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CHAPTER 19 PROBLEM 19.1 Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 it without permission. 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 it without permission. 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 it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83 PROBLEM 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 it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 121 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 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. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 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. 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. 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. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 . 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. 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) 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. 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 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. 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: 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. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 186 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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. 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. 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 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. 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. 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. 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) 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 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. 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 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. 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 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. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 213 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 214 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 216 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218 PROBLEM 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 219 PROBLEM 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 ˙˚ 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. 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 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. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 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. 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 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. 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 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. 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 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. 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 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. 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. 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. 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 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. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244 PROBLEM 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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= 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. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 258