Download Mechanics

CHAPTER 19
PROBLEM 19.1
Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic
motion with an amplitude of 5 mm and a period of 0.1 s.
SOLUTION
Simple harmonic motion.
where
x = xm sin(w nt + f )
wn =
2p
2p
=
= 62.83 rad/s
T
0.1 s
xm = 5 mm = 5 ¥ 10 -3 m
Then
x = 5 ¥ 10 -3 sin(62.83t + f )
Differentiate to obtain velocity and acceleration.
v = x& = w n xm cos(w nt + f )
a = &&
x = - w n2 xm sin(w nt + f )
Maximum velocity.
Maximum acceleration.
| x& max | = w n xm = (62.83)(5 ¥ 10 -3 )
| &&
xmax | = w n2 xm = (62.83)2 (5 ¥ 10 -3 )
x& max = 0.314 m/s &&
xmax = amax = 19.74 m/s2 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.
3
PROBLEM 19.2
Determine the amplitude and maximum velocity of a particle which moves in simple harmonic motion with a
maximum acceleration of 60 m/s2 and a frequency of 40 Hz.
SOLUTION
Simple harmonic motion.
where
x = xm sin(w nt + f )
w n = 2p f n = (2p )( 40) = 80p rad/s
Differentiate to obtain velocity and acceleration.
v = x& = w n xm cos(w nt + f )
vm = w n xm
a = &&
x = - w n2 xm sin(w nt + f )
am = w n2 xm
Use information given to obtain amplitude and maximum velocity.
xm =
am
w n2
=
60 m/s2
(80p )2
= 0.000950 m
vm = w n xm = (80p )(0.000950) = .2387 m/s
xm = 0.950 mm vm = 239 mm/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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
4
PROBLEM 19.3
A particle moves in simple harmonic motion. Knowing that the amplitude is 300 mm and the maximum
acceleration is 5 m/s2, determine the maximum velocity of the particle and the frequency of its motion.
SOLUTION
Simple harmonic motion.
x = xm sin(w nt + f ) xm = 0.300 m
x = (0.300) sin(w nt + f ) ( m)
x& = (0.3)(w n ) cos(w nt + f ) ( m/s)
&&
x = -(0.3)(w n )2 sin(w nt + f ) (m/s)
| am | = (0.3 m/s)(w n )2
Natural frequency.
Maximum velocity.
am = 5 m/s2
| am |
(5 m/s2 )
=
= 16.667 rad/s2
(0.3 m) (0.3 m)
w n = 4.082 rad/s
w n2 =
fn =
wn
2p
fn =
( 4.082 rad/s)
= 0.6497 Hz
(2p rad/cycle)
f n = 0.650 Hz vm = xmw n = (0.3 m)( 4.082 rad/s)
vm = 1.225 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
5
PROBLEM 19.4
A 5 kg block is supported by the spring shown. If the block is moved vertically downward
from its equilibrium position and released, determine (a) the period and frequency of the
resulting motion, (b) the maximum velocity and acceleration of the block if the amplitude
of its motion is 50 mm.
SOLUTION
(a)
Simple harmonic motion.
Natural frequency.
x = xm sin(w nt + f )
wn =
k
m
wn =
(12000 N/m)
15 kg
k = 12000 N/m
w n = 20 2 = 28.284 rad/s
tn =
2p
wn
tn =
2p
= 0.22214 s
28.284
t n = 0.222 s fn =
1
1
=
= 4.50 Hz t n 0.22214
xm = 50 mm = 0.05 m
(b)
x = 0.05 sin (28.284t + f )
Maximum velocity.
vm = xmw n = (0.05)(28.284)
vm = 1.414 m/s Maximum acceleration.
am = xmw n2 = (0.05)(28.284)2
am = 40.0 m/s2 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.
6
PROBLEM 19.5
A 32-kg block is attached to a spring and can move without friction in a slot as
shown. The block is in its equilibrium position when it is struck by a hammer,
which imparts to the block an initial velocity of 250 mm/s. Determine (a) the
period and frequency of the resulting motion, (b) the amplitude of the motion
and the maximum acceleration of the block.
SOLUTION
x = xm sin(w nt + f )
(a)
12 ¥ 103 N/m
k
=
m
32 kg
wn =
w n = 19.365 rad/s
tn =
2p
wn
2p
19.365
t n = 0.324 s
tn =
(b)
At t = 0, x0 = 0,
x&0 = v0 = 250 mm/s
Thus,
x0 = 0 = xm sin(w n (0) + f )
and
f=0
fn =
1
1
=
= 3.08 Hz t n 0.324
x&0 = v0 = xmw n cos(w n (0) + 0) = xmw n
v0 = 0.250 m/s = xm (19.365 rad/s)
xm =
(0.250 m/s)
(19.365 rad/s)
xm = 12.91 ¥ 10 -3 m
xm = 12.91 mm am = xmw n2 = (12.91 ¥ 10 -3 m)(19.365 rad/s)2
am = 4.84 m/s2 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.
7
PROBLEM 19.6
A simple pendulum consisting of a bob attached to a cord oscillates in a vertical
plane with a period of 1.3 s. Assuming simple harmonic motion and knowing that the
maximum velocity of the bob is 400 mm/s, determine (a) the amplitude of the motion
in degrees, (b) the maximum tangential acceleration of the bob.
SOLUTION
(a)
q = q m sin(w nt + f )
Simple harmonic motion.
wn =
(2p )
2p
=
t n (1.3 s)
w n = 4.833 rad/s
q& = q mw n cos(w nt + f )
q& = q w
m
m
n
vm = lq&m = lq mw n
For a simple pendulum,
qm =
vm
lw n
wn =
g
l
l=
Thus,
(1)
9.81 m/s2
g
=
w n2 ( 4.833 rad/s)2
l = 0.4200 m
Amplitude.
qm =
From Eq. (1),
vm
(0.4 m/s)
=
lw n (0.42 m)( 4.833 rad/s)
q m = 0.19706 rad
(b)
Maximum tangential acceleration.
q m = 11.29∞ at = lq&&
Maximum tangential acceleration occurs when q&& is maximum.
q&& = -q mw n2 sin(w nt + f )
q&&max = q mw n2
( at ) max = lq mw n2
( at ) max = (0.4200 m)(0.19706 rad )( 4.833 rad/s)2
( at ) m = 1.933 m/s2 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.
8
PROBLEM 19.7
A simple pendulum consisting of a bob attached to a cord of length l = 800 mm
oscillates in a vertical plane. Assuming simple harmonic motion and knowing that
the bob is released from rest when q = 6∞, determine (a) the frequency of oscillation,
(b) the maximum velocity of the bob.
SOLUTION
(a)
Frequency.
wn =
(9.81 m/s2 )
g
=
l
(0.8 m)
w n = 3.502 rad/s
fn =
(b)
Simple harmonic motion.
where
Maximum velocity.
w n (3.502 rad/s)
=
2p
2p
f n = 0.557 Hz q = q m sin(w nt + f )
q m = 6∞ = 0.10472 rad
q& = q mw n cos(w nt + f )
q&m = q mw n
vm = lq&&m = lq mw n = (0.8 m)(0.10472)(3.502)
vm = 293.4 ¥ 10 -3 m/s
vm = 293 mm/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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9
PROBLEM 19.8
An instrument package A is bolted to a shaker table as shown. The table
moves vertically in simple harmonic motion at the same frequency as
the variable-speed motor which drives it. The package is to be tested
at a peak acceleration of 50 m/s2. Knowing that the amplitude of the
shaker table is 60 mm, determine (a) the required speed of the motor in
rpm, (b) the maximum velocity of the table.
SOLUTION
In simple harmonic motion,
amax = xmaxw n2
50 m/s2 = (0.06 m)w n2
w n2 = (833.33 rad/s)2
w n = 28.8675 rad/s
wn
2p
28.8675
=
2p
= 4.5945 Hz (cycles per second)
fn =
(a)
Motor speed.
(b)
Maximum velocity.
speed = 276 rpm ( 4.5945 rev/s)(60 s/min)
vmax = xmaxw n = (0.06 m)(28.8675 rad/s)
vmax = 1.732 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
10
PROBLEM 19.9
The motion of a particle is described by the equation x = 5 sin 2t + 4 cos 2t , where x is expressed in meters and
t in seconds. Determine (a) the period of the motion, (b) its amplitude, (c) its phase angle.
SOLUTION
x = xm sin(w nt + f )
For simple harmonic motion
Double angle formula (trigonometry): sin( A + B) = (sin A)(cos B) + (sin B)(cos A)
Let
A = w nt , B = f
Then
x = xm sin(w nt + f )
x = xm (sin w nt )(cos f ) + xm (sin f )(cos w nt )
x = ( xm cos f))(sin w nt ) + ( xm sin f )(cos w nt )
x = 5 sin 2t + 4 cos 2t
Given
Comparing,
w n = 2 xm cos f = 5
(1)
xm sinf = 4
(2)
tn =
(a)
(b)
2p
2p
=
=p s
w n (2 rad/s)
t = 3.14 s Squaring Eqs. (1) and (2) and adding,
xm2 cos2 f + xm2 sin 2 f = 42 + 52
xm2 (cos2 f + sin 2 f ) = xm2 = 41 m2
(c)
tanf =
Dividing Eq. (2) by Eq. (1),
4
5
xm = 6.40 m f = 38.7∞ 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
11
PROBLEM 19.10
An instrument package B is placed on the shaking table C as shown. The table is
made to move horizontally in simple harmonic motion with a frequency of 3 Hz.
Knowing that the coefficient of static friction is m s = 0.40 between the package and
the table, determine the largest allowable amplitude of the motion if the package is
not to slip on the table. Given answers in both SI and U.S. customary units.
SOLUTION
Maximum allowable acceleration of B.
+
m s = 0.40
SF = ma :
Fm = mam
m s mg = mam
am = m s g
Simple harmonic motion.
am = 0.40 g
wn
2p
w n = 6p rad/s
f n = 3 Hz =
am = xmw n2
0.40 g = xm (6p rad/s)2
xm = 1.1258 ¥ 10 -3g
Largest allowable amplitude.
SI:
xm = 1.1258 ¥ 10 -3 (9.81) = 11.044 ¥ 10 -3 m
xm = 11.04 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
12
PROBLEM 19.11
A 32-kg block attached to a spring of constant k = 12 kN/m can move without
friction in a slot as shown. The block is given an initial 300-mm displacement
downward from its equilibrium position and released. Determine 1.5 s after
the block has been released (a) the total distance traveled by the block, (b) the
acceleration of the block.
SOLUTION
x = xm sin(w nt + f )
(a)
wn =
k
(12 ¥ 103 N/m)
=
m
(32 kg)
w n = 19.365 rad/s
tn =
(2p )
2p
=
w n (19.365)
t n = 0.3245 s
Initial conditions.
x(0) = 0.3 m, x& (0) = 0
0.3 = xm sin(0 + f )
x& (0) = 0 = xmw n cos( 0 + f )
p
2
xm = 0.3
f=
pˆ
Ê
x(t ) = (0.3)sin Á19.365t + ˜ , t n = 0.3245 s
Ë
2¯
p˘
È
x(1.5 s) = (0.3)sin Í(19.365)(1.5) + ˙ = -0.2147 m
2˚
Î
p˘
È
x& (1.5 s) = (0.3)(19.365) cos Í(19.365)(1.5) + ˙ = 4.057 m/s Ø
2˚
Î
In one cycle, block travels
( 4)(0.3 m) = 1.2 m
To travel 4 cycles, it takes
(4 cyc)(0.3245 s/cyc) = 1.2980 s
at t = 1.5 s. Thus, total distance traveled is
(b)
4(1.2) + 0.6 + (0.3 - 0.2147) = 5.49 m p˘
È
&&
x(1.5) = -(0.3)(19.365)2 sin Í(19.365)(1.5) + ˙ = 80.5 m/s2 Ø 2˚
Î
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.
13
PROBLEM 19.12
A 2 kg block is supported as shown by a spring of constant k = 400 N/m, which can act in
tension or compression. The block is in its equilibrium position when it is struck from below
by a hammer, which imparts to the block an upward velocity of 2.5 m/s. Determine (a) the
time required for the block to move 100 mm upward, (b) the corresponding velocity and
acceleration of the block.
SOLUTION
Simple harmonic motion.
Natural frequency.
x = xm sin(w nt + f )
wn =
k
, k = 400 N/m
m
wn =
400 N/m
2
w n = 10 2 = 14.1421 rad/s
x(0) = 0 = xm sin(0 + f )
f=0
x& (0) = xmw n cos( 0 + 0)
x& (0) = 2.5 m/s
2.5 = xm (14.1421) xm = 0.17678 m
x = (0.17678)sin(14.1421t )( m/s)
(a)
(1)
Time at x = 100 mm (x = 0.1 m)
0.1 = 0.17678 sin(14.1421t )
t=
(b)
sin -1
0.1
( 0.17678
) = 0.04252 s
14.1421
t = 0.046 s
Note: Since w is in rad/s, convert the argument of sin-1 to radians.
Velocity and acceleration.
x& = xmw n cos(w nt )
&&
x = - xmw n2 sin w nt
t = 0.04252
21) cos[(14.1421)(0.04252)]
x& = (0.17678)(14.142
&x = 2.0615 m/s
v = 2.06 m/s ≠ &&
x = -(0.1768)(14.1421) sin[(14.1421)(0.04252)]
2
a = 20.0 m/s2 Ø = -20 m/s2
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 19.13
In Problem 19.12, determine the position, velocity, and acceleration of the block 0.90 s after it has been struck
by the hammer.
SOLUTION
Simple harmonic motion.
Natural frequency.
x = xm sin(w nt + f )
wn =
k
, k = 400 N/m
m
wn =
400 N/m
2 kg
w n = 10 2 = 14.1421 rad/s
x(0) = 0 = xm sin(0 + f )
f=0
x& (0) = xmw n cos( 0 + 0) x& (0) = 2.5 m/s
2.5 = xm (14.1421) xm = 0.17678 m
x = (0.17678)sin (14.1421t )( m/s)
Simple harmonic motion.
x = xm sin(w nt + f )
x& = xmw n cos(w nt + f )
&&
x = - xmw n2 sin(w nt + f )
Note: arguments of sin and cos are in radians.
At 0.90 s:
x = (0.17678)sin[(14.1421)( 0.90)] = 0.02843 m
x& = (0.17678)(14.1421) cos[(14.1421)( 0.90)] = -2.4675 m/s
x = 0.0284 m ≠ v = 2.47 m/s ≠ &&
x = -(0.17678)(14.1421)2 sin[(14.1421)(0.90)] = -5.6859 m/s2 a = 5.69 m/s2 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
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15
PROBLEM 19.14
The bob of a simple pendulum of length l = 800 mm is released from rest when q = + 5∞.
Assuming simple harmonic motion, determine 1.6 s after release (a) the angle q, (b) the
magnitudes of the velocity and acceleration of the bob.
SOLUTION
q = q m sin(w nt + f )
Initial conditions:
q ( 0 ) = 5∞ =
g
9.81 m/s
=
l
0.8 m
w n = 3.502 rad/s
wn =
(5)(p )
180 rad
q& (0) = 0
5p
= q m sin(0 + f )
q ( 0) =
180
q& (0) = 0 = q mw n cos( 0 + f )
p
2
5p
qm =
rad
180
pˆ
5p
Ê
q=
sin Á 3.502t + ˜
Ë
180
2¯
f=
(a)
At t = 1.6 s:
q=
p˘
5p
È
sin Í(3.502)(1.6) + ˙
180
2˚
Î
q = 0.06786 rad = 3.89° (b)
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 19.14 (Continued)
p˘
È
Ê 5p ˆ
(3.502) cos Í(3.502)(1.6) + ˙
q& = q mw n cos(w nt + f ) = Á
˜
Ë 180 ¯
2˚
Î
& 1.6 s) = 0.19223 rad/s
q(
v = lq& = (0.800 m)(0.19223 rad/s) = 0.1538 m/s q&& = -q mw n2 sin(w nt + f )
p˘
Ê 5p ˆ
È
= -Á
(3.502)2 sin Í(3.502)(1.6) + ˙
Ë 180 ˜¯
2˚
Î
q&& = -0.8319 rad/s2
a = ( at )2 + ( an )2
at = lq&& = (0.8 m)( -0.8319 rad/s2 ) = -0.6655 m/s2
a = lq& 2 = (0.8 m)(0.19223 rad/s)2 = 0.02956 m/s2
n
a = (0.6655)2 + (0.02956)2 = 0.6662 m/s2
a = 0.666 m/s2 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.
17
PROBLEM 19.15
A 5-kg collar rests on but is not attached to the spring shown. It is observed that when the collar
is pushed down 180 mm or more and released, it loses contact with the spring. Determine (a) the
spring constant, (b) the position, velocity, and acceleration of the collar 0.16 s after it has been
pushed down 180 mm and released.
SOLUTION
(a)
x = xm sin(w nt + f )
x0 = xm sin(0 + f ) = 0.180 m
x&0 = 0 = xm cos( 0 + f )
p
2
xm = 0.180 m
f=
pˆ
Ê
x = 0.180 sin Á w nt + ˜
Ë
2¯
When the collar just leaves the spring, its acceleration is gØ and v = 0.
pˆ
Ê
x& = (0.180)w n cos Á w nt + ˜
Ë
2¯
pˆ
Ê
v = 0 0 = (0.180)w n cos Á w nt + ˜
Ë
2¯
pˆ p
Ê
ÁË w nt + ˜¯ =
2
2
pˆ
Ê
a = - g = ( -0.180)(w n )2 sin Á w nt + ˜
Ë
2¯
9.81 m/s2
0.180 m
w n = 7.382 rad/s
- g = ( -0.180)((w n2 ) w n =
wn =
k
m
k = mw n2
= (5 kg)(7.382 rad/s)2
= 272.5 N/m
k = 273 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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it without permission.
18
PROBLEM 19.15 (Continued)
w n = 7.382 rad/s
(b)
p˘
È
x = 0.180 sin Í(7.382)t + ˙
2˚
Î
At t = 0.16 s :
Position.
p˘
È
x = 0.180 sin Í(7.382)(0.16) + ˙ = 0.06838 m
2˚
Î
x = 68.4 mm below equilibrium position Velocity.
p˘
È
x& = (0.180)(7.382) cos Í(7.382)(0.16) + ˙
2˚
Î
= -1.229 m/s
v = 1.229 m/s ≠ Acceleration.
p˘
È
&&
x = -(0.180)(7.382)2 sin Í(7.382)(0.16) + ˙
2˚
Î
a = 3.73 m/s2 ≠ = -3.726 m/s2
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 19.16
An 8-kg collar C can slide without friction on a horizontal rod between
two identical springs A and B to which it is not attached. Each spring
has a constant of 600 N/m. The collar is pushed to the left against
spring A, compressing that spring 20 mm, and released in the position
shown. It then slides along the rod to the right and hits spring B. After
compressing that spring 20 mm, the collar slides to the left and hits
spring A, which it compresses 20 mm. The cycle is then repeated.
Determine (a) the period of the motion of the collar, (b) the position
of the collar 1.5 s after it was pushed against spring A and released.
(Note: This is a periodic motion, but not a simple harmonic motion.)
SOLUTION
(a)
For either spring,
tn =
tn =
2p
k
mC
2p
600 N/m
8 kg
t n = 0.7255 s
Complete cycle is 1234, 4321.
Time from 1 to 2 is (t n /4), which is the same as time from 3 to 4, 4 to 3 and 2 to 1.
Thus, the time during which the springs are compressed is 4(t n /4) = t n = 0.7255 s.
Velocity at 2 or 3.
Use conservation of energy.
v1 = 0 T1 = 0 V1 =
1 2 1
kx = (600 N/m)(0.020 m)2
2
2
V1 = 0.120 J
1
1
T2 = mv22 = (8 kg)(v2 )2 T2 = 4v22
2
2
V2 = 0
T1 + V1 = T2 + V2
Time from 2 to 3 is
t2 -3 =
0 + 0.120 = 4v22
v2 = 0.1732 m/s
(0.020 m)
= 0.11545 s
(0.1732 m/s)
and is the same as the time from 3 to 2.
Thus, total time for a complete cycle is
t c = t n + 2t2-3
= 0.7255 + 2(0.11545)
= 0.9564
t c = 0.956 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
20
PROBLEM 19.16 (Continued)
(b)
From (a), in 0.9564, the spring A is again fully compressed. Spring B is compressed the second time in
1.5 cycles or (1.5)(0.9564) = 1.4346 s. At 1.5 s, the collar is still in contact with spring B moving to the
left and is at a distance Dx from the maximum deflection of B equal to
È 2p
˘
Dx = 20 - 20 cos Í
(1.5 - 1.4346) ˙
Î 0.7255
˚
Dx = 20 - 16.877
= 3.123 mm
Thus, collar C is 60 - 3.123 = 56.877 mm from its initial position.
56.9 mm from initial position 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.
21
PROBLEM 19.17
A 35-kg block is supported by the spring arrangement shown. The block is moved
vertically downward from its equilibrium position and released. Knowing that the
amplitude of the resulting motion is 45 mm, determine (a) the period and frequency
of the motion, (b) the maximum velocity and maximum acceleration of the block.
SOLUTION
(a)
Determine the constant k of a single spring equivalent to the three springs
P=k
d = 16d + 8d + 8d
k = 32 kN/m
Natural frequency.
wn =
=
k
m
32 ¥ 103 N/m
35 kg
(1 N = 1 kg ◊ 1 m/s2 )
w n = 30.237 rad/s
tn =
2p
2p
=
= 0.208 s w n 30.23
fn =
(b)
1
= 4.81 Hz tn
x = xm sin(w nt + f ) x0 = 0.045 m = xm
w n = 30.24 rad/s
x = 0.045 sin(30.24tt + f)
x& = (0.045)(30.24) cos(30.24t + f )
vmax = 1.361 m/s &&
x = -(0.045)(30.237)2 sin(30.24t + f )
amax = 41.1 m/s2 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.
22
PROBLEM 19.18
A 35-kg block is supported by the spring arrangement shown. The block is moved vertically
downward from its equilibrium position and released. Knowing that the amplitude of the
resulting motion is 45 mm, determine (a) the period and frequency of the motion, (b) the
maximum velocity and maximum acceleration of the block.
SOLUTION
(a)
Determine the constant k of a single spring equivalent to the two springs shown.
d = d1 + d 2 =
P
P
P
+
=
16 kN/m 16 kN/m k
1 1 1
= +
k 16 16
k = 8 kN/m
Period of the motion.
tn =
2p
fn =
(b)
k
m
=
2p
8 ¥ 103
35
= 0.416 s 1
1
=
= 2.41 Hz t n 0.416
w n = 2p f n = 2p (2.41) = 15.12 rad/s
x = 0.045 sin(15.12t + f )
x& = (0.045)(15.12) cos(15.12t + f )
vmax = 0.680 m/s &&
x = -(0.045)(15.12)2 sin(15.12t + f )
amax = 10.29 m/s2 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.
23
PROBLEM 19.19
A 15 kg block is supported by the spring arrangement shown. If the block is
moved from its equilibrium position 40 mm vertically downward and released,
determine (a) the period and frequency of the resulting motion, (b) the maximum
velocity and acceleration of the block.
SOLUTION
Determine the constant k of a single spring equivalent to the three springs shown.
Springs 1 and 2:
d = d1 + d 2 , and
Hence,
k¢ =
P1 P1 P1
= +
k ¢ k1 k2
k1k2
k1 + k2
where k ¢ is the spring constant of a single spring equivalent of springs 1 and 2.
Springs k ¢ and 3 Deflection in each spring is the same.
So
Now
P = P1 + P2 , and
P = kd , P1 = k ¢d , P2 = k3d
kd = k ¢d + k3d
k = k ¢ + k3 =
k1k2
+ k3
k1 + k2
( 4)(2.4)
+ 3.2 = 4.7 kN/m = 4700 N/m
4 + 2.4
m = 15 kg
k 4700
w n2 = =
= 313.33 w n = 17.7012 rad/s
m
15
k=
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.
24
PROBLEM 19.19 (Continued)
(a)
(b)
Period.
tn =
2p
wn
t n = 0.355 s Frequency.
fn =
wn
2p
f n = 2.82 Hz Simple harmonic motion.
x = 0.04 sin(w t + j ) in m
xm = 40 mm x& = (0.04)(17.7012) cos(w t + f )
vm = x&m = 0.70805 m/s
vm = 0.708 m/s &&
x = -(0.04)(17.7012)2 sin(w t + f )
am = &&
xm = 12.533 m/s2
am = 12.53 m/s2 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.
25
PROBLEM 19.20
A 5-kg block, attached to the lower end of a spring whose upper end is fixed, vibrates with a period of 6.8 s.
Knowing that the constant k of a spring is inversely proportional to its length, determine the period of a 3-kg
block which is attached to the center of the same spring if the upper and lower ends of the spring are fixed.
SOLUTION
Equivalent spring constant.
k ¢ = 2k + 2k = 4 k
For case 1 ,
(Deflection of each spring is the same.)
t n1 = 6.8 s
w n1 =
2p 2p
=
= 0.924 rad/s
t n1 6.8
w n2 =
k
m1
k = m1w n21 = (5)(0.924)2 = 4.2689 N/m
For case 2 ,
w n22 =
4k ( 4)( 4.2689)
=
= 5.6918 (rad/s)2
m2
3
w n2 = 2.3857 rad/s
tn =
2p
2p
=
w n2 2.3857
t n2 = 2.63 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
26
PROBLEM 19.21
A 15 kg block is supported by the spring arrangement shown. The block is moved from its
equilibrium position 20 mm vertically downward and released. Knowing that the period of the
resulting motion is 1.5 s, determine (a) the constant k, (b) the maximum velocity and maximum
acceleration of the block.
SOLUTION
Since the force in each spring is the same, the constant k ¢ of a single equivalent spring is
1
1 1 1
=
+ +
k ¢ 2k k k
(a)
t 1n = 1.5 s =
2p
k¢
m
k¢ =
k
2.5
(See Problem 19.19.)
2
Ê 2p ˆ
; k = Á ˜ ( m)(2.5)
Ë 1.5 ¯
(2p )2
(15 kg)(2.5)
(1.5 s)2
= 657.974 N/m
k=
(b)
k = 658 N/m x = xm sin(w nt + f )
x& = xmw n cos(w nt + f )
vmax = xmw n
wn =
2p
2p
=
= 4.189 rad/s
t n 1.5 s
xm = 20 mm = 0.02 m
vmax = 0.0838 m/s vmax = (0.02 m)( 4.189 rad/s)
&&
x = - xmw n2 cos(w nt + f )
| amax | = xmw n2 = (0.02 m)( 4.189)2
| amax | = 0.351 m/s2 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.
27
PROBLEM 19.22
Two springs of constants k1 and k2 are connected in series to a block A that vibrates
in simple harmonic motion with a period of 5 s. When the same two springs are
connected in parallel to the same block, the block vibrates with a period of 2 s.
Determine the ratio k1 /k2 of the two spring constants.
SOLUTION
Equivalent springs.
k1k2
k1 + k2
Series:
ks =
Parallel:
k p = k1 + k2
ts =
2p 2p
2p
2p
=
=
; tp =
ks
kp
ws
wp
m
m
2
2
Ê ts ˆ
k p k1 + k2 ( k1 + k2 )2
Ê 5ˆ
=
= (k k ) =
=
ÁË ˜¯
Á ˜
1 2
2
ks
k1k2
Ëtp¯
k1 + k2
(6.25)( k1k2 ) = k12 + 2k1k2 + k22
k1 =
( 4.25)k2 m ( 4.25)2 k22 - 4k22
2
k1
= 2.125 m 3.516 = 0.250 or 4.00
k2
k1
= 0.250 or 4.00 k2
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.
28
PROBLEM 19.23
The period of vibration of the system shown is observed to be 0.6 s. After cylinder B has been
removed, the period is observed to be 0.5 s. Determine (a) the weight of cylinder A, (b) the
constant of the spring.
SOLUTION
m1 = mA + 1.5 t 1 = 0.6 s
2p
2p 2p
=
= 3.333p rad/s
t1 =
w1 =
w1
t 1 0.6
k
k = m1w12 = mA + 1.5(3.333p )2
w12 =
m1
m2 = mA t 2 = 0.5 s
2p
2p 2p
=
= 4p rad/s
t2 =
w2 =
w2
t 2 0.5
k
k = m2w 22 = mA ( 4p )2
w 22 =
m2
(a)
(1)
(2)
Equating the expressions found for k in Eqs. (1) and (2):
mA + 1.5(3.333p )2 = mA ( 4p )2
(11.111)( mA + 1.5) = 16mA
4.889mA = 16.6665
mA = 3.40898 kg
W A = mA g = (3.40898)(9.81)
(b)
Eq. (1):
W A = 33.4 N k = 3.40898 + 1.5(3.333p rad/s)2
k = 538.22 N/m
k = 538 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
29
PROBLEM 19.24
The period of vibration of the system shown is observed to be 0.8 s. If
block A is removed, the period is observed to be 0.7 s. Determine (a) the
mass of block C, (b) the period of vibration when both blocks A and B
have been removed.
SOLUTION
m1 = mC + 6 kg t 1 = 0.8 s
W1 =
2p
2p
2p
=
=
rad/s
t 1 0.8 s 0.8
k
Ê 2p ˆ
; k = m1w12 = ( mC + 6) Á
Ë 0.8 ˜¯
m1
m2 = mC + 3 kg t 2 = 0.7 s
2
w12 =
w2 =
2p
2p
2p
=
=
rad/s
t 2 0.7 s 0.7
w 22 =
k
Ê 2p ˆ
; k = m2w 22 = ( mC + 3) Á
Ë 0.7 ˜¯
m2
(1)
2
(2)
Equating the expressions found for k in Eqs. (1) and (2):
2
Ê 2p ˆ
Ê 2p ˆ
= ( mC + 3) Á
( mC + 6) Á
˜
Ë 0.7 ˜¯
Ë 0.8 ¯
2
2
mC + 6 Ê 0.8 ˆ
=Á
˜ ; solve for mC :
mC + 3 Ë 0.7 ¯
w3 =
2p
t3
w 32 =
Ê 2p ˆ
k
; k = mC w 32 = mC Á ˜
mC
Ë t3 ¯
mC = 6.80 kg 2
(3)
Equating expressions for k from Eqs. (2) and (3),
2
2
2
2
Ê 2p ˆ
Ê 2p ˆ
= mC Á ˜
( mC + 3) Á
˜
Ë 0.7 ¯
Ë t3 ¯
Recall mC = 6.8 kg :
Ê 2p ˆ
Ê 2p ˆ
= 6.8 Á ˜
(6.8 + 3) Á
˜
Ë 0.7 ¯
Ë t3 ¯
2
t3
6.8
Ê t3 ˆ
;
= 0.833
ÁË
˜¯ =
0.7
9.8 0.7
t 3 = 0.583 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
30
PROBLEM 19.25
The period of vibration of the system shown is observed to be 0.2 s. After the spring of
constant k2 = 4 kN/m is removed and block A is connected to the spring of constant k1, the
period is observed to be 0.12 s. Determine (a) the constant k1 of the remaining spring, (b) the
weight of block A.
SOLUTION
Equivalent spring constant for springs in series.
ke =
For k1 and k2 ,
t=
2p
ke
mA
=
k1k2
( k1 + k2 )
2p
( k1k2 )
( mA )( k1 + k2 )
2p
For k1 alone,
t¢ =
(a)
( k1 + k2 )( k1 )
t
=
=
t¢
( k1k2 )
k1
mA
k1 + k2
k2
2
Êtˆ
k2 Á ˜ = k1 + k2
Ë t ¢¯
0.2
t
=
= 1.6667
t ¢ 0.12
k2 = 4 kN/m
( 4 kN/m)(1.6667 s)2 = k1 + 4 kN/m
(b)
k1 = 7.1111 kN / m
mA =
k1 = 7.11 kN/m t¢ =
2p
k1
mA
(t ¢ )2 k1
; W A = mA g
(2p )2
k1 = 7.1111 kN/m = 7111.1 N/m
WA =
(9.81 m/s2 ) = 25.4453 N (0.12 s)2 (7111.1 N/m)
(2p )2
W A = 25.4 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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it without permission.
31
PROBLEM 19.26
The 50-kg platform A is attached to springs B and D, each of which has
a constant k = 2 kN/m. Knowing that the frequency of vibration of the
platform is to remain unchanged when a 40-kg block is placed on it and
a third spring C is added between springs B and D, determine the required
constant of spring C.
SOLUTION
Frequency of the original system.
Springs B and D are in parallel.
ke = k B + k D = 2(2 kN/m) = 4 kN/m = 4000 N / m
w n2 =
ke 4000 N/m
=
mA
50 kg
w n2 = 80 ( rad/s)2
Frequency of new system.
Springs A, B, and C are in parallel.
ke¢ = k B + k D + kC = (2)(200) + kc in N/m
(2000 + kc )
ke¢
=
mA + mB (50 kg + 40 kg)
(2000 + kc )
(w n¢ )2 =
90
2
w n = (w n¢ )2
(w n¢ )2 =
(2000 + kc )
90
kC = 5200 N/m = 5.2 kN / m
80 =
kC = 5.20 kN/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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
32
PROBLEM 19.27
From mechanics of materials it is known that when a static load P is applied
at the end B of a uniform metal rod fixed at end A, the length of the rod will
increase by an amount d = PL /AE , where L is the length of the undeformed
rod, A is its cross-sectional area, and E is the modulus of elasticity of the
metal. Knowing that L = 450 mm and E = 200 GPa and that the diameter
of the rod is 8 mm, and neglecting the mass of the rod, determine (a) the
equivalent spring constant of the rod, (b) the frequency of the vertical
vibrations of a block of mass m = 8 kg attached to end B of the same rod.
SOLUTION
(a)
P = ke d
PL
AE
Ê AE ˆ
P=Á
d
Ë L ˜¯
d=
AE
L
p d 2 p (8 ¥ 10 -3 m)2
=
A=
4
4
-5 2
A = 5.027 ¥ 10 m
ke =
L = 0.450 m
E = 200 ¥ 109 N/m2
ke =
(5.027 ¥ 10 -5 m2 )(200 ¥ 109 N/m2 )
(0.450 m)
ke = 22.34 ¥ 106 N/m
(b)
fn =
=
ke = 22.3 MN/m ke
m
2p
22.3¥106
8
2p
= 265.96 Hz
f n = 266 Hz 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.
33
PROBLEM 19.28
From mechanics of materials it is known that for a cantilever
beam of constant cross section, a static load P applied at end
B will cause a deflection d B = PL3 / 3EI , where L is the length
of the beam, E is the modulus of elasticity, and I is the moment
of inertia of the cross-sectional area of the beam. Knowing that
L = 3 m, E = 230 GPa, and I = 5 ¥ 106 mm4, determine (a) the
equivalent spring constant of the beam, (b) the frequency of
vibration of a 250 kg block attached to end B of the same beam.
SOLUTION
(a)
Equivalent spring constant.
P
dB
P = ke d B
ke =
PL3
3EI
Ê 3EI ˆ
P = Á 3 ˜ dB
Ë L ¯
3EI
ke = 3
L
(3)(230 ¥ 109 N/m2 )(5 ¥ 106 ¥ 10 -12 m 4 )
=
(3)3
dB =
ke = 127.778 ¥ 103 N/m
(b)
Natural frequency.
fn =
ke = 127.8 kN/m ke
m
2p
ke = 127.778 ¥ 103 N/m
fn =
(127.778 ¥ 103 N/m )
250 kg
2p
f n = 3.5981 Hz
f n = 3.60 Hz 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.
34
PROBLEM 19.29
A 40-mm deflection of the second floor of a building is measured directly under a newly installed 4000-kg
piece of rotating machinery, which has a slightly unbalanced rotor. Assuming that the deflection of the floor
is proportional to the load it supports, determine (a) the equivalent spring constant of the floor system, (b) the
speed in rpm of the rotating machinery that should be avoided if it is not to coincide with the natural frequency
of the floor-machinery system.
SOLUTION
(a)
Equivalent spring constant.
W = ke d s
W
d
( 4000)(9.81)
=
= 981 ¥ 103 N / m
(0.04 m)
ke =
(b)
Natural frequency.
fn =
=
ke = 981 kN/m ke
m
2p
( 981 ¥ 103 N/m)
(4000 kg)
2p
f n = 2.4924 Hz
1 Hz = 1 cycle/s
= 60 rpm
Speed = (2.4924 Hz )
= 149.5 rpm
(60 rpm)
Hz
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.
35
PROBLEM 19.30
The force-deflection equation for a nonlinear spring fixed at one end is F = 5 x1/ 2 where F is the force,
expressed in newtons, applied at the other end and x is the deflection expressed in meters. (a) Determine the
deflection x0 if a 120-g block is suspended from the spring and is at rest. (b) Assuming that the slope of the
force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant,
determine the frequency of vibration of the block if it is given a very small downward displacement from its
equilibrium position and released.
SOLUTION
(a)
Deflection x0 .
mg = (0.120 kg)(9.81 m/s2 )
mg = 1.177 N
F = mg
= 5 x10/2
Ê 1.177 ˆ
x0 = Á
Ë 5 ˜¯
2
= 0.0554 m
x0 = 55.4 mm Equivalent spring constant.
At x0 ,
5
5
Ê dF ˆ
-1/ 2
= (0.0554)-1/ 2
ÁË ˜¯ = ( x0 )
dx x0 2
2
Ê dF ˆ
ÁË ˜¯ = 10.618 N/m
dx x
0
ke = 10.618 N/m
(b)
Natural frequency.
fn =
=
ke
m
2p
(10.618 N/m )
( 0.120 kg)
2p
f n = 1.4971 Hz
f n = 1.497 Hz 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 19.31
If h = 700 mm and d = 500 mm and each spring has a constant k = 600 N/m,
determine the mass m for which the period of small oscillations is (a) 0.50 s,
(b) infinite. Neglect the mass of the rod and assume that each spring can act in
either tension or compression.
SOLUTION
xs = x
d
h
2 F = 2kxs = 2k
+
d
x
h
SM A = S ( M A )eff : 2 Fd - mgx = -( mx&&)h
Êd ˆ
&&
2k Á x˜ d - mgx = - mxh
Ëh ¯
È 2kd 2 g ˘
&&
x+Í 2 - ˙x = 0
h˚
Î mh
È 2kd 2 g ˘
w n2 = Í 2 - ˙
h˚
Î mh
2
w n2 =
Data:
2k Ê d ˆ
g
ÁË ˜¯ m h
h
(1)
d = 0.5 m
h = 0.7 m
k = 600 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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37
PROBLEM 19.31 (Continued)
(a)
For t = 0.5 s :
t=
2p
2p
; 0.5 =
wn
wn
w n = 4p
2
Eq. (1):
( 4p )2 =
2(600) Ê 0.5 ˆ
9.81
˜¯ ÁË
m
0.7
0.7
m = 3.561 kg
(b)
m = 3.56 kg For t = infinite:
t=
2p
wn
Eq. (1):
0=
2(600) Ê 0.5 ˆ
9.81
˜¯ ÁË
m
0.7
0.7
wn = 0
2
m = 43.69 kg
m = 43.7 kg 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.
38
PROBLEM 19.32
Denoting by d ST the static deflection of a beam under a given load, show that the frequency of vibration of the
load is
f =
1
2p
g
d ST
Neglect the mass of the beam, and assume that the load remains in contact with the beam.
SOLUTION
k=
W
d ST
m=
W
g
w n2
k
= =
m
W
d ST
W
g
=
g
d ST
Fn =
wn
1
=
2p 2p
g
d ST
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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39
PROBLEM 19.33*
Expanding the integrand in Equation (19.19) of Section 19.4 into a series of even powers of sinf and
integrating, show that the period of a simple pendulum of length l may be approximated by the formula
t = 2p
l Ê 1 2 qm ˆ
˜
Á1 + sin
2¯
gË 4
where q m is the amplitude of the oscillations.
SOLUTION
Using the Binomial Theorem, we write
1
1 - sin 2
( )
qm
2
È
˘
Êq ˆ
= Í1 - sin 2 Á m ˜ sin f ˙
Ë 2¯
˚
sin 2 f Î
-1/ 2
q
1
= 1 + sin 2 m sin 2 f + ◊ ◊ ◊ ◊
2
2
Neglecting terms of order higher than 2 and setting sin 2 f = 12 (1 - cos 2f ), we have
tn = 4
l 2p Ï 1 2 q m
Ì1 + sin
2
g Ú0 Ó 2
È1
˘¸
ÍÎ 2 (1 - cos 2f )˙˚ ˝ df
˛
=4
l 2p Ï 1 2 q m 1 2 q m
¸
- sin
cos 2f ˝ df
Ì1 + sin
2 4
2
g Ú0 Ó 4
˛
=4
l
g
=4
˘
l Èp 1 Ê 2 qm ˆ p
+ Á sin
+ 0˙
˜
Í
2¯2
g Î2 4Ë
˚
p /2
È
˘
1 Ê 2 qm ˆ
1 2 qm
Íf + 4 ÁË sin 2 ˜¯ f - 8 sin 2 sin 2f ˙
Î
˚0
t n = 2p
l Ê 1 2 qm ˆ
Á1 + sin
˜ gË 4
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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40
PROBLEM 19.34*
Using the formula given in Problem 19.33, determine the amplitude q m for which the period of a simple
pendulum is 12 percent longer than the period of the same pendulum for small oscillations.
SOLUTION
For small oscillations,
(t n )0 = 2p
l
g
t n = 1.005(t n )0
We want
= 1.005 2p
l
g
Using the formula of Problem 19.33, we write
q ˆ
Ê 1
t n = (t n )0 Á1 + sin 2 m ˜
Ë 4
2¯
= 1.005(t n )0
qm
= 4[1.005 - 1] = 0.022
2
q
sin m = 0.02
2
qm
= 8.130∞
2
sin 2
q m = 16.26∞ 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.
41
PROBLEM 19.35*
Using the data of Table 19.1, determine the period of a simple pendulum of length l = 750 mm (a) for small
oscillations, (b) for oscillations of amplitude q m = 60∞, (c) for oscillations of amplitude q m = 90∞.
SOLUTION
(a)
t n = 2p
l
(Equation 19.18 for small oscillations):
g
0.750 m
t n = 2p
9.81 m/s2
t n = 1.737 s = 1.737 s
(b)
For large oscillations (Eq. 19.20),
lˆ
Ê 2k ˆ Ê
t n = Á ˜ Á 2p
Ë p ¯Ë
g ˜¯
=
For q m = 60∞,
2k
(1.737 s)
p
k = 1.686 (Table 19.1)
2(1.686)(1.737 s)
p
= 1.864 s
t n (60∞) =
(c)
For q m = 90∞,
k = 1.854
t n (60) = 1.864 s tn =
2(1.854)(1.737 s)
= 2.05 s 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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42
PROBLEM 19.36*
Using the data of Table 19.1, determine the length in mm of a simple pendulum which oscillates with a period
of 2 s and an amplitude of 90°.
SOLUTION
For large oscillations (Eq. 19.20),
lˆ
Ê 2k ˆ Ê
t n = Á ˜ Á 2p
Ë p ¯Ë
g ˜¯
for
q m = 90∞
k = 1.854 (Table 19.1)
(2 s) = (2)(1.854)(2)
l=
l
9.81 m/s2
(2 s)2 (9.81 m/s2 )
[( 4)(1.854)]2
= 0.71349 m
l = 713 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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43
PROBLEM 19.37
The 5-kg uniform rod AC is attached to springs of constant
k = 500 N/m at B and k = 620 N/m at C, which can act in tension
or compression. If the end C of the rod is depressed slightly and
released, determine (a) the frequency of vibration, (b) the amplitude
of the motion of Point C, knowing that the maximum velocity of that
point is 0.9 m/s.
SOLUTION
FB = k B ( x B + (d ST ) B )
= k B (0.7q + (d ST ) B )
FC = kC ( xC + (d ST )C )
= kC (1.4q + (d ST )C )
Equation of motion.
+
SM A = ( SM A )eff :
But in equilibrium (q = 0),
+
(0.7)[k B [(0.7q + (d ST ) B ) - mg ] + 1.4[kC (1.4q + (d ST )C )] = - I a - (0.7)( mat )
SM A = 0 = 0.7[k B (d ST ) B - mg ] + 1.4 kC (d ST )C
(1)
(2)
Substituting Equation (2) into Equation (1),
I a + 0.7mat + (0.7)2 k Bq + (1.4)2 kC q = 0
Kinematics (a = q&&):
at = 0.7a = 0.7q&&
[ I + m(0.7)2 ] q&& + [(0.7)2 k B + (1.4)2 kC ]q = 0
1 2
ml
12
1
= (5 kg)(1.4 m)2
12
= 0.8167 kg ◊ m2
I =
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 19.37 (Continued)
(.7)2 m = (0.49 m2 )(5 kg)
= 2.45 kg ◊ m2
(0.7)2 k B + (1.4)2 kC = (0.49 m2 )(500 N/m) + (1.96 m2 )(620 N/m)
= 245 + 1215.2
= 1460.2 N ◊ m
[0.8167 + 2.45] q&& + 1460.2q = 0
(1460.2 N ◊ m)
q&& +
q =0
(3.267 kg ◊ m2 )
q&& + 447q = 0
(a)
Natural frequency.
( N/kg ◊ m = s -2 )
w n = 447 s -2
= 21.14 rad/s
fn =
(b)
Maximum velocity.
Maximum angular velocity.
Maximum velocity at C.
w n 21.14
=
= 3.36 Hz
2p
2p
q = q m sin(w nt + f )
q& = (q m )(w n ) cos(w nt + f )
q&m = q mw n
( x&C ) m = 1.4 q&m
= (1.4 m)(q m )(w n )
(0.9 m/s)
(1.4 m)(21.14 rad/s)
= 0.03041 rad
qm =
Maximum amplitude at C.
( xC ) m = (1.4 m)(q m )
= (1.4 m)(0.03041)
( xC ) m = 0.0426 m
( xC ) m = 42.6 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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45
PROBLEM 19.38
The uniform rod shown weighs 7.5 kg and is attached to a spring of
constant k = 800 N/m If end B of the rod is depressed 10 mm and
released, determine (a) the period of vibration, (b) the maximum
velocity of end B.
SOLUTION
k = 800 N/m
m = 7.5 kg
F = k ( x + d ST )
where
= k (0.5q + d ST )
15
ma = mr a = (7.5)(0.125 m)q&& = q&&
16
1
I a = (7.5)(0.75)2 q&&
12
45 &&
=
q
128
(a)
Equation of motion.
+
SM C = S ( M C )eff : mg (0.125 m) - F (0.5 m) = I a + ma (0.125 m)
mg (0.125) - k (0.5q + d ST )(0.5 m) =
But in equilibrium, we have
+
45 && 15 &&
q + q (0.25)
128
16
mg (0.125) - kd ST (0.5) = 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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46
PROBLEM 19.38 (Continued)
Thus,
È 45 15 ˘ &&
- k(0.5)2 q = Í
+
q
Î128 128 ˙˚
15
-(800)(0.5)2 q = q&&
32
q&& + ( 426.667)q = 0
Natural frequency and period.
w n2 = 426.667
w n = 20.656 rad/s
t=
(b)
At end B.
2p
2p
=
w n 20.656 rad/s
t = 0.304 s xm = 10 mm = 0.01 m
vm = xmw n
= (0.01 m)(20.656 rad/s)
= 0.20656 m/s
vm = 0.207 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
47
PROBLEM 19.39
A 15 kg uniform cylinder can roll without sliding on a 15° incline.
A belt is attached to the rim of the cylinder, and a spring holds the
cylinder at rest in the position shown. If the center of the cylinder
is moved 50 mm down the incline and released, determine (a) the
period of vibration, (b) the maximum acceleration of the center of
the cylinder.
SOLUTION
x A = x0 + x A/0
Spring deflection.
x A/0 = rq
x0
r
x A = 2 x0
q=
FS = k ( x A + d ST ) = k (2 x0 + d ST )
+
But in equilibrium,
SM C = ( SM )eff : - 2rk (2 x0 + d ST ) + rw sin 15∞ = rmx&&0 + I q&&
(1)
x0 = 0
SM C = 0 = -2rkd ST + rw sin 15∞
Substituting Eq. (2) into Eq. (1) and noting that q =
rmx&&0 + I
(2)
&&
x0
x
, q&& = 0
r
r
&&
x0
+ 4rkx0 = 0
r
1
I = mr 2
2
3
mrx&&0 + 4rkx0 = 0
2
Ê8 k ˆ
&&
x =0
x0 + Á
Ë 3 m ˜¯ 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
it without permission.
48
PROBLEM 19.39 (Continued)
Natural frequency.
(a)
Period.
wn =
tn =
(8)(6000 N / m)
8k
=
= 32.6599 s -1
3m
(3)(15 kg)
2p
2p
=
w n 32.6599
t n = 0.1924 s x0 = ( x0 ) m sin(w nt + f )
(b)
At t = 0,
x0 = 0.05 m
x&0 = 0
x&0 = ( x0 ) m w n cos(w nt + f )
t=0
0 = ( x0 ) m w n cos f
Thus,
p
2
t=0
x0 (0) = 0.05 m = ( x0 ) m sin f = ( x0 ) m (1)
f=
( x0 ) m = 0.05 m
&&
x0 = -( x0 ) m w n2 sin(w nt + f )
( a0 ) max = ( &&
x0 ) max
= -( x0 ) m w n2
= -(0.05 m)(32.6599 s -1 )2
= 53.333 m/s2
( a0 ) max = 53.3 m/s2 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.
49
PROBLEM 19.40
A 7.5 kg slender rod AB is riveted to a 6 kg uniform disk as shown. A
belt is attached to the rim of the disk and to a spring which holds the rod
at rest in the position shown. If end A of the rod is moved 20 mm down
and released, determine (a) the period of vibration, (b) the maximum
velocity of end A.
SOLUTION
Equation of motion.
+
SM B = S ( M B )eff : mg
where x = rq + d ST and from statics, mg
L
Ê
cos q - kxr = I ABa + m Á a
Ë
2
Lˆ Ê Lˆ
˜ Á ˜ + I Diska
2¯ Ë 2¯
(1)
L
= kd ST r
2
Assuming small angles (cos q ª 1), Equation (1) becomes
2
Ê
ˆ
L
Ê Lˆ
mg q - kr 2q - krd ST = Á I AB + m Á ˜ + I Disk ˜ a
Ë 2¯
2
Ë
¯
Ê
ˆ
mL2
+
+ I Disk ˜ q&& + kr 2q = 0
I
Á AB
4
Ë
¯
Data:
m = 7.5 kg
mdisk = 6 kg
L = 900 mm = 0.9 m
r = 250 mm = 0.25 m
k = 6 kN/m = 6000 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
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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50
PROBLEM 19.40 (Continued)
1
mL2
12
1
= (7.5)(0.9)2
12
= 0.50625 kg ◊ m 2
I AB =
1
mDisk r 2
2
1
= (6)(0.25)2
2
= 0.1875 kg ◊ m2
I Disk =
1
È
˘ &&
2
2
ÍÎ0.50625 + 4 (7.5)(0.9) + 0.1875˙˚ q + (6000)(0.25) q = 0
2.2125q&& + 375q = 0 or q&& + 169.492q = 0
(a)
Natural frequency and period.
w n2 = 169.492 ( rad/s)2
w n = 13.0189 rad/s
t=
(b)
Maximum velocity.
2p
2p
=
w n 13.0189
vm = w n xm = (13.0189 rad/s)(0.02 m)
t = 0.483 s vm = 260 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
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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51
PROBLEM 19.41
An 8-kg uniform rod AB is hinged to a fixed support at A and
is attached by means of pins B and C to a 12-kg disk of radius
400 mm. A spring attached at D holds the rod at rest in the
position shown. If Point B is moved down 25 mm and released,
determine (a) the period of vibration, (b) the maximum velocity
of Point B.
SOLUTION
(a)
SM A = ( SM A )eff : FS = k (0.6q + d ST )
Equation of motion.
+
0.6( mR g - FS ) + 1.2mD g = ( I R + I D )a + 0.6( mR )( at ) D + 1.2( mD )( at ) B
At equilibrium (q = 0),
FS = kd ST
SM A = 0 = 0.6( mR g - k (d ST )) + 1.2mD g
Substituting Eq. (2) into Eq. (1),
(1)
(2)
( I R + I D )a + 0.6 mR ( at ) D + 1.2mD ( at ) B + (0.6)2 kq = 0
a = q&&
( at ) B = 0.6q&&
( a ) = 1.2q&&
t D
1
1
mR l 2 = (8)(1.2)2
12
12
= 0.960 kg ◊ m
IR =
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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52
PROBLEM 19.41 (Continued)
1
1
mD R2 = (12)(0.4)2 = 0.960 kg ◊ m
2
2
2
2
&&
[0.960 + 0.960 + (0.6) (8) + (1.2) (12)]q + (0.6)2 (800)q = 0
288 N ◊ m
q&& +
q =0
(22.08 kg ◊ m2 )
ID =
(a)
Natural frequency and period.
288
22.08
= 3.6116 rad/s
2p
2p
tn =
=
w n 3.6116
wn =
(b)
t n = 1.740 s Maximum velocity at B.
( v B ) max = (1.2)(q&max )
yB
1.2
0.025
qm =
= 0.02083 rad
1.2
q = q m sin(w nt + f )
q& = q w cos(w t + f )
qm =
m
n
n
q&max = q mw n = (0.02083)(3.612) = 0.0752
24 rad/s
( v B ) max = (1.2)(q&max ) = (1.2 m)(0.07524) rad/s
( v B ) max = 0.09029 m/s
( v B ) max = 90.3 mm/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
teachers and 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 19.42
Solve Problem 19.41, assuming that pin C is removed and that
the disk can rotate freely about pin B.
SOLUTION
(a)
Note: This problem is the same as Problem 19.41, except that the disk does not rotate, so that the effective
moment I Da = 0.
SM A = ( SM A )eff : FS = k (0.60 + d ST )
Equation of motion.
+
(0.6)( mR g - FS ) + 1.2mD g = I Ra + (0.6)( mR )( at ) D + 1.2( mD )( at ) B
At equilibrium (q = 0),
FS = kd ST
+
Substituting Eq. (2) into Eq. (1),
(1)
SM A = 0 = 0.6( mR g - d ST ) + 1.2 mD g
(2)
I Ra + 0.6 mR ( at ) D + 1.2mD ( at ) B + (0.6)2 kq = 0
a = q&&
( at ) B = 0.6q&&
( a ) = 1.2q&&
t D
1
1
I R = mR l 2 = (8)(1.2)2
12
12
= 0.960 kg ◊ 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
teachers and 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 19.42 (Continued)
[0.960 + (0.6)2 (8) + (1.2)2 (12)]q&& + (0.6)2 (800)q = 0
(288 N ◊ m))
q =0
q&& +
21.12 kg ◊ m2
(a)
Natural frequency and period.
288
21.12
= 3.693 rad/s
wn =
tn =
(b)
2p
2p
=
w n 3.693
t n = 1.701 s Maximum velocity at B.
( v B ) max = (1.2)(q& ) max
m B 0.025
=
= 0.02083 rad
1.2 1.20
q = q m sin(w nt + f )
q&& = q w cos(w t + f )
qm =
m
n
n
q&&max = q mw n
( v B ) max
= (0.02083)(3.693)
= 0.076994 rad/s
= (1.2)(q& )
max
= (1.2)(0.07694)
= 0.09233 m/s
( v B ) max = 92.3 mm/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
teachers and 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 19.43
A belt is placed around the rim of a 240-kg flywheel and attached as
shown to two springs, each of constant k = 15 kN/m. If end C of the belt is
pulled 40 mm down and released, the period of vibration of the flywheel is
observed to be 0.5 s. Knowing that the initial tension in the belt is sufficient
to prevent slipping, determine (a) the maximum angular velocity of the
flywheel, (b) the centroidal radius of gyration of the flywheel.
SOLUTION
Denote the initial tension by T0.
Equation of motion.
+
SM 0 = S ( M 0 )eff : - TA r + TB r = I q&&
- (T0 + k r q )r + (T0 - krq )q = I q&&
2
2kr
q&& +
q =0
I
2 kr 2
w n2 =
I
Data:
(a)
m = 240 kg
t = 0.5 s
2p
t=
;
wn
(1)
k = 15 kN/m
r = 0.45 m
2p 2p
wn =
=
= 4p rad/s
t
0.5
Maximum angular velocity. If Point C is pulled down 40 mm and released,
q m = q max
Ê 40 mm ˆ
=Á
Ë 450 mm ˜¯
= 88.89 ¥ 10 -3 rad
q&m = q mw n
= (88.92 ¥ 10 -3 rad )( 4p rad/s)
q&m = 1.117 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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it without permission.
56
PROBLEM 19.43 (Continued)
(b)
Centroidal radius of gyration.
2kr 2
I
2(15 kN/m)(0.45 m)2
( 4p rad/s)2 =
I
I = 38.47 kg ◊ m2
w n2 =
or since
I = mk 2
(240 kg) k 2 = 28.47 kg ◊ m2
k = 0.4004 m
k = 400 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
57
PROBLEM 19.44
A 75-mm-radius hole is cut in a 200-mm-radius uniform disk, which
is attached to a frictionless pin at its geometric center O. Determine
(a) the period of small oscillations of the disk, (b) the length of a simple
pendulum which has the same period.
SOLUTION
Equation of motion.
SM 0 = ( SM 0 )eff :
+
- mH g (0.1)sin q = I Dq - I H q&& - (0.1)2 mH q&&
mD = rtp R2
= ( rtp )(0.2)2
= (0.04) prt
mH = rtp r 2
= ( rtp )(0.075)2
= (0.005625) prt
1
1
mD R2 = (0.04 prt )(.2)2
2
2
-6
= 800 ¥ 10 prt
ID =
1
mH r 2
2
1
= (0.005625 prt )(0.75)2
2
= 15.82 ¥ 10 -6 prt
IH =
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.
58
PROBLEM 19.44 (Continued)
Small angles.
sinq ª q
[(800 ¥ 10 -6 p - 15.82 ¥ 10 -6 p - (0.1)2 (0.005625p )] r t q&&
+ (0.005625) p r t (9.81)(0.1)q = 0
727.9 ¥ 10 -6 q&& + 5.518 ¥ 10 -3 q = 0
(a)
Natural frequency and period.
5.518 ¥ 10 -3
727.9 ¥ 10 -6
= 7.581
w n = 2.753 rad/s
w n2 =
tn =
(b)
2p
wn
tn =
2p
= 2.28 s 2.753
Length and period of a simple pendulum.
t n = 2p
l
g
2
Êt ˆ
l=Á n˜ g
Ë 2p ¯
2
È (2.753) ˘
(9.81 m/s2 )
l=Í
˙
p
2
Î
˚
l = 1.294 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.
59
PROBLEM 19.45
Two small weights w are attached at A and B to the rim of a uniform disk of radius r
and weight W. Denoting by t 0 the period of small oscillations when b = 0, determine
the angle b for which the period of small oscillations is 2t 0 .
SOLUTION
a = q&&
at = ra = rq&&
ID =
1W 2
r
2 g
Equation of motion.
2w
rat + I a
g
wr[sin( b - q ) - sin( b + q )] = -2wr sin q cos b
sin q ª q
SM C = ( SM C )eff : wr sin( b - q ) - wr sin( b + q ) =
Ê 2w 2 W 2 ˆ &&
ÁË g r + 2 g r ˜¯ q + (2 wr cos b ) q = 0
Natural frequency.
wn =
2wg cos b
4 g cos b
=
W
( 4 + Ww )r
2w + 2 r
(
)
b =0
tn =
t0 =
2p
cos b
( 4 + Ww ) r
2p
=
w0
= 2t 0 =
(1)
2p
4g
( 4 + Ww ) r
4p
4g
( 4 + Ww ) r
2
1
Ê 1ˆ
cos b = Á ˜ =
Ë 2¯
4
b = 75.5∞ 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.
60
PROBLEM 19.46
Two 50 g weights are attached at A and B to the rim of a 1.5 kg uniform disk of radius
r = 100 mm . Determine the frequency of small oscillations when b = 60∞.
SOLUTION
a = q&&
at = ra = rq&&
ID =
1W 2
r
2 g
Equation of motion.
2w
rat + I a
g
wr[sin( b - q ) - sin( b + q )] = -2wr sin q cos b
sin q ª q
SM C = ( SM C )eff : wr sin( b - q ) - wr sin( b + q ) =
Ê 2w 2 W 2 ˆ &&
ÁË g r + 2 g r ˜¯ q + (2 wr cos b ) q = 0
Natural frequency.
Data:
wn =
2wg cos b
4 g cos b
=
W
4 + Ww )r
(
2w + 2 r
(
)
(1)
W m
1.5
= =
= 30
w m 50 ¥ 10 -3
r = 100 mm = 0.1 m b = 60∞
wn =
fn =
( 4)(9.81) cos 60∞
= 2.4022 rad/s
( 4 + 30)(0.1)
w n 2.4022
=
2p
2p
f n = 0.382 Hz 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.
61
PROBLEM 19.47
For the uniform square plate of side b = 300 mm, determine (a) the
period of small oscillations if the plate is suspended as shown,
(b) the distance c from O to a Point A from which the plate should
be suspended for the period to be a minimum.
SOLUTION
(a)
Equation of motion.
SM 0 = ( SM 0 )eff : a = q&&
1
I = mb2
6
at = (OG )(a )
OG = b
2
2
Ê 2 ˆ &&
at = Á b
˜q
Ë 2 ¯
+
(OG )(sin q )( mg ) = -(OG )mat - I a
sin q ª q
Ê 2 ˆ Ê 2 ˆ && 1
2ˆ
2 && Ê
Á b 2 ˜ m Á b 2 ˜ q + 6 mb q + Á b 2 ˜ mg q = 0
¯
¯
Ë
¯ Ë
Ë
Ê 2ˆ
Ê 1 1ˆ
(b) Á + ˜ m q&& + Á
˜ mg q = 0
Ë 2 6¯
Ë 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
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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62
PROBLEM 19.47 (Continued)
q&& +
( )gq = 0
2
2
3 2g
or q&& +
=0
4 b
( )b
2
3
Natural frequency and period.
3 2 g 3 2 (9.81)
=
= 34.68
4 b
( 4)(0.3)
= 5.889 rad/s
w n20 =
w n0
t n0 =
(b)
Suspended about A.
2p
w n0
t n0 = 1.067 s Let e = (OG - c)
at = ea
Equation of motion.
+ SM A
= S ( M A )eff :
mge sin q = -emat - I a = -( me 2 + I )a
1 ˆ
Ê
m Á e 2 + b2 ˜ q&& + mgeq = 0
Ë
6 ¯
w n2 =
Frequency and period.
For t n to be minimum,
eg
e + 16 b2
2
t n2 =
2 2
2
4p 2 4p (e + 16 b )
=
eg
w n2
t n2 =
b2 ˆ
4p 2 Ê
+
e
g ÁË
6e ˜¯
d Ê
b2 ˆ
e
+
=0
6e ˜¯
de ÁË
1-
b2
=0
6e 2
b2
=6
e2
e=
b
6
2
b
b= 0.29886b
2
6
c = (0.29886)(300 mm))
c = OG - e =
c = 89.7 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
63
PROBLEM 19.48
A connecting rod is supported by a knife-edge at Point A; the period of its small oscillations
is observed to be 0.87 s. The rod is then inverted and supported by a knife-edge at Point B
and the period of small oscillations is observed to be 0.78 s. Knowing that ra + rb = 250 mm
determine (a) the location of the mass center G, (b) the centroidal radius of gyration k .
SOLUTION
Consider general pendulum of centroidal radius of gyration k .
Equation of motion.
+
SM 0 = S ( M 0 )eff : - mgr sin q = ( mr q&&)r + mk 2q&&
È gr ˘
sin q = 0
q&& + Í 2
Î r + k 2 ˙˚
For small oscillations, sin q ª q , we have
È gr ˘
q&& + Í 2
q =0
Î r + k 2 ˙˚
w n2 =
t=
For rod suspended at A,
gr
r +k2
2
2p
r2 +k2
= 2p
wn
gr
ra2 + k 2
gra
t A = 2p
(
gt 2A ra = 4p 2 ra2 + k 2
For rod suspended at B,
t B = 2p
)
(1)
rb2 + k 2
grb
(
gt B2 rb = 4p 2 rb2 + k 2
)
(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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
64
PROBLEM 19.48 (Continued)
(a)
Value of ra .
Subtracting Eq. (2) from Eq. (1),
(
gt 2A ra - gt B2 rb = 4p 2 ra2 - rb2
)
gt 2A ra - gt B2 rb = 4p 2( ra + rb )( ra - rb )
Applying the numerical data with ra + rb = 250 mm = 0.25 m
(9.81)(0.87)2 ra - (9.81)(0.78)2 rb = 4p 2(0.25)( ra - rb )
7.4252ra - 5.9684rb = 9.8696( ra - rb )
3.9012rb = 2.4444ra
0.25 = ra + 0.6266ra
rb = 0.25 - 0.15369
(b)
rb = 0.6266ra
ra = 0.15369 m
ra = 153.7 mm rb = 0.09631 m
rb = 96.31 mm Centroidal radius of gyration.
From Eq. (1),
4p 2 k 2 = g t2A ra - 4p 2 ra2
= (9.81)(0.87)2 (0.15369) - 4p 2 (0.15369)2 = 0.20867 m2
k = 0.072703 m
k = 72.7 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
65
PROBLEM 19.49
For the uniform equilateral triangular plate of side l = 300 mm, determine
the period of small oscillations if the plate is suspended from (a) one of its
vertices, (b) the midpoint of one of its sides.
SOLUTION
For an equilateral triangle,
h=
3
b,
2
1
3 2
A = bh =
b
2
4
3
1
1 Ê 3 ˆ
3 4
I x = bh3 = l Á
l˜ =
l
36
36 Ë 2 ¯
96
Iy = 2 -
3
3
1 Ê bˆ
1Ê 3 ˆÊ lˆ
3 4
l
hÁ ˜ = Á
l˜ Á ˜ =
12 Ë 2 ¯
6 Ë 2 ¯ Ë 2¯
96
3 4
l
48
I
1
1
k 2 = z = l 2 = (0.300)2 = 0.0075 m2
12
A 12
I z = I x + I4 =
Let r be the distance from the suspension point to the center of mass.
Equation of motion.
where
For small angle q,
Natural frequency:
+
SM 0 = + S( M 0 )eff : - mge sin q = r ( mr q&&) + I q&&
I = mk 2
- mrg q = m ( r 2 + k 2 ) q&&
gr
q&& + 2
q =0
r +k2
gr
w n2 = 2
r +k2
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.
66
PROBLEM 19.49 (Continued)
(a)
Plate is suspended from a vertex.
2
2 3
2 3
h=
l=
(0.3 m)
3
3 2
3 2
= 0.173205 m
r=
r 2 = 0.03 m2
(9.81)(0.173205)
w n2 =
= 45.31
0.03 + 0.0075
w n = 6.7313 rad/s
t=
(b)
2p
wn
t = 0.933 s Plate is suspended at the midpoint of one side.
1
1Ê 3 ˆ 1 3
r= h= Á
l =
(0.3)
3
3 Ë 2 ˜¯ 3 2
= 0.0866025 m
r = 0.0075 m2
(9.81)(0.0866025)
= 56.64
w n2 =
0.0075 + 0.0075
w n = 7.5258 rad/s
2
t=
2p
wn
t = 0.835 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
67
PROBLEM 19.50
A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of
negligible mass, which can rotate freely in a vertical plane about B. Determine the
period of small oscillations (a) if the disk is free to rotate in a bearing at A, (b) if
the rod is riveted to the disk at A.
SOLUTION
1 2
mr
2
m
1
= (0.250)2 m =
2
32
at = la = 0.650a
a = q&&
The disk is free to rotate and is in curvilinear translation.
I =
Ia = 0
Thus,
SM B = S ( M B )eff :
+
(a)
- mgl sin q = lmat
sin q ª q
ml 2q&& - mglq = 0
g 9.81 m/s2
=
0.650 m
I
= 15.092
w n = 3.885 rad/s
w n2 =
tn =
2p
2p
=
w n 3.885
t n = 1.617 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
68
PROBLEM 19.50 (Continued)
When the disk is riveted at A, it rotates at an angular acceleration a.
SM B = S ( M B )eff :
+
(b)
- mgl sinq = I a + lmat
I=
1 2
mr
2
Ê1 2
2 ˆ &&
ÁË mr + ml ˜¯ q + mglq = 0
2
w n2 =
(
gl
2
r
2
+ l2
)
(9.81 m/s2 )(0.650 m)
È 0.2502 + (0.650)2 ˘
Î 2
˚
= 14.053
w n = 3.749 rad/s
=
(
tn =
2p
wn
=
2p
3.749
)
t n = 1.676 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
69
PROBLEM 19.51
A small collar weighing 1 kg is rigidly attached to a 3 kg uniform rod of length L = 1 m
Determine (a) the distance d to maximize the frequency of oscillation when the rod is
given a small initial displacement, (b) the corresponding period of oscillation.
SOLUTION
Equation of motion.
-WR
L
L
sin q - WC d sin q = I Ra + mR ( at ) R + mC d ( at )C
2
2
SM A = ( SM A )eff :
+
sinq ª q
L
L
a = q&&, ( at ) R = a = q&&, ( at )C = da = dq&&
2
2
2
Ê
ˆ
L
ˆ
Ê Lˆ
2 && Ê
Á I R + mR ÁË ˜¯ + mC d ˜ q + ÁË mR g + mC gd ˜¯ q = 0
2
2
Ë
¯
IR =
1
mR L2
12
2
m L2
Ê Lˆ
I R + mR Á ˜ = R
Ë 2¯
3
Ê mR L2
L
ˆ
2 ˆ && Ê
Á 3 + mC d ˜ q + ÁË mR g 2 + mC gd ˜¯ q = 0
Ë
¯
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.
70
PROBLEM 19.51 (Continued)
q&& +
(
(
+
L
2
2
L
3
mC
mR
+
mC
mR
)
d g
d2
)
q =0
mC 1
=
mR 3
q&& +
Natural frequency.
(a)
( 12 + 13 d )
Ê 1 1 2ˆ
ÁË + d ˜¯
3 3
w n2 =
L =1m
q =0
( 12 + 13 d ) g = (1.5 + d ) g
1
3
+ 13 d 2
(1 + d 2 )
To maximize the frequency, we need to take the derivative and set A equal to zero.
( )
2
(1 + d 2 )(1) - (1.5 + d )(2d )
1 d wn
=
=0
g dt
(1 + d 2 )2
d 2 + 3d - 1 = 0
Solve for d.
d = 0.3028 or - 3.3028
d = 0.303 m
w n2 =
d = 303mm (1.5 + 0.3028)(9.81)
(1 + 0.3028)2
w n2 = 16.200
(b)
Period of oscillation.
w n = 4.0249 rad/s
w n = 4.02 rad/s 2p
2p
=
w n 4.0249
t n = 1.561 s tn =
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.
71
PROBLEM 19.52
A compound pendulum is defined as a rigid slab which oscillates about a fixed Point O
called the center of suspension. Show that the period of oscillation of a compound
pendulum is equal to the period of a simple pendulum of length OA, where the distance
from A to the mass center G is GA = k 2 /r . Point A is defined as the center of oscillation
and coincides with the center of percussion defined in Problem 17.66.
SOLUTION
+
SM 0 = S ( M 0 )eff : - W r sin q = I a + mat r
- mgr sinq = mk 2q&& + mr 2q&&
q&& +
gr
r +k2
2
sin q = 0
(1)
For a simple pendulum of length OA = l ,
g
q&& + q = 0
l
Comparing Equations (1) and (2),
l=
r2 +k2
r
(2)
GA = l - r =
k2
Q.E.D. r
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.
72
PROBLEM 19.53
A rigid slab oscillates about a fixed Point O. Show that the smallest period of oscillation
occurs when the distance r from Point O to the mass center G is equal to k .
SOLUTION
See Solution to Problem 19.52 for derivation of
q&& +
gr
r +k2
2
sin q = 0
For small oscillations, sinq ª q and
tn =
2p
r 2 + k 2 2p
= 2p
=
wn
gF
g
r+
k2
r
2
For smallest t n , we must have r + kr as a minimum:
(
2
d r + kr
dr
) =1- k
r
2
2
r2 = k2
=0
r = k Q.E.D. 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.
73
PROBLEM 19.54
Show that if the compound pendulum of Problem 19.52 is suspended from A instead of O, the period of
oscillation is the same as before and the new center of oscillation is located at O.
SOLUTION
Same derivation as in Problem 19.52 with r replaced by R.
Thus,
q&& +
gR
R +k
2
q =0
Length of the equivalent simple pendulum is
R2 + k 2
k2
=R+
R
R
2
k
L = (l - r ) + 2 = l
L=
k
r
Thus, the length of the equivalent simple pendulum is the same as in Problem 19.52. It follows that the period
is the same and that the new center of oscillation is at O.
Q.E.D.
PROPRIETARY MATERIAL. © 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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74
PROBLEM 19.55
The 8-kg uniform bar AB is hinged at C and is attached at A to a spring of
constant k = 500 N/m. If end A is given a small displacement and released,
determine (a) the frequency of small oscillations, (b) the smallest value of
the spring constant k for which oscillations will occur.
SOLUTION
I =
1 2 Ê 1ˆ
ml = Á ˜ (8)(0.250)2
Ë 12 ¯
12
I = 0.04167 kg ◊ m2
a = q&&
at = 0.04a = 0.04q&&
sin q ª q
Equation of motion.
SMC = S ( MC )eff :
&&
-(0.165)2 kq + 0.04 mgq = I q&& + (0.04)2 mA
(0.04167 + 0.01280)q&& + (0.02722k - 0.32 g )q = 0
(a)
Frequency if k = 500 N ◊ m.
0.05447q&& + (10.47)q = 0
(b)
(1)
w
fn = n =
2p
For t n Æ d or w n Æ 0, oscillations will not occur.
From Equation (1),
w n2 =
k=
(
10.47
0.05447
)
2p
f n = 2.21 Hz 0.02722k - 0.32 g
=0
(0.05447)
0.32 g
(0.32)(9.81)
=
0.02722
(0.02722)
k = 115.3 N/m PROPRIETARY MATERIAL. © 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.
75
PROBLEM 19.56
A 20-kg uniform square plate is suspended from a pin located
at the midpoint A of one of its 0.4 m edges and is attached to
springs, each of constant k = 1.6 kN/m If corner B is given a
small displacement and released, determine the frequency of
the resulting vibration. Assume that each spring can act in
either tension or compression.
SOLUTION
a = q&&
b
b
at = a - q&&
2
2
sin q ª q
2
+
Equation of motion.
b
Ê bˆ
SM 0 = S ( M 0 )eff : - mg q + 2kb2q = I a + Á ˜ ma
Ë 2¯
2
2
1
b2 5
Ê bˆ
I + m Á ˜ = mb2 + m = mb2
Ë 2¯
6
4 12
b
5
ˆ
Ê
mb2q&& + Á mg + 2kb2 ˜ q = 0
¯
Ë
12
2
Ê 12 g 24k ˆ
+
q&& + Á
q =0
Ë 10 b 5m ˜¯
Data:
b = 0.4 m; m = 20 kg
k = 1.6 kN/m = 1600 N/m
È (12)(9.81) (24)(1600) ˘
q&& + Í
+
q =0
(5)(20) ˙˚
Î (10)(0.4)
q&& + 413.43 = 0
w n2 = 413.43 w n = 20.333 rad/s
Frequency.
fn =
w n 20.333
=
2p
2p
f n = 3.24 Hz 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.
76
PROBLEM 19.57
Two uniform rods, each of mass m = 12 kg and length L = 800 mm, are
welded together to form the assembly shown. Knowing that the constant of
each spring is k = 500 N/m and that end A is given a small displacement
and released, determine the frequency of the resulting motion.
SOLUTION
SM 0 = S ( M 0 )eff :
+
Equation of motion.
2
2
È
L
Ê Lˆ
Ê Lˆ ˘
ÍmAC g - 2k Á ˜ ˙ q = ( I AC + I BD )a + mAC Á ˜ a
Ë 2¯
Ë 2¯ ˙
2
ÍÎ
˚
mBD = mAC = m
1 2
mL
2
2
L˘
Ê 1 1 ˆ 2 && È Ê L ˆ
+
q
mL
k
+
2
Í Á ˜ - mg ˙ q = 0
˜¯
ÁË
6 4
2 ˙˚
ÍÎ Ë 2 ¯
I BD = I AC = I =
10 2 && È kL2 mgL ˘
6( kL - mg )
2
mL q + Í
˙q = 0 fi wn =
24
2 ˚
5 mL
Î 2
Data:
Frequency.
L = 800 mm = 0.8 m, m = 12 kg, k = 500 N/m
w n2 =
6[(500)(0.8) - (12)(9.81)]
= 35.285
(5)(12)(0.8)
w n = 5.9401 rad/s
fn =
wn
2p
f n = 0.945 Hz 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.
77
PROBLEM 19.58
The rod ABC of total mass m is bent as shown and is supported in
a vertical plane by a pin at B and by a spring of constant k at C. If
end C is given a small displacement and released, determine the
frequency of the resulting motion in terms of m, L, and k.
SOLUTION
L
a = q&& at = q&&
2
1 m 2 mL2
I =
L =
12 2
24
sin q ª q cos q ª 1
Equation of motion.
+
SM B = S ( H B )eff :
But for equilibrium (q = 0),
mg L
mg L
m L
sin q +
cos q - kL( Lq + d ST )cosq = 2 I q&& + 2 at
2 2
2 2
2 2
mgL
mgL
mL2 && mL2 &&
q+
- kL2q - kL2d ST =
q+
q
4
4
12
4
m L
SM B = 0 = g - kL2d ST
2 2
(1)
(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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78
PROBLEM 19.58 (Continued)
Substituting Eq. (2) into Eq. (1),
mL2 &&
Ê mgL
2ˆ
=
q
q
kL
˜¯
ÁË
4
3
(kL - ) q = 0
q&& +
mgL
4
2
mL2
3
3k 3 g
m 4L
k
g
wn = 3
m 4L
w n2 =
Frequency.
fn =
wn
2p
fn =
3 k
g
2p m 4 L
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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79
PROBLEM 19.59
A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of
negligible mass which can rotate freely in a vertical plane about B. If the rod is
displaced 2° from the position shown and released, determine the magnitude of
the maximum velocity of Point A, assuming that the disk (a) is free to rotate in a
bearing at A, (b) is riveted to the rod at A.
SOLUTION
1 2
mr
2
a = q&&
I =
Kinematics:
at = la = lq&&
(a)
The disk is free to rotate and is in curvilinear translation.
Thus,
Equation of motion.
Ia = 0
SM B = ( SM B )eff : - mgl sin q = lmat , sin q ª q
ml 2q&& + mglq = 0
Frequency.
g
l
9.81
=
0.650
= 15.092
w n = 3.8849 rad/s
w n2 =
tn =
2p
= 1.617 s
wn
t n = 1.617 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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it without permission.
80
PROBLEM 19.59 (Continued)
(b)
When the disk is riveted at A, it rotates at an angular acceleration a.
Equation of motion.
SM B = ( SM B )eff : - mgl sin q = I a + lmat , I =
1 2
mr , sin q ª q
2
Ê1 2
2 ˆ &&
ÁË mr + ml ˜¯ q + mglq = 0
2
Frequency.
w n2 =
=
(
gl
r2
2
+ l2
)
(9.81)(0.650)
+ (0.650)2
2
1
2 ( 0.250)
= 14.053
w n = 3.7487 rad/s
tn =
2p
= 1.676 s
wn
t n = 1.676 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
81
PROBLEM 19.60
A 3-kg slender rod is suspended from a steel wire which is known to
have a torsional spring constant K = 2.25 N ◊ m/rad. If the rod is rotated
through 180° about the vertical and released, determine (a) the period of
oscillation, (b) the maximum velocity of end A of the rod.
SOLUTION
K
SM G = S ( M G )eff : - Kq = I q&& q&& + q = 0
I
K
2
2
q&& + w nq = 0 w n =
I
Equation of motion.
m = 3 kg
l = 200 mm = 0.2 m
1
1
2
I = ml 2 = (3) (0.2)
12
12
= 0.01 kg ◊ m2
K = 2.25 N ◊ m/ rad
2.25
w n2 =
= 225
0.01
w n = 15 rad/s
Data:
(a)
tn =
Period of oscillation.
2p 2p
=
w n 15
t n = 0.419 s q = q m sin (w nt + j )
q& = w q cos (w t + j )
Simple harmonic motion:
n m
n
q&m = w nq m
l
1
( v A )m = q&m = lw Aq m
2
2
q m = 180∞ = p radians
(b)
Maximum velocity at end A.
Ê 1ˆ
( v A )m = Á ˜ (0.2)(15)(p )
Ë 2¯
( v A )m = 4.71m/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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
82
PROBLEM 19.61
A homogeneous wire bent to form the figure shown is attached to a pin
support at A. Knowing that r = 220 mm and that Point B is pushed
down 20 mm and released, determine the magnitude of the velocity
of B, 8 s later.
SOLUTION
Determine location of the centroid G.
Let
r = mass per unit length
Then total mass
m = r ( 2r + p r ) = r r ( 2 + p )
Ê 2r ˆ
mgc = 0 + p r r Á ˜ g = 2r 2 r g
Ëp¯
About C:
y=
2r
p
for a semicircle
r r ( 2 + p )c = 2 r 2 r , c =
+
Equation of motion.
2r
(2 + p )
SM 0 = S ( M 0 )eff : a = q&& at = ca = cq&&
- mgc sin q = I a + mcan
sinq ª q
( I + mc )q&& + mgcq = 0
I 0q&& + mgcq = 0
2
Moment of inertia.
I + mc 2 = I 0
I 0 = ( I 0 )semicirc. + ( I 0 )line = msemicirc. r 2 + mline
msemicirc. = rp r
mline = 2r r
r=
( 2r )2
12
m
(2 + p )r
È
r2 ˘
mr 2 È
2˘
p+ ˙
I 0 = r Íp r 2 ◊ r + 2r ◊ ˙ =
Í
3 ˚ (2 + p ) Î
3˚
Î
2˘
2r
mr 2 È
p + ˙ q&& + mg
q =0
Í
(2 + p ) Î
3˚
(2 + 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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83
PROBLEM 19.61 (Continued)
Frequency.
w n2 =
w n2
2g
(2)(9.81)
=
2
p+3 r
p + 23 (0.220)
(
)
= 23.418 s
(
-2
)
w n = 4.8392 rad/s
q = q m sin(w nt + f )
At t = 0,
y B = rq
y B = 20 mm, y& B = 0
y& B = 0 = ( y B ) m w n cos( 0 + f ), f =
p
2
pˆ
Ê
y B = 20 mm = ( y B )m sin Á 0 + ˜ , (y B )m = 20 mm
Ë
2¯
pˆ
Ê
y& B = (20 mm) sin Á w nt + ˜ w n = 4.8392 rad/s
Ë
2¯
pˆ
Ê
y& B = 20w cos Á w nt + ˜ = -(20 mm)w n sin w nt
Ë
2¯
At t = 8 s,
y& B = -(20)( 4.8392)sin[( 4.8392)(8)] = -(96.88)(0.8492)
= -82.2 mm/s
vB = 82.2 mm/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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
84
PROBLEM 19.62
A homogeneous wire bent to form the figure shown is attached to a pin
support at A. Knowing that r = 400 mm and that Point B is pushed down
40 mm and released, determine the magnitude of the acceleration of B,
10 s later.
SOLUTION
Determine location of the centroid G.
Let
r = mass per unit length
Then total mass
m = r ( 2r + p r ) = r r ( 2 + p )
Ê 2r ˆ
mgc = 0 + p r r Á ˜ g = 2r 2 r g
Ëp¯
About C:
y=
2r
p
for a semicircle
r r ( 2 + p )c = 2 r 2 r
+
Equation of motion.
c=
2r
(2 + p )
SM 0 = S ( M 0 )eff : a = q&& at = ca = cq&&
- mgc sin q = I a + mcan
sinq ª q
&&
&&
( I + mc )q + mgcq = 0 I 0q + mgcq = 0
2
Moment of inertia.
I + mc 2 = I 0
I 0 = ( I 0 )semicirc. + ( I 0 )line = msemicirc. r 2 + mline
msemicirrc. = rp r mline = 2 r r r =
( 2r )2
12
m
r (2 + p )
È
2r ◊ r 2 ˘
mr 2 È
2˘
=
I 0 = r Íp r ◊ r 2 +
p+ ˙
˙
Í
3 ˚ (2 + p ) Î
3˚
Î
mr 2 È
2˘
2r
p + ˙ q&& + mg
q =0
Í
(2 + p ) Î
(2 + p )
3˚
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.
85
PROBLEM 19.62 (Continued)
Frequency.
w n2 =
(2)(9.81)
2g
=
2
p+3 r
p + 23 (0.4)
(
)
(
)
w n2 = 12.8799 (rad/s)2 w n = 3.5889 rad/s
q = q m sin(w nt + f )
y B = rq
y B = rq m sin(w nt + f ) = ( y B ) m sin(w nt + f)
At t = 0,
y B = 40 mm
y& B = 0
(t = 0): y& B = 0 = ( y B )m cos(0 + f ) f =
pˆ
Ê
y B = 40 mm = ( y B ) m sin Á 0 + ˜
Ë
2¯
pˆ
Ê
y B = ( 40 mm)sin Á w nt + ˜
Ë
2¯
p
2
( y B ) m = 40 mm
w n = 3.5889 radd/s
pˆ
Ê
y& B = ( 40)(w n ) cos Á w nt + ˜ = -( 40)w n sin w n t
Ë
2¯
( )
pˆ
Ê
&&y B = ( -40) w n2 sin Á w nt + ˜ = 40w n2 cos w nt
Ë
2¯
At t = 10 s,
( aB )t = &&y B = ( -40)(3.5889)2 cos(3.5889)(10) = -122.124 mm/s2
( v B ) = y& B = -( 40)(3.5889)sin [(3.5889)(10)] = -139.465 mm/s
1
1
2
2
È
È v B2 ˘ ˘ 2 È
È ( -139.465)2 ˘ ˘ 2
2
2
2
aB = Í( aB )t + Í ˙ ˙ = Í(122.124) + Í
˙ ˙ = 122.3 mm/s r
400
Í
˙
˙
Í
Î
˚ ˚
Î ˚ ˚
Î
Î
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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86
PROBLEM 19.63
A uniform disk of radius r = 120 mm is welded at its center to two elastic rods of
equal length with fixed ends at A and B. Knowing that the disk rotates through an
8∞ angle when a 500-mN ◊ m couple is applied to the disk and that it oscillates with
a period of 1.3 s when the couple is removed, determine (a) the mass of the disk,
(b) the period of vibration if one of the rods is removed.
SOLUTION
Torsional spring constant.
k=
T 0.5 N ◊ m
=
p
q
(8) 180
( )
k = 3.581 N ◊ m/rad
Equation of motion.
K
SM 0 = S ( M 0 )eff : -Kq = J q&& q&& + q = 0
J
Natural frequency and period.
w n2 =
K
J
Period.
tn =
2p
J
= 2p
wn
K
J=
(a)
Mass of the disk.
m=
(b)
With one rod removed:
Period.
t n2 K
(1.35)2 (3581 N ◊ m/r )
(2p )2
(2p )2
1
1
J = 0.1533 N ◊ m ◊ s2 = mr 2 = m(0.120 m)2
2
2
Mass moment of inertia.
K¢ =
=
(0.1533 N ◊ m ◊ s2 )(2)
(0.120 m)2
m = 21.3 kg K 3.581
=
= 1.791 N ◊ m/rad
2
2
t = 2p
(0.1533 N ◊ m ◊ s2 )
J
= 2p
K¢
1.791 N ◊ m/rad
t = 1.838 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
87
PROBLEM 19.64
A 5 kg uniform rod CD of length l = 0.75 m is welded at C to two
elastic rods, which have fixed ends at A and B and are known to have
a combined torsional spring constant K = 30 N ◊ m/rad. Determine
the period of small oscillation if the equilibrium position of CD is
(a) vertical as shown, (b) horizontal.
SOLUTION
(a)
Equation of motion.
+
l
l
a = q&& at = a = q&&
2
2
l
l
SM C = S( M C )eff : - Kq - ( mg ) sin q = I a + ( mat )
2
2
1
1
- Kq - mgl sinq = I q&& + ml 2q&&
2
4
1 2 ˆ && 1
Ê
ÁË I + ml ˜¯ q + mgl sin q + Kq = 0
4
2
1
Ê 1 2 1 2 ˆ &&
ÁË ml + ml ˜¯ q + Kq + mglq = 0
12
4
2
1 2 && Ê
1
ˆ
ml q + Á K + mgl ˜ q = 0
¯
Ë
3
2
Ê 3K 3 g ˆ
q&& + Á 2 + ˜ q = 0
Ë ml
2l ¯
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.
88
PROBLEM 19.64 (Continued)
K = 30 N ◊ m/rad, m = 5 kg, l = 0.75 m
Data:
È (3)(30)
(3)(9.81) ˘
+
q&& + Í
˙q = 0
2
(2)(0.75) ˚
Î (5)(0.75)
q&& + 51.62q = 0
w n2 = 51.62 w n = 7.1847 rad/s
Frequency.
Period.
(b)
t=
2p
2p
=
w n 7.1847
t n = 0.875 s If the rod is horizontal, the gravity term is not present and the equation of motion is
3K
q&& + 2 q = 0
ml
(3)(30)
3K
w n2 = 2 =
= 32
(5)(0.75)2
ml
w n = 5.9938 rad/s t =
2p
2p
=
w n 5.65685
t n = 1.111 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
89
PROBLEM 19.65
A 1.8-kg uniform plate in the shape of an equilateral triangle is suspended at its
center of gravity from a steel wire which is known to have a torsional constant
K = 35 mN ◊ m/rad. If the plate is rotated 360° about the vertical and then
released, determine (a) the period of oscillation, (b) the maximum velocity of
one of the vertices of the triangle.
SOLUTION
Mass moment of inertia of plate about a vertical axis:
h=
3
b
2
1 3
3b 4
bh =
36
96
3 4
Iz = Ix + I y =
b
48
For area,
Ix = I y =
For mass,
I =
m
( I z )area
A
Ê 4 mˆ Ê 3 4ˆ 1
=Á
b = mb2
Ë 3b ˜¯ ÁË 48 ˜¯ 12
I =
Equation of motion.
1
3 2
A = bh =
b
2
4
1
(1.8)(0.150)2 = 3.375 ¥ 10 -3 kg ◊ m2
12
+ SM G = S ( M G )eff : - Kq = I q&&
K
q&& + q = 0
I
Frequency.
K
35 ¥ 10 -3
=
= 10.37
I 3.375 ¥ 10 -3
w n = 3.2203 rad/s
w n2 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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90
PROBLEM 19.65 (Continued)
(a)
(b)
Period.
t=
2p
2p
=
w n 3.2203
Maximum rotation.
q m = 360∞ = 2p rad
Maximum angular velocity.
q&m = w nq m = (3.2203)(2p ) = 20.234 rad/s
t = 1.951 s Maximum velocity at a vertex.
2
2 3
Ê 2ˆ Ê 3 ˆ
vm = rq&m = hq&m =
b = Á ˜ Á ˜ (0.150)(20.234)
Ë 3¯ Ë 2 ¯
3
3 2
vm = 1.752 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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91
PROBLEM 19.66
A horizontal platform P is held by several rigid bars which are connected
to a vertical wire. The period of oscillation of the platform is found to be
2.2 s when the platform is empty and 3.8 s when an object A of unknown
moment of inertia is placed on the platform with its mass center directly
above the center of the plate. Knowing that the wire has a torsional
constant K = 30 N ◊ m/rad, determine the centroidal moment of inertia
of object A.
SOLUTION
Equation of motion.
SM G = S ( M G )eff : -Kq = I q&&
K
K
q&& + q = 0 w n2 =
&I
I
2p 2p
=
= 2.856 rad/s
t 1 2.2
K
30
I1 = 2 =
= 3.6779 kg ◊ m2
2
w n1 (2.856)
Case 1. The platform is empty.
w n1 =
Case 2. Object A is on the platform.
w n2 =
Moment of inertia of object A.
2p 2p
=
= 1.653 rad/s
t 2 3.8
K
30
I2 = 2 =
= 10.9793 kg ◊ m2
2
w n2 (1.653)
I A = I 2 - I1
I A = 7.30 kg ◊ m2 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.
92
PROBLEM 19.67
A thin rectangular plate of sides a and b is suspended from four vertical
wires of the same length l. Determine the period of small oscillations of
the plate when (a) it is rotated through a small angle about a vertical axis
through its mass center G, (b) it is given a small horizontal displacement
in a direction perpendicular to AB, (c) it is given a small horizontal
displacement in a direction perpendicular to BC.
SOLUTION
(a)
Plate is rotated about vertical axis.
Similarly, for B, D, and C,
r
mg
SM G = S ( M G )eff : - 4T q ◊ r = J q&& T =
l
4
2
mgr
1
q&& =
q = 0 J = m( a2 + b2 )
Jl
12
1
r 2 = ( a2 + b2 )
4
tn =
2p
Jl
= 2p
wn
mgr 2
t n = 2p
(b)
2
2
1
12 m( a + b )l
mg 14 ( a2 + b2 )
= 2p
l
3g
Plate is moved horizontally.
The plate is in curvilinear translation.
sin q ª q
cos q ª 1
lq&& cos q ª lq&& = 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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93
PROBLEM 19.67 (Continued)
SFy = 0 = 4(T cos q ) - mg = 0
+
T=
mg
4
+® SFH = S ( FH )eff : - 4T sin q = ma
lq&& + gq = 0
(c)
t n = 2p
l
g
Since the oscillation about axes parallel to AB (and CD) is independent of the length of the sides of
the plate, the period of vibration about axes parallel to BC (and AD) is the same.
t n = 2p
l
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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94
PROBLEM 19.68
A circular disk of radius r = 0.8 m is suspended at its center C from wires AB and BC
soldered together at B. The torsional spring constants of the wires are K1 = 100 N ◊ m/rad
for AB and K 2 = 50 N ◊ m/rad for BC. If the period of oscillation is 0.5 s about the axis
AC, determine the mass of the disk.
SOLUTION
Spring constant. Let T be the torque carried by the wires.
q B /A =
T
K1
q C /B =
T
K2
Ê 1
1 ˆ
T
q C /A = q B /A + q C /B = Á
+
T=
˜
K eq
Ë K1 K 2 ¯
1
1
1
=
+
K eq K1 K 2
Equation of motion.
K eq =
SM C = S ( M C )eff : - K eqq = I q&&
q&& +
w n2 =
But t n = 0.5 s:
K1 K 2
(100)(50)
=
= 33.333 N ◊ m/rad
K1 + K 2 100 + 50
wn =
K eq
I
K eq
I
or
q =0
I =
K eq
w n2
2p 2p
=
= 12.566 rad/s
t n 0.5
Then
I =
33.333
= 0.21109 kg ◊ m2
(12.566)2
For a circular disk,
I =
1 2
mr
2
m=
2 I (2)(0.21109)
=
r2
(0.8)2
m = 0.660 kg 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.
95
PROBLEM 19.69
Determine the period of small oscillations of a small particle which
moves without friction inside a cylindrical surface of radius R.
SOLUTION
Datum at 2 :
Position j
T1 = 0
V1 = WR(1 - cos q m )
Small oscillations:
(1 - cos q m ) = 2 sin 2
V1 =
q m q m2
ª
2
2
WRq m2
2
Position k
vm = Rq&m T2 =
1 2 1
mvm = mR2q&m2
2
2
V2 = 0
Conservation of energy.
T1 + Y1 = T2 + V2
0 + WR
mgR
q m2 1
= mR2q&m2 + 0 q&m = w nq m
2 2
W = mg
q m2 1
= mR2w n2q m2
2 2
g
Wn =
R
tn =
Period of oscillations.
2p
R
= 2p
wn
g
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96
PROBLEM 19.70
A 400 g sphere A and a 300 sphere C are attached to the ends of a rod AC of negligible weight
which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations
of the rod.
SOLUTION
Datum at j:
Position j
T1 = 0
V1 = WC hC - WA hA
hC = BC (1 - cos q m )
hA = BA(1 - cos q m )
Small angles.
1 - cos q m =
q m2
2
V1 = [( mC g )( BC ) - ( mA g )( BA)]
q m2
2
Êq2 ˆ
V1 = [(0.3 kg)(0.2 m) - (0.4 kg)(0.125 m) ] Á m ˜ (9.81 m/s2 )
Ë 2¯
V1 = (0.0981)
Position k
1
1
mC ( v0 )2m + mA ( v A )2m , ( vC ) m = 0.2q&m
2
2
1
1
T2 = mC (0.2)2 q&m2 + mA (0.125)2 q&m2 ( v A ) m = 0.125q&m
2
2
1
T2 = ÈÎ(0.3)( 0.2)2 + (0.4)(0.125)2 ˘˚Wn2q m2 q&m2 = w nq m
2
1
T2 = (0.01825)w 2n q m2
2
V2 = 0 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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97
PROBLEM 19.70 (Continued)
Conservation of energy.
Period of oscillations.
T1 + V1 = T2 + V2 : 0 + 0.0981
tn =
2p
2p
=
wn
5.3753
q m2 0.01825 2 2
=
w nq m
2
2
(0.0981)
= 5.3753
w n2 =
(0.01825)
t n = 2.71 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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it without permission.
98
PROBLEM 19.71
A 1.8-kg collar A is attached to a spring of constant 800 N/m and can slide
without friction on a horizontal rod. If the collar is moved 70 mm to the left
from its equilibrium position and released, determine the maximum velocity
and maximum acceleration of the collar during the resulting motion.
SOLUTION
Datum at j:
Position j
T1 = 0 V1 =
1 2
kxm
2
Position k
1 2
mv2 V2 = 0 v2 = x&m
2
x&m = w n xm
T2 =
1
1
0 + kxm2 = mx&m2 + 0
2
2
k 800 N/m
1 2 1
2 2
2
kxm = mw n xm w n = =
m
2
1.8 kg
2
T1 + V1 = T2 + V2
w n2 = 444.4 s -2 w n = 21.08 rad/s
x&m = w n xm = (21.08 s -1 )(0.070 m) = 1.476 m/s &&
xm = w n2 xm = (21.08 s -2 )(0.070 m) = 31.1 m/s2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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it without permission.
99
PROBLEM 19.72
A 1.5-kg collar A is attached to a spring of constant 1 kN/m and can slide without friction on a horizontal
rod. The collar is at rest when it is struck with a mallet and given an initial velocity of 1 m/s. Determine the
amplitude of the resulting motion and the maximum acceleration of the collar.
SOLUTION
m = 1.5 kg k = 1 kN/m = 1000 N/m
Position j Immediately after collar is struck:
x = 0, v = 1 m/s
1
1
V1 = 0, T1 = mv 2 = (1.5)(1)2 = 0.75 N ◊ m
2
2
Position k v = 0, x is maximum, i.e., x = xm
V2 =
Conservation of energy.
1 2 1
kxm = (1000) xm2 = 500 xm2 T2 = 0
2
2
T1 + V1 = T2 + V2
0.75 + 0 = 0 + 500 xm2
Amplitude of motion.
xm = 0.03873 m
Maximum force:
Fm = kxm = (1000)(0.03873) = 38.73 N
Maximum acceleration.
am =
xm = 38.7 mm Fm 38.73
=
m
1.5
am = 25.82 m/s2
am = 25.8 m/s2 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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100
PROBLEM 19.73
A uniform rod AB can rotate in a vertical plane about a horizontal axis at C located at a distance
c above the mass center G of the rod. For small oscillations, determine the value of c for which
the frequency of the motion will be maximum.
SOLUTION
Find w n as a function of c.
Datum at k:
Position j
T1 = 0 V1 = mgh
V1 = mgc(1 - cos q m )
1 - cos q m = 2 sin 2
V1 = mgc
Position k
T2 =
q m q m2
ª
2
2
q m2
2
1 &2
ICq m
2
1 2
ml + mc 2
12
ˆ
1 Ê l2
V2 = 0
T2 = m Á + c 2 ˜ q&m2
2 Ë 12
¯
I C = I + mc 2 =
T1 + V1 = T2 + V2
0 + mgc
Ê l2
ˆ q& 2
q m2
= m Á + c2 ˜ m + 0
2
Ë 12
¯ 2
q&m = w nq m
Ê l2
ˆ
gc = m Á + c 2 ˜ w n2
Ë 12
¯
gc
w n2 = 2
2
l
12 + c
(
Maximum c, when
g
dw n2
=0=
dc
l2
- c2 = 0
12
(
)
l2
12
)
+ c 2 - 2c 2 g
(
2
l
12
+ c2
)
=0
c=
l
12
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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101
PROBLEM 19.74
A homogeneous wire of length 2 l is bent as shown and allowed to
oscillate about a frictionless pin at B. Denoting by t 0 the period of small
oscillations when b = 0, determine the angle b for which the period of
small oscillations is 2t 0 .
SOLUTION
We denote by m the mass of half the wire.
Position j Maximum deflections:
l
l
T1 = 0, V1 = - mg cos(q m - b ) - mg cos(q m + b )
2
2
l
= - mg (cos q m cos b + sin q m sin b + cos q m cos b - sin q m sin b )
2
V1 = - mgl cos b cos q m
For small oscillations,
1
cos q m ª 1 - q m2
2
1
V1 = - mgl cos b + mgl cos b q m2
2
Position k Maximum velocity:
T2 =
Thus,
1 &2
I Bq m
2
but
ˆ
Ê1
I B = 2 Á ml 2 ˜
¯
Ë3
1Ê2
ˆ
T2 = Á ml 2 ˜ q&m2
¯
2Ë3
ˆ
Êl
V2 = -2mg Á cos b ˜ = - mgl cos b
¯
Ë2
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 19.74 (Continued)
T1 + V1 = T2 + V2
Conservation of energy.
1
1
- mgl cos b + mgl cos b q m2 = ml 2q&m2 - mgl cos b
2
3
Setting q&m = q mw n ,
1
1
mgl cos b q m2 = ml 2 q m2 w n2
2
3
w n2 =
3g
2p
2l
cos b t =
= 2p
wn
2 l
3g cos b
But for b = 0,
For t = 2t 0 ,
t 0 = 2p
2p
(1)
2l
3g
Ê
2l
2l ˆ
= 2 Á 2p
3g cos b
3g ˜¯
Ë
Squaring and reducing,
1
1
= 4 cos b =
cos b
4
b = 75.5∞ 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.
103
PROBLEM 19.75
The inner rim of an 40 kg flywheel is placed on a knife edge, and
the period of its small oscillations is found to be 1.26 s. Determine
the centroidal moment of inertia of the flywheel.
SOLUTION
Datum at j:
Position j
T1 =
1 &2
J 0q m V1 = 0
2
Position k
T2 = 0 V2 = mgh
h = r (1 - cos q m ) = r 2 sin 2
ªr
qm
2
q m2
2
V2 = mgr
q m2
2
Conservation of energy.
T1 + V1 = T2 + V2 :
q2
1 &2
I 0 q m + 0 = 0 + mgr m
2
2
q&m = w nq m
For simple harmonic motion,
I 0w n2q m2 = mgr q m2 w n2 =
Moment of inertia.
I 0 = I + mr 2
I =
I + mr 2 =
(t ) (mgr) - mr
2
n
4p 2
2
=
mgr
J0
t n2 =
4p 2 ( 4p 2 ) I 0
=
mgr
w n2
(t ) (mgr)
2
n
4p 2
(1.26)2 ( 40)((9.81) Ê 0.35 ˆ
Ê 0.35 ˆ
˜
˜¯ - ( 40) ÁË
ÁË
2
2
2 ¯
4p
I = 2.7615 - 1.225
2
I = 1.537 kg ◊ m2 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.
104
PROBLEM 19.76
A connecting rod is supported by a knife edge at Point A; the period of its small oscillations
is observed to be 1.03 s. Knowing that the distance ra is 150 mm, determine the centroidal
radius of gyration of the connecting rod.
SOLUTION
Position j Displacement is maximum.
T1 = 0, V1 = mgra (1 - cos q m ) ª
1
mgraq m2
2
Position k Velocity is maximum.
( vG ) m = rq q&m
1 2 1 &2 1 2 &2 1
mvG + I q m = mra q m + mk 2q&m2
2
2
2
2
1
= m ra2 + k 2 q&m2
2
V2 = 0
T2 =
(
For simple harmonic motion,
Conservation of energy.
)
q&m = w nq m
T1 + V1 = T2 + V2
(
)
1
1
0 + mgraq m2 = m ra2 + k 2 w n2 q m2 + 0
2
2
gr
gr
w n2 = 2 a 2 or k 2 = 2a - ra2
ra + k
wn
Data:
t n = 1.03 s w n =
2p
2p
=
= 6.1002 rad/s
t n 1.03
ra = 150 mm = 0.15 m
k2 =
g = 9.81 m/s2
(9.81)(0.15)
- (0.15)2 = 0.039543 - 0.0225 = 0.017043 m2
2
(6.1002)
k = 0.13055 m
k = 130.6 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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it without permission.
105
PROBLEM 19.77
The rod ABC of total mass m is bent as shown and is supported in a
vertical plane by a pin at B and a spring of constant k at C. If end C is
given a small displacement and released, determine the frequency of the
resulting motion in terms of m, L, and k.
SOLUTION
Position j
T1 =
1 &2
1
I Bq m V1 = k (d ST )2
2
2
m
m L
1
gh - g sin q m + k ( Lq m + d ST )2
2
2 2
2
L
L 2 qm
h = (1 - cos q m ) = 2 sin
2
2
2
2
m L
m L
1
q mL
hª
V2 = - g q m2 - g q m + k ( L q m + d ST )2
4
2 4
2 2
2
sin q m ª q m
Position k
T2 = 0 V2 = -
Conservation of energy.
T1 + V1 = T2 + V2
m Lq 2 mg L
1 &2 1 2
1
2
˘
q m + k ÈÍ L2q m2 + d ST
I Bq m + k d ST = 0 - g m T + 2 Ld STq m ˙
˚
2
2
2
4
2 2
2 Î
(1)
When the rod is in equilibrium,
+
SM B = 0 =
m L
g - kLd ST
2 2
(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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106
PROBLEM 19.77 (Continued)
Substituting Eq. (2) into Eq. (1)
mgL ˆ 2 &
Ê
I Bq&m2 = Á kL2 ˜ q m q m = w nq m
Ë
4 ¯
mgL ˆ 2
Ê
I Bw n2q m2 = Á kL2 ˜ qm
Ë
4 ¯
w n2 =
kL2 -
mgL
4
IB
2
Ê 1 m 2 ˆ mL
IB = 2Á
L˜=
Ë3 2 ¯
3
w n2
=
kL2 mL2
3
mgL
4
=
3k 3g
m 4L
Frequency of oscillation.
w
3 k
g
f&n = n =
2p 2p m 4 L
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 19.78
A 7.5 kg uniform cylinder can roll without sliding on an incline and
is attached to a spring AB as shown. If the center of the cylinder
is moved 10 mm down the incline and released, determine (a) the
period of vibration, (b) the maximum velocity of the center of the
cylinder.
SOLUTION
(a)
Position j
T1 = 0 V1 =
1
k (d ST + rq m )2
2
Position k
1 &2 1
J q m + mvm2
2
2
1
V2 = mgh + k (d ST )2
2
T2 =
Conservation of energy.
1
1
1
1
T1 + V1 = T2 + V2 : 0 + k (d ST + rq m )2 = J q&m2 + mvm2 + mgh + k (d ST )2
2
2
2
2
2
2 2
2
2
2
&
kd ST + 2kd ST rq m + kr q m = J q m + mvm + 2mgh + kd ST
(1)
When the disk is in equilibrium,
+
Also,
SM c = 0 = mg sin b r - kd ST r
h = r sin b q m
Thus,
mgh - kd ST r = 0
(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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108
PROBLEM 19.78 (Continued)
Substituting Eq. (2) into Eq. (1)
kr 2q m2 = J q&m2 + mvm2
q&m = w nq m vm = rq&m = rw nq m
kr 2q m2 = ( J + mr 2 )q m2 w n2
w n2 =
w n2 =
kr 2
J + mr 2
J =
kr 2
1 2
mr + mr 2
2
=
2k
3m
tn =
(b)
1 2
mr
2
2p
=
wn
2p
2p
=
= 0.702 s 80
2 (900 N/m)
3 (7.5 kg)
Maximum velocity.
vm = rq&m
q&m = q mw n
vm = rq mw n
rq m =
10
= 0.01 m
1000
vm = (0.01 m/s) 80 = 0.0894 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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109
PROBLEM 19.79
Two uniform rods, each of weight m = 600 g and length l = 200 mm, are
welded together to form the assembly shown. Knowing that the constant of
each spring is k = 120 N/m and that end A is given a small displacement and
released, determine the frequency of the resulting motion.
SOLUTION
Mass and moment of inertia of one rod.
I =
m = 600 g = 0.6 kg
1 2 1
ml = (0.6)(0.2)2 = 2 ¥ 10 -3 kg ◊ m2
12
12
Approximation.
sin q m ª tan q m ª q m
1 - cos q m = 2 sin 2
Spring constant:
Position j
qm 1 2
ª qm
2 2
k = 120 N/m
ˆ 1 Êl ˆ
Ê1
T1 = 2 Á I q&m2 ˜ + m Á q&m ˜
¯ 2 Ë2 ¯
Ë2
2
Ê 1ˆ
Ê 1ˆ
= (2) Á ˜ (2 ¥ 10 -3 )q&m2 + Á ˜ (0.6) 0.1q&m
Ë 2¯
Ë 2¯
-3 & 2
= 5 ¥ 10 q
(
)
2
m
V1 = 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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110
PROBLEM 19.79 (Continued)
Position k
T2 = 0
V2 = ª-
Wl
Ê 1ˆ Ê l ˆ
(1 - cos q m ) + 2 Á ˜ k Á q m ˜
Ë 2¯ Ë 2 ¯
2
2
(0.6)(9.81)(0.2) Ê 1 2 ˆ
Ê 0.2 ˆ
Ê 1ˆ
qm ˜
ÁË q m ˜¯ + (2) ÁË ˜¯ (120) ÁË
¯
2
2
2
2
2
ª 1.4943q m2
Conservation of energy.
T1 + V1 = T2 + V2
5 ¥ 10 -3 q&m2 + 0 = 0 + 1.4943q m2
q& = 17.2876q
m
Simple harmonic motion.
m
q&m = w nq m
w n = 17.2876 rad/s
Frequency.
fn =
wn
2p
f n = 2.75 Hz 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 19.80
A slender 8-kg rod AB of length l = 600 mm is connected to two
collars of negligible mass. Collar A is attached to a spring of
constant k = 1.2 kN/m and can slide on a vertical rod, while collar
B can slide freely on a horizontal rod. Knowing that the system is
in equilibrium and that q = 40°, determine the period of vibration
if collar B is given a small displacement and released.
SOLUTION
Vertical rod.
y = l sin q
d y = l cos qdq
d y = l cos qdq&
x = -l cos q
d x = -l sin qdq
d x& = -l sin qdq&
y=
y
2
x=
x
2
Position j (Maximum velocity, dq& m)
1
1
J (dq& )2m + m ÈÎ(d x&m )2 + (d y )2m ˘˚
2
2
2
2
ÈÊ l
1Ê 1
ˆ ˘
Êl
ˆ
ˆ
T1 = Á ml 2 ˜ (dq& )2m + 1m ÍÁ sin q ˜ + Á cos q ˜ ˙ (dq& )2m
¯ ˙
Ë2
¯
¯
2 Ë 12
ÍÎË 2
˚
1
1
1
È 1 1˘
T1 = ml 2 Í + ˙ (dq&m )2 = ml 2 (dq&m )2
2
12
4
2
3
Î
˚
1
V1 = k (d ST )2 + mg y
2
T1 =
Position k (Zero velocity, maximum dq m )
T2 = 0
1
V2 = k (d y + d ST )2 + mg ( y - d ym )
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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112
PROBLEM 19.80 (Continued)
T1 + V1 = T2 + V2 :
Conservation of energy.
1 2 1 & 2 1 2
1
ml + (dq m ) + kd ST + mgy = 0 + (d y + d ST )2 + mg ( y - d ym )
2
3
2
2
1
2
2
+ mgy = k (d y 2 + 2d yd ST + d ST
) + mg ( y - d y x )
ml 2 (dq&m )2 + k d ST
3
(1)
But when the rod is in equilibrium,
SM B = mg
x
- kd ST x = 0 mg = 2kd ST
2
(2)
+
Substituting Eq. (2) into Eq. (1),
1
ml 2 (dq&m )2 = kd ym2 d ym = l cos q dq m
3
1
ml 2 (dq&m )2 = kl 2 cos2 q (dq m )2
3
For simple harmonic motion,
dq = dq m sin(w nt + f )
dq&m = dq mw n
1
m(dq m )2 w n2 = k cos2 q (dq m )2
3
3(1200 N/m)
k
cos2 40∞
w n2 = 3 cos2 q =
m
8kg
w n2 = 264.07
Period of vibration.
tn =
2p
2p
=
wn
264.07
t n = 0.387 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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113
PROBLEM 19.81
A slender rod AB of length l = 600 mm and negligible mass is
connected to two collars, each of mass 8 kg. Collar A is attached to
a spring of constant k = 1.2 kN/m and can slide on a vertical rod,
while collar B can slide freely on a horizontal rod. Knowing that
the system is in equilibrium and that q = 40∞, determine the period
of vibration if collar A is given a small displacement and released.
SOLUTION
Horizontal rod.
Y = l sin q
dY = l cos qdq
d = l cos qdq&
g
x = l cos q
d x = -l sin qdq
d = -l sin qdq&
x
Position j (Maximum velocity, dq&m )
1
1
m(d y& m )2 + m(d x&m )2
2
2
1
T1 = m[(l cos q )2 + (l sin q )2 ](dq&m )2
2
1 2 & 2
T1 = ml (dq m )
2
V1 = 0
T1 =
Position k (Zero velocity, maximum dq )
T2 = 0
1
V2 = kd ym2
2
Conservation of energy.
T1 + V1 = T2 + V2
1 2 & 2
1
ml (dq m ) + 0 = kd ym2
2
2
d ym = l cos qdq m
2
ml (dq&m )2 = kl 2 cos2 q (dq m )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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114
PROBLEM 19.81 (Continued)
Simple harmonic motion.
dq = dq m sin (w nt + f )
dq&m = dq mw n
ml 2 (dq m )2 w n2 = kl 2 cos2 q (dq m )2
k
cos2 q
m
1200
N/m
cos2 40 = 88.02 s -2
w n2 =
8 kg
w n2 =
Period of vibration.
tn =
2p
2p
=
= 0.6697 s
wn
88.02
t n = 0.670 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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115
PROBLEM 19.82
A 3-kg slender rod AB is bolted to a 5-kg uniform disk. A spring of
constant 280 N/m is attached to the disk and is unstretched in the
position shown. If end B of the rod is given a small displacement and
released, determine the period of vibration of the system.
SOLUTION
r = 0.08 m
l = 0.3 m
Position j
Position k
1
1
J diskq&m2 + ( JA ) rod q&m2
2
2
V1 = 0
1
Jdisk = m0 r 2
2
1
( JA ) rod = mAB l 2
3
T1 =
T2 = 0
l
1
k ( rq m )2 + mAB g (1 - cos q m )
2
2
2
q
q
1 - cos q m = 2 sin 2 m ª m
2
2
mAB g 2l 2
1
V2 = kr 2q m2 +
qm
2
2
V2 =
()
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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116
PROBLEM 19.82 (Continued)
Conservation of energy.
T1 + V1 = T2 + V2 :
1Ê1
1
1 2 2 1
l 2
2
2 ˆ &2
ÁË m0 r + mAB l ˜¯ q m + 0 = 0 + kr q m + mAB g q m
2
2 2
3
2
2
For simple harmonic motion,
q&m = w nq m
1
lˆ 2
Ê 2
Ê1
2
2ˆ 2 2
ÁË m0 t + mAB l ˜¯ w nq m = ÁË kr + mAB g ˜¯ q m
2
3
2
w n2 =
w n2 =
Period of vibration.
kr 2 + mAB gl
2
1
2 m0 r
+ 13 mAB l 2
(280 N/m)(0.08 m)2 + (3 kg)(9.81 m/s2 )
1
2 (5
( 02.3 m)
kg)(0.08 m)2 + 13 (3 kg)(0.300 m)2
w n2 =
6.207
= 58.55
0.106
tn =
2p
2p
=
wn
58.55
t n = 0.821 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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it without permission.
117
PROBLEM 19.83
A 400-g sphere A and a 300-g sphere C are attached to the ends of a 600-g rod AC, which can
rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the
rod.
SOLUTION
Position j
1
1
1
1
mA (0.125)2 + mC (0.2q&m )2 + mAC (0.0375q&m )2 + I AC q&m2
2
2
2
2
1
I AC = mAC (0.325)2
12
1È
1
˘
T1 = Í0.4(0.125)2 + 0.3(0.2)2 + 0.6(0.0375)2 + (0.6)(0.325)2 ˙ q&m2
2Î
12
˚
1
T1 = (0.024375)q&m2 ( N ◊ m)
2
V1 = 0
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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118
PROBLEM 19.83 (Continued)
Position k
T2 = 0
V2 = -WA 0.125(1 - cos q m ) + WC 0.2(1 - cos q m ) + WAC 0.0375(1 - cos q m )
1 - cos q m = 2 sin 2 q m/ 2 ª
q m2
2
V2 = [ -(0.4)(9.81)(0.125) + (0.3)(9.81)(0.2) + (0.6)(9.81)(0.0375)]
V2 =
Conservation of energy.
q m2
( N ◊ m)
2
0.318825q m2
2
T1 + V1 = T2 + V2:
1
0.318825 2
(0.024375)q&m2 + 0 = 0 +
qm
2
2
Simple harmonic motion.
q&m = w nq m
0.318825
= 13.08
0.024375
2p
2p
=
tn =
13.08
wn
w n2 =
t n = 1.737 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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119
PROBLEM 19.84
Three identical rods are connected as shown. If b = 43 l , determine the
frequency of small oscillations of the system.
SOLUTION
l = length of each rod
m = mass of each rod
Kinematics:
Position j
l
vm = q&m
2
( v BE ) m = bq&m
T1 = 0
l
V1 = -2mg cos q m - mgb cos q m
2
V1 = - mg (l + b) cos q m
Position k
l
- mgb
2
= - mg (l + b)
V2 = -2mg
1
È1
˘ 1
T2 = 2 Í I q&m2 + mvm2 ˙ + m( v BE )2m
2
2
Î
˚ 2
2
1
1
Êl ˆ
= ml 2q&m2 + m Á q&m ˜ + m(bq&m )2
¯
Ë
12
2
2
1 ˆ
Ê1
T2 = Á l 2 + b2 ˜ mq&m2
Ë3
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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120
PROBLEM 19.84 (Continued)
Conservation of energy.
1 ˆ
Ê1
T1 + V1 = T2 + V2 : 0 - mg (l + b) cos q m = Á l 2 + b2 ˜ mq&m2 - mg (l + b)
Ë3
2 ¯
1 ˆ
Ê1
mg (l + b)(1 - cos q m ) = Á l 2 + b2 ˜ mq&m2
Ë3
2 ¯
For small oscillations,
But for simple harmonic motion,
1
(1 - cos q m ) = q m2
2
1
1
1 ˆ
Ê
mg (l + b)q m2 = Á l 2 + b2 ˜ mq&m2
Ë
2
3
2 ¯
1
1 ˆ
Ê1
mg (l + b)q m2 = Á l 2 + b2 ˜ m(w nq m )2
q&m = w nq m :
Ë
2
3
2 ¯
1
l+b
w n2 = g 1 2 1 2
2 3l + 2b
w n2 = 3g
or
For b =
3
l , we have
4
w n2 = 3g
= 3g
l+b
2l + 3b2
(1)
2
l + 43 l
2l 2 + 3
( 43 l )2
7
4l
59 2
16 l
= 1.4237
w n = 1.1932
g
l
g
l
wn
2p
1.1932 g
=
2p
l
fn =
f n = 0.1899
g
l
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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121
PROBLEM 19.85
An 800-g rod AB is bolted to a 1.2-kg disk. A spring of constant
k = 12 N/m is attached to the center of the disk at A and to the wall
at C. Knowing that the disk rolls without sliding, determine the
period of small oscillations of the system.
SOLUTION
Position j
2
1
1
1
1
ˆ
Êl
T1 = ( I G ) AB q&m2 + mAB Á - r ˜ q&m + ( I G )disk q&m2 + m diskk r 2q&m2
Ë2 ¯
2
2
2
2
1
1
( I G ) AB = ml 2 = (0.8)(0.6)2 = 0.024 kg ◊ m2
12
12
2
Êl
ˆ
mAB Á - r ˜ = (0.8)(0.3 - 0.25)2 = 0.002 kg ◊ m2
Ë2 ¯
1
1
mdisk r 2 = (1.2)(0.25)2 = 0.0375 kg ◊ m 2
2
2
2
2
mdisk r = 1.2(0.25) = 0.0750 kg ◊ m 2
( I G )disk =
1
[0.024 + 0.002 + 0.0375 + 0.0750]q&m2
2
1
T1 = [0.1385]q&m2
2
V1 = 0
T1 =
Position k
T2 = 0
1
l
V2 = k ( rq m )2 + mAB g (1 - cos q m )
2
2
2
q
q
1 - cos q m = 2 sin 2 m ª m (smalll angles)
2
2
2
1
Ê 0.6 m ˆ q m
V2 = (12 N/m)(0.25 m)2 q m2 + (8 kg)(9.81 m/s2 ) Á
˜
Ë 2 ¯ 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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122
PROBLEM 19.85 (Continued)
1
V2 = [0.750 + 2.354]q m2
2
1
= (3.104) q m2 N ◊ m
2
T1 + V1 = T2 + V2
q& 2 = w 2q 2
m
n m
1
1
(0.1385)q m2 w n2 + 0 = 0 + (3.104)q m2
2
2
(
3
.
104 N ◊ m)
w n2 =
(0.1385 kg ◊ m2 )
= 22.41 s -2
tn =
2p
2p
=
= 1.327 s wn
22.41
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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123
PROBLEM 19.86
Two uniform rods AB and CD, each of length l and mass m, are attached to gears
as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4 m,
determine the period of small oscillations of the system.
SOLUTION
Kinematics:
2rq A = rqC
2q A = qC
2q& A = q&C
Let
q A = qm
2q m = (qC ) m
2q&m = (q&C ) m
Position j
T1 =
1 &2 1
1
1
I Aq m + I C (2q&m )2 + I ABq&m2 + I CD (2q&m )2
2
2
2
2
2
1
1
ˆ
Êl
Êl ˆ
+ mAB Á q&m ˜ + mCD Á 2q&m ˜
¯
Ë
¯
Ë
2
2
2
2
2
1
I A = ( 4m)(2r )2 = 8mr 2
2
1
1
I C = ( m)( r )2 = mr 2
2
2
1 2
1
I AB = ml
I CD = ml 2
12
12
2
È
˘
Êr ˆ
l2 l2 l2
1
T1 = m Í8r 2 + Á ˜ 4 + + + + l 2 ˙
2 ÍÎ
12 3 4
Ë 2¯
˙˚
5 ˘
1 È
T1 = m Í10r 2 + l 2 ˙ q&m2
3 ˚
2 Î
V1 = 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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124
PROBLEM 19.86 (Continued)
Position k
T1 = 0
l
mgl
(1 - cos 4q m )
V1 = mg (1 - cos q m ) +
2
2
For small angles,
q m q m2
ª
2
2
2
1 - cos 4q m = 2 sin 2q m = 2q m2
1 - cos q m = 2 sin 2
V1 =
Êq2
ˆ 1
5q 2
1
mgl Á m + 2q m2 ˜ = mgl m
2
2
Ë 2
¯ 2
q&m2 = w n2q m2
T1 + V1 = T2 + V2
5q 2
1
1 È 2 5 2˘ 2 2
m Í10r + l ˙ w nq m + 0 = 0 + mgl m
3 ˚
2
2
2 Î
w n2 =
=
5
2
2
gl
10r + 53 l 2
3gl
12r 2 + 2l 2
tn =
2p
12r 2 + 2l 2
= 2p
3gl
wn
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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125
PROBLEM 19.87
Two uniform rods AB and CD, each of length l and mass m, are attached to gears
as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4m,
determine the period of small oscillations of the system.
SOLUTION
Kinematics:
Let
2rq A = rqC
2q& A = q&C
2q A = qC
2q m = (qC ) m
q A = qm
&
&
2q m = (qC ) m
2
Position j
1
1
1
1
1
1
ˆ
Êl
Êl ˆ
T1 = I Aq&m2 + I C (2q&m )2 + I ABq&m2 + I CD (2q&m )2 + mAB Á q&m ˜ + mCD Á 2q&m ˜
¯
Ë
¯
Ë
2
2
2
2
2
2
2
2
2
1
I A = ( 4m)(2r )2 = 8mr 2
2
1
1
I C = ( m)( r 2 ) = mr 2
2
2
1 2
1
I AB = ml
I CD = ml 2
12
12
2
È
˘
Êr ˆ
l2 l2 l2
1
T1 = m Í8r 2 + Á ˜ 4 + + + + l 2 ˙ q&m2
2 ÍÎ
12 3 4
Ë 2¯
˙˚
1 È
5 ˘
V1 = 0
T1 = m Í10r 2 + l 2 ˙ q&m2
2 Î
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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126
PROBLEM 19.87 (Continued)
Position k
T2 = 0
l
mgl
(1 - cos 2q m )
V2 = - mg (1 - cos q m ) +
2
2
For small angles,
q m q m2
ª
2
2
2
1 - cos 2q m = 2 sin q m ª 2q m2
1 - cos q m = 2 sin 2
l q m2 mgl 2
2q m
+
2
2 2
1
3
= mgl q m2
2
2
T1 + V1 = T2 + V2
q&m = w nq m
V2 = - mg
1 È 2 5 2˘ 2 2
1
3
m 10 r + l ˙ q mw n + 0 = 0 + mgl q m2
2
2
2 ÍÎ
3 ˚
3
2 gl
w n2 =
10 r 2 + 53 l 2
=
9 gl
60r + 10l 2
2
tn =
2p
60r 2 + 10l 2
= 2p
9 gl
wn
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127
PROBLEM 19.88
A 5-kg uniform rod CD is welded at C to a shaft
of negligible mass which is welded to the centers of
two 10-kg uniform disks A and B. Knowing that the
disks roll without sliding, determine the period of
small oscillations of the system.
SOLUTION
W A = WB = Wdisk
Position j
1
1
T1 = 2( I A )disk q&m2 + (2mdisk )( rq&m )2
2
2
2
Position k
1
1
ˆ
Êl
+ I CDq&m2 + mCD Á - r ˜ q&m2
Ë
2
2
2 ¯
1
1
( I A )disk = mdisk r 2 = (10)(0.5)2 = 1.25 kg ◊ m2
2
2
1
1
I CD = mCD l 2 = (5)(1.5)2 = 0.9375 kg ◊ m2
12
12
1
T1 = [2.5 + 5 + 0.9375 + 0.3125]q&m2
2
1
T1 = (8.75)q&m2
V1 = 0
2
T2 = 0
l
V2 = WCD (1 - cos q m )
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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128
PROBLEM 19.88 (Continued)
Small angles:
q m q m2
ª
2
2
2
q
1
V2 = WCD l m
2
2
q2
1
= (5)(9.81)(1.5) m
2
2
1
= (36.7875) q m2
2
1 - cos q m = 2 sin 2
Conservation of energy and simple harmonic motion.
T1 + V1 = T2 + V2
q& = w q
m
n m
1
1
(8.75)w n2q m2 + 0 = (36.7875)q m2
2
2
36.7875
= 4.2043
w n2 =
8.75
Period of oscillations.
tn =
2p
=
wn
2p
4.2043
t n = 3.06 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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129
PROBLEM 19.89
Four bars of the same mass m and equal length l are connected by
pins at A, B, C, and D and can move in a horizontal plane. The
bars are attached to four springs of the same constant k and are in
equilibrium in the position shown (q = 45∞). Determine the period of
vibration if corners A and C are given small and equal displacements
toward each other and released.
SOLUTION
Kinematics:
BC
l
xG = cos q
2
l
yG = sin q
2
l
d xG = - sin q dq
2
l
d x&G = - sin q dq&
2
l
d yG = cos q dq
2
l
d yG = cos q dq&
2
xC = l cos q d xC
d x&C
x B = l sin q d x B
d x& B
= -l sin q dq
= -l sin q dq&
= l cos q dq
= l cosq dq&
yC = 0
y B = l sin q d y B = l cos q dq
d y& B = l cos q dq&m
The kinetic energy is the same for all four bars.
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 19.89 (Continued)
Position j
1
È1
˘
T1 = 4 Í I (dq&m )2 + m[(d x&G )2m + (d y&G )2m ]˙
2
2
Î
˚
1 2
I = ml
12
È1 1
˘
T1 = 2ml 2 Í + (sin 2 q m + cos2 q m )˙ d q&m2
Î12 4
˚
2
T1 = ml 2
3
V1 = 0
Position k
T2 = 0
1
1
V2 = (2) k (d xm )2 + (2) k (d ym )2
2
2
2
2
2
2
V2 = k (l sin q + l cos q ) d q m2
= k l 2d q m2
Conservation of energy.
Simple harmonic motion.
T1 + V1 = T2 + V2
2 2 & 2
ml (dq m ) + 0 = 0 + k l 2 (dq m )2
3
dq& = w dq
m
n
m
Ê3 kˆ
w n2 = Á
Ë 2 m ˜¯
tn =
Period of vibration.
2p
wn
t n = 2p
2m
3k
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 19.90
The 10-kg rod AB is attached to two 4-kg disks as shown. Knowing that the
disks roll without sliding, determine the frequency of small oscillations of
the system.
SOLUTION
Position k
Position j
r = 0.15 m
Masses and moments of inertia.
mA = mB = 4 kg
1
1
I A = I B = mA rA2 = ( 4)(0.15)2
2
2
2
= 0.045 kg ◊ m
mAB = 10 kg
Kinematics:
vm = rAq&m = 0.15 q&m
v = 0.05 q&
AB
Position j (Maximum displacement)
m
T1 = 0
V1 = -W AB (0.1 cos q m ) = -(10)(9.81)(0.1 cos q m )
= -9.81 cos q m
Position k (Maximum speed)
1
1
1
1
1
mA vm2 + I Aq&m2 + mB vm2 + I Bq&m2 + mAB v 2AB
2
2
2
2
2
1
È1
˘ 1
= 2 Í ( 4)(0.15q&m )2 + (0.045)q&m2 ˙ + (10)(0.05q&m )2
2
2
Î
˚ 2
= 0.1475 q& 2
T2 =
m
V2 = -W
WAB (0.1) = -(10)(9.81)(0.1) = -9.81
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132
PROBLEM 19.90 (Continued)
Conservation of energy.
T1 + V1 = T2 + V2
-9.81cos q m = 0.1475 q&m2 - 9.81
q&m2 = 66.5085(1 - cos q m )
Ê1 ˆ
ª 66.5085 Á q m2 ˜
Ë2 ¯
= 33.2542q m2
q m = 5.7666 q m
Simple harmonic motion.
Frequency.
q&m = w nq m
fn =
w n 5.7666
=
2p
2p
w n = 5.7666 rad/s
f n = 0.918 Hz 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 19.91
An inverted pendulum consisting of a sphere of mass m and a rigid bar ABC of
length l and negligible weight is supported by a pin and bracket at C. A spring
of constant k is attached to the bar at B and is undeformed when the bar is in the
vertical position shown. Determine (a) the frequency of small oscillations, (b) the
smallest value of a for which these oscillations will occur.
SOLUTION
(a)
Position j
1
1
m(lq&m )2 = ml 2q&m2
2
2
V1 = 0
Position k
T2 = 0
1
V2 = k ( a sin q m )2 - ml (1 - cos q m )
2
For small angles,
T1 =
sin q m ª q m
1 - cos q m = 2 sin 2
qm
2
q m2
2
q2
q2
1
V2 = ka2 m - ml m
2
2
2
1
= [ka2 - ml ]q m2
2
ª
T1 + V1 = T2 + V2
1
1 2 &2
ml q m + 0 = 0 + [ka2 - ml ]q m2
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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134
PROBLEM 19.91 (Continued)
q&m = w nq m
m=
m
g
m 2 2 2
l w nq m = ka2 - ml
g
w n2 =
=
(b)
ka2 - ml
( )l
m
g
2
g È ka2 ˘
- 1˙
Í
l Î ml
˚
fn =
1
1
wn =
2p
2p
g Ê ka2 ˆ
- 1˜ l ÁË ml
¯
fn = 0
ka2
- 1⬎ 0
ml
a⬎
ml
k
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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135
PROBLEM 19.92
For the inverted pendulum of Problem 19.91 and for given values of k, a, and l,
it is observed that f = 1.5 Hz when m = 1 kg and that f = 0.8 Hz when m = 2 kg.
Determine the largest value of W for which small oscillations will occur.
SOLUTION
See Solution to Problem 19.91 for the frequency in terms of W, k, a, and l.
fn =
1
2p
g Ê ka2 ˆ
- 1˜
l ÁË Wl
¯
f n = 1.5 Hz W = (1) (g) N 1.5 =
1
2p
f n = 0.8 Hz W = (2) ( g) N 0.8 =
2
Dividing Eq. (1) by Eq. (2) and squaring,
Ê 1.5 ˆ
ÁË
˜ =
0.8 ¯
(
(
ka2
gl
2
ka
2 gl
g Ê ka2 ˆ
- 1˜
l ÁË gl
¯
1
2p
) = 2(ka - gl)
- 1) ( ka - 2 gl )
-1
g Ê ka2 ˆ
-1
l ÁË 2 gl ˜¯
(1)
(2)
2
2
225 2
( ka - 2 gl ) = 2( ka2 - gl )
64
ka2 = 3.3196 gl
fn =
1
2p
g Ê 3.3196 gl ˆ
- 1˜
Á
¯
l Ë Wl
fn = 0
3.3196 g
-1 = 0
W
W = 32.6 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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136
PROBLEM 19.93
A uniform rod of length L is supported by a ball-and-socket joint
at A and by a vertical wire CD. Derive an expression for the
period of oscillation of the rod if end B is given a small horizontal
displacement and then released.
SOLUTION
Position j (Maximum deflection)
Looking from above:
Horizontal displacement of C:
xC = bq m
Looking from right:
fm =
xC b
= qm
h h
yC = h(1 - cos fm ) ª
1 2
hfm
2
2
yC =
1 Êb ˆ
1 b2 2
h Á qm ˜ =
qm
2 Ëh ¯
2 h
ym = yG =
ym =
We have
1 Ê
L 1 b2 2 ˆ
AG
- yC = 2 Á
qm ˜
AC
b Ë2 h
¯
1 bL 2
◊ qm
4 h
T1 = 0
V1 = mgym =
1 mgbL 2
qm
4 h
Position k (Maximum velocity)
Looking from above:
T2 =
1 &2 1
I q m + mvm2
2
2
1Ê 1
1 ÊL ˆ
ˆ
= Á mL2 ˜ q&m2 + m Á q& ˜
¯
Ë
2 12
2 Ë2 ¯
1
T2 = mL2q&m2
6
V2 = 0
Conservation of energy.
T1 + V1 = T2 + V2 : 0 +
2
1 mgbL 2 1 2 & 2
q m = mL q m
4 h
6
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137
PROBLEM 19.93 (Continued)
But for simple harmonic motion,
q&m = w nq m
1 mgbL 2 1 2
q m = mL (w nq m )2
4 h
6
3 bg
w n2 =
2 hL
Period of vibration.
tn =
2p
wn
t n = 2p
2hL
3bg
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138
PROBLEM 19.94
A 2-kg uniform rod ABC is supported by a pin at B and is
attached to a spring at C. It is connected at A to a 2-kg block
DE, which is attached to a spring and can roll without friction.
Knowing that each spring can act in tension or compression,
determine the frequency of small oscillations of the system
when the rod is rotated through a small angle and released.
SOLUTION
Position j
1
1
1
T1 = ( I R )q&m2 + ( mRC 2 )q&m + mE ( x&1 )2m
2
2
2
1
I R = mR l 2
12
1
= (2 kg)(00.9 m)2
12
= 0.135 kg ◊ m2
ml C 2 = 2 kg(0.15 m)2 = 0.0225 kg ◊ m2
x = 0.6q m ( x& )2 = (2 kg)(0.6q& m)2
1
E
1 m
m
x&1 = 0.6q& mE ( x&1 )2m = 0.72q m2 m2
1
T1 = [0.135 + 0.0225 + 0.72]q& 2 m
2
1
T1 = (0.9)q&m
2
V1 = 0
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139
PROBLEM 19.94 (Continued)
Position k
T2 = 0
1
1
V2 = k ( x1 )2m + k ( x2 )2m - mR gc(1 - cos q m )
2
2
( x1 )m = 0.6q m x2 = 0.3q m
1 - cos q m = 2 sin 2
q
m ª q m2
2
1
V2 = [(50 N/m)(0.6 m)2q m2 + (50 N/m)(0.3 m)2q m2
2
- (2 kg)(9.81 m/s2 )(0.15)q m2
1
V2 = [18 + 4.5 - 2.943]q 2 m
2
1
V2 = (19.55)q m2
2
Conservation of energy.
Simple harmonic motion.
T1 + V1 = T2 + V2 :
q&m = w n2q m2
w n2 =
Frequency of oscillations.
fn =
1
1
(0.9)q&m2 + 0 = 0 + (19.55)q m2
2
2
(0.9)(w n2 )q m2 = (19.55)q m2
19.55
= 21.73 s -2
0.9
wn
21.73
=
2p
2p
f n = 0.742 Hz PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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140
PROBLEM 19.95
A 750-g uniform arm ABC is supported by a pin at B and is attached
to a spring at A. It is connected at C to a 1.5-kg mass m which is
attached to a spring. Knowing that each spring can act in tension or
compression, determine the frequency of small oscillations of the
system when the weight is given a small vertical displacement and
released.
SOLUTION
Position j
kC = 300 N/m
k A = 400 N/m
2
T1 =
1
1
Êl ˆ
I BC (q&m )2 + mBC Á BC ˜ q&m2
Ë 2 ¯
2
2
2
1
1
1
Êl ˆ
+ I BAq&m2 + mBA Á AB ˜ q&m2 + my& m2
Ë 2 ¯
2
2
2
2
1
1
1
Êl ˆ
2
2
2
+ mBC lBC
= mBC lBC
I BC + mB Á BC ˜ = mBC lBC
Ë 2 ¯
12
4
3
2
1
1
1
Êl ˆ
2
2
2
I BA + mBA Á AB ˜ = mAB l AB
+ mAB l AB
= mAB l AB
Ë 2 ¯
12
4
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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141
PROBLEM 19.95 (Continued)
3
mBC = mABC = 0.45 kg
5
2
mBA = mABC = 0.3 kg
5
1
1
2
mBC lBC = (0.45 kg)(0.3 m))2 = 0.0135 kg ◊ m2
3
3
1
1
2
mBAlBA = (0.3)(0.2 m)2 = 0.004 kg ◊ m2
3
3
y& = l q&
m
my& m2
=
=
BC m
2 &2
qm =
mlBC
0.135q&m2
(1.5 kg)(0.3 m)2q&m2
1
T1 = [0.0135 + 0.004 + 0.135]q&m2
2
q& 2
= 0.1525 m
2
1
1
V1 = kC (d ST )C2 + k A (d ST )2A
2
2
T2 = 0
Position k
Ê
W ˆ
V2 = WlBC q m + WBC lBC Á 2q m - l BA ˜ (1 - cos q m )
AB
Ë
2 ¯
1
1
+ kC (lBC q m + (d ST )C )2 + k A (l ABq m + (d ST ) A )2
2
2
when the system is in equilibrium (q = 0).
SM B = 0 = WlBC + WBC
1 - cos q m - 2 sin 2
lBC
- kC (d ST )C lBC - k A (d ST ) A l AB
2
(1)
q m q m2
=
2
2
È
Êl ˆ
V2 = ÍWlBC + WBC Á BC ˜
Ë 2 ¯
ÍÎ
2
È
˘
Ê l AB ˆ Ê q m ˆ ˘ 1
2
2
˙ + kC lBC q m
˙ q m - ÍWBA Á
˜
Á
˜
Ë
¯
2 Ë 2 ¯ ˙˚ 2
˙˚
ÍÎ
1
- kC (d ST )C lBC q m + kC (d ST )C2
2
1 2 2
1
+ k Al ABq m - k A (d ST ) A l q m + k A (d ST )2A
AB
2
2
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142
PROBLEM 19.95 (Continued)
Taking Equation (1) into account,
2
È
1
Êl ˆq
2
V2 = - ÍWBA Á AB ˜ m + kC lBC
q m2
¯
Ë
2
2
2
ÍÎ
1
1
1
˘
2
q m2 + k A (d ST )2A ˙
+ kC (d ST )C2 + k Al AB
2
2
2
˚
V2 =
1È
Ê 0.2 ˆ
2
2˘ 2
-(0.3)(9.81) Á
˜¯ + (300)(0.3) + 400(0.2) ˙ q m
Í
Ë
2Î
2
˚
1
1
+ kC (d ST )C2 + k A (d ST )2A
2
2
1
1
1
V2 = [-0.2943 + 27 + 16]q m2 + kC (d ST )C2 + k A (d ST )2A
2
2
2
1
1
1
V2 = [42.7057]q m2 + kC (d ST )C2 + k A (d ST )2A
2
2
2
Conservation of energy.
T1 + V1 = T2 + V2 :
Simple harmonic motion.
1
1
1
(0.1525)q&m2 + kC (d ST )C2 + k A (d ST )2A
2
2
2
1
1
1
= 0 + ( 42.7057)q m2 + kC (d ST )C2 + k A (d ST )2A
2
2
2
q&m = w nq m
0.1525w n2q m2 = 42.7057q m2
w n2 =
Frequency.
fn =
42.7057
= 280.037 (raad/s)2
0.1525
wn
280.037
=
2p
2p
f n = 2.66 Hz PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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143
PROBLEM 19.96*
Two uniform rods AB and BC, each of mass m and length l, are
pinned together at A and are pin-connected to small rollers at B
and C. A spring of constant k is attached to the pins at B and C, and
the system is observed to be in equilibrium when each rod forms an
angle b with the vertical. Determine the period of small oscillations
when Point A is given a small downward deflection and released.
SOLUTION
xC = l sin b
d xC = l cos bdb
l
xG = cos b
2
l
d x&G = - sin bdb&
2
l
yG = sin b
2
l
d yG = cos bdb
2
l
d y&G = cos bdb&
2
Position j
1
È1
˘
T1 = 2 Í I (db& m )2 + m(d x&m )C2 + (d y& m )G2 ˙
2
2
Î
˚
2
˘
È1
ml
= Í ml 2 +
(sin 2 b + cos2 b ) ˙ db m2
4
˚
Î12
1
= ml 2db& m
3
V1 = 0
Position k
T2 = 0
1
V2 = k (2d xm )2
2
1 2
= ( 4l cos2b )
2
= 2l 2 cos2b
db& = w db
m
n
m
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144
PROBLEM 19.96* (Continued)
T1 + V1 = T2 + V2
Conservation of energy.
Period of oscillations.
1 2 2 2
ml w n db n + 0 = 0 + 2kl 2 cos2bdb n2
3
6k
w n2 =
cos2b
m
tn =
2p
2p
=
wn
6 mk cos2 b
( )
tn =
2p
cos b
m
6k
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145
PROBLEM 19.97*
As a submerged body moves through a fluid, the particles of the fluid flow around
the body and thus acquire kinetic energy. In the case of a sphere moving in an ideal
fluid, the total kinetic energy acquired by the fluid is 14 rVv 2 , where r is the mass
density of the fluid, V is the volume of the sphere, and v is the velocity of the sphere.
Consider a 500-g hollow spherical shell of radius 80 mm, which is held submerged
in a tank of water by a spring of constant 500 N/m. (a) Neglecting fluid friction,
determine the period of vibration of the shell when it is displaced vertically and
then released. (b) Solve Part a, assuming that the tank is accelerated upward at the
constant rate of 8 m/s2.
SOLUTION
This is not a damped vibration. However, the kinetic energy of the fluid must be included.
(a)
Position k
T2 = 0
1
V2 = kxm2
2
Position j
T1 = Tspere + Tfluid =
1
1
ms vm2 + rVvm2
2
4
V1 = 0
Conservation of energy and simple harmonic motion.
T1 + V1 = T2 + V2 :
1
1
1
ms vm2 + rVvm2 + 0 = 0 + kxm2
2
4
2
vm = lm = xmw n
1Ê
1 ˆ 2 2 1 2
Á ms + rV ˜¯ xmw n = kxm
2Ë
2
2
k
w n2 =
1
ms + rV
2
500
N/m
w n2 =
Ê1 ˆ
(0.5 kg) + Á rV ˜
Ë2 ¯
1
1 Ê 1000 kg ˆ Ê 4
3ˆ
rV = Á
˜ Á p (0.08 m) ˜¯ = 1.0723 kg
2
2 Ë m3 ¯ Ë 3
w n2 = 318 s -2
Period of vibration.
(b)
tn =
2p
2p
=
wn
318
t n = 0.352 s Acceleration does not change mass.
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146
PROBLEM 19.98*
A thin plate of length l rests on a half cylinder of radius r. Derive an expression
for the period of small oscillations of the plate.
SOLUTION
( r sin q m )sin q m ª rq m2
r (1 - cos q m ) ª r
Position j (Maximum deflection)
q m2
2
T1 = 0
V1 = Wym
= mgr
Position k (q = 0):
q m2
2
1 &2
I qm
2
1Ê 1 ˆ
= Á ˜ ml 2q&m2
2 Ë 12 ¯
T2 =
q&m = w nq m
1Ê 1 ˆ
T2 = Á ˜ ml 2w n2q m2
2 Ë 12 ¯
T1 + V1 = T2 + V2
1
1Ê 1 ˆ
0 + mgrq m2 = Á ˜ ml 2w n2q m2
2
2 Ë 12 ¯
12 gr
w n2 = 2
l
tn =
l2
2p
= 2p
wn
12 gr
tn =
pl
3gr
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147
PROBLEM 19.99
A 50-kg block is attached to a spring of constant k = 20 kN/m and can
move without friction in a vertical slot as shown. It is acted upon by a
periodic force of magnitude P = Pm sin w f t , where w f = 18 rad/s. Knowing
that the amplitude of the motion is 3 mm, determine the value of Pm .
SOLUTION
Equation of motion.
The steady state response is
mx&& + kx = Pm sinw f t
xm =
( )
Pm
k
1-
or
Data:
( )
wf 2
wn
È Êw f ˆ2˘
˙
Pm = kxm Í1 - Á
ÍÎ Ë w n ˜¯ ˙˚
k = 20 ¥ 103 N/m
m = 50 kg
wn =
wf
wn
=
k
20 ¥ 103
=
= 20 rad/s
m
50
18 rad/s
= 0.9
20 rad/s
xm = 3 mm = 3 ¥ 10 -3 m
Pm = (20 ¥ 103 )(3 ¥ 10 -3 )[1 - (0.9)2 ]
Pm = 11.40 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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148
PROBLEM 19.100
A 5 kg collar can slide on a frictionless horizontal rod and is attached to a
spring of constant 500 N/m. It is acted upon by a periodic force of magnitude
P = Pm sin w f t , where Pm = 15 N. Determine the amplitude of the motion of
the collar if (a) w f = 5 rad/s, (b) w f = 10 rad/s.
SOLUTION
Eq. (19.33):
xm =
( )
Pm
k
1-
( )
wf 2
wn
Pm = 15 N, k = 500 N/m
m = 5 kg
k
500
=
= 10 rad/s
m
5
Pm
5N
=
= 0.03 m
k
500 N/m
wn =
Amplitude of vibration.
0.03
= 0.04 m
1 - (10)2
(a)
w f = 5 rad/s
xm =
(b)
w f = 10 rad/s
w f = w n fi xm | xm | = 0.04 m (in phase) xm PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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149
PROBLEM 19.101
A 5-kg collar can slide on a frictionless horizontal rod and is attached to
a spring of constant k. It is acted upon by a periodic force of magnitude
P = Pm sin w f t , where Pm = 10 N and w f = 5 rad/s. Determine the value of
the spring constant k knowing that the motion of the collar has an amplitude
of 150 mm and is (a) in phase with the applied force, (b) out of phase with
the applied force.
SOLUTION
Eq. (19.33):
Pm
k
xm =
1xm =
k=
( )
wf
wn
2
k
m
Pm
k - mw 2f
Pm
+ mw 2f
xm
Pm = 10 N,
Data:
w n2 =
m = 5 kg
w f = 5 rad/s
k=
=
(a)
(In phase)
(Out of phase)
Pm
+ 125
xm
xm = 150 mm = 0.15 m
k=
(b)
Pm
+ (5)(5)2
xm
10
+ 125
0.15
k = 191.7 N/m xm = -150 mm = -0.15 m
k=
10
+ 125
-0.15
k = 58.3 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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150
PROBLEM 19.102
A collar of mass m which slides on a frictionless horizontal rod is attached
to a spring of constant k and is acted upon by a periodic force of magnitude
P = Pm sin w f t . Determine the range of values of w f for which the amplitude
of the vibration exceeds two times the static deflection caused by a constant
force of magnitude Pm .
SOLUTION
wn =
Circular natural frequency.
k
m
For forced vibration, the equation of motion is
mx&& + kx = pm sin (w f t + j )
The amplitude of vibration is
pm
k
xm =
1-
=
d ST
( ) (1 - )
wf
wn
2
wf 2
wn
For w f ⬍ w n and xm = 2 d ST , we have
2 d ST =
1-
For
2
Êwf ˆ
1
=
or 1 - Á
˜
2
Ë wn ¯
d ST
( )
wf 2
wn
1
1k
w 2f = w n2 =
2
2m
wf =
k
2m
k
⬍ w f ⱕ wn ,
2m
| xm | exceeds 2d ST
(1)
For w f ⬎ w n and xm = 2d ST , we have
2d ST =
d ST
(w f - w n ) - 1
2
3
3k
w 2f = w n2 =
2
2m
For
wn ⱕ w f ⱕ
3k
2m
From Eqs. (1) and (2),
or
w 2f
w n2
-1 =
wf =
1
2
3k
2m
(2)
| xm | exceeds 2d ST
Range:
3k
k
⬍w f ⬍
2m
2m
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151
PROBLEM 19.103
An 8-kg uniform disk of radius 200 mm is welded to a vertical shaft with a
fixed end at B. The disk rotates through an angle of 3∞ when a static couple
of magnitude 50 N ◊ m is applied to it. If the disk is acted upon by a periodic
torsional couple of magnitude T = Tm sin w f t , where Tm = 60 N ◊ m, determine
the range of values of w f for which the amplitude of the vibration is less than
the angle of rotation caused by a static couple of magnitude Tm .
SOLUTION
Mass moment of inertia:
I =
Torsional spring constant:
K=
Natural circular frequency:
wn =
For forced vibration,
qm =
1 2 1
mr = (8)(0.200)2 = 0.16 kg ◊ m2
2
2
T
q
T = 50 N ◊ m
q = 3∞ = 0.05236 rad
0.50
K=
0.05236
= 954.93 N ◊ m/rad
K
954.93
=
= 77.254 rad/s
I
0.16
Tm
K
1-
qST
=
( )
wf 2
wn
1-
( )
wf 2
wn
For the amplitude |q m | to be less than qST, we must have w f ⬎ w n .
Then
|q m | =
qST
( )
wf 2
wn
⬍ qST
-1
2
Êwf ˆ
ÁË w ˜¯ - 1 ⬎ 1
n
2
Êwf ˆ
ÁË w ˜¯ ⬎ 2
n
w f ⬎ 2w n = ( 2 )(77.254)
w f ⬎109.3 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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152
PROBLEM 19.104
For the disk of Problem 19.103 determine the range of values of w f for which
the amplitude of the vibration will be less than 3.5∞.
SOLUTION
Mass moment of inertia:
I =
Torsional spring constant:
K=
Natural circular frequency:
wn =
For forced vibration,
1 2 1
mr = (8)(0.200)2 = 0.16 kg ◊ m2
2
2
T
q
T = 50 N ◊ m, q = 3∞ = 0.05236 rad
50
K=
= 954.93 N ◊ m/rad
0.05236
K
954.93
=
= 77.254 rad/s
I
0.16
Tm
K
qm =
1-
( )
wf
wn
2
=
qST
1-
Tm = 60 N ◊ m
qST =
( )
wf 2
wn
Tm
60
=
= 0.062832 rad
K 954.93
|q m | ⬍ 3.5∞ = 0.061087 rad ⬍ qST
For the amplitude |q m | to be less than qST , we must have w f ⬎ w n .
|q m | =
qST
=
0.062832
( ) -1 ( ) -1
wf
wn
2
wf 2
wn
⬍ 0.061087
2
Êwf ˆ
0.062832
ÁË w ˜¯ - 1 ⬎ 0.061087 = 1.02857
n
2
Êwf ˆ
ÁË w ˜¯ ⬎ 2.02857
n
w f ⬎ 2.02857w n
w f ⬎110.0 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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153
PROBLEM 19.105
An 8-kg block A slides in a vertical frictionless slot and is connected to a
moving support B by means of a spring AB of constant k = 1.6 kN/m. Knowing
that the displacement of the support is d = d m sin w f t , where d m = 150 mm,
determine the range of values of w f for which the amplitude of the fluctuating
force exerted by the spring on the block is less than 120 N.
SOLUTION
Natural circular frequency:
wn =
Eq. (19.33' ):
xm =
k
1.6 ¥ 103
=
= 14.1421 rad/s
m
8
dm
1-
Spring force:
( )
wf 2
wn
È
Í
1
Fm = - K ( xm - d m ) = - kd m Í1 Í 1- wf
wn
ÍÎ
( )
= kd m
( )
1- ( )
wf 2
wn
wf 2
wn
( )
= (1.6 ¥ 10 )(0.150)
1- ( )
wf 2
wn
3
wf 2
wn
Limit on spring force:
˘
˙
2˙
˙
˙˚
( )
= 240
1- ( )
wf 2
wn
wf 2
wn
| Fm | ⬍120 N
( )
240
1- ( )
wf 2
wn
wf 2
wn
⬍ 120 or
( )
1- ( )
wf 2
wn
wf 2
wn
⬍
1
2
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154
PROBLEM 19.105 (Continued)
In phase motion.
( )
1- ( )
wf 2
wn
wf
wn
2
⬍
1
2
2
Êwf ˆ
1 1Êwf ˆ
ÁË w ˜¯ ⬍ 2 - 2 ÁË w ˜¯
n
n
2
3Êwf ˆ
1
⬍
Á
˜
2 Ë wn ¯
2
wf ⬍
Out of phase motion.
wf
wn
2
⬍
1
wn
3
1
3
w f ⬍ 8.16 rad/s ( ) ⬍1
( ) -1 2
wf 2
wn
wf 2
wn
2
2
Êwf ˆ
1Êwf ˆ
1
ÁË w ˜¯ ⬍ 2 ÁË w ˜¯ - 2
n
n
2
Êwf ˆ
1
ÁË w ˜¯ ⬍ - 2
n
No solution for w f .
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155
PROBLEM 19.106
Rod AB is rigidly attached to the frame of a motor running at a constant
speed. When a collar of mass m is placed on the spring, it is observed to
vibrate with an amplitude of 15 mm. When two collars, each of mass m,
are placed on the spring, the amplitude is observed to be 18 mm. What
amplitude of vibration should be expected when three collars, each of
mass m, are placed on the spring? (Obtain two answers.)
SOLUTION
(a)
One collar:
( xm )1 = 15 mm
(w n )12 =
k
m
(b)
Two collars:
( xm )2 = 18 mm
(w n )22 =
k
1
= (w n )12
2m 2
Êwˆ
Êwˆ
ÁË w ˜¯ = 2 ÁË w ˜¯
n 2
n 1
(c)
Three collars:
( xm )3 = unknown, (w n )32 =
1
k
= (w n )12 ,
3m 3
Êwˆ
Êwˆ
ÁË w ˜¯ = 3 ÁË w ˜¯
n 3
n 1
We also note that the amplitude d m of the displacement of the base remains constant.
Referring to Section 19.7, Figure 19.9, we note that, since ( xm )2 ⬎ ( xm )1 and
w
(w n ) 2
⬎
w
,
(w n )1
we must
have (ww ) ⬍ 1 and ( xm )1 ⬎ 0. However, (ww ) may be either ⬍ 1 or ⬎ 1, with ( xm )2 being correspondingly
n 1
n 2
either ⬎ 0 or ⬍ 0.
1.
Assuming ( xm )2 ⬎ 0 :
For one collar,
( xm )1 =
dm
2
j
+ 15 mm =
2
j
+ 18 mm =
Êwˆ
1- Á ˜
Ë wn ¯1
dm
(1)
2
Êwˆ
1- Á ˜
Ë wn ¯1
For two collars,
( xm ) 2 =
dm
Êwˆ
1- Á ˜
Ë wn ¯ 2
dm
2
Êwˆ
1 - 2Á ˜
Ë wn ¯1
(2)
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156
PROBLEM 19.106 (Continued)
Dividing Eq. (2) by Eq. (1), member by member:
2
Êwˆ
1- Á ˜
Ë wn ¯1
2
Êwˆ
1
1.2 =
; we find Á ˜ =
2
Ë wn ¯1 7
Êwˆ
1 - 2Á ˜
Ë wn ¯1
Substituting into Eq. (1),
1 ˆ 90
Ê
mm
d m = (15 mm) Á1 - ˜ =
Ë
7¯
7
For three collars,
Ê 90 ˆ
ÁË ˜¯ mm 90
7
( xm )3 =
mm,
=
=
2
4
Ê 1ˆ
Êwˆ
1
3
ÁË ˜¯
1 - 3Á ˜
7
Ë wn ¯1
dm
2.
( xm )3 = 22.5 mm Assuming ( xm )2 ⬍ 0 :
For two collars, we have
-18 mm =
dm
2
Êwˆ
1 - 2Á ˜
Ë wn ¯1
(3)
Dividing Eq. (3) by Eq. (1), member by member:
2
-1.2 =
2
Êwˆ
1- Á ˜
Ë wn ¯1
2
Êwˆ
1 - 2Á ˜
Ë wn ¯1
2
Êwˆ
Êwˆ
-1.2 + 2.4 Á ˜ = 1 - Á ˜
Ë wn ¯1
Ë wn ¯1
2
Êwˆ
2.2 1.1
ÁË w ˜¯ = 3.4 = 1.7
n 1
Substitute into Eq. (1),
1.1ˆ
9
Ê
mm
d m = (15 mm) Á1 ˜ =
Ë
1.7 ¯ 1.7
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157
PROBLEM 19.106 (Continued)
For three collars,
Ê 9ˆ
ÁË ˜¯
9 mm
1.7
( xm )3 =
=
=
,
2
- 1.6
Ê 1.1ˆ
Êwˆ
1 - 3Á ˜
1 - 3Á ˜
Ë 1.7 ¯
Ë wn ¯1
dm
( xm )3 = - 5.63 mm (out of phase)
Points corresponding to the two solutions are indicated below:
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158
PROBLEM 19.107
A cantilever beam AB supports a block which causes a static deflection
of 50 mm at B. Assuming that the support at A undergoes a vertical
periodic displacement d = d m sin w f t , where d m = 12 mm, determine
the range of values of w f for which the amplitude of the motion of
the block will be less than 25 mm. Neglect the weight of the beam
and assume that the block does not leave the beam.
SOLUTION
For the static condition.
mg = kd ST
Natural circular frequency.
wn =
k
g
=
d ST
m
g = 9.81 m/s2 , d ST = 50 mm = 0.05 m
wn =
9.81
= 14.007 rad/s
0.05
dm
( xm ) B =
From Eqs. (19.31 and 19.33' ):
1-
wf 2
wn
| xm |B ⬍ 0.025 m d m = 0.012 m
Conditions:
In phase motion.
( )
0.012
1-
( )
wf 2
wn
⬍ 0.025
2
Êwf ˆ
0.012
⬎
1- Á
˜
0.025
Ë wn ¯
Êwf ˆ
0.52 ⬎ Á
Ë w n ˜¯
Out of phase motion.
0.012
2
( )
( )
Ê 0.012 ˆ
- 1⬎ Á
Ë 0.025 ˜¯
wf 2
wn
w f ⬍10.10 rad/s ⬍ 0.025
-1
wf 2
wn
w f ⬍ 0.52w n
2
Êwf ˆ
ÁË w ˜¯ - 1 ⬎ 0.48
n
2
Êwf ˆ
ÁË w ˜¯ ⬎ 1.48
n
w f ⬎ 1.48w n
w f ⬎17.04 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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it without permission.
159
PROBLEM 19.108
A variable-speed motor is rigidly attached to a beam BC. When the
speed of the motor is less than 600 rpm or more than 1200 rpm, a
small object placed at A is observed to remain in contact with the
beam. For speeds between 600 and 1200 rpm the object is observed
to “dance” and actually to lose contact with the beam. Determine the
speed at which resonance will occur.
SOLUTION
Let m be the unbalanced mass and r the eccentricity of the unbalanced mass. The vertical force exerted on the
beam due to the rotating unbalanced mass is
P = mr w 2f sin w f t = Pm sin w f t
Then from Eq. 19.33,
Pm
k
xm =
1-
( )
wf 2
wn
mr w 2f
k
=
1-
( )
wf 2
wn
For simple harmonic motion, the acceleration is
am =
-w 2f xm
mr w 4f
k
=
1-
( )
wf 2
wn
When the object loses contact with the beam is | am |. Acceleration is greater than g. Let w1 = 600 rpm = 62.832 rad/s.
| am |1 =
where
Let
U=
mr w14
k
1-
( )
w1 2
wn
=
mrw n4U 4
k
2
1-U
(1)
w1
.
wn
w 2 = 1200 rpm = 125.664 rad/s = 2w1
| am | 2 =
mr w 24
k
w 2
( )
2
wn
-1
=
mrw n4 ( 2U )4
k
2
4U - 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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160
PROBLEM 19.108 (Continued)
Dividing Eq. (1) by Eq. (2),
1=
4U 2 - 1
16(1 - U 2 )
20U 2 = 17
w1
17
=
wn
20
U=
or 16 - 16U 2 = 4U 2 - 1
17
20
wn =
20
w 1 = 1.08465 w1
17
w n = (1.08465)(600 rpm)
w n = 651 rpm 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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161
PROBLEM 19.109
An 8-kg block A slides in a vertical frictionless slot and is connected to a
moving support B by means of a spring AB of constant k = 120 N/m. Knowing
that the acceleration of the support is a = am sin w f t , where am = 1.5 m/s2
and w f = 5 rad/s, determine (a) the maximum displacement of block A, (b) the
amplitude of the fluctuating force exerted by the spring on the block.
SOLUTION
(a)
Support motion.
a = d&& = am sin w f t
Êa ˆ
d = - Á m2 ˜ sin w f t
Ëwf ¯
dm =
- am
w 2f
=-
1.5 m/s2
(5 rad/ss)2
= -0.06 m
From Equation (19.31 and 19.33¢ ):
xm =
dm
w 2f
1 - w2
w n2 =
k 120 N/m
=
= 15( rad/s)2
m
8 kg
n
xm =
(b)
-0.06
= 0.09 m
25
1 - ( 15
)
xm = 90.0 mm x is out of phase with d for w f = 5 rad/s.
Thus,
Fm = k ( xm + d m ) = 120 N/m (0.09 m + 0.06 m)
= 18 N
Fm = 18.00 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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162
PROBLEM 19.110
A 20 g ball is connected to a paddle by means of an elastic cord AB of
constant k = 7.5 N/m. Knowing that the paddle is moved vertically
according to the relation d = d m sin w f t , where d m = 200 mm, determine
the maximum allowable circular frequency w f if the cord is not to become
slack.
SOLUTION
k (d - x ) = mx&&
SF = ma
From Equation (19.31 and 19.33¢ ):
Data:
xm =
Êkˆ
&&
x+Á ˜ x =d
Ë m¯
dm
Ê1 - w 2f ˆ
Ë w n2 ¯
m = 20 kg = 0.02 kg
k = 7.5 N/m
wn =
d m = 200 mm = 0.2 m
k
m
7.5
0.02
= 19.3649 rad/s
=
The cord becomes slack if xm - d m exceeds d ST , where
d ST =
m 0.02 ¥ 9.81
=
= 0.02616 m
k
7.5
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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163
PROBLEM 19.110 (Continued)
0.2
Then
1-
( )
wf 2
wn
- 0.2 ⬍ 0.02616
2
Êwf ˆ
Êwf ˆ
0.2 - 0.2 + 0.2 Á
⬍ 0.02616 - 0.02616 Á
˜
Ë wn ¯
Ë w n ˜¯
2
2
Êw f ˆ
0.22616 Á
⬍ 0.02616
Ë w n ˜¯
wf
wn
⬍
0.02616
= 0.3401
0.22616
Maximum allowable circular frequency.
w f ⬍ (0.3401)(19.3649 rad/s)
w f ⬍ 6.59 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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164
PROBLEM 19.111
A simple pendulum of length l is suspended from a collar C, which is forced
to move horizontally according to the relation xC = d m sin w f t . Determine the
range of values of w f for which the amplitude of the motion of the bob is less
than d m . (Assume that d m is small compared with the length l of the pendulum.)
SOLUTION
x = xC + l sin q
Geometry.
sin q =
+
+
x - xC
l
SFy = ma y ª 0 : T cosq - mg = 0 T ª mg
-T sinq = mx&&
SFx = max :
mx&& +
mg ( x - xC )
=0
l
g
g
&&
x + x = xC
l
l
Using the given motion of xC,
&&
x+
Circular natural frequency.
g
g
x = d m sin w f t
l
l
wn =
g
l
&&
x + w n2 x = w n2d m sin w f t
The steady state response is
dm
xm =
1xm2 =
Consider
( )
wf 2
wn
d m2
È
Í1 Î
( ) ˙˚
wf
wn
2 ˘2
ⱕ d m2
xm2 = d m2 .
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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165
PROBLEM 19.111 (Continued)
2
Then
2
4
È Êw f ˆ2˘
Êwf ˆ
Êwf ˆ
˙ = 1- 2Á
Í1 - Á
+Á
=1
Ë w n ˜¯
Ë w n ˜¯
ÍÎ Ë w n ˜¯ ˙˚
2
2
Êwf ˆ
Êwf ˆ
ÁË w ˜¯ = 0 and ÁË w ˜¯ = 2
n
n
Êwf ˆ
ÁË w ˜¯ = 0 and
n
For
For
Then
0⬍
wf
wn
wf
wn
wf
wn
= 2
⬍ 2 , | xm | ⬎ d m
⬎ 2, | xm | ⬍ d n
w f ⬎ 2 wn =
2g
l
wf ⬎
2g
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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166
PROBLEM 19.112
The 1.2-kg bob of a simple pendulum of length l = 600 mm is suspended
from a 1.4-kg collar C. The collar is forced to move according to the relation
xC = d m sin w f t , with an amplitude dm = 10 mm and a frequency ff = 0.5 Hz.
Determine (a) the amplitude of the motion of the bob, (b) the force that must
be applied to collar C to maintain the motion.
SOLUTION
(a)
SFx = max
-T sin q = mx&&
SFy = T cos q - mg = 0
For small angles cos q ª 1. Acceleration in the y direction is second order and is neglected.
T = mg
mx&& = - mg sin q
x - xc
l
mg
g
mg
mx&& +
x = xc =
d m sin w f t
l
l
l
g
w n2 =
l
2
&x& + w n x = w n2d m sin w f t
sin q =
From Equation (19.33¢ ):
xm =
dm
w2
1 - w 2f
n
So
w 2f = (2p f n )2 = 4p 2 (0.5)2 = p 2 s -2
w n2 =
xm =
g 9.81 m/s2
=
= 16.35 s -2
l
0.600 m
10 ¥ 10 -3 m
2
1 - 16p.35
= 0.02523 m
xm = 25.2 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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167
PROBLEM 19.112 (Continued)
(b)
+
ac = &&
xc = -d mw 2f sin w f t
SFx = mc ac
F - T sinq = mc ac
From Part (a):
Thus,
T = mg , sinq =
x - xc
l
È x - xc ˘
xc
F = - mg Í
˙ + mc &&
Î l ˚
= - mw n2 x + mw n2 xc + mc &&
xc
= - mw n2 xm sin w f t + mw n2d m sin w f t - mcw 2f d m sin w f t
= [-(1.2)(16.35)(0.02523) + (1.2)((16.35)(10 ¥ 10 -3 ) - (1.4)p 2 (10 ¥ 10 -3 )]sin p t
= -0.437 sin p t
F = -0.437 sin p t ( 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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168
PROBLEM 19.113
A motor of mass M is supported by springs with an equivalent spring constant k. The unbalance of its rotor is
equivalent to a mass m located at a distance r from the axis of rotation. Show that when the angular velocity of
the motor is w f , the amplitude xm of the motion of the motor is
xm =
r
( Mm ) ( ww
1-
)
2
f
n
( )
wf 2
wn
k
.
M
where w n =
SOLUTION
Rotor
Motor
+
SF = ma Pm sinw f t - kx = Mx&&
Mx&& + kx = Pm sin w f t
&&
x+
P
k
x = m sin w f t
M
m
k
w n2 =
M
From Equation (19.33):
Pm
k
xm =
1-
But
( )
wf 2
wn
2
Pm mrw f
=
k
k
k = Mw n2
Pm
Ê m ˆ Êwf ˆ
= rÁ ˜ Á
Ë M ¯ Ë w n ˜¯
k
2
xm =
Thus,
r
( Mm ) ( ww
1-
)
2
f
n
( )
wf 2
wn
Q.E.D. 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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169
PROBLEM 19.114
As the rotational speed of a spring-supported 100-kg motor is increased, the amplitude of the vibration due to
the unbalance of its 15-kg rotor first increases and then decreases. It is observed that as very high speeds are
reached, the amplitude of the vibration approaches 3.3 mm. Determine the distance between the mass center of
the rotor and its axis of rotation. (Hint: Use the formula derived in Problem 19.113.)
SOLUTION
Use the equation derived in Problem 19.113 (above).
xm =
r
( Mm ) ( ww
1-
For very high speeds,
thus,
1
( )
wf
wn
2
)
2
f
n
( )
wf 2
wn
=
r
( Mm )
1
( )
wf 2
wn
® 0 and xm ®
-1
rm
,
M
Ê 15 ˆ
3.3 mm = r Á
Ë 100 ˜¯
r = 22 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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170
PROBLEM 19.115
A motor of mass 200 kg is supported by springs having a total constant of 240 kN/m. The unbalance of the rotor
is equivalent to a 30-g mass located 200 mm from the axis of rotation. Determine the range of allowable values
of the motor speed if the amplitude of the vibration is not to exceed 1.5 mm.
SOLUTION
Let M = mass of motor, m = unbalance mass, r = eccentricity
M = 200 kg
m = 30 g = 0.03 kg
r = 200 mm = 0.2 m K = 240 kN/m = 240 ¥ 103 N/m
Natural circular frequency:
240 ¥ 103 N/m
k
=
= 34.641 rad/s
M
200
rm (0.2)(0.03)
=
M
200
= 0.3 ¥ 10 -5 m = 0.03 mm
wn =
From the derivation given in Problem 19.113,
xm
( rmM ) ( ww
)
=
1- ( )
2
f
( )
=
1- ( )
wf 2
wn
0.03
n
wf 2
wn
wf 2
wn
mm
In phase motion with | xm | ⬍1.5 mm
( )
1- ( )
0.03
wf 2
wn
wf 2
wn
⬍ 1.5
2
Êwf ˆ
Êwf ˆ
0.03 Á
⬍ 1.5 - 1.5 Á
˜
Ë wn ¯
Ë w n ˜¯
2
2
Êwf ˆ
⬍ 1.5
1.53 Á
Ë w n ˜¯
wf
wn
⬍
1.5
= 0.99015
1.53
w f ⬍ (0.99015)(34.641) = 34.30 rad/s
w f ⬍ 328 rpm 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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171
PROBLEM 19.115 (Continued)
Out of phase motion with | xm | = 1.5 mm
( ) ⬍ 1.5
( ) -1
0.03
wf 2
wn
wf 2
wn
2
2
Êwf ˆ
Êwf ˆ
.
0.03 Á
1
5
⬍
ÁË w ˜¯ - 1.5
Ë w n ˜¯
n
Êwf ˆ
1.5 ⬍ 1.47 Á
Ë w n ˜¯
wf
wn
⬎
2
1.5
= 1.01015
1.47
w f ⬎ (1.01015)(34.641) = 34.992 rad/s
w f ⬎ 334 rpm 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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172
PROBLEM 19.116
As the rotational speed of a spring-supported motor is slowly increased from 300 to 500 rpm, the amplitude of
the vibration due to the unbalance of its rotor is observed to increase continuously from 1.5 to 6 mm. Determine
the speed at which resonance will occur.
SOLUTION
Let m¢ be the mass of rotor of the motor and m the total mass of the motor. Let e be the distance from the rotor
axis to the mass center of the rotor. The magnitude of centrifugal force due to unbalance of the rotor is
Pm = m¢ew 2f
and in rotation, the unbalanced force is
Pm sin w f t = m¢ ew 2f sin w f t
The equation of motion of the spring mounted motor is
mx&& + kx = m¢ ew 2f sin w f t = Pm sin w f t
The steady state response is
Pm
k
xm =
1xm =
Case 1.
( )
wf 2
wn
m¢ ew 2f
k
1
◊
1-
( )
wf 2
wn
w f = w1 = 300 rpm =
Cw 2f
=
1-
( )
wf 2
wn
2p (300)
= 10p rad/s
60
| xm | = x1 = 1.5 mm
Case 2.
w f 1 = w 2 = 500 rpm =
2p (500) 50p
=
rad/s
60
3
| xm | = x2 = 6 mm
È
w 22 Í1 +
x2
6
Î
=
=4=
x1 1.5
2È
w1 Í1 +
Î
2
2
È
È Êw ˆ ˘
Êw ˆ ˘
4w12 Í1 - Á 2 ˜ ˙ = w 22 Í1 - 1Á 1 ˜ ˙
Ë wn ¯ ˙
ÍÎ
ÍÎ Ë w n ¯ ˙˚
˚
( ) ˘˙˚
( ) ˘˙˚
w1 2
wn
w2 2
wz
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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173
PROBLEM 19.116 (Continued)
Multiplying by w n2 and transposing terms,
( 4w12 - w 22 )w n2 - 3w12w 22 = 0
w n2 =
3w12w 22
4w12
- w 22
=
(3)(986.96)(2741.56)
= 6729.3 ( rad/s)2
3947.84 - 2741.56
w n = 82.032 rad/s
w n = 783 rpm 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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174
PROBLEM 19.117
A100 kg motor is bolted to a light horizontal beam. The unbalance of
its rotor is equivalent to a mass of 60 g located 100 mm from the axis of
rotation. Knowing that resonance occurs at a motor speed of 400 rpm,
determine the amplitude of the steady-state vibration at (a) 800 rpm,
(b) 200 rpm, (c) 425 rpm.
SOLUTION
From Problem 19.113:
xm =
r
( Mm ) ( ww
1-
n
( )
wn
wf
)
2
f
2
Resonance at 400 rpm means that w n = 400 rpm.
Ê 60 ¥ 10 -3 ˆ
Ê mˆ
r Á ˜ = (100 mm) Á
˜ = 0.06 mm
ËM¯
Ë 100 ¯
2
(a)
2
Êwf ˆ
Ê 800 ˆ
=
˜ =4
ÁË
ÁË w ˜¯
400 ¯
n
xm =
0.06( 4)
1- 4
xm = - 0.080 mm 2
(b)
2
Êwf ˆ
1
Ê 200 ˆ
=
=
˜
Á
ÁË w ˜¯
Ë 400 ¯
4
n
xm =
0.06
( 14 )
xm = 0.020 mm 1 - 14
2
(c)
2
Êwf ˆ
Ê 425 ˆ
=
˜ = 1.1289
ÁË
ÁË w ˜¯
400 ¯
n
xm =
0.06(1.1289)
1 - 1.1289
xm = - 0.525 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
175
PROBLEM 19.118
A 180-kg motor is bolted to a light horizontal beam. The unbalance of
its rotor is equivalent to a 28-g mass located 150 mm from the axis of
rotation, and the static deflection of the beam due to the weight of the
motor is 12 mm. The amplitude of the vibration due to the unbalance can
be decreased by adding a plate to the base of the motor. If the amplitude
of vibration is to be less than 60 mm for motor speeds above 300 rpm,
determine the required mass of the plate.
SOLUTION
M1 = 180 kg, m = 28 ¥ 10 -3 kg
r = 150 mm = 0.150 m
Before the plate is added,
Equivalent spring constant:
k=
W1 M1 g
=
d ST d ST
k=
(180)(9.81)
= 147.15 ¥ 103 N/m
-3
12 ¥ 10
Let M2 be the mass of motor plus the plate.
k
M2
Natural circular frequency.
wn =
Forcing frequency:
w f = 300 rpm = 31.416 rad/s
2
w 2f M 2 (31.416)2 M 2
Êwf ˆ
=
=
= 0.006707 M 2
ÁË w ˜¯
k
147.15 ¥ 103
n
From the derivation in Problem 19.113,
xm
( )( )
=
rm
M2
1For out of phase motion with
wf 2
wn
( )
wf 2
wn
xm = -60 ¥ 10 -6 m,
-60 ¥ 10 -6
È ( 0.150 )( 28¥10-3 ) ˘ (0.006707 M )
2
M2
Í
˚˙
=Î
1 - 0.006707 M 2
-60 ¥ 10 -6 + (60 ¥ 10 -6 )(0.006707) M 2 = 28.170 ¥ 10 -6
402.49 ¥ 10 -9 M 2 = 88.170 ¥ 10 -6
M 2 = 219.10 kg
Added mass:
DM = M 2 - M1 = 219.10 - 180
DM = 39.1 kg 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.
176
PROBLEM 19.119
The unbalance of the rotor of a 200-kg motor is equivalent to a 100-g mass
located 150 mm from the axis of rotation. In order to limit to 1-N the amplitude
of the fluctuating force exerted on the foundation when the motor is run at
speeds of 100 rpm and above, a pad is to be placed between the motor and
the foundation. Determine (a) the maximum allowable spring constant k of
the pad, (b) the corresponding amplitude of the fluctuating force exerted on
the foundation when the motor is run at 200 rpm.
SOLUTION
Mass of motor.
M = 200 kg
Unbalance mass.
m = 100 g = 0.1 kg
Eccentricity.
r = 150 mm = 0.15 m
Equation of motion:
Mx&& + kx = Pm sin w f t = mrw 2f sin w f t
( - Mw 2f + k ) xm = mrw 2f
xm =
Fm = kxm =
Transmitted force.
| Fm | =
For out of phase motion,
(a)
mrw 2f
k - Mw 2f
kmrw 2f
k - M w 2f
kmrw 2f
(1)
Mw 2f - k
Required value of k.
Solve Eq. (1) for k.
| Fm | ( Mw 2f - k ) = kmrw 2f
k ( mrw 2f + | Fm |) = | Fm | M w 2f
k=
Data:
| Fm | = 1 N
| Fm | Mw 2f
mrw 2f + | Fm |
w f = 100 rpm = 10.472 rad/s
k=
(1)(200)(10.472)2
= 8292.3 N/m
(0.1)(0.15)(10.472)2 + 1
k = 8.292 kN/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.
177
PROBLEM 19.119 (Continued)
(b)
Force amplitude at 200 rpm.
From Eq. (1),
w f = 20.944 rad/s
| Fm | =
(8292.3)(0.1)(0.15)(20.944)2
(200)(20.944)2 - 8292.3
| Fm | = 0.687 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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178
PROBLEM 19.120
A 180-kg motor is supported by springs of total constant 150 kN/m. The unbalance of the rotor is equivalent to
a 28-g mass located 150 mm from the axis of rotation. Determine the range of speeds of the motor for which
the amplitude of the fluctuating force exerted on the foundation is less than 20 N.
SOLUTION
Equation derived in Problem 19.113:
xm
( rmM ) ( ww
)
=
1- ( )
n
wf 2
wn
Fm = kxm =
Force transmitted:
2
f
w
( kmr
M )( w
1-
)
2
f
n
( )
wf 2
wn
k = 150 kN/m = 150 ¥ 103 N/m
r = 150 mm = 0.150 m
Data:
m = 28 g = 28 ¥ 10 -3 kg
M = 180 kg
krm (150 ¥ 103 )(0.150)(28 ¥ 10 -3 )
=
= 3.5 N
M
180
k
150 ¥ 103
=
= 28.868 rad/s
M
180
wn =
In phase motion with | Fm | ⬍ 20 N.
( )
1- ( )
3.5
wf 2
wn
wf 2
wn
⬍ 20
2
Êwf ˆ
Êwf ˆ
3.5 Á
⬍ 20 - 20 Á
˜
Ë wn ¯
Ë w n ˜¯
2
2
Êwf ˆ
23.5 Á
⬍ 20
Ë w n ˜¯
wf
wn
⬍
wf ⬍
20
23.5
20
(28.868) = 26.63 rad/s
23.5
w f ⬍ 254 rpm 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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179
PROBLEM 19.120 (Continued)
Out of phase motion with | Fm | ⬍ 20 N.
( ) ⬍ 20
( ) -1
3.5
wf 2
wn
wf 2
wn
2
2
Êwf ˆ
Êwf ˆ
3.5 Á
⬍ 20 Á
- 20
˜
Ë wn ¯
Ë w n ˜¯
Êwf ˆ
20 ⬍ 16.5 Á
Ë w n ˜¯
wf
wn
⬎
20
16.5
wf ⬎
2
20
(28.868) = 31.78 rad/s
16.5
w f ⬎ 304 rpm 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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180
PROBLEM 19.121
A vibrometer used to measure the amplitude of vibrations consists
essentially of a box containing a mass-spring system with a known
natural frequency of 120 Hz. The box is rigidly attached to a surface,
which is moving according to the equation y = d m sin w f t . If the
amplitude zm of the motion of the mass relative to the box is used
as a measure of the amplitude d m of the vibration of the surface,
determine (a) the percent error when the frequency of the vibration
is 600 Hz, (b) the frequency at which the error is zero.
SOLUTION
Ê dm ˆ
x=Á
w 2f ˜ sin w f t
ÁË 1 - w 2 ˜¯
n
y = d m sin w f t
z = relative motion
È dm
˘
z = x - y = Í w 2f - d m ˙ sin w f t
Í1 - w 2
˙
n
Î
˚
w2
È 1
˘ d m w 2f
1
n
2
˙=
zm = d m Í w f
2
Í1 - w 2
˙ 1- wf
n
Î
˚
w n2
w 2f
(a)
( 600 )2 = 25 = 1.0417
( )2 24
zm
w n2
120
=
2 =
d m 1 - w f 1 - 600
120
w2
n
Error = 4.17%
(b)
w 2f
w n2
zm
=1=
w2
dm
1- f
w n2
1= 2
ff =
w 2f
w n2
2
2
fn =
(120) = 84.853 Hz
2
2
f n = 84.9 Hz 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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181
PROBLEM 19.122
A certain accelerometer consists essentially of a box containing a
mass-spring system with a known natural frequency of 2200 Hz. The
box is rigidly attached to a surface, which is moving according to the
equation y = d m sin w f t . If the amplitude zm of the motion of the mass
relative to the box times a scale factorw n2 is used as a measure of the
maximum acceleration am = d mw 2f of the vibrating surface, determine
the percent error when the frequency of the vibration is 600 Hz.
SOLUTION
Ê dm ˆ
x=Á
w 2f ˜ sin w f t
ÁË 1 - w 2 ˜¯
n
y = d m sin w f t
z = relative motion
È dm
˘
z = x - y = Í w 2f - d m ˙ sin w f t
Í1 - w 2
˙
n
Î
˚
w2
È 1
˘ d m w 2f
1
n
˙=
zm = d m Í w 2f
2
w
1
Í w2
˙ 1- f
n
Î
˚
w n2
The actual acceleration is
am = -w 2f d m
zmw n2 .
The measurement is proportional to
Then
zmw n2 zm Ê w n ˆ
=
Á
˜
am
dm Ë w f ¯
1
=
1=
2
( )
wf 2
wn
1
1-
600 2
( 2200
)
= 1.0804
Error = 8.04% 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.
182
PROBLEM 19.123
Figures (1) and (2) show how springs can be used to support a
block in two different situations. In Figure (1), they help decrease
the amplitude of the fluctuating force transmitted by the block to
the foundation. In Figure (2), they help decrease the amplitude
of the fluctuating displacement transmitted by the foundation
to the block. The ratio of the transmitted force to the impressed
force or the ratio of the transmitted displacement to the impressed
displacement is called the transmissibility. Derive an equation for
the transmissibility for each situation. Give your answer in terms
of the ratio w f /w n of the frequency wf of the impressed force
or impressed displacement to the natural frequency w n of the
spring-mass system. Show that in order to cause any reduction in
transmissibility, the ratio w f /w n must be greater than 2.
SOLUTION
(1)
From Equation (19.33):
Pm
k
xm =
1-
Force transmitted:
( )
wf 2
wn
È
( PT ) m = kxm = k Í
Í1 ÍÎ
˘
˙
2
˙
˙˚
( )
wf
wn
Transmissibility =
Thus,
(2)
Pm
k
( PT ) m
=
Pm
1
1-
( )
1
wf 2
wn
From Equation (19.33¢ ):
Displacement transmitted:
dm
xm =
1-
For
( PT )m
Pm
or
xm
dm
( )
2
wf
wn
1
to be less than 1,
1-
( )
wf 2
wn
1⬍ 1 2
Transmissibility =
xm
=
dm
1-
( )
wf 2
wn
⬍1
( )
wf 2
wn
Êwf ˆ
ÁË w ˜¯ ⬎ 2
n
wf
wn
⬎ 2 Q.E.D. 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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183
PROBLEM 19.124
A 30-kg disk is attached with an eccentricity e = 0.15 mm to the midpoint of a
vertical shaft AB, which revolves at a constant angular velocity w f . Knowing that
the spring constant k for horizontal movement of the disk is 650 kN/m, determine
(a) the angular velocity w f at which resonance will occur, (b) the deflection r of
the shaft when w f = 1200 rpm.
SOLUTION
G describes a circle about the axis AB of radius r + e.
Thus,
an = ( r + e)w 2f
Deflection of the shaft is
F = kr
Thus,
F = man
kr = m( r + e)w 2f
w n2 =
k
m
kr =
k
m=
w n2
k
w n2
( r + e)w 2f
w2
r=
e w 2f
n
w 2f
1 - w2
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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184
PROBLEM 19.124 (Continued)
(a)
Resonance occurs when
w f = w n , i.e., r Æ d
wn =
k
m
(650 ¥ 1000)
30
= 147.196 rad/s
= 1405.6 rpm
=
(b)
r=
w n = w f = 1406 rpm 1200 2
(1405
.6 )
2
1200
1 - ( 1405
.6 )
(0.15 mm)
= 0.02964 mm
r = 0.403 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
185
PROBLEM 19.125
A small trailer and its load have a total mass of 250 kg. The trailer is
supported by two springs, each of constant 10 kN/m, and is pulled over
a road, the surface of which can be approximated by a sine curve with an
amplitude of 40 mm and a wavelength of 5 m (i.e., the distance between
successive crests is 5 m and the vertical distance from crest to trough is
80 mm). Determine (a) the speed at which resonance will occur, (b) the
amplitude of the vibration of the trailer at a speed of 50 km/h.
SOLUTION
k = 2(10 ¥ 103 N/m)
Total spring constant
= 20 ¥ 103 N/m
k 20 ¥ 103 N/m
=
= 80 s -2
m
250 kg
l =5m
w n2 =
(a)
d m = 40 mm = 40 ¥ 10 -3 m
y = d m sin
x = vt
x
l
l
Ê vˆ
y = d m sin Á ˜ t , t =
Ë l¯
v
so
y = 40 ¥ 10 -3 sinw f t
From Equation (19.33¢ ):
xm =
Resonance:
wf =
and
where
2p
,
t
2p v 2p v
wf =
=
l
5
wf =
dm
Ê1 - w 2f ˆ
Ë w n2 ¯
2p v
= w n = 80 s -1 ,
5
v = 7.1176 m/s
(b)
Amplitude at
v = 25.6 km/h v = 50 km/h = 13.8889 m/s
2p (13.8889)
wf =
= 17.4533 rad/s
5
w 2f = 304.60 s -2
xm =
40 ¥ 10 -3
= -14.246 ¥ 10 -3 m
1 - 30480.62
xm = -14.25 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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it without permission.
186
PROBLEM 19.126
Block A can move without friction in the slot as shown and is acted upon by a
vertical periodic force of magnitude P = Pm sin w f t , where w f = 2 rad/s and
Pm = 20 N. A spring of constant k is attached to the bottom of block A and to
a 22-kg block B. Determine (a) the value of the constant k which will prevent
a steady-state vibration of block A, (b) the corresponding amplitude of the
vibration of block B.
SOLUTION
In steady state vibration, block A does not move and therefore, remains in its original equilibrium position.
Block A:
+
SF = 0
kx = - Pm sinw f t
Block B:
+
(1)
SF = mB &&
x
mB &&
x + kx = 0
x = xm sinw nt
w n2 = k /mB
From Eq. (1):
kxm sin w nt = - Pm sin w f t
w n = w f = 2 rad/s
kxm = - Pm
k
mB
wn =
k = mBw n2
k = (22)(2)2
(a)
Required spring constant.
(b)
Corresponding amplitude of vibration of B.
k = 88.0 N/m k xm = - Pm
Pm
k
20 N
xm = 88 N/m
xm = -
xm = -0.227 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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it without permission.
187
PROBLEM 19.127
Show that in the case of heavy damping (c ⬎ cc ), a body never passes through its position of equilibrium O
(a) if it is released with no initial velocity from an arbitrary position or (b) if it is started from O with an
arbitrary initial velocity.
SOLUTION
Since c ⬎ cc , we use Equation (19.42), where
l1 ⬍ 0, l2 ⬍ 0
x = c1e l1t + c2 e l2t
dx
v=
= c1l1e l1t + c2 l2 e l2t
dt
(a)
t = 0, x = x0 ,
(1)
(2)
v = 0:
x0 = c1 + c2
From Eqs. (1) and (2):
0 = c1l1 + c2 l2
l2
x0
l2 - l1
- l1
c2 =
x0
l2 - x1
c1 =
Solving for c1 and c2 ,
x=
Substituting for c1 and c2 in Eq. (1),
x2
Èl2 e l1t - l1e l2t ˘
˚
l2 - l1 Î
For x = 0: when t , we must have
l1e l2t - l2 e l1t = 0
l2
= e( l2 - l1 )t
l1
(3)
Recall that
l1 ⬍ 0, l2 ⬍ 0. Choosing l1 and l2 so that l1 ⬍ l2 ⬍ 0, we have
0⬍
l2
⬍ 1 and l2 - l1 ⬎ 0
l1
Thus a positive solution for t ⬎ 0 for Equation (3) cannot exist, since it would require that e raised to a
positive power be less than 1, which is impossible. Thus, x is never 0.
The x - t curve for this
case is as shown.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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188
PROBLEM 19.127 (Continued)
(b)
t = 0, x = 0, v = v0 :
Equations (1) and (2) yield
0 = c1 + c2
v0 = c1l1 + c2 l2
Solving for c1 and c2 ,
v0
l2 - l1
v2
c2 =
l2 - l1
c1 = -
v0
[e l2t - e l1t ]
l2 - l1
Substituting into Eq. (1),
x=
For x = 0,
t πa
e l2t = e l1t
For c ⬎ cc , l1 - l2 ; thus, no solution can exist for t, and x is never 0.
The x - t curve for this
case is as shown.
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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189
PROBLEM 19.128
Show that in the case of heavy damping (c ⬎ cc ), a body released from an arbitrary position with an arbitrary
initial velocity cannot pass more than once through its equilibrium position.
SOLUTION
Substitute the initial conditions, t = 0, x = x0 , v = v0 in Equations (1) and (2) of Problem 19.127.
x0 = c1 + c2
( v0 - l2 x0 )
l2 - l1
( v0 - l1 x0 )
c2 =
l2 - l1
c1 = -
Solving for c1 and c2 ,
x=
And substituting in Eq. (1)
For x = 0, t :
v0 = c1l1 + c2 l2
1
È( v - l1 x0 )e l2t - ( v0 - l2 x0 )e l1t ˘˚
l2 - l1 Î 0
( v0 - l1 x0 )e l2t = ( v0 - l2 x0 )e l1t
e( l2 - l1 )t =
t=
( v0 - l2 x0 )
( v0 - l1 x0 )
v - l2 x0
1
ln 0
( l2 - l1 ) v0 - l1 x0
This defines one value of t only for x = 0, which will exist if the natural log is positive,
i.e., if
v0 - l2 x0
v0 - l1 x0
⬎ 1. Assuming l1 ⬍ l2 ⬍ 0,
this occurs if v0 ⬍ l1 x0 .
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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190
PROBLEM 19.129
In the case of light damping, the displacements x1, x2 , x3, shown in Figure 19.11 may be assumed equal to
the maximum displacements. Show that the ratio of any two successive maximum displacements xn and xn+1 is
constant and that the natural logarithm of this ratio, called the logarithmic decrement, is
ln
xn
2p (c /cc )
=
xn+1
1 - (c /cc )2
SOLUTION
For light damping,
Equation (19.46):
- c t
x = x0 e ( 2 m ) sin(w 0 t + f )
At given maximum displacement,
t = t n , x = xn
sin(w 0 t n + f ) = 1
- c t
xn = x0 e ( 2m ) n
t = t n+1 , x = xn+1
At next maximum displacement,
sin(w 0 t n+1 + f ) = 1
- c t
xn+1 = x0 e ( 2m ) n + 1
But
w D t n+1 - w D t n = 2p
t n+1 - t n =
2p
wD
-
Ratio of successive displacements:
=e
Thus,
From Equations (19.45) and (19.41):
ln
t
- 2cm (tn - tn+1 )
=e
+
c 2p
2m wD
xn
cp
=
xn+1 mw D
(1)
Ê cˆ
wD = wn 1 - Á ˜
Ë cc ¯
2
cc
Ê cˆ
1- Á ˜
Ë c¯
2m
2
wD =
Thus,
c
xn
x e 2m n
= 0-ct
xn+1 x0 e 2 m n+1
x
cp 2m
ln n =
xn+1 m cc
1
Ê cˆ
1- Á ˜
Ë cc ¯
2
2p
x
ln n =
xn+1
1-
( )
( )
c
cc
c
cc
2
Q.E.D. 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.
191
PROBLEM 19.130
In practice, it is often difficult to determine the logarithmic decrement of a system with light damping defined in
Problem 19.129 by measuring two successive maximum displacements. Show that the logarithmic decrement
can also be expressed as (1/k ) ln ( xn /xn+ k ), where k is the number of cycles between readings of the maximum
displacement.
SOLUTION
As in Problem 19.129, for maximum displacements xn and xn+ k at t n and t n+ k , sin(w 0 t n + f ) = 1
- c t
xn = x0 e ( 2 m ) n
and sin(w nt n+ k + f ) = 1.
- c (t )
xn+ k = x0 e ( 2 m ) n+k
-c
xn
x0 e( 2 m ) n
t -t
=
= e 2 m ( n n+ k )
-c
t
)
(
n
k
+
xn + k x e 2 m
0
-c
Ratio of maximum displacements:
But
t
w D t n+ k - w D t n = k (2p )
tn - tn+ k = k
2p
wD
xn
c Ê 2k p ˆ
=+
xn + k
2m ÁË w D ˜¯
Thus,
ln
xn
cp
=k
xn + k
mw D
(2)
But from Problem 19.129, Equation (1):
log decrement = ln
xn
cp
=
xn+1 mw D
log decrement =
Comparing with Equation (2),
x
1
ln n Q.E.D. k xn + k
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.
192
PROBLEM 19.131
In a system with light damping (c ⬍ cc ), the period of vibration is commonly defined as the time interval
t d = 2p /w d corresponding to two successive points where the displacement-time curve touches one of
the limiting curves shown in Figure 19.11. Show that the interval of time (a) between a maximum positive
displacement and the following maximum negative displacement is 12 t d , (b) between two successive zero
displacements is 12 t d , (c) between a maximum positive displacement and the following zero displacement is
greater than 14 t d .
SOLUTION
Equation (19.46):
(a)
- c t
x = x0 e ( 2 m ) sin(w D t + f )
Maxima (positive or negative) when x& = 0:
- c t
Ê -c ˆ - c t
x& = x0 Á ˜ e ( 2 m ) sin(w D t + f ) + x0w D e ( 2 m ) cos(w D t + f )
Ë 2m ¯
Thus, zero velocities occur at times when
x& = 0, or tan(w D t + f ) =
2mw D
c
(1)
The time to the first zero velocity, t1 , is
Ètan -1
t1 = Î
(
2 mw D
c
) - f ˘˚
wD
(2)
The time to the next zero velocity where the displacement is negative is
Ètan -1
t1¢ = Î
(
2 mw D
c
) - f + p ˘˚
wD
(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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193
PROBLEM 19.131 (Continued)
Subtracting Eq. (2) from Eq. (3),
t1¢ - t1 =
(b)
p ◊t D t D
p
Q.E.D.
=
=
wD
2p
2
Zero displacements occur when
sin(wq t + f ) = 0 or at intervals of
w D t + f = p , 2p np
Thus,
Time between
(t1 )0 = (p - f )
wD
and (t1¢)0 =
0 ¢s = (t1¢)0 - (t1 )0 =
(2p - f )
wD
2p - p pt D t D
=
=
Q.E.D.
wD
2p
2
Plot of Equation (1)
(c)
The first maxima occurs at 1:
(w D t1 + f )
The first zero occurs at
(w D (t1 )0 + f ) = p
From the above plot,
(w D (t1 )0 + f ) - (w D t1 + f ) ⬎
or
(t1 )0 - t1 ⬎
p
2w D
p
2
(t1 )0 - t1 ⬎
tD
Q.E.D.
4
Similar proofs can be made for subsequent maximum and minimum.
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194
PROBLEM 19.132
The block shown is depressed 30 mm from its equilibrium position and released.
Knowing that after 10 cycles the maximum displacement of the block is 1.25 mm,
determine (a) the damping factor c/c, (b) the value of the coefficient of viscous
damping. (Hint: See Problems 19.129 and 19.130.)
SOLUTION
From Problems 19.130 and 19.129:
Ê 1 ˆ Ê xn ˆ
=
ÁË ˜¯ ln Á
k
Ë xn + k ˜¯
2p
1-
c
cc
( )
c
cc
2
where k = number of cycles = 10
(a)
First maximum is
Thus, n = 1
x1 = 30 mm
x1
30
=
= 2.4
x1 + 10 1.25
1
ln 2.4 = 0.08755
10
2p cc
c
=
2
Ê cˆ
1- Á ˜
Ë cc ¯
2
Damping factor.
2
Ê cˆ
Ê 2p ˆ Ê c ˆ
1- Á ˜ = Á
Ë 0.08755 ˜¯ ÁË cc ˜¯
Ë cc ¯
2
2
2
˘
Ê c ˆ ÈÊ 2p ˆ
Í
ÁË c ˜¯ ÁË 0.08755 ˜¯ + 1˙ = 1
c
ÍÎ
˙˚
2
1
Ê cˆ
ÁË c ˜¯ = 5150 + 1
(
)
c
= 0.0001941
c
= 0.01393 cc
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195
PROBLEM 19.132 (Continued)
(b)
k
m
Critical damping coefficient.
cc = 2 m
or
cc = 2 km
( Eq. 19.41)
cc = 2 (175 N/m)(4 kg)
cc = 52.915 N ◊ s/m
From Part (a),
c
= 0.01393
cc
c = (0.01393)(52.915)
c = 0.737 N ◊ s/m Coefficient of viscous damping.
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196
PROBLEM 19.133
A loaded railroad car of mass 15,000 kg is rolling at a constant velocity v0 when it couples with a spring and
dashpot bumper system (Figure 1). The recorded displacement-time curve of the loaded railroad car after
coupling is as shown (Figure 2). Determine (a) the damping constant, (b) the spring constant. (Hint: Use the
definition of logarithmic decrement given in Problem 19.129.)
SOLUTION
m = 15000 kg
Mass of railroad car:
The differential equation of motion for the system is
mx&& + cx& + kx = 0
For light damping, the solution is given by Eq. (19.44):
- c t
x = e ( 2 m ) (c1 sin w d t + c2 cos w d t )
From the displacement versus time curve,
t d = 0.41 s
wd =
2p
2p
=
= 15.325 rad/s
t d 0.41
At the first peak, x1 = 12.5 mm and t = t1 .
At the second peak, x2 = 3 mm and t = t1 + t d .
Forming the ratio
x2
x1
x2 e ( 2 m ) 1 d
- c td
=
= e ( 2m )
c
t
x1
e ( 2m ) 1
ct d
x1
= e 2m
x2
-
,
c
( t +t )
(1)
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197
PROBLEM 19.133 (Continued)
(a)
Damping constant.
From Eq. (1):
Êx ˆ
ct d
= ln Á 1 ˜
2m
Ë x2 ¯
c=
2m x1
ln
td
x2
(2)(15000) 12.5
ln
0.41
3
3
= 104.423 ¥ 10 N ◊ s/m
=
(b)
c = 104.4 kN ◊ s/m Spring constant.
Equation for w d :
w d2 =
k Ê c ˆ
-Á ˜
m Ë 2m ¯
k = mw d2 +
2
c2
4m
= (15000)(15.325)2 +
= 3.7046 ¥ 106 N/m
(104.423 ¥ 103 )2
( 4)(15000)
k = 3.70 ¥ 106 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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198
PROBLEM 19.134
A 4-kg block A is dropped from a height of 800 mm onto a 9-kg block B
which is at rest. Block B is supported by a spring of constant k = 1500 N/m
and is attached to a dashpot of damping coefficient c = 230 N ◊ s/m. Knowing
that there is no rebound, determine the maximum distance the blocks will
move after the impact.
SOLUTION
Velocity of Block A just before impact.
v A = 2 gh
= 2(9.81)(0.8)
= 3.962 m/s
Velocity of Blocks A and B immediately after impact.
Conservation of momentum.
mA v A + mB v B = ( mA + mB )v ¢
( 4)(3.962) + 0 = ( 4 + 9)v ¢
v ¢ = 1.219 m/s
x&0 = +1.219 m/s Ø = x&0
Static deflection (Block A):
mA g
k
( 4)(9.82)
=1500
= -0.02619 m
x0 = -
x = 0, Equivalent position for both blocks:
cc = 2 km
= 2 (1500)(13)
= 279.3 N ◊ s/m
Since c < cc , Equation (19.44):
x = e ( 2 m ) [c1 sin w D t + c2 cos w D t ]
230
c
=
2m (2)(13)
-
c
t
= 8.846 s -1
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199
PROBLEM 19.134 (Continued)
w D2
Expression for w D :
k Ê c ˆ
= -Á ˜
m Ë 2m ¯
2
1500 Ê 230 ˆ
wD =
13 ÁË (2)(13) ˜¯
2
= 6.094 rad/s
x = e -8.846t (c1 sin 6.094t + c2 cos 6.094t )
x0 = -0.02619 m
(t = 0) x&0 = +1.219 m/s
Initial conditions:
’
x0 = -0.02619 = e 0 [c1 (0) + c2 (1)]
c2 = -0.02619
x& (0) = -8.846e( -8.846)0 [c1 (0) + ( -0.02619)(1)]
+ e( -8.846)( 0 ) [6.094c1 (1) + c2 (0)] = 1.219
1.219 = ( -8.846)( -0.02619) + 6.094c1
c1 = 0.16202
x = e -8.846t (0.16202 sin 6.094t - 0.02619 cos 6.094t )
x& = 0
Maximum deflection occurs when
x& = 0 = -8.846e -8.846tm (0.16202 sin 6.094t m - 0.02619 cos 6.094t m )
+ e -8.846tm [6.094][0.1620 cos 6.094t m + 0.02619 sin 6.094t m ]
0 = [( -8.846)(0.16202) + (6.094)(0.02619)]sin 6.094t m
+ [( -8.846)( -0.02619) + (6.094)(0.1620)]cos 6.094t m
0 = -1.274 sin 6.094t + 1.219 cos 6.094t
1.219
tan 6.094t =
= 0.957
1.274
tan -1 0.957
= 0.1253 s
6.094
xm = e -(8.846)( 0.1253) [0.1620 sin(6.094)(0.1253)
Time at maximum deflection = t m =
- 0.02619 cos(6.0944)(0.1253)]
xm = (0.3301)(0.1120 - 0.0189) = 0.307 m
Blocks move, static deflection + xm
Total distance = 0.02619 + 0.307 = 0.0569 m = 56.9 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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200
PROBLEM 19.135
Solve Problem 19.134, assuming that the damping coefficient of the dashpot
is c = 300 N ◊ s/m.
SOLUTION
Velocity of Block A just before impact.
v A = 2 gh
= 2(9.81)(0.8)
= 3.962 m/s
Velocity of Blocks A and B immediately after impact.
Conservation of momentum.
mA v A + mB v B = ( mA + mB )v ¢
( 4)(3.962) + 0 = ( 4 + 9)v ¢
v ¢ = 1.219 m/s
x&0 = +1.219 m/s Ø = x&0
Static deflection (Block A).
mA g
k
( 4)(9.82)
=1500
= -0.02619 m
x0 = -
x = 0, Equivalent position for both blocks:
cc = 2 km
= 2 (1500)(13)
= 279.3 N ◊ s/m
Since c ⬎ cc , the system is heavily damped.
Eq. 19.42:
x = c1e l1t + c2 e l2t
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201
PROBLEM 19.135 (Continued)
2
l=-
Eq. 19.40:
c
k
Ê c ˆ
± Á ˜ ¯
Ë
2m
2m
m
2
300
1500
Ê 300 ˆ
=± Á
˜¯ Ë
26
26
13
= -11.538 ± 4.2213
l1 = -15.751 s -1
l2 = -7.325 s -1
x = c1e -15.751t + c2 e -7.325t
x& = -15.751c1e -11.751t - 7.325c2 e -7.325t
x0 = -0.02619 m
x&0 = 1.219 m/s
Initial conditions:
Then
-0.02619 = c1 + c2
1.219 = -15.751c1 - 7.325c2
Solving the simultaneous equations,
Then
c1 = -0.1219 m
c2 = 0.09571 m
x = -0.1219e -15.751t + 0.09571e -7.325t
x& = -(15.751)( -0.1219)e
-15.751t
( m)
- (7.325)(0.09571)e -7.325t
= 1.92e -15.751t - 0.70108e -7.325t
The maximum value of x occurs when x& = 0
0 = 1.92e -15.751tm - 0.70108e -7.325tm
1.92e -15.751tm = 0.70108e -7.325tm
1.92
= e(15.751- 0.7325)tm
0.70108
2.7386 = e8.426tm
8.426t m = ln(2.7386)
t m = 0.11956 s
xm = -0.1219e -(15.751)( 0.11956) + 0.09571e -( 7.325)( 0.11956)
= -0.018542 + 0.039867
= 0.02132 m
Total deflection = static deflection + xm
= 0.02619 + 0.02132
= 0.004751 m
Total deflection = 47.5 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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202
PROBLEM 19.136
The barrel of a field gun weighs 750 kg and is returned into firing position after recoil by a recuperator of
constant c = 18 kN ◊ s/m Determine (a) the constant k which should be used for the recuperator to return the
barrel into firing position in the shortest possible time without any oscillation, (b) the time needed for the barrel
to move back two-thirds of the way from its maximum-recoil position to its firing position.
SOLUTION
(a)
A critically damped system regains its equilibrium position in the shortest time.
c = cc
= 18000
= 2m
k
m
Then
(b)
( ) =( )
k=
cc 2
2
18000 2
2
m
750
= 108 kN/m For a critically damped system, Equation (19.43):
x = (c1 + c2t )e -w nt
We take t = 0 at maximum deflection x0 .
Thus,
x& (0) = 0
x(0) = x0
Using the initial conditions,
x(0) = x0 = (c1 + 0)e 0 ,
so c1 = x0
x = ( x0 + c2t )e -w nt
and
x& = -w n ( x0 + c2t )e -w nt + c2 e -w nt
x& (0) = 0 = -w n x0 + c2 , so c2 = w n x0
Thus,
x = x0 (1 + w nt )e -w nt
For
x=
x0
1
,
= (1 + w nt )e -w nt ,
3
3
We obtain
wn =
108 ¥ 103
= 12 rad / s -1
(750)
From solution of (1)
w nt = 2.289,
t=
2.289
= 0.19075 s
12
with w n =
k
m
(1)
t = 0.1908 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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203
PROBLEM 19.137
A uniform rod of mass m is supported by a pin at A and a spring
of constant k at B and is connected at D to a dashpot of damping
coefficient c. Determine in terms of m, k, and c, for small oscillations,
(a) the differential equation of motion, (b) the critical damping
coefficient cc.
SOLUTION
In equilibrium, the force in the spring is mg.
sin q ª q cos q ª 1
l
d yB = q
2
d yC = lq
For small angles,
(a)
SM A = ( SM A )eff
Newton’s Law:
+
mgl Ê l
l
ˆl
- Á k q + mg ˜ - clq& l = I a + mat
¯2
Ë 2
2
2
a = q&&
Kinematics:
l
l
at = a = q&&
2
2
2
2
È
Ê lˆ
Ê lˆ ˘
2
Í I + m Á ˜ ˙ q&& + cl q& + k Á ˜ q = 0
Ë 2¯
Ë 2¯ ˙
ÍÎ
˚
2
1
Ê lˆ
I + m Á ˜ = ml 2
Ë 2¯
3
(b)
Ê 3k ˆ
Ê 3c ˆ
q&& + Á ˜ q& + Á
q =0 Ë 4m ˜¯
Ë m¯
Substituting q = e lt into the differential equation obtained in (a), we obtain the characteristic equation,
3k
Ê 3c ˆ
=0
l2 + Á ˜ l +
Ë m¯
4m
and obtain the roots
l=
-3c
m
m
( 3mc )2 - ( 3mk )
2
The critical damping coefficient, cc , is the value of c, for which the radicand is zero.
2
Thus,
3k
Ê 3cc ˆ
ÁË
˜¯ =
m
m
cc =
km
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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204
PROBLEM 19.138
A 2-kg uniform rod is supported by a pin at O and a spring
at A, and is connected to a dashpot at B. Determine (a) the
differential equation of motion for small oscillations, (b) the
angle that the rod will form with the horizontal 5 s after end B
has been pushed 25 mm down and released.
SOLUTION
sin q ª q , cosq ª 1
Small angles:
d y A = (0.2 m )q = 0.2 q
d yC = (0.2 m )q = 0.2 q
d y B = (0.6 m )q = 0.6 q
(a)
SM 0 = ( SM 0 )eff
Newton’s Law:
+
- (0.2 m ) Fs + (0.2 m ) (2)(9.81) - (0.6 m ) FD
= I a + (0.2 m ) mat
(1)
Fs = k (d y A + (d ST ) A ) = k (0.2 q + (d ST ) A )
F = cd y& = c0.6 q&
D
B
4m
1
1
I = ml 2 = m(0.8)2 =
12
12
75
a = q&&,
Kinematics:
at = (0.2 m ) a = 0.2q&&
È 4m
2 ˘ &&
2 &
ÍÎ 75 + (0.2) m˙˚ q + (0.6) c q + 0.2k ((0.2q + (d ST ) A ) - (0.2)(2)(9.81) = 0
Thus, from Eq. (1),
(2)
SM 0 = 0
But in equilibrium,
+
k (d ST ) A (0.2) - (2)(9.81)(0.2) = 0, 0.2k (d ST ) A = (0.2)(2)(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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205
PROBLEM 19.138 (Continued)
Equation (2) becomes
Ê 7 ˆ && Ê 9 & ˆ Ê 1
ˆ
ÁË ˜¯ mq + ÁË cq ˜¯ ÁË kq ˜¯ = 0
75
25
25
7
Ê 7ˆ
m = Á ˜ (2) = 0.18667
Ë 75 ¯
75
9
Ê 9ˆ
c = Á ˜ (8) = 2.88
Ë 25 ¯
25
k 50
=
=2
25 25
(b)
0.18667q&& + 2.88q& + 2q = 0 Substituting e lt into the above differential equation,
0.18667l 2 + 2.88l + 2 = 0
l = -0.7289, - 14.699i
Since the roots are negative and real, the system is heavily damped. (Eq. 19.46):
q = c1e -0.7289t
+ c1e -14.699t
q& = -(0.7289c1 )e -0.7289t
- (14.699c2 )e -14.699t
Initial conditions.
q 0 = sin -1
(3)
25
= 0.04168 rad
600
q0 = 0
0.04168 = c1 + c2
0 = -0.07289c1 - 14.699c2
Solving for c1 and c2:
c1 = 0.04385, c2 = -2.1747 ¥ 10 -3
At t = 5,
q = 0.04385e -( 0.7289)(5) - 2.1747 ¥ 10 -3 e
= 0.1146 rad = 0.657∞
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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206
PROBLEM 19.139
A 500-kg machine element is supported by two springs, each of constant 50 kN/m. A periodic force of 150 N
amplitude is applied to the element with a frequency of 2.8 Hz. Knowing that the coefficient of damping is
1.8 kN ◊ s/m, determine the amplitude of the steady-state vibration of the element.
SOLUTION
Eq. (19.52):
Total spring constant:
xm =
Pm
(k - mw )
2 2
f
+ (cw f )2
k = (2)(50 kN/m)
= 100 ¥ 103 N/m
w f = 2p f f = 2p (2.8) = 5.6p rad/s
m = 500 kg
Pm = 150 N
c = 1800 N ◊ s/m
xm =
=
150
[100, 000 - (500)(5.6p )2 ]2 + [1800(5.6p )2 ]
150
2.9982 ¥ 109 + 1.0028 ¥ 109
= 2.3714 ¥ 10 -3 m = 2.3714 mm
xm = 2.37 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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it without permission.
207
PROBLEM 19.140
In Problem 19.139, determine the required value of the constant of each spring if the amplitude of the steadystate vibration is to be 1.25 mm.
SOLUTION
w f = 2p f f = (2p )(2.8) = 5.6p rad/s
Pm = 150 N
W
= 500 kg
g
c = 1800 N ◊ s/m
m=
xm = 1.25 mm ¥ 1.25 ¥ 10 -3 m
Eq. (19.52):
Solve for k.
xm =
Pm
( k - mw 2f )2 + (cw f )2
ÊP ˆ
( k - mw 2f )2 + (cw f )2 = Á m ˜
Ë xm ¯
2
2
k=
mw 2f
ÊP ˆ
+ Á m ˜ - (cw f )2
Ëx ¯
m
2
150
ˆ
Ê
= (500)(5.6p )2 + Á
- [(1800)(5.6p )]2
Ë 1.25 ¥ 10 -3 ˜¯
= 154.755 ¥ 103 + 1.44 ¥ 1010 - 1.00281 ¥ 109
= 154.755 ¥ 103 + 115.746 ¥ 103
= 270.501 ¥ 103 N/m
The above calculated value is the equivalent spring constant for two springs.
For each spring, the spring constant is
k
.
2
k
= 135.3 kN/m 2
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208
PROBLEM 19.141
In the case of the forced vibration of a system, determine the range of values of the damping factor c/cc for
which the magnification factor will always decrease as the frequency ratio w f /w n increases.
SOLUTION
From Eq. (19.53)¢ :
Magnification factor:
Find value of
xm
Pm
k
1
=
È
Í1 Î
( ) ˙˚
2 ˘2
wf
wn
+ È2
ÎÍ
( )(
c
cc
wf
wn
)˘˚˙
2
wf
x
c
for which there is no maximum for Pm as
increases.
m
wn
cc
k
( )
d( )
È Ê
- Í2 Á 1 Î Ë
xm 2
d
Pm
k
wf 2
wn
=
ÏÔ È
Ì Í1 ÔÓ Î
( ) ˆ˜¯ (-1) + 4
wf 2
wn
( ) ˙˚
wf
wn
2 ˘2
È
+ Í2
Î
( )(
c
cc
wf
wn
˘
c2
cc2 ˙
˚
2
2 ˘¸
) ˙˚Ô˝Ô˛
=0
wf 2
c2
-2 + 2 ÊÁ w ˆ˜ + 4 2 = 0
Ë n¯
cc
2
2
Ê w f ˆ =1- 2 c
ÁË w n ˜¯
c2
c
wf
x
c2 1
For 2 ⱖ , there is no maximum for Pm and the magnification factor will decrease as
increases.
m
wn
2
cc
( )
k
1
c
ⱖ
cc
2
c
ⱖ 0.707 cc
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209
PROBLEM 19.142
Show that for a small value of the damping factor c/cc , the maximum amplitude of a forced vibration occurs
when w f ª w n and that the corresponding value of the magnification factor is 12 (c /cc ).
SOLUTION
From Eq. (19.53' ):
Magnification factor =
Find value of
wf
wn
for which
xm
Pm
k
xm
Pm
k
1
=
È
Í1 Î
( ) ˙˚
2 ˘2
wf
wn
+ È2
ÎÍ
( )(
c
cc
wf
wn
)˘˚˙
2
is a maximum.
( )
0=
d( )
d
Pm
k
wf 2
wn
2
È È
Í2 Í1 Î Î
xm 2
=-
ÏÔ È
Ì Í1 ÓÔ Î
( ) ˘˙˚ (-1) + 4 ( )˘˙˚
( )
wf 2
wn
2
wf 2˘
˙
wn
+ È2
˚ ÍÎ
c2
cc2
( )(
c
cc
wf
wn
)
2¸
˘ Ô
˙˚ ˝
˛Ô
2
2
Êwf ˆ
Ê cˆ
- 2 + 2Á
+ 4Á ˜ = 0
˜
Ë wn ¯
Ë cc ¯
For small
For
c
,
cc
wf
wn
wf
wn
xm
Pm
k
ª1
w f ª wn
=1
=
xm
1
[1 - 1]2 + ÈÎ2 ( cc )1˘˚
c
2
( )
Pm
k
=
1 cc
2 c
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210
PROBLEM 19.143
A 50-kg motor is directly supported by a light horizontal beam, which
has a static deflection of 6 mm due to the mass of the motor. The
unbalance of the rotor is equivalent to a mass of 100 g located 75 mm
from the axis of rotation. Knowing that the amplitude of the vibration of
the motor is 0.8 mm at a speed of 400 rpm, determine (a) the damping
factor c/cc , (b) the coefficient of damping c.
SOLUTION
k=
Spring constant:
Natural undamped circular frequency:
m
mg (50)(9.81)
=
=
= 81.75 ¥ 103 N ◊ m
-3
d ST d ST
6 ¥ 10
k
81.75 ¥ 103
=
= 40.435 rad/s
50
m
wn =
m¢ = 100 g = 0.100 kg
r = 75 mm = 0.075 m
Unbalance:
Forcing frequency:
w f = 400 rpm = 41.888 rad/s
Unbalance force:
Pm = m¢ rw 2f = (0.100)(0.075)( 41.888)2 = 13.1595 N
d ST =
Static deflection:
Pm
13.1595
=
= 0.16097 ¥ 10 -3 m
3
k
81.75 ¥ 10
xm = 0.8 mm = 0.8 ¥ 10 -3 m
Amplitude:
wf
Frequency ratio:
wn
=
41.888
= 1.0359
40.435
xm =
Eq. (19.53):
d ST
È
Í1 Î
( ) ˙˚
wf
wn
2 ˘2
+ È2
ÍÎ
( ) ( )˘˙˚
c
cc
wf
wn
2
2
È Ê w f ˆ 2 ˘ È Ê c ˆ Ê w f ˆ ˘2 Ê d ˆ 2
ST
˙ + Í2
Í1 - Á
˙ =
ÍÎ Ë w n ˜¯ ˙˚ ÍÎ ÁË cc ˜¯ ÁË w n ˜¯ ˙˚ ÁË xm ˜¯
2
È Ê cˆ
˘
È 0.16097 ¥ 10 -3 ˘
[1 - (1.0359)2 ]2 + Í2 Á ˜ (1.0359) ˙ = Í
˙
-3
ÍÎ Ë cc ¯
˙˚
˚
Î 0.8 ¥ 10
2
2
Ê cˆ
0.0053523 + 4.2924 Á ˜ = 0.040486
Ëc ¯
c
2
Ê cˆ
ÁË c ˜¯ = 0.008185
c
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211
PROBLEM 19.143 (Continued)
(a)
Damping factor.
c
= 0.090472
cc
Critical damping factor.
cc = 2 km
c
= 0.0905 cc
= 2 (81.75 ¥ 103 )(50)
= 4.0435 ¥ 103 N ◊ s/m
(b)
Coefficient of damping.
Ê cˆ
c = Á ˜ cc
Ë cc ¯
= (0.090472)( 4.0435 ¥ 103 )
c = 366 N ◊ s/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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212
PROBLEM 19.144
A 15-kg motor is supported by four springs, each of constant 45 kN/m. The unbalance of the motor is equivalent
to a mass of 20 g located 125 mm from the axis of rotation. Knowing that the motor is constrained to move
vertically, determine the amplitude of the steady-state vibration of the motor at a speed of 1500 rpm, assuming
(a) that no damping is present, (b) that the damping factor c /cc is equal to 1.3.
SOLUTION
Mass of motor:
m = 15 kg
Equivalent spring constant:
k = ( 4)( 45 ¥ 103 )
= 180 ¥ 103 lb/ft
k
180 ¥ 103
=
15
m
= 109.545 rad/s
Undamped natural circular frequency:
wn =
Forcing frequency:
w f = 1500 rpm = 157.08 rad/s
wf
Frequency ratio:
wn
=
157.08
= 1.43394
109.545
Unbalance:
m¢ = 20 g = 20 ¥ 10 -3 kg
r = 125 mm = 0.125 m
Unbalance force:
Pm = m¢ rw 2f = (20 ¥ 10 -3 )(0.125)(157.08)2 = 61.685 N
Pm
61.685
=
k 180 ¥ 103
= 0.3427 ¥ 10 -3 m
d ST =
Static deflection:
xm =
Eq. (19.53):
(a)
No damping.
È
ÍÎ1 -
( )
2
wf 2˘
˙˚
wn
d ST
È
ÍÎ1 -
( )
2
wf 2˘
˙˚
wn
+ ÈÍ2
Î
( ) ( )˘˚˙
c
cc
wf
wn
2
= [1 - (1.43394)2 ]2 = 1.11522
xm =
d ST
1.11522
0.3427 ¥ 10 -3
1.05618
= 0.324 ¥ 10 -3 m
=
xm = 0.324 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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213
PROBLEM 19.144 (Continued)
(b)
c
= 1.3 ÈÍ1 cc
Î
( )
2
wf 2˘
˙˚
wn
+ È2
ÎÍ
( ) ( )˘˚˙
c
cc
wf
wn
2
= 1.11522 + [(2)(1.3)(1.43394)]2 = 15.015
xm =
d ST
15.015
=
0.3427 ¥ 10 -3
3.8749
= 0.0884 ¥ 10 -3 m
xm = 0.0884 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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214
PROBLEM 19.145
A 100-kg motor is supported by four springs, each of constant 90 kN/m,
and is connected to the ground by a dashpot having a coefficient of damping
c = 6500 N ◊ s/m. The motor is constrained to move vertically, and the
amplitude of its motion is observed to be 2.1 mm at a speed of 1200 rpm.
Knowing that the mass of the rotor is 15 kg, determine the distance between
the mass center of the rotor and the axis of the shaft.
SOLUTION
Rotor
Equation (19.52):
xm =
Pm
(k - mw )
2 2
f
È (1200)(2p ) ˘
w 2f = Í
˙˚
60
Î
+ (cw f )2
2
w 2f = 15, 791 s -2
k = 4(90 ¥ 103 N/m)
= 360 ¥ 103
Pm = m¢ ew 2f
= (15 kg)e(15, 791 s -2 )
= 236, 865e
2.1 ¥ 10 -3 m =
236, 865e
[(360 ¥ 10 ) - (100)(15, 791)]2 + (6500)2 (15, 791)
3
= 0.16141e
e = 13.01 ¥ 10 -3 m
e = 13.01 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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215
PROBLEM 19.146
A counter-rotating eccentric mass exciter consisting of two rotating 400-g masses
describing circles of 150-mm radius at the same speed, but in opposite senses, is
placed on a machine element to induce a steady-state vibration of the element and to
determine some of the dynamic characteristics of the element. At a speed of
1200 rpm, a stroboscope shows the eccentric masses to be exactly under their
respective axes of rotation and the element to be passing through its position of static
equilibrium. Knowing that the amplitude of the motion of the element at that speed
is 15 mm, and that the total mass of the system is 140 kg, determine (a) the combined
spring constant k, (b) the damping factor c /cc .
SOLUTION
Forcing frequency:
w f = 1200 rpm = 125.664 rad/s
Unbalance of one mass:
m = 400 g = 0.4 kg
r = 150 mm = 0.15 m
Shaking force:
P = 2 mrw 2f sin w f t
= (2)(0.4)(0.15)(125.664)2 sin w f t
= 1.89497 ¥ 103 siin w f t
Pm = 1.89497 ¥ 103 N
Total mass:
M = 140 kg
By Eqs. (19.48) and (19.52), the vibratory response of the system is
x = xm sin(w f t - j )
xm =
where
tanj =
and
p
,
2
Combined spring constant.
Since j = 90∞ =
(a)
Pm
(
k - M w 2f
)
2
(1)
+ (cw f )2
cw f
k - M w 2f
(2)
tan j and k - M w 2f = 0.
k = M w 2f
= (140)(125.664)2
= 2.2108 ¥ 106 N/m
k = 2.21 Mn/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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216
PROBLEM 19.146 (Continued)
The observed amplitude is
From Eq. (1):
xm = 15 mm = 0.015 m
1
c=
wf
=
2
Ê Pm ˆ
Pm
2
ÁË x ˜¯ - ( k - M w f ) = w x
m
f m
1.89497 ¥ 103
(125.664)(0.015)
= 1.00531 ¥ 103 N ◊ s/m
Critical damping coefficient:
cc = 2 kM
= 2 (2.2108 ¥ 106 )(140)
= 35.186 ¥ 103 N ◊ s/m
(b)
Damping factor.
c 1.00531 ¥ 103
=
cc 35.186 ¥ 103
c
= 0.0286 cc
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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217
PROBLEM 19.147
A simplified model of a washing machine is shown.
A bundle of wet clothes forms a mass mb of 10 kg in the
machine and causes a rotating unbalance. The rotating
mass is 20 kg (including mb) and the radius of the
washer basket e is 25 cm. Knowing the washer has an
equivalent spring constant k = 1000 N/m and damping
ratio z = c/ cc = 0.05 and during the spin cycle the drum
rotates at 250 rpm, determine the amplitude of the
motion and the magnitude of the force transmitted to
the sides of the washing machine.
SOLUTION
wf =
Forced circular frequency:
(2p )(250)
= 26.18 rad/s
60
System mass:
m = 20 kg
Spring constant:
k = 1000 N/m
k
1000
=
= 7.0711 rad/s
20
m
wn =
Natural circular frequency:
cc = 2 km = 2 (1000)(20) = 282.84 N ◊ s/m
Critical damping constant:
Ê cˆ
c = Á ˜ c = (0.05)(141.42) = 14.1421 N ◊ s/m
Ë cc ¯
Damping constant:
Pm = mb ew 2f
Unbalance force:
Pm = (10 kg)(0.25 m)(26.18 rad/s)2 = 1713.48 N
The differential equation of motion is
mx&& + cx& + kx = Pm sinw f t
The steady state response is
x = xm sin(w f t - j )
where
xm =
=
=
x& = w f xm cos(w f t - j )
Pm
(k - mw )
2 2
f
+ (cw f )2
1713.48
[1000 - (20)(26.18)2 ]2 + [(141421)(26.18)]2
1713.48
( -12, 707.8) + (370.24)
2
2
=
1713.48
= 0.13, 478 m
12, 713.2
xm = 134.8 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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218
PROBLEM 19.147 (Continued)
(a)
Amplitude of vibration.
x = 0.13478 sin(w f t - j )
x& = (26.18)(0.13478) cos(w f t - j )
= 3.5285 cos(w f t - j )
Spring force:
kx = (1000)(0.13478)sin(w f t - j )
= 134.78 sin(w f t - j )
Damping force:
cx& = (14.1421)(3.5285) cos(w f t - j )
= 49.901cos(w f t - j )
(b)
Total force:
F = 134.78 sin(w f t - j ) + 49.901cos(w f t - j )
Let
F = Fm cos y sin(w f t - j ) + Fm sin j sin(w f t - j )
= Fm sin(w f t - j + y )
Maximum force.
Fm2 = Fm2 cos2 y + Fm2 sin 2 y
= (134.78)2 + ( 49.901)2
= 20, 656
Fm = 143.7 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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219
PROBLEM 19.148
A machine element is supported by springs and is connected to a dashpot as
shown. Show that if a periodic force of magnitude P = Pm sin wf t is applied to the
element, the amplitude of the fluctuating force transmitted to the foundation is
( ) ( )˘˚˙
È
˘ È
˘
Í1 - ( ) ˙ + Í2 ( ) ( ) ˙
Î
˚ Î
˚
1 + È2
ÎÍ
Fm = Pm
c
cc
2
wf 2
wn
wf
wn
c
cc
2
2
wf 2
wn
SOLUTION
x = xm sin(w f t - f )
From Equation (19.48), the motion of the machine is
The force transmitted to the foundation is
Springs:
Fs = kx = kxm sin(w f t - f )
Dashpot:
FD = cx& = cxmw f cos(w f t - f )
FT = xm [k sin(w f - f ) + cw f cos(w f t - f )]
or recalling the identity,
A sin y + B cos y =
sin y =
cos y =
A2 + B 2 sin( y + f )
B
A2 + B 2
A
A2 + B 2
FT = È xm k 2 + (cw f )2 ˘ sin(w f t - f + y )
Î
˚
Thus, the amplitude of FT is
From Equation (19.53):
Fm = xm k 2 + (cw f )2
xm =
Substituting for xm in Equation (1),
È
Í1 Î
( )
wf
wn
Fm =
w n2 =
(1)
Pm
k
2 ˘2
È
˙ + ÍÎ2
˚
( )
2
c wf ˘
cc w n ˙
˚
Pm 1 +
È
Í1 Î
( )˚
2
wf 2˘
˙
wn
( )
cw f 2
k
+ È2
ÍÎ
( ) ( )˘˙˚
c
cc
(2)
wf
w n2
k
m
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220
PROBLEM 19.148 (Continued)
and Equation (19.41),
cc = 2mw n
m=
cw f
k
=
cw n
2
cw f
Ê c ˆ Êwf ˆ
2
=
ÁË c ˜¯ ÁË w ˜¯
mw n2
c
n
Fm =
Substituting in Eq. (2),
Pm 1 + È2
ÍÎ
È
Í1 Î
( ) ˙˚
wf
wn
2 ˘2
( ) ( )˘˙˚
c
cc
+ È2
ÎÍ
wf
wn
2
( ) ( )˘˚˙
c
cc
wf
wn
Q.E.D. 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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221
PROBLEM 19.149
A 100-kg machine element supported by four springs, each of constant k = 200 N/m,
is subjected to a periodic force of frequency 0.8 Hz and amplitude 100 N.
Determine the amplitude of the fluctuating force transmitted to the foundation
if (a) a dashpot with a coefficient of damping c = 420 N · s/m is connected to the
machine element and to the ground, (b) the dashpot is removed.
SOLUTION
Forcing frequency:
w f = 2p f f = (2p )(0.8) = 1.6p rad/s
Exciting force amplitude:
Pm = 100 N
Mass:
m = 100 kg
Equivalent spring constant:
k = ( 4)(200 N/m) = 800 N/m
Natural frequency:
Frequency ratio:
k 800
m 100
= 2.8284
wn =
wf
wn
=
1.6p
2.8284
= 1.77715
Critical damping coefficient:
cc = 2 km
= 2 (800)(100)
= 565.685 N ◊ s/m
From the derivation given in Problem 19.148, the amplitude of the force transmitted to the foundation is
Fm =
Pm 1 + È2
ÍÎ
È
Í1 Î
( ) ˙˚
wf
wn
2 ˘2
( ) ( )˘˙˚
c
cc
+ È2
ÎÍ
wf
wn
2
( ) ( )˘˚˙
c
cc
wf
wn
(1)
2
2
Êwf ˆ
= 1 - (1.77715)2 = -2.1583
1- Á
˜
w
Ë 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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222
PROBLEM 19.149 (Continued)
(a)
Fm when c = 420 N ◊ s/m:
c
420
=
cc 565.685
= 0.74263
Ê c ˆ Êwf ˆ
= (2)(0.74263)(1.77715)
2Á ˜ Á
Ë cc ¯ Ë w n ˜¯
= 2.63894
From Eq. (1):
Fm =
=
(b)
Fm when c = 0:
100 1 + (2.63894)2
( -2.1583)2 + (2.63894)2
100 7.964
11.6223
Pm
Fm =
1-
( )
wf
wn
2
=
Fm = 82.8 N 100
2.1583
Fm = 46.3 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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223
PROBLEM 19.150*
For a steady-state vibration with damping under a harmonic force, show that the mechanical energy dissipated
per cycle by the dashpot is E = p cxm2 w f , where c is the coefficient of damping, xm is the amplitude of the
motion, and w f is the circular frequency of the harmonic force.
SOLUTION
Energy is dissipated by the dashpot.
From Equation (19.48), the deflection of the system is
x = xm sin(w f t - j )
FD = cx&
FD = cxmw f cos(w f t - j )
The force on the dashpot.
The work done in a complete cycle with
cf =
2p
wf
E=Ú
2p / w f
0
Fd dx (i.e., force ¥ distance)
dx = xmw f cos(w f t - j )dt
E=Ú
cos2 (w D t - j ) =
2p / w f
0
cxm2 w 2f cos2 (w f t - j )dt
[1 - 2 cos(w f t - f )]
E = cxm2 w 2f Ú
2
2p / w f
1 - 2 cos(w f t - f )
2
0
dt
2p / w f
cxm2 w 2f È 2 sin(w f t - f ) ˘
E=
Ít ˙
2 ÍÎ
wf
˙˚0
0
˘
cxm2 w 2f È 2p
2
(sin(2p - j ) - sin j )˙ E = p cxm2 w f Q.E.D. E=
Í
2 ÍÎ w f w f
˙˚
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224
PROBLEM 19.151*
The suspension of an automobile can be approximated by the
simplified spring-and-dashpot system shown. (a) Write the differential
equation defining the vertical displacement of the mass m when the
system moves at a speed v over a road with a sinusoidal cross section
of amplitude d m and wavelength L. (b) Derive an expression for the
amplitude of the vertical displacement of the mass m.
SOLUTION
(a)
+
d2x
Ê dx dd ˆ
SF = ma : W - k (d ST + x - d ) - c Á - ˜ = m 2
Ë dt dt ¯
dt
Recalling that W = kd ST , we write
m
d2x
dt
2
+c
dx
dd
+ kx = kd + c
dt
dt
(1)
Motion of wheel is a sine curve, d = d m sin w f t . The interval of time needed to travel a distance L at a
speed v is t =
Thus,
L
, which is the period of the road surface.
v
2p 2p 2p v
wf =
= L =
tf
L
v
and
d = d m sin w f t
dd d m 2p
= L cos w f t
dt
v
Thus, Equation (1) is
m
dx
d2x
+ c + kx = ( k sin w f t + cw f cos w f t )d m dt
dt
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225
PROBLEM 19.151* (Continued)
(b)
From the identity
A sin y + B cos y =
sin j =
cos j =
A2 + B 2 sin( y + y )
B
A2 + B 2
A
A2 + B 2
We can write the differential equation
d2x
dx
+ c + kx = d m k 2 + (cw f )2 sin(w f t + y )
2
dt
dt
cw f
y = tan -1
k
The solution to this equation is analogous to Equations 19.47 and 19.48, with
m
Pm = d m k 2 + (cw f )2
x = xm sin(w f t - j + y ) (where analogous to Equations (19.52)) xm =
d k 2 + (cw f )2
(k -
)
2
mw 2f
tanj =
+ (cw f )
2
cw f
k - mw 2f
tany =
cw f
k
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226
PROBLEM 19.152*
Two blocks A and B, each of mass m, are supported as shown by three springs of
the same constant k. Blocks A and B are connected by a dashpot, and block B is
connected to the ground by two dashpots, each dashpot having the same coefficient
of damping c. Block A is subjected to a force of magnitude P = Pm sin w f t . Write
the differential equations defining the displacements x A and x B of the two blocks
from their equilibrium positions.
SOLUTION
Since the origins of coordinates are chosen from the equilibrium position, we may omit the initial spring
compressions and the effect of gravity
For load A,
+
SF = ma A : Pm sin w f t + 2k ( x B - x A ) + c( x& B - x& A ) = mx&&A
(1)
+
SF = maB : - 2k ( x B - x A ) - c( x& B - x& A ) - kx B - 2cx& B = mx&&B
(2)
For load B,
mx&&A + c( x& A - x& B ) + 2k ( x A - x B ) = Pm sin w f t Rearranging Equations (1) and (2), we find:
mx&&B + 3cx& B - cx& A + 3kx B - 2kx A = 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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227
PROBLEM 19.153
Express in terms of L, C, and E the range of values of the resistance R
for which oscillations will take place in the circuit shown when switch
S is closed.
SOLUTION
For a mechanical system, oscillations take place if c ⬍ cc (lightly damped).
But from Equation (19.41),
cc = 2m
k
= 2 km
m
Therefore,
c ⬍ 2 km
From Table 19.2:
c®R
(1)
m®L
1
(2)
C
Substituting in Eq. (1) the analogous electrical values in Eq. (2), we find that oscillations will take place if
k®
Ê 1ˆ
R ⬍ 2 Á ˜ ( L)
ËC¯
R⬍2
L
C
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228
PROBLEM 19.154
Consider the circuit of Problem 19.153 when the capacitor C is removed. If switch S is closed at time t = 0,
determine (a) the final value of the current in the circuit, (b) the time t at which the current will have reached
(1 - 1/e) times its final value. (The desired value of t is known as the time constant of the circuit.)
SOLUTION
Electrical system
Mechanical system
The mechanical analogue of closing a switch S is the sudden application of a constant force of magnitude P to
the mass.
(a)
Final value of the current corresponds to the final velocity of the mass, and since the capacitance is zero,
the spring constant is also zero
+
Final velocity occurs when
SF = ma: P - C
d2x
=0
dt 2
dx
=0
P -C
dt final
vfinal =
From Table 19.2:
dx
d2x
=m 2
dt
dt
(1)
dx
= vfinal
dt final
P
C
v ® L, P ® E, C ® R
Lfinal =
Thus,
E
R
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229
PROBLEM 19.154 (Continued)
(b)
Rearranging Equation (1), we have
m
Substitute
Thus,
At t = 0,
d2x
dt
2
+C
dx
=P
dt
P
dx
= Ae - lt + ;
dt
C
P˘
È
m ÈÎ- Al e - lt ˘˚ + C Í Ae - lt + ˙ = P
C˚
Î
c
- ml + C = 0 l =
m
dx
p
= Ae -( c /m)t +
dt
c
dx
P
= 0 0 = A+
C
dt
v=
From Table 19.2:
A= -
P
C
dx p
= È1 - e -( c /m)t ˘˚
dt c Î
v ® c, p ® E, c ® R, m ® L
L=
Ê E ˆ Ê 1ˆ
For L = Á ˜ Á1 - ˜ ,
Ë R¯ Ë e¯
d2x
= - Al e - lt
dt
E
È1 - e -( R /L )t ˘
˚
RÎ
Ê Rˆ
ÁË ˜¯ t = 1
L
t=
L
R
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230
PROBLEM 19.155
Draw the electrical analogue of the mechanical system shown. (Hint: Draw the loops
corresponding to the free bodies m and A.)
SOLUTION
We note that both the spring and the dashpot affect the motion of Point A. Thus, one loop in
the electrical circuit should consist of a capacitor k fi 1c and a resistance (c fi R).
(
)
The other loop consists of ( Pm sin w f t ® Em sin w f t ), an inductor (m® L) and the resistor
(c ® R).
Since the resistor is common to both loops, the circuit is
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231
PROBLEM 19.156
Draw the electrical analogue of the mechanical system shown. (Hint: Draw the
loops corresponding to the free bodies m and A.)
SOLUTION
Loop 1 (Point A):
k1 ®
1
1
, k2 ® , c1 ® R1
c1
c2
Loop 2 (Mass m):
k2 ®
1
, m ® L, c2 ® R2
c2
With k2 ® c1 common to both loops,
2
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232
PROBLEM 19.157
Write the differential equations defining (a) the displacements of the mass m and of the
Point A, (b) the charges on the capacitors of the electrical analogue.
SOLUTION
Mechanical system.
SF = 0:
+
Point A:
+
(a)
SF = ma : c
c
d
( x A - xm ) + kx A = 0 dt
Mass m:
d2x
d
( xm - x A ) - Pm sinw f t = - m 2m
dt
dt
m
(b)
d 2 xm
dt
2
+c
d
( xm - x A ) = Pm sinw f t dt
Electrical analogue.
From Table 19.2:
m®L
c®R
1
C
x®q
k®
P®E
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233
PROBLEM 19.157 (Continued)
Substituting into the results from Part (a), the analogous electrical characteristics,
L
d 2 qm
dt 2
R
d
Ê 1ˆ
( q A - qm ) + Á ˜ qn = 0 ËC¯
dt
+R
d
( qm - qA ) = Em sinw f t dt
Note: These equations can also be obtained by summing the voltage drops around the loops in the circuit of
Problem 19.155.
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234
PROBLEM 19.158
Write the differential equations defining (a) the displacements of the mass m and of
the Point A, (b) the charges on the capacitors of the electrical analogue.
SOLUTION
(a)
Mechanical system.
+
k1 x A + c1
Point A:
Mass m:
(b)
+
SF = 0
dx A
+ k 2 ( x A - xm ) = 0
dt
SF = ma : k2 ( x A - xm ) - c2
d2x
dxm
= m 2m
dt
dt
c1
m
dx A
+ ( k1 + k2 ) x A - k2 xm = 0 dt
d 2 xm
dt
2
+ c2
dxm
+ k 2 ( xm - x A ) = 0 dt
Electrical analogue.
Substituting into the results from Part (a) using the analogous electrical characteristics from Table 19.2
(see left),
R1
L
dqA Ê 1
1ˆ
1
qm = 0 + Á + ˜ qA dt Ë C1 C2 ¯
C2
d 2 qm
dt
2
+ R2
dqm
1
+
( qm - qA ) = 0 dt
C2
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235
PROBLEM 19.159
A thin square plate of side a can oscillate about an axis AB
located at a distance b from its mass center G. (a) Determine the
period of small oscillations if b = 12 a. (b) Determine a second
value of b for which the period of small oscillations is the same
as that found in Part a.
SOLUTION
Let the plate be rotated through angle q about the axis as shown.
+
SMAB = S ( M AB )eff : mgb sin q = - I a - ( mat )(b)
a = q&&
Kinematics:
at = ba = bq&&
sin q ª 0
I =
Moment of inertia:
1
ma2
12
( I + mb2 )q&& + mgbq = 0
Then
Ê1 2
2 ˆ &&
ÁË a + b ˜¯ q + gbq = 0
12
q&& +
Natural circular frequency:
(a)
b=
1
a:
2
Period of vibration:
12 gb
a + 12b2
2
q =0
wn =
12 gb
a + 12b2
wn =
16 ga
3g
=
2
2
a
a + 3a
tn =
2
2
2p
wn
t n = 2p
2a
3g
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236
PROBLEM 19.159 (Continued)
(b)
Another value of b giving the same period:
12 gb
3g
=
2
2
a
a + 12b
12b
3
=
2
2
2a
a + 12b
24ab = 3a2 + 36b2
wn =
2
36b2 - 24ab + 3a2 = 0
b = 0.5a and b = 0.16667a
b = 0.1667a PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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237
PROBLEM 19.160
A 150-kg electromagnet is at rest and is holding 100 kg of scrap steel when
the current is turned off and the steel is dropped. Knowing that the cable
and the supporting crane have a total stiffness equivalent to a spring of
constant 200 kN/m, determine (a) the frequency, the amplitude, and the
maximum velocity of the resulting motion, (b) the minimum tension which
will occur in the cable during the motion, (c) the velocity of the magnet
0.03 s after the current is turned off.
SOLUTION
m1 = 150 kg m2 = 100 kg k = 200 ¥ 103 N/m
Data:
From the first two sketches,
Subtracting Eq. (2) from Eq. (1),
T0 + kxm = ( m1 + m2 ) g
(1)
T0 = m1 g
(2)
kxm = m2 g
xm =
Natural circular frequency:
wn =
m2 g (100)(9.81)
=
= 4.905 ¥ 10 -3 m = 4.91 mm
3
k
200 ¥ 10
k
200 ¥ 103
=
= 36.515 rad/s
m1
150
w n 36.515
=
2p
2p
Natural frequency:
fn =
Maximum velocity:
vm = w n xm = (36.515)( 4.905 ¥ 10 -3 ) = 0.1791 m/s
(a)
f n = 5.81 Hz
amplitude xm = 4.91 mm Resulting motion:
frequency
f n = 5.81 Hz maximum velocity vm = 0.1791 m/s PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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238
PROBLEM 19.160 (Continued)
(b)
Minimum value of tension occurs when x = - xm .
Tmin = T0 - kxm
= m1 g - m2 g
= ( m1 - m2 ) g
= (50)(9.81)
The motion is given by
Tmin = 491 N x = xm sin(w nt + j )
x& = w n xm cos(w nt + j )
x0 = - xm or sinj = -1
x&0 = 0 or cosj = 0
Initally,
j=-
p
2
pˆ
Ê
x& = w n xm cos Á w nt - ˜
Ë
2¯
(c)
Velocity at t = 0.03 s.
w nt = (36.515)(0.03) = 1.09545 rad
w nt - j = -0.47535 rad
cos(w nt - j ) = 0.88913
x& = (36.515)( 4.905 ¥ 10 -3 )(0.88913)
x& = 0.1592 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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239
PROBLEM 19.161
Disks A and B weigh 15 kg and 6 kg, respectively, and a small 2.5-kg block C
is attached to the rim of disk B. Assuming that no slipping occurs between the
disks, determine the period of small oscillations of the system.
SOLUTION
Small oscillations:
Position j
h = rB (1 - cos q m ) ª
rBq& B = rAq& A
rBq B2
2
ˆ
1
1
1 Êr
T1 = mC ( rBq&m )2 + I Bq&m2 + I A Á B q&m ˜
2
2
2 Ë rA ¯
2
mB rB2
2
m r2
IA = A A
2
2 ˆ Ê r ˆ2˘
1È
m r2
T1 = ÍmC rB2 + B B + ÊÁ mA rA ˜ Á B ˜ ˙ q&m2
2 ÍÎ
Ë 2 ¯ Ë rA ¯ ˙˚
2
1 ÈÊ
m
m ˆ˘
T1 = ÍÁ mC + B + A ˜ ˙ rB2q&m2
2 ÎË
2
2 ¯˚
IB =
V1 = 0
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240
PROBLEM 19.161 (Continued)
T2 = 0e
V2 = mC gh
m rq& 2
= C m
2
Position k
Conservation of energy and simple harmonic motion.
T1 + V1 = T2 + V2
q& = w q
m
n m
mC grBq m2
1 ÈÊ
mB mA ˆ ˘ 2 2 2
w
q
+
=
+
r
0
0
+
+
m
Á
˜
B
n
m
C
2
2 ÍÎË
2
2 ¯ ˙˚
mC
g
w n2 =
( mB + mA ) ˘ rB
È
ÍÎmC +
˙˚
2
(2.5)(9.81)
w n2 =
(15 + 6) ˘
È
(0.15)
ÍÎ2.5 +
2 ˙˚
w n2 = 12.5769 s -2
Period of small oscillations.
tn =
2p
2p
=
wn
12.5769
t n = 1.772 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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241
PROBLEM 19.162
A period of 6.00 s is observed for the angular oscillations of a 120-g gyroscope rotor
suspended from a wire as shown. Knowing that a period of 3.80 s is obtained when a
30 mm-diameter steel sphere is suspended in the same fashion, determine the centroidal
radius of gyration of the rotor. (Density of steel = 7800 kg/m3.)
SOLUTION
+
SM = S ( M )eff : - Kq = I q&&
K
q&& + q = 0
I
K
w n2 =
I
I
K
t = 2p
K=
I =
For the sphere,
Volume:
Mass:
r=
4p 2 I
t2
Kt 2
4p 2
(1)
(2)
(3)
d 30
=
= 15 mm = 0.015 m
2 2
4
4
Vs = p r 3 = p (0.015)3
3
3
= 1.41372 ¥ 10 -5 m3
ms = SVs
= (7800 kg/m3 )(1.41372 ¥ 10 -5 )
= 0.11027 kg
Moment of inertia:
Period:
2
2
ms r 2 = (0.11027)(0.015)2
5
5
= 9.9243 ¥ 10 -6 kgm2
I =
t s = 3.80 s
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242
PROBLEM 19.162 (Continued)
From Eq. (2):
K=
4p 2(9.9243 ¥ 10 -6 )
(3.80)2
= 2.7133 ¥ 10 -5 N ◊ m/rad
For the rotor,
m = 120g = 0.12 kg
t = 6.00 s
From Eq. (3):
I = (2.7133 ¥ 10 -5 )
(6.00)2
4p 2
= 2.4742 ¥ 10 -5 kg ◊ m2
Radius of gyration.
I = mk 2
k =
I
m
2.4742 ¥ 10 -5
0.12
= 0.014359 m
=
k = 14.36 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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243
PROBLEM 19.163
A 1.5-kg block B is connected by a cord to a 2-kg block A, which is suspended from a spring
of constant 3 kN/m. Knowing that the system is at rest when the cord is cut, determine (a) the
frequency, the amplitude, and the maximum velocity of the resulting motion, (b) the minimum
tension that will occur in the spring during the motion, (c) the velocity of block A 0.3 s after the
cord has been cut.
SOLUTION
Before the cord is cut, the tension in the spring is
T0 = ( mA + mB ) g
= (2 + 1.5)(9.81)
= 34.335 N
The elongation of the spring is
T0¢ 34.335
=
k 3 ¥ 103
= 11.445 ¥ 10 -3 m
e0 =
After the cord is cut, the tension in the equilibrium position is
T0¢ = mA g
= (2.0)(9.81)
= 19.62 N
The corresponding elongation is
T0¢ 19.62
=
k 3 ¥ 103
= 6.54 ¥ 10 -3 m
e0¢ =
Let x be measured downward from the equilibrium position.
T = T0¢ + kx
(a)
Newton’s Second Law after the cord is cut.
mx&& = mA g - T
mA &&
x + mA g - kx - T0 = - kx
mA &&
x + kx = 0
wn =
fn =
3 ¥ 103
k
=
= 38.7298 rad/s
2
mA
w n 38.7298
=
2p
2p
f n = 6.16 Hz 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.
244
PROBLEM 19.163 (Continued)
The resulting motion is
Initial condition:
x = xm sin (w nt + j )
x& = w n xm cos (w t + j )
x0 = e0 - e0¢
= 4.905 ¥ 10 -3 m
x&0 = 0
0.75375 ¥ 10 -3 = xm sin j
0 = w n xm cos j
p
2
xm = 4.905 m
j=
xm = 4.91 mm pˆ
Ê
x = 4.905 ¥ 10 -3 sin Á w nt + ˜ ( m)
Ë
2¯
pˆ
Ê
x& = 0.18997 cos Á w nt + ˜ (m/s)
Ë
2¯
(b)
x&m = 0.1900 m/s Minimum tension occurs when x is minimum.
Tmin = T0¢ - kxm
= 19.62 - (3 ¥ 103 )( 4.905 ¥ 10 -3 )
(c)
Tmin = 4.91 N Velocity when t = 0.3 s.
w nt - j = (38.7298)(0.3) +
p
2
= 13.1897 radians
x& = (0.18997) cos (13.1897)
= (0.18997)(0.8119)
x& = 0.1542 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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245
PROBLEM 19.164
Two rods, each of mass m and length L, are welded together to form the
assembly shown. Determine (a) the distance b for which the frequency
of small oscillations of the assembly is maximum, (b) the corresponding
maximum frequency.
SOLUTION
Position j
V1 = 0
1
2
2
ÈmvCD
+ mv AB
+ I CDq&m2 + I ABq&m2 ˘˚
2Î
= bq&m
T1 =
vCD
Ê Lˆ
v AB = Á ˜ q&m
Ë 2¯
I CD = I AB
1
mL2
12
2
1 È
1
1 ˘
Ê Lˆ
T1 = m Íb2 + Á ˜ + L2 + L2 ˙ q&m2
Ë 2¯
2 ÍÎ
12
12 ˙˚
5L2 ˆ & 2
mÊ
= Á b2 +
qm
2Ë
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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246
PROBLEM 19.164 (Continued)
Position k
V2 = mgb(1 - cos q m ) + mg
L
(1 - cos q m )
2
2
Ê q ˆ aq
1 - cos q m = 2 sin 2 Á m ˜ ª m
Ë 2¯
2
Small angles:
q m2 Ê
Áb +
2 Ë
V2 = mg
Lˆ
˜
2¯
T2 = 0
Conservation of energy and simple harmonic motion.
T1 + V1 = T2 + V2
mg È
1 È 2 5 2 ˘ &2
L˘
m Íb + L ˙ q m + 0 = 0 +
b + ˙ q m2
Í
2 Î
12 ˚
2 Î
2˚
&
q =w q
m
w n2
(a)
n m
=
(
g b+
(b
+
2
L
2
)
5 2
12 L
(1)
)
Distance b for maximum frequency.
Maximum w n2
when
dw n2
=0
db
(
)
(
b2 + 125 L2 g - g b +
dw n2
=
2
db
b2 + 5 L2
(
12
L
2
)
) (2b) = 0
Ê 5ˆ
- b2 - Lb + Á ˜ L2 = 0
Ë 12 ¯
b=
(b)
- L m L2 +
( 1220 ) L2
2
Corresponding maximum frequency.
= 0.316 L, 1.317 L
b = 0.316 L From Eq. (1) and the answer to Part (a):
w n2 =
g[0.316 + 0.5]
ÈÎ(0.316)2 + 125 ˘˚ L
= 1.580
fn =
wn
2p
= 0.200
g
L
=
1.580
2p
g
Hz
L
g
L
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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247
PROBLEM 19.165
As the rotating speed of a spring-supported motor is slowly increased from 200 to 500 rpm, the amplitude
of the vibration due to the unbalance of the rotor is observed to decrease steadily from 8 mm to 2.5 mm.
Determine (a) the speed at which resonance would occur, (b) the amplitude of the steady-state vibration at a
speed of 100 rpm.
SOLUTION
Since the amplitude decreases with increasing speed over the range w1 = 200 rpm to w 2 = 500 rpm, the motion
is out of phase with the force.
)
x =
=
=
1- ( )
1- ( )
1- ( )
( )( )
|x | =
( ) -1
mrw 2f
k
Pm
k
m
wf 2
wn
mr
M
m
Let u =
wf 2
wn
( mrM ) ( ww
2
f
n
wf 2
wn
wf 2
wn
wf 2
wn
w1
, where w1 = 200 rpm = 20.944 rad/s .
wn
| xm |1
mr
u2
(
M)
=
(1)
u2 - 1
w f = w 2 = 500 rpm
At
wf
wn
=
| xm |2 =
500
u = 2.5u
200
( mrM ) (6.25u2 )
6.25u 2 - 1
(2)
Dividing Eq. (1) by Eq. (2),
| xm |1
8 mm
(6.25u 2 - 1)u 2
= 2
=
= 3.2
2
| xm |2 (u - 1)(6.25u ) 2.5 mm
6.25u 4 - u 2 = 20u 4 - 20u 2
13.75u 4 - 19u 2 = 0 u =
(a)
Natural frequency.
wn =
19
= 1.17551
13.75
w1 20.944
=
= 17.817 rad/s
u 1.17551
w n = 170.1 rpm 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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248
PROBLEM 19.165 (Continued)
At 200 rpm,
From Eq. (1):
| xm |1 = 8 mm = 8 ¥ 10 -3 m, u = 1.17551
8 ¥ 10
-3
mr
(1.17551)2
(
M)
=
(1.17551)2 - 1
19 mr
=
5.25 M
mr
= 2.2105 ¥ 10 -3 m
M
(b)
Amplitude at 100 rpm.
w f = 10.472 rad/s ⬍ w n
wf
wn
=
10.472
= 0.58775
17.817
The motion is in phase with the force.
xm
)
=
1- ( )
( mrM ) ( ww
2
f
n
wf 2
wn
xm =
(2.2105 ¥ 10 -3 )(0.58775)2
1 - (0.58775)2
= 1.167 ¥ 10 -3 m
xm = 1.167 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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it without permission.
249
PROBLEM 19.166
The compressor shown has a mass of 250 kg and operates at 2000 rpm. At this operating condition, undesirable
vibration occurs when the compressor is attached directly to the ground. To reduce the vibration of the concrete
floor that is resting on clay soil, it is proposed to isolate the compressor by mounting it on a square concrete
block separated from the rest of the floor as shown. The density of concrete is 2400 kg/m3 and the spring
constant for the soil is found to be 80 ¥ 106 N/m. The geometry of the compressor leads to choosing a block
that is 1.5 m by 1.5 m. Determine the depth h that will reduce the force transmitted to the ground by 75%.
SOLUTION
Forced circular frequency corresponding to 2000 rpm:
wf =
(2p )(2000)
= 209.44 rad/s
60
Natural circular frequency for compressor attached directly to the ground:
w1 =
80 ¥ 106 N/m
k
=
= 565.68 rad/s
m1
250 kg
Let P be the force due to the compressor unbalance and F1 be the force transmitted to the ground.
F1 = kx1 = k
F1
=
P
P
k
1-
1
1-
( )
wf
wn
2
( )
=
wf
wn
2
P
=
1-
( )
1
1-
.44 2
( 209
565.6
68 )
wf 2
wn
= 1.15951
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.
250
PROBLEM 19.166 (Continued)
Modify the system by making the mass much larger so that the natural frequency of modified system is much
less than the forcing frequency. Let w 2 be the new natural circular frequency. For the ground force to be
reduced by 75%,
F2
F
= 0.25 1 = (0.25)(1.15951) = 0.28988 =
P
P
1
( )
wf 2
wn
-1
2
wf
Êwf ˆ
1
ÁË w ˜¯ = 1 + 0.28988 = 4.4497 w = 2.109
2
2
w 22 =
w 2f
4.4497
k
w 22 =
m2
m2 =
=
(209.44)2
= 9.8579 ¥ 103 ( rad/s)2
4.4497
80 ¥ 106 N/m
k
= 8115 kg
=
w 22 9.8579 ¥ 103 ( rad/s)2
Required properties of attached concrete block:
mass = m2 - m1 = 8115 - 250 = 7865 kg
7865 kg
mass
=
= 3.277 m3
volume =
density 2400 kg/m3
area = 1.5 m ¥ 1.5 m = 2.25 m2
depth =
volume 3.277 m3
=
= 1.456 m
area
2.225 m2
h = 1.456 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
251
PROBLEM 19.167
If either a simple or a compound pendulum is used to determine experimentally the acceleration
of gravity g, difficulties are encountered. In the case of the simple pendulum, the string is not truly
weightless, while in the case of the compound pendulum, the exact location of the mass center
is difficult to establish. In the case of a compound pendulum, the difficulty can be eliminated
by using a reversible, or Kater, pendulum. Two knife edges A and B are placed so that they
are obviously not at the same distance from the mass center G, and the distance l is measured
with great precision. The position of a counterweight D is then adjusted so that the period of
oscillation t is the same when either knife edge is used. Show that the period t obtained is equal
to that of a true simple pendulum of length l and that g = 4p 2l/t 2.
SOLUTION
From Problem 19.52, the length of an equivalent simple pendulum is:
lA = r +
k2
r
and
lB = R +
k2
R
But
tA = tB
2p
lA
l
= 2p B
g
g
Thus,
l A = lB
For
l A = lB
k2
k2
=R+
r
R
2
2
2
r R + k R = r R + k 2r
r+
r R[ r - R ] = k 2 [ r - R ]
(r - R) = 0
Thus,
or
rR =k2
k2
R
k2
R=
r
r=
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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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252
PROBLEM 19.167 (Continued)
Thus,
AG = GA¢ and
That is,
A = A¢ and
Noting that
l A = lB = l
t = 2p
or
g=
BG = GB ¢
B = B¢
l
g
4p 2 l
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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253
PROBLEM 19.168
A 400-kg motor supported by four springs, each of constant 150 kN/m, is
constrained to move vertically. Knowing that the unbalance of the rotor is
equivalent to a 23-g mass located at a distance of 100 mm from the axis of
rotation, determine for a speed of 800 rpm (a) the amplitude of the fluctuating
force transmitted to the foundation, (b) the amplitude of the vertical motion
of the motor.
SOLUTION
Total mass:
M = 400 kg
Unbalance:
m¢ = 23 g = 0.023 kg
r = 100 mm = 0.100 m
Forcing frequency:
w f = 800 rpm
= 83.776 rad/s
Spring constant:
k = ( 4)(150 ¥ 103 N/m)
= 600 ¥ 103 N/m
Natural frequency:
wn =
k
m
600 ¥ 103
400
= 38.730 rad/s
=
Frequency ratio:
Unbalance force:
wf
wn
= 2.1631
Pm = m¢ rw 2f
= (0.023)(0.100)(83.776)2
= 16.1424 N
Static deflection:
Pm
16.1424
=
k
600 ¥ 103
= 26.904 ¥ 10 -6 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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254
PROBLEM 19.168 (Continued)
Amplitude of vibration. From Eq. (19.33):
Pm
k
xm =
1=
( )
wf 2
wn
26.904 ¥ 10 -6
1 - (2.1631)2
= -7.313 ¥ 10 -6 m
(a)
Transmitted force.
(b)
Amplitude of motion.
Fm = - kxm = - (600 ¥ 103 )( -7.313 ¥ 10 -6 )
| xm | = 7.313 ¥ 10 -6 m
Fm = 4.39 N | xm | = 0.00731 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
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
255
PROBLEM 19.169
Solve Problem 19.168, assuming that a dashpot of constant c = 6500 N ◊ s/m
is introduced between the motor and the ground.
PROBLEM 19.168 A 400-kg motor supported by four springs, each of
constant 150 kN/m, is constrained to move vertically. Knowing that the
unbalance of the rotor is equivalent to a 23-g mass located at a distance of
100 mm from the axis of rotation, determine for a speed of 800 rpm (a) the
amplitude of the fluctuating force transmitted to the foundation, (b) the
amplitude of the vertical motion of the motor.
SOLUTION
Total mass:
M = 400 kg
Unbalance:
m = 23 g = 0.023 kg
r = 100 mm = 0.100 m
Forcing frequency:
w f = 800 rpm
= 83.776 rad/s
Spring constant:
Natural frequency:
Frequency ratio:
Viscous damping coefficient:
Critical damping coefficient:
( 4)(150 ¥ 103 N/m) = 600 ¥ 103 N/m
k
600 ¥ 103
=
400
m
= 38.730 rad/s
wn =
wf
wn
= 2.1631
c = 6500 N ◊ s/m
cc = 2 kM = 2 (600 ¥ 103 )( 400)
= 30, 984 N ◊ s/m
Damping factor:
c
= 0.20978
cc
Unbalance force:
Pm = mrw 2f = (0.023)(0.100)(83.776)2
= 16.1424 N
Static deflection:
Pm
16.1424
=
k
600 ¥ 103
= 26.904 ¥ 10 -6 m
d ST =
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.
256
PROBLEM 19.169 (Continued)
Amplitude of vibration. Use Eq. (19.53).
xm =
1 ÈÍ1 Î
Pm
k
2
2˘
( ) ˙˚
wf
wn
+ ÈÍ2
Î
( ) ( )˘˙˚
c
cc
wf
wn
2
2
Êwf ˆ
= 1 - (2.1631)2 = -3.679
1- Á
Ë w n ˜¯
where
Ê c ˆ Êwf ˆ
= (2)(0.20978)(2.1631) = 0.90755
2Á ˜ Á
Ë cc ¯ Ë w n ˜¯
and
xm =
26.904 ¥ 10 -3
( -3.679)2 + (0.90755)2
= 7.1000 ¥ 10 -6 m
Resulting motion:
x = xm sin (w f t - j )
x& = w f xm cos (w f t - j )
Spring force:
Fs = kx = kxm sin (w f t - j ) = 4.26 sin (w f t - j )
Damping force:
Fd = cx& = cw f xm cos (w f t - j ) = 3.8663 cos (w f t - j )
Let
Fs = Fm cos y sin (w f t - j ) and
Total force:
F = Fm cos y sin (w f t - j ) + Fm sin y cos (w f t - j )
Fd = Fm sin y cos (w f t - j )
= Fm sin (w f t - j + y )
(a)
Force amplitude.
Fm = ( Fm cos y )2 + ( Fm sin y )2
Fm = ( kxm )2 + (cw f xm )2
= ( 4.26)2 + (3.8663)2
(b)
Fm = 5.75 N xm = 0.00710 mm Amplitude of vibration.
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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257
PROBLEM 19.170
A small ball of mass m attached at the midpoint of
a tightly stretched elastic cord of length l can slide
on a horizontal plane. The ball is given a small
displacement in a direction perpendicular to the
cord and released. Assuming the tension T in the
cord to remain constant, (a) write the differential
equation of motion of the ball, (b) determine the
period of vibration.
SOLUTION
(a)
Differential equation of motion.
SF = ma : 2T sin q = - mx&&
+
sin q ª tan q =
For small x,
x
()
l
2
=
2x
l
Ê 2x ˆ
mx&& + (2T ) Á ˜ = 0 Ë l ¯
Ê 4T ˆ
mx&& + Á ˜ x = 0
Ë l ¯
Natural circular frequency.
w n2 =
4T
ml
wn = 2
(b)
Period of vibration.
tn =
T
ml
2p
wn
tn = p
ml
T
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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258