vG)0 t + 12 aG t2

and

use

United

+ B A;

not

instructors

vG = 1.5 ft>s

States

t = 0.1553 s

for

work

student

the

by

= 0 + 0 + 12 (9.66)(0.1553)2

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This

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is

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the

(including

= 0.116 ft = 1.40 in.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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6 rad/s

Ans.

17–111. A long strip of paper is wrapped into two rolls, each having a mass of 8 kg. Roll A is pin supported about its center whereas roll B is not centrally supported. If B is brought into contact with A and released from rest, determine the initial tension in the paper between the rolls and the angular acceleration of each roll. For the calculation, assume the rolls to be approximated by cylinders.

A 90 mm 90 mm

SOLUTION For roll A. T(0.09) = 12 (8)(0.09)2 aA

c + ©MA = IA a;

(1)

For roll B a + ©MO = ©(Mk)O;

8(9.81)(0.09) = 12 (8)(0.09)2 aB + 8aB (0.09)

(2)

+ c ©Fy = m(aG)y ;

T - 8(9.81) = - 8aB

(3)

Kinematics:

laws

or

aB = aO + (aB>O)t + (aB>O)n

permitted.

Dissemination

copyright

(4)

Wide

in

teaching

Web)

c aBd = c aOd + caB (0.09) d + [0] T T T aB = aO + 0.09aB A+TB

is

of

the

not

instructors

States

World

also,

learning.

aO = aA (0.09)

and

use

United

on

(5)

the

by

A+TB

the

(including

for

work

student

Solving Eqs. (1)–(5) yields:

Ans.

assessing

is

work

solely

of

protected

aA = 43.6 rad>s2

Ans.

part

the

is and

any

courses

T = 15.7 N

This

provided

integrity

of

and

work

this

aB = 43.6 rad>s2

aO = 3.92 m>s2 sale

will

their

destroy

of

aB = 7.85 m>s2

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

B

*17–112. v

The circular concrete culvert rolls with an angular velocity of v = 0.5 rad>s when the man is at the position shown. At this instant the center of gravity of the culvert and the man is located at point G, and the radius of gyration about G is kG = 3.5 ft. Determine the angular acceleration of the culvert. The combined weight of the culvert and the man is 500 lb.Assume that the culvert rolls without slipping, and the man does not move within the culvert.

4 ft O 0.5 ft

SOLUTIONS Equations of Motion: The mass moment of inertia of the system about its mass 500 (3.52) = 190.22 slug # ft2. Writing the moment equation of center is IG = mkG2= 32.2 motion about point A, Fig. a, + ©MA = ©(Mk)A ; - 500(0.5) = -

500 500 (a ) (4) (a ) (0.5) - 190.22a (1) 32.2 G x 32.2 G y

Kinematics: Since the culvert rolls without slipping, a0 = ar = a(4) : laws

or

Applying the relative acceleration equation and referrring to Fig. b, in

teaching

Web)

aG = aO + a * rG>O - v2rG>A Dissemination

Wide

copyright

(aG)xi - (aG)y j = 4ai + ( -ak) * (0.5i) - (0.52)(0.5i)

not

instructors

States

World

permitted.

(aG)xi - (aG)y j = (4a - 0.125)i - 0.5aj

and

use

United

on

learning.

is

of

the

Equation the i and j components,

(2) work

student

the

by

(aG)x = 4a - 0.125

(3)

solely

of

protected

the

(including

for

(aG)y = 0.5a

this

assessing

is

work

Subtituting Eqs. (2) and (3) into Eq. (1), provided

integrity

of

and

work

500 500 (4a - 0.125)(4) (0.5a)(0.5) - 190.22a 32.2 32.2 any

Ans.

sale

will

their

destroy

of

and

a = 0.582 rad>s2

courses

part

the

is

This

-500(0.5) = -

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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G

17–113. v0

The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the initial angular acceleration of the disk and the acceleration of its mass center. The coefficient of kinetic friction between the disk and the floor is mk.

r

SOLUTION Equations of Motion. Since the disk slips, the frictional force is Ff = mkN. The mass 1 moment of inertia of the disk about its mass center is IG = mr2. We have 2 + c ©Fy = m(aG)y;

N - mg = 0

N = mg

+ ©F = m(a ) ; ; x G x

mk(mg) = maG

aG = mkg ;

+ ©MG = IGa;

1 -mk(mg)r = a mr2 ba 2

a =

Ans.

2mkg r

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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17–114. v0

The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the time before it starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is mk.

r

SOLUTION Equations of Motion: Since the disk slips, the frictional force is Ff = mkN. The mass 1 moment of inertia of the disk about its mass center is IG = mr2. 2 + c ©Fy = m(aG)y; N - mg = 0 N = mg + ©F = m(a ) ; ; x G x

mk(mg) = maG

+ ©MG = IGa;

1 -mk(mg)r = - a m r2 b a 2

aG = mkg 2mkg r

a =

Kinematics: At the instant when the disk rolls without slipping, vG = vr. Thus, or

vG = (vG)0 + aGt teaching

Web)

vr mkg

in

(1) Wide

permitted.

Dissemination

t =

laws

vr = 0 + mkgt copyright

+ B A;

on

learning.

is

of

the

not

instructors

States

World

and

the

by

and

use

United

v = v0 + at

work

student

2mkg bt r

the

(including

for

(2)

assessing

is

work

solely

of

v = v0 + a -

protected

(a + )

provided

integrity

of

and

work

this

Solving Eqs. (1) and (2) yields v0r 3mkg

the part any

courses sale

will

their

destroy

of

t =

is

This

1 v 3 0

and

v =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

17–115. ω

The 16-lb bowling ball is cast horizontally onto a lane such that initially v = 0 and its mass center has a velocity v = 8 ft>s. If the coefficient of kinetic friction between the lane and the ball is mk = 0.12, determine the distance the ball travels before it rolls without slipping. For the calculation, neglect the finger holes in the ball and assume the ball has a uniform density.

8 ft/s G 0.375 ft

SOLUTION 16 a 32.2 G

+ ©F = m(a ) ; : x G x

0.12NA =

+ c ©Fy = m(aG)y ;

NA - 16 = 0

a + ©MG = IG a;

2 16 0.12NA(0.375) = c a b (0.375)2 da 5 32.2

Solving, NA = 16 lb;

aG = 3.864 ft>s2;

a = 25.76 rad>s2

laws

or

When the ball rolls without slipping v = v(0.375), Web)

v = v0 + ac t in

teaching

(a+)

States

World

permitted.

Dissemination

Wide

copyright

v = 0 + 25.76t 0.375

learning.

is

of

the

not

instructors

v = 9.660t by

and

use

United

on

v = v0 + ac t

the

+ B A;

the

(including

for

work

student

9.660t = 8 - 3.864t

integrity

of

and

work

provided

the part

1 2 a t 2 c

is

s = s0 + v0 t +

This

+ B A;

this

assessing

is

work

solely

of

protected

t = 0.592 s

destroy

of

and

any

courses

1 (3.864)(0.592)2 2 sale

will

their

s = 0 + 8(0.592) s = 4.06 ft

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

*17–116. The uniform beam has a weight W. If it is originally at rest while being supported at A and B by cables, determine the tension in cable A if cable B suddenly fails. Assume the beam is a slender rod.

A

B

SOLUTION + c ©Fy = m(aG)y; c + ©MA = IAa;

Wa

L 1 W W L L b = c a b L2 d a + a b aa b g 4 4 12 g 4 1 =

L 1 L a + ba g 4 3

L b. 4 or

12 g a b 7 L

teaching

Web)

a =

laws

Since aG = a a

L –– 4

W TA - W = - a G g

in

g W 12 L W L (a)a b = W a ba ba b g g 7 4 L 4 States

World

permitted.

Dissemination

Wide

copyright

TA = W -

not

instructors

4 W 7

United

on

learning.

is

of

the

Ans.

the

by

and

use

TA =

the

(including

for

work

student

Also, W a g G

work

solely

of

protected

integrity provided

L 1 W b = c a b L2 d a 4 12 g

part

the

is

and

any

courses

TA a

This

c + ©MG = IG a;

of

and

work

this

assessing

TA - W = -

is

+ c ©Fy = m(aG)y ;

will

their

destroy

of

L a 4

sale

Since a G =

TA =

1 W a b La 3 g

W L 1 W a b La - W = - a b a g 4 3 g a =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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12 g a b 7 L

TA =

g 1 W 12 a bLa b a b 3 g 7 L

TA =

4 W 7

Ans.

L –– 2

L –– 4

17–117. A cord C is wrapped around each of the two 10-kg disks. If they are released from rest, determine the tension in the fixed cord D. Neglect the mass of the cord.

D

A 90 mm C

SOLUTION For A: B

c + a MA = IA aA;

90 mm

1 T(0.09) = c (10)(0.09)2 d aA 2

(1)

1 T(0.09) = c (10)(0.09)2 d aB 2

(2)

For B: a + a MB = IBaB;

10(9.81) - T = 10aB

(3) or

+ T a Fy = m(aB)y;

teaching

Web)

laws

aB = aP + (aB>P)t + (aB>P)n

(4)

Dissemination

Wide

copyright

in

(+ T)aB = 0.09aA + 0.09aB + 0

not

instructors

States

World

permitted.

Solving.

use

United

on

learning.

is

of

the

aB = 7.85 m>s2

for

work

student

the

by

and

aA = 43.6 rad>s2

solely

of

protected

the

(including

aB = 43.6 rad>s2 work

T = 19.6 N work

this

assessing

is

Ans.

any sale

will

their

destroy

of

and

= 118 N

courses

part

the

is

This

provided

integrity

of

and

Ay = 10(9.81) + 19.62

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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17–118. The 500-lb beam is supported at A and B when it is subjected to a force of 1000 lb as shown. If the pin support at A suddenly fails, determine the beam’s initial angular acceleration and the force of the roller support on the beam. For the calculation, assume that the beam is a slender rod so that its thickness can be neglected.

1000 lb 5

3

4

B 8 ft

SOLUTION + ; a Fx = m(aG)x ;

500 4 1000 a b = (a ) 5 32.2 G x

+ T a Fy = m(aG)y ;

3 500 1000 a b + 500 - By = (a ) 5 32.2 G y

a + a MB = a (Mk)B;

3 500 1 500 500(3) + 1000 a b (8) = (aG)y(3) + c a b (10)2 d a 5 32.2 12 32.2

aB = aG + aB>G - aBi = - (aG)x i - (aG)yj + a(3)j or

(aG)y = a(3)

laws

(+ T)

Web)

a = 23.4 rad>s2

Wide

copyright

in

teaching

Ans.

By = 9.62 lb States

World

permitted.

Dissemination

Ans.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

By 7 0 means that the beam stays in contact with the roller support.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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A 2 ft

17–119. The 30-kg uniform slender rod AB rests in the position shown when the couple moment of M = 150 N # m is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers.

A

0.75 m

SOLUTION M ⫽ 150 N⭈m

Equations of Motion: Here, the mass moment of inertia of the rod about its mass 1 2 1 center is IG = ml = (30)(1.52) = 5.625 kg # m2. Writing the moment 12 12 equations of motion about the intersection point A of the lines of action of NA and

0.75 m

NB and using, Fig. a, B

+ ©MA = ©(Mk)A;

-150 = 30(aG)x(0.75) - 5.625a (1)

5.625a - 22.5(aG)x = 150

Kinematics: Applying the relative acceleration equation to points A and G, Fig. b,

laws

or

aG = aA + a * rG>A - v2rG>A teaching

Web)

(aG)xi + (aG)y j = - aAj + ( - ak) * (-0.75j) - 0

Dissemination

Wide

copyright

in

(aG)xi + (aG)y j = - 0.75ai - aA j

of

the

not

instructors

States

World

permitted.

Equating the i components,

(2)

by

and

use

United

on

learning.

is

(aG)x = - 0.75a

for

work

student

the

Substituting Eq. (2) into Eq. (1),

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

a = 6.667 rad>s2 = 6.67 rad>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

*17–120. The 30-kg slender rod AB rests in the position shown when the horizontal force P = 50 N is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers.

A

SOLUTION 1.5 m

Equations of Motion: Here, the mass moment of inertia of the rod about its mass 1 2 1 ml = (30)(1.52) = 5.625 kg # m2. Writing the moment equations center is IG = 12 12 of motion about the intersection point A of the lines of action of NA and NB and using, Fig. a, + ©MA = ©(Mk)A;

B

-50(0.15) = 30(aG)x(0.75) - 5.625a

P ⫽ 50 N

(1)

5.625a - 22.5(aG)x = 75

Kinematics: Applying the relative acceleration equation to points A and G, Fig. b,

or

aG = aA + a * rG>A - v2rG>A

teaching

Web)

laws

(aG)xi + (aG)y j = - aA j + ( - ak) * (-0.75j) - 0

Wide

copyright

in

(aG)xi + (aG)y j = - 0.75ai - aA j

(2) and

use

United

on

learning.

is

of

the

(aG)x = - 0.75a

work

student

the

by

Substituting Eq. (2) into Eq. (1),

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

a = 3.333 rad>s2 = 3.33 rad>s2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

not

instructors

States

World

permitted.

Dissemination

Equating the i components,

18–1. At a given instant the body of mass m has an angular velocity V and its mass center has a velocity vG. Show that its kinetic energy can be represented as T = 12IICv2, where IIC is the moment of inertia of the body determined about the instantaneous axis of zero velocity, located a distance rG>IC from the mass center as shown.

IC rG/IC

G vG

SOLUTION T =

1 1 my2G + IG v2 2 2 1 1 m(vrG>IC)2 + IG v2 2 2

=

1 A mr2G>IC + IG B v2 2

=

1 I v2 2 IC

However mr2G>IC + IG = IIC Q.E.D.

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

=

where yG = vrG>IC

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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V

18–2. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, and the wheel is rotated until the torque M = 25 N # m is developed, determine the maximum angular velocity of the wheel if it is released from rest.

0.5 m O

M

SOLUTION Kinetic Energy and Work: The mass moment of inertia of the wheel about point O is IO = mRr 2 + 2 ¢

1 m l2 ≤ 12 r

= 5(0.52) + 2c

1 (2)(12) d 12

= 1.5833 kg # m2 Thus, the kinetic energy of the wheel is

or

1 1 I v2 = (1.5833) v2 = 0.79167 v2 2 O 2

teaching

Web)

laws

T =

permitted.

Dissemination

Wide

copyright

in

Since the wheel is released from rest, T1 = 0. The torque developed is M = ku = 2u. Here, the angle of rotation needed to develop a torque of M = 25 N # m is

learning.

is

of

the

not

instructors

States

World

u = 12.5 rad

2u = 25

for

work

student

the

by

and

use

United

on

The wheel achieves its maximum angular velocity when the spacing is unwound that M is when the wheel has rotated u = 12.5 rad. Thus, the work done by q is the

(including

12.5 rad work

solely

of

protected

2u du

L0

this

assessing

L

Mdu =

is

UM =

is

part

the

= 156.25 J

This

0

will sale

Principle of Work and Energy:

their

destroy

of

and

any

courses

= u †

provided

integrity

of

and

work

12.5 rad 2

T1 + © u 1 - 2 = T2 0 + 156.25 = 0.79167 v2 v = 14.0 rad/s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

18–3. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, so that the torque on the center of the wheel is M = 12u2 N # m, where u is in radians, determine the maximum angular velocity of the wheel if it is rotated two revolutions and then released from rest.

0.5 m O

M

SOLUTION Io = 2c

1 (2)(1)2 d + 5(0.5)2 = 1.583 12

T1 + ©U1 - 2 = T2 4p

0 +

L0

2u du =

1 (1.583) v2 2

(4p)2 = 0.7917v2 v = 14.1 rad/s

sale

will

their

destroy

of

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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*18–4. The 50-kg flywheel has a radius of gyration of k0 = 200 mm about its center of mass. If it is subjected to a torque of M = (9u1>2) N # m, where u is in radians, determine its angular velocity when it has rotated 5 revolutions, starting from rest.

M ⫽ (9 u1/2) N⭈m

O

SOLUTION Kinetic Energy and Work: The mass moment inertia of the flywheel about its mass center is IO = mkO2= 50(0.22) = 2 kg # m2. Thus, the kinetic energy of the flywheel is T =

1 1 I v2 = (2)v2 = v2 2 O 2

Since the wheel is initially at rest, T1 = 0 . Referring to Fig. a, W, Ox, and Oy do no work while M does positive work. When the wheel rotates u = (5 rev) ¢

2p rad ≤ = 10p, the work done by M is 1 rev 10p or teaching

Web)

10p 0

Wide Dissemination

= 6u3>2 `

9u1>2du laws

L0

in

L

Mdu =

copyright

UM =

not

instructors

States

World

permitted.

= 1056.52 J United

on

learning.

is

of

the

Principle of Work and Energy: the

by

and

use

T1 + ©U1 - 2 = T2

the

(including

for

work

student

0 + 1056.52 = v2

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this

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protected

v = 32.5 rad>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

18–5. The spool has a mass of 60 kg and a radius of gyration kG = 0.3 m. If it is released from rest, determine how far its center descends down the smooth plane before it attains an angular velocity of v = 6 rad>s. Neglect friction and the mass of the cord which is wound around the central core.

0.3 m

SOLUTION

0.5 m

G

T1 + ©U1 - 2 = T2 0 + 60(9.81) sin 30°(s) =

1 1 C 60(0.3)2 D (6)2 + (60) C 0.3(6) D 2 2 2

A 30⬚

s = 0.661 m

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of

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permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

18–6. Solve Prob. 18–5 if the coefficient of kinetic friction between the spool and plane at A is mk = 0.2.

0.3 m

SOLUTION

0.5 m

G

sG sA = 0.3 (0.5 - 0.3) A

sA = 0.6667sG +a©Fy

= 0;

30

NA - 60(9.81) cos 30° = 0 NA = 509.7 N T1 + ©U1 - 2 = T2 1 C 60(0.3)2 D (6)2 2 or

0 + 60(9.81) sin 30°(sG) - 0.2(509.7)(0.6667sG) =

teaching

Web)

laws

1 (60) C (0.3)(6) D 2 2 in

+

Wide

copyright

sG = 0.859 m

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of

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not

instructors

States

World

permitted.

Dissemination

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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18–7. v

The double pulley consists of two parts that are attached to one another. It has a weight of 50 lb and a radius of gyration about its center of kO = 0.6 ft. If it rotates with an angular velocity of 20 rad>s clockwise, determine the kinetic energy of the system. Assume that neither cable slips on the pulley.

20 rad/s

0.5ft

1 ft O

SOLUTION T =

1 1 1 I v2 + mA v2A + mB v2B 2 O O 2 2

T =

1 50 1 20 1 30 a (0.6)2 b (20)2 + a b C (20)(1) D 2 + a b C (20)(0.5) D 2 2 32.2 2 32.2 2 32.2

B 30 lb A 20 lb

= 283 ft # lb

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on

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of

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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*18–8. v ⫽ 20 rad/s

The double pulley consists of two parts that are attached to one another. It has a weight of 50 lb and a centroidal radius of gyration of kO = 0.6 ft and is turning with an angular velocity of 20 rad> s clockwise. Determine the angular velocity of the pulley at the instant the 20-lb weight moves 2 ft downward.

0.5 ft

1 ft O

SOLUTION Kinetic Energy and Work: Since the pulley rotates about a fixed axis, vA = vrA = v(1) and vB = vrB = v(0.5). The mass moment of inertia of the B 30 lb

50 ≤ (0.62) = 0.5590 slug # ft2. Thus, the 32.2

pulley about point O is IO = mkO 2 = ¢

A 20 lb

kinetic energy of the system is T =

1 1 20 1 30 (0.5590)v2 + ¢ ≤ [v(1)]2 + ¢ ≤ [v(0.5)]2 2 2 32.2 2 32.2 or

=

1 1 1 I v2 + mAvA2 + mBvB2 2 O 2 2

teaching

Web)

laws

= 0.7065v2

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

Thus, T1 = 0.7065(202) = 282.61 ft # lb. Referring to the FBD of the system shown in Fig. a, we notice that Ox, Oy, and Wp do no work while WA does positive work and WB does negative work. When A moves 2 ft downward, the pulley rotates

by

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SA SB = rA rB

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SB 2 = 1 0.5

work

this

assessing

is

work

SB = 2(0.5) = 1 ft c

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the

is

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provided

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Thus, the work of WA and WB are

destroy

of

and

any

courses

UWA = WA SA = 20(2) = 40 ft # lb

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UWB = - WB SB = - 30(1) = - 30 ft # lb Principle of Work and Energy: T1 + U1 - 2 = T2 282.61 + [40 + ( -30)] = 0.7065 v2 v = 20.4 rad>s

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Ans.

18–9. 400 mm

If the cable is subjected to force of P = 300 N, and the spool starts from rest, determine its angular velocity after its center of mass O has moved 1.5 m. The mass of the spool is 100 kg and its radius of gyration about its center of mass is kO = 275 mm. Assume that the spool rolls without slipping.

P ⫽ 300 N

O 200 mm

SOLUTION Kinetic Energy and Work: Referring to Fig. a, we have vO = vrO>IC = v(0.4) The mass moment of inertia of the spool about its mass center is IO = mkO2 = 100(0.2752) = 7.5625 kg # m2. Thus, the kinetic energy of the spool is 1 1 mvO2+ IOv2 2 2 1 1 = (100)[v(0.4)]2 + (7.5625)v2 2 2

T =

= 11.78125v2

of

the

not

instructors

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by

and

use

United

on

learning.

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UP = Psp = 300(2.25) = 675 J

(including

for

work

student

Principle of Work and Energy:

is

work

solely

of

protected

the

T1 + ©U1 - 2 = T2

part

the

is sale

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destroy

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any

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v = 7.57 rad>s

This

provided

integrity

of

and

work

this

assessing

0 + 675 = 11.78125v2

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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Ans.

permitted.

Dissemination

Wide

copyright

in

teaching

work P does positive work. When the center O of the spool moves to the right by rP>IC 0.6 SO = 1.5 m, P displaces sP = s = ¢ ≤ (1.5) = 2.25 m. Thus, the work done rO> IC O 0.4 by P is

Web)

laws

or

Since the spool is initially at rest, T1 = 0. Referring to Fig. b, W, N, and Ff do no

18–10. The two tugboats each exert a constant force F on the ship. These forces are always directed perpendicular to the ship’s centerline. If the ship has a mass m and a radius of gyration about its center of mass G of kG, determine the angular velocity of the ship after it turns 90°. The ship is originally at rest.

F

G

SOLUTION Principle of Work and Energy: The two tugboats create a couple moment of p M = Fd to rotate the ship through an angular displacement of u = rad. The mass 2 moment of inertia about its mass center is IG = mk2G. Applying Eq. 18–14, we have

–F

T1 + a U 1 - 2 = T2 0 + Mu =

1 I v2 2 G or

1 p b = A mk2G B v2 2 2 in

teaching

Web)

laws

0 + Fda

Ans.

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and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

pFd 1 kG A m

copyright

v =

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d

18–11. At the instant shown, link AB has an angular velocity vAB = 2 rad>s. If each link is considered as a uniform slender bar with a weight of 0.5 lb>in., determine the total kinetic energy of the system.

vAB

A

2 rad/s

3 in. 4 in.

C

SOLUTION vBC =

6 = 1.5 rad>s 4

B 5 in.

vC = 1.5(4 22) = 8.4853 in.>s 45

rIC - G = 2(2)2 + (4)2 = 4.472 vG = 1.5(4.472) = 6.7082 in.>s

3 2 4 2 1 4(0.5) 6.7082 2 1 1 4(0.5) 1 1 3(0.5) c a b a b d (2)2 + c da b + c a b a b d(1.5)2 2 3 32.2 12 2 32.2 12 2 12 32.2 12 laws

T =

8.4853 = 1.697 rad>s 5

or

vDC =

teaching

Web)

5 2 1 1 5(0.5) c a b a b d (1.697)2 = 0.0188 ft # lb 2 3 32.2 12

in

Ans.

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United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

+

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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D

*18–12. Determine the velocity of the 50-kg cylinder after it has descended a distance of 2 m. Initially, the system is at rest. The reel has a mass of 25 kg and a radius of gyration about its center of mass A of kA = 125 mm.

A

SOLUTION T1 + ©U1 - 2 = T2 0 + 50(9.81)(2) =

2 1 v [(25)(0.125)2] ¢ ≤ 2 0.075

+

1 (50) v2 2

v = 4.05 m>s

sale

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of

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the

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on

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of

the

not

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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75 mm

18–13. The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of kA = 6 in. If pulley B attached to the motor is subjected to a torque of M = 40(2 - e -0.1u) lb # ft, where u is in radians, determine the velocity of the 200-lb crate after it has moved upwards a distance of 5 ft, starting from rest. Neglect the mass of pulley B.

7.5 in.

M

SOLUTION Kinetic Energy and Work: Since the wheel rotates about a fixed axis , vC = vrC = v(0.375). The mass moment of inertia of A about its mass center is 50 b A 0.52 B = 0.3882 slug # ft2. Thus, the kinetic energy of the IA = mkA 2 = a 32.2 system is

1 1 200 (0.3882)v2 + a b C v(0.375) D 2 2 2 32.2

teaching

Web)

=

laws

1 1 IA v2 + mC vC 2 2 2 or

T = TA + TC =

Wide

copyright

in

= 0.6308v2

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

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United

on

learning.

is

of

the

not

instructors

States

World

integrity

provided

part

the

is

This

any

courses and

destroy

of

0

will

= 2280.93 ft # lb

33.33 rad

their

= c 40 A 2u + 10e - 0.1u B d 2

of

and

work

40 A 2 - e - 0.1u B du

L0

sale

L

this

33.33 rad

MduB =

UWC = - WC sC = - 200(5) = - 1000 ft # lb Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [2280.93 - 1000] = 0.6308v2 v = 45.06 rad>s Thus, vC = 45.06(0.375) = 16.9 ft>s c

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Ans.

permitted.

Dissemination

Since the system is initially at rest, T1 = 0. Referring to Fig. b, Ax, Ay, and WA do no work, M does positive work, and WC does negative work. When crate C moves 5 ft sC 5 = 13.333 rad. Then, = upward, wheel A rotates through an angle of uA = r 0.375 rA 0.625 b (13.333) = 33.33 rad u = a pulley B rotates through an angle of uB = rB A 0.25 . Thus, the work done by M and WC is UM =

A

3 in. B

4.5 in.

18–14. The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of kA = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb # ft, determine the velocity of the 200-lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

7.5 in.

M

SOLUTION Kinetic Energy and Work: Since the wheel at A rotates about a fixed axis, vC = vrC = v(0.375). The mass moment of inertia of wheel A about its mass center 50 b A 0.52 B = 0.3882 slug # ft2. Thus, the kinetic energy of the is IA = mkA 2 = a 32.2 system is

1 1 200 (0.3882)v2 + a b C v(0.375) D 2 2 2 32.2

Web)

=

laws

1 1 IA v2 + mC vC 2 2 2

or

T = TA + TC =

Wide

copyright

in

teaching

= 0.6308v2

of

the

is

2p rad b = 10p rad, the wheel rotates through an angle of 1 rev rB 0.25 uA = uB = a b (10p) = 4p. Thus, the crate displaces upwards through a rA 0.625 distance of sC = rC uA = 0.375(4p) = 1.5p ft. Thus, the work done by M and WC is assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

uB = (5 rev)a

this

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of

and

work

provided

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will

Principle of Work and Energy:

their

destroy

of

and

any

courses

part

the

is

This

UWC = -WC sC = - 200(1.5p) = -300p ft # lb

T1 + ©U1 - 2 = T2 0 + [500p - 300p] = 0.6308v2 v = 31.56 rad>s Thus, vC = 31.56(0.375) = 11.8 ft>s c

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Ans.

not

instructors

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permitted.

Dissemination

Since the system is initially at rest, T1 = 0. Referring to Fig. b, Ax, Ay, and WA do no work, M does positive work, and WC does negative work. When pulley B rotates

UM = MuB = 50(10p) = 500p ft # lb

A

3 in. B

4.5 in.

18–15. The 50-kg gear has a radius of gyration of 125 mm about its center of mass O. If gear rack B is stationary, while the 25-kg gear rack C is subjected to a horizontal force of P = 150 N, determine the speed of C after the gear’s center O has moved to the right a distance of 0.3 m, starting from rest.

C

P ⫽ 150 N 150 mm O B

SOLUTION Kinetic Energy and Work: Referring to Fig. a, vC vC v = = = 3.333vC rC>IC 0.3 Then, vO = vrO>IC = (3.333vC)(0.15) = 0.5vC The mass moment of inertia of the gear about its mass center is IO = mkO2 = 50(0.1252) = 0.78125 kg # m2. Thus, the kinetic energy of the system is T = TA + TC

teaching

Web)

laws

or

1 1 1 = c mAvO 2 + IOv 2 d + mC vC2 2 2 2

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permitted.

Dissemination

Wide

copyright

in

1 1 1 = c (50)(0.5vC)2 + (0.78125)(3.333vC)2 d + (25) vC2 2 2 2

and

use

United

on

learning.

is

of

the

not

instructors

= 23.090vC 2

for

work

student

the

by

Since the system is initially at rest, T1 = 0. Referring to Fig. b, WC, WA, F, and N do of

protected

the

(including

no work, while P does positive work. When the center O of the gear travels to the

part

the

is

This

provided

integrity

of

and

work

this

assessing

is

work

solely

right through a distance of sO = 0.3 m, P displaces horizontally through a distance rC>IC 0.3 sO = a b(0.3) = 0.6 m. Thus, the work done by P is of sC = rO>IC 0.15

sale

will

Principle of Work and Energy:

their

destroy

of

and

any

courses

UP = PsD = 150(0.6) = 90 J

T1 + ©U1-2 = T2 0 + 90 = 23.090vC 2 vC = 1.97 m>s

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*18–16. Gear B is rigidly attached to drum A and is supported by two small rollers at E and D. Gear B is in mesh with gear C and is subjected to a torque of M = 50 N # m. Determine the angular velocity of the drum after C has rotated 10 revolutions, starting from rest. Gear B and the drum have 100 kg and a radius of gyration about their rotating axis of 250 mm. Gear C has a mass of 30 kg and a radius of gyration about its rotating axis of 125 mm.

300 mm A

E 150 mm

SOLUTION Kinetic Energy and Work: Since gear B is in mesh with gear C and both gears rotate rB 0.2 b v = 1.333vA. The mass moment of the about fixed axes, vC = a b vA = a rC 0.15 A drum and gear C about their rotating axes are IA = mAk2 = 100(0.252) = 6.25 kg # m2 and IC = mCk2 = 30(0.1252) = 0.46875 kg # m2. Thus, the kinetic energy of the system is

1 1 (6.25)vA2 + (0.46875)(1.333vA)2 2 2

or

=

in

teaching

Web)

1 1 I v 2 + ICvC 2 2 A A 2 laws

=

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permitted.

Dissemination

Wide

copyright

= 3.5417vA2

student

the

by

and

use

United

on

learning.

is

of

the

protected

the

(including

for

work

UM = 50(20p) = 1000p J assessing

is

work

solely

of

Principle of Work and Energy:

and

any

courses

part

the

is

This

provided

integrity

of

and

work

this

T1 + ©U1-2 = T2

sale

will

their

destroy

of

vA = 29.8 rad>s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Ans.

not

instructors

Since the system is initially at rest, T1 = 0. Referring to Fig. a, M does positive work. 2p rad b = 20p, the work done by M is When the gear C rotates u = (10 rev)a 1 rev

D C

M ⫽ 50 N⭈m

T = TA + TC

0 + 1000p = 3.5417vA2

B

200 mm

18–17. The center O of the thin ring of mass m is given an angular velocity of v0. If the ring rolls without slipping, determine its angular velocity after it has traveled a distance of s down the plane. Neglect its thickness.

v0 s r O

SOLUTION u

T1 + ©U1-2 = T2 1 1 (mr2 + mr2)v0 2 + mg(s sin u) = (mr2 + mr2)v2 2 2 g v = v0 2 + 2 s sin u A r

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laws

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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18–18. If the end of the cord is subjected to a force of P = 75 lb, determine the speed of the 100-lb block C after P has moved a distance of 4 ft, starting from rest. Pulleys A and B are identical, each of which has a weight of 10 lb and a radius of gyration of k = 3 in. about its center of mass.

4 in. B

A 4 in. P = 75 lb

SOLUTION C

Kinetic Energy and Work: Referring to Fig. a, we have vD = vArD>IC = vA(0.6667) (vG)A = vC = vArC>IC = vA(0.3333) Since pulley B rotates about a fixed axis, its angular velocity is vB =

vA(0.6667) vD = = 2vA rB 0.3333

The mass moment of inertia of pulleys A and B about their resperctive mass centers or

10 3 2 ≤ ¢ ≤ = 0.01941 slug # ft2. Thus, the kinetic 32.2 12 teaching

Web)

laws

are (IA)G = (IB)G = mk2 = ¢

Dissemination

Wide

copyright

in

enegry of the system is

not

instructors

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World

permitted.

T = TA + TB + TC

the

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on

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the

1 1 1 1 = B mA(vG)A2 + (IG)AvAR2 + (IG)BvB2+ mCvC2 2 2 2 2

the

(including

for

work

student

1 10 1 1 ¢ ≤ [vA(0.3333)]2 + (0.01941)vA2R + (0.01941)(2vA)2 2 32.2 2 2 is

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= B

this

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1 100 ¢ ≤ [vA(0.3333)]2 2 32.2

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= 0.2383vA2

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Since the system is initially at rest, T1 = 0. Referring to Fig. b, R 1, R 2, and WB do no work, P does positive work, and WA and WC do negative work. When P moves sD = 4 ft downward, the center of the pulley moves upward through a distance of rC>IC 0.3333 sC = s = (4) = 2 ft. Thus, the work done by WA, WC, and P is rD>IC D 0.6667 UWA = - WAsC = - 10(2) = - 20 ft # lb UWC = - WCsC = - 100(2) = - 200 ft # lb UP = PsD = 75(4) = 300 ft # lb Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [- 20 + ( - 200) + 300] = 0.2383vA2 vA = 18.32 rad>s Thus, vC = 18.32(0.3333) = 6.11 ft>s c © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

18–19. When u = 0°, the assembly is held at rest, and the torsional spring is untwisted. If the assembly is released and falls downward, determine its angular velocity at the instant u = 90°. Rod AB has a mass of 6 kg, and disk C has a mass of 9 kg.

C 450 mm B k ⫽ 20 N⭈m/rad u

SOLUTION

A

Kinetic Energy and Work: Since the rod rotates about a fixed axis, (vG)AB = vrGAB = v(0.225) and (vG)C = vrGC = v(0.525). The mass moment of the rod 1 1 and the disk about their respective mass centers are (IAB)G = ml2 = (6)(0.452) 12 12 1 1 = 0.10125 kg # m2 and (IC)G = mr2 = (9)(0.0752) = 0.0253125 kg # m2. Thus, 2 2 the kinetic energy of the pendulum is 1 1 T = © mvG 2 + IGv 2 2 2

or

1 1 1 1 = c (6)[v(0.225)]2 + (0.10125)v2 d + c (9)[v(0.525)]2 + (0.0253125)v2 d 2 2 2 2

teaching

Web)

laws

= 1.4555v2 in

1 1 I v2, where IO = c (6)(0.452) 2 O 12 Dissemination

Wide

copyright

This result can also be obtained by applying T =

and work

student

the

by

1 1 IOv2 = (2.9109)v2 = 1.4555v2 2 2

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for

T =

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1 + 6(0.2252) d + c (9)(0.0752) + 9(0.5252) d = 2.9109 kg # m2. Thus, 2

part

the

is

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protected

the

Since the pendulum is initially at rest, T1 = 0. Referring to Fig. a, Ox and Oy do no work, WC and WAB do positive work, and M does negative work. When u = 90°, WAB and WC displace vertically through distances of hAB = 0.225 m and hC = 0.525 m. Thus, the work done by WAB, WC, and M is

destroy

of

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any

courses

UWAB = WABhAB = 6(9.81)(0.225) = 13.24 J

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UWC = WChC = 9(9.81)(0.525) = 46.35 J p>2

UM = -

L

Mdu = -

L0

20udu = - 24.67 J

Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [13.24 + 46.35 + ( - 24.67)] = 1.4555v2 v = 4.90 rad>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Ans.

75 mm

*18–20. If P = 200 N and the 15-kg uniform slender rod starts from rest at u = 0°, determine the rod’s angular velocity at the instant just before u = 45°.

600 mm A 45°

u

P ⫽ 200 N B

SOLUTION Kinetic Energy and Work: Referring to Fig. a, rA>IC = 0.6 tan 45° = 0.6 m Then rG>IC = 30.32 + 0.62 = 0.6708 m

Thus,

teaching

Web)

laws

or

(vG)2 = v2rG>IC = v2(0.6708)

of

the

not

instructors

States

World

permitted.

Dissemination

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copyright

in

1 The mass moment of inertia of the rod about its mass center is IG = ml2 12 1 = (15)(0.62) = 0.45 kg # m2. Thus, the final kinetic energy is 12

by

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1 1 m(vG)22 + IG v2 2 2 2

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=

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for

1 1 (15)[w2(0.6708)]2 + (0.45) v2 2 2 2

work

student

the

T2 =

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and

work

this

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= 3.6v2 2

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Since the rod is initially at rest, T1 = 0. Referring to Fig. b, NA and NB do no work, while P does positive work and W does negative work. When u = 45°, P displaces through a horizontal distance sP = 0.6 m and W displaces vertically upwards through a distance of h = 0.3 sin 45°, Fig. c. Thus, the work done by P and W is UP = PsP = 200(0.6) = 120 J UW = - Wh = - 15(9.81)(0.3 sin 45°) = - 31.22 J Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [120 - 31.22] = 3.6v22 v2 = 4.97 rad>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Ans.

18–21. A yo-yo has a weight of 0.3 lb and a radius of gyration kO = 0.06 ft. If it is released from rest, determine how far it must descend in order to attain an angular velocity v = 70 rad>s. Neglect the mass of the string and assume that the string is wound around the central peg such that the mean radius at which it unravels is r = 0.02 ft.

SOLUTION r

vG = (0.02)70 = 1.40 ft>s T1 + ©U1 - 2 = T2 0 + (0.3)(s) =

1 0.3 1 0.3 a b(1.40)2 + c(0.06)2 a b d(70)2 2 32.2 2 32.2

s = 0.304 ft

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on

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is

of

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States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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O

18–22. If the 50-lb bucket is released from rest, determine its velocity after it has fallen a distance of 10 ft. The windlass A can be considered as a 30-lb cylinder, while the spokes are slender rods, each having a weight of 2 lb. Neglect the pulley’s weight.

B

3 ft

4 ft 0.5 ft

SOLUTION Kinetic Energy and Work: Since the windlass rotates about a fixed axis, vC = vArA vC vC = or vA = = 2vC. The mass moment of inertia of the windlass about its rA 0.5 mass center is IA =

C

2 1 30 1 2 a b A 0.52 B + 4 c a b A 0.52 B + A 0.752 B d = 0.2614 slug # ft2 2 32.2 12 32.2 32.2

Thus, the kinetic energy of the system is

or

T = TA + T C 1 1 I v 2 + m C vC 2 2 A 2

=

1 1 50 (0.2614)(2vC)2 + a bv 2 2 2 32.2 C instructors

States

World

permitted.

Dissemination

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copyright

in

teaching

Web)

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=

and

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is

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= 1.2992vC 2

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the

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Since the system is initially at rest, T1 = 0. Referring to Fig. a, WA, Ax, Ay, and RB do no work, while WC does positive work. Thus, the work done by WC, when it displaces vertically downward through a distance of sC = 10 ft, is

vC = 19.6 ft>s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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any will sale

0 + 500 = 1.2992vC 2

their

destroy

of

and

T1 + ©U1-2 = T2

courses

part

the

is

Principle of Work and Energy:

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provided

integrity

of

and

work

this

assessing

is

work

solely

UWC = WCsC = 50(10) = 500 ft # lb

Ans.

A

0.5 ft

18–23. The combined weight of the load and the platform is 200 lb, with the center of gravity located at G. If a couple moment of M = 900 lb # ft is applied to link AB, determine the angular velocity of links AB and CD at the instant u = 60°. The system is at rest when u = 0°. Neglect the weight of the links.

1 ft

G

M ⫽ 900 lb⭈ft

4 ft

A 2 ft

B

C

3 ft u D

SOLUTION Kinetic Energy and Work: Since the weight of the links are negligible and the crate and platform undergo curvilinear translation, the kinetic energy of the system is T =

1 1 200 b v 2 = 3.1056vG2 mvG2 = a 2 2 32.2 G

Since the system is initially at rest, T1 = 0. Referring to Fig. a, Bx, By, Dx, and Dy do no work while M does positive work and W does negative work. When u = 60°, W displaces upward through a distance of h = 4 sin 60° ft = 3.464 ft. Thus, the work done by M and W is

laws

or

p UM = Mu = 900 a b = 300p ft # lb 3

in

teaching

Web)

UW = - Wh = - 200(3.464) = - 692.82 ft # lb

Wide

copyright

Principle of Work and Energy: States

World

permitted.

Dissemination

T1 + ©U1-2 = T2

not

instructors

0 + [300p - 692.82] = 3.1056vG2 by

and

use

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on

learning.

is

of

the

vG = 8.966 ft >s

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for

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student

the

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the

vG 8.966 = = 2.24 rad>s r 4

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is

vAB = vCD =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Ans.

*18–24. 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. If a constant torque M = 60 lb # ft is applied to the dumping wheel, determine the angular velocity of the tub when it has rotated u = 90°. Originally the tub is at rest when u = 0°.

θ 0.8 ft

SOLUTION G

T1 + ©U1 - 2 = T2 M

1 70 1 70 p )(1.3)2 d (v)2 + [ ] (0.8v)2 0 + 60( ) - 70(0.8) = c ( 2 2 32.2 2 32.2 v = 3.89 rad>s

sale

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the

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on

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of

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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18–25. 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. If a constant torque M = 60 lb # ft is applied to the tub, determine its angular velocity when it has rotated u = 45°. Originally the tub is at rest when u = 0°.

u 0.8 ft

SOLUTION

G

Kinetic Energy and Work: The mass moment of inertia of the tub about point O is M

IO = mkG2+ mrG2 =

70 70 (1.32) + (0.82) 32.2 32.2

= 5.0652 slug # ft2 Thus, kinetic energy of the tub is T =

1 1 I v2 = (5.0652)v2 = 2.5326 v2 2 O 2

in

teaching

Web)

laws

or

Initially, the tub is at rest. Thus, T1 = 0. Referring to the FBD of the tub, Fig. a, we notice that Ox and Oy do no work while M does positive work and W does negative work. Thus, the work done by M and W are

instructors

States

World

permitted.

Dissemination

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copyright

p UM = Mu = 60 a b = 15p ft # lb 4

use

United

on

learning.

is

of

the

not

UW = - Wh = - 70 [0.8 (1 - cos 45°)] = - 16.40 ft # lb

student

the

by

and

Principle of Work and Energy:

the

(including

for

work

T1 + U1 - 2 = T2

assessing

is

work

solely

of

protected

0 + [15p + ( - 16.40)] = 2.5326 v2

Ans.

sale

will

their

destroy

of

and

any

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the

is

This

provided

integrity

of

and

work

this

v = 3.48 rad>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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18–26. Two wheels of negligible weight are mounted at corners A and B of the rectangular 75-lb plate. If the plate is released from rest at u = 90°, determine its angular velocity at the instant just before u = 0°.

A 3 ft 1.5 ft

u

SOLUTION

B

Kinetic Energy and Work: Referring Fig. a, (vG)2 = vrA>IC = v a 20.752 + 1.52 b = 1.677v2 The mass moment of inertia of the plate about its mass center is 1 1 75 b (1.52 + 32) = 2.1836 slug # ft2. Thus, the final IG = m(a2 + b2) = a 12 12 32.2 kinetic energy is

or

1 75 1 a b (1.677v2)2 + IG (2.1836)v22 2 32.2 2

teaching

Web)

=

1 1 m(vG)22 + v22 2 2 laws

T2 =

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copyright

in

= 4.3672v2 2

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is

of

the

for

work

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UW = Wh = 75(1.677) = 125.78 ft # lb

solely

of

protected

the

(including

Principle of Work and Energy:

this

assessing

is

work

T1 + ©U1-2 = T2

the

is

This

provided

integrity

of

and

work

0 + 125.78 = 4.3672v2 2

sale

will

their

destroy

of

and

any

courses

part

v2 = 5.37 rad>s

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Ans.

not

instructors

States

World

permitted.

Dissemination

Since the plate is initially at rest, T1 = 0. Referring to Fig. b, NA and NB do no work, while W does positive work. When u = 0°, W displaces vertically through a distance of h = 20.752 + 1.52 = 1.677 ft, Fig. c. Thus, the work done by W is

18–27. P ⫽ 25 lb

The 100-lb block is transported a short distance by using two cylindrical rollers, each having a weight of 35 lb. If a horizontal force P = 25 lb is applied to the block, determine the block’s speed after it has been displaced 2 ft to the left. Originally the block is at rest. No slipping occurs.

1.5 ft

SOLUTION T1 + ©U1-2 = T2 0 + 25(2) =

vB 2 vB 2 1 35 1 1 35 1 100 a b (vB)2 + 2 B a ba b + a a b (1.5)2 b a b R 2 32.2 2 32.2 2 2 2 32.2 3 vB = 5.05 ft>s

sale

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and

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of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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1.5 ft

*18–28. The hand winch is used to lift the 50-kg load. Determine the work required to rotate the handle five revolutions. The gear at A has a radius of 20 mm.

rB = 130 mm B

SOLUTION

100 mm

20(uA) = uB (130)

A

When uA = 5 rev. = 10 p uB = 4.8332 rad Thus load moves up s = 4.8332(0.1 m) = 0.48332 m U = 50(9.81)(0.48332) = 237 J

provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

United

on

learning.

is

of

the

not

instructors

States

World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

any

courses

part

the

is

This

destroy

of

and

will

their

sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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375 mm

18–29. A motor supplies a constant torque or twist of M = 120 lb # ft to the drum. If the drum has a weight of 30 lb and a radius of gyration of kO = 0.8 ft, determine the speed of the 15-lb crate A after it rises s = 4 ft starting from rest. Neglect the mass of the cord.

M = 120 lb · ft

1.5 ft O

SOLUTION Free Body Diagram: The weight of the crate does negative work since it acts in the opposite direction to that of its displacement sw. Also, the couple moment M does positive work as it acts in the same direction of its angular displacement u. The reactions Ox, Oy and the weight of the drum do no work since point O does not displace.

A

s

teaching

Web)

laws

or

Kinematic: Since the drum rotates about point O, the angular velocity of the vA vA drum and the speed of the crate can be related by vD = = = 0.6667 vA . rD 1.5 When the crate rises s = 4 ft, the angular displacement of the drum is given by s 4 u = = = 2.667 rad . rD 1.5

is

of

the

not

instructors

States

World student

the

by

and

use

United

on

learning.

T1 + a U1 - 2 = T2

the

(including

for

work

1 1 I v2 + mC v2C 2 O 2 is

work

solely

of

protected

0 + Mu - WC sC =

integrity

of

and

work

this

assessing

1 15 1 (0.5963)(0.6667vA)2 + a bvA2 2 2 32.2 part

the

is

This

provided

0 + 120(2.667) - 15(4) =

any

courses

Ans.

sale

will

their

destroy

of

and

vA = 26.7 ft s

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permitted.

Dissemination

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copyright

in

Principle of Work and Energy: The mass moment of inertia of the drum about point 30 O is IO = mk2O = a b (0.82) = 0.5963 slug # ft2. Applying Eq. 18–13, we have 32.2

18–30. M P ⫽ 750 N

Motor M exerts a constant force of P = 750 N on the rope. If the 100-kg post is at rest when u = 0°, determine the angular velocity of the post at the instant u = 60°. Neglect the mass of the pulley and its size, and consider the post as a slender rod.

C

A 4m

SOLUTION

3m

Kinetic Energy and Work: Since the post rotates about a fixed axis, vG = vrG = v (1.5). The mass moment of inertia of the post about its mass center is 1 IG = (100)(32) = 75 kg # m2. Thus, the kinetic energy of the post is 12 T = =

u B

1 1 mvG2 + IGv2 2 2 1 1 (100)[v(1.5)]2 + (75)v2 2 2

= 150v2

in

teaching

Web)

laws

or

1 This result can also be obtained by applying T = IBv2, where IB = 2 1 (100)(32) + 100 (1.52) = 300 kg # m2. Thus, 12 1 1 I v2 = (300)v2 = 150v2 2 B 2

World

permitted.

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copyright

T =

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P displaces sP = A¿C - AC, where AC = 24 2 + 32 - 2(4)(3) cos 30° = 2.053 m

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and A¿C = 24 2 + 32 = 5 m. Thus, sP = 5 - 2.053 = 2.947 m. Also, W displaces vertically upwards through a distance of h = 1.5 sin 60° = 1.299 m. Thus, the work done by P and W is

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this

UP = PsP = 750(2.947) = 2210.14 J

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T1 + ©U1-2 = T2

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Principle of Work and Energy:

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UW = - Wh = - 100 (9.81)(1.299) = - 1274.36 J

0 + [2210.14 - 1274.36] = 150v2 v = 2.50 rad>s

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Ans.

not

instructors

States

Since the post is initially at rest, T1 = 0. Referring to Fig. a, Bx, By, and R C do no work, while P does positive work and W does negative work. When u = 60° ,

18–31. The uniform bar has a mass m and length l. If it is released from rest when u = 0°, determine its angular velocity as a function of the angle u before it slips.

u

O

2l — 3

SOLUTION Kinetic Energy and Work: Before the bar slips, the bar rotates about the fixed axis passing through point O. The mass moment of inertia of the bar about this axis is 1 l 2 1 IO = ml2 + ma b = ml2. Thus, the kinetic energy of the bar is 12 6 9 1 1 1 1 I v2 = a ml2 bv2 = ml2v2 2 O 2 9 18

T =

Initially, the bar is at rest. Thus, T1 = 0. Referring to the FBD of the bar, Fig. a, we notice that N and Ff do no work while W does positive work which is given by

teaching

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or

mgl l UW = Wh = mg a sin ub = sin u 6 6

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in

Principle of Work and Energy:

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T1 + U1-2 = T2

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mgl 1 sin u = ml2v2 6 18 by

0 +

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the

3g sin u l

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sin u

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C l

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v =

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v2 =

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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Ans.

l — 3

*18–32. The uniform bar has a mass m and length l. If it is released from rest when u = 0°, determine the angle u at which it first begins to slip. The coefficient of static friction at O is ms = 0.3.

θ

O

2l — 3

SOLUTION T1 + ©U1-2 = T2 1 1 l l 0 + m g ( sin u) = [ m l2 + m ( )2]v2 6 2 12 6 v =

3 g sin u l

l l 1 m g cos u( ) = [ m l2 + m( )2] a 6 12 6 3 g cos u 2l laws

a =

or

a + ©MO = IO a;

A

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Web)

3 g sin u l )( ) l 6

in

ms N - m g sin u = m(

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copyright

+ ©Fn = m(aG)n ;

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permitted.

Dissemination

msN = 1.5 m g sin u

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3 g cos u l )( ) 2l 6

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- N + m g cos u = m(

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N = 0.75 m g cos u

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1.5 tan u 0.75

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ms =

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will sale

u = 8.53°

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0.3 = 2 tan u

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l — 3

18–33. The two 2-kg gears A and B are attached to the ends of a 3-kg slender bar. The gears roll within the fixed ring gear C, which lies in the horizontal plane. If a 10-N # m torque is applied to the center of the bar as shown, determine the number of revolutions the bar must rotate starting from rest in order for it to have an angular velocity of vAB = 20 rad>s. For the calculation, assume the gears can be approximated by thin disks.What is the result if the gears lie in the vertical plane?

C 400 mm

A

SOLUTION

M = 10 N · m

Mu = T2 1 1 1 10u = 2a IGv2gear b + 2 a mgear b (0.200vAB)2 + IABv2AB 2 2 2 1 (2)(0.150)2 = 0.0225 kg # m2, 2 1 200 IAB = (3)(0.400)2 = 0.0400 kg # m2 and vgear = v , 12 150 AB 2 200 b v2AB + 2(0.200)2v2AB + 0.0200v2AB 10u = 0.0225 a 150 in

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laws

or

Using mgear = 2 kg, IG =

Dissemination

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copyright

When vAB = 20 rad>s,

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permitted.

u = 5.60 rad

is

= 0.891 rev, regardless of orientation

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on

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sale Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

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150 mm

150 mm

Energy equation (where G refers to the center of one of the two gears):

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

B

18–34. A ball of mass m and radius r is cast onto the horizontal surface such that it rolls without slipping. Determine its angular velocity at the instant u = 90°, if it has an initial speed of vG as shown.

SOLUTION

G

vG . r

Kinetic Energy and Work: Since the ball rolls without slipping, vG = vr or v =

2 2 mr . Thus, the 5

The mass moment of inertia of the ball about its mass cener is IG = kinetic energy of the ball is T =

1 1 mvG2+ IGv2 2 2

=

vG 2 1 1 2 mvG2+ a mr2 b a b r 2 2 5

=

7 mv 2 10 G

7 mvG2. Referring to the FBD of the 10 ball, Fig. a, we notice that N does no work while W does negative work. When laws

or

Initially, the ball has a speed of vG. Thus, T1 =

copyright

in

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u = 90°, h = R (1 - cos 90°) = R. Thus, Dissemination

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UW = - Wh = - mgR instructors

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Principle of Work and Energy:

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T1 + U1-2 = T2

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7 7 mvG2 + (- mgR) = m(vG)22 10 10

the

(including

1 (7vG2 - 10gR ) A7

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(vG)2 =

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(vG)2 10 = vG2 gR>r r A 7

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v2 =

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u R

r

vG

18–35. A ball of mass m and radius r is cast onto the horizontal surface such that it rolls without slipping. Determine the minimum speed vG of its mass center G so that it rolls completely around the loop of radius R + r without leaving the track. u R r G

SOLUTION + T ©Fy = m(aG)y ;

mg = m a

v2 b R

v2 = gR T1 + ©U1 - 2 = T2 gR 1 2 2 v2G 1 1 2 1 a mr b a 2 b + mv2G - mg(2R) = a mr2 b a 2 b + m(gR) 2 5 2 2 5 2 r r 1 1 1 2 1 vG + v2G = 2gR + gR + gR 5 2 5 2 or

3 gR A7

vG = 3

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Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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vG

*18–36. At the instant shown, the 50-lb bar rotates clockwise at 2 rad>s. The spring attached to its end always remains vertical due to the roller guide at C. If the spring has an unstretched length of 2 ft and a stiffness of k = 6 lb>ft, determine the angular velocity of the bar the instant it has rotated 30° clockwise.

C

4 ft

k

A B

SOLUTION

2 rad/s 6 ft

Datum through A. T1 + V1 = T2 + V2 1 1 1 50 1 1 50 c a b(6)2 d(2)2 + (6)(4 - 2)2 = c a b (6)2 dv2 2 3 32.2 2 2 3 32.2 +

1 (6)(7 - 2)2 - 50(1.5) 2

v = 2.30 rad>s

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copyright

in

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Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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18–37. At the instant shown, the 50-lb bar rotates clockwise at 2 rad>s. The spring attached to its end always remains vertical due to the roller guide at C. If the spring has an unstretched length of 2 ft and a stiffness of k = 12 lb>ft, determine the angle u, measured from the horizontal, to which the bar rotates before it momentarily stops.

C

4 ft

k

A B

SOLUTION

2 rad/s 6 ft

T1 + V1 = T2 + V2 1 1 1 1 50 c a b (6)2 d (2)2 + (12)(4 - 2)2 = 0 + (12)(4 + 6 sin u - 2)2 - 50(3 sin u) 2 3 32.2 2 2 61.2671 = 24(1 + 3 sin u)2 - 150 sin u 37.2671 = - 6 sin u + 216 sin2 u Set x = sin u, and solve the quadratic equation for the positive root: sin u = 0.4295 u = 25.4°

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Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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18–38. The spool has a mass of 50 kg and a radius of gyration kO = 0.280 m. If the 20-kg block A is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity v = 5 rad>s. Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

0.3 m 0.2 m O

SOLUTION vA = 0.2v = 0.2(5) = 1 m>s System:

A

T1 + V1 = T2 + V2 [0 + 0] + 0 =

1 1 (20)(1)2 + [50(0.280)2](5)2 - 20(9.81) s 2 2

s = 0.30071 m = 0.301 m

Ans.

Block: laws in

teaching

Web)

1 (20)(1)2 2

Dissemination

Wide

copyright

0 + 20(9.81)(0.30071) - T(0.30071) =

or

T1 + ©U1 - 2 = T2

permitted.

Ans.

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T = 163 N

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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18–39. The spool has a mass of 50 kg and a radius of gyration kO = 0.280 m. If the 20-kg block A is released from rest, determine the velocity of the block when it descends 0.5 m.

0.3 m 0.2 m O

SOLUTION Potential Energy: With reference to the datum established in Fig. a, the gravitational potential energy of block A at position 1 and 2 are V1 = (Vg)1 = WAy1 = 20 (9.81)(0) = 0

A

V2 = (Vg)2 = - WAy2 = - 20 (9.81)(0.5) = - 98.1 J vA vA = = 5vA. rA 0.2 Here, the mass moment of inertia about the fixed axis passes through point O is Kinetic Energy: Since the spool rotates about a fixed axis, v =

IO = mkO2 = 50 (0.280)2 = 3.92 kg # m2. Thus, the kinetic energy of the system is

or laws in

teaching

Web)

1 1 (3.92)(5vA)2 + (20)vA2 = 59vA2 2 2

Wide Dissemination

=

1 1 I v2 + mAvA2 2 O 2

copyright

T =

instructors

States

World

permitted.

Since the system is at rest initially, T1 = 0

United

on

learning.

is

of

the

not

Conservation of Energy: the

by

and

use

T1 + V1 = T2 + V2

the

(including

for

work

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0 + 0 = 59vA2 + (-98.1)

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vA = 1.289 m>s

Ans.

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this

= 1.29 m>s

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*18–40. G

An automobile tire has a mass of 7 kg and radius of gyration kG = 0.3 m. If it is released from rest at A on the incline, determine its angular velocity when it reaches the horizontal plane. The tire rolls without slipping.

0.4 m A 5m 30° 0.4 m B

SOLUTION nG = 0.4v Datum at lowest point. T1 + V1 = T2 + V2 0 + 7(9.81)(5) =

1 1 (7)(0.4v)2 + [7 (0.3)2]v2 + 0 2 2

v = 19.8 rad s

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copyright

in

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laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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18–41. The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant the rod becomes horizontal, i.e., u = 0°. The system is released from rest when u = 45°.

C

3 ft

SOLUTION

u B

T1 + V1 = T2 + V2

0.8 ft A

vC 2 1 1 4 1 1 b (3)2 d a b + a b(vC)2 + 0 0 + 4(1.5 sin 45°) + 1(3 sin 45°) = c a 2 3 32.2 3 2 32.2 vC = 13.3 ft>s

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18–42. The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant u = 30°. The system is released from rest when u = 45°.

C

3 ft

SOLUTION

u B

vB = 0.8vA vBC

0.8 ft

vB vC vG = = = 1.5 2.598 1.5

A

Thus, vB = vG = 1.5vBC

vC = 2.598vBC

vA = 1.875 vBC

laws

or

T1 + V1 = T2 + V2

in

teaching

Web)

0 + 4(1.5 sin 45°) + 1(3 sin 45°) 1 1 20 1 20 c a b (0.8)2 d (1.875vBC)2 + a b (1.5 vBC)2 2 2 32.2 2 32.2 instructors

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1 1 4 1 4 + c a b(3)2 dv2BC + a b (1.5vBC)2 2 12 32.2 2 32.2

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1 1 b (2.598vBC)2 + 4(1.5 sin 30°) + 1(3 sin 30°) + a 2 32.2

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the

is

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vBC = 1.180 rad>s

courses

vC = 2.598(1.180) = 3.07 ft>s sale

will

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destroy

of

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Ans.

18–43. The door is made from one piece, whose ends move along the horizontal and vertical tracks. If the door is in the open position, u = 0°, and then released, determine the speed at which its end A strikes the stop at C. Assume the door is a 180-lb thin plate having a width of 10 ft.

5 ft

3 ft

SOLUTION T 1 + V1 = T2 + V2 0 + 0 =

1 1 180 1 180 c a b (8)2 d v2 + a b (1v)2 - 180(4) 2 12 32.2 2 32.2

v = 6.3776 rad>s vC = v(5) = 6.3776(5) = 31.9 m>s

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sale © 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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A

C

B

u

*18–44. Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10 kg and a radius of gyration of 125 mm about its center of mass. Gear B and drum C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm.

100 mm 150 mm C

A

B 200 mm

SOLUTION Potential Energy: With reference to the datum shown in Fig. a, the gravitational potential energy of block D at position (1) and (2) is V1 = (Vg)1 = WD(yD)1 = 50 (9.81)(0) = 0

D

V2 = (Vg)2 = - WD(yD)2 = - 50(9.81)(2) = - 981 J vD vD = = 10vD. rD 0.1

Kinetic Energy: Since gear B rotates about a fixed axis, vB =

rB 0.2 bv = a b(10vD) = 13.33vD. rA B 0.15 The mass moment of inertia of gears A and B about their mass centers are IA = mAkA2 = 10(0.1252) = 0.15625 kg # m2 and IB = mBkB2 = 30(0.152) = 0.675 kg # m2.Thus, the kinetic energy of the system is laws

or

Also, since gear A is in mesh with gear B, vA = a

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1 1 1 (0.15625)(13.33vD)2 + (0.675)(10vD)2 + (50)vD2 2 2 2

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=

1 1 1 I v 2 + IBvB2 + mDvD2 2 A A 2 2 copyright

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0 + 0 = 72.639vD2 - 981

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vD = 3.67 m>s

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18–45. The disk A is pinned at O and weighs 15 lb. A 1-ft rod weighing 2 lb and a 1-ft-diameter sphere weighing 10 lb are welded to the disk, as shown. If the spring is orginally stretched 1 ft and the sphere is released from the position shown, determine the angular velocity of the disk when it has rotated 90˚.

k = 4 lb/ft

1 ft

1 ft

SOLUTION T1 + V1 = T2 + V2 [0 + 0 + 0] +

1 1 1 15 1 1 2 1 2 (4)(1)2 = [ ( )(2)2]v2 + [ ( )(1)2]v2 + ( )(v )2R 2 2 2 32.2 2 12 32.2 2 32.2 G +

1 2 10 1 10 1 p [ ( )(0.5)2]v2 + ( )(v )2s - 2(2.5) - 10(3.5) + (4)(1 + 2( ))2 2 5 32.2 2 32.2 G 2 2

Since (vG)S = 3.5v

laws

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(vG)R = 2.5v

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v = 1.73 rad s

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Substituting and solving, yields

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O

2 ft

A

18–46. The disk A is pinned at O and weighs 15 lb. A 1-ft rod weighing 2 lb and a 1-ft-diameter sphere weighing 10 lb are welded to the disk, as shown. If the spring is originally stretched 1 ft and the sphere is released from the position shown, determine the angular velocity of the disk when it has rotated 45°.

k ⫽ 4 lb/ft

1 ft

1 ft

SOLUTION Potential Energy: From the geometry shown in Fig. a, we obtain (yG1)2 = 2.5 sin 45° ft = 1.7678 ft and (yG2)2 = 3.5 sin 45° = 2.4749 ft. With reference to the datum set in Fig. a, the initial and final gravitational potential energy of the system is (Vg)1 = W1(yG1)1 + W2(yG2)1 = 2(0) + 10(0) = 0 (Vg)2 = - W1(yG1)2 - W2(yG2)2 = - 2(1.7678) - 10(2.4749) = - 28.284 ft # lb p (2) = 2.5708 ft. 4

The initial and final stretch of the spring is s1 = 1 ft and s2 = 1 +

laws

or

Thus the initial and final elastic potential energy of the system are 1 1 ks 2 = (4)(12) = 2 ft # lb 2 1 2

(Ve)2 =

1 1 ks 2 = (4)(2.5708)2 = 13.218 ft # lb 2 2 2

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Kinetic Energy: The mass moment of inertia of the disk assembly about the fixed

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2 10 10 1 2 2 a b (0.52) + a b (3.52) + a b(12) + a b (2.52) 5 32.2 32.2 12 32.2 32.2 integrity

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= 5.1605 slug # ft2

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IO =

T =

1 1 I v2 = (5.1605)v2 = 2.5802v2 2 O 2

Since the system is at rest initally, T1 = 0 Conservation of Energy: T1 + V1 = T2 + V2 0 + (0 + 2) = 2.5802v2 + (- 28.284 + 13.218) v = 2.572 rad>s = 2.57 rad>s

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O

2 ft

Ans.

A

18–47. At the instant the spring becomes undeformed, the center of the 40-kg disk has a speed of 4 m>s. From this point determine the distance d the disk moves down the plane before momentarily stopping. The disk rolls without slipping.

k = 200 N/m

0.3 m

SOLUTION

30°

Datum at lowest point. T1 + V1 = T2 + V2 4 2 1 1 1 1 c (40)(0.3)2 d a b + (40)(4)2 + 40(9.81)d sin 30° = 0 + (200)d2 2 2 0.3 2 2 100d2 - 196.2d - 480 = 0 Solving for the positive root d = 3.38 m

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*18–48. A chain that has a negligible mass is draped over the sprocket which has a mass of 2 kg and a radius of gyration of kO = 50 mm. If the 4-kg block A is released from rest from the position s = 1 m, determine the angular velocity of the sprocket at the instant s = 2 m.

100 mm O

s⫽1m

SOLUTION T1 + V1 = T2 + V2 0 + 0 + 0 =

1 1 (4)(0.1 v)2 + C 2(0.05)2 D v2 - 4(9.81)(1) 2 2

A

v = 41.8 rad>s

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18–49. Solve Prob. 18–48 if the chain has a mass of 0.8 kg>m. For the calculation neglect the portion of the chain that wraps over the sprocket.

100 mm O

s

1m

SOLUTION T1 + V1 = T2 + V2 1 1 (4)(0.1 v)2 + C 2(0.05)2 D v2 2 2

0 - 4(9.81)(1) - 2 C 0.8(1)(9.81)(0.5) D = +

A

1 (0.8)(2)(0.1 v)2 - 4(9.81)(2) - 0.8(2)(9.81)(1) 2 v = 39.3 rad>s

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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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or transmission in any form or by any means, electronic, mechanical,

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18–50. The compound disk pulley consists of a hub and attached outer rim. If it has a mass of 3 kg and a radius of gyration kG = 45 mm, determine the speed of block A after A descends 0.2 m from rest. Blocks A and B each have a mass of 2 kg. Neglect the mass of the cords.

100 mm 30 mm

SOLUTION B

T1 + V1 = T2 + V2 [0 + 0 + 0] + [0 + 0] =

1 1 1 [3(0.045)2]v2 + (2)(0.03v)2 + (2)(0.1v)2 - 2(9.81)sA + 2(9.81)sB 2 2 2

sA sB = 0.03 0.1

u =

sB = 0.3 sA Set sA = 0.2 m, sB = 0.06 m

or

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laws

v = 14.04 rad>s vA = 0.1(14.04) = 1.40 m>s

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A

18–51. A spring having a stiffness of k = 300 N>m is attached to the end of the 15-kg rod, and it is unstretched when u = 0°. If the rod is released from rest when u = 0°, determine its angular velocity at the instant u = 30°. The motion is in the vertical plane.

B

u k

300 N/m

SOLUTION

0.6 m

Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the rod at positions (1) and (2) is

A

A Vg B 1 = W(yG)1 = 15(9.81)(0) = 0 A Vg B 2 = - W(yG)2 = - 15(9.81)(0.3 sin 30°) = -22.0725 J Since the spring is initially unstretched, (Ve)1 = 0. When u = 30°, the stretch of the spring is sP = 0.6 sin 30° = 0.3 m. Thus, the final elastic potential energy of the spring is

or

1 1 ks 2 = (300) A 0.32 B = 13.5 J 2 P 2 laws

A Ve B 2 =

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V1 = (Vg)1 + (Ve)1 = 0 + 0 = 0

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V2 = (Vg)2 + (Ve)2 = - 22.0725 + 13.5 = -8.5725 J

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1 1 (15) C v2 A 0.3 B D 2 + A 0.45 B v2 2 2 2 sale

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1 1 m(vG)2 2 + IGv2 2 2 2

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Kinetic Energy: Since the rod is initially at rest, T1 = 0. From the geometry shown in Fig. b, rG>IC = 0.3 m. Thus, (VG)2 = v2rG>IC = v2 (0.3). The mass moment of inertia 1 1 of the rod about its mass center is IG = ml2 = (15) A 0.62 B = 0.45 kg # m2. Thus, 12 12 the final kinetic energy of the rod is

= 0.9v2 2 Conservation of Energy: T1 + V1 = T2 + V2 0 + 0 = 0.9v2 2 - 8.5725 v2 = 3.09 rad>s

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Ans.

*18–52. The two bars are released from rest at the position u. Determine their angular velocities at the instant they become horizontal. Neglect the mass of the roller at C. Each bar has a mass m and length L.

B L A

C

SOLUTION Potential Energy: Datum is set at point A. When links AB and BC is at their L initial position, their center of gravity is located 2 sin u above the datum. Their L gravitational potential energy at this position is mg a sin u b . Thus, the initial and 2 final potential energies are V1 = 2 a

mgL sin u b = mgL sin u 2

V2 = 0

Kinetic Energy: When links AB and BC are in the horizontal position, then vB = vAB L laws

or

which is directed vertically downward since link AB is rotating about fixed point A. Link

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BC is subjected to general plane motion and its instantaneous center of zero velocity is

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1 1 (I ) v2AB + (IBC)C v2BC 2 AB A 2

1 1 1 1 a mL2 bv2 + a mL2 b v2 2 3 2 3

=

1 mL2 v2 3

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Conservation of Energy: Applying Eq. 18–18, we have T1 + V1 = T2 + V2 0 + mgL sin u =

1 mL2 v2 + 0 3

vAB = vBC = v =

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3g sin u L

Ans.

not

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The mass moment inertia for link AB and BC about point A and C is 1 L 2 1 (IAB)A = (IBC)C = mL2 + m a b = mL2. Since links AB and CD are at rest 2 2 3 initially, the initial kinetic energy is T1 = 0. The final kinetic energy is given by

permitted.

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located at point C. Thus, vB = vBCrB>IC or vABL = vBCL, hence vAB = vBC = v.

L

18–53. The two bars are released from rest at the position u = 90°. Determine their angular velocities at the instant they become horizontal. Neglect the mass of the roller at C. Each bar has a mass m and length L.

B L A

SOLUTION Potential Energy: With reference to the datum established in Fig. a, the gravitational potential energy of the system at position 1 and 2 are (V1)g = WAB(yGAB)1 + WBC(yGBC) 1 = mg a

L L b + mga b = mg L 2 2

(V2)g = WAB(yGAB)2 + WBC(yGBC)2 = 0 Kinetic Energy: Since the system is at rest initially, T1 = 0. Referring to the kinematic diagram of the system at position 2 shown in Fig. b, vB = vAB rAB = vBC rGBC>IC;

vAB(L) = vBC(L) vAB = vBC laws

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vGBC = vBC rGBC>IC = vBC a

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1 1 1 1 1 L 2 a mL2 b vBC 2 + a mL2 b vBC2 + m cvBC a b d 2 3 2 12 2 2 provided

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1 mL2 vBC2 3 Conservation of Energy:

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=

T1 + V1 = T2 + V2 0 + mg L =

1 mL2 vBC 2 3

vAB = vBC =

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3g BL

Ans.

permitted.

Dissemination

The mass moment of inertia of bar AB about the fixed axis passing through A is 1 IA = mL2 and the mass moment of inertia of bar BC about its mass center is 3 1 IGBC = mL2. Thus, the kinetic energy of the system is 12

L u

u C

18–54. If the 250-lb block is released from rest when the spring is unstretched, determine the velocity of the block after it has descended 5 ft. The drum has a weight of 50 lb and a radius of gyration of kO = 0.5 ft about its center of mass O.

0.375 ft k ⫽ 75 lb/ft 0.75 ft O

SOLUTION Potential Energy: With reference to the datum shown in Fig. a, the gravitational potential energy of the system when the block is at position 1 and 2 is (Vg)1 = W(yG)1 = 250(0) = 0 (Vg)2 = - W(yG)2 = - 250(5) = - 1250 ft # lb When the block descends sb = 5 ft, the drum rotates through an angle of sb 5 u = = = 6.667 rad. Thus, the stretch of the spring is x = s + s0 = rb 0.75 rspu + 0 = 0.375(6.667) = 2.5 ft. The elastic potential energy of the spring is 1 2 1 kx = (75)(2.52) = 234.375 ft # lb 2 2 laws

or

(Ve)2 =

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Since the spring is initially unstretched, (Ve)1 = 0. Thus,

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V2 = (Vg)2 + (Ve)2 = - 1250 + 234.375 = - 1015.625 ft # lb

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V1 = (Vg)1 + (Ve)1 = 0

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1 1 IOv2 + mbvb2 2 2

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1 1 250 (0.3882)(1.333vb)2 + a bv 2 2 2 32.2 b their

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Kinetic Energy: Since the drum rotates about a fixed axis passing through point O, vb vb v = = = 1.333vb. The mass moment of inertia of the drum about its mass rb 0.75 50 center is IO = mkO 2 = a 0.52 b = 0.3882 slug # ft2. 32.2

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will

= 4.2271vb2 Since the system is initially at rest, T1 = 0. T1 + V1 = T2 + V2

0 + 0 = 4.2271vb 2 - 1015.625 vb = 15.5 ft>s

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T

Ans.

18–55. The 6-kg rod ABC is connected to the 3-kg rod CD. If the system is released from rest when u = 0°, determine the angular velocity of rod ABC at the instant it becomes horizontal.

A 0.5 m

B 0.3 m

SOLUTION

C

Potential Energy: When rod ABC is in the horizontal position, Fig. a, 0.3 u = sin - 1 a b = 48.59°. With reference to the datum in Fig. a, the initial and final 0.4 gravitational potential energy of the system is

u 0.4 m D

V1 = (Vg)1 = W1(yG1)1 + W2(yG2)1 = 6(9.81)(0.8) + 3(9.81)(0.2) = 52.974 J V2 = (Vg)2 = W1(yG1)2 + W2(yG2)2

or

= 6(9.81)(0.4 cos 48.59°) + 3(9.81)(0.2 cos 48.59°) = 19.466 J

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1 1 m (v ) 2 + IG1(vABC)22 2 1 G1 2 2

this

2 1 1 (6)c (vABC)2(0.4) d + (0.32)(vABC)2 2 2 2 part

the

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= 0.64vABC2

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Conservation of Energy: T1 + V1 = T2 + V2 0 + 52.974 = 0.64vABC 2 + 19.466 (vABC)2 = 7.24 rad>s

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Ans.

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Kinetic Energy: Since the system is initially at rest, T1 = 0. Referring to Fig. b, (vG1)2 = (vABC)2 rG1>IC = (vABC)2(0.4). Since point C is at the IC(vC)2 = 0. Then, (vC)2 0 vCD = = = 0. The mass moment of inertia of rod ABC about its mass rC 0.4 1 center is IG1 = (6)(0.82) = 0.32 kg # m2. Thus, the final kinetic energy of the 12 system is

*18–56. If the chain is released from rest from the position shown, determine the angular velocity of the pulley after the end B has risen 2 ft. The pulley has a weight of 50 lb and a radius of gyration of 0.375 ft about its axis. The chain weighs 6 lb/ft.

0.5 ft

4 ft

SOLUTION

6 ft

Potential Energy: (yG1)1 = 2 ft, (yG 2)1 = 3 ft, (yG1)2 = 1 ft, and (yG2)2 = 4 ft. With reference to the datum in Fig. a, the gravitational potential energy of the chain at position 1 and 2 is

B

V1 = (Vg)1 = W1(yG1)1 - W2(yG2)1

A

= - 6(4)(2) - 6(6)(3) = - 156 ft # lb V2 = (Vg)2 = - W1(yG1)2 + W2(yG2)2 = - 6(2)(1) - 6(8)(4) = - 204 ft # lb

in

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laws

or

Kinetic Energy: Since the system is initially at rest, T1 = 0. The pulley rotates about a fixed axis, thus, (VG1)2 = (VG2)2 = v2 r = v2(0.5). The mass moment of inertia of 50 the pulley about its axis is IO = mkO 2 = (0.3752) = 0.2184 slug # ft2. Thus, the 32.2 final kinetic energy of the system is

permitted.

Dissemination

Wide

copyright

1 1 1 IOv2 2 + m1(VG1)2 2 + m2 (VG2)2 2 2 2 2 instructors

States

World

T =

the

not

1 1 6(2) 1 6(8) 1 6(0.5)(p) (0.2184)v2 2 + c d [v2(0.5)]2 + c d[v2(0.5)]2 + c d[v2(0.5)]2 2 2 32.2 2 32.2 2 32.2 by

and

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=

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= 0.3787v2 2

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Conservation of Energy:

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work

this

T1 + V1 = T2 + V2

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the

is

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of

and

v2 = 11.3 rad >s

provided

0 + (-156) = 0.3787v22 = ( -204)

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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18–57. If the gear is released from rest, determine its angular velocity after its center of gravity O has descended a distance of 4 ft. The gear has a weight of 100 lb and a radius of gyration about its center of gravity of k = 0.75 ft. 1 ft O

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the gear at position 1 and 2 is V1 = (Vg)1 = W(y0)1 = 100(0) = 0 V2 = (Vg)2 = - W1(y0)2 = - 100(4) = - 400 ft # lb Kinetic Energy: Referring to Fig. b, we obtain vO = vrO / IC = v(1).The mass moment 100 (0.752) = 1.7469 kg # m2. 32.2

of inertia of the gear about its mass center is IO = mkO 2 = Thus,

or

1 1 mvO2 + IOv2 2 2 laws

T =

in

teaching

Web)

1 100 1 a b [v (1)]2 + (1.7469)v2 2 32.2 2

Wide

copyright

=

instructors

States

World

permitted.

Dissemination

= 2.4262v2

use

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on

learning.

is

of

the

not

Since the gear is initially at rest, T1 = 0.

student

the

by

and

Conservation of Energy:

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T1 + V1 = T2 + V2

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0 + 0 = 2.4262v2 - 400

Ans.

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this

v = 12.8 rad>s

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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18–58. When the slender 10-kg bar AB is horizontal it is at rest and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the bar is momentarily stopped when it has rotated clockwise 90°.

A

B

1.5 m

SOLUTION T1 + V1 = T2 + V2 0 + 0 = 0 +

1 1.5 (k)(3.3541 – 1.5 )2 – 98.1 a b 2 2 k = 42.8 N>m

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laws

or

Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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photocopying, recording, or likewise. For information regarding permission(s), write to:

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k

1.5 m

C

18–59. When the slender 10-kg bar AB is horizontal it is at rest and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the bar is momentarily stopped when it has rotated clockwise 45°.

A

B

1.5 m

SOLUTION Potential Energy: From the geometry shown in Fig. a, we obtain (yG)2 = 0.75 sin 45° = 0.5303 m and CB¿ = 232 + 1.52 - 2(3)(1.5) cos 45° = 2.2104. With reference to the datum established in Fig. a, the initial and final gravitational potential energy of the system is (Vg)1 = WAB(yG)1 = 0 (Vg)2 = - WAB(yG)2 = - 10(9.81)(0.5303) = - 52.025 J

or

Initially, the spring is unstretched. Thus, (Ve)1 = 0. At the final position, the spring stretches S = CB¿ - CB = 2.2104 - 1.5 = 0.7104 m. Then (Ve)1 = 0 and (Ve)2 = 1 2 1 ks = k (0.7104 2) = 0.2524k. 2 2

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laws

V1 = (Ve)1 + (Vg)1 = 0

learning.

is

of

the

not

instructors

States

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Kinetic Energy: Since the bar is at rest initially and stops momentarily at the final position, T1 = T2 = 0.

the

by

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Conservation of Energy:

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T1 + V1 = T2 + V2

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the

0 + 0 = 0 + 0.2524k - 52.025

Ans.

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assessing

k = 206.15 N>m = 206 N>m

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

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permitted.

Dissemination

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copyright

in

V2 = (Ve)2 + (Vg)2 = 0.2524k - 52.025

k

1.5 m

C

*18–60. If the 40-kg gear B is released from rest at u = 0°, determine the angular velocity of the 20-kg gear A at the instant u = 90°. The radii of gyration of gears A and B about their respective centers of mass are kA = 125 mm and kB = 175 mm. The outer gear ring P is fixed.

P

0.15 m A

u

B

SOLUTION

0.2 m

Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of gear B at positions (1) and (2) is V1 = (Vg)1 = WB(yGB)1 = 40(9.81)(0) = 0 V2 = (Vg)2 = - WB(yGB)2 = - 40(9.81)(0.35) = - 137.34 J Kinetic Energy: Referring to Fig. b, vP = vArA = vA(0.15). Then, vB =

vP =

rP>IC

teaching

Web)

laws

or

vA(0.15) = 0.375vA. Subsequently, vGB = vBrGB>IC = (0.375vA)(0.2) = 0.075vA. 0.4 The mass moments of inertia of gears A and B about their mass centers are and IA = mAkA2 = 20(0.1252) = 0.3125 kg # m2 IB = mBkB2 = 40(0.1752) = 2 # 1.225 kg m . Thus, the kinetic energy of the system is

Wide

copyright

in

T = TA + TB 1 1 1 I v 2 + c mBvGB2 + IBvB 2d 2 A A 2 2

=

1 1 1 (0.3125)vA2 + c (40)(0.075vA)2 + (1.225)(0.375vA)2d 2 2 2 for

work

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=

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the

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= 0.3549vA2

work

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assessing

is

work

Since the system is initially at rest, T1 = 0.

part

the

is

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provided

integrity

of

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Conservation of Energy:

of

and

any

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T1 + V1 = T2 + V2

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destroy

0 + 0 = 0.3549vA2 - 137.34 vA = 19.7 rad>s

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Ans.

18–61. A uniform ladder having a weight of 30 lb is released from rest when it is in the vertical position. If it is allowed to fall freely, determine the angle u at which the bottom end A starts to slide to the right of A. For the calculation, assume the ladder to be a slender rod and neglect friction at A. 10 ft

SOLUTION

u

Potential Energy: Datum is set at point A. When the ladder is at its initial and final position, its center of gravity is located 5 ft and (5 cos u ) ft above the datum. Its initial and final gravitational potential energy are 30(5) = 150 ft # lb and 30(5 cos u ) = 150 cos u ft # lb, respectively. Thus, the initial and final potential energy are V1 = 150 ft # lb

A

V2 = 150 cos u ft # lb

or

Kinetic Energy: The mass moment inertia of the ladder about point A is 1 30 30 IA = a b (102) + a b (52) = 31.06 slug # ft2. Since the ladder is initially at 12 32.2 32.2 rest, the initial kinetic energy is T1 = 0. The final kinetic energy is given by

teaching

Web)

laws

1 1 I v2 = (31.06)v2 = 15.53v2 2 A 2 in

T2 =

Dissemination

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copyright

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

is

of

the

not

instructors

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permitted.

T1 + V1 = T2 + V2

by

and

use

United

on

learning.

0 + 150 = 15.53v2 + 150 cos u

the

(including

for

work

student

the

v2 = 9.66(1 - cos u)

part

the

is

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is

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solely

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protected

Equation of Motion: The mass moment inertia of the ladder about its mass center is 1 30 IG = a b (102) = 7.764 slug # ft2. Applying Eq. 17–16, we have 12 32.2

any

courses

-30 sin u(5) = - 7.764a - a

30 b [a(5)](5) 32.2

destroy

of

and

+ ©MA = ©(Mk)A;

sale

will

their

a = 4.83 sin u c

+ ©Fx = m(aG)x;

Ax = -

30 [9.66(1 - cos u)(5)] sin u 32.2 +

Ax = -

30 [4.83 sin u(5)] cos u 32.2

30 (48.3 sin u - 48.3 sin u cos u - 24.15 sin u cos u) 32.2

= 45.0 sin u (1 - 1.5 cos u) = 0 If the ladder begins to slide, then Ax = 0 . Thus, for u>0, 45.0 sin u (1 - 1.5 cos u) = 0 u = 48.2°

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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Ans.

18–62. The 50-lb wheel has a radius of gyration about its center of gravity G of kG = 0.7 ft. If it rolls without slipping, determine its angular velocity when it has rotated clockwise 90° from the position shown. The spring AB has a stiffness k = 1.20 lb/ft and an unstretched length of 0.5 ft. The wheel is released from rest.

3 ft 0.5 ft

B

0.5 ft G 1 ft

SOLUTION T1 + V1 = T2 + V2 0 +

1 1 50 1 50 (1.20)[2(3)2 + (0.5)2 - 0.5]2 = [ (0.7)2]v2 + ( )(1v)2 2 2 32.2 2 32.2 1 (1.20)(0.9292 - 0.5)2 2

+

v = 1.80 rad s

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the

(including

for

work

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and

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on

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is

of

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

k = 1.20 lb/ft

A

18–63. The uniform window shade AB has a total weight of 0.4 lb. When it is released, it winds up around the spring-loaded core O. Motion is caused by a spring within the core, which is coiled so that it exerts a torque M = 0.3(10 - 3)u lb # ft, where u is in radians, on the core. If the shade is released from rest, determine the angular velocity of the core at the instant the shade is completely rolled up, i.e., after 12 revolutions. When this occurs, the spring becomes uncoiled and the radius of gyration of the shade about the axle at O is kO = 0.9 in. Note: The elastic potential energy of the torsional spring is Ve = 12ku2, where M = ku and k = 0.3(10 - 3) lb # ft>rad.

M B

O

3 ft

A

SOLUTION T1 + V1 = T2 + V2 0 - (0.4)(1.5) +

1 0.4 0.9 2 2 1 (0.3)(10 - 3)(24p)2 = a ba b v 2 2 32.2 12

v = 85.1 rad>s

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destroy

of

and

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of

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is

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solely

of

protected

the

(including

for

work

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by

and

use

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on

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of

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permitted.

Dissemination

Wide

copyright

in

teaching

Web)

laws

or

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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O

*18–64. The motion of the uniform 80-lb garage door is guided at its ends by the track. Determine the required initial stretch in the spring when the door is open, u = 0°, so that when it falls freely it comes to rest when it just reaches the fully closed position, u = 90°. Assume the door can be treated as a thin plate, and there is a spring and pulley system on each of the two sides of the door.

k C u 8 ft 8 ft

A

SOLUTION sA + 2 s s = l ¢sA = - 2¢ss 8 ft = - 2¢ss ¢ss = -4 ft T1 + V1 = T2 + V2 1 1 0 + 2 c (9)s2 d = 0 - 80(4) + 2 c (9)(4 + s)2 d 2 2

laws

or

9s2 = - 320 + 9(16 + 8s + s2) s = 2.44 ft

sale

will

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destroy

of

and

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courses

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the

is

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provided

integrity

of

and

work

this

assessing

is

work

solely

of

protected

the

(including

for

work

student

the

by

and

use

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on

learning.

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of

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World

permitted.

Dissemination

Wide

copyright

in

teaching

Web)

Ans.

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photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

B

9 lb/ft

18–65. The motion of the uniform 80-lb garage door is guided at its ends by the track. If it is released from rest at u = 0°, determine the door’s angular velocity at the instant u = 30°. The spring is originally stretched 1 ft when the door is held open, u = 0°. Assume the door can be treated as a thin plate, and there is a spring and pulley system on each of the two sides of the door.

k ⫽ 9 lb/ft C u 8 ft 8 ft

A

SOLUTION vG = 4v sA + 2ss = l ¢sA = - 2¢ss 4 ft = - 2¢ss ¢ss = - 2 ft T1 + V1 = T2 + V2

laws

or

1 80 1 1 80 1 b (4v)2 + c a b(8)2 dv2 - 80(4 sin 30°) 0 + 2c (9)(1)2 d = a 2 2 32.2 2 12 32.2

Ans.

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of

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the

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for

work

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the

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on

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is

of

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not

instructors

States

World

Dissemination

v = 1.82 rad>s

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1 + 2 c (9)(2 + 1)2 d 2

B

18–66. The end A of the garage door AB travels along the horizontal track, and the end of member BC is attached to a spring at C. If the spring is originally unstretched, determine the stiffness k so that when the door falls downward from rest in the position shown, it will have zero angular velocity the moment it closes, i.e., when it and BC become vertical. Neglect the mass of member BC and assume the door is a thin plate having a weight of 200 lb and a width and height of 12 ft. There is a similar connection and spring on the other side of the door.

12 ft

C

D

CD2 - 11.591CD + 32 = 0 Selecting the smaller root: CD = 4.5352 ft T1 + V1 = T2 + V2

teaching

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laws

or

1 0 + 0 = 0 + 2 c (k)(8 - 4.5352)2 d - 200(6) 2 k = 100 lb/ft

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is

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the

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of

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permitted.

Dissemination

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in

Ans.

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6 ft

2 ft

(2)2 = (6)2 + (CD)2 - 2(6)(CD) cos 15°

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

15

7 ft

SOLUTION

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

A

B

1 ft

18–67. Determine the stiffness k of the torsional spring at A, so that if the bars are released from rest when u = 0°, bar AB has an angular velocity of 0.5 rad/s at the closed position, u = 90°. The spring is uncoiled when u = 0°. The bars have a mass per unit length of 10 kg>m.

B 4m

3m

u C A k

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the system at its open and closed positions is

A Vg B 1 = WAB (yG1)1 + WBC (yG2)1 = 10(3)(9.81)(1.5) + 10(4)(9.81)(1.5) = 1030.5 J

A Vg B 2 = WAB (yG1)2 + WBC (yG2)2 = 10(3)(9.81)(0) + 10(4)(9.81)(0) = 0

teaching

Web)

laws

or

Since the spring is initially uncoiled, (Ve)1 = 0. When the panels are in the closed p position, the coiled angle of the spring is u = rad. Thus, 2 in

p 2 1 2 1 p2 ku = ka b = k 2 2 2 8

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(Ve)2 =

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V1 = A Vg)1 + (Ve)1 = 1030.5 + 0 = 1030.5 J

the

(including

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p2 p2 k = k 8 8 is

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V2 = A Vg)2 + (Ve)2 = 0 +

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Kinetic Energy: Since the system is initially at rest, T1 = 0. Referring to Fig. b, (vB)2 1.5 (vB)2 = (vAB)2 rB = 0.5(3) = 1.5 m>s. Then, (vBC)2 = = 0.375 rad>s. = rB>IC 4 Subsequently, (vG)2 = (vBC)2 rG2>IC = 0.375(2) = 0.75 m>s. The mass moments of inertia of AB about point A and BC about its mass center are (IAB)A =

1 2 1 ml = [10(3)] A 32 B = 90 kg # m2 3 3

and (IBC)G2 =

1 1 ml2 = [10(4)] A 42 B = 53.33 kg # m2 12 12

Thus, T2 = =

1 1 1 (I ) (v ) 2 + c mBC(vG2)2 + (IBC)G2 (vBC)2 2 d 2 AB A AB 2 2 2 1 1 1 (90) A 0.52 B + c [10(4)] A 0.752 B + (53.33) A 0.375 2 B d 2 2 2

= 26.25 J

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18–67. continued Conservation of Energy: T1 + V1 = T2 + V2 0 + 1030.5 = 26.25 +

p2 k 8

k = 814 N # m>rad

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the

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Dissemination

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or

Ans.

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*18–68. The torsional spring at A has a stiffness of k = 900 N # m>rad and is uncoiled when u = 0°. Determine the angular velocity of the bars, AB and BC, when u = 0°, if they are released from rest at the closed position, u = 90°. The bars have a mass per unit length of 10 kg>m.

B 4m

3m

u C A k

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the system at its open and closed positions is

A Vg B 1 = WAB (yG1)1 + WBC (yG2)1 = 10(3)(9.81)(0) + 10(4)(9.81)(0) = 0

A Vg B 2 = WAB (yG1)2 + WBC (yG2)2 = 10(3)(9.81)(1.5) + 10(4)(9.81)(1.5) = 1030.05 J p rad. 2

When the panel is in the closed position, the coiled angle of the spring is u =

or

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p 2 1 2 1 ku = (900)a b = 112.5p2 J 2 2 2

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The spring is uncoiled when the panel is in the open position (u = 0°). Thus,

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(Ve)2 = 0 the

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V1 = A Vg)1 + (Ve)1 = 0 + 112.5p2 = 112.5p2 J

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V2 = A Vg)2 + (Ve)2 = 1030.05 + 0 = 1030.05 J

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Kinetic Energy: Since the panel is at rest in the closed position, T1 = 0. Referring to Fig. b, the IC for BC is located at infinity. Thus, Ans.

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(vBC)2 = 0

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(vG)2 = (vB)2 = (vAB)2 rB = (vAB)2 (3) The mass moments of inertia of AB about point A and BC about its mass center are (IAB)A =

1 2 1 ml = [10(3)] A 32 B = 90 kg # m2 3 3

and (IBC)G2 =

1 1 ml2 = [10(4)] A 42 B = 53.33 kg # m2 12 12

Thus, T2 = =

1 1 (IAB)A(vAB)2 2 + mBC(vG2)2 2 2 1 1 (90)(vAB)2 2 + [10(4)] C (vAB)2 (3) D 2 2 2

= 225(vAB)2 2

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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

18–68. continued Conservation of Energy: T1 + V1 = T2 + V2 0 + 112.5p2 = 225(vAB)2 2 + 1030.05 (vAB)2 = 0.597 rad>s

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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical,

photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.