Transcription
SolutionsManual for Engineering Mechanics Dynamics 14th Edition by Hibbeler IBSN 9780134116990 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they 4th-edition-by-hibbeler-iNo portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Full Download:22–1.A spring is stretched 175 mm by an 8-kg block. If the block isdisplaced 100 mm downward from its equilibrium positionand given a downward velocity of 1.50 m s, determine thedifferential equation which describes the motion. Assumethat positive displacement is downward. Also, determine theposition of the block when t 0.22 s.SOLUTION mg - k(y yst) my T ΣFy may;where kyst mgk y y 0mHencep kBmBWhere k 8(9.81)0.175448.46 7.4878 y (7.487)2y 06 448.46 N m y 56.1y 0Ans.The solution of the above differential equation is of the form:y A sin pt B cos pt(1)#v y Ap cos pt - Bp sin pt(2)At t 0, y 0.1 m and v v0 1.50 m sFrom Eq. (1)0.1 A sin 0 B cos 0B 0.1 mv01.50 0.2003 mp7.487From Eq. (2)v0 Ap cos 0 - 0Hencey 0.2003 sin 7.487t 0.1 cos 7.487tAt t 0.22 s,y 0.2003 sin [7.487(0.22)] 0.1 cos [7.487(0.22)]A Ans. 0.192 m1190Ans: y 56.1 y 0y 0 t 0.22 s 0.192 mFull all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–2.A spring has a stiffness of 800 N m. If a 2-kg block isattached to the spring, pushed 50 mm above its equilibriumposition, and released from rest, determine the equationthat describes the block’s motion. Assume that positivedisplacement is downward.SOLUTIONp k800 20AmA 2x A sin pt B cos ptx - 0.05 mwhen t 0,-0.05 0 B;B -0.05v Ap cos pt - Bp sin ptv 0 when t 0,0 A(20) - 0;A 0Thus,Ans.x - 0.05 cos (20t)Ans:x -0.05 cos (20t)1191
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–3.A spring is stretched 200 mm by a 15-kg block. If the blockis displaced 100 mm downward from its equilibriumposition and given a downward velocity of 0.75 m s,determine the equation which describes the motion. What isthe phase angle? Assume that positive displacement isdownward.SOLUTIONk 15(9.81)F 735.75 N m y0.2vn k735.75 7.00AmA 15y A sin vn t B cos vn ty 0.1 m when t 0,0.1 0 B;B 0.1v A vn cos vn t - Bvn sin vn tv 0.75 m s when t 0,0.75 A(7.00)A 0.107Ans.y 0.107 sin (7.00t) 0.100 cos (7.00t)f tan - 1 aB0.100b tan - 1 ab 43.0 A0.107Ans.Ans:y 0.107 sin (7.00t) 0.100 cos (7.00t)f 43.0 1192
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–4.When a 20-lb weight is suspended from a spring, the springis stretched a distance of 4 in. Determine the naturalfrequency and the period of vibration for a 10-lb weightattached to the same spring.SOLUTIONk vn t 20412 60 lb ftk60 13.90 rad s10AmA 32.2Ans.2p 0.452 svnAns.Ans:vn 13.90 rad st 0.452 s1193
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–5.When a 3-kg block is suspended from a spring, the spring isstretched a distance of 60 mm. Determine the naturalfrequency and the period of vibration for a 0.2-kg blockattached to the same spring.SOLUTIONk vn 3(9.81)F 490.5 N m x0.060k490.5 49.52 49.5 rad sAmA 0.2f vn49.52 7.88 Hz2p2pt 11 0.127 sf7.88Ans.Ans.Ans:vn 49.5 rad st 0.127 s1194
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–6.An 8-kg block is suspended from a spring having a stiffnessk 80 N m. If the block is given an upward velocity of0.4 m s when it is 90 mm above its equilibrium position,determine the equation which describes the motion and themaximum upward displacement of the block measuredfrom the equilibrium position. Assume that positivedisplacement is measured downward.SOLUTIONvn k80 3.162 rad sAmA8y - 0.4 m s,x - 0.09 m at t 0x A sin vn t B cos vn t- 0.09 0 BB -0.09y Avn cos vn t - Bvn sin vn t-0.4 A(3.162) - 0A -0.126Thus,x - 0.126 sin (3.16t) - 0.09 cos (3.16t) mAns.C 2A2 B2 2( -0.126)2 ( - 0.09) 0.155 mAns.Ans:x 5 - 0.126 sin (3.16t) - 0.09 cos (3.16t) 6 mC 0.155 m1195
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–7.A 2-lb weight is suspended from a spring having a stiffnessk 2 lb in. If the weight is pushed 1 in. upward from itsequilibrium position and then released from rest, determinethe equation which describes the motion. What is theamplitude and the natural frequency of the vibration?SOLUTIONk 2(12) 24 lb ftvn k24 19.66 19.7 rad s2AmA 32.2y -1,12Ans.y 0 at t 0From Eqs. 22–3 and 22–4,-1 0 B12B -0.08330 Avn 0A 0C 2A2 B2 0.0833 ft 1 in.Ans.y (0.0833 cos 19.7t) ftAns.Position equation,Ans:vn 19.7 rad sC 1 in.y (0.0833 cos 19.7t) ft1196
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–8.A 6-lb weight is suspended from a spring having a stiffnessk 3 lb in. If the weight is given an upward velocity of20 ft s when it is 2 in. above its equilibrium position,determine the equation which describes the motion and themaximum upward displacement of the weight, measuredfrom the equilibrium position. Assume positive displacementis downward.SOLUTIONk 3(12) 36 lb ftvn k36 13.90 rad sAmA 632.2t 0,y - 20 ft s,1y - ft6From Eq. 22–3,-1 0 B6B -0.167From Eq. 22–4,-20 A(13.90) 0A - 1.44Thus,Ans.y [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ftFrom Eq. 22–10,C 2A2 B2 2(1.44)2 (- 0.167)2 1.45 ftAns.Ans:y [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ftC 1.45 ft1197
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–9.A 3-kg block is suspended from a spring having a stiffnessof k 200 N m. If the block is pushed 50 mm upwardfrom its equilibrium position and then released from rest,determine the equation that describes the motion. What arethe amplitude and the frequency of the vibration? Assumethat positive displacement is downward.SOLUTIONvn k200 8.16 rad sAmA 3Ans.x A sin vn t B cos vn tx -0.05 m when t 0,- 0.05 0 B;B -0.05v Ap cos vn t - Bvn sin vn tv 0 when t 0,0 A(8.165) - 0;A 0Hence,x - 0.05 cos (8.16t)Ans.C 2A2 B2 2(0)2 ( - 0.05) 0.05 m 50 mmAns.Ans:vn 8.16 rad sx - 0.05 cos (8.16t)C 50 mm1198
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–10.The uniform rod of mass m is supported by a pin at A and aspring at B. If B is given a small sideward displacement andreleased, determine the natural period of vibration.ALSolution1Equation of Motion. The mass moment of inertia of the rod about A is IA mL2.3Referring to the FBD. of the rod, Fig. a,L1a ΣMA IAa ;-mg a sin u b - (kx cos u)(L) a mL2 ba23BkHowever; x L sin u. Then- mgL1sin u - kL2 sin u cos u mL2a23Using the trigonometry identity sin 2u 2 sin u cos u,- mgLKL21sin u sin 2u mL2a223 Here since u is small sin u u and sin 2u 2u. Also a u . Then the aboveequation becomes mgL1mL2 u a kL2 bu 032 3mg 6kLu u 02mLComparing to that of the Standard form, vn t A3mg 6kL. Then2mL2p2mL 2p vnA 3mg 6kLAns.Ans:t 2p11992mLA 3mg 6kL
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–11.While standing in an elevator, the man holds a pendulumwhich consists of an 18-in. cord and a 0.5-lb bob. If the elevatoris descending with an acceleration a 4 ft s2, determine thenatural period of vibration for small amplitudes of swing.a4 ft/s2SOLUTIONSince the acceleration of the pendulum is (32.2 - 4) 28.2 ft s2Using the result of Example 22–1,We havevn t g28.2 4.336 rad sAlA 18 122p2p 1.45 s vn4.336Ans.Ans:t 1.45 s1200
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–12.Determine the natural period of vibration of the uniformbar of mass m when it is displaced downward slightly andreleased.OkL—2L—2SolutionEquation of Motion. The mass moment of inertia of the bar about O is I0 1mL2.12Referring to the FBD of the rod, Fig. a,a ΣM0 I0a ;However, y L1- ky cos ua b a mL2 ba212Lsin u. Then2-ka1LLsin u b cos u a b mL2a2212Using the trigonometry identity sin 2u 2 sin u cos u, we obtain1kL2mL2a sin 2u 0128 Here since u is small, sin 2u 2u. Also, a u . Then the above equation becomes 1kL2mL2 u u 0124 3ku 0u mComparing to that of the Standard form, vn t 3k. ThenAmm2p 2p vnA 3kAns.Ans:t 2p1201mA 3k
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–13.The body of arbitrary shape has a mass m, mass center at G,and a radius of gyration about G of kG. If it is displaced aslight amount u from its equilibrium position and released,determine the natural period of vibration.OduGSOLUTIONa MO IO a; -mgd sin u C mk2G md2 D u u gdk2G d2sin u 0However, for small rotation sin u Lu. Hence u gd2kG d2u 0From the above differential equation, vn t 2p vn2pgdA k2G d2gdB k2G 2p d2.k2G d2C gdAns.Ans:t 2p1202k 2G d 2C gd
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–14.The 20-lb rectangular plate has a natural period of vibrationt 0.3 s, as it oscillates around the axis of rod AB.Determine the torsional stiffness k, measured in lb # ft rad,of the rod. Neglect the mass of the rod.AkBSolutionT kuΣMz Iza ;- ku u k(4.83)u 0t 2p2k(4.83)4 ft 1 20ab(2)2 u12 32.22 ft 0.3k 90.8 lb # ft rad Ans.Ans:k 90.8 lb # ft rad1203
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–15.A platform, having an unknown mass, is supported by foursprings, each having the same stiffness k. When nothing ison the platform, the period of vertical vibration is measuredas 2.35 s; whereas if a 3-kg block is supported on theplatform, the period of vertical vibration is 5.23 s. Determinethe mass of a block placed on the (empty) platform whichcauses the platform to vibrate vertically with a period of5.62 s. What is the stiffness k of each of the springs?kkSolution T ΣFy may; mtg - 4k(y yts) mty Where 4k yts mtg4k y y 0mtHence P t 4kA mt2pmt 2pPA 4kFor empty platform mt mP, where mP is the mass of the platform.2.35 2pmP A 4k(1)When 3-kg block is on the platform mt mP 3.5.23 2pmP 3 A 4k(2)When an unknown mass is on the platform mt mP mB.5.62 2pmP mB A4k(3)Solving Eqs. (1) to (3) yields :k 1.36 N m mB 3.58 kg Ans.mP 0.7589 kgAns:k 1.36 N mmB 3.58 kg1204
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–16.A block of mass m is suspended from two springs having astiffness of k1 and k2, arranged a) parallel to each other, andb) as a series. Determine the equivalent stiffness of a singlespring with the same oscillation characteristics and theperiod of oscillation for each case.k1k2k1k2SOLUTION(a) When the springs are arranged in parallel, the equivalent spring stiffness isAns.keq k1 k2The natural frequency of the system isvn keqCm (a)k1 k2B m(b)Thus, the period of oscillation of the system ist 2p vn2pk1 k2B m 2pmAns.C k1 k2(b) When the springs are arranged in a series, the equivalent stiffness of the systemcan be determined by equating the stretch of both spring systems subjected tothe same load F.FFF k1k2keq111 k1k2keqk2 k11 k1k2keqkeq k1k2k1 k2Ans.The natural frequency of the system isk1k2bk2 k1vn mCmSkeqaThus, the period of oscillation of the system ist 2p vn2pk1k2bak2 k1mS 2pCm(k1 k2)k1k2Ans.Ans:keq k1 k2mt 2pA k1 k2k1k2keq k1 k2m(k1 k2)t 2pAk1k21205
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–17.The 15-kg block is suspended from two springs having adifferent stiffness and arranged a) parallel to each other,and b) as a series. If the natural periods of oscillation of theparallel system and series system are observed to be 0.5 sand 1.5 s, respectively, determine the spring stiffnesses k1and k2.k1k2k1k2SOLUTIONThe equivalent spring stiffness of the spring system arranged in parallel isA keq B P k1 k2 and the equivalent stiffness of the spring system arranged in aseries can be determined by equating the stretch of the system to a single equivalentspring when they are subjected to the same load.FFF k1k2(keq)S(a)(b)k2 k11 k1k2A keq B SA keq B S k1k2k1 k2Thus the natural frequencies of the parallel and series spring system are(vn)P (vn)S A keq B PD mA keq B SD m Bk1 k215 k1k2 k1 k2U15 k1k2D 15 A k1 k2 BThus, the natural periods of oscillation aretP tS 2p15 2p 0.5(vn)PB k1 k2(1)15 A k1 k2 B2p 2p 1.5(vn)Sk1k2D(2)Solving Eqs. (1) and (2),k1 2067 N m or 302 N mAns.k2 302 N m or 2067 N mAns.Ans:k1 2067 N mk2 302 N mor vice versa1206
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–18.The uniform beam is supported at its ends by two springsA and B, each having the same stiffness k. When nothing issupported on the beam, it has a period of vertical vibrationof 0.83 s. If a 50-kg mass is placed at its center, the periodof vertical vibration is 1.52 s. Compute the stiffness of eachspring and the mass of the beam.ABkkSOLUTIONt 2pmAkmt2 k(2p)2(0.83)2(2p)2(1.52)2(2p)2 mB2k(1) mB 502k(2)Eqs. (1) and (2) becomemB 0.03490kmB 50 0.1170kAns.mB 21.2 kgAns.k 609 N mAns:mB 21.2 kgk 609 N m1207
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–19.The slender rod has a mass of 0.2 kg and is supported at Oby a pin and at its end A by two springs, each having astiffness k 4 N m. The period of vibration of the rod canbe set by fixing the 0.5-kg collar C to the rod at anappropriate location along its length. If the springs areoriginally unstretched when the rod is vertical, determinethe position y of the collar so that the natural period ofvibration becomes t 1 s. Neglect the size of the collar.OyC600 mmSOLUTIONk A kMoment of inertia about O:IO 1(0.2)(0.6)2 0.5y2 0.024 0.5y23Each spring force Fs kx 4x.a MO IO a;- 2(4x)(0.6 cos u) - 0.2(9.81)(0.3 sin u) - 0.5(9.81)(y sin u) (0.024 0.5y2) u - 4.8x cos u - (0.5886 4.905y) sin u (0.024 0.5y2) uHowever, for small displacement x 0.6u, sin u L u and cos u 1. Hence 3.4686 4.905yu u 00.024 0.5y2From the above differential equation, p t 1 3.4686 4.905yB 0.024 0.5y2.2pp2p3.4686 4.905yB 0.024 0.5y219.74y 2 - 4.905y - 2.5211 0Ans.y 0.503 m 503 mmAns:y 503 mm1208
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–20.A uniform board is supported on two wheels which rotatein opposite directions at a constant angular speed. If thecoefficient of kinetic friction between the wheels and boardis m, determine the frequency of vibration of the board if itis displaced slightly, a distance x from the midpoint betweenthe wheels, and released.xABddSOLUTIONFreebody Diagram: When the board is being displaced x to the right, the restoringforce is due to the unbalance friction force at A and B C (Ff)B 7 (Ff)A D .Equation of Motion:a MA (MA)k ;NB (2d) - mg(d x) 0NB c Fy m(aG)y ;NA mg(d x)- mg 02dNA F m(a ) ;:xG xmcKinematics: Since a mg(d - x)2dmg(d x)mg(d-x)d - mcd ma2d2dmgx 0a d(1)d2x## x, then substitute this value into Eq.(1), we havedt2##x From Eq.(2), vn 2 mg(d x)2dmgx 0d(2)mgmg, thus, vn . Applying Eq. 22–4, we havedA df vn1 2p2pmgdAns.Ans:f 1209mg12p A d
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–21.If the wire AB is subjected to a tension of 20 lb, determinethe equation which describes the motion when the 5-lbweight is displaced 2 in. horizontally and released from rest.A6 ftSolutionL′ K L6 ft ΣF m a ; - 2T x mx dxxLB2T x x 0LmP 2(20)2T 6.55 rad s5A LmA 6 ( 32.2)x A sin pt B cos ptx 11ft at t 0, Thus B 0.16766v A p cos pt - B p sin ptv 0 at t 0, Thus A 0So thatAns.x 0.167 cos 6.55t Ans:x 0.167 cos 6.55t1210
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–22.The bar has a length l and mass m. It is supported at its endsby rollers of negligible mass. If it is given a small displacementand released, determine the natural frequency of vibration.RABlSOLUTIONMoment of inertia about point O:IO 1 2l2 21ml ma R2 b ma R2 - l2 b1246Bc MO IOa;mg aBR2 -l21 b u - ma R2 - l2 bu461 3g(4R 2 - l2)2u u 06R 2 - l21From the above differential equation, vn 3g(4R2 - l2)2D6R2 - l2.Ans.Ans:vn 12113g(4R2 - l 2)1 2C6R2 - l 2
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–23.The 20-kg disk, is pinned at its mass center O and supportsthe 4-kg block A. If the belt which passes over the disk isnot allowed to slip at its contacting surface, determine thenatural period of vibration of the system.OSolutionk 30050 N/mmmk 200 N/mEquation of Motion. The mass moment of inertia of the disk about its mass11center O is I0 mr 2 (20) ( 0.32 ) 0.9 kg # m2. When the disk undergoes a22small angular displacement u, the spring stretches further by s ru 0.3u. Thus,Athe total stretch is y yst 0.3u. Then Fsp ky 200(yst 0.3u). Referring tothe FBD and kinetic diagram of the system, Fig. a,a ΣM0 Σ(mk)0; 4(9.81)(0.3) - 200(yst 0.3u)(0.3) 0.90a 4[a(0.3)](0.3)(1)11.772 - 60yst - 18u 1.26a When the system is in equilibrium, u 0 . Thena ΣM0 0;4(9.81)(0.3) - 200(yst)(0.3) 060yst 11.772Substitute this result into Eq. (1), we obtain- 18u 1.26aa 14.2857u 0 Since a u , the above equation becomes u 14.2857u 0Comparing to that of standard form, vn 214.2857 3.7796 rad s.Thus,t 2p2p 1.6623 s 1.66 s vn3.7796Ans.Ans:t 1.66 s1212
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–24.The 10-kg disk is pin connected at its mass center. Determinethe natural period of vibration of the disk if the springs havesufficient tension in them to prevent the cord from slippingon the disk as it oscillates. Hint: Assume that the initialstretch in each spring is dO.150 mmk 80 N/mOk 80 N/mSolutionEquation of Motion. The mass moment of inertia of the disk about its mass center O11is I0 Mr 2 (10) ( 0.152 ) 0.1125 kg # m2. When the disk undergoes a small22angular displacement u, the top spring stretches further but the stretch of the springis being reduced both by s ru 0.15u. Thus, (Fsp)t Kxt 80(d0 - 0.15u) and(Fsp)b 80(d0 - 0.15u). Referring to the FBD of the disk, Fig. a,a ΣM0 I0a; - 80(d0 0.15u)(0.15) 80(d0 - 0.15u)(0.15) 0.1125a- 3.60u 0.1125aa 32u 0 Since a u , this equation becomes u 32u 0Comparing to that of standard form, vn 232 rad s. Thent 2p2p 1.1107 s 1.11 s vn232Ans.Ans:t 1.11 s1213
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–25.If the disk in Prob. 22–24 has a mass of 10 kg, determine thenatural frequency of vibration. Hint: Assume that the initialstretch in each spring is dO.150 mmk 80 N/mOk 80 N/mSolutionEquation of Motion. The mass moment of inertia of the disk about its mass center O11is I0 mr 2 (10) ( 0.152 ) 0.1125 kg # m2 when the disk undergoes a small22angular displacement u, the top spring stretches but the bottom spring compresses,both by s ru 0.15u. Thus, (Fsp)t (Fsp)b ks 80(0.15u) 12u. Referring tothe FBD of the disk, Fig. a,a ΣM0 I0a; - 12u(0.3) 0.1125a- 3.60u 0.1125aa 32u 0 Since a u , this equation becomes u 32u 0Comparing to that of Standard form, vn 232 rad s. Thenf vn232 0.9003 Hz 0.900 Hz2p2pAns:f 0.900 Hz1214
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–26.A flywheel of mass m, which has a radius of gyration aboutits center of mass of kO, is suspended from a circular shaftthat has a torsional resistance of M Cu. If the flywheelis given a small angular displacement of u and released,determine the natural period of oscillation.LSOLUTIONEquation of Motion: The mass moment of inertia of the wheel about point O isIO mkO 2. Referring to Fig. a, a MO IO a;-Cu mkO 2u u OuCu 0mkO 2Comparing this equation to the standard equation, the natural circular frequency ofthe wheel isvn 1 CC kO A mA mkO 2Thus, the natural period of the oscillation ist 2pm 2pkOvnACAns.Ans:t 2pkO1215mAC
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–27.k 5 lb/ftThe 6-lb weight is attached to the rods of negligible mass.Determine the natural frequency of vibration of the weightwhen it is displaced slightly from the equilibrium positionand released.2 ftSOLUTIONOTO is the equilibrium force.TO 3 ft6(3) 9 lb2Thus, for small u,c MO IO a;6(3) - C 9 5(2)u D (2) a 6b a3u b(3)32.2Thus, u 11.926u 0vn 211.926 3.45 rad/sAns.Ans:vn 3.45 rad s1216
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–28.The platform AB when empty has a mass of 400 kg, centerof mass at G1, and natural period of oscillation t1 2.38 s.If a car, having a mass of 1.2 Mg and center of mass at G2, isplaced on the platform, the natural period of oscillationbecomes t2 3.16 s. Determine the moment of inertia ofthe car about an axis passing through G2.O1.83 m2.50 mG2G1ABSOLUTIONFree-body Diagram: When an object arbitrary shape having a mass m is pinned at Oand being displaced by an angular displacement of u, the tangential component of itsweight will create the restoring moment about point O.Equation of Motion: Sum moment about point O to eliminate Ox and Oy.(1)- mg sin u(l) IOaa MO IOa : d2u u and sin u u if u is small, then substituting these2dtvalues into Eq. (1), we haveKinematics: Since a -mglu IOuFrom Eq. (2), v2n t ormgl u u 0IO(2)mglmgl, thus, vn , Applying Eq. 22–12, we haveB IOIOIO2p 2pB mglvn(3)When the platform is empty, t t1 2.38 s, m 400 kg and l 2.50 m.Substituting these values into Eq. (3), we have2.38 2p(IO)pC 400(9.81)(2.50)(IO)p 1407.55 kg # m2When the car is on the platform, t t2 3.16 s, m 400 kg 1200 kg 1600 kg.2.50(400) 1.83(1200)andl 1.9975 mIO (IO)C (IO)p (IO)C 16001407.55. Substituting these values into Eq. (3), we have3.16 2p(IO)C 1407.55(I ) 6522.76 kg # m2D 1600(9.81)(1.9975) O CThus, the mass moment inertia of the car about its mass center is(IG)C (IO)C - mCd2 6522.76 - 1200(1.832) 2.50(103) kg # m2Ans.Ans:(IG)C 2.50(103) kg # m21217
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–29.The plate of mass m is supported by three symmetricallyplaced cords of length l as shown. If the plate is given aslight rotation about a vertical axis through its center andreleased, determine the natural period of oscillation.llSolution120 1ΣMz Iza -3(T sin f)R mR2 u2l120R120sin f K f 6Tf 0u RmΣFz 0 3T cos f - mg 0f 0, T u mgR, f u3l6 mg Raba u b 0Rm 3l 2gu u 0lt l2p 2p vnA 2gAns.Ans:t 2p1218lA 2g
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–30.Determine the differential equation of motion of the 3-kg blockwhen it is displaced slightly and released. The surface is smoothand the springs are originally unstretched.k500 N/mk500 N/m3 kgSOLUTIONT V const.T 1#(3)x22V 11(500)x 2 (500)x222#T V 1.5x 2 500x 2# #1.5(2x) x 1000xx 0 3x 1000x 0 x 333x 0Ans.Ans: x 333x 01219
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–31.Determine the natural period of vibration of the pendulum.Consider the two rods to be slender, each hav
The solution of the above differential equation is of the form: y A sin pt B cos pt (1) . Solutions Manual for Engineering Mechanics Dynamics 14th Edition by Hibbeler IBSN 9780134116990 . Solutions Manual for Engineering Mechanics Dynamics 14th Edition by Hibbeler IBSN 9780134116990
