CHAPTER 17
PROBLEM 17.1
It is known that 1500 revolutions are required for the 3000 kg flywheel to coast to rest from an angular velocity
of 300 rpm. Knowing that the radius of gyration of the flywheel is 1 m determine the average magnitude of the
couple due to kinetic friction in the bearings.
SOLUTION
Angular velocity:
Ê 2p ˆ
w 0 = 300 rpm Á ˜
Ë 60 ¯
= 10 p rad
w2 = 0
Moment of inertia:
I = mk 2 = (3000 kg) (1 m)2
= 3000 kg ◊ m2
Kinetic energy:
Work:
Principle of work and energy:
1
I w 02
2
1
= (3000)(10 p )2
2
= 1.48044 ¥ 106 N ◊ m
T2 = 0
T1 =
U1Æ2 = - M q
= - M (1500 rev)(2p rad/rev )
= -9424.8 M
T1 + U1Æ2 = T2
1.48044 ¥ 106 - 9424.8 M = 0
M = 157.08 N ◊ m
Average friction couple:
M = 157.1 N ◊ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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3
PROBLEM 17.2
The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off.
The 50-kg rotor, which has a centroidal radius of gyration of 180 mm, then coasts to rest. Knowing that the
kinetic friction of the rotor produces a couple of magnitude 3.5 N ◊ m, determine the number of revolutions that
the rotor executes before coming to rest.
SOLUTION
Ê 2p ˆ
w 0 = 3600 rpm Á ˜
Ë 60 ¯
Angular velocity:
= 120p rad/s
w2 = 0
I = mk 2
Moment of inertia:
= (50 kg)(0.180 m)2
= 1.620 kg ◊ m 2
1 2
I w0
2
1
= (1.620)(120p )2
2
= 115.12 kJ ,
T2 = 0
T1 =
Kinetic energy:
U1Æ2 = - M q
Work:
= -(3.5 N ◊ m) q
Principle of work and energy:
T1 + U1Æ2 = T2 : 115.12 kJ - (3.5 N ◊ m) q = 0
q = 32.891 ¥ 103 rad
Rotation angle:
q = 5230 rev PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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4
PROBLEM 17.3
Two disks of the same material are attached to a shaft as shown. Disk A is of radius r
and has a thickness b, while disk B is of radius nr and thickness 3b. A couple M of
constant magnitude is applied when the system is at rest and is removed after the
system has executed 2 revolutions. Determine the value of n which results in the
largest final speed for a point on the rim of disk B.
SOLUTION
m = r(p r 2t )
1
I = mr 2
2
1
= prtr 4
2
For any disk:
Moment of inertia.
1
I A = p r b r4
2
1
I B = p r(3b)( nr )4
2
È1
˘
= 3n4 Í prbr 4 ˙
Î2
˚
Disk A:
Disk B:
I total
= 3n4 I A
= IA + IB
= (1 + 3n4 ) I A
T1 = 0 U1Æ2 = M q = M ( 4p rad )
Work-energy.
T2 =
1
I total w 22
2
1
T1 + U1Æ2 = T2 : 0 + M ( 4p ) = (1 + 3n4 ) I Aw 22
2
8p M
w 22 =
(1 + 3n4 ) I A
For Point D on rim of disk B
vD = ( nr )w 2
or
vD2 = n2 r 2w 22 =
n2
8p Mr 2
◊
IA
1 + 3n4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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5
PROBLEM 17.3 (Continued)
Value of n for maximum final speed.
For maximum
vD :
d Ê n2 ˆ
=0
dn ÁË 1 + 3n4 ˜¯
1
[n2 (12n3 ) - (1 + 3n4 )(2n)] = 0
(1 + 3n4 )2
12n5 - 2n - 6n5 = 0
2n(3n4 - 1) = 0
Ê 1ˆ
n = 0 and n = Á ˜
Ë 3¯
0.25
= 0.7598
n = 0.760 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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6
PROBLEM 17.4
Two disks of the same material are attached to a shaft as shown. Disk A has a mass
of 15 kg and a radius r = 125 mm. Disk B is three times as thick as disk A. Knowing
that a couple M of magnitude 20 N ◊ m is to be applied to disk A when the system is
at rest, determine the radius nr of disk B if the angular velocity of the system is to be
600 rpm after 4 revolutions.
SOLUTION
For any disk:
m = r(p r 2t )
1
I = mr 2
2
1
= prtr 4
2
Moment of inertia.
Disk A:
Disk B:
1
I A = prbr 4
2
1
I B = pr(3b)( nr )4
2
È1
˘
= 3n4 Í prbr 4 ˙
Î2
˚
= 3n4 I A
I total = I A + I B = (1 + 3n4 ) I A
Angular velocity:
Rotation:
Kinetic energy:
Work:
(1)
w1 = 0
w 2 = 600 rpm
= 20p rad/s
q = 4 rev = 8 p rad
T1 = 0
1
T2 = I total w 22
2
U1Æ2 = M q
= (20 N ◊ m)(8p rad )
= 502.65 J
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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7
PROBLEM 17.4 (Continued)
Principle of work and energy:
T1 + U1Æ2 = T2
1
I total (20p )2
2
= 0.25465 kg ◊ m2
0 + 502.65 =
I total
But,
From (1)
1
mA rA2
2
1
= (15 kg)(0.125 m)2
2
= 0.117188 kg ◊ m2
IA =
0.25465 = (1 + 3n4 )(0.117188)
n4 = 0.390998
Radius of disk B:
n = 0.79076
rB = nrA = (0.79076)(125 mm)
rB = 98.8 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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8
PROBLEM 17.5
The flywheel of a punching machine has a mass of 300 kg and a radius of gyration of 600 mm. Each punching
operation requires 2500 J of work. (a) Knowing that the speed of the flywheel is 300 rpm just before a punching,
determine the speed immediately after the punching. (b) If a constant 25-N ◊ m couple is applied to the shaft of
the flywheel, determine the number of revolutions executed before the speed is again 300 rpm.
SOLUTION
I = mk 2
Moment of inertia.
= (300 kg)(0.6 m)2
= 108 kg ◊ m 2
w1 = 300 rpm
= 10p rad/s
Kinetic energy. Position 1.
1 2
I w1
2
1
= (108)(10p )2
2
= 53.296 ¥ 103 J
T1 =
T2 =
Position 2.
1 2
I w 2 = 54w 22
2
U1Æ2 = -2500 J
Work.
Principle of work and energy for punching.
T1 + U1Æ2 = T2 : 53.296 ¥ 103 - 2500 = 54w 22
w 22 = 940.66
(a)
w 2 = 30.67 rad/s
w 2 = 293 rpm Principle of work and energy for speed recovery.
T2 + U 2Æ1 = T1
U 2Æ1 = 2500 J
M = 25 N ◊ m
U 2Æ1 = M q 2500 = 25q q = 100 rad
q = 15.92 rev (b)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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9
PROBLEM 17.6
The flywheel of a small punching machine rotates at 360 rpm. Each punching operation requires 2250 N ◊ m
of work and it is desired that the speed of the flywheel after each punching be not less than 95 percent of the
original speed. (a) Determine the required moment of inertia of the flywheel. (b) If a constant 27 N ◊ m couple
is applied to the shaft of the flywheel, determine the number of revolutions that must occur between two
successive punchings, knowing that the initial velocity is to be 360 rpm at the start of each punching.
SOLUTION
Angular speeds.
Work.
w1 = 360 rpm = 12p rad/s
w 2 = 0.95 w1
w 2 = 11.4p rad/s
U1Æ2 = 2250 N ◊ m
Principle of work and energy for punching
T1 + U1Æ2 = T2
1 2
1
I w1 + U1Æ2 = I w 22
2
2
(a)
Solving for I
I=
=
-2U1Æ2
w 22 - w12
-(2)( -2250)
= 32.475 kg ◊ m2
2
2
(12p ) - (11.4p )
I = 32.5 kg ◊ m2 Principle of work and energy for speed recovery.
T2 + U 2Æ3 = T3
But,
T3 = T1
U 2Æ3 = T3 - T2
= T1 - T2
= -U1Æ2
= 2250 N ◊ m
Work,
(b)
U 2Æ 3 = M q
U 2Æ 3
M
2250
=
27
= 83.33 rad
q=
q = 13.26 rev PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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10
PROBLEM 17.7
Disk A is of constant thickness and is at rest when it is placed in contact
with belt BC, which moves with a constant velocity v. Denoting by m k
the coefficient of kinetic friction between the disk and the belt, derive an
expression for the number of revolutions executed by the disk before it attains
a constant angular velocity.
SOLUTION
Work of external friction force on disk A.
Only force doing work is F. Since its moment about A is M = rF , we have
U1Æ2 = M q
= rFq
= r ( m k mg )q
Kinetic energy of disk A.
Angular velocity becomes constant when
v
r
T1 = 0
1
T2 = I w 22
2
w2 =
1Ê1
ˆ Ê vˆ
= Á mr 2 ˜ Á ˜
¯ Ë r¯
2Ë2
=
mv 2
4
T1 + U1Æ2 = T2 : 0 + r m k mg q =
mv 2
4
2
Principle of work and energy for disk A.
Angle change.
q=
v2
rad
4r mk g
q=
v2
rev 8p r m k g
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11
PROBLEM 17.8
Disk A, of weight 5 kg and radius r = 150 mm, is at rest when it is placed
in contact with belt BC, which moves to the right with a constant speed
v = 12 m/s. Knowing that m k = 0.20 between the disk and the belt, determine
the number of revolutions executed by the disk before it attains a constant
angular velocity.
SOLUTION
Work of external friction force on disk A.
Only force doing work is F. Since its moment about A is M = rF , we have
U1Æ2 = M q
= rFq
= r ( m k mg )q
Kinetic energy of disk A.
Angular velocity becomes constant when
v
r
T1 = 0
1
T2 = I w 22
2
w2 =
1Ê1
ˆ Ê vˆ
= Á mr 2 ˜ Á ˜
¯ Ë r¯
2Ë2
=
mv 2
4
T1 + U1- 2 = T2 : 0 + r m k mgq =
mv 2
4
2
Principle of work and energy for disk A.
q=
Angle change
v2
rad
4 r mk g
q=
v2
rev
8 p r mk g
r = 0.15 m
m k = 0.20
Data:
v = 12 m/s
q=
(12 m/s)2
8p (0.15 m)(0.20)(9.81 m/s2 )
q = 19.47 rev PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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12
PROBLEM 17.9
Each of the gears A and B has a mass of 2.4 kg and a radius of
gyration of 60 mm, while gear C has a mass of 12 kg and a radius
of gyration of 150 mm. A couple M of constant magnitude 10 N ◊ m
is applied to gear C. Determine (a) the number of revolutions of
gear C required for its angular velocity to increase from 100 to
450 rpm, (b) the corresponding tangential force acting on gear A.
SOLUTION
Moments of inertia.
Gears A and B:
I A = I B = mk 2 = (2.4)(0.06)2 = 8.64 ¥ 10 -3 kg ◊ m2
Gear C:
I C = (12)(0.15)2 = 270 ¥ 10 -3 kg ◊ m2
Kinematics.
rAw A = rBw B = rC w C
200
w C = 2.5wC
80
q A = q B = 2.5qC
wA = wB =
Kinetic energy.
Position 1.
1 2
Iw :
2
10
w C = 100 rpm = p rad/s
3
25
w A = w B = 250 rpm = p rad/s
3
T=
2
Gear A:
1
Ê 25p ˆ
= 2.9609 J
(T1 ) A = (8.64 ¥ 10 -3 ) Á
Ë 3 ˜¯
2
Gear B:
1
Ê 25p ˆ
= 2.9609 J
(T1 ) B = (8.64 ¥ 10 -3 ) Á
Ë 3 ˜¯
2
Gear C:
1
Ê 10p ˆ
= 14.8044 J
(T1 )C = (270 ¥ 10 -3 ) Á
Ë 3 ˜¯
2
System:
T1 = (T1 ) A + (T1 ) B + (T1 )C = 20.726 J
2
2
Position 2.
w C = 450 rpm = 15p rad/s
w A = w B = 37.5p rad/s
Gear A:
1
(T2 ) A = (8.64 ¥ 10 -3 )(37.5p )2 = 59.957 J
2
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13
PROBLEM 17.9 (Continued)
1
(T2 ) B = (8.64 ¥ 10 -3 )(37.5p )2 = 59.957 J
2
1
(T2 )C = (270 ¥ 10 -3 )(15p )2 = 299.789 J
2
Gear B:
Gear C:
T2 = (T2 ) A + (T2 ) B + (T2 )C = 419.7 J
System:
U1Æ2 = M qC = 10qC
Work of couple.
Principle of work and energy for system.
T1 + U1Æ2 = T2 : 20.726 + 10qC = 419.7
qC = 39.898 radians
(a)
qC = 6.35 rev Rotation of gear C.
Rotation of gear A.
q A = (2.5)(39.898)
= 99.744 radians
Principle of work and energy for gear A.
(T1 ) A + M Aq A = (T2 ) A : 2.9609 + M A (99.744) = 59.957
M A = 0.57142 N ◊ m
(b)
Tangential force on gear A.
Ft =
M A 0.57142
=
rA
0.08
Ft = 7.14 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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14
PROBLEM 17.10
Solve Problem 17.9, assuming that the 10-N ◊ m couple is applied
to gear B.
PROBLEM 17.9 Each of the gears A and B has a mass of 2.4 kg and
a radius of gyration of 60 mm, while gear C has a mass of 12 kg and
a radius of gyration of 150 mm. A couple M of constant magnitude
10 N ◊ m is applied to gear C. Determine (a) the number of revolutions
of gear C required for its angular velocity to increase from 100 to
450 rpm, (b) the corresponding tangential force acting on gear A.
SOLUTION
Moments of inertia.
Gears A and B:
I A = I B = mk 2 = (2.4)(0.06)2 = 8.64 ¥ 10 -3 kg ◊ m2
Gear C:
I C = (12)(0.15)2 = 270 ¥ 10 -3 kg ◊ m2
Kinematics.
rAw A = rBw B = rC w C
200
w C = 2.5wC
80
q A = q B = 2.5qC
wA = wB =
Kinetic energy.
Position 1.
T=
1 2
Iw :
2
10
p rad/s
3
25
w A = w B = 250 rpm = p rad/s
3
w C = 100 rpm =
2
Gear A:
1
Ê 25p ˆ
= 2.9609 J
(T1 ) A = (8.64 ¥ 10 -3 ) Á
Ë 3 ˜¯
2
Gear B:
1
Ê 25p ˆ
= 2.9609 J
(T1 ) B = (8.64 ¥ 10 -3 ) Á
Ë 3 ˜¯
2
Gear C:
1
Ê 10p ˆ
= 14.8044 J
(T1 )C = (270 ¥ 10 -3 ) Á
Ë 3 ˜¯
2
System:
T1 = (T1 ) A + (T1 ) B + (T1 )C = 20.726 J
2
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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15
PROBLEM 17.10 (Continued)
w C = 450 rpm = 15 p rad/s
Position 2.
w A = w B = 37.5 p rad/s
Gear A:
1
(T2 ) A = (8.64 ¥ 10 -3 )(37.5p )2 = 59.957 J
2
Gear B:
1
(T2 ) B = (8.64 ¥ 10 -3 )(37.5p )2 = 59.957 J
2
Gear C:
1
(T2 )C = (270 ¥ 10 -3 )(15p )2 = 299.789 J
2
System:
T2 = (T2 ) A + (T2 ) B + (T2 )C = 419.7 J
Work of couple.
U1Æ2 = M q B = 10q B
Principle of work and energy for system.
T1 + U1Æ2 = T2 : 20.726 + 10q B = 419.7
q B = 39.898 radians
(a)
39.898
= 15.959 radians
2.5
Rotation of gear C.
qC =
Rotation of gear A.
q A = q B = 39.898 radians
qC = 2.54 rev Principle of work and energy for gear A.
(T1 ) A + M Aq A = (T2 ) A : 2.9609 + M A (39.898) = 59.957
M A = 1.4285 N ◊ m
(b)
Tangential force on gear A.
Ft =
M A 1.4285
=
rA
0.08
Ft = 17.86 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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16
PROBLEM 17.11
The double pulley shown weighs 15 kg and has a centroidal
radius of gyration of 160 mm. Cylinder A and block B are
attached to cords that are wrapped on the pulleys as shown.
The coefficient of kinetic friction between block B and the
surface is 0.25. Knowing that the system is released from rest
in the position shown, determine (a) the velocity of cylinder A
as it strikes the ground, (b) the total distance that block B moves
before coming to rest.
SOLUTION
Let v A = speed of block A,v B = speed of block B, w = angular speed of pulley.
rA = 0.25 m,
rB = 0.15 m
v A = rAw = 0.25 w
v B = rBw = 0.15 w
Kinematics.
s A = rAq = 0.25 q
sB = rBq = 0.15 q
(a)
s A = 0.9 m
Cylinder A falls to ground.
Ê 0.9 ˆ
= 0.54 m
sB = (0.15) Á
Ë 0.25 ˜¯
U1Æ2 = ( mA g ) s A = (12.5)(9.81)(0.9)
Work of weight A:
Normal contact force acting on block B:
F f = m k N = (0.25)(98.1) = 24.525 N
Friction force on block B:
U1Æ2 = - F f sB = -(24.525)(0.54) = -13.2435 N ◊ m
Work of friction force:
Total work:
Kinetic energy:
N = mB g = (10)(9.81) = 98.1 N
T1 = 0;
U1Æ2 = 110.3625 - 13.2435 = 97.119 N ◊ m
1
1
1
T2 = mA v 2A + I w 2 + mB v B2
2
2
2
1
1
1
mA rA2w 2 + mC k 2w 2 + mB rB2w 2
2
2
2
1
= ÈÎ(12.5) (0.25)2 + (15) (0.16)2 + (10) (0.15)2 ˘˚ w 2
2
= 0.695125w 2
T2 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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17
PROBLEM 17.11 (Continued)
Principle of work and energy.
T1 + U1Æ2 = T : 0 + 97.119 = 0.695125w 2
w = 11.820 rad/s2
v A = (0.25)(11.820)
Velocity of cylinder A:
(b)
vA = 2.96 m/s Ø Block B comes to rest.
1 2 1
I w + mB v B2 ;
T4 = 0 :
2
2
1
1
T3 = mC k 2w 2 + mB rB2w 2
2
2
1
= ÈÎ(15)(0.16)2 + (10)(0.15)2 ˘˚ (11.820)2
2
= 42.5424 N ◊ m
T3 =
For block B and pulley C.
Work of friction force:
U 3Æ 4 = - F f sB¢ = -24.525 sB¢
Principle of work and energy.
T3 + U 3Æ 4 = T4 : 42.5424 - 24.525sB¢ = 0
sB¢ = 1.7347 m
Total distance for block B. d = sB + sB¢ : d = 0.54 + 1.7347 = 2.2747 m
d = 2.27 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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18
PROBLEM 17.12
The 160 mm-radius brake drum is attached to a larger flywheel that is
not shown. The total mass moment of inertia of the flywheel and drum
is 20 kg ◊ m2 and the coefficient of kinetic friction between the drum and
the brake shoe is 0.35. Knowing that the initial angular velocity of the
flywheel is 360 rpm counterclockwise, determine the vertical force P that
must be applied to the pedal C if the system is to stop in 100 revolutions.
SOLUTION
w1 = 360 rpm = 12p rad/s
w2 = 0
1
T1 = I w12
2
1
= (20)(12p )2
2
= 14, 212.23 N ◊ m
T2 = 0
Kinetic energy.
q = (100)(2p ) = 628.32 rad
M D = F f r = F f (0.16 m)
Work.
U1Æ2 = - M Dq = - F f (0.16)(628.32)
= -100.531F f
Principle of work and energy.
T1 + U1Æ2 = T2 : 14, 212.23 - 100.531 F f = 0
F f = 141.371 N
Kinetic friction force.
F f = mk N
N=
Statics.
+
Ff
mk
=
141.371
= 403.918 N
0.35
SM A = 0 : (180 mm) P + ( 40 mm) F f - (200 mm) N = 0
180P + ( 40)(141.371) - (200)( 403.918) = 0
P = 417.38 N
P = 417 N Ø PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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19
PROBLEM 17.13
Solve Problem 17.12, assuming that the initial angular velocity of the
flywheel is 360 rpm clockwise.
PROBLEM 17.12 The 160-mm-radius brake drum is attached to a
larger flywheel that is not shown. The total mass moment of inertia of
the flywheel and drum is 20 kg ◊ m2 and the coefficient of kinetic friction
between the drum and the brake shoe is 0.35. Knowing that the initial
angular velocity of the flywheel is 360 rpm counterclockwise, determine
the vertical force P that must be applied to the pedal C if the system is to
stop in 100 revolutions.
SOLUTION
w1 = 360 rpm = 12p rad/s
w2 = 0
1
T1 = I w12
2
1
= (20)(12p )2
2
= 14, 212.23 N ◊ m
T2 = 0
Kinetic energy.
q = (100)(2p ) = 628.32 rad
M D = F f r = F f (0.16)
Work.
U1Æ2 = - M Dq = - F f (0.16)(628.32)
= -100.531F f
Principle of work and energy.
T1 + U1Æ2 = T2 : 14, 212.23 - 100.531F f = 0
F f = 141.371 N
F f = mk N : N =
Kinetic friction.
Statics.
+
Ff
mk
=
141.371
= 403.918 N
0.35
+ SM A = 0 : (180 mm) P - ( 40 mm) F f - (200 mm) N = 0
180 P - ( 40)(141.371) - (200)( 403.918) = 0
P = 480.21 N
P = 480 N Ø PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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20
PROBLEM 17.14
The gear train shown consists of four gears of the same thickness
and of the same material; two gears are of radius r, and the other
two are of radius nr. The system is at rest when the couple M0
is applied to shaft C. Denoting by I0 the moment of inertia of a
gear of radius r, determine the angular velocity of shaft A if the
couple M0 is applied for one revolution of shaft C.
SOLUTION
Mass and moment of inertia:
m = r(p r 2 )t = rp tr 2
For a disk of radius r and thickness t:
1 2 1
1
mr = ( rp + r 2 )r 2 = rp tr 4
2
2
2
1
I = rp t ( nr )4 I = n4 I 0
2
I0 =
For a disk of radius nr and thickness t,
Kinematics:
If for shaft A we have w A
Then, for shaft B we have w B = w A /n
And, for shaft C we have w C = w A /n2
Principle of work-energy:
Couple M 0 applied to shaft C for one revolution. q = 2p radians, T1 = 0,
U1- 2 = M 0q = M 0 (2p radians) = 2p M 0
1
1
1
T2 = ( I shaft A )w 2A + ( I shaft B )w B2 + ( I shaft C )w C2
2
2
2
2
=
1
1
1
Êw ˆ
Êw ˆ
I 0 w 2A + ( I 0 + n4 I 0 ) Á A ˜ + ( n4 I 0 ) Á 2A ˜
Ën ¯
Ë n ¯
2
2
2
=
1
1ˆ
Ê
I 0w 2A Á n2 + 2 + 2 ˜
Ë
2
n ¯
1
1ˆ
Ê
I 0w 2A Á n + ˜
Ë
n¯
2
2
=
1
1ˆ
Ê
I 0w 2A Á n + ˜
Ë
n¯
2
4p M 0
1
w 2A =
2
I0
n+ 1
2
2
T1 + U1- 2 = T2 : 0 + 2p M 0 =
Angular velocity.
(
n
)
wA =
p M0
2n
n + 1 I0
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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21
PROBLEM 17.15
The three friction disks shown are made of the same
material and have the same thickness. It is known that disk A
weighs 6 kg and that the radii of the disks are rA = 200 mm,
rB = 150 mm, and rC = 100 mm. The system is at rest when a
couple M0 of constant magnitude 7.5 N ◊ m is applied to disk A.
Assuming that no slipping occurs between disks, determine the
number of revolutions required for disk A to reach an angular
velocity of 150 rpm.
SOLUTION
Kinematics.
Denote velocity of perimeter by v.
Mass and moment of inertia of disks.
Denote mass density of material by r and thickness of disks by t.
Then mass of a disk is
m = ( volume) r = (p r 2t ) r
and
I =
Kinetic energy:
1 2 prt 4
mr =
r
2
2
1
T = S Iw2
2
1 Ê prt ˆ 4 2
4 2
4 2
T= Á
˜ ÈrA w A + rB w B + rC w C ˘˚
2Ë 2 ¯ Î
2
2
˘
1 Ê prt ˆ È 4 2
4 Ê rA ˆ
2
4 Ê rA ˆ
ÍrA w A + rB Á ˜ w A + rC Á ˜ w 2A ˙
= Á
˜
2Ë 2 ¯ Í
Ë rB ¯
Ë rC ¯
˙˚
Î
1 Ê prtw 2A ˆ 2 2
T= Á
rA ÈrA + rB2 + rC2 ˘˚
2 Ë 2 ˜¯ Î
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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22
PROBLEM 17.15 (Continued)
Recall that
Eq. (1) as:
Data:
mA = p rA2t r and write
1
T = p rA2t r rA2 + rB2 + rC2
4
È Ê r ˆ2 Ê r ˆ2˘
1
T = ( mA )rA2 Í1 + Á B ˜ + Á C ˜ ˙ w 2A
4
Ë rA ¯ ˙
ÍÎ Ë rA ¯
˚
(
)(
)
Ê 2p ˆ
w A = 150 rpm Á ˜ = 5p rad/s
Ë 60 ¯
mA = 6 kg, rA = 0.2 m, rB = 0.15 m, rC = 0.1 m
M = 7.5 N ◊ m
Work:
U1- 2 = M q = (7.5 N ◊ m)q
Principle of work and energy for system:
T1 + U1Æ2 = T2
0 + 7.5q =
È Ê 0.15 ˆ 2 Ê 0.1ˆ 2 ˘
1
2
(6)(0.2)2 Í1 + Á
˜¯ ˙ (5p )
˜¯ + ÁË
Ë
4
0
.
2
0
.
2
ÍÎ
˙˚
9 1˘
È
7.5q = 0.06 Í1 + + ˙ (25p )2
Î 16 4 ˚
7.5q = 26.833
Ê rev ˆ
q = 3.5777 rad Á
Ë 2p rad ˜¯
= 0.5694 rev
q = 0.569 rev PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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23
PROBLEM 17.16
A slender 4-kg rod can rotate in a vertical plane about a pivot at B. A spring
of constant k = 400 N/m and of unstretched length 150 mm is attached to
the rod as shown. Knowing that the rod is released from rest in the position
shown, determine its angular velocity after it has rotated through 90°.
SOLUTION
Position 1:
Spring:
Gravity:
Unstretched
Length
67
4 4
8
x1 = CD - ( 150 mm ) = 370 - 150 = 220 mm = 0.22 m
1
1
Ve = kx12 = ( 400 N/m)(0.22 m)2 = 9.68 J
2
2
Vg = Wh = mgh = ( 4 kg)(9.81 m/s2 )(0.18 m) = 7.063 J
V1 = Ve + Vg = 9.68 J + 7.063 J = 16.743 J
Kinetic energy:
T1 = 0
Position 2:
Spring:
x2 = 230 mm - 150 mm = 80 mm = 0.08 m
1
1
Ve = kx22 = ( 400 N/m)(0.08 m)2 = 1.28 J
2
2
Gravity:
Vg = Wh = 0
V2 = Ve + Vg
= 1.28 J
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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24
PROBLEM 17.16 (Continued)
Kinetic energy:
v2 = rw 2 = (0.18 m)w 2
1
1
I = mL2 = ( 4 kg)(0.6 m)2 = 0.12 kg ◊ m 2
12
12
1
1
T2 = mv22 + I w 22
2
2
1
1
= ( 4 kg)(0.18w 2 )2 + (0.12)w 22
2
2
2
T2 = 0.1248w 2
Conservation of energy:
T1 + V1 = T2 + V2
0 + 16.743 J = 0.1248w 22 + 1.28 J
w 22 = 123.9
w 2 = 11.131 rad/s
w 2 = 11.13 rad/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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25
PROBLEM 17.17
A slender 4-kg rod can rotate in a vertical plane about a pivot at B. A spring
of constant k = 400 N/m and of unstretched length 150 mm is attached to
the rod as shown. Knowing that the rod is released from rest in the position
shown, determine its angular velocity after it has rotated through 90°.
SOLUTION
Position 1:
Spring:
Gravity:
Unstretched
Length
67
4 4
8
x1 = CD - ( 150 mm ) = 370 - 150 = 220 mm = 0.22 m
1
1
Ve = kx12 = ( 400 N/m)(0.22 m)2 = 9.68 J
2
2
Vg = Wh = mgh = ( 4 kg)(9.81 m/s2 )( -0.22 m) = -7.063 J
V1 = Ve + Vg = 9.68 J - 7.063 J = 2.617 J
Kinetic energy:
T1 = 0
Position 2:
Spring:
x2 = 230 mm - 150 mm = 80 mm = 0.08 m
1
1
Ve = kx22 = ( 400 N/m)(0.08 m)2 = 1.28 J
2
2
Gravity:
Vg = Wh = 0
V2 = Ve + Vg
= 1.28 J
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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26
PROBLEM 17.17 (Continued)
Kinetic energy:
v2 = rw 2 = (0.18 m)w 2
1
1
I = mL2 = ( 4 kg)(0.6 m)2 = 0.12 kg ◊ m 2
12
12
1
1
T2 = mv22 + I w 22
2
2
1
1
= ( 4 kg)(0.18w 2 )2 + (0.12)w 22
2
2
2
T2 = 0.1248w 2
Conservation of energy:
T1 + V1 = T2 + V2
0 + 2.617 J = 0.1248 w 22 + 1.28 J
w 22 = 10.713
w 2 = 3.273 rad/s
w 2 = 3.27 rad/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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27
PROBLEM 17.18
A slender rod of length l and weight W is pivoted at one end as shown. It is
released from rest in a horizontal position and swings freely. (a) Determine
the angular velocity of the rod as it passes through a vertical position and
determine the corresponding reaction at the pivot, (b) Solve Part a for
W = 10 N and l = 1 m.
SOLUTION
v1 = 0
w1 = 0
T1 = 0
l
v2 = w 2
2
1
1
T2 = mr22 + I w 22
2
2
Position 1:
Position 2:
2
1 Êl ˆ
1Ê 1
ˆ
m Á w 2 ˜ + Á ml 2 ˜ w 22
¯
2 Ë2 ¯
2 Ë 12
1
T2 = ml 2w 22
6
l
U1Æ2 = mg
2
=
Work:
Principle of work and energy:
T1 + U1Æ2 = T2
0 + mg
(a)
l 1 2 2
= ml w 2
2 6
Expressions for angular velocity and reactions.
3g
l
l 2 l 3g 3
a = w2 = ◊
= g
2
2 l
2
w 22 =
w2 =
3g
l
SF = S ( F )eff : A - W = ma
+
3
A - mg = m g
2
5
A = mg
2
5
A= W ≠
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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28
PROBLEM 17.18 (Continued)
(b)
Application of data:
W = 10 N, l = 1 m
3g 3g
=
= 29.43 rad 2 /s2
w 22 =
l
1
5
5
A = W = (10 N ) = 25 N
2
2
w 2 = 5.42 rad/s
A = 25.0 N ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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29
PROBLEM 17.19
A slender rod of length l is pivoted about a Point C located at a distance b from
its center G. It is released from rest in a horizontal position and swings freely.
Determine (a) the distance b for which the angular velocity of the rod as it
passes through a vertical position is maximum, (b) the corresponding values
of its angular velocity and of the reaction at C.
SOLUTION
Position 1.
v = 0,
Elevation:
h=0
w =0
T1 = 0
V1 = mgh = 0
v2 = bw 2
1
I = ml 2
12
1
1
T2 = mv22 + I w 22
2
2
1 Ê
1 ˆ
= m Á b2 + l 2 ˜ w 22
2 Ë
12 ¯
Position 2.
h = -b
Elevation:
V2 = - mgb
Principle of conservation of energy.
T1 + V1 = T2 + V2 : 0 + 0 =
w 22 =
(a)
1 Ê 2 1 2ˆ 2
m Á b + l ˜ w 2 - mgb
2 Ë
12 ¯
2 gb
b + 121 l 2
2
Value of b for maximumw 2 .
(
)
b2 + 121 l 2 - b (2b )
ˆ
d Ê
b
=0
Á
˜=
2
db Ë b2 + 121 l 2 ¯
b2 + 121 l 2
(b)
Angular velocity.
w 22 =
(
)
b2 =
1 2
l
12
b=
l
12
l
12
l2
l2
12 + 12
2g
= 12
w 2 = 121/4
g
l
g
l
w 2 = 1.861
g
l
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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30
PROBLEM 17.19 (Continued)
an = bw 22
Reaction at C.
l
g
12
l
12
=
=g
SFy = man : C y - mg = mg
+
C y = 2mg
+
SM C = mbat + I a :
0 = ( mb2 + I )a
+
a = 0, at = 0
SFx = mat :
C x = - mat = 0
C = 2mg ≠
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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31
PROBLEM 17.20
An 80-kg gymnast is executing a series of full-circle swings
on the horizontal bar. In the position shown he has a small
and negligible clockwise angular velocity and will maintain his
body straight and rigid as he swings downward. Assuming that
during the swing the centroidal radius of gyration of his body
is 0.4 m determine his angular velocity and the force exerted on
his hands after he has rotated through (a) 90°, (b) 180°.
SOLUTION
Position 1. (Directly above the bar).
Elevation:
h1 = 1 m
Potential energy:
V1 = mgh1 = (80 kg)(9.81 m/s2 )(1 m) = 784.8 N ◊ m
Speeds:
w1 = 0, v1 = 0
Kinetic energy:
T1 = 0
(a)
Position 2. (Body at level of bar after rotating 90∞).
Elevation:
h2 = 0.
Potential energy:
V2 = 0
Speeds:
v2 = (1 m) w 2 .
1
1
T2 = mv22 + mk 2w 22
2
2
1
1
T2 = (80)w 22 + (80)(0.4)2w 22
2
2
2
= 46.4w 2
Kinetic energy:
Principle of conservation of energy.
T1 + V1 = T2 + V2 : 0 + 784.8 = 46.4w 22
w 22 = 16.9138
w 2 = 4.11264 rad/s
w 2 = 4.11 rad/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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32
PROBLEM 17.20 (Continued)
at = (1)(a )
Kinematics:
an = (1)(w 22 ) = 16.9138 m/s2 ¨
+
SM 0 = S ( M 0 )eff : (80)(9.81)(1) = (80)(1)(a )(1) + (80)(0.4)2 (a )
a = 8.4569 rad/s2
+
SFx = man :
+
SFy = - mat :
at = 8.4569 m/s2
Rx = (80)(16.9138) = 1353.1 N
Ry - (80)(9.81) = -(80)(8.4569) ≠
Ry = 108.248 N ≠
(b)
R = 1357 N
4.57∞ Position 3. (Directly below bar after rotating 180∞).
Elevation:
h3 = -1 m.
Potential energy:
V3 = Wh3 = (80)(9.81)( -1) = -784.8 N ◊ m
Speeds:
v3 = (1)w 3 .
Kinetic energy:
T3 = 46.4w 32
Principle of conservation of energy.
T1 + V1 = T3 + V3 : 0 + 784.8 = 46.4w 32 - 784.8
w 32 = 33.828 rad 2 /s2
Kinematics:
From
w 3 = 5.82 rad/s
an = (1)(33.828) = 33.828 m/s ≠
2
SM 0 = S ( M 0 )eff
a = 0,
and
at = 0
SFx = 0,
Rx = 0
SFy = man : Ry - (80)(9.81) = (80)(33.828)
+
Ry = 3491.04 N
R = 3490 N ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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33
PROBLEM 17.21
Two identical slender rods AB and BC are welded together to
form an L-shaped assembly. The assembly is pressed against a
spring at D and released from the position shown. Knowing that
the maximum angle of rotation of the assembly in its subsequent
motion is 90° counterclockwise, determine the magnitude of the
angular velocity of the assembly as it passes through the position
where rod AB forms an angle of 30° with the horizontal.
SOLUTION
Moment of inertia about B.
Position 2.
1
1
I B = mAB l 2 + mBC l 2
3
3
q = 30∞
V2 = W AB ( hAB )2 + WBC ( hBC )2
l
ˆ
Ê l
= WAB sin 30∞ + WBC Á - cos 30∞˜
¯
Ë
2
2
1
1
T2 = I Bw 22 = ( mAB + mBC )l 2w 22
2
6
Position 3.
q = 90∞
V3 = WAB
l
T3 = 0
2
Conservation of energy.
T2 + V2 = T3 + V3 :
l
l
l
1
( mAB + mBC )l 2w 22 + WAB sin 30∞ - WBC cos 30∞ = 0 + W AB
6
2
2
2
1
W
- sin 30∞) + WBC cos 30∞
(
3
w 22 = ◊ AB
l
mAB + mBC
3g
[1 - sin 30∞ + cos 30∞]
2 l
9.81
g
= 2.049 = 2.049
= 50.25
l
0.4
=
w 2 = 7.09 rad/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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34
PROBLEM 17.22
A collar with a mass of 1 kg is rigidly attached at a distance
d = 300 mm from the end of a uniform slender rod AB. The rod has
a mass of 3 kg and is of length L = 600 mm. Knowing that the rod
is released from rest in the position shown, determine the angular
velocity of the rod after it has rotated through 90°.
SOLUTION
Kinematics.
Rod
Collar
L
w
2
vC = dw
vR =
Position 1.
w =0
T1 = 0 V1 = 0
Position 2.
T2 =
1
1
1
mR vR2 + I Rw 2 + mC vC2
2
2
2
2
1
1Ê 1
1
ˆ
ÊL ˆ
mR Á w ˜ + Á mR L2 ˜ w 2 + mC d 2w 2
¯
Ë2 ¯
2
2
2 Ë 12
1
1
= mR L2w 2 + mC d 2w 2
6
2
L
V2 = -WC d - WR
2
=
1
L
ˆ
Ê1
T1 + V1 = T2 + V2 : 0 + 0 = Á mR L2 + mC d 2 ˜ w 2 - WC d - WR
¯
Ë6
2
2
w2 =
3(2WC d + WC L)
3mC d + mR L
2
2
=
3g (2mC d + mR L)
Data:
mC = 1 kg, d = 0.3 m, mR = 3 kg, L = 0.6 m
From Eq. (1),
È (2)(1)(0.3) + 3(0.6) ˘
w 2 = 3(9.81) Í
2
2˙
Î 3(1)(0.3) + 3(0.6) ˚
= 52.32
(1)
3mC d 2 + mR L2
w = 7.23 rad/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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35
PROBLEM 17.23
A collar with a mass of 1 kg is rigidly attached to a slender rod AB
of mass 3 kg and length L = 600 mm. The rod is released from
rest in the position shown. Determine the distance d for which the
angular velocity of the rod is maximum after it has rotated 90°.
SOLUTION
Kinematics.
Rod
Collar
L
w
2
vC = dw
vR =
Position 1.
w =0
T1 = 0 V1 = 0
Position 2.
T2 =
1
1
1
mR vR2 + I Rw 2 + mC vC2
2
2
2
2
1
1
1Ê 1
ˆ
ÊL ˆ
mR Á w ˜ + Á mR L2 ˜ w 2 + mC d 2w 2
¯
Ë2 ¯
2
2
2 Ë 12
1
1
= mR L2w 2 + mC d 2w 2
6
2
L
V2 = -WC d - WR
2
L
1
1
ˆ
Ê
T1 + V1 = T2 + V2 : 0 + 0 = Á mR L2 + mC d 2 ˜ w 2 - WC d - WR
¯
Ë6
2
2
=
w2 =
Let
x=
3(2WC d + WC L)
3mC d + mR L
2
2
=
3g (2mC d + mR L)
3mC d 2 + mR L2
(1)
d
.
L
mR
mC
3g
w2 =
◊
L 3x 2 + mR
mC
2x +
Data:
mC = 1 kg, mR = 3 kg
Lw 2 2 x + 3
= 2
3g
3x + 3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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36
PROBLEM 17.23 (Continued)
Lw 2 / 3g is maximum. Set its derivative with respect to x equal to zero.
d Ê Lw 2 ˆ (3x 2 + 3)(2) - (2 x + 3)(6 x )
=
=0
dx ÁË 3g ˜¯
(3x 2 + 3)2
-6 x 2 - 18 x + 6 = 0
Solving the quadratic equation
x = -3.30 and
x = 0.30278
d = 0.30278L
= (0.30278)(0.6)
= 0.1817 m
d = 181.7 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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37
PROBLEM 17.24
A 20-kg uniform cylindrical roller, initially at rest, is acted upon by a 90-N
force as shown. Knowing that the body rolls without slipping, determine
(a) the velocity of its center G after it has moved 1.5 m, (b) the friction force
required to prevent slipping.
SOLUTION
Since the cylinder rolls without slipping, the point of contact with the ground is the instantaneous center.
v = rw
T1 = 0
Kinematics:
Position 1. At rest.
s = 1.5 m vG = v
Position 2.
T2 =
w=
vG
r
1
1
mv 2 + I w 2
2
2
2
1 2 1 Ê 1 2 ˆ Ê vG ˆ
mvG + Á mr ˜ Á ˜
¯Ë r ¯
2
2Ë2
3
3
= mvG2 = (20)vG2 = 15vG2
4
4
=
U1Æ2 = Ps = (90)(1.5) = 135 J.
Work:
(a)
F f does no work.
Principle of work and energy.
T1 + U1Æ2 = T2 : 0 + 135 = 15vG2
vG2 = 9
(b)
Since the forces are constant,
vG = 3.00 m/s Æ
aG = a = constant
aG =
=
vG2
2s
9
(2)(1.5)
= 3 m/s2
+
SFx = ma : P - F f = ma
F f = P - ma
= 90 - (20)(3)
F f = 30.0 N ¨ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
38
PROBLEM 17.25
A rope is wrapped around a cylinder of radius r and mass m as shown. Knowing that
the cylinder is released from rest, determine the velocity of the center of the cylinder
after it has moved downward a distance s.
SOLUTION
Point C is the instantaneous center.
v = rw w =
Position 1. At rest.
v
r
T1 = 0
Position 2. Cylinder has fallen through distance s.
T2 =
Work.
U1Æ2
1
1
mv 2 + I w 2
2
2
1
1Ê1
ˆÊvˆ
= mv 2 + Á mr 2 ˜ Á ˜
¯Ë r¯
Ë
2
2 2
3
= mv 2
4
= mgs
2
Principle of work and energy.
3
mv 2
4
4 gs
v2 =
3
T1 + U1Æ2 = T2: 0 + mgs =
v=
4 gs
Ø
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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39
PROBLEM 17.26
Solve Problem 17.25, assuming that the cylinder is replaced by a thin-walled pipe of
radius r and mass m.
PROBLEM 17.25 A rope is wrapped around a cylinder of radius r and mass m as
shown. Knowing that the cylinder is released from rest, determine the velocity of the
center of the cylinder after it has moved downward a distance s.
SOLUTION
Point C is the instantaneous center.
v = rw w =
Position 1. At rest.
v
r
T1 = 0
Position 2. Cylinder has fallen through distance s.
T2 =
=
Work.
U1Æ2
1
1
mv 2 + I w 2
2
2
1
1
Êvˆ
mv 2 + ( mr 2 ) Á ˜
Ë r¯
2
2
2
= mv 2
= mgs
Principle of work and energy.
T1 + U1Æ2 = T2 : 0 + mgs = mv 2
v 2 = gs
v = gs Ø PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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40
PROBLEM 17.27
The mass center G of a 3-kg wheel of radius R = 180 mm is located at a distance
r = 60 mm from its geometric center C. The centroidal radius of gyration of the wheel
is k = 90 mm. As the wheel rolls without sliding, its angular velocity is observed
to vary. Knowing that w = 8 rad/s in the position shown, determine (a) the angular
velocity of the wheel when the mass center G is directly above the geometric center C,
(b) the reaction at the horizontal surface at the same instant.
SOLUTION
v1 = ( BG )w1
= (0.18)2 + (0.06)2 (8)
= 8 0.036 m/s
v2 = 0.24w 2
m = 3 kg
k = 0.09 m
Position 1.
V1 = 0
1
1
T1 = mv12 + I w12
2
2
1
1
= (3)(8 0.036 )2 + (3)(0.09)2 (8)2
2
2
= 4.2336 J
Position 2.
V2 = Wh
= mgh
= (3)(9.81)(0.06)
= 1.7658 J
1
1
T2 = mv22 + I w 22
2
2
1
1
= (3)(0.24w 2 )2 + (3)(0.09)2 w 22
2
2
= 0.09855w 22
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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41
PROBLEM 17.27 (Continued)
(a)
Conservation of energy.
T1 + V1 = T2 + V2
4.2336 J + 0 = 0.09855w 22 + 1.7658 J
w 22 = 25.041
w 2 = 5.004 rad/s
(b)
w 2 = 5.00 rad/s Reaction at B.
man = m(CG )w 22
+
= (3 kg)(0.06 m)(5.00 rad/s)2
= 4.5 N Ø
SFy = ma y : N - mg = - man
N - (3)(9.81) = -4.5
N = 24.9 N ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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42
PROBLEM 17.28
A collar B, of mass m and of negligible dimension, is attached to the rim of a
hoop of the same mass m and of radius r that rolls without sliding on a horizontal
surface. Determine the angular velocity w1 of the hoop in terms of g and r when B
is directly above the center A, knowing that the angular velocity of the hoop is 3w1
when B is directly below A.
SOLUTION
The point of contact with ground is the instantaneous center.
Position 1. Point B is at the top.
w = w1 v B = 2rw1 v A = rw1
1
1
1
T1 = mv B2 + mv 2A + I w 2
2
2
2
1
1
1
= m(2rw1 )2 + m( rw1 )2 + ( mr 2 )w12
2
2
2
2 2
= 3mr w1
Position 2. Point B is at the bottom.
w = w 2 v B = 0 v A = rw 2
1
1
1
T2 = mv B2 + mv 2A + I w 22
2
2
2
1
1
= 0 + m( rw 2 )2 + ( mr 2 )w 22
2
2
2 2
= mr w 2
= mr 2 (3w1 )2
= 9mr 2w12
U1Æ2 = mg ( Dh) = 2mgr
Work.
Principle of work and energy.
T1 + U1Æ2 = T2 : 3mr 2w12 + 2mgr = 9mw12
w12 =
g
3r
w1 = 0.577
g
r
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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43
PROBLEM 17.29
A half section of pipe of mass m and radius r is released from rest in the position
shown. Knowing that the pipe rolls without sliding, determine (a) its angular
velocity after it has rolled through 90°, (b) the reaction at the horizontal surface at
the same instant. [Hint: Note that GO = 2r /p and that, by the parallel-axis theorem,
I = mr 2 - m(GO )2 .]
SOLUTION
w1 = 0 v1 = 0 T1 = 0
Position 1.
Position 2.
2ˆ
Ê
v2 = ( AG )w 2 = r Á1 - ˜ w 2
Ë p¯
Kinematics:
2
4ˆ
Ê
Ê 2r ˆ
I = mr 2 - m(0.6)2 = mr 2 - m Á ˜ = mr 2 Á1 - 2 ˜
Ëp¯
Ë p ¯
Moment of inertia:
T2 =
Kinetic energy:
1
1
mv22 + I w 22
2
2
2
=
2ˆ
1
4ˆ
1 Ê
Ê
m Á1 - ˜ r 2w 22 + mr 2 Á1 - 2 ˜ w 22
Ë p ¯
2 Ë p¯
2
=
4 4ˆ Ê
4 ˆ˘
1 2 ÈÊ
mr ÍÁ1 - + 2 ˜ + Á1 - 2 ˜ ˙
2
ÎË p p ¯ Ë p ¯ ˚
1 2Ê
4ˆ
mr Á 2 - ˜
Ë
p¯
2
2r 2
= W (OG ) = mg = mgr
p p
= T2
=
Work:
Principle of work and energy:
U1Æ2
T1 + U1Æ2
0 + mg
2r 1 2 Ê
4ˆ
= mr Á 2 - ˜ w 22
Ë
p 2
p¯
g
2
g
w 22 =
◊ = 1.7519
2
r
p 1- p r
(
)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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44
PROBLEM 17.29 (Continued)
(a)
Angular velocity.
(b)
Reaction at A.
Kinematics:
w 2 = 1.324
g
r
Since O moves horizontally, ( a0 ) y = 0
an = (0.6)w 22
=
2r Ê
gˆ
ÁË1.7519 ˜¯
p
r
= 1.1153g ≠
+
Kinetics:
SFy = S ( Fy )eff : A - mg = 1.1153mg
A = 2.12mg ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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45
PROBLEM 17.30
Two uniform cylinders, each of mass m = 7 kg and radius r = 100 mm are
connected by a belt as shown. Knowing that the initial angular velocity of
cylinder B is 30 rad/s counterclockwise, determine (a) the distance through
which cylinder A will rise before the angular velocity of cylinder B is reduced
to 5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.
SOLUTION
v AB = rw B
Kinematics.
Point C is the instantaneous center of cylinder A.
v AB 1
= wB
2r 2
1
v A = rw A = rw B
2
1W 2
r
I =
2 g
wA =
Moment of inertia.
Kinetic energy.
1
1 Ê 1 W 2ˆ 2 1 W 2 2
I w B2 = Á
r wB =
r wB
2
2 Ë 2 g ˜¯
4 g
Cyl B:
1 2 1
1W
mv A + I w 2A =
2
2
2 g
Cyl A:
Total:
(a)
2
1 Ê 1 W 2ˆ Ê 1 ˆ
ˆ
Ê1
r Á wB ˜
ÁË rw B ˜¯ + Á
2
2 Ë 2 g ˜¯ Ë 2 ¯
=
3W 2 2
r wB
16 g
T=
7W 2 2
r wB
16 g
2
Distance h that cylinder A will rise.
Conservation of energy for system.
T1 + V1 = T2 + V2 :
7W 2
7W 2
r (w B )22 + Wh
r (w B )12 + 0 =
16 g
16 g
h=
7 r2
[(w B )12 - (w B )2 ]2
16 g
2
Ê 7 ˆ (0.1)
(302 - 52 )
=Á ˜
Ë 16 ¯ (9.81)
= 0.3902 m
h = 0.390 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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it without permission.
46
PROBLEM 17.30 (Continued)
(b)
Tension in belt between the cylinders.
When cylinder A moves up a distance h, the belt moves up a distance 2h.
U1Æ2 = P (2h) - Wh
Work:
Principle of work and energy for cylinder A.
T1 + U1Æ2 = T2 :
3W 2
3W 2
r (w B )12 + 2 Ph - Wh =
r (w B )22
16 g
16 g
1
3 Wr 2
È(w B )12 - (w B )22 ˘
P= W˚
2
32 gh Î
1
3
= W- W
2
14
2
2
P = 19.62 N = W = (7)(9.81)
7
7
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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47
PROBLEM 17.31
Two uniform cylinders, each of mass m = 7 kg and radius r = 100 mm
are connected by a belt as shown. If the system is released from rest,
determine (a) the velocity of the center of cylinder A after it has moved
through 1 m, (b) the tension in the portion of belt connecting the two
cylinders.
SOLUTION
v AB = rw B
Kinematics.
Point C is the instantaneous center of cylinder A.
v AB 1
= wB
2r 2
1
v A = rw A = rw B
2
wA =
Moment of inertia.
I =
1W 2
r
2 g
Kinetic energy.
Cyl B:
1
1 Ê 1 W 2ˆ 2 1 W 2 2
I w B2 = Á
r wB =
r wB
2
2 Ë 2 g ˜¯
4 g
Cyl A:
1 2 1
1W Ê1
1 Ê 1 W 2ˆ Ê 1 ˆ
ˆ
mv A + I w 2A =
r Á wB ˜
ÁË rw B ˜¯ + Á
2
2
2 g 2
2 Ë 2 g ˜¯ Ë 2 ¯
2
Total:
=
3W 2 2
r wB
16 g
T=
7W 2 2
r wB
16 g
T1 = 0
Position 1.
Rest
Position 2.
Cylinder has fallen through. d = 1 m.
2
V1 = 0
7W 2 2
r wB
16 g
V2 = -Wd
T2 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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48
PROBLEM 17.31 (Continued)
T1 + V1 = T2 + V2
7W 2 2
0+0=
r w B - Wd
16 g
Principle of conservation of energy:
r = 0.1 m
16 gd
w B2 =
7 r2
16 (9.81)(1)
= ◊
7 (0.1)2
= 2242.29
w B = 47.3528 rad/s
where
(a)
Velocity of the center of cylinder A.
1
rw B
2
1
= (0.1)( 47.3528)
2
vA =
(b)
vA = 2.37 m/s Ø Tension in belt between the cylinders.
When cylinder A moves down a distance d, the belt moves down a distance 2d.
P
U1Æ2 = (2d ) + Wd
Work:
2
Principle of work and energy for cylinder A:
T1 + U1Æ2 = T2 : 0 - P (2d ) + Wd =
3W 2 2
r wB
16 g
1
3 Wr 2w B2
P= W2
32 g d
3 (7)(9.81)(0.1)2 (2242.29)
1
P = (7)(9.81) 32
(9.81)(1)
2
= 34.335 - 14.715 = 19.620
P = 19.62 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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49
PROBLEM 17.32
The 5-kg rod BC is attached by pins to two uniform disks as
shown. The mass of the 150-mm-radius disk is 6 kg and that
of the 75-mm-radius disk is 1.5 kg. Knowing that the system is
released from rest in the position shown, determine the velocity
of the rod after disk A has rotated through 90°.
SOLUTION
T1 = 0
Position 1.
Position 2.
Kinematics.
v B = v AB
vC = v AB
Kinetic energy.
T2 =
vB
v AB
=
v A = 2v B = 2v AB
BE 0.075 m
v
v AB
wC = C =
w AB = 0
CF 0.075 m
wA =
1
1
1
1
1
mA v A + I Aw 2A + mAB v 2AB + mB v B2 + I Bw B2
2
2
2
2
2
1È
1
ˆ
2
2Ê v
2
Í(2 m/s)(2v AB ) + (6 kg)(0.15 m) Á AB ˜ + (5 kg)v AB
Ë 0.075 ¯
2 ÍÎ
2
2
=
1
Ê v ˆ
+ (1.5 kg)( v AB ) + (1.5 kg)(0.075 m) Á AB ˜
Ë 0.075 ¯
2
2
2˘
˙
˙˚
1
[24 + 12 + 5 + 1.5 + 0.75] v 2AB
2
T2 = 21.625 v 2AB
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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50
PROBLEM 17.32 (Continued)
Work:
U1Æ2 = W AB (0.1125 m - 0.075 m)
U1Æ2
Principle of work and energy:
= (5 kg)(9.81)(0.0375 m)
= 1.8394 J
T1 + U1Æ2 = T2
0 + 1.8394 J = 21.625 v 2AB
v 2AB = 0.08506
v AB = 0.2916 m
vAB = 292 mm/s Æ Velocity of the rod.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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51
PROBLEM 17.33
The 9-kg cradle is supported as shown by two uniform disks that roll
without sliding at all surfaces of contact. The mass of each disk is
m = 6 kg and the radius of each disk is r = 80 mm. Knowing that the
system is initially at rest, determine the velocity of the cradle after it
has moved 250 mm.
SOLUTION
IA = IB =
Moments of inertia.
Wr 2
2g
wA = wB =
Kinematics.
vC
r
v A = v B = vC
T1 = 0
Kinetic energy.
1
1
1
1
1
mA v 2A + I Aw 2A + mB v B2 + I Bw B2 + mC vC2
2
2
2
2
2
1È
1
1
˘
= Ím + m + m + m + mC ˙ vC2
2Î
2
2
˚
1
= (3m + mC )vC2
2
1
= [(3)(6) + 9] vC2
2
= 13.5vC2
T2 =
Work.
U1Æ2 = Fs = (30 N )(0.25 m) = 7.5 J
Principle of work and energy.
T1 + U1Æ2 = T2 : 0 + 7.5 = 13.5vC2
vC = 0.745 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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52
PROBLEM 17.34
The 9-kg cradle is supported as shown by two uniform disks that roll
without sliding at all surfaces of contact. The mass of each disk is
m = 6 kg and the radius of each disk is r = 80 mm. Knowing that the
system is initially at rest, determine the velocity of the cradle after it
has moved 250 mm.
SOLUTION
IA = IB =
Moments of inertia.
Wr 2
2g
vC
r
v A = vB = 0
wA = wB =
Kinematics.
T1 = 0
Kinetic energy.
1
1
1
1
1
mA v 2A + I Aw 2A + mB v B2 + I Bw B2 + mC vC2
2
2
2
2
2
1
1
1È
˘ 2
= Í0 + m + 0 + m + mC ˙ vC
2
2
2Î
˚
1
= ( m + mC )vC2
2
1
= (6 + 9)vC2
2
= 7.5vC2
T2 =
Work.
U1Æ2 = Fs = (30 N )(0.25 m) = 7.5 J
Principle of work and energy.
T1 + U1Æ2 = T2 : 0 + 7.5 = 7.5vC2
vC = 1.000 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
53
PROBLEM 17.35
The 9-kg cradle is supported as shown by two uniform disks that roll
without sliding at all surfaces of contact. The mass of each disk is
m = 6 kg and the radius of each disk is r = 80 mm. Knowing that the
system is initially at rest, determine the velocity of the cradle after it
has moved 250 mm.
SOLUTION
IA = IB =
Moments of inertia.
Wr 2
2g
wA = wB =
Kinematics.
vC
2r
v A = vB =
1
vC
2
T1 = 0
Kinetic energy.
1
1
1
1
1
mA v 2A + I Aw 2A + mB v B2 + I Bw B2 + mC vC2
2
2
2
2
2
1 È1
1
1
1
˘
= Í m + m + m + m + mC ˙ vC2
8
2 Î4
8
4
˚
1
= (0.75m + mC )vC2
2
1
= [(0.75)(6) + 9]vC2
2
= 6.75vC2
T2 =
Work.
U1Æ2 = Fs = (30 N )(0.25 m) = 7.5 J
Principle of work and energy.
T1 + U1Æ2 = T2 : 0 + 7.5 = 6.75vC2
vC = 1.054 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
54
PROBLEM 17.36
The motion of the slender 10-kg rod AB is guided by collars of negligible
mass that slide freely on the vertical and horizontal rods shown. Knowing
that the bar is released from rest when q = 30°, determine the velocity of
collars A and B when q = 60°.
SOLUTION
Position 1:
T1 = 0
V1 = W (0.3 m)
= -(10 kg)(9.8)(0.3)
= -29.43 J
Position 2:
V2 = -W (0.5196 m)
= -(10 kg)(9.81)(0.5196)
= -50.974 J
1
1
T2 = mv22 + I w 22
2
2
1
1Ê1
ˆ
= (10 kg)(0.6w 2 )2 + Á (10 kg)(1.2 m)2 w 22 ˜
¯
2
2Ë2
= 2.4w 22
Principle of conservation of energy.
T1 + V1 = T2 + V2
0 - 29.43 J = 2.4w 22 - 50.974
w 22 = 8.9768
Velocity of collars when
w 2 = 2.996 rad/s
q = 60∞
v A = ( AC )w 2 = (2 ¥ 0.5196 m)(2.996 rad/s)
vA = 3.11 m/s Æ v B = ( BC )w 2 = (2 ¥ 0.3 m)(2.996 rad/s)
vB = 1.798 m/s Ø PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
55
PROBLEM 17.37
The motion of the slender 10-kg rod AB is guided by collars of negligible
mass that slide freely on the vertical and horizontal rods shown. Knowing
that the bar is released from rest when q = 20°, determine the velocity of
collars A and B when q = 90°.
SOLUTION
Position 1:
Position 2:
T1 = 0
V1 = -W (0.2052 m)
= - mg (0.2052)
V2 = -W (0.6 m) = - mg (0.6)
1
1
T2 = mv 22 + I w 22
2
2
1
1Ê 1
ˆ
= m(0.6w 2 )2 + Á m(1.2)2 ˜ w 22
¯
2
2 Ë 12
T2 = 0.24mw 22
Principle of conservation of energy.
T1 + V1 = T2 + V2
0 - 0.2052mg = 0.24mw 22 - 0.6mg
w 22 = 1.645 g = 1.645(9.81) = 16.137
w 2 = 4.017 rad/s
Velocity of collars when
q = 90∞
v A = ( AB)w 2 = (1.2 m)( 4.017 rad/s)
vB = 0
vA = 4.82 m/s Æ vB = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
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56
PROBLEM 17.38
The ends of a 4.5 kg rod AB are constrained to move along slots cut in a
vertical plane as shown. A spring of constant k = 600 N/m is attached to
end A in such a way that its tension is zero when q = 0 . If the rod is
released from rest when q = 0, determine the angular velocity of the rod
and the velocity of end B when q = 30∞.
SOLUTION
Moment of inertia. Rod.
Position 1.
1
mL2
12
q1 = 0
v1 = 0
w1 = 0
h1 = elevation above slot.
h1 = 0
e1 = elongation of spring.
e1 = 0
I =
1
1
mv12 + I w12 = 0
2
2
1 2
V1 = ke1 + Wh1 = 0
2
T1 =
Position 2.
q = 30∞
e2 + L cos 30∞ = L
e2 = L(1 - cos 30∞)
L
1
h2 = - sin 30∞ = - L
2
4
1
1
1
V2 = ke22 + Wh2 = k L2 (1 - cos 30∞)2 - WL
2
4
2
Kinematics. Velocities at A and B are directed as shown. Point C is the instantaneous center of rotation. From
geometry, b = L2 .
L
w
2
v B = ( L cos 30∞)w
1
1
T2 = mv 2 + I w 2
2
2
v = bw =
2
1 ÊL ˆ
1Ê 1
ˆ
m Á w ˜ + Á mL2 ˜
¯
¯
Ë
Ë
2
2
2 12
1W 2 2
=
Lw
6 g
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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57
PROBLEM 17.38 (Continued)
Conservation of energy.
1W 2 2 1 2
1
L w + kL (1 - cos 30∞)2 - WL
6 g
2
4
3g 3 kg
w2 =
(1 - cos 30∞)2
L
W
T1 + V1 = T2 + V2 : 0 + 0 =
Data:
W = ( 4.5)(9.81) N
g = 9.81 m/s2
L = 0.6 m
k = 600 N/m
(3)(9.81) 3(600)(9.81)
(1 - cos 30∞)2
(0.6)
( 4.5)(9.81)
= 41.8703
w2 =
v B = (0.6)(cos 30∞)(6.4707)
w = 6.47 rad/s vB = 3.36 m/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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58
PROBLEM 17.39
The ends of a 4.5 kg rod AB are constrained to move along slots cut in a
vertical plate as shown. A spring of constant k = 600 N/m is attached to
end A in such a way that its tension is zero when q = 0. If the rod is
released from rest when q = 50∞, determine the angular velocity of the
rod and the velocity of end B when q = 0.
SOLUTION
L
w2
2
v B = Lw 2
x1 = L - L cos 50∞
= 0.6 m(1 - cos 50∞)
= 0.21433 m
v2 =
Position 1.
Position 2.
1
L
sin 50∞ + kx12
2
2
1
Ê 0.6 ˆ
sin 50∞ + (600)(0.21433)2
V1 = -( 4.5)(9.81) Á
Ë 2 ˜¯
2
= -10.1451 + 13.7812
= 3.6361 N ◊ m
T1 = 0
V1 = -W
V2 = (Vg )2 + (Ve )2 = 0
T2 =
1
1
mv22 + I w 22
2
2
2
1 ÊL ˆ
1Ê 1
ˆ
m Á w 2 ˜ + Á mL2 ˜ w 22
¯
2 Ë2 ¯
2 Ë 12
1
1
= mL2w 22 = ( 4.5 kg)(0.6 m)2 w 22 = 0.27w 22
6
6
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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59
PROBLEM 17.39 (Continued)
Conservation of energy:
T1 + V1 = T2 + V2
0 + 3.6361 N ◊ m = 0.27w 22
w 22 = 13.467 rad 2 /s2
w 2 = 3.6697 rad/s
Velocity of B:
w 2 = 3.67 rad/s
v B = Lw 2 = (0.6 m)(3.6697 rad/s)
= 2.2018 m/s
vB = 2.20 m/s ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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60
PROBLEM 17.40
The motion of the uniform rod AB is guided by small wheels of negligible
mass that roll on the surface shown. If the rod is released from rest when
q = 0, determine the velocities of A and B when q = 30∞.
SOLUTION
q =0
v A = vB = 0
w =0
T1 = 0
V1 = 0
Position 1.
q = 30∞
Position 2.
Kinematics. Locate the instantaneous center C. Triangle ABC is equilateral.
v A = v B = Lw
vG = Lw cos 30∞
I =
Moment of inertia.
Kinetic energy.
T2 =
Potential energy.
1 2
ml
12
1 2 1
1
1Ê 1
ˆ
mvG + I w 2 : T2 = m( Lw cos 30∞)2 + Á mL2 ˜ w 2
¯
2
2
2
2 Ë 12
5
= ml 2w 2
12
1
L
V2 = - mg sin 30∞ = - mgL
2
4
Conservation of energy.
T1 + V1 = T2 + V2 : 0 + 0 =
5
1
mL2w 2 - mgL
12
4
w 2 = 0.6
g
L
w = 0.775
g
L
v A = 0.775 gL
v B = 0.775 gL
vA = 0.775 gL ¨ vB = 0.775 gL
60∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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61
PROBLEM 17.41
The motion of a slender rod of length R is guided by pins at A and B which
slide freely in slots cut in a vertical plate as shown. If end B is moved
slightly to the left and then released, determine the angular velocity of the
rod and the velocity of its mass center (a) at the instant when the velocity
of end B is zero, (b) as end B passes through Point D.
SOLUTION
The rod AB moves from Position 1, where it is nearly vertical, to Position 2, where vB = 0.
In Position 2, vA is perpendicular to both CA and AB, so CAB is a straight line of length 2L and slope angle 30∞.
In Position 3 the end B passes through Point D.
T1 = 0 V1 =
Position 1:
v
R
= mg
h1
2
Position 2: Since instantaneous center is at B,
1
Rw 2
2
1
1
T2 = mv22 + I w 22
2
2
v2 =
2
1 Ê1
1Ê 1
ˆ
ˆ
= m Á Rw 2 ˜ + Á mR2 ˜ w 22
¯
Ë
¯
Ë
2
2
2 12
1
= mR2w 22
6
R
V2 = Wh2 = mg
4
Position 3:
V3 = 0
Since both vA and vB are horizontal,
w3 = 0
1
T3 = mv22
2
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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62
PROBLEM 17.41 (Continued)
(a)
From 1 to 2: Conservation of energy
1
1
1
T1 + V1 = T2 + V2 : 0 + mgR = mR2w 22 + mgR
2
6
4
3
g
w 22 =
2R
3g
w2 =
2R
v2 =
(b)
1
1 3
3
Rw 2 =
gR =
gR
2
2 2
8
w 2 = 1.225
vR = 0.612 gR
g
R
60∞ From 1 to 3: Conservation of energy
w3 = 0 From Eq. (1) we have
1
1
T1 + V1 = T3 + V3 : 0 + mgR = mv32
2
2
2
v3 = gR
v3 = gR Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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63
PROBLEM 17.42
Two uniform rods, each of mass m and length L, are connected to form
the linkage shown. End D of rod BD can slide freely in the horizontal
slot, while end A of rod AB is supported by a pin and bracket. If end
D is moved slightly to the left and then released, determine its velocity
(a) when it is directly below A, (b) when rod AB is vertical.
SOLUTION
Moments of inertia.
I AB = I BD =
1
mL2
12
1
I A = mL2
3
Position 1. At rest as shown.
T1 = 0
V1 = mghAB + mghBC
Ê Lˆ
= 0 + mg Á - ˜
Ë 2¯
1
= - mgL
2
(a)
Position 2. In Position 2, Point A is the instantaneous center of both AB and BD. Since Point B is
common to both bars,
w AB = w BD
= w2
vD = lw 2
vG = lw cos 30∞
Kinetic energy.
T2 =
1Ê 1
1
1
1
1Ê1
1
ˆ
ˆ
I Aw 22 + mvG2 + I BDw 22 : T2 = Á mL2 ˜ w 22 + m(lw 2 cos 30∞)2 + Á mL2 ˜ w 22
¯
¯
2 Ë 12
2
2
2
2Ë3
2
7
= mL2w 22
12
Potential energy.
V2 = mghAB + mghBD
ˆ
Ê 3L
ˆ
Ê L
V2 = mg Á - sin 30∞˜ + mg Á - sin 30∞˜
¯
Ë 2
¯
Ë 2
= - mgL
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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64
PROBLEM 17.42 (Continued)
Conservation of energy.
1
7
T1 + V1 = T2 + V2 : 0 - mgL = mL2w 22 - mgL
2
12
6g
w 22 =
7L
g
w 2 = 0.926
L
vD = Lw 2 = 0.926 gL
(b)
vD = 0.926 gL ¨ Position 3. In Position 3, bar BD is in translation.
vD = v B = Lw AB = Lw 3
Kinetic energy.
T3 =
1
1
1Ê1
1
2
ˆ
I Aw 32 + mv B2 : T3 = Á mL2 ˜ w 32 + m( Lw 3 )2 = mL2w 32
¯
2
2
2Ë3
2
3
Potential energy.
Ê Lˆ
V3 = mg Á - ˜ + mg ( - L)
Ë 2¯
3
= - mgL
2
Conservation of energy.
1
2
3
T1 + V1 = T3 + V3 : 0 - mgL = mL2w 32 - mgL
2
3
2
3g
w 32 =
2L
g
w 3 = 1.225
L
vD = Lw 3
= 1.225 gL
vD = 1.225 gL ¨ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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65
PROBLEM 17.43
The uniform rods AB and BC have a mass of 1.2 kg and 2 kg,
respectively, and the small wheel at C is of negligible weight. If the
wheel is moved slightly to the right and then released, determine the
velocity of pin B after rod AB has rotated through 90∞.
SOLUTION
Moments of inertia.
Bar AB:
Bar BC:
mAB = 1.2 kg
mBC = 2 kg
1
1
I A = mAB L2AB = (1.2)(0.45)2 = 0.081 kg ◊ m2
3
3
I =
1
1
mBC L2BC = (2)(0.75)2 = 0.09375 kg ◊ m2
12
12
Position 1. As shown with bar AB vertical. Point G is the midpoint of BC.
V1 = W AB hAB + WBC hBC
Ê 0.45 ˆ
Ê 0.45 ˆ
+ (2)(9.81) Á
= (1.2)(9.81) Á
Ë 2 ˜¯
Ë 2 ˜¯
= 7.0632 N ◊ m
Bar BC is at rest.
w BC = 0
v = vG = v B = vC = 0
w AB =
vB
=0
LAB
T1 = 0
Position 2. Bar AB is horizontal.
V2 = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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66
PROBLEM 17.43 (Continued)
Kinematics.
w AB =
vB
v
= B
LAB 0.45
w BC =
vB
v
= B
LBC 0.75
1
vB
2
1
1
1 2
T2 = I Aw 2AB + mBC v 2 + Iw BC
2
2
2
v=
2
2
1 Ê1 ˆ
1
1
Ê v ˆ
Ê v ˆ
= (0.081) Á B ˜ + (2) Á v B ˜ + (0.09375) Á B ˜
¯
Ë
Ë 0.75 ¯
Ë
¯
0.45
2
2
2
2
2
= 0.53333v B2
Conservation of energy.
T1 + V1 = T2 + V2 : 0 + 7.0632 = 0.53333v B2
v B = 3.6392 m/s
vB = 3.64 m/s Ø PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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67
PROBLEM 17.44
The uniform rods AB and BC have a mass 1.2 kg and 2 kg,
respectively, and the small wheel at C is of negligible weight.
Knowing that in the position shown the velocity of wheel C is 2 m/s
to the right, determine the velocity of pin B after rod AB has rotated
through 90°.
SOLUTION
Moments of inertia.
Bar AB:
Bar BC:
1
1
I A = mAB L2AB = (1.2)(0.45)2 = 0.081 kg ◊ m 2
3
3
I =
1
1
mBC LBC 2 = (2)(0.75)2 = 0.09375 kg ◊ m2
12
12
Position 1. As shown with bar AB vertical. Point G is the midpoint of BC.
V1 = W AB hAB + WBC hBC
Ê 0.45 ˆ
Ê 0.45 ˆ
+ (2)(9.81) Á
= (1.2)(9.81) Á
Ë 2 ˜¯
Ë 2 ˜¯
= 7.0632 N ◊ m
Kinematics, Bar BC is in translation.
w BC = 0
v = vG = v B = vC = 2 m/s
vB
2
40
=
=
rad/s
LAB 0.45 9
1 2
1
1
T1 = IAw 2AB + mBC v 2 + Iw BC
2
2
2
w AB =
2
1
1
Ê 40 ˆ
= (0.081) Á ˜ + (2)(2)2 + 0
Ë 9¯
2
2
= 4.8 N ◊ m
Position 2. Bar AB is horizontal.
V2 = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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68
PROBLEM 17.44 (Continued)
Kinematics.
w AB =
vB
v
= B
LAB 0.45
w BC =
vB
v
= B
LBC 0.75
1
vB
2
1
1
1 2
T2 = I Aw 2AB + mBC v 2 + I w BC
2
2
2
v=
2
2
1 Ê1 ˆ
1
1
Ê v ˆ
Ê v ˆ
= (0.081) Á B ˜ + (2) Á v B ˜ + (0.09375) Á B ˜
Ë 0.75 ¯
Ë 0.45 ¯
2 Ë2 ¯
2
2
2
= 0.53333 v B2
Conservation of energy.
T1 + V1 = T2 + V2 : 7.0632 + 4.8 = 0.53333v B2 + 0
v B = 4.7163 m/s
vB = 4.72 m/s Ø PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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69
PROBLEM 17.45
The 4-kg rod AB is attached to a collar of negligible mass at A and to a flywheel
at B. The flywheel has a mass of 16 kg and a radius of gyration of 180 mm. Knowing
that in the position shown the angular velocity of the flywheel is 60 rpm clockwise,
determine the velocity of the flywheel when Point B is directly below C.
SOLUTION
Moments of inertia.
Rod AB:
Flywheel:
1
mAB L2AB
12
1
= ( 4 kg)(0.72 m)2
12
= 0.1728 kg ◊ m2
I AB =
I C = mk 2
= (16 kg)(0.18 m)2
= 0.5184 kg ◊ m2
Position 1. As shown.
w = w1
0.12
b = 19.471∞
0.72
1
h1 = (0.72) cos b = 0.33941 m
2
V1 = W AB h1
= ( 4)(9.81)(0.33941)
= 13.3185 J
sin b =
Kinematics.
Bar AB is in translation.
v B = rw1 = 0.24w1
w AB = 0,
v = vB
1
1
1
mAB v 2 + IABw 2AB + I C w12
2
2
2
1
1
= ( 4)(0.24w1 )2 + 0 + (0.5184)w12
2
2
2
= 0.3744w1
T1 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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70
PROBLEM 17.45 (Continued)
Position 2. Point B is directly below C.
1
LAB - r
2
1
= (0.72) - 0.24
2
= 0.12 m
V2 = WAB h2
h2 =
= ( 4)(9.81)(0.12)
= 4.7088 J
Kinematics.
v B = rw 2 = 0.24w 2
vB
= 0.33333w 2
0.72
1
v = v B = 0.12w 2
2
1
1
1
T2 = mAB v 2 + IABw 2AB + I C w 22
2
2
2
1
1
1
= ( 4)(0.12w 2 )2 + (0.1728)(0.33333w 2 )2 + (0.5184)w 22
2
2
2
2
= 0.2976w 2
w AB =
Conservation of energy.
T1 + V1 = T2 + V2 : 0.3744w12 + 13.3185 = 0.2976w 22 + 4.7088
Angular speed data:
w1 = 60 rpm = 2p rad/s
Solving Equation (1) for w 2 ,
w 2 = 8.8655 rad/s
w 2 = 84.7 rpm
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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71
PROBLEM 17.46
If in Problem 17.45 the angular velocity of the flywheel is to be the same in the
position shown and when Point B is directly above C, determine the required value
of its angular velocity in the position shown.
SOLUTION
Moments of inertia.
Rod AB:
Flywheel:
I AB = mAB L2AB
1
= 4 kg (0.72 m)2
12
= 0.1728 kg ◊ m2
I C = mk 2
= 16 kg(0.18 m)2
= 0.5184 kg ◊ m2
Position 1. As shown.
w = w1
0.12
0.72
b = 19.471∞
1
h1 = 0.72 cos b = 0.33941 m
2
V1 = W AB h1 = ( 4)(9.81)(0.33941)
= 13.3185
sin b =
Kinematics.
v B = rw 1 = 0.24w 1
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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72
PROBLEM 17.46 (Continued)
Bar AB is in translation.
w AB = 0, v = v B
1
1
1
mAB v 2 + I ABw 2AB + I C w12
2
2
2
1
1
= ( 4)(0.24w 1)2 + 0 + (0.55901)w 12
2
2
2
= 0.3744w1
T1 =
Position 2. Point B is directly above C.
1
LAB + r
2
1
= (0.72) + 0.24
2
= 0.6 m
V2 = W AB h2
h2 =
= ( 4)(9.81)(0.6)
= 23.544 J
Kinematics.
v B = rw 2 = 0.24w 2
vB
= 0.33333w 2
0.72
1
v = v B = 0.12w 2
2
1 W AB 2 1
1
T2 =
v + I ABw 2AB + I C w 22
2 g
2
2
1 8
1
=
(0.500w 2 )2 + (0.186335)(0.33333w 2 )2
2
2 32.2
1
+ (0.55901)w 22
2
= 0.320913w 22
w AB =
Conservation of energy.
Angular speed data:
Then,
T1 + V1 = T2 + V2 : 0.3744w12 + 13.3135 = 0.2976w 22 + 23.544
(1)
w 2 = w1
0.0760w12 = + 0.4105
w1 = 11.602 rad/s
w1 = 110.8 rpm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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73
PROBLEM 17.47
The 80-mm-radius gear shown has a mass of 5 kg and a centroidal radius
of gyration of 60 mm. The 4-kg rod AB is attached to the center of the gear
and to a pin at B that slides freely in a vertical slot. Knowing that the
system is released from rest when q = 60∞, determine the velocity of the
center of the gear when q = 20∞.
SOLUTION
Kinematics.
vA = v A ¨
vB = v B Ø
Point D is the instantaneous center of rod AB.
vA
L cos q
v B = ( L sin q )w AB = v A tan q
vA
L
vG = w AB =
2
2 cos q
w AB =
Gear A effectively rolls with slipping, with Point C being the contact point.
vC = 0
vA
.
r
Potential energy: Use the level of the center of gear A as the datum.
Angular velocity of gear
wA =
1
ˆ
ÊL
V = -WAB Á cos q ˜ = - mAB gL cos q
¯
Ë2
2
Kinetic energy:
Masses and moments of inertia:
1
1
1
1
mA v 2A + I Aw 2A + mAB vq2 + I ABw 2AB
2
2
2
2
mA = 5 kg, mAB = 4 kg
T=
I A = mA k 2 = (5)(0.060)2 = 0.018 kg ◊ m2
1
1
I AB = mAB L2 = ( 4)(0.320)2 = 0.03413 kg ◊ m2
12
12
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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74
PROBLEM 17.47 (Continued)
Conservation of energy:
Position 1:
Position 2:
T1 + V1 = T2 + V2
q = 60∞
v A = 0 T1 = 0
1
V1 = - ( 4)(9.81)(0.320) cos 60∞
2
= -3.1392 J
q = 20∞ v A = ?
2
1
1
1 Ê vA ˆ
Ê v ˆ
T2 = (5)v 2A + (0.018) Á A ˜ + ( 4) Á
˜
Ë 0.080 ¯
2
2
2 Ë 2 cos 20∞ ¯
vA
1
Ê
ˆ
+ (0.03413) Á
Ë 0.320 cos 20∞ ˜¯
2
2
2
= (2.5 + 1.40625 + 0.56624 + 0.18875)v 2A
= 4.66124 v 2A
1
V2 = ( 4)(9.81)(0.320) cos 20∞
2
= -5.8998 J
Conservation of energy:
T1 + V1 = T2 + V2
0 - 3.1392 = 4.66124v 2A - 5.8998
v 2A = 0.59225 m2 /s2
v A = 0.770 m/s
vA = 0.770 m/s ¨ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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75
PROBLEM 17.48
The motor shown rotates at a frequency of 22.5 Hz and runs
a machine attached to the shaft at B. Knowing that the motor
develops 3 kW, determine the magnitude of the couple exerted
(a) by the motor on pulley A, (b) by the shaft on pulley B.
SOLUTION
Ê 2p rad ˆ
w A = 22.5 Hz Á
Ë cycle ˜¯
= 45p rad/s
rAw A = rBw B : (0.03 m)( 45p rad/s) = (0.180 m)w B
w B = 7.5p rad/s
(a)
Power = M Aw A
Pulley A:
3000 W = M A ( 45p rad/s)
M A = 21.2 N ◊ m
(b)
Power = M Bw B
Pulley B:
3000 W = M B (7.5p rad/s)
M B = 127.3 N ◊ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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76
PROBLEM 17.49
Knowing that the maximum allowable couple that can be applied to a shaft is 2000 N · m determine the
maximum power (m kW) that can be transmitted by the shaft at (a) 180 rpm, (b) 480 rpm.
SOLUTION
M = 2000 N ◊ m
(a)
Ê 2p ˆ
w = 180 rpm Á ˜
Ë 60 ¯
= 6p rad/s
Power = M w
= (2000 N ◊ m)(6p rad/s) = 37699.1 W
= 37.699 kW
37.7 kW (b)
Ê 2p ˆ
w = 480 rpm Á ˜
Ë 60 ¯
= 16p rad/s
Power = Mw
= (2000 N ◊ m)(16p rad/s) = 100531 W
= 100.531 kW
100.5 kW PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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77
PROBLEM 17.50
Three shafts and four gears are used to form a gear train which
will transmit 7.5 kW from the motor at A to a machine tool at F.
(Bearings for the shafts are omitted in the sketch.) Knowing that
the frequency of the motor is 30 Hz, determine the magnitude of
the couple which is applied to shaft (a) AB, (b) CD, (c) EF.
SOLUTION
w AB = 30 Hz
= 30(2p ) rad/s
= 60p rad/s
rB = 75 mm
rC = 180 mm
Kinematics.
Gears B and C.
rBw AB = rC w CD : (75 mm)(60p rad/s) = (180 mm)(w CD )
w CD = 25p rad/s
Gears D and E.
rDw CD
rD = 75 mm
rE = 180 mm
= rE w EF : (75 mm)(25p rad/s) = (180 mm)(w EF )
w EF = 10.4167p rad/s
Power = 7.5 kW
(a)
Shaft AB.
Power = M ABw AB : 7500 W = M AB (60p rad/s)
M AB = 39.8 N ◊ m (b)
Shaft CD.
Power = M CDw CD: 7500 W = M CD (25p rad/s)
M CD = 95.5 N ◊ m (c)
Shaft EF.
Power = M EF w EF : 7500 W = M EF (10.4167p rad/s)
M EF = 229 N ◊ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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78
PROBLEM 17.51
The shaft-disk-belt arrangement shown is used to transmit 2.4 kW from Point A
to Point D. Knowing that the maximum allowable couples that can be applied
to shafts AB and CD are 25 N ◊ m and 80 N ◊ m, respectively, determine the
required minimum speed of shaft AB.
SOLUTION
Power.
2.4 kW = 2400 W
M AB ⬍ 25 N ◊ m
P = M ABw AB
P
2400
min w AB =
=
= 96 rad/s
max M AB
25
M CD ⬍ 80 N ◊ m
P = M CDw CD
min w CD =
Kinematics.
P
2400
=
= 30 rad/s
max M CD
80
rAw AB = rC w CD
min w AB =
rC
(min w CD )
rA
Ê 120 ˆ
(30)
=Á
Ë 30 ˜¯
= 120 rad/s
Choose the larger value for min w AB .
minw AB = 120 rad/s
minw AB = 1146 rpm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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79
PROBLEM 17.52
The rotor of an electric motor has a mass of 25 kg and a radius of gyration of 180 mm. It is observed that
4.2 min is required for the rotor to coast to rest from an angular velocity of 3600 rpm. Determine the average
magnitude of the couple due to kinetic friction in the bearings of the motor.
SOLUTION
t = 4.2 min = 252 s
Time.
I G = mk 2
Moment of inertia.
= (25)(0.18 m)2
= 0.81 kg ◊ m2
w1 = 3600 rev/min
= 376.99 rad/s
w2 = 0
Angular velocities.
Principle of impulse and momentum.
Syst. Momenta1 +
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
I Gw1 - Mt = 0
Moments about A.
I Gw1 (0.81 kg ◊ m )(376.99 rad/s)
=
t
252 s
2
Average magnitude of couple.
M=
M = 1.212 N ◊ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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80
PROBLEM 17.53
A 2000-kg flywheel with a radius of gyration of 700 mm is allowed to coast to rest from an angular velocity of
450 rpm. Knowing that kinetic friction produces a couple of magnitude 16 N · m determine the time required
for the flywheel to coast to rest.
SOLUTION
I = mk 2
Moment of inertia.
= (2000 kg)(0.7 m)2
= 980 kg ◊ m2
Ê 2p ˆ
w 1 = 450 rpm Á ˜
Ë 60 ¯
Angular velocities.
= 47.125 rad/s
M = 16 N ◊ m
Principle of impulse and momentum.
Syst. Momenta1
+
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
Required time.
+
Moments about A:
I w1 - Mt = 0
Iw1
M
(980 kg ◊ m 2 )( 47.125 rad/s)
=
16 N ◊ m
= 2886.4 s
t=
Ê min ˆ
t = 2886.4 s Á
Ë 60 s ˜¯
t = 48 min 6 sec PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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81
PROBLEM 17.54
Two disks of the same thickness and same material are attached to a shaft
as shown. The 4-kg disk A has a radius rA = 100 mm and disk B has a
radius rB = 150 mm. Knowing that a couple M of magnitude 2.5 N · m is
applied to disk A when the system is at rest, determine the time required
for the angular velocity of the system to reach 960 rpm.
SOLUTION
2
Êr ˆ
mB = Á B ˜ mA
Ë rA ¯
Mass of disk B.
2
Ê 0.15 m ˆ
=Á
( 4 kg)
Ë 0.1 m ˜¯
= 9 kg
I = IA + IB
1
1
= ( 4 kg)(0.1 m)2 + (9 kg)(0.15 m)2
2
2
2
= 0.12125 kg ◊ m
Moment of inertia.
Angular velocity.
Ê 2p ˆ
w 2 = 960 rpm Á ˜ = 100.53 rad/s
Ë 60 ¯
Moment.
M = 2.5 N ◊ m
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about C:
Required time.
+
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
0 + Mt = I w 2
I w2
M
(0.12125 kg ◊ m 2 )(100.53 rad/s)
=
2.5 N ◊ m
t = 4.8757 s
t=
t = 4.88 s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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82
PROBLEM 17.55
Two disks of the same thickness and same material are attached to a shaft
as shown. The 3-kg disk A has a radius rA = 100 mm, and disk B has a
radius rB = 125 mm. Knowing that the angular velocity of the system is to
be increased from 200 rpm to 800 rpm during a 3-s interval, determine the
magnitude of the couple M that must be applied to disk A.
SOLUTION
2
Êr ˆ
mB = Á B ˜ mA
Ë rA ¯
Mass of disk B.
2
Ê 125 mm ˆ
=Á
3 kg
Ë 100 mm ˜¯
= 4.6875 kg
I = IA + IB
1
1
= (3 kg)(0.1 m)2 + ( 4.6875 kg)(0.125 m)2
2
2
2
= 0.05162 kg ◊ m
Moment of inertia.
Ê 2p ˆ
w1 = 200 rpm Á ˜ = 20.944 rad/s
Ë 60 ¯
Angular velocities.
Ê 2p ˆ
w 2 = 800 rpm Á ˜ = 83.776 rad/s
Ë 60 ¯
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about B:
Couple M.
+
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
I w 1 + Mt = I w 2
M=
=
I
(w 2 - w 1)
t
0.05162 kg ◊ m2
(83.776 rad/s - 20.944 rad/s)
3s
M = 1.081 N ◊ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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83
PROBLEM 17.56
A cylinder of radius r and weight W with an initial counterclockwise angular
velocity w 0 is placed in the corner formed by the floor and a vertical wall.
Denoting by m k the coefficient of kinetic friction between the cylinder and the
wall and the floor derive an expression for the time required for the cylinder to
come to rest.
SOLUTION
Since the cylinder’s centre does not move, the center is the instantaneous center. Hence at both the contacts, the
contact points move w.r.t. ground and wall and hence friction force at each contact = mkN.
1
I = mr 2 , W = mg
For the cylinder
2
Principle of impulse and momentum.
Syst. Momenta1 +
Linear momentum
+
Linear momentum
+
:
:
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
0 + N B t - FAt = 0
N B = FA
0 + N At + FB t - Wt = 0
N A + FB = N A + m k N B
= N A + m FA + N A + m k2 N A = W
W
NA =
1 + m k2
mW
FA = m k N A = k 2
1 + mk
mW
NB = k 2
1 + mk
FB =
+
Moments about G:
m k2W
1 + m k2
I w 0 - FA rt - FB rt = 0
t=
I w0
(1 + m k2 ) I w 0
=
( FA + FB )r m k (1 + m k )Wr
t=
1 + m k2
rw 0
2 m k (1 + m k ) g
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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84
PROBLEM 17.57
A 3-kg cylinder of radius r = 125 mm with an initial counterclockwise angular
velocity w 0 = 90 rad/s is placed in the corner formed by the floor and a vertical
wall. Knowing that the coefficient of kinetic friction is 0.10 between the cylinder
and the wall and the floor, determine the time required for the cylinder to come
to rest.
SOLUTION
Since the cylinder’s centre does not move, the center is the instantaneous center. Hence at both the contacts, the
contact points move w.r.t. ground and wall and hence friction force at each contact = mkN.
1
For the cylinder
I = mr 2 , W = mg
2
Principle of impulse and momentum.
Syst. Momenta1
Linear momentum
+
+
Linear momentum
:
:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
0 + N B t - FAt = 0
N B = FA
0 + N At + FB t - Wt = 0
N A + FB = N A + m k N B
= N A + m FA + N A + m k2 N A = W
W
NA =
1 + m k2
mW
FA = m k N A = k 2
1 + mk
m kW
NB =
1 + m k2
FB =
+
Moments about G:
m k2W
1 + m k2
I w 0 - FA rt - FB rt = 0
t=
=
rw 0
1 + m k2
2 m k (1 + m k ) g
1 + (0.1)2 (0.125)(90)
2(0.1)(1 + 0.1)
9.81
t = 5.26 s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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85
PROBLEM 17.58
A disk of constant thickness, initially at rest, is placed in contact with a belt that
moves with a constant velocity v. Denoting by m k the coefficient of kinetic
friction between the disk and the belt, derive an expression for the time required
for the disk to reach a constant angular velocity.
SOLUTION
1 2
mr
2
v
w2 =
r
I =
Moment of inertia.
Final state of constant angular velocity.
+
Syst. Momenta1 +
y components:
+
Moments about A:
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
0 + Nt - mgt = 0 N = mg
0 + m k Ntr = I w 2
t=
1
mr 2 vr
I w2
v
= 2
=
m k mgr m k mgr 2m k g
t=
v
2m k g
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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86
PROBLEM 17.59
Disk A, of weight 2.5 kg and radius r = 100 mm is at rest when it is placed in
contact with a belt which moves at a constant speed v = 15 m/s. Knowing that
m k = 0.20 between the disk and the belt, determine the time required for the
disk to reach a constant angular velocity.
SOLUTION
1 2
mr
2
v
w2 =
r
I =
Moment of inertia.
Final state of constant angular velocity.
+
Syst. Momenta1
y components:
+
Moments about A:
Data:
+
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
0 + Nt - mgt = 0 N = mg
0 + m k Ntr = I w 2
t=
1
mr 2 vr
I w2
v
= 2
=
m k mgr m k mgr 2m k g
t=
v
2mk g
v = 15 m/s
m k = 0.20
t=
15
(2)(0.20)(9.81)
t = 3.82 s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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87
PROBLEM 17.60
The 350-kg flywheel of a small hoisting engine has a radius of gyration of
600 mm. If the power is cut off when the angular velocity of the flywheel
is 100 rpm clockwise, determine the time required for the system to come
to rest.
SOLUTION
v B = rw
Kinematics.
Syst. Momenta1 +
+
Moments about A:
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
I Aw1 + mB ( v B )1 r - WB tr = I Aw 2 + mB ( v B )2 r
( mA k 2 + mB r 2 )w1 - mB gtr = ( mA k 2 + mB r 2 )w 2
t=
Data:
( mA k 2 + mB r 2 )(w1 - w 2 )
mB rg
mA = 350 kg
k = 600 mm = 0.6 m
mB = 120 kg
r = 225 mm = 0.225 m
w 1 = 100 rpm = 10.472 rad/s
w2 = 0
t=
[(350)(0.6)2 + (120)(0.225)2 ](10.472 - 0)
t = 5.22 s (120)(0.225)(9.81)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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88
PROBLEM 17.61
In Problem 17.60, determine the time required for the angular velocity of the
flywheel to be reduced to 40 rpm clockwise.
PROBLEM 17.60 The 350-kg flywheel of a small hoisting engine has
a radius of gyration of 600 mm. If the power is cut off when the angular
velocity of the flywheel is 100 rpm clockwise, determine the time required
for the system to come to rest.
SOLUTION
v B = rw
Kinematics.
Syst. Momenta1
+
Moments about A:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
I Aw1 + mB ( v B )1 r - WB tr = I Aw 2 + mB ( v B )2 r
( mA k 2 + mB r 2 )w1 - mB gtr = ( mA k 2 + mB r 2 )w 2
t=
Data:
( mA k 2 + mB r 2 )(w1 - w 2 )
mB rg
mA = 350 kg
k = 600 mm = 0.6 m
mB = 120 kg
r = 225 mm = 0.225 m
w 1 = 100 rpm = 10.472 rad/s
w 2 = 40 rpm = 4.189 rad/s
t=
[(350)(0.6)2 + (120)(0.225)2 ](10.472 - 4.189)
(120)(0.225)(9.81)
t = 3.13 s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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89
PROBLEM 17.62
A tape moves over the two drums shown. Drum A weighs 0.6 kg and has
a radius of gyration of 20 mm, while drum B weighs 1.75 kg and has a
radius of gyration of 30 mm. In the lower portion of the tape the tension
is constant and equal to TA = 4 N. Knowing that the tape is initially at
rest, determine (a) the required constant tension TB if the velocity of the
tape is to be v = 3 m/s after 0.24 s, (b) the corresponding tension in the
portion of tape between the drums.
SOLUTION
Kinematics.
Drums A and B rotate about fixed axes. Let v be the tape velocity in m/s.
Moments of inertia.
v = rAw A = 0.025 w A
w A = 40 v
v = rBw B = 0.04 w B
w B = 25 v
I A = mA k A2 = (0.6 kg)(0.02 m)2 = 2.4 ¥ 10 -4 kg ◊ m 2
I B = mB k B2 = (1.75 kg)(0.03 m)2 = 15.75 ¥ 10 -4 kg ◊ m2
State 1.
t=0
v=0
wA = wB = 0
State 2.
t = 0.24 s,
v = 3 m/s
w A = ( 40)(3) = 120 rad/s
w B = (25)(3) = 75 rad/s
Drum A.
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
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90
PROBLEM 17.62 (Continued)
+
Moments about A:
0 + rATAB t - rATAt = I Aw A
0 + (0.025)(TAB t ) - (0.025)( 4)(0.24) = (2.4 ¥ 10 -4 )(120)
TAB t = 2.112 N ◊ s
Drum B.
Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2
+
(a)
(b)
Moments about B:
0 + rBTB t - rBTAB t = I Bw B
0 + (0.04)(TB t ) - (0.04)(2.112) = (15.75 ¥ 10 -4 )(75)
TB t = 5.065125 N ◊ s
T t 5.065125
TB = B =
0.24
t
T t 2.112
TAB = AB =
0.24
t
TB = 21.1 N TAB = 8.80 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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91
PROBLEM 17.63
Disk B has an initial angular velocity w 0 when it is brought into contact
with disk A which is at rest. Show that the final angular velocity of disk B
depends only on w 0 and the ratio of the masses mA and mB of the two disks.
SOLUTION
Let Points A and B be the centers of the two disks and Point C be the contact point between the two disks.
Let w A and w B be the final angular velocities of disks A and B, respectively, and let vC be the final velocity at
C common to both disks.
Kinematics: No slipping
vC = rAw A = rBw B
Moments of inertia. Assume that both disks are uniform cylinders.
IA =
1
mA rA2
2
IB =
1
mB rB2
2
Principle of impulse and momentum.
Disk A
Disk B
Syst. Momenta1 + Syst. Ext. Imp. 1Æ2
Disk A:
Moments about A:
=
Syst. Momenta 2
0 + rA Ft = I Aw A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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92
PROBLEM 17.63 (Continued)
Ft =
I Aw A 1 mA rA2 vC
=
rA
2 rA2
1
mA vC
2
1
= mA rBw B
2
=
Disk B:
Moments about B:
I Bw 0 - rB Ft = I Bw B
1
ˆ 1
Ê1
mB rB2w 0 - rB Á mA rBw B ˜ = mrB2w B
¯ 2
Ë2
2
wB =
w0
m
1 + mA
B
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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93
PROBLEM 17.64
The 4 kg disk A has a radius rA = 150 mm and is initially at rest. The 5 kg
disk B has a radius rB = 200 mm and an angular velocity w 0 of 900 rpm
when it is brought into contact with disk A. Neglecting friction in the
bearings, determine (a) the final angular velocity of each disk, (b) the
total impulse of the friction force exerted on disk A.
SOLUTION
Let Points A and B be the centers of the two disks and Point C be the contact point between the two disks.
Let w A and w B be the final angular velocities of disks A and B, respectively, and let vC be the final velocity at
C common to both disks.
vC = rAw A = rBw B
Kinematics: No slipping
Moments of inertia. Assume that both disks are uniform cylinders.
IA =
1
mA rA2
2
IB =
1
mB rB2
2
Principle of impulse and momentum.
Disk A
Disk B
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
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94
PROBLEM 17.64 (Continued)
Disk A:
Moments about A:
0 + rA Ft = I Aw A
I w
Ft = A A
rA
=
1 mA rA2 vC 1
= mA vC
2 rA2
2
1
mA rBw B
2
I Bw 0 - rB Ft = I Bw B
=
Disk B:
Moments about B:
1
ˆ 1
Ê1
mB rB2w 0 - rB Á mA rBw B ˜ = mrB2w B
¯ 2
Ë2
2
w0
wB =
m
1 + mA
B
Data:
mA = 4 kg
mA 4
= = 0.8
mB 5
rB = 0.2 m
rB 200 mm 4
=
=
rA 150 mm 3
w 0 = 900 rpm = 30p rad/s
(a)
w0
1 + 0.8
30p
=
1.8
= 52.3599 rad/s
wB =
wA =
rB
wB
rA
Ê 4ˆ
= Á ˜ (52.3599)
Ë 3¯
= 69.8132 rad/s
(b)
Ê 1ˆ
Ft = Á ˜ ( 4)(0.2)(52.3599)
Ë 2¯
w A = 667 rpm
w B = 500 rpm
Ft = 20.9 N ◊ s ≠
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95
PROBLEM 17.65
Show that the system of momenta for a rigid slab in plane motion reduces to a single vector, and express the
distance from the mass center G to the line of action of this vector in terms of the centroidal radius of gyration k
of the slab, the magnitude v of the velocity of G, and the angular velocity w.
SOLUTION
Syst. Momenta
=
Components parallel to mv:
mv = X
Moments about G:
I w = ( mv ) d
d=
I w mk 2w
=
mv
mv
Single Vector
X = mv d=
k 2w
v
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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96
PROBLEM 17.66
Show that, when a rigid slab rotates about a fixed axis through O perpendicular
to the slab, the system of the momenta of its particles is equivalent to a single
vector of magnitude mrw, perpendicular to the line OG, and applied to a Point P
on this line, called the center of percussion, at a distance GP = k 2/r from the
mass center of the slab.
SOLUTION
v = rw
Kinematics. Point O is fixed.
System momenta.
X = mv = mr w
Components parallel to mv:
Moments about G:
X = mr w (GP ) X = I w
(GP )m r w = mk 2w
(GP ) =
k2
r
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97
PROBLEM 17.67
Show that the sum HA of the moments about a Point A of the momenta of the particles of a rigid slab in plane
motion is equal to IA w, where w is the angular velocity of the slab at the instant considered and IA the moment
of inertia of the slab about A, if and only if one of the following conditions is satisfied: (a) A is the mass center
of the slab, (b) A is the instantaneous center of rotation, (c) the velocity of A is directed along a line joining
Point A and the mass center G.
SOLUTION
Kinematics.
Let
w = wk
and
rG/ A = rG/ A
Then,
vG/ A = w ¥ rG/ A = (w rG/ A )
q
Where
b = q + 90∞
Also
v = vA + vG/ A
Define
h = rG/ A ¥ vG/ A
b
h = ( rG/ A )( vG/ A )k = ( rG/ A )2 w k = ( rG/ A )2 w
System momenta. Moments about A:
H A = rG/ A ¥ mv + I w
= rG/ A ¥ m( vA + vG/ A ) + I w
= rG/ A ¥ mvA + mrG/ A ¥ vG/ A + I w
= rG/ A ¥ mvA + mh + I w
= rG/ A ¥ mvA + mrG2/ Aw + I w
= rG/ A ¥ mvA + ( m rG2/ A + I ) w
The first term on the right hand side is equal to zero if
(a)
rG/ A = 0
or
(b)
vA = 0
or
(c)
(A is the mass center)
(A is the instantaneous center of rotation)
rG/ A is perpendicular to vA .
In the second term,
mrG2/ A + I = I A
H A = I Aw
by the parallel axis theorem. Thus,
when one or more of the conditions (a), (b) or (c) is satisfied.
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98
PROBLEM 17.68
Consider a rigid slab initially at rest and subjected to an impulsive force F contained in the
plane of the slab. We define the center of percussion P as the point of intersection of the
line of action of F with the perpendicular drawn from G. (a) Show that the instantaneous
center of rotation C of the slab is located on line GP at a distance GC = k 2 /GP on
the opposite side of G. (b) Show that if the center of percussion were located at C the
instantaneous center of rotation would be located at P.
SOLUTION
(a)
Locate the instantaneous center C corresponding to center of percussion P. Let dP = GP.
Syst. Momenta1
+
Components parallel to FDt :
Moments about G:
Syst. Ext. Imp. 1Æ2
Syst. Momenta 2
0 + F Dt = mv
0 + dP ( F Dt ) = I w
v
I
=
w md P
Eliminate F Dt to obtain
=
k2
dP
GC = dC =
Kinematics. Locate Point C.
(b)
=
v k2
=
w dP
GC =
k2
GP
Place the center of percussion at P ¢ = C. Locate the corresponding instantaneous center C ¢. Let
dP ¢ = GP ¢ = GC = dC .
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
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99
PROBLEM 17.68 (Continued)
Components parallel to FDt :
Moments about G:
0 + F Dt = mv ¢
0 + dP ¢ ( F Dt ) = I w ¢
Eliminate F Dt to obtain
Kinematics. Locate Point C ¢.
Using dC = dP ¢ =
v¢
I
k2
=
=
w ¢ md P ¢ d P ¢
GC ¢ = dC ¢ =
k2
gives
dP
v¢ k 2 k 2
=
=
w ¢ dP ¢ dC
dC ¢ = dP or GC ¢ = GP Thus Point C ¢ coincides with Point P.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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100
PROBLEM 17.69
A wheel of radius r and centroidal radius of gyration k is released from rest on
the incline shown at time t = 0. Assuming that the wheel rolls without sliding,
determine (a) the velocity of its center at time t, (b) the coefficient of static
friction required to prevent slipping.
SOLUTION
Kinematics. Rolling motion. Instantaneous center at C.
v = vG = rw
w=
v
r
I = mk 2
Moment of inertia.
Kinetics.
Syst. Momenta1
moments about C:
+
Syst. Ext. Imp. 1Æ2
Syst. Momenta 2
0 + ( mgt )r sin b = mvr + I w
( mgr sin b )t = mrv +
(a)
=
mk 2 v
r
v=
Velocity of Point G.
+
r 2 gt sin b
r2 + k 2
b components parallel to incline:
0 + mgt sin b - Ft = mv
mr 2 gt sin b
r2 + k 2
k 2 mgt sin b
Ft = mgt sin b =
r2 + k 2
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101
PROBLEM 17.69 (Continued)
+
components normal to incline:
0 + Nt - mgt cos b = 0
Nt = mgt cos b
(b)
Required coefficient of static friction.
F
N
Ft
=
Nt
ms ³
=
k 2 mgt sin b
( r 2 + k 2 )mgt cos b
ms ³
k 2 tan b
r2 + k 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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102
PROBLEM 17.70
A flywheel is rigidly attached to a 40 mm-radius shaft that rolls without sliding
along parallel rails. Knowing that after being released from rest the system
attains a speed of 150 mm/s in 30 s, determine the centroidal radius of gyration
of the system.
SOLUTION
Kinematics. Rolling motion. Instantaneous center at C.
v = vG = rw
I = mk 2
Moment of inertia.
Kinetics.
Syst. Momenta1
Moments about C:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + ( mgt )r sin b = mvr + I w
Ê
k2ˆ
mgtr sin b = m Á r + ˜ v
r ¯
Ë
Solving for k 2,
Data:
Ê gt sin b ˆ
k 2 = r2 Á
- 1˜
¯
Ë v
r = 40 mm = 0.04 m
g = 9.81 m/s2
t = 30 s
v = 150 mm/s = 0.15 m/s
È (9.81)(30)sin 15∞ ˘
- 1˙
k 2 = (0.04)2 Í
0.15
Î
˚
= 0.81088 m2
k 2 = 0.90049 m
k = 900 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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103
PROBLEM 17.71
The double pulley shown has a mass of 3 kg and a radius of gyration of
100 mm. Knowing that when the pulley is at rest, a force P of magnitude 24 N
is applied to cord B, determine (a) the velocity of the center of the pulley after
1.5 s, (b) the tension in cord C.
SOLUTION
rC = 0.150 m
For the double pulley,
rB = 0.080 m
k = 0.100 m
Principle of impulse and momentum.
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
Syst. Momenta 2
v = rC w
Kinematics. Point C is the instantaneous center.
Moments about C:
=
0 + Pt ( rC + rB ) - mgtrC = I w + mvrC
= mk 2w + m( rC w )rC
w=
=
Pt ( rC + rB ) - mgtrC
(
m k 2 + rC2
)
(24)(1.5)(0.230) - (3)(9.81)(1.5) (0.150)
3 (0.1002 + 0.1502 )
= 17.0077 rad/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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104
PROBLEM 17.71 (Continued)
(a)
v = (0.150)(17.0077) = 2.55115 m/s
Linear components:
v = 2.55 m/s ≠ 0 + Pt - mgt + Qt = mv
+
mv
+ mg - P
t
(3)(2.55115)
=
+ (3)(9.81) - 24
1.5
Q=
(b)
Q = 10.53 N Tension in cord C.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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105
PROBLEM 17.72
Two uniform cylinders, each of mass m = 7 kg and radius r = 100 mm
are connected by a belt as shown. If the system is released from rest
when t = 0, determine (a) the velocity of the center of cylinder B at
t = 3s, (b) the tension in the portion of belt connecting the two cylinders.
SOLUTION
v AB = rw B
Kinematics.
Point C is the instantaneous center of cylinder A.
v AB 1
= wB
2r 2
1
v A = rw A = rw B
2
1
I = mr 2
2
wA =
Moment of inertia.
(a)
Velocity of the center of A.
Cyl. B:
Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2
+
0 + Ptr = Iw B
Moments about B:
(1)
Cyl. A:
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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106
PROBLEM 17.72 (Continued)
Moments about C:
0 - 2 Ptr + mgtr = mv A r + I wA
Ê1 ˆ
ˆ
Ê1
0 - 2 I w B - mgtr = m Á rw B ˜ r + I Á w B ˜
Ë2 ¯
¯
Ë2
1 2ˆ
Ê5
ÁË I + mr ˜¯ w B = mgrt
2
2
Ê 5 mr 2 1 2 ˆ
Á 2 2 + 2 mr ˜ w B = mgrt
Ë
¯
7
rw B = gt
4
4 gt
wB =
7 r
1
2
2
v A = rw B = gt = (9.81)(3)
2
7
7
= 8.4086 m/s
(b)
(2)
v A = 8.41 m/s Ø Tension in the belt.
From (1) and (2)
Ê 4 gt ˆ
Ptr = I Á
Ë 7 r ˜¯
P=
4 Ig 4
4
= mg = (7)(9.81) = 39.24 N
2
7r
7
7
P = 39.2 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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107
PROBLEM 17.73
Two uniform cylinders, each of mass m = 7 kg and radius r = 100 mm
are connected by a belt as shown. Knowing that at the instant shown the
angular velocity of cylinder A is 30 rad/s counterclockwise, determine
(a) the time required for the angular velocity of cylinder A to be reduced to
5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.
SOLUTION
v AB = rw B
Kinematics.
Point C is the instantaneous center of cylinder A.
v AB 1
= wB
2r 2
1
v A = rw A = rw B
2
1W 2
r
I =
2 g
wA =
Moment of inertia.
(a)
Required time.
Cyl. B:
Syst. Momenta1
Moments about B:
+ Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
I (w B )1 - Ptr = I (w B )2
Ptr = I [(w B )1 - (w B )2 ]
1
= mr 2 [(w B )1 - (w B )2 ]
2
(1)
Cyl. A:
Syst. Momenta1 +
Syst. Ext. Imp. 1Æ2
= Syst. Momenta 2
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108
PROBLEM 17.73 (Continued)
Moments about C:
I (w A )1 + m( v A )1 r + 2 Ptr - mgtr = I (w A )2 + m( v A )2 r
1 2
mr [(w A )1 - (w A )2 + mr[(w A )1 - (w A )2 ]r + 2 Ptr - mgtr = 0
2
˘
3 2 ÈÊ 1 ˆ 1
Ï1
¸
mr ÍÁ w B ˜ - (w B )2 ˙ + 2 Ì mr 2 [(w B )1 - (w B )2 ]˝ - mgtr = 0
Ë
¯
2
2
2
2
Ó
˛
˚
Î
1
7 2
mr [(w B )1 - (w B )2 ] - mgtr = 0
4
t=
Data:
From Equation (2),
(b)
7r[(w B )1 - (w B )2 ]
4g
m = 7 kg
r = 0.1 m
(7)(0.1)(30 - 5)
t=
= 0.56612
( 4)(9.81)
(2)
t = 0.446 s Tension in belt between cylinders.
From Equation (1) and (2)
1
4g
P = ◊ m◊
2
7
1
( 4)(9.81)
= (7)
= 19.62 N
2
7
P = 19.62 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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109
PROBLEM 17.74
A 240-mm-radius cylinder of mass 8 kg rests on a 3-kg carriage. The
system is at rest when a force P of magnitude 10 N is applied as shown
for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage
and neglecting the mass of the wheels of the carriage, determine the
resulting velocity of (a) the carriage, (b) the center of the cylinder.
SOLUTION
1
mA r 2
2
1
= (8 kg)(0.24 m)2
2
= 0.2304 kg ◊ m2
I =
Moment of inertia.
Cylinder alone:
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + 0 = I w - mA v A r
Moments about C:
0 = 0.2304w - (8)(0.24) v A
or
(1)
Cylinder and carriage:
+
Syst. Momenta1
or
Horizontal components:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + Pt = mA v A + mB v B
0 + (10)(1.2) = 8 v A + 3v B
(2)
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110
PROBLEM 17.74 (Continued)
Kinematics.
v A = v B - rw
v A = v B - 0.24w
(3)
Solving Equations (1), (2) and (3) simultaneously gives
w = 5.68 rad/s
(a)
Velocity of the carriage.
vB = 2.12 m/s Æ (b)
Velocity of the center of the cylinder.
vA = 0.706 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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111
PROBLEM 17.75
A 240-mm-radius cylinder of mass 8 kg rests on a 3-kg carriage. The
system is at rest when a force P of magnitude 10 N is applied as shown for
1.2 s. Knowing that the cylinder rolls without sliding on the carriage and
neglecting the mass of the wheels of the carriage, determine the resulting
velocity of (a) the carriage, (b) the center of the cylinder.
SOLUTION
1
mA r 2
2
1
= (8 kg)(0.24 m)2
2
= 0.2304 kg ◊ m2
I =
Moment of inertia.
Cylinder alone:
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2
0 + Ptr = I w + mA v A r
Moments about C:
0 + (10)(1.2)(0.24) = 0.2304w + (8)(0.24)v A
or
(1)
Cylinder and carriage:
+
Syst. Momenta1
or
Horizontal components:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + Pt = mA v A + mB v B
0 + (10)(1.2) = 8 v A + 3v B
(2)
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112
PROBLEM 17.75 (Continued)
Kinematics.
v A = v B + rw
v A = v B + 0.24w
(3)
Solving Equations (1), (2) and (3) simultaneously gives
w = 2.21 rad/s
(a)
Velocity of the carriage.
v B = 0.706 m/s Æ (b)
Velocity of the center of the cylinder.
v A = 0.235 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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113
PROBLEM 17.76
In the gear arrangement shown, gears A and C are attached to rod ABC,
which is free to rotate about B, while the inner gear B is fixed. Knowing
that the system is at rest, determine the magnitude of the couple M which
must be applied to rod ABC, if 2.5 s later the angular velocity of the rod is
to be 240 rpm clockwise. Gears A and C have a mass of 1.25 kg each and
may be considered as disks of radius 50 mm; rod ABC weighs 2 kg.
SOLUTION
Kinematics of motion
Let w ABC = w
v A = vC = ( BC )w = 2rw
Since gears A and C roll on the fixed gear B,
w A = wC =
vC 2rw
=
= 2w
r
r
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about D:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + (Qt )r = mC vC r + I C wC
1
(Qt )r = mC (2rw )r + mC r 2 (2w )
2
Qt = 3mC rw
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
(1)
Syst. Momenta 2
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114
PROBLEM 17.76 (Continued)
+
Moments about B:
Mt - Qt ( 4r ) = I ABC w
1
mABC ( 4r )2w
12
4
Mt - 4(Qt )r = mABC r 2w
3
Mt - 4(Qt )r =
(2)
Substitute for (Qt) from (1) into (2):
4
mABC r 2w
3
4
Mt = r 2w ( mABC + 9mC )
3
Mt - 4(3mC rw )r =
(3)
Couple M.
Data:
t = 2.5 s
r = 50 mm = 0.05 m
mABC = 2 kg
mC = 1.25 kg
w = 240 rpm
= 8p rad/s
Eq. (3):
M (2.5 s) =
4
(0.05)2 (8p ) (2 + 9(1.25))
3
2.5 M = 1.1100
M = 0.444 N ◊ m
M = 0.444 N ◊ m
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115
PROBLEM 17.77
A sphere of radius r and mass m is placed on a horizontal floor with no linear velocity but
with a clockwise angular velocity w 0 . Denoting by m k the coefficient of kinetic friction
between the sphere and the floor, determine (a) the time t 1 at which the sphere will start
rolling without sliding, (b) the linear and angular velocities of the sphere at time t 1 .
SOLUTION
I =
Moment of inertia. Solid sphere.
2 2
mr
5
Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 =
Nt1 - Wt1 = 0
+
+
y components:
+
x components:
Moments about G:
Syst. Momenta 2
N = W = mg
(1)
Ft1 = mv2
(2)
I w 0 - Ft1r = I w 2
(3)
Since F = m k N = m k mg , Equation (2) gives
m k mgt1 = mv2
v2 = m k gt1
or
(4)
Using the value for I in Equation (3),
2 2
2
mr w 0 - m k mgt1r = mr 2w 2
5
5
w2 = w0 -
or
(a)
5 m k gt1
2 r
(5)
Time t1 at which sliding stops.
v2 = rw
From kinematics,
5
m k gt1 = rw 0 - m k gt1
2
t1 =
2 rw 0
7 mk g
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116
PROBLEM 17.77 (Continued)
(b)
Linear and angular velocities.
From Equation (4),
From Equation (6),
v2 = m k g
w2 =
2 rw 0
7 mk g
v2 2
= w0
r 7
v2 =
2
rw 0 Æ 7
2
w 2 = w0
7
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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117
PROBLEM 17.78
A sphere of radius r and mass m is projected along a rough horizontal surface
with the initial velocities shown. If the final velocity of the sphere is to be zero,
express (a) the required magnitude of w 0 in terms of v0 and r, (b) the time
required for the sphere to come to rest in terms of v0 and coefficient of kinetic
friction m k .
SOLUTION
I =
Moment of inertia. Solid sphere.
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2
Nt - Wt = 0
+
+
y components:
+
x components:
Moments about G:
2 2
mr
5
mv0 - Ft = 0
N = W = mg
(1)
Ft = mv0
(2)
I w 0 - Ftr = 0
(3)
2 2
mr w 0 - mv0 r = 0
5
(a)
Solving for w 0 ,
(b)
Time to come to rest.
From Equation (2),
t=
mv0
mv0
=
F
m k mg
w0 =
5 v0
2 r
t=
v0
mk g
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118
PROBLEM 17.79
A 1.25-kg disk of radius 100 mm is attached to the yoke BCD by means
of short shafts fitted in bearings at B and D. The 0.75 kg yoke has a
radius of gyration of 75 mm about the x axis. Initially the assembly is
rotating at 120 rpm with the disk in the plane of the yoke (q = 0). If
the disk is slightly disturbed and rotates with respect to the yoke until
q = 90∞, where it is stopped by a small bar at D, determine the final
angular velocity of the assembly.
SOLUTION
I C = mkC2 = (0.75 kg)(0.075 m)2 = 4.21875 ¥ 10 -3 kg ◊ m2
Moment of inertia of yoke:
Moment of inertia of disk about x axis:
q = 0: I A =
1 2
mr
4
1
(1.25)(0.1)2 = 3.125 ¥ 10 -3 kg ◊ m2
4
1 2
q = 90∞ : I A = mr
2
1
= (1.25)(0.1)2 = 6.25 ¥ 10 -3 kg ◊ m2
2
=
Total moment of inertia about the x axis:
q = 0 : ( I x )1 = I C + I A
= 7.34375 ¥ 10 -3 kg ◊ m2
q = 90∞ : ( I x )2 = I C + I A
= 10.46875 ¥ 10 -3 kg ◊ m2
Angular momentum about the x axis:
q = 0 : H1 = ( I x )1w1
= 7.34375 ¥ 10 -3 w1
q = 90∞ : H 2 = ( I x )2 w 2
= 10.46875 ¥ 10 -3w 2
Conservation of angular momentum.
H1 = H 2 : 7.34375 ¥ 10 -3 w 1 = 10.46875 ¥ 10 -3 w 2
w 2 = 0.7015w 1 = (0.7015)(120 rpm)
w 2 = 84.2 rpm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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119
PROBLEM 17.80
Two panels A and B are attached with hinges to a rectangular plate
and held by a wire as shown. The plate and the panels are made of
the same material and have the same thickness. The entire assembly
is rotating with an angular velocity w0 when the wire breaks.
Determine the angular velocity of the assembly after the panels
have come to rest against the plate.
SOLUTION
Geometry and kinematics:
Panels in up position
Panels in down position
3
v2 = b w 0
2
v0 = bw 0
Let r = mass density, t = thickness
Plate:
mplate = rt (2b)( 4b) = 8rtb2
1
(8rtb2 )[(2b)2 + ( 4b)2 ]
12
160
rtb 4
=
12
40
=
rtb 4
3
I plate =
Each panel:
Panel in up position
mpanel = rt (b)(2b) = 2rtb2
1
( I panel )0 = (2 rt b2 )(2b)2
12
8
2
= r t b 4 = rt b 4
12
3
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120
PROBLEM 17.80 (Continued)
1
(2 rt b2 )[b2 + (2b)2 ]
12
10
= rt b 4
12
5
= rt b 4
6
( I panel )1 =
Panel in down position
Conservation of angular momentum about the vertical spindle.
Initial momenta
+
Final momenta
Moments about C:
È
Ê 3b ˆ ˘
I platew 0 + 2[( I panel )0 w 0 + mpanel v0 (b)] = I platew1 + 2 Í( I panel )1w1 + mpanel v1 Á ˜ ˙
Ë 2 ¯˚
Î
40
È5
È2
ˆ Ê 3 ˆ˘
˘ 40
Ê3
rtb 4w 0 + 2 Í rtb 4w 0 + (2 rtb2 )(bw 0 )b ˙ =
rtb 4w 0 + 2 Í rtb 4w1 + 2 rtb2 Á bw 0 ˜ Á b˜ ˙
¯ Ë 2 ¯˚
Ë
2
3
Î3
˚ 3
Î6
È 40 4
˘
È 40 10
˘
4
4
ÍÎ 3 + 3 + 4 ˙˚ rt b w 0 = ÍÎ 3 + 6 + 9˙˚ rtb w1
56
w 0 = 24w 1
3
56
w1 =
(3)(24)
7
w1 = w 2 9
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121
PROBLEM 17.81
A 1.6-kg tube AB can slide freely on rod DE which in turn can rotate freely in
a horizontal plane. Initially the assembly is rotating with an angular velocity
w = 5 rad/s and the tube is held in position by a cord. The moment of inertia of the
rod and bracket about the vertical axis of rotation is 0.30 kg ◊ m2 and the centroidal
moment of inertia of the tube about a vertical axis is 0.0025 kg ◊ m2 . If the cord
suddenly breaks, determine (a) the angular velocity of the assembly after the tube
has moved to end E, (b) the energy lost during the plastic impact at E.
SOLUTION
Let Point C be the intersection of axle C and rod DE. Let Point G be the mass center of tube AB.
Masses and moments of inertia about vertical axes.
mAB = 1.6 kg
I AB = 0.0025 kg ◊ m2
I DCE = 0.30 kg ◊ m 2
State 1.
State 2.
1
( rG /A )1 = (125)
2
= 62.5 mm
w1 = 5 rad/s
( rG /A )2 = 500 - 62.5
= 437.5 mm
w = w2
Kinematics.
( vG )q = vq = rG /C w
Syst. Momenta1 +
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
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122
PROBLEM 17.81 (Continued)
Moments about C:
I ABw1 + I DCE w1 + mAB ( vq )1 ( rG /C )1 + 0 = I ABw 2 + I DCE w 2 + mAB ( vq )2 ( rG /C )2
È I AB + I DCE + mAB ( rG /G )12 ˘ w1 = È I AB + I DCE + mAB ( rG /C )22 ˘ w 2
Î
˚
Î
˚
[0.0025 + 0.30 + (1.6)(0.0625)2 ](5) = [0.0025 + 0.30 + (1.6)(0.4375)2 ]w
w2
(0.30875)(5) = 0.60875w 2
w 2 = 2.5359 rad/s
(a)
Kinetic energy.
(b)
2.54 rad/s Angular velocity after the plastic impact.
1
1
1
I ABw 2 + I DCE w 2 + mAB v 2
2
2
2
1
1
1
T1 = (0.0025)(5)2 + (0.30)(5)2 + (1.6)(0.0625)2 (5)2
2
2
2
= 3.859375 J
1
1
1
T2 = (0.0025)(2.5359)2 + (0.30)(2.5359)2 + (1.6)(0.4375)2 (2.5359)2
2
2
2
= 1.9573 J
T=
T1 - T2 = 1.902 J Energy lost.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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123
PROBLEM 17.82
Two 0.4-kg balls are to be put successively into the center C of the slender
2-kg tube AB. Knowing that when the first ball is put into the tube the initial
angular velocity of the tube is 8 rad/s and neglecting the effect of friction,
determine the angular velocity of the tube just after (a) the first ball has left
the tube, (b) the second ball has left the tube.
SOLUTION
Conservation of angular momentum about C.
I w1 = I w 2 + mrvq
= I w 2 + mr ( rw 2 )
I
w1
I + mr 2
= Cw1
w2 =
Data:
I =
(1)
1
mtube L2
12
1
(2)(1)2
12
1
= kg ◊ m2
6
(one ball) m = 0.4 kg
r = 500 mm = 0.5 m
1
I + mr 2 = + (0.4)(0.5)2
6
4
=
kg ◊ m2
15
=
C=
( 16 ) = 0.625
(154 )
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124
PROBLEM 17.82 (Continued)
(a)
First ball moves through the tube.
w1 = 8 rad/s
By Equation (1),
(b)
w 2 = (0.625)(8)
w 2 = 5.00 rad/s Second ball moves through the tube.
w1 = 5.00 rad/s
By Equation (1),
w 2 = (0.625)(5.00)
w 2 = 3.13 rad/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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125
PROBLEM 17.83
A 3-kg rod of length 800 mm can slide freely in the 240-mm cylinder DE, which
in turn can rotate freely in a horizontal plane. In the position shown the assembly
is rotating with an angular velocity of magnitude w = 40 rad/s and end B of the
rod is moving toward the cylinder at a speed of 75 mm/s relative to the cylinder.
Knowing that the centroidal mass moment of inertia of the cylinder about a
vertical axis is 0.025 kg ◊ m2 and neglecting the effect of friction, determine the
angular velocity of the assembly as end B of the rod strikes end E of the cylinder.
SOLUTION
Kinematics and geometry.
v1 = (0.04 m)w1 = (0.4 m)(40 rad/s)
v1 = 1.6 m/s
v2 = (0.28 m)w 2
Initial position
Conservation of angular momentum about C.
+
Moments about C:
I AB =
Final position
1
(3 kg)(0.8 m)2 = 0.16 kg ◊ m2
12
I ABw1 + mv1 (0.04 m) + I DE w1 = I ABw 2 + mv2 (0.028 m) + I DE w 2
(0.16 kg ◊ m 2 )( 40 rad/s) + (8 kg)(1.6 m/s) (0.04 m) + (0.025 kg ◊ m2 )(40 rad/s)
= (0.16 kg ◊ m2 )w 2 + (3 kg)(0.28w 2 )(0.28) + (0.025 kg ◊ m2 )w 2
(6.4 + 0.192 + 1.00) = (0.16 + 0.2352 + 0.025)w 2
7.592 = 0.4202w 2 ; w 2 = 18.068 rad/s
w 2 = 18.07 rad/s Angular velocity.
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126
PROBLEM 17.84
In the helicopter shown, a vertical tail propeller is used to prevent
rotation of the cab as the speed of the main blades is changed.
Assuming that the tail propeller is not operating, determine the final
angular velocity of the cab after the speed of the main blades has
been changed from 180 to 240 rpm. (The speed of the main blades is
measured relative to the cab, and the cab has a centroidal moment of
inertia of 1000 kg ◊ m2 . Each of the four main blades is assumed to
be a slender 4.2-m rod of mass 25 kg.)
SOLUTION
Let W be the angular velocity of the cab and w be the angular velocity of the blades relative to the cab. The
absolute angular velocity of the blades is W + w .
w1 = 180 rpm = 6p rad/s
w 2 = 240 rpm = 8p rad/s
Moments of inertia.
Cab:
I C = 1000 kg ◊ m2
Blades:
ˆ
Ê1
I B = 4 Á mL2 ˜
¯
Ë3
Ê 1ˆ
= ( 4) Á ˜ (25)( 4.2)2
Ë 3¯
= 588 kg ◊ m2
Assume W1 = 0.
Conservation of angular momentum about shaft.
I B (w1 + W1 ) + I C W1 = I B (w 2 + W2 ) + I C W2
W2 = -
I B (w 2 - w1 )
IC + I B
(588)(8p - 6p )
(588 + 1000)
= -2.3265 rad/s
=-
W2 = -22.2 rpm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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127
PROBLEM 17.85
Assuming that the tail propeller in Problem 17.84 is operating and that the
angular velocity of the cab remains zero, determine the final horizontal
velocity of the cab when the speed of the main blades is changed from
180 to 240 rpm. The cab weighs 625 kg and is initially at rest. Also determine
the force exerted by the tail propeller if the change in speed takes place
uniformly in 12 s.
SOLUTION
Let W be the angular velocity of the cab and w be the angular velocity of the blades relative to the cab. The
absolute angular velocity of the blades is W + w .
w1 = 180 rpm = 6p rad/s
w 2 = 240 rpm = 8p rad/s
Moments of inertia.
Cab:
I C = 1000 kg ◊ m2
Blades:
Ê 1ˆ
ˆ
Ê1
I B = 4 Á mL2 ˜ = ( 4) Á ˜ (25)( 4.2)2
Ë 3¯
¯
Ë3
= 588 kg ◊ m2
The cab does not rotate.
W1 = W2 = 0
Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2
Moments about shaft:
I B (w1 + W1 ) + I C W1 + Frt = I B (w 2 + W2 ) + I C W2
Frt = I B (w 2 - w1 )
= (588)(8p - 6p )
= 3694.51 N ◊ m ◊ s
Frt 3694.51
Ft =
=
= 738.903 N ◊ s
r
5
Linear components:
(a)
Assume v1 = 0.
(b)
Force.
mv1 + Ft = mv2
Ft
738.903
v2 - v1 =
=
m 625 + ( 4)(25)
= 1.0192 m/s
v2 = 1.019 m/s F=
Ft 738.903
=
t
12
F = 61.6 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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128
PROBLEM 17.86
The 4-kg disk B is attached to the shaft of a motor mounted on plate A, which can
rotate freely about the vertical shaft C. The motor-plate-shaft unit has a moment
of inertia of 0.20 kg ◊ m2 with respect to the axis of the shaft. If the motor is
started when the system is at rest, determine the angular velocities of the disk and
of the plate after the motor has attained its normal operating speed of 360 rpm.
SOLUTION
Moments of inertia. motor-plate-shaft:
I C = 0.20 kg ◊ m2
1
mB rB2
2
1
= ( 4 kg)(0.090 m)2
2
= 0.0162 kg ◊ m2
Disk:
IB =
Kinematics.
v B = rB/C w C = 0.090w C
Kinetics.
Syst. Momenta1
Moments about C:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + 0 = I C w C + mB v B - I Bw B
I C w C + mB ( rB/C w C )rB/C - I Bw B = 0
( I C + mB rB2/C )w C = I Bw B
[0.20 + ( 4)(0.090)2 ]w C = 0.0162w B
w C = 0.069707
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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129
PROBLEM 17.86 (Continued)
Angular velocity of motor.
w M = w B + wC
= 1.06971w B
w M = 360 rpm
wM
wB =
1.06971
360 rpm
=
1.06971
= 336.54 rpm
w B = 337 rpm w C = (0.069707)(336.54 rpm)
w C = 32.5 rpm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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130
PROBLEM 17.87
The circular platform A is fitted with a rim of 200-mm inner radius and can
rotate freely about the vertical shaft. It is known that the platform-rim unit has
a mass of 5 kg and a radius of gyration of 175 mm with respect to the shaft. At
a time when the platform is rotating with an angular velocity of 50 rpm, a 3-kg
disk B of radius 80 mm is placed on the platform with no velocity. Knowing
that disk B then slides until it comes to rest relative to the platform against the
rim, determine the final angular velocity of the platform.
SOLUTION
I A = mA k 2
Moments of inertia.
= (5 kg)(0.175 m)2
= 0.153125 kg ◊ m2
1
mB rB2
2
1
= (3 kg)(0.08 m)2
2
= 9.6 ¥ 10 -3 kg ◊ m2
IB =
Syst. Momenta,
State 1 Disk B is at rest.
State 2 Disk B moves with platform A.
Kinematics.
In State 2,
v B = (0.12 m)w 2
Syst. Momenta,
Principle of conservation of angular momentum.
+
Moments about D:
I Aw1 = I Aw 2 + I Bw 2 + mB v B (0.12 m)
(0.153125 kg ◊ m2 )w1 = (0.153125 kg ◊ m2 )w 2
+ (9.6 ¥ 10 -3 kg ◊ m2 )w 2 + (3 kg)(0.12 m)2w 2
0.153125w 2 = 0.20593w1
w 2 = 0.7436w1
= 0.7436(50 rpm)
w 2 = 37.2 rpm Final angular velocity
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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131
PROBLEM 17.88
A small 2-kg collar C can slide freely on a thin ring of mass 3 kg and radius
250 mm. The ring is welded to a short vertical shaft, which can rotate freely in a
fixed bearing. Initially the ring has an angular velocity of 35 rad/s and the collar
is at the top of the ring (q = 0) when it is given a slight nudge. Neglecting the
effect of friction, determine (a) the angular velocity of the ring as the collar passes
through the position q = 90∞, (b) the corresponding velocity of the collar relative
to the ring.
SOLUTION
IR =
Moment of inertia of ring.
1
mR R2
2
Position 1
Position 1.
Position 2.
Position 2
q = 0.
vC = 0
q = 90∞
( vC ) y = v y = Rw 2
Conservation of angular momentum about y axis for system.
I Rw1 = I Rw 2 + mC v y R
1
1
mR R2w1 = mR R2w 2 + mC R2w 2
2
2
2
mR R w1 = ( mR + 2mC ) R2w 2
w2 =
mR
w1
mR + 2mC
(1)
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132
PROBLEM 17.88 (Continued)
Potential energy. Datum is the center of the ring.
V1 = mC gR V2 = 0
1
1Ê1
ˆ
I Rw12 = Á mR R2 ˜ w 12
¯
Ë
2
2 2
1
= mR R2w 12
4
1
1
T2 = I Rw 22 + mC v x2 + v 2y
2
2
1
1
1
= mR R2w 22 + mC R2 v22 + mC v 2y
4
2
2
Principle of conservation of energy:
T1 =
Kinetic energy:
(
)
T1 + V1 = T2 + V2
1
1 ˆ
1
Ê1
mR R2w12 + mC gR = Á mR + mC ˜ R2w 22 + mC v 2y
¯
Ë
4
4
2
2
(2)
mC = 2 kg
Data:
mR = 3 kg
R = 0.25 m
w1 = 35 rad/s
(a)
Angular velocity.
From Eq. (1),
(b)
w2 =
3 kg
(35 rad/s)
3 kg + 2(2 kg)
w 2 = 15.00 rad/s Velocity of collar relative to ring.
1
From Eq. (2),
(3 kg)(0.25 m)2 (35 rad/s)2 + (2 kg)(9.81 m/s2 )(0.25 m)
4
1
1
È1
˘
= Í (3 kg) + (2 kg)˙ (0.25 m)2 (15 rad/s)2 + (2 kg)v 2y
2
Î4
˚
2
57.422 + 4.905 = 24.609 + v 2y
v 2y = 37.716
v y = 6.14 m/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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133
PROBLEM 17.89
Collar C has a mass of 8 kg and can slide freely on rod AB, which in
turn can rotate freely in a horizontal plane. The assembly is rotating
with an angular velocity w of 1.5 rad/s when a spring located between
A and C is released, projecting the collar along the rod with an initial
relative speed vr = 1.5 m/s. Knowing that the combined mass moment
of inertia about B of the rod and spring is 1.2 kg ◊ m2 , determine (a) the
minimum distance between the collar and Point B in the ensuing
motion, (b) the corresponding angular velocity of the assembly.
SOLUTION
vq = rw
Kinematics.
Moments of inertia.
I = I B + mC r 2
= 1.2 + 8r 2
Conservation of angular momentum.
( H B )1 = ( H B )2 : I Bw1 + mC ( vq )1r1 = I Bw 2 + mC ( vq )2 r2
( I B + mC r12 )w1 = ( I B + mC r22 )w 2
w2 =
I1w1 = I 2w 2
I1w1
I2
Potential energy.
V1 = 0 V2 = 0
Kinetic energy.
T=
Position 1. Just after spring is released.
r = r1
vr = ( vr )1
w = w1
I = I1
Position 2. Distance r is minimum.
or
(1)
1
1
1
mC vr2 + mC vq2 + I Bw 2
2
2
2
1
1
= mC vr2 + I w 2
2
2
r = r2
( vr ) = ( vr ) = 0
w = w2
I = I2
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134
PROBLEM 17.89 (Continued)
Conservation of energy.
T1 + V1 = T2 + V2 :
1
1
1
mC ( vr )12 + I1w12 + 0 = 0 + I 2w 22
2
2
2
2
ÊIw ˆ
mC ( vr )12 = I 2 Á 1 1 ˜ - I1w12
Ë I2 ¯
ÊI
ˆ
= I1 Á 1 - 1˜ w12
Ë I2 ¯
(2)
r1 = 600 mm = 0.6 m
w1 = 15 rad/s
(vr )1 = 1.5 m
Data.
I1 = 1.2 + (8)(0.6)2
= 4.08 kg ◊ m2
Equation (2):
ÊI
ˆ
(8)(1.5)2 = 4.08 Á 1 - 1˜ (1.5)2
Ë I2 ¯
I1
= 2.9608
I2
I2 =
(a)
(b)
Equation (1):
4.08
= 1.378 = 1.2 + 8r22
2.9608
r2 = 0.14917 m
r2 = 149.2 mm 4.08
(1.5)
1.378
w 2 = 4.44 rad/s w2 =
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135
PROBLEM 17.90
In Problem 17.89, determine the required magnitude of the initial
relative speed vr if during the ensuing motion the minimum distance
between collar C and Point B is to be 300 mm.
SOLUTION
vq = rw
Kinematics.
I = I B + mC r 2
Moments of inertia.
= 1.2 + 8r 2
Conservation of angular momentum.
( H B )1 = ( H B )2 : I Bw1 + mC ( vq )1r1 = I Bw 2 + mC ( vq )2 r2
( I B + mC r12 )w1 = ( I B + mC r22 )w 2
w2 =
I1w1 = I 2w 2
I1w1
I2
Potential energy.
V1 = 0 V2 = 0
Kinetic energy.
T=
Position 1. Just after spring is released.
r = r1
vr = ( vr )1
w = w1
I = I1
Position 2. Distance r is minimum.
or
(1)
1
1
1
mC vr2 + mC vq2 + I Bw 2
2
2
2
1
1
= mC vr2 + I w 2
2
2
r = r2
( vr ) = ( vr ) = 0
w = w2
I = I2
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136
PROBLEM 17.90 (Continued)
Conservation of energy.
T1 + V1 = T2 + V2 :
1
1
1
mC ( vr )12 + I1w12 + 0 = 0 + I 2w 22
2
2
2
2
ÊIw ˆ
mC ( vr )12 = I 2 Á 1 1 ˜ - I1w12
Ë I2 ¯
ÊI
ˆ
= I1 Á 1 - 1˜ w12
Ë I2 ¯
Data.
(2)
r1 = 600 mm = 0.6 m
w1 = 1.5 rad/s
r2 = 300 mm = 0.3 m
I1 = 1.2 + (8)(0.6)2 = 4.08 kg ◊ m2
I 2 = 1.2 + (8)(0.3)2 = 1.92 kg ◊ m2
Equation (2):
Ê 4.08 ˆ
8( vr )12 = 4.08 Á
- 1 (1.5)2
Ë 1.92 ˜¯
( vr )12 = 1.29094 m2 /s2
( vr )1 = 1.136 m/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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137
PROBLEM 17.91
A 3 kg collar C is attached to a spring and can slide on rod AB, which in turn
can rotate in a horizontal plane. The mass moment of inertia of rod AB with
respect to end A is 0.5 kg ◊ m2. The spring has a constant k = 3000 N/m and an
undeformed length of 250 mm. At the instant shown the velocity of the collar
relative to the rod is zero and the assembly is rotating with an angular velocity
of 12 rad/s. Neglecting the effect of friction, determine (a) the angular velocity of
the assembly as the collar passes through a point located 180 mm from end A of
the rod, (b) the corresponding velocity of the collar relative to the rod.
SOLUTION
Potential energy of spring: undeformed length = 250 mm = 0.25 m
Position 1:
Position 2:
D = 0.65 m - 0.25 m = 0.4 m
1
1
V1 = k D 2 = (3000)(0.4)2
2
2
= 240 N ◊ m
D = 0.30806 - 0.25 = 0.05806 m
1
1
V2 = k D 2 = (3000)(0.05806)2
2
2
= 5.0564 N ◊ m
Kinematics:
Kinetics: Since moments of all forces about shift at A are zero, ( H A )1 = ( H A )2
I Rw1 + mC ( v0 )r1 = I Rw C + mC ( v0 )2 r2
(I
R
)
(
)
+ mC r12 w1 = I R + mC r23 w 2
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138
PROBLEM 17.91 (Continued)
I R = 0.5 kg ◊ m2 , mC = 3 kg
Data:
r1 = 0.6 m, r2 = 0.18 m, w1 = 12 rad/s
È0.5 + (3)(0.6)2 ˘ (12 rad/s) = È0.5 + (3)(0.18)2 ˘ w 2
Î
˚
Î
˚
18.96 = 0.5972w 2 ; w 2 = 31.748 rad/s
(a)
w 2 = 31.7 rad/s Angular velocity.
Kinetic energy.
T1 =
1
1
1
I Aw12 + mC ( vD )12 + mC ( vR )12
2
2
2
1
1
= (0.5)(12)2 + (3)(0.6)2 (12)2 + 0
2
2
= 36 + 77.76
T1 = 113.76 N ◊ m
1
1
1
I Rw 22 + m( v B )22 + m2 ( vR )22
2
2
2
1
= (0.5)(31.748)2
2
1
1
+ (3)(0.18)2 (31.748)2 + (3)( vR )22
2
2
2
T2 = 300.9695 + 1.5 ( vr )2
T2 =
Principle of conservation of energy:
Recall:
T1 + V1 = T2 + V2
V1 = 240 N ◊ m and V2 = 5.0564 N ◊ m
113.76 + 240 = 300.9695 + 1.5( vr )22 + 5.0564
47.7341 = 1.5 vr 22
(b)
( vr )2 = 5.64 m/s Velocity of collar relative to rod.
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139
PROBLEM 17.92
A uniform rod AB, of mass 7.5 kg and length 1 m is attached to the 12.5-kg
cart C. Knowing that the system is released from rest in the position shown and
neglecting friction, determine (a) the velocity of Point B as rod AB passes through
a vertical position (b) the corresponding velocity of cart C.
SOLUTION
vC = vA
Kinematics
vC Æ = v AB Æ + (0.5 m)w ¨
vC = v AB - 0.5w
(1)
AB = 1 m
mC = 12.5 kg, mAB = 7.5 kg
Mass.
Kinetics
Linear momentum
+
0 = mC vC + mAB v AB
v AB = -
Substitute into Eq. (1):
mC
(12.5 kg)
5
vC = vC , v AB = - vC
mAB
(7.5 kg)
3
5
vC = - vC - 0.5w
3
8
vC = - 0.5w
3
Substitute into Eq. (2):
(2)
vC = - 0.1875w (3)
5
v AB = - ( - 0.1875w )
3
v AB = 0.3125w
(4)
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140
PROBLEM 17.92 (Continued)
Kinetic and potential energies.
T1 = 0
V1 = WAB b = (7.5)(9.81)(0.5)(1 - cos 30∞)
= 4.9286 N ◊ m
V2 = 0
1
1
1
2
T2 = mC vC2 + mAB v AB
+ I ABw 2
2
2
2
1
1È 1
1
˘
= (12.5)( - 0.1875w )2 + (7.5)(0.3125w )2 + Í (7.5)(1)2 ˙ w 2
2
2
2 Î12
˚
= w 2 (0.21973 + 0.36621 + 0.3125)
= 0.89844w 2
Conservation of energy:
T1 + V1 = T2 + V2
0 + 4.9286 = 0.89844w 2
w 2 = 5.4857 w = 2.3422 rad/s
(b)
Velocity of C: Eq. (3)
vC = - 0.1875w
(a)
Velocity of B:
v B = vC + (1)w Æ = (0.43916 m/s ¨) + [(1)(2.3422)] Æ]
v B = 0.43916 ¨ + 2.3422 Æ
vC = 0.439 m/s ¨ v B = 1.903 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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141
PROBLEM 17.93
In Problem 17.83, determine the velocity of rod AB relative to cylinder DE as end B of the rod strikes the end
E of the cylinder.
SOLUTION
Kinematics and geometry.
v1 = (0.04 m)w1 = (0.4 m)(40 rad/s)
v1 = 1.6 m/s
v2 = (0.28 m)w 2
Initial position
Final position
Conservation of angular momentum about C.
+
Moments about C:
I AB =
1
(3 kg)(0.9 m)2 = 0.16 kg ◊ m2
12
I ABw1 + mv1 (0.04 m) + I DE w1 = I ABw 2 + mv2 (0.28 m) + I DE w 2
(0.16 kg ◊ m2 )( 40 rad/s) + (8 kg)(1.6 m/s) (0.04 m) + (0.025 kg ◊ m 2 )(40 rad/s)
= (0.16 kg ◊ m2 )w 2 + (3 kg)(0.28w 2 )(0.28) + (0.025 kg ◊ m2 )w 2
(6.4 + 0.192 + 1.00) = (0.16 + 0.2352 + 0.025)w 2
7.592 = 0.4202w 2 ; w 2 = 18.068 rad/s: w 2 = 18.07 rad/s
Conservation of energy
( vr ) = 0.075 m/s
V1 = V2 = 0
1
1
1
1
T1 = I DE w12 + I ABw12 + mAB v12 + mAB ( vr )12
2
2
2
2
1
1
= (0.025 kgg ◊ m2 )( 40 rad/s)2 + (0.16 kg ◊ m2 )( 40 rad/s)2
2
2
1
1
+ (3 kg)(1.6 m/s)2 + (3 kg)(0.075 m/s)2
3
2
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142
PROBLEM 17.93 (Continued)
T1 = 20 J + 120 J + 3.84 J + 0.008 J = 151.85 J
v2 = (0.28 m)w 2 + (0.28 m)(18.068 rad/s) = 5.059 m/s
1
1
1
1
T2 = I DE w 22 + I ABw 22 + mAB v22 + mAB ( vr )22
2
2
2
2
1
= (0.025 kg ◊ m 2 )(18.068 rad/s)2
2
1
+ (0.16 kg ◊ m 2 )(18.068 rad/s)2
2
1
1
= (3 kg)(5.059 m/s)2 + (3 kg)( vr )22
2
2
T2 = 4.081 J + 26.116 J + 38.391 J + 1.5( vr )22
T2 = 68.587 J + 1.5( vr )22
T1 + V1 = T2 + V2 : 151.85 J + 0 = 68.587 J + 1.5( vr )22
83.263 = 1.5( vr )22
( vr )2 = 7.45 m/s Velocity of rod relative to cylinder.
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143
PROBLEM 17.94
In Problem 17.81 determine the velocity of the tube relative to the rod as the
tube strikes end E of the assembly.
PROBLEM 17.81 A 1.6-kg tube AB can slide freely on rod DE which in turn
can rotate freely in a horizontal plane. Initially the assembly is rotating with
an angular velocity w = 5 rad/s and the tube is held in position by a cord. The
moment of inertia of the rod and bracket about the vertical axis of rotation is
0.30 kg ◊ m2 and the centroidal moment of inertia of the tube about a vertical
axis is 0.0025 kg ◊ m2 . If the cord suddenly breaks, determine (a) the angular
velocity of the assembly after the tube has moved to end E, (b) the energy lost
during the plastic impact at E.
SOLUTION
Let Point C be the intersection of axle C and rod DE. Let Point G be the mass center of tube AB.
Masses and moments of inertia about vertical axes.
mAB = 1.6 kg,
I AB = 0.0025 kg ◊ m2 ,
State 1.
1
( rG/ A )1 = (125) = 62.5 mm,
2
State 2.
( rG/ A )2 = 500 - 62.5 = 437.5 mm,
Kinematics.
I DCE = 0.30 kg ◊ m2
w1 = 5 rad/s, ( vr )1 = 0
w = w2 ,
vr = ( vr ) 2 = 0
( vG )q = vq = rG/C w
Syst. Momenta1 +
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
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144
PROBLEM 17.94 (Continued)
Moments about C:
I ABw1 + I DCE w1 + mAB ( vq )1 ( rG/C )1 + 0 = I ABw 2 + I DCE w 2 + mAB ( vq )2 ( rG/C )2
È I AB + I DCE + mAB ( rG/C )12 ˘ w1 = È I AB + I DCE + mAB ( rG/C )22 ˘ w 2
Î
˚
Î
˚
[0.0025 + 0.30 + (1.6)(0.0625)2 ](5) = [0.0025 + 0.30 + (1.6)(0.4375)2 ]w
w2
(0.30875)(5) = 0.60875w 2
Kinetic energy.
w 2 = 2.5359 rad/s
1
1
1
I ABw 2 + I DCE w 2 + mAB v 2
2
2
2
1
1
1
= I ABw 2 + I DCE w 2 + mAB rG2/C w 2 + vr2
2
2
2
1
1
1
T1 = (0.0025)(5)2 + (0.3)(5)2 + (1.6)(0.0625)2 + 0 = 3.859375 J
2
2
2
1
1
T2 = (0.0025)(2.5359)2 + (0.30)(2.5359)2
2
2
1
1
+ (1.6)(0.0625)2 (2.5359)2 + (1.6)( vr )22
2
2
2
= 1.95737 + 0.8( vr )2
T=
(
Work.
)
The work of the bearing reactions at C is zero. Since the sliding contact between the rod and the tube
is frictionless, the work of the contact force is zero.
U1Æ2 = 0
Principle of work and energy.
T1 + U1Æ2 = T2
3.859375 + 0 = 1.95737 + 0.8( vr )22
( vr )2 = 1.542 m/s Velocity of the tube relative to the rod.
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145
PROBLEM 17.95
The 3-kg steel cylinder A and the 5-kg wooden cart B are at rest
in the position shown when the cylinder is given a slight nudge,
causing it to roll without sliding along the top surface of the cart.
Neglecting friction between the cart and the ground, determine
the velocity of the cart as the cylinder passes through the lowest
point of the surface at C.
SOLUTION
Kinematics (when cylinder is passing C )
+
v B = vC = rw - v A
w=
v A + vB
r
Principle of impulse and momentum.
Syst. of Momenta1
+
x components:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
mA v A - mB v B = 0
3 v A = 5 v B ; v B = 0.6 v A
Work:
Kinetic energy:
U1Æ2 = W A (0.15 m) = (3 kg)(9.81 m/s2 )(0.15 m) = 4.4145 N ◊ m; T1 = 0
1
1
1
mA v A2 + I w 2 + mB v32
2
2
2
v A + v B v A + 0.6v A 1.6v A
v B = 0.6v A ; w =
=
=
r
r
r
2
2
1
1
1 Ê 3 kg r ˆ Ê 1.6v A ˆ
2
T2 = (3 kg) v A2 + Á
ÁË
˜¯ + (5 kg)(0.6v A )
˜
2
2
2Ë 2 ¯
r
T2 =
( )
= 1.5 v A2 + 1.92 v A2 + 0.9 v A2 = 4.32 v A2
Principle of work and energy:
T1 + U1Æ2 = T2
0 + 4.4145 = 4.32 v A2
v A2 = 1.021875 v A = 1.01088 m/s Æ
v B = 0.6vA = 0.6(1.01088)
v B = 0.607 m/s ¨ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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146
PROBLEM 17.96
A bullet weighing 40 gm is fired with a horizontal velocity of 550 m/s into
the lower end of a slender 7.5-kg bar of length L = 800 mm. Knowing that
h = 300 mm and that the bar is initially at rest, determine (a) the angular velocity
of the bar immediately after the bullet becomes embedded, (b) the impulsive
reaction at C, assuming that the bullet becomes embedded in 0.001 s.
SOLUTION
L = 800 mm = 0.8 m m = 7.5 kg
1
1
I = mL2 = (7.5)(0.8)2 = 0.4 kg ◊ m2
12
12
Bar:
m0 = 40 g = 0.04 kg
Bullet:
h = 300 mm = 0.3 m
Support location:
v B = ( L - h)w = (0.8 - 0.3)w = 0.5w
Kinematics.
ˆ
ÊL
vG = Á - h˜ w = (0.4 - 0.3)w = 0.1w
¯
Ë2
Kinetics.
Syst. Momenta1
Moments about C:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
ˆ
ÊL
m0 v0 ( L - h) = m0 v B ( L - h) + mvG Á - h˜ + I w
¯
Ë2
(0.04)(550)(0.5) = (0.04)(0.5w )(0.5) + (7.5)(0.1w )(0.1) + (0.4w )
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147
PROBLEM 17.96 (Continued)
11 = 0.01w + 0.075w + 0.4w = 0.485w
(a)
or
w = 22.6804
w = 22.7 rad/s
v B = (0.5)(22.6804) = 11.3402 m/s
vG = (0.1)(22.6804) = 2.26804 m/s
+
Horizontal components:
- m0 v0 + C ( Dt ) = - m0 v B - mvG : C ( Dt ) = m0 ( v0 - v B ) - mv0
C ( Dt ) = (0.04)(550 - 11.3402) - (7.5)(2.26804)
= 4.53609 N ◊ s
(b)
C=
C Dt 4.53609
=
Dt
0.001
C = 4540 N Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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148
PROBLEM 17.97
In Problem 17.96, determine (a) the required distance h if the impulsive reaction
at C is to be zero, (b) the corresponding angular velocity of the bar immediately
after the bullet becomes embedded.
SOLUTION
L = 800 mm = 0.8 m m = 0.75 kg
1
1
I = mL2 = (7.5)(0.8)2 = 0.4 kg ◊ m2
12
12
Bar:
Bullet:
m0 = 0.04 kg
Kinematics.
v B = ( L - h)w = (0.8 - h)w
ˆ
ÊL
vG = Á - h˜ w = (0.4 - h)w
¯
Ë2
Kinetics.
Syst. Momenta1
moment about B:
+
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
Ê Lˆ
0 + 0 = I w - mvG Á ˜
Ë 2¯
0 + 0 = 0.4w - (7.5)(0.4 - h)w (0.4)
Divide by w
0 = 0.4 - 3(0.4 - h)
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149
PROBLEM 17.97 (Continued)
h=
(a)
4
m = 0.26667 m
15
h = 267 mm 4ˆ
8
Ê
v B = Á 0.8 - ˜ w = w
¯
Ë
15
15
4ˆ
2
Ê
vG = Á 0.4 - ˜ w = w
Ë
15 ¯
15
+
Horizontal components: m0 v0 + 0 = mvG + m0 v B
Ê8 ˆ
Ê2 ˆ
(0.04)(550) + 0 = (7.5) Á w ˜ + (0.04) Á w ˜
Ë 15 ¯
Ë 15 ¯
(b)
w = 21.541
w = 21.5 rad/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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150
PROBLEM 17.98
A 45-g bullet is fired with a velocity of 400 m/s at q = 30∞ into a 9-kg
square panel of side b = 200 mm. Knowing that h = 150 mm and that the
panel is initially at rest, determine (a) the velocity of the center of the panel
immediately after the bullet becomes embedded, (b) the impulsive reaction
at A, assuming that the bullet becomes embedded in 2 ms.
SOLUTION
1
1
mB = 0.045 kg mP = 9 kg I G = mP b2 = (9)(0.200)2 = 0.06 kg ◊ m2
6
6
Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity w = w .
b
vG = w Æ .
2
Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.
Syst. Momenta1
(a)
+
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
Moments about A:
b
Ê bˆ
( mB v0 cos 30∞)h + mB v0 sin 30∞ Á ˜ + 0 = I Gw + mP vG
Ë 2¯
2
b
1
ˆ
ˆ Ê
Ê
mB v0 Á h cos 30∞ + sin 30∞˜ = Á I G + mP b2 ˜ w
¯
¯ Ë
Ë
2
4
(0.045)( 400)(0.150 cos 30∞ + 0.100 sin 30∞)
1
È
˘
= Í0.06 + (9)(0.2)2 ˙ w = 0.15w
4
Î
˚
w = 21.588 rad/s
v B = (0.100)(21.556) = 2.1556 m/s
vG = 2.16 m/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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151
PROBLEM 17.98 (Continued)
(b)
+
Linear momentum:
mB v0 cos 30∞ + Ax ( Dt ) = mP vG
(0.045)( 400 cos 30∞) + Ax (0.002) = (9)(2.1556)
A X = 1920 N
Linear momentum:
A x = 1920 N Æ
- mB v0 sin 30∞ + Ay ( Dt ) = 0
- (0.045)( 400)sin 30∞ + Ay (0.002) = 0
Ay = 4500 N
A y = 4500 N ≠
` A = 4892 N = 4.892 kW
tan b =
4500
1920
b = 66.9∞
A = 4.87 kN
66.9∞
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152
PROBLEM 17.99
A 45-g bullet is fired with a velocity of 400 m/s at q = 5∞ into a 9-kg square
panel of side b = 200 mm. Knowing that h = 150 mm and that the panel
is initially at rest, determine (a) the required distance h if the horizontal
component of the impulsive reaction at A is to be zero, (b) the corresponding
velocity of the center of the panel immediately after the bullet becomes
embedded.
SOLUTION
1
1
mB = 0.045 kg mP = 9 kg I G = mP b2 = (9)(0.200)2 = 0.06 kg ◊ m2
6
6
Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity w = w .
b
vG = w Æ .
2
Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.
AX ( Dt ) = 0.
Also
Syst. Momenta1
+
Linear momentum:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
Êb ˆ
mB v0 cos 5∞ + 0 = mP vG = mP Á w ˜
Ë2 ¯
(0.045)( 400 cos 5∞) = (9)(0.100)w w = 19.9239 rad/s
vG = (0.100)(19.9239) = 1.99239 m/s
+
Moments about A:
(1)
b
b
= I Gw + mP vG
2
2
b
1
ˆ
ˆ Ê
Ê
mB v0 Á b cos 5∞ + sin 5∞˜ = Á I G + mP b2 ˜ w
¯
¯ Ë
Ë
2
4
( mB v0 cos 5∞)h + ( mB v0 sin 5∞)
1
È
˘
(0.045)( 400)( h cos 5∞ + 0.100 sin 5∞) = Í0.06 + (9)(0.100)2 ˙ (19.9239)
4
Î
˚
17.9315h + 1.5688 = 2.9886
h = 0.07918 m
(a)
(b)
h = 79.2 mm vG = 1.992 m/s Æ From Eq. (1),
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153
PROBLEM 17.100
A 8-kg wooden panel is suspended from a pin support at A and is initially
at rest. A 2-kg metal sphere is released from rest at B and falls into a
hemispherical cup C attached to the panel at a point located on its top edge.
Assuming that the impact is perfectly plastic, determine the velocity of the
mass center G of the panel immediately after the impact.
SOLUTION
Mass and moment of inertia
ms = 2 kg mP = 8 kg
1
1
I = mP (0.5 m)2 = (8)(0.5)2 = 0.3333 kg ◊ m2
6
6
Velocity of sphere at C.
( vC )1 = 2 gy = 2(9.81 m/s2 )(0.25 m) = 2.2147 m/s
Impact analysis.
Kinematics
Immediately after impact in terms of w 2
v2 = 0.25w 2
( vC )2 = 0.2w 2
Principle of impulse and momentum.
Syst. Momenta1 +
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
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154
PROBLEM 17.100 (Continued)
+
Moments about A:
ms ( vC )1 (0.2 m) + 0 = ms ( vC )2 (0.2 m) + I w 2 + mP v2 (0.25 m)
(2 kg)(2.2147 m/s)(0.2 m) = (2 kg)(0.2w 2 )(0.2 m) + 0.33333w 2 + (8 kg)(0.25 m)2w 2
0.88589 = (0.08 + 0.33333 + 0.500)w 2
w 2 = 0.96995 rad/s w 2 = 0.96995 rad/s
Velocity of the mass center
v2 = (0.25 m)w 2 = (0.25 m)(0.96995 rad/s)
v2 = 0.24249 m/s
v2 = 242 mm/s Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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155
PROBLEM 17.101
An 8-kg wooden panel is suspended from a pin support at A and is initially
at rest. A 2-kg metal sphere is released from rest at B¢ and falls into a
hemispherical cup C¢ attached to the panel at the same level as the mass
center G. Assuming that the impact is perfectly plastic, determine the
velocity of the mass center G of the panel immediately after the impact.
SOLUTION
mS = 2 kg
Mass and moment of inertia.
mP = 8 kg
1
I = mP (0.5 m)2
6
1
= (8)(0.5)2
6
= 0.3333 kg ◊ m2
Velocity of sphere at C ¢.
( vC ¢ )1 = 2 gy
= 2(9.81 m/s2 )(0.5 m)
= 3.1321 m/s
Impact analysis.
Kinematics
Immediately after impact in terms of w 2 .
AC ¢ = (0.2)2 + (0.25)2 = 0.32016 m
( vC ¢ )2 = AC ¢w 2
= 0.32016w 2
v2 = 0.25 w 2
q
(perpendicular to AC .)
Principle of impulse and momentum.
Syst. Momenta1
+
Syst. Ext. Imp.1-2 =
Syst. Momenta2
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156
PROBLEM 17.101 (Continued)
+
Moments about A:
mS ( vC ¢ )1 (0.2 m) + 0 = mS ( vC ¢ )2 ( AC ¢ ) + I w 2 + mP v2 (0.25 m)
(2 kg)(3.1321 m/s)(0.2 m) = (2 kg)(0.32016w 2 )(0.32016 m)
+ 0.3333w 2 + (8 kg)(0.25 m)2 w 2
1.25284 = (0.2050 + 0.3333 + 0.500)w 2
w 2 = 1.2066 rad/s
Velocity of the mass center.
v2 = (0.25 m)w 2
= (0.25 m)(1.2066 rad/s)
= 0.3016 m/s
v2 = 302 mm/s ¨ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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157
PROBLEM 17.102
The gear shown has a radius R = 150 mm and a radius of gyration
k = 125 mm. The gear is rolling without sliding with a velocity v1
of magnitude 3 m/s when it strikes a step of height h = 75 mm.
Because the edge of the step engages the gear teeth, no slipping
occurs between the gear and the step. Assuming perfectly plastic
impact, determine the angular velocity of the gear immediately
after the impact.
SOLUTION
Kinematics. Just before impact, the contact point with the rack is the instantaneous center of rotation of the
gear.
v1 = Rw1 ¨
Just after impact, Point S is the instantaneous center of rotation
v 2 = Rw 2
q
( perpendicular to GS )
Principle of impulse and momentum.
+
Moments about S:
mv2 ( R - h) + I w1 = mv2 R + I w 2
m( Rw1 )( R - h) + mk 2w1 = m( Rw 2 ) R + mk 2w 2
[ R( R - h) + k 2 ]w1 = ( R2 + k 2 )w 2
w2 =
Data:
R2 + k 2 - Rh
w1
R2 + k 2
Rh ˘
È
w 2 = Í1 - 2
w1
Î R + k 2 ˙˚
(1)
R = 150 mm, k = 125 mm, v1 = 3 m/s, h = 75 mm
w1 =
v1
3 m/s
=
= 20 rad/s
R 0.150 m
Angular velocity.
From (1),
È
(150)(75) ˘
w 2 = Í1 (20 rad/s) = 0.7049(20)
2
2 ˙
Î (150 + 125 ) ˚
w 2 = 14.10 rad/s
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158
PROBLEM 17.103
A uniform slender rod AB of mass m is at rest on a frictionless
horizontal surface when hook C engages a small pin at A.
Knowing that the hook is pulled upward with a constant
velocity v0, determine the impulse exerted on the rod (a) at A,
(b) at B. Assume that the velocity of the hook is unchanged
and that the impact is perfectly plastic.
SOLUTION
vB = 0.
At the given instant,
Moment of inertia.
I =
1
mL2
12
Kinematics. (Rotation about B).
w=
v0
L
vG =
1
L
w = v0
2
2
Kinetics.
Syst. Momenta1
Moments about B:
Moments about A:
Syst. Ext. Imp.1Æ2
0 + L Ú Adt = mvG
Ú Adt =
(a)
(b)
+
Syst. Momenta2
L
+ Iw
2
mvG mLw mv0 mv0 mv0
+
=
+
=
2
12
4
12
3
0 + L Ú Bdt = mvG
Ú Bdt =
=
Ú Adt =
mv0
≠
3
Ú Bdt =
mv0
≠
6
L
- Iw
2
mvG mLw mv0 mv0 mv0
=
=
2
12
4
12
6
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159
PROBLEM 17.104
A uniform slender bar of length L and mass m is supported
by a frictionless horizontal table. Initially the bar is spinning
about its mass center G with a constant angular velocity w 1 .
Suddenly latch D is moved to the right and is struck by end A
of the bar. Assuming that the impact of A and D is perfectly
plastic, determine the angular velocity of the bar and the
velocity of its mass center immediately after the impact.
SOLUTION
I =
1
mL2
12
Before impact.
( vA )1 =
L
w1 Ø
2
Impact condition.
1
( vA )2 = -e( vA )1 = eLw1 ≠
2
Moment of inertia.
v2 = ( v A )2 +
Kinematics after impact.
1
1
L
w 2 = eLw1 + Lw 2
2
2
2
Principle of impulse-momentum at impact.
Syst. Momenta1
Moments about D:
For perfectly plastic impact,
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
L
2
1
ˆL
Ê1
I w1 = I w 2 + m Á eLw1 + Lw 2 ˜
¯2
Ë2
2
1
1
1
1
mL2w1 = mL2w 2 + mL2 ew1 + mL2w 2
4
4
12
12
1
w 2 = (1 - 3e)w1
4
1
1
1 Ê 1ˆ
v2 = Lew1 + L Á ˜ (1 - 3e)w1 = (1 + e)w1 L
8
2
2 Ë 4¯
I w1 + 0 = I w 2 + mv2
1
w 2 = w1 4
1
v2 = Lw1 ≠ 8
e=0
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160
PROBLEM 17.105
Solve Problem 17.104, assuming that the impact of A and D
is perfectly elastic.
PROBLEM 17.104 A uniform slender bar of length L and
mass m is supported by a frictionless horizontal table.
Initially the bar is spinning about its mass center G with a
constant angular velocity w 1 . Suddenly latch D is moved
to the right and is struck by end A of the bar. Assuming that
the impact of A and D is perfectly plastic, determine the
angular velocity of the bar and the velocity of its mass center
immediately after the impact.
SOLUTION
I =
1
mL2
12
Before impact.
( vA )1 =
L
w1 Ø
2
Impact condition.
1
( vA )2 = -e( vA )1 = eLw1 ≠
2
Moment of inertia.
v2 = ( vA )2 +
Kinematics after impact.
1
1
L
w 2 = eLw1 + Lw 2
2
2
2
Principle of impulse-momentum at impact.
Syst. Momenta1
Moments about D:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
L
2
1
ˆL
Ê1
I w1 = I w 2 + m Á eLw1 + Lw 2 ˜
¯2
Ë2
2
1
1
1
1
mL2w1 = mL2w 2 + mL2 ew1 + mL2w 2
4
4
12
12
1
w 2 = (1 - 3e)w1
4
1
1
1 Ê 1ˆ
v2 = Lew1 + L Á ˜ (1 - 3e)w1 = (1 + e)w1 L
8
2
2 Ë 4¯
I w1 + 0 = I w 2 + mv2
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161
PROBLEM 17.105 (Continued)
For perfectly elastic impact,
e =1
w2 =
v2 =
1
1
(1 - 3)w1 = - w1
4
2
1
w 2 = w1
2
1
1 Ê 1 ˆ 1
Lw1 + L Á - w1 ˜ = Lw1
2
2 Ë 2 ¯ 4
v2 =
1
Lw1 ≠ 4
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162
PROBLEM 17.106
A uniform slender rod of length L is dropped onto rigid supports at A
and B. Since support B is slightly lower than support A, the rod strikes
A with a velocity v1 before it strikes B. Assuming perfectly elastic
impact at both A and B, determine the angular velocity of the rod and
the velocity of its mass center immediately after the rod (a) strikes
support A, (b) strikes support B, (c) again strikes support A.
SOLUTION
I =
Moment of inertia.
(a)
1
mL2
12
First Impact at A.
Syst. Momenta1
+
Syst. Ext. Imp.1Æ2 =
Syst. Momenta2
e = 1: ( v A )2 = v1 ≠
Condition of impact:
L
L
w - ( v A )2 = w - v1
2
2
L
L
mv1 + 0 = mv2 + I w 2
2
2
v2 =
Kinematics:
Moments about A:
ˆ
ˆL Ê1
ÊL
= m Á w - v1 ˜ + Á mL2 ˜ w 2
¯
¯ 2 Ë 12
Ë2
v2 =
1
L Ê 3v1 ˆ
Á
˜ - v1 = v1
2Ë L ¯
2
w2 =
3v1
L
v2 =
1
v1 Ø 2
( v B )2 = Lw - ( v A )2 = 3v1 - v1 = 2v1 Ø
(b)
Impact at B.
Syst. Momenta1
Condition of impact.
Kinematics:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
e = 1: ( v B )3 = 2v1 ≠
v3 = ( v B )2 -
L
L
w = 2v1 - w
2
2
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163
PROBLEM 17.106 (Continued)
- mv2
Moments about B:
L
L
+ I w 2 + 0 = mv3 - I w 3
2
2
L ˆL Ê1
Ê
ˆ Ê 3v ˆ
ˆ
Ê1 ˆ L Ê 1
- m Á v1 ˜ + Á mL2 ˜ Á 1 ˜ + 0 = m Á 2v1 - w 3 ˜ - Á mL2 ˜ w 3
Ë
¯Ë L ¯
¯
Ë 2 ¯ 2 Ë 12
2 ¯ 2 Ë 12
v3 = 2v1 -
L Ê 3v1 ˆ 1
˜ = v1
Á
2Ë L ¯ 2
w3 =
3v1
L
v3 =
1
v1 ≠ 2
( v A )3 = Lw - ( v B )3 = 3v1 - 2v1 = v1 Ø
(c)
Second Impact at A.
Syst. Momenta3
Condition of impact.
Kinematics:
Moments about A:
+
Syst. Ext. Imp.3Æ4
=
Syst. Momenta4
e = 1: ( v A )4 = v1 ≠
v4 = ( v A ) 4 +
mv3
L
L
w 4 = v1 + w 4
2
2
L
L
+ I w 3 + 0 = mv4 + I w 4
2
2
L ˆL Ê1
Ê
ˆ Ê 3v ˆ
ˆ
Ê1 ˆ L Ê 1
m Á v1 ˜ + Á mL2 ˜ Á 1 ˜ + 0 = m Á v1 + w 4 ˜ + Á mL2 ˜ w 4
¯
Ë
¯Ë L ¯
¯
Ë
Ë 2 ¯ 2 Ë 12
2
2
12
v4 = v1 + 0
w4 = 0 v 4 = v1 ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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164
PROBLEM 17.107
A uniform slender rod AB is at rest on a frictionless horizontal
table when end A of the rod is struck by a hammer which delivers
an impulse that is perpendicular to the rod. In the subsequent
motion, determine the distance b through which the rod will
move each time it completes a full revolution.
SOLUTION
I =
Moment of inertia.
1
mL2
12
Kinetics.
Syst. Momenta1
+
Syst. Ext. Imp.1Æ2 =
0 + 0 + I w + mv
Moments about A:
Syst. Momenta2
L
2
2
2 I w 2 121 mL w 1
=
= wL
v=
mL
mL
6
Motion after impact.
q = wt
t=
q 2p
=
w w
ˆ 2p
Ê1
b = vt = Á w L˜
Ë6 ¯ w
b=
p
L 3
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165
PROBLEM 17.108
A uniform sphere of radius r rolls down the incline shown without slipping.
It hits a horizontal surface and, after slipping for a while, it starts rolling
again. Assuming that the sphere does not bounce as it hits the horizontal
surface, determine its angular velocity and the velocity of its mass center
after it has resumed rolling.
SOLUTION
I =
Moment of inertia. Solid sphere.
2 2
mr
5
Kinematics.
Before
After
Before impact (rolling).
v1 = rw1
After slipping has stopped.
v2 = rw 2
Kinetics.
Syst. Momenta1
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
I w1 + mv1r cos b + 0 = I w 2 + mv2 r
Moments about C:
I w1 + mr 2w1 cos b = I w 2 + mr 2w 2
w2 =
2
2
2
I + mr 2 cos b
5 mr + mr cos b
w
=
w1
1
2
2
2
I + mr 2
5 mr + mr
v2 = rw 2 =
2 + 5 cos b
rw 1
7
1
w 2 = (2 + 5 cos b )w1
7
1
v2 = (2 + 5 cos b )v1 ¨ 7
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166
PROBLEM 17.109
The slender rod AB of length L forms an angle b with the vertical as it strikes the
frictionless surface shown with a vertical velocity v1 and no angular velocity.
Assuming that the impact is perfectly elastic, derive an expression for the angular
velocity of the rod immediately after the impact.
SOLUTION
1
mL2
12
Moment of inertia.
I =
Perfectly elastic impact.
e =1
[( v A ) y ]2 = -e[( v A ) y ]1 = ev1 ≠
v A = ( v A ) x i + ( v A ) y j = ( v A ) x i + v1 j
Kinetics.
Syst. Momenta1
+
+
0 + 0 = mv x
horizontal components:
Kinematics.
vG = vA + vG/ A
mv1
=
Syst. Momenta2
mv x = 0
ÈL
[v y ≠] = [v1 ≠] + [( v A ) x Æ] + Í w
Î2
v y = v1 -
Velocity components :
Moments about A:
Syst. Ext. Imp.1Æ2
˘
b˙
˚
L
w sin b
2
L
L
sin b + 0 = - mv y sin b + I w
2
2
L
1
ˆL
ÊL
mv1 sin b = m Á w sin b - v1 ˜ sin b + mL2w
¯2
Ë2
2
12
1 2 2 ˆ
Ê1
2
ÁË mL + mL sin b ˜¯ w = mv1 L sin b
12
4
w=
12 sin b v1
1 + 3 sin 2 b L
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167
PROBLEM 17.110
Solve Problem 17.109, assuming that the impact between rod AB and the
frictionless surface is perfectly plastic.
SOLUTION
1
mL2
12
Moment of inertia.
I =
Perfectly plastic impact.
e=0
[( v A ) y ]2 = -e( v A ) y1 = 0
v A = (v A ) x i + (v A ) y j = (v A ) x i
Kinetics.
Syst. Momenta1
+
vG = v A + vG/ A
=
Syst. Momenta2
mv x = 0
ÈL
[v y ≠] = [( v A ) x Æ] + + Í w
Î2
˘
b˙
˚
L
v y = - w sin b
2
Velocity components ≠:
Moments about A:
Syst. Ext. Imp.1Æ2
0 + 0 = mv x
horizontal components:
Kinematics.
+
mv1
L
L
sin b + 0 = - mv y sin b + I w
2
2
1
L
ˆL
ÊL
mv1 sin b = m Á w sin b ˜ sin b + mL2w
¯
Ë
2
2
2
12
1 2 2 ˆ
1
Ê1
2
ÁË mL + mL sin b ˜¯ w = mv1 L sin b
12
4
2
w=
6 sin b v1
1 + 3 sin 2 b L
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168
PROBLEM 17.111
A uniformly loaded rectangular crate is released from rest in the
position shown. Assuming that the floor is sufficiently rough
to prevent slipping and that the impact at B is perfectly plastic,
determine the smallest value of the ratio a/b for which corner A
will remain in contact with the floor.
SOLUTION
We consider the limiting case when the crate is just ready to rotate about B. At that instant the velocities must
be zero and the reaction at corner A must be zero. Use the principle of impulse and momentum.
Syst. Momenta1
+
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
Moments about B:
I w1 + ( mv1 )2
Note:
sin f =
b
a
- ( mv1 ) y + 0 = 0
2
2
b
a +b
2
2
, cos f =
(1)
a
a + b2
2
1 2
a + b2 w1
2
1
m 2
b
= mbw 1
( mv1 ) x = ( mv1 )sin f =
a + b2 w 1
2
2
2
2
a +b
v1 = ( AG )v1 =
Thus:
Also,
( mv1 ) y = ( mv1 ) cos f =
I =
From Eq. (1)
1
maw
2
1
m( a2 + b2 )
12
b 1
a
1
1
m( a2 + b2 )w1 + ( mbw1 ) - ( maw1 ) = 0
12
2
2 2
2
1 2
1
mb w1 - ma2 v1 = 0
3
6
a2
=2
b2
a
= 2
b
a
= 1.414 b
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169
PROBLEM 17.112
A uniform slender rod AB of length L is falling freely
with a velocity v0 when cord AC suddenly becomes
taut. Assuming that the impact is perfectly plastic,
determine the angular velocity of the rod and the
velocity of its mass center immediately after the cord
becomes taut.
SOLUTION
Immediately after impact
tan q =
1
q = 26.565∞
2
Due to constraint of inextensible cord,
vA = vA
q
vG = vA + vG /A
Kinematics:
= vA
q+
L
wØ
2
(1)
Principle of impulse and momentum.
Syst. Momenta1
Components
q:
+ Syst. Ext. Imp.1Æ2
mv0 cos q + 0 = mv A +
=
Syst. Momenta2
mL
w cos q
2
ˆ
Ê1
v A + Á cos q ˜ ( Lw ) = v0 cos q
¯
Ë2
+
Moments about A:
mv0
(2)
L
L Ê mL ˆ Ê L ˆ
w˜ Á ˜ + I w
+ 0 = ( mv A cos q ) + Á
2
2 Ë 2 ¯ Ë 2¯
L
Ê 1 2 1 2ˆ
ˆ
ÊL
ÁË cosq ˜¯ v A + ÁË L + L ˜¯ w = v0
2
4
12
2
(3)
Solving Eqs. (2) and (3) simultaneously (with q = 26.565∞)
vA = 0.55902v0
Lw = 0.75v0
w = 0.750
Angular velocity after impact.
v0
L
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170
PROBLEM 17.112 (Continued)
Velocity of mass center.
vG = vA + vG /A
vG = 0.55902v0
1
q + (0.75v0 ) Ø
2
= 0.25v0 ¨ + 0.5v0 Ø + 0.375v0 Ø
= 0.25v0 ¨ + 0.875v0 Ø
v B = 0.910v0
74.1∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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171
PROBLEM 17.113
A uniform slender rod AB of length L is falling freely with a velocity v0
when cord AC suddenly becomes taut. Assuming that the impact is perfectly
plastic, determine the angular velocity of the rod and the velocity of its mass
center immediately after the cord becomes taut.
SOLUTION
Immediately after impact,
tan q =
1
q = 26.565∞
2
Due to constraint of inextensible cord,
vA = vA
q
vG = vA + vG /A
Kinematics.
= vA
q+
L
wÆ
2
Principle of impulse and momentum.
Syst. Momenta1
Components
q:
+ Syst. Ext. Imp.1Æ2
mv0 cos q + 0 = mvA -
=
Syst. Momenta2
mL
w sin q
2
ˆ
Ê1
vA - Á sin q ˜ ( Lw ) = v0 cos q
¯
Ë2
(2)
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172
PROBLEM 17.113 (Continued)
+
Moments about A:
0 + 0 = -( mv A sin q )
L Ê mL ˆ Ê L ˆ
+Á
w ˜ Á ˜ + Iw
2 Ë 2 ¯ Ë 2¯
1 ˆ
Ê1
ˆ
ÊL
- Á sinq ˜ v A + Á L2 + L2 ˜ w = 0
Ë4
¯
Ë2
12 ¯
(3)
Solving Eqs. (2) and (3) simultaneously (with q = 26.565∞),
vA = 1.05527v0
Lw = 0.70588v0
w = 0.706
Angular velocity after impact.
Velocity of mass center.
v0
L
vG = vA + vG /A
vG = 1.055227v0
1
q + (0.70588)v0 Æ
2
= 0.47059v0 ¨ + 0.94118v0 Ø + 0.35294v0 Æ
= 0.11765v0 ¨ + 0.94118v0 Ø
vG = 0.949v0
82.9∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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173
PROBLEM 17.114
A slender rod of length L and mass m is released from rest in the position
shown. It is observed that after the rod strikes the vertical surface it
rebounds to form an angle of 30° with the vertical. (a) Determine the
coefficient of restitution between knob K and the surface. (b) Show that
the same rebound can be expected for any position of knob K.
SOLUTION
For analysis of the downward swing of the rod before impact and for the upward swing after impact use the
principle of conservation of energy.
Before impact.
V1 = 0
V2 = -W
L
L
= - mg
2
2
T1 = 0
2
T2 =
1
1
1
1Ê 1
1 Ê1 ˆ
ˆ
I w 22 + mv22 = Á mL2 ˜ v22 + m Á w 2 ˜ = mL2w 22
¯
6
2
2
2 Ë 12
2 Ë2 ¯
1
L
g
T1 + V1 = T2 + V2 : 0 = mL2w 22 = - mg ; w 22 = 3
L
6
2
w 2 = 1.73205
g
L
After impact.
V3 = -W
L
L
= - mg
2
2
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174
PROBLEM 17.114 (Continued)
L
cos 30∞
2
1
1
T3 = I w 32 + mv32
2
2
V4 = -W
1Ê 1
1 Ê1 ˆ
ˆ
= Á mL2 ˜ w 32 + m Á w 3 ˜
¯
2 Ë 12
2 Ë2 ¯
1
= mL2w 32
6
T4 = 0
1 2 2
L
L
T3 + V3 = T4 + V4 :
mL w 3 - mq = 0 - mg cos 30∞
6
2
2
w 32 = (1 - cos 30∞)
g
L
2
w 3 = 0.63397
g
L
Analysis of impact.
Let r be the distance BK.
Before impact,
( v k )3 = bw 2 Æ = 1.73205b
g
Æ
L
After impact,
( v k )4 = bw 3 ¨ = 0.63397b
g
¨
L
Coefficient of restitution.
(b)
e=
|( vk )4 n |
|(vk )3n |
e=
0.63397
1.73205
e = 0.366 Clearly the answer is independent of b.
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175
PROBLEM 17.115
The uniform rectangular block shown is moving along a
frictionless surface with a velocity v1 when it strikes a
small obstruction at B. Assuming that the impact between
corner A and obstruction B is perfectly plastic, determine
the magnitude of the velocity v1 for which the maximum
angle q through which the block will rotate is 30∞.
SOLUTION
Let m be the mass of the block.
a = 250 mm = 0.25 m
b = 125 mm = 0.125 m
Dimensions:
Moment of inertia about the mass center.
Let d be one half the diagonal.
I =
1
m( a2 + b2 )
12
d=
1 2
5
a + b2 =
m = 0.13975 m
2
16
v1 = v1 Æ, w1 = 0
Kinematics. Before impact
After impact, the block is rotating about corner at B.
w2 = w2
v2 = dw 2
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about B:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
mv1b
+ 0 = I w 2 + mdv2
2
1
1
mv1b = m( a2 + b2 )w 2 + md 2w 2
2
12
1
= m( a2 + b2 )w 2
3
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176
PROBLEM 17.115 (Continued)
w2 =
Angular velocity after impact
3v1b
(1)
2( a2 + b2 )
The motion after impact is a rotation about corner B.
Position 2 (immediately after impact).
v2 = dw 2
Position 3 (q = 30∞).
b
= tan -1 0.5 = 26.565∞
a
h = d sin( b + 30∞) = 0.13975 sin 56.565∞ = 0.11663 m
w3 = 0
v3 = 0
b = tan -1
Potential energy:
V2 =
Kinetic energy:
T2 =
mgb
V3 = mgh
2
1 2 1
I w 2 + mv22 =
2
2
1
= m( a2 + b2 )w 22
6
1
( I + md 2 )w 22
2
T3 = 0
Principle of conservation of energy:
T2 + V3 = T3 + V3
1
mgb
m( a2 + b2 )w 22 +
= 0 + mgh
6
2
3g (2h - b) (3)(9.81)(0.23325 - 0.125)
w 22 = 2
=
(0.25)2 + (0.125)2
( a + b2 )
= 40.7794 (rad/s)2
w 2 = 6.3859 rad/s
Magnitude of initial velocity.
Solving Eq. (1) for v1
v1 =
v1 =
2( a2 + b2 )w 2
3b
(2) ÈÎ(0.25)2 + (0.125)2 ˘˚ (6.3859)
(3)(0.125)
v1 = 2.66 m/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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177
PROBLEM 17.116
A slender rod of mass m and length L is released from rest in the position
shown and hits edge D. Assuming perfectly plastic impact at D, determine
for b = 0.6 L, (a) the angular velocity of the rod immediately after the impact,
(b) the maximum angle through which the rod will rotate after the impact.
SOLUTION
For analysis of the falling motion before impact use the principle of conservation of energy.
L
T1 = 0, V1 = mg
Position 1:
4
V2 = 0
Position 2:
2
1 ÊL ˆ
1Ê 1
ˆ
m Á w 2 ˜ + Á mL2 ˜ w 22
¯
2 Ë2 ¯
2 Ë 12
1
T2 = mL2w 22
6
L 1
3g
T1 + V1 = T2 + V2 : 0 + mg = mL2w 22 w 2 =
4 6
2L
T2 =
Analysis of impact. Kinematics
Before impact, rotation is about Point A.
v2 =
L
w2
2
After impact, rotation is about Point D.
v3 =
L
w3
10
Principle of impulse-momentum.
Syst. Momenta2
+
Moments about D:
+
Syst. Ext. Imp.2Æ3
=
Syst. Momenta3
Ê Lˆ
Ê Lˆ
I w 2 - mv2 Á ˜ = I w 3 + mv3 Á ˜
Ë 10 ¯
Ë 10 ¯
(1)
1
1
ÊL ˆ L
ÊL ˆ L
= mL2w 3 + m Á w 3 ˜
mL2w 2 - m Á w 2 ˜
Ë 10 ¯ 10
Ë 2 ¯ 10 12
12
1 ˆ
1ˆ
Ê1
Ê1
ÁË - ˜¯ w 2 = ÁË +
˜ w3
12 100 ¯
12 20
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178
PROBLEM 17.116 (Continued)
(a)
Angular velocity. w 3 =
5
5 3g
w2 =
14
14 2 L
w 3 = 0.437
g
L
For analysis of the rotation about Point D after the impact use the principle of conservation of energy.
Position 3.
(Just after impact)
v3 =
L
w 3 V3 = 0
10
2
T3 =
1 ÊL ˆ
1Ê 1
14
ˆ
m Á w 3 ˜ + Á mL2 ˜ w 32 =
mL2w 32
¯
Ë
¯
Ë
2 10
2 12
300
2
Ê 5 2g ˆ
14
mgL
=
mL2 Á
=
˜
300
112
Ë 14 2 L ¯
q = maximum rotation angle.
Position 4.
h¢ =
L
sin q
10
mgL
sin q
10
v4 = 0, w 4 = 0, T4 = 0
mgL
mgL
+0=0+
sinq
10
112
10
sinq =
112
V4 = mgh¢ =
T3 + V3 = T4 + V4 ;
(b)
Maximum rotation angle.
q = 5.12∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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179
PROBLEM 17.117
A 30-g bullet is fired with a horizontal velocity of 350 m/s into the 8-kg wooden
beam AB. The beam is suspended from a collar of negligible mass that can
slide along a horizontal rod. Neglecting friction between the collar and the rod,
determine the maximum angle of rotation of the beam during its subsequent
motion.
SOLUTION
m¢ = 30 g = 0.03 kg
Mass of bullet.
m = 8 kg
Mass of beam AB.
m¢
= 0.00375 m¢ = b m
m
Since b is so small, the mass of the bullet will be neglected in comparison with that of the beam in determining
the motion after the impact.
1
Moment of inertia.
I = mL2
12
b=
Mass ratio.
Impact kinetics.
Syst. Momenta2
+
linear components:
Moments about B:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
- b mv0 + 0 = mv2
v2 = b v0
L
0 + 0 = I w - mv2
2
mv2 L 12mb v0 L
=
2I
2mL2
6b v0
w=
L
w=
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180
PROBLEM 17.117 (Continued)
Motion during rising. Position 2. Just after the impact.
V2 = - mg
T2 =
=
L
2
(datum at level A)
1
1
mv22 + I w 22
2
2
1
1Ê 1
ˆ Ê 6b v0 ˆ
m( b v0 )2 + Á mL2 ˜ Á
¯ Ë L ˜¯
2
2 Ë 12
2
= 2b 2 mv02
w = 0,
Position 3.
Syst. Momenta2
+
q = qm .
Syst. Ext. Imp.2Æ3
V3 = - mg
=
Syst. Momenta3
L
cosq m
2
1
mv32
2
mv2 + 0 = mv3
T3 =
+
linear components:
v3 = v2 = b v0
Conservation of energy.
L 1
L
= m( b v0 )2 - mg cos q m
2 2
2
2 2
3b v0
= 1 - cos q m
gL
T2 + V2 = T3 + V3 : 2b 2 mv02 - mg
cos q m = 1 =1-
3b 2 v02
gL
(3)(0.00375)2 (350)2
(9.81)(1.2)
= 0.56099
q m = 55.9∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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181
PROBLEM 17.118
For the beam of Problem 17.117, determine the velocity of the 30-g bullet for
which the maximum angle of rotation of the beam will be 90∞.
PROBLEM 17.117 A 30-g bullet is fired with a horizontal velocity of 350 m/s
into the 8-kg wooden beam AB. The beam is suspended from a collar of negligible
weight that can slide along a horizontal rod. Neglecting friction between the
collar and the rod, determine the maximum angle of rotation of the beam during
its subsequent motion.
SOLUTION
m¢ = 30 g = 0.03 kg
Mass of bullet.
Mass of beam AB.
m = 8 kg
Mass ratio.
b=
m¢
= 0.00375 m¢ = b m
m
Since b is so small, the mass of the bullet will be neglected in comparison with that of the beam in determining
the motion after the impact.
1
Moment of inertia.
I = mL2
12
Impact Kinetics.
Syst. Momenta1
+
linear components:
Moments about B:
+
Syst. Ext. Imp.1Æ2
- b mv0 + 0 = mv2
=
Syst. Momenta2
v2 = b v0
0 + 0 = I w - mv2
L
2
mv2 L 12mb v0 L
=
2I
2mL2
6b v0
w=
L
w=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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182
PROBLEM 17.118 (Continued)
Motion during rising. Position 2. Just after the impact.
V2 = - mg
T2 =
=
L
2
(datum at level A)
1
1
mv22 + I w 22
2
2
1
1Ê 1
ˆ Ê 6b v0 ˆ
m( b v0 )2 + Á mL2 ˜ Á
¯ Ë L ˜¯
2 Ë 12
2
2
= 2b 2 mv02
w = 0,
Position 3.
Syst. Momenta2
+
q = q m& .
Syst. Ext. Imp.2Æ3
V3 = - mg
T3 =
+
linear components:
=
Syst. Momenta3
L
cosq m
2
1
mv32
2
mv2 + 0 = mv3
v3 = v2 = b v0
Conservation of energy.
L 1
L
= m( b v0 )2 - mg cos q m
2 2
2
1
b v0 =
gL(1 - cos q m )
3
T2 + V2 = T3 + V3 : 2b 2 mv02 - mg
Ê 1ˆ
= Á ˜ (9.81)(1.2)(1 - cos 90∞)
Ë 3¯
= 1.98091 m/s
v0 =
1.98091
0.00375
v0 = 528 m/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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183
PROBLEM 17.119
A uniformly loaded square crate is released from rest with
its corner D directly above A; it rotates about A until its
corner B strikes the floor, and then rotates about B. The floor
is sufficiently rough to prevent slipping and the impact at B
is perfectly plastic. Denoting by w 0 the angular velocity of
the crate immediately before B strikes the floor, determine
(a) the angular velocity of the crate immediately after B strikes
the floor, (b) the fraction of the kinetic energy of the crate lost
during the impact, (c) the angle q through which the crate will
rotate after B strikes the floor.
SOLUTION
Let m be the mass of the crate and c be the length of an edge.
1
1
I = m(c 2 + c 2 ) = mc 2
Moment of inertia
12
6
Syst. Momenta1
Kinematics:
Moments about B:
+
Syst. Ext. Imp.1Æ2
=
Syst. Momenta2
1
2cw 0
2
1
v = rG/Bw =
2cw
2
v0 = rG/ Aw 0 =
I w 0 + 0 = I w + rG/B mv
1 2
1
ˆ 2
ˆ Ê1
Ê1
mc w 0 + 0 = mc 2w + Á
2c ˜ m Á
2cw ˜ = mc 2w
¯ 3
¯ Ë2
Ë2
6
6
(a)
1
w = w0
4
Solving for w,
Kinetic Energy.
Before impact:
T1 =
1 2 1
I w 0 + mv02
2
2
1Ê1
1 Ê1
ˆ
ˆ
= Á mc 2 ˜ w 02 + m Á
2cw 0 ˜
¯
¯
2Ë6
2 Ë2
1
= mc 2w 02
3
2
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184
PROBLEM 17.119 (Continued)
After impact:
T2 =
1 2 1
1Ê1
1 Ê1
ˆ
ˆ
I w + mv 2 = Á mc 2 ˜ w 2 + m Á
2cw ˜
¯
¯
2
2
2Ë6
2 Ë2
2
2
1
1
1
Ê1 ˆ
= mc 2w 2 = mc 2Á w 0 ˜ =
mc 2w 0
Ë4 ¯
48
3
3
(b)
Fraction of energy lost:
1
T1 - T2 13 - 48
1
= 1 =116
T1
3
Conservation of energy during falling.
T0 + V0 = T1 + V1
(1)
Conservation of energy during rising.
T3 + V3 = T2 + V2
(2)
T0 = 0, T3 = 0 T2 =
Conditions:
15
16
1
T1
16
Ê1 ˆ
ˆ
Ê1
V0 = mg Á
2c˜ V1 = V2 = mg Á c˜ V3 = mgh3
Ë2 ¯
¯
Ë2
(
)
T1 = V0 - V1 =
From Equation (2),
1
T2 = V3 - V2 = mgh3 - mgc
2
h3 - 12 c
1
2
(c)
1
2
From Equation (1),
From geometry,
1
2 - 1 16
=
h3 =
2 - 1 mgc
È1 1
h3 = Í +
Î 2 16
(
)
˘
2 -1 ˙c
˚
1
2c sin (q + 45∞)
2
Equating the two expressions for h3 ,
sin ( 45∞ + q ) =
1
2
+ 161
(
1
2
45∞ + q = 46.503∞
)
2 -1
2
q = 1.50∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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185
PROBLEM 17.120
A uniform slender rod AB of length L = 800 mm is placed
with its center equidistant from two supports that are located
at a distance b = 200 mm from each other. End B of the rod is
raised a distance h0 = 100 mm and released; the rod then rocks
on the supports as shown. Assuming that the impact at each
support is perfectly plastic and that no slipping occurs between
the rod and the supports, determine (a) the height h1 reached
by end A after the first impact, (b) the height h2 reached by
end B after the second impact.
SOLUTION
I =
Moment of inertia.
1
mL2
12
Let q be the angle of inclination of the bar and h the elevation of the center of gravity.
Position 0
Position¢
bh0
T0 = 0
L+b
Position 0.
V0 = mgh0 = mg
Position¢.
1
v ¢ = bw ¢ V ¢ = 0
2
2
1
1
1 Ê1
1Ê 1
ˆ
ˆ
m( v ¢ )2 + I (w ¢ )2 = m Á bw ¢˜ + Á mL2 ˜ (w ¢ )2
¯
Ë
¯
Ë
2
2
2
2
2 12
1
=
m( L2 + 3b2 )(w ¢ )2
24
T¢ =
Conservation of energy.
T0 + V0 = T ¢ + V ¢ : 0 +
(w ¢ )2 =
where
C1 =
mgbh0
1
m( L2 + 3b2 )(w ¢ )2
=
L + b 24
24 gbh0
( L + b)( L2 + 3b2 )
24 gb
( L + b)( L2 + 3b2 )
= C1h0
(1)
(2)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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186
PROBLEM 17.120 (Continued)
Impact at D.
Syst. Momenta¢
+
Syst. Ext. Imp.¢Æ≤ =
1
v ¢ = bw ¢
2
Kinematics.
Moments about D:
Syst. Momenta≤
1
v ¢¢ = bw ¢¢
2
b
b
mv ¢ - I w ¢ + 0 = mv ¢¢ + I w ¢¢
2
2
ˆb 1
Ê1
ˆb 1
Ê1
- m Á bw ¢˜ + mL2w ¢ + 0 = m Á bw ¢¢˜ + mL2w ¢¢
¯ 2 12
Ë2
¯ 2 12
Ë2
where
w ¢¢ =
L2 - 3b2
w ¢ = C2w ¢
L2 + 3b2
(3)
C2 =
L2 - 3b2
L2 + 3b2
(4)
Position 1 Maximum elevation of end A.
Conservation of energy.
T1 + V1 = T ¢¢ + V ¢
Result
(w ¢¢ )2 = C1h1
h1 =
Data:
(w ¢¢ )2 C22 (w ¢ )2 C22C1
=
=
h1 = C22 h1
C1
C1
C1
L = 800 mm = 0.8 m, b = 200 mm = 0.2 m
(0.8)2 - 3(0.2)2 13
=
= 0.68421
(0.8)2 + 3(0.2)2 19
169
C22 =
= 0.468144
361
C2 =
(a)
h1 = 0.468144 h0 = (0.468144)(100 mm) = 46.8144 mm
h1 = 46.8 mm (b)
h2 = 0.468144 h1 = (0.468144)( 46.8144) = 21.916 mm
h2 = 21.9 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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187
PROBLEM 17.121
A small plate B is attached to a cord that is wrapped around a uniform 4 kg disk
of radius R = 200 mm. A 1.5 kg collar A is released from rest and falls through
a distance h = 300 mm before hitting plate B. Assuming that the impact is
perfectly plastic and neglecting the weight of the plate, determine immediately
after the impact (a) the velocity of the collar, (b) the angular velocity of the disk.
SOLUTION
The collar A falls a distance h. From the principle of conservation of energy.
v1 = 2 gh
Impact analysis. Kinematics. Plastic impact.
e=0
Collar A and plate B move together. The cord is inextensible.
v2 = R w
or w 2 =
v2
R
Let m = mass of collar A and M = mass of disk
I =
Moment of inertia of disk:
1
MR2
2
Principle of impulse and momentum.
I w1 = 0
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2 =
Syst. Momenta 2
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188
PROBLEM 17.121 (Continued)
+
Moments about C:
m v1 R = I w 2 + m v2 R
(1)
1
Êv ˆ
MR2 Á 2 ˜ + m v2 R
Ë R¯
2
1
m v1 = Mv2 + m v2
2
2m
v1
v2 =
2m + M
m v1 R =
m = 1.5 kg
M = 4 kg
h = 300 mm = 0.3 m
R = 200 mm = 0.2 m
Data:
v1 = 2(9.81)(0.3)
= 2.4261 m/s
(a)
Velocity of A.
v2 =
(2)(1.5)
3
v1 = v1
[(2)(1.5) + 4]
7
3
v2 = (2.4261) = 1.03976 m/s
7
(b)
Angular velocity.
w2 =
1.03976
= 5.1988 rad/s
0.2
v 2 = 1.040 m/s Ø w 2 = 5.20 rad/s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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189
PROBLEM 17.122
Solve Problem 17.121, assuming that the coefficient of restitution between A
and B is 0.8.
PROBLEM 17.121 A small plate B is attached to a cord that is wrapped
around a uniform 4-kg disk of radius 200 mm. A 1.5 kg collar A is released
from rest and falls through a distance h = 300 mm before hitting plate B.
Assuming that the impact is perfectly plastic and neglecting the weight of the
plate, determine immediately after the impact (a) the velocity of the collar,
(b) the angular velocity of the disk.
SOLUTION
mD = 4 kg
R = 0.2 m
1
1
I D = mD R2 = ( 4)(0.2)2 = 0.08 kg ◊ m2
2
2
mA = 1.5 kg
h = 300 mm = 0.3 m
Collar A falls through distance h. Use conservation of energy.
T1 = 0
V1 = W A h
1
T2 = mA v 2A
2
V2 = 0
1
mA v 2A + 0
2
2mA h
= 2 gh
v 2A =
WA
= (2)(9.81)(0.3)
T1 + V1 = T2 + V2 : 0 + W A h =
= 5.886 m2 /s2
vA = 2.4261 m/s Ø
Impact. Neglect the mass of plate B. Neglect the effect of weight during the duration of the impact.
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190
PROBLEM 17.122 (Continued)
Kinematics.
v ¢B = Rw Ø = 0.2w ¢ Ø
w¢ = w
Conservation of momentum.
+
mA v A R + 0 = mA v A¢ R + I Dw ¢ + mB v B¢ R
Moments about D:
(1.5)(2.4261)(0.2) = (1.5)(0.75)v A¢ + (0.08 w ¢ ) + 0
(1)
1.125v A¢ + 0.08 w ¢ = 0.72783
Coefficient of restitution.
v B¢ - v A¢ = e( v A - v B )
0.2w ¢ - v A¢ = 0.8(2.4261 - 0)
(2)
- v A¢ + 0.2w ¢ = 1.94088
Solving Eqs. (1) and (2) simultaneously
(a)
Velocity of A.
v A¢ = - 0.0318 m/s
vA¢ = 0.0318 m/s ≠ (b)
Angular velocity.
w ¢ = 9.545 rad/s
w¢ = 9.55 rad/s
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191
PROBLEM 17.123
A slender rod AB is released from rest in the position shown. It
swings down to a vertical position and strikes a second and identical
rod CD which is resting on a frictionless surface. Assuming that
the coefficient of restitution between the rods is 0.5, determine the
velocity of rod CD immediately after the impact.
SOLUTION
I =
Moment of inertia.
1
mL2 for each rod.
12
Rod AB swings to vertical position.
Position 1
Position 2
Position 1.
V1 = 0
T1 = 0
Position 2.
V2 = - mg
L
2
T2 =
1
1
mv22 + I w 22
2
2
2
1 ÊL ˆ
1Ê 1
ˆ
m Á w 2 ˜ + Á mL2 ˜ w 22
¯
2 Ë2 ¯
2 Ë 12
1
= mL2w 22
6
1
L
T1 + V1 = T2 + V2 : 0 + 0 = mL2w 22 - mg
6
2
=
Conservation of energy.
w2 =
v2 =
3g
L
L 3g
Æ
2 L
( v B )2 = Lw = 3gL
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192
PROBLEM 17.123 (Continued)
( vC )3 - ( v B )3 = e( v B )2
Impact condition:
( vC )3 - Lw 3 = e 3gL
( vC )3 = Lw 3 + e 3gL
Principle of impulse-momentum at impact.
Syst. Momenta 2
Moments about B:
mv2
+
Syst. Ext. Imp. 2Æ3
= Syst. Momenta3
L
L
+ I w 2 + 0 = mv3 + I w 3 + m( vC )3 L
2
2
Ê L 3g ˆ L 1
ÊL ˆ L 1
2 3g
2
mÁ
˜ 2 + 12 mL L + 0 = m ÁË 2 w 3 ˜¯ 2 + 12 mL w 3 + m Lw 3 + e 3gL L
2
L
Ë
¯
(
)
Ê 1 3 ˆ 3g
w 3 = Á - e˜
Ë4 4 ¯ L
Ê1 3 ˆ
( vC )3 = Á - e˜ 3gL + e 3gL
Ë4 4 ¯
For e = 0.5
( vC )3 =
1
(11 + e) 3gL
4
( vC )3 =
1
(1 + 0.5) 3gL
4
( vC )3 = 0.650 gL Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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193
PROBLEM 17.124
Solve Problem 17.123, assuming that the impact between the rods
is perfectly elastic.
PROBLEM 17.123 A slender rod AB is released from rest in the
position shown. It swings down to a vertical position and strikes
a second and identical rod CD which is resting on a frictionless
surface. Assuming that the coefficient of restitution between the
rods is 0.5, determine the velocity of rod CD immediately after
the impact.
SOLUTION
I =
Moment of inertia.
1
mL2
12
for each rod.
Rod AB swings to vertical position.
Position 1
Position 2
Position 1
V1 = 0
T1 = 0
Position 2
V2 = - mg
L
2
T2 =
1
1
mv22 + I w 22
2
2
2
1 ÊL ˆ
1Ê 1
ˆ
m Á w 2 ˜ + Á mL2 ˜ w 22
¯
2 Ë2 ¯
2 Ë 12
1
= mL2w 22
6
=
Conservation of energy.
1
L
T1 + V1 = T2 + V2 : 0 + 0 = mL2w 22 - mg
6
2
w2 =
v2 =
3g
L
L 3g
Æ
2 L
( v B )2 = Lw = 3gL
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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194
PROBLEM 17.124 (Continued)
-( vC )3 - ( v B )3 = e( v B )2
Impact condition:
( vC )3 - Lw 3 = e 3gL
( vC )3 = Lw 3 + e 3gL
Principle of impulse-momentum at impact.
Syst. Momenta2
Moments about B:
+
mv2
Syst. Ext. Imp.2Æ3 =
Syst. Momenta3
L
L
+ I w 2 + 0 = mv3 + I w 3 + m( vC )3 L
2
2
Ê L 3g ˆ L 1
3g
ÊL ˆ L 1
mÁ
+ mL2
+ 0 = m Á w 3 ˜ + mL2w 3 + m Lw 3 + e 3gL L
˜
Ë 2 ¯ 2 12
L
Ë 2 L ¯ 2 12
(
)
Ê 1 3 ˆ 3g
w 3 = Á - e˜
Ë4 4 ¯ L
Ê1 3 ˆ
( vC )3 = Á - e˜ 3gL + e 3gL
Ë4 4 ¯
For perfectly elastic impact, e = 1.
( vC )3 =
1
(11 + e) 3gL
4
( vC )3 =
1
(1 + 1) 3gL
4
( vC )3 = 0.866 gL Æ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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195
PROBLEM 17.125
The plank CDE has a mass of 15 kg and rests on a small pivot
at D. The 55-kg gymnast A is standing on the plank at C when
the 70-kg gymnast B jumps from a height of 2.5 m and strikes the
plank at E. Assuming perfectly plastic impact and that gymnast A
is standing absolutely straight, determine the height to which
gymnast A will rise.
SOLUTION
I =
Moment of inertia.
1
1
mP (2 L)2 = mP L2
12
3
( v )1 = 2 gh1
Velocity of jumper at E.
(1)
Principle of impulse-momentum.
Syst. Momenta1
Kinematics:
Moments about D:
+
Syst. Ext. Imp. 1Æ2
vC = Lw
=
Syst. Momenta 2
vD = Lw
mE v1 L + 0 = mE vE L + mC vC L + I w
1
= mE L2w + mC L2w + mP L2w
3
mE
v1
w=
mE + mC + 13 mP L
vC = Lw =
mE v1
mE + mC + 13 mP
vC2
2g
Gymnast (flier) rising.
hC =
Data:
mE = mB = 70 kg
mC = mA = 55 kg
(2)
(3)
mP = 15 kg
h1 = 2.5 m
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196
PROBLEM 17.125 (Continued)
From Equation (1)
v1 = (2)(9.81)(2.5)
= 7.0036 m/s
(70)(7.0036)
70 + 55 + 5
= 3.7712 m/s
From Equation (2)
vC =
From Equation (3)
h2 =
(3.7712)2
(2)(9.81)
= 0.725 m
h2 = 725 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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197
PROBLEM 17.126
Solve Problem 17.125, assuming that the gymnasts change places so
that gymnast A jumps onto the plank while gymnast B stands at C.
PROBLEM 17.125 The plank CDE has a mass of 15 kg and rests
on a small pivot at D. The 55-kg gymnast A is standing on the plank
at C when the 70-kg gymnast B jumps from a height of 2.5 m and
strikes the plank at E. Assuming perfectly plastic impact and that
gymnast A is standing absolutely straight, determine the height to
which gymnast A will rise.
SOLUTION
I =
Moment of inertia.
1
1
mP (2 L)2 = mP L2
3
12
( v )1 = 2 gh1
Velocity of jumper at E.
(1)
Principle of impulse-momentum.
Syst. Momenta1
Kinematics:
Moments about D:
+
Syst. Ext. Imp. 1Æ2
vC = Lw
=
Syst. Momenta 2
vD = Lw
mE v1 L + 0 = mE vE L + mC vC L + I w
1
= mE L2w + mC L2w + mP L2w
3
mE
v1
w=
mE + mC + 13 mP L
vC = Lw =
Gymnast (flier) rising.
hC =
mE v1
mE + mC + 13 mP
vC2
2g
(2)
(3)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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198
PROBLEM 17.126 (Continued)
Data:
mE = mA = 55 kg
mC = mB = 70 kg
mP = 15 kg
h1 = 2.5 m
From Equation (1)
v1 = (2)(9.81)(2.5)
= 7.0036 m/s
(55)(7.0036)
55 + 70 + 5
= 2.9631 m/s
From Equation (2)
vC =
From Equation (3)
h2 =
(2.9631)2
(2)(9.81)
= 0.447 m
h2 = 447 mm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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199
PROBLEM 17.127
Member ABC has a mass of 2.4 kg and is attached to a pin support
at B. An 800-g sphere D strikes the end of member ABC with
a vertical velocity v1 of 3 m/s. Knowing that L = 750 mm and
that the coefficient of restitution between the sphere and member
ABC is 0.5, determine immediately after the impact (a) the angular
velocity of member ABC, (b) the velocity of the sphere.
SOLUTION
mD = 0.800 kg
L = 0.750 m
1
L = 0.1875 m
4
mAC = 2.4 kg
Let Point G be the mass center of member ABC.
1
mAC L2
12
1
= (2.4)(0.750)2
12
= 0.1125 kg ◊ m2
IG =
w¢ = w ¢
Kinematics after impact.
,
vG¢ =
L
w ¢ ≠,
4
v ¢A =
L
w¢ Ø
4
Conservation of momentum.
+
Moments about B:
L
L
L
+ 0 = mD vD¢ + I Gw ¢ + mAC vG¢
2
2
4
2
È
L
L
Ê Lˆ ˘
mD vD = mD vD¢ + Í I G + mAD Á ˜ ˙ w ¢
Ë 4¯ ˙
4
4 ÍÎ
˚
mD vD
(0.800)(3)(0.1875) = (0.800)(0.1875)vD¢ + [0.1125 + (2.4)(0.1875)2 ]w ¢
0.45 = 0.15vD¢ + 0.196875 w ¢
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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200
PROBLEM 17.127 (Continued)
L
w¢
4
= - e( v D - v A )
vD¢ - v A¢ = vD¢ -
Coefficient of restitution.
vD¢ - 0.1875w ¢ = -(0.5)(3 - 0)
(2)
Solving Eqs. (1) and (2) simultaneously.
(a)
Angular velocity.
w¢ = 3
w ¢ = 3.00 rad/s
(b)
Velocity of D.
vD¢ = -0.9375
vD¢ = 0.938 m/s ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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201
PROBLEM 17.128
Member ABC has a mass of 2.4 kg and is attached to a pin support
at B. An 800-g sphere D strikes the end of member ABC with
a vertical velocity v1 of 3 m/s. Knowing that L = 750 mm and
that the coefficient of restitution between the sphere and member
ABC is 0.5, determine immediately after the impact (a) the
angular velocity of member ABC, (b) the velocity of the sphere.
SOLUTION
Let M be the mass of member ABC and I its moment of inertia about B.
where
1
M ( 2 L)2
12
L = 750 mm = 0.75 m
Let m be the mass of sphere D.
m = 800 g = 0.8 kg
M = 2.4 kg I =
Impact kinematics and coefficient of restitution.
( v1 sin q )e = Lw 2 - ( vD )n : ( vD ) n = Lw 2 - ( v1 sin q ) e
(1)
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about E:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
mv1 L sin q = I w 2 + m( vD )n L
1
M (2 L)2 w 2 + m[ Lw 2 - ( v1 sin q )e]L
12
1
mv1 sin q = MLw 2 - mLw 2 - m( v1 sin q )e
3
v
ˆ
Ê1
m(1 + e) 1 sin q = Á M + m˜ w 2
¯
Ë
L
3
mv1 L sin q =
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202
PROBLEM 17.128 (Continued)
(a)
Angular velocity.
w2 =
(3)(1 + e)mv1 sin q
M + 3m
(3)(1.5)(0.8)(3)sin 60∞
(2.4 + 2.4)(0.75)
= 2.5981
w2 =
(b)
w 2 = 2.60 rad/s
Velocity of D.
From Eq. (1),
( vD )n = (0.75)(2.5981) - (3 sin 60∞)(0.5)
= 0.64976 m/s
( vD )t = v1 cos 60∞
= 3 cos 60∞
= 1.5 m/s
( v D ) n = 0.64976 m/s
( v D )t = 1.5 m/s
30∞
30∞
vD = (0.64976)2 + (1.5)2
= 1.63468 m/s
0.64976
tan q =
1.5
q = 23.4∞
q = 30∞ = 53.4∞
vD = 1.635 m/s
53.4∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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203
PROBLEM 17.129
A slender rod CDE of length L and mass m is attached to a pin
support at its midpoint D. A second and identical rod AB is
rotating about a pin support at A with an angular velocity w1
when its end B strikes end C of rod CDE. Denoting by e the
coefficient of restitution between the rods, determine the angular
velocity of each rod immediately after the impact.
SOLUTION
Rod AB.
w1 = w1
Kinematics.
w2 = w2
( v AB )1 =
L
w1 Ø
2
( v AB )2 =
L
w2 Ø
2
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about A:
+
I w1 + m( v AB )1
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
L
L
- ( F Dt ) L = I (w AB )2 + m( v AB )2
2
2
L
L
1
1
Ê Lˆ L
mL2w1 + m Á w1 ˜ - ( F Dt ) L = mL2 (w AB )2 + m (w AB )2
Ë 2¯ 2
12
12
2
2
1
1 2
mL w1 - ( F Dt ) L = mL2 (w AB )2
3
3
1
F D t = mL2 [w1 - (w AB )2 ]
3
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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204
PROBLEM 17.129 (Continued)
Rod CE.
Principle of impulse and momentum.
Syst. Momenta1
+
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
L
= I (w CE )2
2
L 1
( F Dt ) = mL2 (w CE )2
2 12
( F Dt )
Moments about D:
Substitute for ( F D t ) from (1)
L 1
1 2
mL [w1 - (w AB )2 ] = mL2 (w CE )2
3
2 12
1
w1 - (w AB )2 = (w CE )2
2
(2)
Condition of impact. e = coefficient of restitution.
( v B )1 e = ( vC )2 - ( v B )2
Lw1e =
L
(w CE )2 - L(w AB )2
2
1
(w AB )2 = (w CE )2 - w1e
2
From Eq. (2)
È1
˘ 1
w1 - Í (w CE )2 - w1e ˙ = (w CE )2
˚ 2
Î2
w1 (1 + e) = (w CE )2
From Eq. (3)
(3)
1
1
1
(w AB )2 - w1 (1 + e) - w1e = w1 + w1e - w1e
2
2
2
(w CE )2 = w1 (1 + e)
1
(w AB )2 = w1 (1 - e)
2
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205
PROBLEM 17.130
The 2.5-kg slender rod AB is released from rest in the position shown
and swings to a vertical position where it strikes the 1.5-kg slender rod
CD. Knowing that the coefficient of restitution between the knob K
attached to rod AB and rod CD is 0.8, determine the maximum angle
q m through which rod CD will rotate after the impact.
SOLUTION
Let
Let
Then
Moments of inertia.
b=
mCD 1.5 kg
=
= 0.6
mAB 2.5 kg
m = mAB .
mCD = b m.
I AB = I =
I CD =
1
mL2
12
1
b mL2
12
Rod AB falls to vertical position.
Position 0.
V0 = 0
Position 1.
V1 = - mg
T0 = 0
L
2
L
(w AB )1
2
1
1
T1 = m( v AB )12 + I (w AB )12
2
2
1 2
= mL (w AB )12
6
( v AB )1 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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206
PROBLEM 17.130 (Continued)
Conservation of energy.
1
1
T0 + V0 = T1 + V1: 0 + 0 = mL2 (w AB )12 - mgL
6
2
(w AB )12 =
3g
L
(1)
Impact.
Syst. Momenta1 + Syst. Ext. Imp. 1®2 = Syst. Momenta 2
L
(w AB )1
2
L
( v AB )2 = (w AB )2
2
( v AB )1 =
Kinematics
Moments about B:
L
Ê Lˆ
+ I (w AB )1 - b Ú Kdt = m ( v AB )2 Á ˜ + I (w AB )2
Ë 2¯
2
m( v AB )1
1 2
1
mL (w AB )1 - b Ú Kdt = mL2 (w AB )2
3
3
Syst. Momenta1
Kinematics
Moments about C:
+
Syst. Ext. Imp. 1®2
( vCD )2 =
=
(2)
Syst. Momenta 2
L
(w CD )2
2
0 + b Ú Kdt = b m( vCD )2
L
+ b I (w CD )2
2
1
b Ú Kdt = b mL2 (w CD )2
3
(3)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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207
PROBLEM 17.130 (Continued)
Add Equations (1) and (2) to eliminate b Ú Kdt .
1 2
1
1
mL (w AB )1 = mL2 (w AB )2 + b mL2 (w CD )2
3
3
3
b (w CD )2 + (w AB )2 = (w AB )1
Condition of impact.
(4)
b(w CD )2 - b(w AB )2 = eb(w AB )1
(w CD )2 - (w AB )2 = e(w AB )1
(5)
Add Equations (4) and (5) to eliminate (w AB )2 .
(1 + b )(w CD )2 = (1 + e)(w AB )1
Ê 1+ e ˆ
(w CD )2 = Á
(w AB )1
Ë 1 + b ˜¯
(6)
Rod CD rises to maximum height.
Position 2.
V2 = - b mg
L
2
1
1
b m( vCD )22 + b I (w CD )22
2
2
1
= b mL2 (w CD )22
6
T2 =
Position 3.
V3 = - b mg
L
cos q m
2
T3 = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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208
PROBLEM 17.130 (Continued)
Conservation of energy.
T2 + V2 = T3 + V3 :
1
1
1
b mL2 (w CD )22 - b mgL = 0 - b mgL cos q m
6
2
2
L
w CD )22
1 - cos q m =
(w
3g
2
=
L Ê 1+ e ˆ
(w AB )12
3g ÁË 1 + b ˜¯
=
L Ê 1 + e ˆ 3g
3g ÁË 1 + b ˜¯ L
2
Ê 1+ e ˆ
cos q m = 1 - Á
Ë 1 + b ˜¯
2
Ê 1 + 0.8 ˆ
=1- Á
Ë 1 + 0.6 ˜¯
2
q m = 105.4∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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209
PROBLEM 17.131
Sphere A of mass m and radius r rolls without slipping with
a velocity v1 on a horizontal surface when it hits squarely an
identical sphere B that is at rest. Denoting by m k the coefficient
of kinetic friction between the spheres and the surface, neglecting
friction between the spheres, and assuming perfectly elastic
impact, determine (a) the linear and angular velocities of each
sphere immediately after the impact, (b) the velocity of each
sphere after it has started rolling uniformly.
SOLUTION
I =
Moment of inertia.
2 2
mr
5
Analysis of impact. Sphere A.
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
Kinematics: Rolling without slipping in Position 1.
v1
r
I w1 + 0 = I w A
wA =
Moments about G:
+
w A = w1 =
Linear components:
v1
r
mv1 - Ú Pdt = mv A
(1)
Analysis of impact. Sphere B.
+
Syst. Momenta1
Linear components:
+ Syst. Ext. Imp. 1Æ2
0 + Ú Pdt = mv B
=
Syst. Momenta 2
(2)
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210
PROBLEM 17.131 (Continued)
Add Equations (1) and (2) to eliminate Ú Pdt.
mv1 = mv A + mv B
Condition of impact. e = 1.
or v B + v A = v1
(3)
v B - v A = ev1 = v1
(4)
Solving Equations (4) and (5) simultaneously,
Moments about G:
(a)
v A = 0,
v B = v1
0 + 0 = I wB
wB = 0
vA = 0; w A =
Velocities after impact.
v1
r
; vB = v1 Æ; w B = 0 Motion after Impact. Sphere A.
Syst. Momenta1
+
Condition of rolling without slipping:
Syst. Ext. Imp. 1®2
=
Syst. Momenta 2
v A¢ = w A¢ r
I w A + 0 + I w A¢ + mv A¢ r
Moments about C:
Ê 2 2ˆ
Ê 2 2 ˆ Ê v1 ˆ
ÁË mr ˜¯ ÁË ˜¯ + 0 = ÁË mr ˜¯ w A¢ + m( rw A¢ )r
r
5
5
2 v1
7 r
2
v A¢ = v1
7
w A¢ =
Motion after impact. Sphere B.
Syst. Momenta1
+
Syst. Ext. Imp. 1®2
=
Syst. Momenta 2
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211
PROBLEM 17.131 (Continued)
Condition of rolling without slipping:
Moments about C:
v B¢ = rw B¢
mv B r + 0 = I w B¢ + mv B¢ r
ˆ
Ê2
mv1r + 0 = Á mr 2 ˜ w B¢ + m( rw B¢ )r
¯
Ë5
5 v1
7 r
5
v B¢ = v1
7
w B¢ =
(b)
v ¢A =
Final Rolling Velocities.
2
5
v1 Æ; v ¢B = v1 Æ 7
7
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212
PROBLEM 17.132
A small rubber ball of radius r is thrown against a rough floor with a
velocity v A of magnitude υ0 and a backspin w A of magnitude w 0 . It
is observed that the ball bounces from A to B, then from B to A, then
from A to B, etc. Assuming perfectly elastic impact, determine the
required magnitude w 0 of the backspin in terms of υ0 and r.
SOLUTION
I =
Moment of inertia.
2 2
mr Ball is assumed to be a solid sphere.
5
Impact at A.
Syst. Momenta1
+
Syst. Ext. Imp. 1®2
=
Syst. Momenta 2
For the velocity of the ball to be reversed on each impact,
v A¢ = v A = v0
w A¢ = w A = w 0
This is consistent with the assumption of perfectly elastic impact.
Moments about C:
mv A r cos 60∞ - I w A + 0 = I w A¢ - mv A¢ r cos 60∞
2
2
mv0 r cos 60∞ - mr 2w 0 + 0 = mr 2w 0 - mv0 r cos 60∞
5
5
2
rw 0 = v0 cos 60∞
5
w0 =
5 v0
4 r
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213
PROBLEM 17.133
In a game of pool, ball A is rolling without slipping with a velocity v0 as it hits
obliquely ball B, which is at rest. Denoting by r the radius of each ball and by m k
the coefficient of kinetic friction between the balls, and assuming perfectly elastic
impact, determine (a) the linear and angular velocity of each ball immediately
after the impact, (b) the velocity of ball B after it has started rolling uniformly.
SOLUTION
I =
Moment of inertia.
(a)
2 2
mr
5
Impact analysis.
Ball A:
Syst. Momenta1
+
Syst. Ext. Imp. 1®2
w0 =
Kinematics of rolling:
=
Syst. Momenta 2
v0
r
(1)
mv0 sin q + 0 = m( v A ) y
(2)
Moments about y axis:
I w 0 cos q + 0 = I w A cos b
(3)
Moments about x axis:
- I w 0 sin q + 0 = - I w A sin b
(4)
+
mv0 cos q - Ú Pdt = m( v A ) x
Linear components:
Linear components:
+
Ball B:
+
Syst. Momenta1
Linear components:
Linear components:
+
Syst. Ext. Imp. 1®2
=
Syst. Momenta 2
0 + Ú Pdt = m( v B ) x
(5)
0 + 0 = m( v B ) y
(6)
+
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214
PROBLEM 17.133 (Continued)
Moments about y axis:
0 + 0 = I w B cos g
(7)
Moments about x axis:
0 + 0 = I w B sin g
(8)
Adding Equations (1) and (5) to eliminate Ú Pdt ,
mv0 cos q + 0 = m( v A ) x + m( v B ) x
( v B ) x + ( v A ) x = v0 cos q
or
(9)
e = 1 ( v B ) x - ( v A ) x = ev0 cos q = v0 cos q
Condition of impact.
(10)
Solving Equations (9) and (10) simultaneously,
( v A ) x = 0, ( v B ) x = v0 cos q
( v A ) y = v0 sin q , ( v B ) y = 0
From Equations (2) and (6),
v A = ( v0 sin q ) j v B = ( v0 cos q )i From Equations (3) and (4) simultaneously,
b = q , w A = w0 =
v0
r
wA =
v0
( - sin q i + cos q j) r
From Equations (7) and (8) simultaneously,
wB = 0
(b)
wB = 0 Subsequent motion of ball B.
Syst. Momenta1
+
Syst. Ext. Imp. 1®2
Kinematics of rolling without slipping.
Moments about C:
=
Syst. Momenta 2
v B¢ = rw B¢
mv B r + 0 = I w B¢ + mv B¢ r
2
= mr 2w B¢ + m( rw B¢ )r
5
5 v B¢ 5 v1 cos q
=
w B¢ =
r
7 r
7
5
v B¢ = v1 cosq
7
5
v ¢B = ( v0 cos q )i 7
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215
PROBLEM 17.134
Each of the bars AB and BC is of length L = 400 mm and mass m = 1 kg. Determine the
angular velocity of each bar immediately after the impulse QDt = (1.5 N ◊ s)i is applied at C.
SOLUTION
Principle of impulse and momentum.
Bar BC:
Syst. Momenta1
+
+
Syst. Ext. Imp. 1®2
=
Syst. Momenta 2
L
2
L
ˆL
Ê
+ m Á Lw AB + w BC ˜
¯2
Ë
2
0 + (Q Dt ) L = I w BC + mv BC
Moments about B:
(Q Dt ) L =
1
mL2w BC
12
1
1
mLw AB + mLw BC
2
3
Q Dt - Bx Dt = mv BC
+
Q Dt =
x components:
(1)
1
ˆ
Ê
Q Dt - Bx Dt = m Á Lw AB + w BC ˜
¯
Ë
2
(2)
Bar AB:
Syst. Momenta1
+
Syst. Ext. Imp. 1®2
=
Syst. Momenta 2
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216
PROBLEM 17.134 (Continued)
+
Moments about A:
( Bx Dt ) L =
Add Eqs. (2) and (3):
Subtract Eq. (1) from Eq. (4):
Substitute for w BC in Eq. (1):
1
mL2w AB
12
1
Bx Dt = mLw AB
3
4
1
Q Dt = mLw AB + mLw BC
3
2
5
1
0 = mLw AB + mLw BC
6
6
(4)
w BC = -5w AB
(5)
(3)
1
1
mLw AB + mL( -5w AB )
2
3
7
= - mLw AB
6
Q Dt =
w AB = Substituting into Eq. (5):
L
2
ˆL
Ê1
+ m Á w AB ˜
¯2
Ë2
0 + ( Bx Dt ) L = I w AB + mv AB
6 Q Dt
7 mL
(6)
Ê 6 Q Dt ˆ
w BC = -5 Á Ë 7 mL ˜¯
w BC =
30 Q Dt
7 mL
Given data:
L = 400 mm = 0.4 m
Q Dt = 1.5 N ◊ s
m = 1 kg
Angular velocity of bar AB.
w AB = -
Angular velocity of bar BC.
w BC =
6 Q Dt
6 (1.5)
=7 mL
7 (1)(0.4)
30 Q Dt
(30)(1.5)
=
7 mL (7)(1)(0.4)
(7)
w AB = 3.21 rad/s
w BC = 16.07 rad/s
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217
PROBLEM 17.135
The motion of the slender 250-mm rod AB is guided by pins at A and B that
slide freely in slots cut in a vertical plate as shown. Knowing that the rod has
a mass of 2 kg and is released from rest when q = 0, determine the reactions
at A and B when q = 90∞.
SOLUTION
Let Point G be the mass center of rod AB.
m = 2 kg
L = 0.025 m
1
I G = mL2 = 0.0104667 kg ◊ m2
12
Kinematics.
q = 90∞
AD = R = 0.125 m
AB = L = 0.25 m
R 1
b = 30∞
sin b = =
L 2
L
AG = = 0.125 m
2
BG = 0.125 m
Point E is the instantaneous center of rotation of bar AB.
L
vG = w = 0.125 w
2
v A = ( L cos 30∞)w = 0.21651w
v B = ( L sin 30∞)w = 0.125w
Use principle of conservation of energy to obtain the velocities when q = 90∞:
Use level A as the datum for potential energy.
Position 1.
q = 0 T1 = 0
L
V1 = - mg
2
= -(2)(9.81)(0.125)
= -2.4525 J
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218
PROBLEM 17.135 (Continued)
Position 2.
q = 90∞
1
1
I Gw 2 + mvG2
2
2
1
1
= (0.0104667)w 2 + (0.125w )2
2
2
= 0.0208583w 2
T2 =
L
ˆ
Ê
V2 = - mg Á R + cos b ˜
¯
Ë
2
= -(2)(9.81)(0.125 + 0.125 cos 30∞)
= -4.5764 J
T1 + V1 = T2 + V2 : 0 - 2.4525 = 0.0208583w 2 - 4.5764
w 2 = 101.826 rad 2 /s2
w = 10.091 rad/s
v A = (0.21651)(10.091)
= 2.1848 m/s
vG = (0.125)(10.091)
= 1.2614 m/s
More kinematics:
For Point A moving in the curved slot,
v 2A
j
R
(2.1847)2
= ( aC ) x i +
j
0.125
= ( aC ) x i + 38.1833j
a A = ( aC ) x i +
For the rod AB,
a = ak , v B = v B j
rA/B = - L sin 30∞i + L cos 30∞ j
= -0.125i + 0.21651j
1
rG /B = rA/B
2
= -0.06255i + 0.108253j
a A = a B + a + rA/B - w 2 rA/B
= aB j + a k ¥ ( -0.125i + 0.21651j)
- (10.091)2 ( -0.125i + 0.21651j)
= aB j - 0.125a j - 0.21651a i + 12.7285i - 22.0468 j
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219
PROBLEM 17.135 (Continued)
Matching vertical components of a A
38.1833 = aB - 0.125a - 22.0468
aB = 0.125a + 60.2301
a G = a B + a G /B
= a B + a k ¥ rG /B - w 2 rG /B
= (0.125a + 60.2324) j + a k ¥ ( -0.0625i + 0.108253j)
- (10.091)2 ( -0.0625i + 0.108253j)
= 0.125a j + 60.2301j - 0.0625a j - 0.108253a i
+ 6.3643i - 11.0232 j
aG = ( -0.108253a + 6.3643)i + (0.0625a + 49.2069)j
Kinetics:
Use rod AB as a free body.
+
- mg
SM E = S ( M E )eff :
L
sin bk = I G a + rG /E ¥ aG - (2)(9.81)(0.125)sin 30∞k
2
= 0.0104667a + (0.0625i + 0.108253j)( maG )
-1.22625 = 0.0104667a + 0.03125a + 4.7730
0.0417167a = -5.999 R
a = -143.808 rad/s2
aG = ( -21.933 m/s2 )i + ( 40.2189 m/s2 ) j
+
SFx = S ( Fx )eff = m( aG ) x : -B = (2)( -21.932) = 43.864 N
B = 43.9 N Æ SFy = S ( Fy )eff = m( aG ) y : A - mg = (2)( 40.4289) = 80.4378
+
A = (2)(9.81) + 80.4378 = 100.058
A = 100.1 N ≠ Check by considering
+
SM G = SM G eff :
SM G = (0.0625) A - 0.108253B = 1.5052 N ◊ m
S( M G )eff = I G ( -a ) = (0.0104667)(143.808) = 1.5052 N ◊ m
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220
PROBLEM 17.136
A uniform disk of constant thickness and initially at rest is placed
in contact with the belt shown, which moves at a constant speed
n = 25 m/s. Knowing that the coefficient of kinetic friction between
the disk and the belt is 0.15, determine (a) the number of revolutions
executed by the disk before it reaches a constant angular velocity,
(b) the time required for the disk to reach that constant angular velocity.
SOLUTION
F f = m k N = 0.15 N
Kinetic friction.
+≠ SFy = N cos 25∞ - F f sin 25∞ - mg = 0
(cos 25∞ - m k sin 25∞)N = mg
mg
cos 25∞ - 0.15 sin 25∞
= 1.18636mg
F f = (0.15)(1.18636)mg
N=
= 0.177954mg
v
r
1
I = mr 2
2
w2 =
Final angular velocity.
Moment of inertia.
(a)
Principle of work and energy.
T1 + W1Æ2 = T2 : T1 = 0
W1Æ2 = F f rq = 0.177954mgrq
2
1
1Ê1
1
ˆ Ê vˆ
T2 = I w 22 = Á mr 2 ˜ Á ˜ = mv 2
¯
Ë
Ë
¯
r
2
2 2
4
1 2
0 + 0.177954 mgrq = mv
4
v2
q = 1.40486
gr
=
(1.40486)(25)2
(9.81)(0.120)
= 745.87 radians
q = 118.7 rev PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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221
PROBLEM 17.136 (Continued)
(b)
Principle of impulse-momentum.
Syst. Momenta1
Moments about A:
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta 2
0 + F f tr = I w 2
t=
=
I w2
Ff r
(
2
1
2 mr
)( )
v
r
0.177954mgr
v
= 2.8097
g
(2.8097)(25)
=
9.81
t = 7.16 s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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222
PROBLEM 17.137
Solve Problem 17.136, assuming that the direction of motion of the belt
is reversed.
SOLUTION
While slipping occurs:
SFy = 0 : N cos b + m k N sin b - mg = 0
mg
cos b + m k sin b
v
w=
r
M F = Fr = Moment of F about A.
+
N=
For cylinder slipping occurs until
Work:
(1)
U1Æ2 = M F q ¥ Frq
= m k Nrq
2
Kinetic energy:
Principle of work and energy:
T1 = 0 : T2 =
1
1Ê1
1
ˆ Ê vˆ
I w 2 = Á mr 2 ˜ Á ˜ = mv 2
¯ Ë r¯
2
2Ë2
2
T1 + U1Æ2 = T2
0 + m k Nrq =
q=
=
q=
1 2
mr
4
1 mv 2 1
◊
4 mk r N
1 mv 2 cos b + m k sin q
◊
4 mk r
mg
1 v2
◊
(cos b + m k sinq
q)
4 m k rg
(2)
Principle of impulse-momentum
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223
PROBLEM 17.137 (Continued)
+
Moments about A:
Ftr = I w
m k Ntr =
Substituting for N:
1 2 Ê vˆ
mv Á ˜
Ë r¯
2
Ê
ˆ
mg
1
mk Á
tr = mrv
2
Ë cos b + m k sin a ˜¯
1 v
t= ◊
(cos b + m k sin b )
2 mk g
(3)
m k = 0.15
Data:
b = 25∞
v = 25 m/s
r = 0.12 m
(a)
From Eq. (2),
q=
(25 m/s)2
[cos 25∞ + (0.15)sin 25∞]
4(0.15)(0.12 m)(9.81 m/s2 )
q = 858.05 rad
(b)
From Eq. (3),
t=
q = 136.6 revolutions 25 m/s
[cos 25∞ + (0.15)sin 25∞] t = 3.24 s 2(0.15)(9.81 m/s2 )
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224
PROBLEM 17.138
A uniform slender rod is placed at corner B and is given a slight
clockwise motion. Assuming that the corner is sharp and becomes
slightly embedded in the end of the rod, so that the coefficient of static
friction at B is very large, determine (a) the angle b through which
the rod will have rotated when it loses contact with the corner, (b) the
corresponding velocity of end A.
SOLUTION
T1 = 0
Position 1.
V1 = mgh1 =
mgL
2
mgL cos b
2
1 2 1 Ê 1 2ˆ 2
T2 = Iw 2 = Á mL ˜ w 2
¯
2
2Ë3
V2 = mgh2 =
Position 2.
Principle of conservation of energy.
T1 + V1 = T2 + V2
0+
mgL 1 Ê 1 2 ˆ 2 mgL cos b
= Á mL ˜ w 2 +
¯
2
2Ë3
2
3g
w 22 =
(1 - cos b )
L
(1)
Normal acceleration of mass center.
an =
+
L 2 3
w 2 = g (1 - cos b )
2
2
SF = + SFeff = man
3
mg (1 - cos b )
2
5
3
cos b =
cos b = 0.6
2
2
mg cos b =
(a)
(b)
Angle b.
b = 53.1∞ 3g
g
g
(1 - 0.6) = 1.2
w 2 = 1.09545
L
L
L
From (1)
w 22 =
Velocity of end A
v A = tw 2
vA = 1.095 gL
53.1∞ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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225
PROBLEM 17.139
A 35-g bullet B is fired with a velocity of 400 m/s into the side of a
3-kg square panel suspended as shown from a pin at A. Knowing that
the panel is initially at rest, determine the components of the reaction at
A after the panel has rotated 90∞.
SOLUTION
Masses and moment of inertia.
mB = 0.035 kg
mP = 3 kg
b = 500 mm = 0.5 m
1
1
I G = mP b2 = (3)(0.5)2 = 0.125 kg ◊ m2
6
6
Note: The mass of the bullet is neglected in comparison with that of the plate after impact.
Analysis of impact: Use principle of impulse and momentum.
Kinematics:
+
After impact the plate rotates about the pin at A.
0.5
b
vG =
w=
w
2
2
b
b
Moments about A:
mG v0 + 0 = I Gw + m p vG
2
2
1
1
mG v0 b = ( I G + mP b2 )w
2
2
1
1
È
˘
(0.035)( 400)(0.5) = Í0.125 + (3)(0.5)2 ˙ w
2
2
Î
˚
w = 7 rad/s
vG =
0.5
(7) = 2.4749 m/s
2
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226
PROBLEM 17.139 (Continued)
1
1
I Gw 2 + mP vG2
2
2
1
1
T1 = (0.125)(7)2 + (3)(2.4749)2
2
2
= 12.25 J
T1 =
Corresponding kinetic energy.
Plate rotates through 45∞.
q = 0∞
Position 1:
Use Point A as the datum for potential energy
V1 = - mP g
b
2
0.5
2
= -10.4051 J
= -(3)(9.81)
q = 90∞
Position 2:
V2 = 0 since G is at level A.
1
1
T2 = I Gw 22 + mP ( vG )22
2
2
1 Ê 0.5 ˆ
1
w2
= (0.125)w 22 + (3) Á
2 Ë 2 ˜¯
2
2
= 0.25w 22
Principle of conservation of energy:
T1 + V1 = T2 + V2
12.25 J - 10.4051 J = 0.25w 22 + 0
w 22 = 7.3796 (rad/s2 )
w 2 = 2.7165 rad/s
Analysis at 90∞ rotation.
Kinematics:
a =a
0.5
b
a=
a
2
2
b 2
( aG ) n =
w
2
(0.5)(7.3796)
=
2
( aG )t =
(aG )t = 0.35355a Ø
(aG )n = 2.6091 m/s2 Æ
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227
PROBLEM 17.139 (Continued)
Kinematics: Use the free body diagram of the plate.
b
b
= I Ga + mP ( aG )t
+ SM A = S ( M A )eff : mP g
2
2
1
ˆ
Ê
= Á I G + mP b2 ˜ a
¯
Ë
2
(3)(9.81)(0.5) È
1
˘
= Í0.125 + (3)(0.5)2 ˙ a
2
2
˚
Î
a = 20.810 rad/s2
+
(aG )t = 7.3574 m/s2 Ø
SFx = S ( Fx )eff : Ax = mP ( aG )n = (3)(2.6091)
A x = 7.83 N Æ SFy = S ( Fy )eff : Ay - mP g = - mP ( aG ) y
+
Ay - (3)(9.81) = (3)( -7.3574)
A y = 7.35 N ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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228
PROBLEM 17.140
A square block of mass m is falling with a velocity v1 when it strikes a small obstruction
at B. Assuming that the impact between corner A and the obstruction B is perfectly
plastic, determine immediately after the impact (a) the angular velocity of the block
(b) the velocity of its mass center G.
SOLUTION
1
I = mb2
6
Moment of inertia.
Kinematics.
Before impact, block is translating.
v1 = v1 Ø w1 = 0
After impact, block is rotating about edge A.
v2 =
b
w2
2
45∞
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about A.
(a)
Angular velocity.
(b)
Velocity of the mass center.
+
Syst. Ext. Imp. 1Æ2 =
Syst. Momenta 2
b
= I w 2 + mv2 ( AG )
2
1
ˆÊ b ˆ
Ê b
w2 ˜ Á
= mb2w 2 + m Á
Ë 2 ¯ Ë 2 ˜¯
6
2
= mb2w 2
3
3v
w2 = 1
4 b
mv1
v2 =
b
3
v1
w2 =
2
4 2
w 2 = 0.750
v2 = 0.530v1
v1
b
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229
PROBLEM 17.141
Solve Problem 17.140, assuming that the impact between corner A and the obstruction
B is perfectly elastic.
SOLUTION
1
I = mb2
6
Moments of inertia.
Kinematics.
Before impact, block is translating.
v1 = v1 Ø w1 = 0
After impact with e = 1:
vA = v1 ≠
v2 = vA + vG/ A
È b
= [v1 ≠] + Í w 2
Î 2
45∞]
(1)
Principle of impulse and momentum.
Syst. Momenta1 +
+
Syst. Ext. Imp. 1Æ2 =
Syst. Momenta2
Moments about A:
mv1
b
b
= Iw1 - mv1 + m
2
2
1 2
I = mb ( Ag ) =
6
b
b
w2
2
2
b
2
2
b 1 2
b
Ê b ˆ
= mb w 2 - mv1 + m Á
w
Ë 2 ˜¯ 2
2 b
2
2
mv1b = b2w 2
3
mv1
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230
PROBLEM 17.141 (Continued)
(a)
Angular velocity.
(b)
Velocity of the mass center.
From Eq. (1),
w2 =
3 v1
2 b
w 2 = 1.500
È b 3 v1
v2 = [v1 ≠] + Í
◊
Î 2 2 b
v
b
˘
45∞ ˙
˚
È 3
˘ È 3
˘
= [v1 ≠] + Í
v1 sin 45∞ Ø˙ + Í
v1 cos 45∞ Æ ˙
Î2 2
˚ Î2 2
˚
˘ È3
˘ È1
˘ È3
È3
˘
= [v1 ≠] + Í v1 Ø˙ + Í v Æ ˙ = Í v1 ≠˙ Í v1 Æ ˙
4
4
4
4
˚ Î
˚ Î
˚Î
Î
˚
v2 = 0.791v1
18.4° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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231
PROBLEM 17.142
A 3-kg bar AB is attached by a pin at D to a 4-kg square plate, which can rotate
freely about a vertical axis. Knowing that the angular velocity of the plate is
120 rpm when the bar is vertical, determine (a) the angular velocity of the plate
after the bar has swung into a horizontal position and has come to rest against
pin C, (b) the energy lost during the plastic impact at C.
SOLUTION
Moments of inertia about the vertical centroidal axis.
1
1
mL2 = ( 4)(0.500)2 = 0.083333 kg ◊ m2
12
12
Square plate.
I =
Bar AB vertical.
I = approximately zero
Bar AB horizontal.
I =
Position 1. Bar AB is vertical.
I1 = 0.083333 kg ◊ m2
1
1
mL2 = (3)(0.500)2 = 0.0625 kg ◊ m2
12
12
w1 = 120 rpm = 4p rad/s
Angular velocity.
Angular momentum about the vertical axis.
( H O )1 = I1w1 = (0.083333)( 4p ) = 1.04720 kg ◊ m2 /s
1
1
I1w12 = (0.083333)( 4p )2 = 6.5797 J
2
2
Kinetic energy.
T1 =
Position 2. Bar AB is horizontal.
I 2 = 0.145833 kg ◊ m2
( H O )2 = I 2w 2 = 0.145833w 2
( H O )1 = ( H O )2 :
Conservation of angular momentum.
1.04720 = 0.145833w 2 w 2 = 7.1808 rad/s
(a)
Final angular velocity.
(b)
Loss of energy.
w 2 = 68.6 rpm T1 - T2 = T1 -
1
1
I 2w 22 = 6.5797 - (0.145833)(0.71808)2
2
2
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232
PROBLEM 17.143
A 300 ¥ 400 mm-rectangular plate is suspended by pins at A and B. The
pin at B is removed and the plate swings freely about pin A. Determine
(a) the angular velocity of the plate after it has rotated through 90°,
(b) the maximum angular velocity attained by the plate as it swings freely.
SOLUTION
Let m be the mass of the plate.
a = 0.4 m b = 0.3 m
Dimensions:
Moment of inertia about A
1
I A = m( a2 + b2 )
3
Position 1. Initial position.
w1 = 0
Position 2. Plate has rotated about A through 90°.
Position 3. Mass center is directly below pivot A.
Potential energy. Use level A as datum.
V1 = -
mab
mga
V2 = V3 = - mgd
2
2
1 2
a + b2 = 0.25 m
2
where
d=
Kinetic energy.
T1 = 0 T2 =
1
1
I Aw 22 T3 = I Aw 32
2
2
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233
PROBLEM 17.143 (Continued)
(a)
90° rotation. Conservation of energy.
mgb 1 1
mga
= ◊ m( a2 + b2 )w 22 +
2
2 3
2
g
a
b
3
)(
9
.
81
)(
0
.
4
0
.
3
)
3
(
)
(
= 11.772( rad/s)2
w 22 = 2
=
È(0.4)2 + (0.3)2 ˘
a + b2
Î
˚
T1 + V1 = T2 + V2 : 0 -
(b)
w 2 = 3.43 rad/s
w 3 = 4.85 rad/s
w is maximum. Conservation of energy.
mgb 1 1
= ◊ m( a2 + b2 )w 32 - mgd
2
2 3
g (6d + 3b) (9.81)(1.5 - 0.9)
= 23.544 (rad/s)2
w 32 =
=
2
2
2
2
È(0.4) + (0.3) ˘
a +b
Î
˚
T1 + V1 = T3 + V3 : 0 -
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234
PROBLEM 17.144
Disks A and B are made of the same material and are of the same thickness;
they can rotate freely about the vertical shaft. Disk B is at rest when it is
dropped onto disk A, which is rotating with an angular velocity of 500 rpm.
Knowing that disk A has a mass of 10 kg, determine (a) the final angular
velocity of the disks, (b) the change is kinetic energy of the system.
SOLUTION
mA = 10 kg rA = 150 mm = 0.15 m
Disk A:
IA =
1
1
9
mA rA2 = (10)(0.15)2 =
kg ◊ m2
2
2
80
rB = 100 mm = 0.1 m
Disk B:
2
2
Êr ˆ
40
Ê 100 ˆ
=
mB = mA Á B ˜ = (10) Á
kg
˜
Ë 150 ¯
9
Ë rA ¯
IB =
1
1
1 Ê 40 ˆ
kg ◊ m2
mB rB2 = Á ˜ (0.1)2 =
45
2
2Ë 9 ¯
Principle of impulse and momentum.
Syst. Momenta1
+
Moments about B:
+
=
Syst. Momenta2
I Aw1 + 0 + 0 = I Aw 2 + I Bw 2
w2 =
Initial angular velocity of disk A:
Syst. Ext. Imp. 1Æ2
IA
81
w1 = w1 = 0.83505w1
IA + IB
97
w1 = 500 rpm = 52.36 rad/s
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235
PROBLEM 17.144 (Continued)
(a)
Final angular velocity of system:
w 2 = (0.83505)(52.36)
w 2 = 43.723 rad/s
Initial kinetic energy:
T1 =
w 2 = 418 rpm 1
I Aw12
2
1Ê 9 ˆ
T1 = Á ˜ (52.36)2 = 154.213 N ◊ m
2 Ë 80 ¯
Final kinetic energy:
1
T2 = ( I A + I B )w 22
2
1Ê 9
1ˆ
T2 = Á + ˜ ( 43.723)2 = 128.774 N ◊ m
2 Ë 80 45 ¯
(b)
Change in energy:
T2 - T1 = - 25.439 N ◊ m
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236
PROBLEM 17.145
At what height h above its center G should a billiard ball of radius r be
struck horizontally by a cue if the ball is to start rolling without sliding?
SOLUTION
I =
Moment of inertia.
2 2
mr
5
Principle of impulse and momentum.
Syst. Momenta1
+
Syst. Ext. Imp. 1Æ2
=
Syst. Momenta2
+
Kinematics. Rolling without sliding. Point C is the instantaneous center of rotation.
Linear components:
0 + P Dt = mv2
= mrw 2
Moments about G:
0 + hP Dt = I w 2
ˆ
Ê2
0 + h( mrw 2 ) = Á mr 2 ˜ w 2
¯
Ë5
h=
2
r
5
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237
PROBLEM 17.146
A large 1.5 kg sphere with a radius r = 100 mm is thrown into a light
basket at the end of a thin, uniform rod weighing 1 kg and length
L = 250 mm as shown. Immediately before the impact the angular
velocity of the rod is 3 rad/s counterclockwise and the velocity of
the sphere is 0.5 m/s down. Assume the sphere sticks in the basket.
Determine after the impact (a) the angular velocity of the bar and
sphere, (b) the components of the reactions at A.
SOLUTION
Let Point G be the mass center of the sphere and Point C be that of the rod AB.
Rod AB:
Sphere:
mAB = 1 kg
1
1
I AB = mAB L2 = (1)(0.25)2
12
12
1
kg ◊ m2
=
192
= 5.2083 ¥ 10 -3 kg ◊ m2
mS = 1.5 kg
IG =
3
2
2
mS r 2 = (1.5)(0.1)2 =
= 6 ¥ 10 -3 kg ◊ m2
5
5
500
Impact. Before impact, bar AB is rotating about A with angular velocity w 0 = w 0
(w 0 = 3 rad/s) and the
sphere is falling with velocity v 0 = v0 Ø ( v0 = 0.5 m/s). After impact, the rod and the sphere move together,
rotating about A with angular velocity w = w .
Geometry.
R = L2 + r 2 = (0.25)2 + (0.1)2 = 0.26926 m
tan q =
r
0.1
=
q = 21.8∞
L 0.25
L
w 0 = (0.125)(3) = 0.375 m/s≠
2
L
vC = w ¢ Ø, vG = Rw ¢ q
After impact,
2
Principle of impulse and momentum. Neglect mass of the rod and sphere over the duration of the impact.
Kinematics: Before impact,
vC =
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238
PROBLEM 17.146 (Continued)
(a)
+
Moments about A:
mS v0 L - I ABw 0 - mAB vC
or
L
L
+ 0 = I Gw ¢ + mS vG¢ R + I ABw ¢ + mAB vC
2
2
mS v0 L - I ABw 0 - mAB vC
1
L Ê
ˆ
= Á I G + mS R2 + I AB + mAB L2 ˜ w ¢
¯
2 Ë
4
(1)
Ê 1 ˆ
(1.5)(0.5)(0.25) - Á
(3) - (1)(0.375)(0.125)
Ë 192 ˜¯
1 Ê 1ˆ
È 3
˘
=Í
+ (1.5)(0.26926)2 +
+ Á ˜ (1)(0.25)2 ˙
¯
Ë
500
192
4
Î
˚
0.125 = 0.13558w ¢ w ¢ = 0.92193 rad/s
w¢ = 0.922 rad/s
Normal accelerations at C and G.
(a C ) n =
L
(w ¢ )2 = (0.125)(0.92193)2 = 0.10624 m/s2 ¨
2
( aG ) n = R(w ¢ )2 = (0.26926)(0.92193)2 = 0.22886 m/s2
21.8°
Tangential accelerations at C and G. a = a
( aC )t =
(b)
L
a = 0.125 a Ø ( aG )t = Ra = 0.26926 a
2
21.8°
Kinetics. Use bar AB and the sphere as a free body.
+
W AB
SM A = S ( M A )eff :
L
L
+ WS L = I ABa + mAB ( aC )t + I Ga + mS ( aG )t R
2
2
1
ˆ
Ê
= Á I AB + mAB L2 + I G + mS R2 ˜ a
¯
Ë
4
3
È 1 Ê 1ˆ
˘
(1)(9.81)(0.125) + (1.5)(9.81)(0.25) = Í
+ Á ˜ (1))(0.25)2 +
+ (1.5)(0.26926)2 ˙ a
¯
Ë
4
500
Î192
˚
4.905 = 0.13558 a
a = 36.1766 rad/s2
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239
PROBLEM 17.146 (Continued)
+
(aC )t = (0.125)(36.1766) = 4.5221 m/s2 Ø, (aG )t = (0.26926)(36.1766) = 9.7409 m/s2
21.8°
SFx = S ( Fx )eff :
Ax = - mAB ( aC ) n - mS ( aG ) n cos 21.8∞ + mS ( aG )t sin 21.8∞
Ax = - (1)(0.106244) - (1.5)(0.22886) cos 21.8∞ + (1.5)(9.7409)sin 21.8∞
Ax = 5.2137 N
A x = 5.21 N Æ SFy = S ( Fy )eff : Ay - W AB - WS = - mAB ( aC )t - mS ( aG )t cos 21.8∞ - mS ( aG )n sin 21.8∞
+
Ay - (1)(9.81) - (1.5)(9.81) = -(1)( 4.5221) - (1.5)(9.7409) cos 21.8∞∞ - (1.5)(0.22886)sin 21.8∞
Ay = 6.309 N
A y = 6.31 N ≠ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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240