Chapter 21

download Chapter 21

of 67

  • date post

    19-Aug-2014
  • Category

    Documents

  • view

    433
  • download

    14

Embed Size (px)

Transcript of Chapter 21

Problem 21.1 InActiveExample21.1, supposethatthe pulley has radius R = 100 mmand its momentof inertia is I = 0.005 kg-m2. The mass m = 2 kg,andthespringconstant is k = 200N/m. If themassis displaced downward fromits equilibriumpositionand released, whatare the periodand frequencyoftheresulting vibration?Rk mSolution: From Active Example 21.1 we have =

km +IR2=

(200 N/m)(2 kg) +(0.005 kg-m2)(0.1 m)2= 8.94 rad/sThus =2=28.94 rad/s = 0.702 s, f =1= 1.42 Hz. = 0.702 s, f = 1.42 Hz.Problem 21.2 InActiveExample 21.1, supposethatthe pulley has radius R = 4 cm and its moment of inertiaisI = 0.06 g- . Thesuspendedobject weighs5 N,2andthespringconstant is k = 10 N/m. Thesystemisinitially at rest in its equilibrium position. Att = 0, thesuspended object is given a downward velocity of 1 m/s.Determine the position of the suspended object relativeto its equilibrium position as a function of time.Rk mSolution: From Active Example 21.1 we have =

km +IR2=

(10 N/m)

5 N2

+(0.06 kg-m2)( )2= rad/sThe general solution isx =Asin t + B cos t , v =Acos t Bsin t .Putting in the initial conditions, we havex(t = 0) = B = 0 B = 0,v(t = 0) = A = (1 ft/s) A =1 m/srad/s = .Thus the equation isx = sin 0.51t ( ).727k m9.81 m/s 0.04 m0.510.511.96 m1.96 mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 21.3 Themassm = 4 kg.The springis un-stretchedwhenx = 0. Theperiodof vibrationof themass is measured and determined to be 0.5 s. The massis displaced to the position x = 0.1 m and released fromrest att = 0. Determine its position att = 0.4 s.kxmSolution: Knowing the period, we can nd the natural frequency =2=20.5 s = 12.6 rad/s.The general solution isx =Asin t + B cos t , v =Acos t Bsin t .Putting in the initial conditions, we havex(t = 0) = B = 0.1 m B = 0.1 m,v(t = 0) =A = 0 A = 0.Thus the equation isx =(0.1 m) cos(12.6 rad/st )At the timet = 0.4 s, we ndx = 0.0309 m.Problem 21.4 Themassm = 4 kg.The springis un-stretched whenx = 0. The frequency of vibration of themass is measured and determined to be 6 Hz. The massis displaced to the position x = 0.1 m and given a veloc-itydx/dt = 5 m/s att = 0. Determine the amplitude ofthe resulting vibration.kxmSolution: Knowingthefrequency, wecanndthenatural fre-quency = 2f = 2(6 Hz) = 37.7 rad/s.The general solution isx =Asin t + B cos t , v =Acos t Bsin t .Putting in the initial conditions, we havex(t = 0) = B = 0.1 m B = 0.1 m,v(t = 0) =A = 5 m/s A =5 m/s37.7 rad/s = 0.133 m.The amplitude of the motion is given by

A2+ B2=

(0.1 m)2+ (0.133 m)2= 0.166 m.Amplitude = 0.166 m.728c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 21.5 The mass m = 4 kg and the spring con-stant is k = 64 N/m. For vibrationof thespring-massoscillator relative to its equilibriumposition, determine(a) the frequency in Hz and (b) the period.m20kxSolution: Sincethevibrationisaroundtheequilibriumposition,we have(a) =

km=

64 N/m4 kg= 4 rad/s

1 cycle2rad

= 0.637 Hz(b) =2=24 rad/s = 1.57 sProblem 21.6 The mass m = 4 kg and the spring con-stant is k = 64 N/m. The springis unstretchedwhenx = 0. At t = 0, x = 0andthemasshasavelocityof2 m/s down the inclined surface. What is the value ofxatt = 0.8 s?Solution: The equation of motion ismd2xdt2 + kx = mg sin 20d2xdt2 +

km

x =g sin 20Putting in the numbers we haved2xdt2 + (16s2)x = 3.36 m/s2The solution to this nonhomogeneous equation isx = Asin([4 s1]t ) + B cos([4 s1]t ) + 0.210 mUsing the initial conditions we have0 =B + 0.210 m2 m/s = A(4 s1)A = 0.5 m, B = 0.210 mThe motion isx = (0.5 m) sin([4 s1]t ) (0.210 m) cos([4 s1]t ) + 0.210 mAtt = 0.8 s we have x = 0.390 mProblem 21.7 Suppose that in a mechanical designcourseyouareaskedtodesignapendulumclock,andyoubeginwiththependulum.Themassofthediskis2 kg. Determine the length L of the bar so that the periodof small oscillationsof thependulumis1 s. For thispreliminary estimate, neglect the mass of the bar.50 mmLSolution: Givenm = 2 kg, r = 0.05 mFor small angles the equation of motion is

12mr2+ m[r + L]2

d2dt2 + mg(L + r) = 0d2dt2 +

2g[L + r]r2+ [r + L]2

= 0The period is = 2

r2+ 2(r + L)22g(L + r)Set = 1 s and solve to nd L = 0.193 m729c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 21.8 Themassof thediskis2 kgandthemass of the slender bar is 0.4 kg. Determine the lengthLofthebarsothattheperiodofsmalloscillationsofthe pendulum is 1 s.Strategy: Draw a graph of the value of the period forarangeoflengthsLtoestimatethevalueof Lcorre-sponding to a period of 1 s.50 mmLSolution: We havemd = 2 kg, mb = 0.4 kg, r = 0.05 m, = 1 s.The moment of inertia of the system about the pivot point isI =13mbL2+ 12mdr2+ md(L + r)2.The equation of motion for small amplitudes is

13mbL2+ 12mdr2+ md(L + r)2

+

mbL2 + md[L + r]

g = 0Thus, the period is given by = 2

13mbL2+ 12mdr2+ md(L + r)2

mb +L2 + md[L + r]

gThis is a complicated equation to solve. You can draw the graph andget anapproximatesolution, or youcanusearoot solver inyourcalculator or on your computer.Using a root solver, we nd L = 0.203 m.Problem 21.9 The spring constant is k = 785 N/m.The spring is unstretched whenx = 0. Neglect the massof the pulley, that is, assume that the tension in the ropeisthesameonbothsidesofthepulley.Thesystemisreleased from rest with x = 0. Determine x as a functionof time.k20 kgx4 kgSolution: We have the equationsT (4 kg)(9.81 m/s2) (785 N/m)x = (4 kg) xT (20 kg)(9.81 m/s2) = (20 kg) xIf we eliminate the tensionTfrom these equations, we nd(24 kg) x + (785 N/m)x = (16 kg)(9.81 m/s2) x + (5.72 rad/s)2x = 6.54 m/s2.The solution of this equation isx =Asin t + B cos t + (0.200 m), v =Acos t Bsin t .Using the initial conditions, we havex(t = 0) = B + (0.200 m) B = 0.200 m,v(t = 0) = A = 0 A = 0.Thus the equation isx = (0.200 m)(1 cos[5.72 rad/st ]).730c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 21.10 Thespringconstant is k = 785 N/m.The spring is unstretched withx = 0. The radius of thepulley is 125 mm, and moment of inertia about its axisis I = 0.05 kg-m2. The system is released from rest withx = 0. Determinexas a function of time.k20 kgx4 kgSolution: Let T1bethetensionintheropeontheleftside, andT2be the tension in the rope on the right side. We have the equationsT1 (4 kg)(9.81 m/s2) (785 N/m)x = (4 kg) xT2 (20 kg)(9.81 m/s2) = (20 kg) x(T2 T1)(0.125 m) =(0.05 kg-m2) x(0.125 m)If we eliminate the tensionsT1andT2from these equations, we nd(27.2 kg) x + (785 N/m)x = (16 kg)(9.81 m/s2) x + (5.37 rad/s)2x = 5.77 m/s2.The solution of this equation isx =Asin t + B cos t + (0.200 m), v = Acos t Bsin t .Using the initial conditions, we havex(t = 0) = B + (0.200 m) B = 0.200 m,v(t = 0) = A = 0 A = 0.Thus the equation isx = (0.200 m)(1 cos [5.37 rad/st ]).Problem 21.11 A bungee jumper who weighs 711.7Nleapsfromabridgeaboveariver.Thebungeecordhas an unstretched length of 18.3 m, anditstretchesadditional 12.2 m before the jumper rebounds. Model thecordas alinear spring. Whenhis motionhas nearlystopped, what are the period and frequency of his verticaloscillations? (You cant model the cord as a linear springduring the early part of his motion. Why not?)Solution: Use energy to nd the spring constantT1 = 0, V1 = 0, T2 = 0, V2 = ( )( ) + 12k( )2T1 + V1 = T2 + V2k = /f =12

km=12

( )/(2)= 0.319 Hz, =1f= 3.13 sThe cord cannot be modeled as a linear spring during the early motionbecause it is slack and does not support a compressive load.731an711.7 N 30.5 m 12.2 m29 .9N m291.9 N/m711.7 N 9.81 m/s1c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 21.12 Thespringconstant is k = 800 N/m,and the spring is unstretched whenx = 0. The mass ofeachobject is 30 kg. Theinclinedsurfaceis smooth.Neglect the mass of thepulley. Thesystem isreleasedfrom rest withx = 0.(a) Determine the frequency and period of the resultingvibration.(b) What is the value ofxatt = 4 s?k20xSolution: The equations of motion areT mg sin kx = m x, T mg = m x.If we eliminateT , we nd2m x + kx = mg(1 sin ), x +k2mx =12g(1 sin ), x + 12

800 N/m30 kg

x =12(9.81 m/s2)(1 sin 20), x + (3.65 rad/s)2x = 3.23 m/s2.(a) The natural frequency, frequency, and period are = 3.65 rad/s, f =2= 0.581 Hz, =1f= 1.72 s.f = 0.581 s, = 1.72 s.(b) The solution to the differential equation isx =Asin t + B cos t + 0.242 m, v = Acos t Bsin t .Putting in the initial conditions, we havex(t = 0) =B + 0.242 m = 0 B = 0.242 m,v(t = 0) =A = 0 A = 0.Thus the equatio