91962_09_s19_p0779-0826 6/8/09 4:38 PM Page 779© 2010 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.•19–1.The rigid body (slab) has a mass mand rotates withmvGan angular velocity Vabout an axis passing through thefixed point O.Show that the momenta of all the particlesVcomposing the body can be represented by a single vectorhaving a magnitude mvvGand acting through point P,calledGrPP/Gthe center of percussion,which lies at a distancerP>G=k2G>rG>Ofrom the mass center G.Here is kGtherradius of gyration of the body,computed about an axisG/OGperpendicular to the plane of motion and passing through G.OH O=(rG>O+rP>G) myG=rG>O (myG)+IG v, where IG=mk2Gr G>O (myG)+rP>G (myG)=rG>O (myG)+(mk2G) v2r =kGP>GyG>v However, yyGG=vrG>O or rG>O=v2r P>=kGGrQ.E.D.G>O19–2.At a given instant,the body has a linear momentumL=mvGand an angular momentum HG=IGVcomputedabout its mass center.Show that the angular momentum ofmvGthe body computed about the instantaneous center of zerovelocity ICcan be expressed as HGIC=IICV,where IIC IGVrepresents the body’s moment of inertia computed aboutthe instantaneous axis of zero velocity.As shown,the ICislocated at a distance rG>ICaway from the mass center G.H IC=rG>IC (myG)+IG v, rG/IC where yG=vrG>IC =r+IICG>IC (mvrG>IC)G v =(IG+mr2G>IC) v =IIC vQ.E.D.19–3.Show that if a slab is rotating about a fixed axisperpendicular to the slab and passing through its mass centerVG,the angular momentum is the same when computed aboutPany other point P.GSince yG=0,the linear momentum L=myG=0.Hence the angular momentumabout any point PisHP=IG vSince vis a free vector,so is HP.Q.E.D.77991962_09_s19_p0779-0826 6/8/09 4:39 PM Page 780© 2010 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.*19–4.The pilot of a crippled jet was able to control hisplane by throttling the two engines.If the plane has a weightof 17000 lb and a radius of gyration of kG=4.7 ftabout themass center G,determine the angular velocity of the planeand the velocity of its mass center Gin t=5 sif the thrust inGeach engine is altered to T1=5000 lband T2=800 lbasTshown.Originally the plane is flying straight at 1200 ft>s.21.25 ftNeglect the effects of drag and the loss of fuel.(a +) T11.25 ft (HG)1+©LMG dt=(HG)20 +5000(5)(1.25)-800(5)(1.25)=ca17 00032.2b(4.7)2dv =2.25 rad>sAns. a:+b vm(vGx)1+©LFx dt=m(vGx)2 a17 00032.2b(1200)+5800(5)=a17 00032.2b(vG)2( vG)2=1.25A103B ft>sAns.•19–5.The assembly weighs 10 lb and has a radius ofgyration kG=0.6 ftabout its center of mass G.The kineticenergy of the assembly is 31 ft#lbwhen it is in the positionshown.If it rolls counterclockwise on the surface without0.8 ft1 ftslipping,determine its linear momentum at this instant.GKinetic Energy:Since the assembly rolls without slipping,then v=yG=yG1 ft= 0.8333yG.rG>IC1.2T =12 my21G+2 IG2 v31=12a10110 32.2b y2G+2 c32.2A0.62Bd(0.8333yG)2y G=12. ft>sLinear Momentum:Applying Eq.19–7,we haveL=my10G=32.2 (12.)=3.92 slug#ft>sAns.19–6.The impact wrench consists of a slender 1-kg rod ABwhich is 580 mm long,and cylindrical end weights at Aand Bthat each have a diameter of 20 mm and a mass of 1 kg.Thisassembly is free to rotate about the handle and socket,whichare attached to the lug nut on the wheel of a car.If the rod ABCis given an angular velocity of 4 rad>sand it strikes the bracketBCon the handle without rebounding,determine the angularimpulse imparted to the lug nut.1A300 mmI= (1)(0.6-0.02)2+2c1 (1)(0.01)2+1(0.3)2d=0.2081 kg#m2300 mmaxle122LMdt=Iaxle v=0.2081(4)=0.833 kg#m2>sAns.78091962_09_s19_p0779-0826 6/8/09 4:39 PM Page 781© 2010 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.19–7.The space shuttle is located in “deep space,”where thezeffects of gravity can be neglected.It has a mass of 120Mg,acenter of mass at G,and a radius of gyration (kG)x=14 mabout the xaxis.It is originally traveling forward atv = 3 km/sv=3 km>swhen the pilot turns on the engine at A,creatinga thrust T=600(1-e-0.3t) kN,where tis in seconds.GDetermine the shuttle’s angular velocity 2 s later.AT 2 mxy(a+) (HG)1+©LMG dt=(HG)220 +0.3 tL600A1030BA1-e-B(2) dt=C120A103B(14)2Dv1200A103Bct+10.3 e-0.3 td2=120A1030B(14)2 vv =0.0253 rad>sAns.*19–8.The 50-kg cylinder has an angular velocity of30rad>swhen it is brought into contact with the horizontalAV ϭ 30 rad/ssurface at C.If the coefficient of kinetic friction is mC=0.2,20Њdetermine how long it will take for the cylinder to stopspinning.What force is developed in link ABduring this500 mm200 mmBtime? The axle through the cylinder is connected to twosymmetrical links.(Only ABis shown.) For the computation,neglect the weight of thelinks.Principle of Impulse and Momentum:The mass moment inertia of the cylinder aboutCits mass center is I1G= (50)A0.22B=1.00 kg#m22.Applying Eq.19–14,we have2mAyGyBt 1+©LF dt=mAytyGy1B2( +c) 0+N(t)+2FAB sin 20° (t)-50(9.81)(t)=0(1)t2m AyGxB1+©LFdt=mAytx Gx1B2 A:+B 0+0.2N(t)-2FAB cos 20°(t)=0(2)t2I G v1+©LMdt=ItG G v21(a +) -1.00(30)+[0.2N(t)](0.2)=0(3)Solving Eqs.(1),(2),and (3) yields FAB=48.7 N t=1. sAns. N=457.22 N78191962_09_s19_p0779-0826 6/8/09 4:40 PM Page 782© 2010 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.•19–9.If the cord is subjected to a horizontal force of75 mmP=150 N,and the gear rack is fixed to the horizontal plane,determine the angular velocity of the gear in 4 s,starting fromP ϭ 150 Nrest.The mass of the gear is 50 kg,and it has a radius ofgyration about its center of mass Oof .k150 mmO=125 mmOKinematics:Referring to Fig.a,vO=vrO>IC=v(0.15)Principle of Angular Impulse and Momentum:The mass moment of inertia of the gear about its mass center is IO=mkO 2=50A0.1252B=0.78125 kg#m2.From Fig.b,t2c + IP v1+©LMtP dt=IP v210 +150(4)(0.225)=0.78125v+50[v(0.15)](0.15)v =70.8 rad>sAns.78291962_09_s19_p0779-0826 6/8/09 4:40 PM Page 783© 2010 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.19–10.If the cord is subjected to a horizontal force of75 mmP=150 N,and gear is supported by a fixed pin at O,determine the angular velocity of the gear and the velocityP ϭ 150 Nof the 20-kg gear rack in 4 s,starting from rest.The mass ofthe gear is 50kg and it has a radius of gyration of150 mmkO=125 mm.Assume that the contact surface betweenOthegear rack and the horizontal plane is smooth.Principle of Impulse and Momentum:The mass moment of inertia of the gear about its mass center is IO=mkO 2=50A0.1252free-body diagram of the gear shown in Fig.aB,=0.78125 kg#m2.Referring to the t2a + IO v1+©LM vtO dt=IO210 +F(4)(0.15)-150(4)(0.075)=-0.78125vAF =75-1.302vA(1)Since the gear rotates about the fixed axis,vP=vA rP=vA(0.15).Referring to thefree-body diagram of the gear rack shown in Fig.b,t2 A;+B mv1+©LFvtxdt=m210 +F(4)=20[vA (0.15)]F =0.75vA(2)Equating Eqs.(1) and (2),0.75vA=75-1.302vAvA=36.548 rad>s=36.5 rad>sAns.Then,v=36.548(0.15)=5.48 m>sAns.78391962_09_s19_p0779-0826 6/8/09 4:40 PM Page 784© 2010 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.19–11.A motor transmits a torque of M=0.05 N#mtothe center of gear A.Determine the angular velocity of eachof the three (equal) smaller gears in 2 s starting from rest.The smaller gears (B) are pinned at their centers,and themasses and centroidal radii of gyration of the gears aremA ϭ 0.8 kg given in the figure.M ϭ 0.05 N и m kA ϭ 31 mm 40 mmGear A:A(c +) (HA)1+©LMA dt=(HA)2B0 -3(F)(2)(0.04)+0.05(2)=[0.8(0.031)2]mB ϭ 0.3 kg vA20 mmkB ϭ 15 mm Gear B:(a +)(HB)1+©LMB dt=(HB)20 +(F)(2)(0.02)=[0.3(0.015)2] vBSince ,0.04vA=0.02 vBor vB=2 vA,then solving,F=0.214 NvA=63.3 rad>svB=127 rad>sAns.78491962_09_s19_p0779-0826 6/8/09 4:41 PM Page 785© 2010 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.*19–12.The 200-lb flywheel has a radius of gyration about1 ftits center of gravity Oof kO=0.75 ft.If it rotatescounterclockwise with an angular velocity of 1200 rev>minOvbefore the brake is applied,determine the time required forthe wheel to come to rest when a force of P=200 lbisapplied to the handle.The coefficient of kinetic frictionPbetween the belt and thewheel rim is mABk=0.3.(Hint:Recall from the statics text that the relation of the tensionCin the belt is given by TB=TC emb,where bis the angle ofcontact in radians.) 1.25 ft2.5 ftEquilibrium:Writing the moment equation of equilibrium about point Aandreferring to the free-body diagram of the arm brake shown in Fig.a,a+©MA=0;TB(1.25)-200(3.75)=0 TB=600 lbUsing the belt friction formula,TB=TC emb600=TC e0.3(p)TC=233.80 lbPrinciple of Angular Impulse and Momentum:The mass moment of inertia of the wheel about its mass center is I=mk200OO 2=abA0.752B=3.494 slug#32.2ft2,and the initial angular velocity of the wheel is vrev1=a1200 minba2p rad1 revba1 min60 sb= 40p rad>s.Applying the angular impulse and momentum equation about point Ousing the free-body diagram of the wheel shown in Fig.b,t2a + IO v1+©LMdt=ItO O v213.494(40p)+233.80(t)(1)-600(t)(1)=0t =1.20 sAns.78591962_09_s19_p0779-0826 6/8/09 4:41 PM Page 786© 2010 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.•19–13.The 200-lb flywheel has a radius of gyration about1 ftits center of gravity Oof kO=0.75 ft.If it rotatesvcounterclockwise with a constant angular velocity ofO1200 rev>minbefore the brake is applied,determine therequired force Pthat must be applied to the handle to stopthe wheel in 2 s.The coefficient of kinetic friction betweenPthe belt and the wheel rim is mk=0.3.(Hint:Recall from theABstatics text that the relation of the tension in the belt is givenCby TB=TC emb,where bis the angle of contact in radians.)1.25 ft2.5 ftPrinciple of Angular Impulse and Momentum:The mass moment of inertia of the wheel about its mass center is Imk200O=O 2=abA0.752B=3.494 slug#ft232.2,andthe initial angular velocity of the wheel is vrev1=a1200 minba2p rad1 revba1 min60 sb= 40p rad>s.Applying the angular impulse and momentum equation about point Ousing the free-body diagram shown in Fig.a,t2a + IO v1+©LM vtO dt=IO213.494(40p)+TC (2)(1)-TB (2)(1)=0T B-TC=219.52(1)Using the belt friction formula,TB=TC embTB=TC e0.3(p)(2)Solving Eqs.(1) and (2),TC=140.15 lb TB=359.67 lbEquilibrium:Using this result and writing the moment equation of equilibriumabout point Ausing the free-body diagram of the brake arm shown in Fig. b,a +©MA=0;359.67(1.25)-P(3.75)=0P =120 lbAns.78691962_09_s19_p0779-0826 6/8/09 4:42 PM Page 787© 2010 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.19–14.The 12-kg disk has an angular velocity ofv=20 rad>s.If the brake ABCis applied such that thePmagnitude of force Pvaries with time as shown,determinethe time needed to stop the disk.The coefficient of kinetic500 mmB500 mmfriction at Bis mCk=0.4.Neglect the thickness of the brake.200 mm400 mmP (N)5A2t (s)Equation of Equilibrium:Since slipping occurs at B,the friction Ff=mk NB=0.4NB.From FBD(a),the normal reaction NBcan be obtained directed by summingmoments about point A.a +©MA=0; NB (0.5)-0.4NB (0.4)-P(1)=0N B=2.941PThus,the friction Ff=0.4NB=0.4(2.941P)=1.176P.Principle of Impulse and Momentum:The mass moment inertia of the cylinder about its mass center is I1O= (12)A0.22we have2B=0.240 kg#m2.Applying Eq.19–14,t2I O v1+©LMdt=ItO O v21t(a +) -0.240(20)+c-a1.176LPdtb(0.2)d=0(1)0tHowever,LPdtis the area under the P-tgraph.Assuming t72 s,then0tLPdt=12 (5)(2)+5(t-2)=(5t-5) N#s0Substitute into Eq.(1) yields -0.240(20)+[-1.176(5t-5)(0.2)]=0t =5.08 sAns.Since t=5.08 s72 s,the above assumption is correct.78791962_09_s19_p0779-0826 6/8/09 4:42 PM Page 788© 2010 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.19–15.The 1.25-lb tennis racket has a center of gravity atGand a radius of gyration about Gof .krpG=0.625 ftDetermine the position Pwhere the ball must be hit so that‘no sting’ is felt by the hand holding the racket,i.e.,the1 fthorizontal force exerted by the racket on the hand is zero.GPPrinciple of Impulse and Momentum:Here,we will assume that the tennis racket isinitially at rest and rotates about point Awith an angular velocity of vimmediately after it is hit by the ball,which exerts an impulse of LFdton the racket,Fig.a.The mass moment of inertia of the racket about its mass center is I1.25G=a32.2bA0.6252= 0.01516 slug#ft2.Since the racket about point A,B(vG)=vrG=v(1).Referring to Fig.b,2 ;+ t m(vG)1+©LF=m(vtx dtG)210 +1.25L Fdt=a32.2bCv(1)D L Fdt=0.03882v(1)andt2a + (HA)1+©LMtA dt=(HA)210 +¢0.01516v+1.25LFdt≤rP=32.2 Cv(1)D(1) 0.05398vLFdt=r(2)PEquating Eqs.(1) and (2) yields0.03882v=0.05398vrP rP=1.39 ftAns.781962_09_s19_p0779-0826 6/8/09 4:42 PM Page 7© 2010 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.*19–16.If the boxer hits the 75-kg punching bag with animpulse of I=20 N#s,determine the angular velocity ofAthe bag immediately after it has been hit.Also,find thelocation dof point B,about which the bag appears to rotate.1 mTreat the bag as a uniform cylinder.Bd1 m1.5 mI ϭ 20 N и sPrinciple of Impulse and Momentum:The mass moment of inertia of the bag aboutits mass center is I=11G12mA3r2+h2B=12(75)c3A0.252B+1.52d=15.23 kg#m2.Referring to the impulse and momentum diagrams of the bag shown in Fig.a,2 A:+B 0.5 m m(vtG)1+©LtG)210 +20=75vG Fx dt=m (vvG=0.2667 m>sandt2a + IGv1+©LMG dt=IG vt210 +20(0.25)=15.23vv =0.3282 rad>s=0.328 rad>sAns.Kinematics:Referring to Fig.b,vG=vrG>IC0.2667=0.3282(0.75+d)d=0.0625 mAns.791962_09_s19_p0779-0826 6/8/09 4:43 PM Page 790© 2010 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.•19–17.The 5-kg ball is cast on the alley with a backspinof v0 =10 rad>s,and the velocity of its center of mass Oisv0 ϭ 10 rad/s100 mmv0=5 m>s.Determine the time for the ball to stop back v0 ϭ 5 m/sspinning,and the velocity of its center of mass at thisOinstant.The coefficient of kinetic friction between the balland the alley is mk=0.08.Principle of Impulse and Momentum:Since the ball slips,Ff=mkN=0.08N.The mass moment of inertia of the ball about its mass center is25 mr2=25 (5)A0.12B=0.02 kg#m2Referring to Fig.a,t2 A+cB IO=mc(vO)yd+© 1LFty dt=mc(vO)y1 d20 +N(t)-5(9.81)t=0N=49.05 Nt2a + (HA)1+©LM)tA dt=(HA210.02(10)-5(5)(0.1)+0=-5(vO)2 (0.1)( vO)2=4.6 m>sAns. A:+B mC(vt2O)xD1+©LFtx dt=mC(vO)x1D25(5)-0.08(49.05)(t)=5(4.6)t =0.510 sAns.79091962_09_s19_p0779-0826 6/8/09 4:56 PM Page 791© 2010 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.19–18.The smooth rod assembly shown is at rest when itis struck by a hammer at Awith an impulse of 10 N#s.zDetermine the angular velocity of the assembly and themagnitude of velocity of its mass center immediately after ithas been struck.The rods have a mass per unit length of0.2 m6 kg>m.0.2 mA0.2 m30Њ0.2 mPrinciple of Impulse and Momentum:The total mass of the assembly x10 N и syis m=3[6(0.4)]=7.2 kg.The mass moment of inertia of the assembly about itsmass center is IG=1 [6(0.4)]A0.42B+2c1[6(0.4)]A0.42B+6(0.4)A0.22Bd= 0.288 kg#m21212Referring to Fig. b,t2( +T)m(vx)1+©LFtxdt=m( vx)210 +10 cos 30°=7.2(vG)x(vG)x=1.203 m>s A:+B m(vt2y)1+©LFtydt=m( vy)210 +10 sin 30°=7.2(vG)y(vG)y=0.6944 m>sThus,the magnitude of vGisvG=2(vG)2x+(vG)2y=21.2032+0.69442=1.39 m>sAns.Also t2a +IG v1+©LM dt=ItGG v210 +[-10 cos 30°(0.2)-10 sin 30°(0.2)]=-0.288vv =9.49 rad>sAns.79191962_09_s19_p0779-0826 6/8/09 4:56 PM Page 792© 2010 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.19–19.The flywheel Ahas a mass of 30 kg and a radius ofMgyration of kC=95 mm.Disk Bhas a mass of 25 kg,ispinned at D,and is coupled to the flywheel using a belt125 mmB125 mmwhich is subjected to a tension such that it does not slip at itsADcontacting surfaces.If a motor supplies a counterclockwiseCtorque or twist to the flywheel,having a magnitude ofM=(12t) N # m,where tis in seconds,determine theangular velocity of the disk 3 s after the motor is turned on.Initially,the flywheel is at rest.Principle of Impulse and Momentum:The mass moment inertia of the flywheelabout point Cis IC=30(0.0952)=0.27075 kg#m2.Applying Eq.19–14 to theflywheel [FBD(a)],we havet2I C v1+©LM dt=ItCC v213 s(a +) 0+L12t dt+[T02 (3)](0.125)-T1 (3)](0.125)=0.27075v54.0+0.375T2-0.375T1=0.27075v(1)The mass moment inertia of the disk about point Dis I1D= (25)A0.1252= 0.1953125 kg#m2.Applying Eq.19–14 to the disk [FBD(b)],we have2Bt2 v1+©LM vtD dt=ID21(a +) ID 0+CT1 (3)D(0.125)-CT2 (3)D(0.125)=0.1953125v0.375T2-0.375T1=-0.1953125v(2)Substitute Eq.(2) into Eq.(1) and solving yieldsv=116 rad>sAns.79291962_09_s19_p0779-0826 6/8/09 4:56 PM Page 793© 2010 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.*19–20.The 30-lb flywheel Ahas a radius of gyrationabout its center of 4in.Disk Bweighs 50 lb and is coupled toM ϭ (50t) lbиftthe flywheel by means of a belt which does not slip at itscontacting surfaces.If a motor supplies a counterclockwise6 in.9 in.torque to the flywheel of M=(50t) lb #f t,where tis inseconds,determine the time required for the disk to attainAan angular velocity of 60 rad>sstarting from rest.BPrinciple of Impulse and Momentum:The mass moment inertia of the flywheel about point Cis I3042C= a12b=0.1035 slug#ft232.2.The angular velocity of theflywheel is vrB0.75A=r vB=0.5 (60)=90.0 rad>s.Applying Eq.19–14 to theAflywheel [FBD(a)],we havet2I C v1+©LMtC dt=IC v21t(a +) 0+Ldt+C0 50t LT2 (dt)D(0.5)-CLT1(dt)D(0.5)=0.1035(90)25t2+0.5L(T2-T1)dt= 9.317(1)The mass moment inertia of the disk about point Dis I1D= a502232.2b(0.752)= 0.4367 slug#ft.Applying Eq.19–14 to the disk [FBD(b)],we havet2I D v1+©LMtD dt=ID v21(a +) 0+CLT1 (dt)D(0.75)-CLT2 (dt)D(0.75)=0.4367(60) L(T2-T1)dt=-34.94(2)Substitute Eq.(2) into Eq.(1) and solving yieldst=1.04 sAns.79391962_09_s19_p0779-0826 6/8/09 4:56 PM Page 794© 2010 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.•19–21.For safety reasons,the 20-kg supporting leg of asign is designed to break away with negligible resistance atBwhen the leg is subjected to the impact of a car.Assumingthat the leg is pinned at Aand approximates a thin rod,determine the impulse the car bumper exerts on it,if afterAvCthe impact the leg appears to rotate clockwise to amaximum angle of u2 mmax=150°.uC0.25 mBt2(+b)IA v1+©LMtA dt=IAv210 +I(1.75)=c13 (20)(2)2dv2v 2=0.065625IT2+V2=T3+V312 c13 (20)(2)2d(0.065625I)2+20(9.81)(-1)=0+20(9.81)(1 sin 60°)I=79.8 N#sAns.79491962_09_s19_p0779-0826 6/8/09 4:56 PM Page 795© 2010 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.19–22.The slender rod has a mass mand is suspended atits end Aby a cord.If the rod receives a horizontal blowgiving it an impulse Iat its bottom B,determine the locationyof the point Pabout which the rod appears to rotateduring the impact.APlyPrinciple of Impulse and Momentum:(a +) t2IG v1+© LM dt=ItG G v210 +Ial12b=c12ml2dvm(yt2 a:+bAx)1+©LF IBI=16 mlvtx dt=m(yAx)210 +16 mlv=mvG yG=l6 vKinematics:Point Pis the IC.yB=v yUsing similar triangles,lv y6 vy= y=2y-l 3 lAns.279591962_09_s19_p0779-0826 6/8/09 4:56 PM Page 796© 2010 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.19–23.The 25-kg circular disk is attached to the yoke bymeans of a smooth axle A.Screw Cis used to lock the diskM ϭ (5t2) N и mto the yoke.0.3 m#If the yoke is subjected to a torque ofM=(5t2) Nm,where tis in seconds,and the disk isunlocked,determine the angular velocity of the yoke whenCt=3 s,starting from rest.Neglect the mass of the yoke.0.15 mAPrinciple of Angular Momentum:Since the disk is not rigidly attached to the yoke,only the linear momentum of its mass center contributes to the angular momentumabout point O.Here,the yoke rotates about the fixed axis,thus vA=vrOA=v(0.3).Referring to Fig.a,t2a + (HO)1+©LMtO dt=(HO)213 s0 +L 5t2dt=25Cv(0.3)D(0.3)05t33 s 3=2.25v0v =20 rad>sAns.79691962_09_s19_p0779-0826 6/8/09 4:56 PM Page 797© 2010 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.*19–24.The 25-kg circular disk is attached to the yoke bymeans of a smooth axle A.Screw Cis used to lock the diskM ϭ (5t2) N и mto the yoke.If the yoke is subjected to a torque of0.3 mM=(5t2) N#m,where tis in seconds,and the disk islocked,determine the angular velocity of the yoke whenCt=3 s,starting from rest.Neglect the mass of the yoke.0.15 mAPrinciple of Angular Momentum:The mass moment of inertia of the disk about itsmass center is I11A= mr2= (25)A0.15222about a fixed axis,vB=0.28125 kg#m2.Since the yoke rotatesA=vrOA=v(0.3).Referring to Fig.a,t2a + (HO)1+©LM )tO dt=(HO213 s0 +L 5t2dt=0.28125v+25Cv(0.3)D(0.3)05t333 s =2.53125v0v =17.8 rad>sAns.79791962_09_s19_p0779-0826 6/8/09 4:56 PM Page 798© 2010 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.•19–25.If the shaft is subjected to a torque ofM=(15t2) N#m,where tis in seconds,determine theangular velocity of the assembly when t=3 s,starting from1 m1 mM ϭ (15t2) N и mrest.Rods ABand BCeach have a mass of 9 kg.CABPrinciple of Impulse and Momentum:The mass moment of inertia of the rods about their mass center is I1212=1G= ml12 (9)A12B=0.75 kg#m2.Since the assembly rotates about the fixed axis,(vG)AB=v(rG)AB=v(0.5)and (vG)BC=v(rG)BC=va212+(0.5)2b=v(1.118).Referring to Fig.a,t2c + (Hz)1+©LMtz dt=(Hz)213 s0 +L15t2dt=9Cv(0.5)D(0.5)+0.75v+9Cv(1.118)D(1.118)+0.75v03 s5 t3=15v0v =9 rad>sAns.791962_09_s19_p0779-0826 6/8/09 4:56 PM Page 799© 2010 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.19–26.The body and bucket of a skid steer loader has aweight of 2000 lb,and its center of gravity is located at G.Each of the four wheels has a weight of 100 lband a radiusof gyration about its center of gravity of 1 ft.If the enginesupplies a torque of M=100 lb#ftto each of the rear drivewheels,determine the speed of the loader in t=10 s,1.25 ftG1.25 ftstarting from rest.The wheels roll without slipping.2 ftM2 ft1 ftPrinciple of Impulse and Momentum.The mass moment of inertia of the wheels about their mass center are I100A=IB=2mk2=2abA12B=6.211 slug#ft2Since the wheels roll without slipping,v=vr=v32.2.1.25=0.8v.From Figs.a,b,and c,t2a + (HC)1+©LMtC dt=(HC)210 +2(100)(10)-A100x(10)(1.25)=6.211(0.8v)+2ca32.2bvd(1.25)A x=160-1.019v(1)and t2c +(HD)1+©LMdt=(HtD D)210 +B100x(10)(1.25)=6.211(0.8v)+2ca32.2bvd(1.25) B x=1.019v(2)From Fig.d,t2 a;+bmC(vG)xD1+©LFdt=mC(vtx G)x1D20 +A2000x(10)-Bx(10)=a32.2bv(3)Substituting Eqs.(1) and (2) into Eq.(3),(160-1.019v)(10)-1.019v(10)=a200032.2bvv=19.4 ft>sAns.79991962_09_s19_p0779-0826 6/8/09 4:56 PM Page 800© 2010 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.19–27.The body and bucket of a skid steer loader has aweight of 2000 lb,and its center of gravity is located at G.Each of the four wheels has a weight of 100 lb and a radiusof gyration about its center of gravity of 1 ft.If the loaderattains a speed of 20 ft>sin 10 s,starting from rest,determine the torque Msupplied to each of the rear drive1.25 ftG1.25 ftwheels.The wheels roll without slipping.2 ftM2 ft1 ftPrinciple of Impulse and Momentum:The mass momentum of inertia of the wheels about their mass centers are I2=2a100A=IB=2mk32.2bA12B=6.211 slug#ft2.Since the wheels roll without slipping,v=v20=16 rad>s.From Figs.a,b,and c,r=1.25t2a + (HC)1+©LMHtC dt=(C)210 +2M(10)-A100x(10)(1.25)=6.211(16)+2c32.2 (20)d(1.25)A x=1.6M-20.37(1)and t2c +(HD)1+©LM)tD dt=(HD210 +B+2c100x(10)(1.25)=6.211(16)32.2 (20)d(1.25) B x=20.37 lb(2)From Fig.d,t2 a;+bmC(vG)xD1+©LFtx dt=mC(vG)x1D20 +A2000x (10)-Bx(10)=32.2 (20)(3)Substituting Eqs.(1) and (2) into Eq.(3),(1.6M-20.37)(10)-20.37(10)=200032.2 (20)M=103 lb#ftAns.80091962_09_s19_p0779-0826 6/8/09 4:56 PM Page 801© 2010 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.*19–28.The two rods each have a mass mand a length l,and lie on the smooth horizontal plane.If an impulse Iislapplied at an angle of 45°to one of the rods at midlength asAl/2l/2shown,determine the angular velocity of each rod just afterthe impact.The rods are pin connected at B.BI45ЊCBar BC:(a +) (HG)1+©LMG dt=(HG)20 +LBly dt a2b=IG vBC(1)( +c) m(vGy)1+©LFy dt=m(vGy)20 -LBy dt+I sin 45°=m(vG)y(2)Bar AB :(a +)(HG¿)1+©LMG¿ dt=(HG¿)20 +LBdt aly 2b=IG vAB(3) A+cB m(vGy)1+©LFy dt=m(vGy)20 +LBy dt=m(vG¿)y(4) vB=vG¿+vB>G¿=vG+vB>GA+cBv(vllBy=G¿)y+vAB a2b=(vG)y-vBC a2b(5)80191962_09_s19_p0779-0826 6/8/09 4:56 PM Page 802© 2010 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.*19–28.ContinuedEliminate LBy dtfrom Eqs.(1) and (2),from Eqs.(3) and (4),and between Eqs.(1) and (3).This yieldsIG vBC=l2 (I sin 45°-m(vG)y)m(v)lGy a2b=IG vABvBC=vABSubstituting into Eq.(5),1m a2lbIlvAB2IlG vAB+vAB a2b=-cIGambalbd+m sin 45°-vAB a2ba4mlbI+vIGvABAB l=m sin 45°a41Imlba12 ml2bvAB+vAB l=m sin 45°43 v I=IABm sin 45°vAB=vBC=3422 aImlbAns.80291962_09_s19_p0779-0826 6/8/09 4:56 PM Page 803© 2010 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.•19–29.The car strikes the side of a light pole,which isCdesigned to break away from its base with negligibleresistance.From a video taken of the collision it is observedthat the pole was given an angular velocity of 60rad>swhen ACwas vertical.The pole has a mass of 175 kg,acenter of mass at G,and a radius of gyration about an axisGperpendicular to the plane of the pole assembly and passingthrough Gof kG=2.25 m.Determine the horizontalimpulse which the car exerts on the pole at the instant ACisessentially vertical.4 m(a +) (HG)1+©LMG dt=(HG)2B0.5 m0 +cLF dtd(3.5)=175(2.25)2 (60)A0.5 m LF dt=15.2 kN#sAns.80391962_09_s19_p0779-0826 6/8/09 4:56 PM Page 804© 2010 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.19–30.The frame of the roller has a mass of 5.5 Mg and acenter of mass at G.The roller has a mass of 2 Mg andaradius of gyration about its mass center of kA=0.45 m.Ifa torque of M=600 N#mis applied to the rear wheels,determine the speed of the compactor in t=4 s,startingfrom rest.No slipping occurs.Neglect the mass of thedriving wheels.GMAD0.6 m0.86 m0.5 mDriving Wheels:(mass is neglected)a +©MD=0; B1.95 m1.10 mC600-FC (0.5)=0 Frame and driving wheels: a;+b FC=1200 Nm(vGx)1+©LFx dt=m(vGx)20 +1200(4)-Ax (4)=5500vGA x=12 00-1375vG(1)Roller:G=vA=0.6v(a +) v (HB)1+©LMB dt=(HB)20 +AvGx (4)(0.6)=C2000(0.45)2Da0.6b+C2000(vG)D(0.6)A x=781.25vG(2)Solving Eqs.(1) and (2):Ax=435 NvG=0.557 m>sAns.80491962_09_s19_p0779-0826 6/8/09 4:56 PM Page 805© 2010 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.19–31.The 200-kg satellite has a radius of gyration aboutzthe centroidal zaxis of kz=1.25 m.Initially it is rotatingwith a constant angular velocity of V0=51500 k6 rev>min./10)If the two jets Aand Bare fired simultaneously and produceT ϭ (5e(–t) kNa thrust of T=(5e—0.1t) kN,where tis in seconds,determine1.5 mBthe angular velocity of the satellite,five seconds after firing.Principle of Angular Impulse and Momentum:The mass moment of inertia of thesatellite about its centroidal zaxis is Iz=mk2z=200A1.252B=312.5 kg#m2.TheAinitial angular velocity of the satellite is vrev2x1.5 my1=a1500 minbap rad1 min1 revba60 sb= 50p rad>s.Applying the angular impulse and momentum equation about the zaxis,T ϭ (5e(–t/10)) kNt2Iz v1+©LM =Itzdtz v215 s312.5(50p)-B2L5000e-0.1t(1.5)dtR=312.5v0215625p+A150 000e-0.1tB5 s=312.5v02Thus,v2=[-31.8k] rad>sAns.*19–32.If the shaft is subjected to a torque ofM=(30e—0.1t) N#m,where tis in seconds,determine thezangular velocity of the assembly when t=5 s,starting fromrest.The rectangular plate has a mass of 25 kg.Rods ACand0.6 mBChave the same mass of 5 kg.A0.6 mM ϭ (30e(Ϫ0.1t)) N и mC0.6 mPrinciple of Angular Impulse and Momentum:The mass moment of inertia of the 0.2 mxBassembly about the zaxis is I2c11z=3 (5)A0.62Bd+c12 (25)A0.62= 8.70 kg#m2.Using the free-body diagram of the assembly shown in Fig.B+25(0.6 sin 60°)2da,yt2a + Iz v1+©LM dt=I tzzv215 s0 +L30e-0.1tdt=8.70v02 A-300e-0.1tB5 s=8.70v02Thus,v 2=13.6 rad>sAns.80591962_09_s19_p0779-0826 6/8/09 4:57 PM Page 806© 2010 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.•19–33.The 75-kg gymnast lets go of the horizontal bar inGa fully stretched position A,rotating with an angularBvelocity of vA=3 rad>s.Estimate his angular velocitywhen he assumes a tucked position B.Assume the gymnast750 mmvA ϭ 3 rad/sat positions Aand Bas a uniform slender rod and a uniformcircular disk,respectively.GA1.75 mConservation of Angular Momentum:Other than the weight,there is no externalimpulse during the motion.Thus,the angular momentum of the gymnast isconserved about his mass center G.The mass moments of inertia of the gymnast atthe fully-stretched and tucked positions are (I1A)G= ml2=1 (75)A1.752= 19.14 kg#m2and .(I)111212BBG=2 mr2=2 (75)A0.3752B=5.273 kg#m2Thus,(HA)G=(HB)G19.14(3)=5.273vBvB=10.9 rad>sAns.19–34.A 75-kg man stands on the turntable Aand rotates az6-kg slender rod over his head.If the angular velocity of the1 m1 mrod is vr=5 rad>smeasured relative to the man and theturntable is observed to be rotating in the opposite directionwith an angular velocity of vt=3 rad>s,determine the radiusof gyration of the man about the zaxis.Consider the turntableas a thin circular disk of 300-mm radius and 5-kg mass.Conservation of Angular Momentum:The mass moment of inertia of the rod about the zaxis is (I1r)z=12 ml2=112 (6)A22B=2 kg#m2and the mass moment of Ainertia of the man and the turntable about the zaxis is (I1m)z= (5)A0.32+75k2B+75k22z= 0.225z.Since no external angular impulse acts on the system,the angularmomentum of the system is conserverved about the zaxis.AHzB1=AHzB20=2(vr)-A0.225+75k2zB(3)vr=vm+lmvr=-3+5=2 rad>s2(2)=A0.225+75k2zB3kz=0.122 mAns.80691962_09_s19_p0779-0826 6/8/09 4:57 PM Page 807© 2010 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.19–35.A horizontal circular platform has a weight ofz300lb and a radius of gyration kz=8 ftabout the zaxispassing through its center O.The platform is free to rotateabout the zaxis and is initially at rest.A man having aweight of 150 lb begins to run along the edge in a circularpath of radius 10ft.If he maintains a speed of 4 ft>srelativeto the platform,determine the angular velocity of theOplatform.Neglect friction.10 ftvm=vp+vm>pa:+b(a +) vm=-10v+4(Hz)1=(Hz)20 =-a30015032.2b(8)2 v+a32.2b(-10v+4)(10)v =0.175 rad>sAns.*19–36.A horizontal circular platform has a weight ofz300lb and a radius of gyration kz=8 ftabout the zaxispassing through its center O.The platform is free to rotateabout the zaxis and is initially at rest.A man having aweight of 150 lb throws a 15-lb block off the edge of theplatform with a horizontal velocity of 5 ft>s,measuredrelative to the platform.Determine the angular velocity ofOthe platform if the block is thrown (a) tangent to theplatform,along the +taxis,and (b) outward along a radial10 ftline,or +naxis.Neglect the size of the man.na)( Htz)1=(Hz)20 +0=a1532.2b(v)(10)-a300150b32.2b(8)2 v-a32.2b(10v)(10)vm+vb>m a:+b v b=228vvb=vb=-10v+5228v=-10v+5v =0.0210 rad>sAns.b)( Hz)1=(Hz)20 +0=0-a30032.2b(8)2 v-a15032.2b(10v)(10)v =0Ans.80791962_09_s19_p0779-0826 6/8/09 4:57 PM Page 808© 2010 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.•19–37.The man sits on the swivel chair holding two 5-lbzweights with his arms outstretched.If he is rotating at3rad>sin this position,determine his angular velocity when3 rad/sthe weights are drawn in and held 0.3 ft from the axis ofrotation.Assume he weighs 160 lb and has a radius of2.5 ft2.5 ftgyration about kz=0.55 ftthe zaxis.Neglect the mass of hisarms and the size of the weights for the calculation.Mass Moment of Inertia:The mass moment inertia of the man and the weightsabout zaxis when the man arms are fully stretched is(Iz)1=a16032.2bA0.552B+2c532.2 A2.52Bd=3.444 slug#ft2The mass moment inertia of the man and the weights about zaxis when the weightsare drawn in to a distance 0.3 ft from zaxis(Iz)2=a16032.2bA0.552B+2c532.2 A0.32Bd=1.531 slug#ft2Conservation of Angular Momentum:Applying Eq.19–17,we have (Hz)1=(Hz)2 3.444(3)=1.531(vz)2 (vz)2=6.75 rad>sAns.19–38.The satellite’s body Chas a mass of 200 kg and azradius of gyration about the zaxis of kz=0.2 m.If thesatellite rotates about the zaxis with an angular velocity of5 rev>s,when the solar panels A and Bare in a position of0.5 mu0.5 mu=0°,determine the angular velocity of the satellite when0.4 mthe solar panels are rotated to a position of u=90°.Consider each solar panel to be a thin plate having a massBof 30 kg.Neglect the mass of the rods.xACyConservation of Angular Momentum:When and u=0°u=90°,the massmomentum of inertia of the satellite are( I1z)1=200A0.22B+2c (30)A0.52+0.42B+30A0.75212Bd =43.8 kg#m2( I1z)2=200A0.22B+2c12 (30)A0.52B+30A0.752Bd =43 kg#m2Thus,(Hz)1=(Hz)2(Iz)1 v1=(Iz)2 v243.8(5)=43v2v2=5.09 rev>sAns.801962_09_s19_p0779-0826 6/8/09 4:57 PM Page 809© 2010 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.19–39.A 150-lb man leaps off the circular platform with av ϭ 5 ft/svelocity of vm>p=5 ft>s,relative to the platform.m/pDetermine the angular velocity of the platform afterwards.Initially the man and platform are at rest.The platformweighs 300 lb and can be treated as a uniform circular disk.10 ft8 ftKinematics:Since the platform rotates about a fixed axis,the speed of point Ponthe platform to which the man leaps is vP=vr=v(8).Applying the relativevelocity equation, A+cBConservation of Angular Momentum: v m=vP+vm>Pvm=-v(8)+5(1)As shown in Fig.b,the impulse L Fdtgenerated during the leap is internal to the system.Thus,angular momentum of the system is conserved about the axis perpendicular to the page passing throughpoint O.The mass moment of inertia of the platform about this axis is I1O= mr213002= 2 a32.2bA102B=465.84 slug#ft2Then(HO)1=(HO)20=a15032.2 vmb(8)-465.84vvm=12.5v(2)Solving Eqs.(1) and (2) yieldsv=0.244 rad>sAns.vm=3.05 ft>s80991962_09_s19_p0779-0826 6/8/09 4:57 PM Page 810© 2010 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.*19–40.The 150-kg platform can be considered as acircular disk.Two men,Aand B,of 60-kg and 75-kg mass,Arespectively,stand on the platform when it is at rest.If theyvA/p = 1.5 m/sstart to walk around the circular paths with speeds of3 mvA>p=1.5 m>sand vB>p =2 m>s,measured relative to theBplatform,determine the angular velocity of the platform.vB/p = 2 m/s2 m2.5 mKinematics:Since the platform rotates about a fixed axis,the speed of points PandP¿on the platform at which men Band Aare located is vP=vrP=v(2.5)andvP¿=vrP¿=v(2).Applying the relative velocity equation,v B=vP+vB>P A+TB vB=-v(2.5)+2(1)and v A=vP¿+vA>P¿ A+TBvB=v(2)+1.5(2)Conservation of Angular Momentum:As shown in Fig.b,the impulses LFA dtand FB dtare internal to the system.Thus,angular momentum of the system is conserved about the axis perpendicular to the page passing through point LO.The mass moment of inertia of the platform about this axis is I=1 mr2=1O (150)A32= 675 kg#m2.Then22B(HO)1=(HO)20=75vB (2.5)-60vA (2)-675v(3)Substituting Eqs.(1) and (2) into Eq.(3),0=75(-2.5v+2)(2.5)-60(2v+1.5)(2)-675vv=0.141 rad>sAns.81091962_09_s19_p0779-0826 6/8/09 4:59 PM Page 811© 2010 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.•19–41.Two children Aand B,each having a mass of 30 kg,zbsit at the edge of the merry-go-round which rotates atv = 2 rad>s.Excluding the children,the merry-go-roundbhas a mass of 180 kg and a radius of gyration k0.75 m0.75 mz=0.6 m.Determine the angular velocity of the merry-go-round if Atjumps off horizontally in the -ndirection with a speed of2m>s,measured relative to the merry-go-round.What is thennBmerry-go-round’s angular velocity if Bthen jumps offAthorizontally in the -tdirection with a speed of 2 m>s,measured relative to the merry-go-round? Neglect frictionand the size of each child.V ϭ 2 rad/sMass Moment of Inertia:The mass moment inertia of the merry-go-round about thezaxis when both children are still on it is(Iz)1=180A0.62B+2C30A0.752BD=98.55 kg#m2The mass moment inertia of the merry-go-round about zaxis when child Ajumps off(Iz)2=180A0.62B+30A0.752B=81.675 kg#m2The mass moment inertia of the merry-go-round about zaxis when both childrenjump off(Iz)3=180A0.62B+0=.80 kg#m2Conservation of Angular Momentum:When child Ajumps off in the –ndirection,applying Eq.19–17,we have (Hz)1=(Hz)2 (Iz)1 v1=(Iz)2 v2 98.55(2)=81.675v2 v2=2.413 rad>s=2.41 rad>sAns.Subsequently,when child Bjumps off from the merry-go-round in the –tdirection,applying Eq.19–17,we have (Hz)2=(Hz)3 (Iz)2 v2=(Iz)3 v3-(mB yB)(0.75) 81.675(2.413)=.80v3-30yB (0.75)(1)Relative Velocity:The speed of a point located on the edge of the merry-go-round atthe instant child Bjumps off is yM=v3 (0.75).yB=-yM+yB>M=-v3 (0.75)+2(2)Substituting Eq.(2) into Eq.(1) and solving yieldsv3=2.96 rad>sAns.81191962_09_s19_p0779-0826 6/8/09 4:59 PM Page 812© 2010 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.19–42.A thin square plate of mass mrotates on thePsmooth surface with an angular velocity v1.Determine itsnew angular velocity just after the hook at its corner strikesV1the peg Pand the plate starts to rotate about Pwithoutarebounding.aMass Moment of Inertia:The mass moment inertia of the thin plate about the zaxispassing through its mass center is(Iz)G=1 (m) Aa2+a2B=16 ma212The mass moment inertia of the thin plate about zaxis passing through peg Pis(I1z)P= (m)Aa2+a2B+mB aab2+aab2R2=2 ma212D223Conservation of Angular Momentum:Applying Eq.19–17,we have HG=HP a16 ma2bv21=a3 ma2bv2 v2=14 v1Ans.81291962_09_s19_p0779-0826 6/8/09 4:59 PM Page 813© 2010 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.19–43.A ball having a mass of 8 kg and initial speed ofvv21=0.2 m>srolls over a 30-mm-long depression.Assumingthat the ball rolls off the edges of contact first A,then B,without slipping,determine its final velocity v2when itv1 ϭ 0.2 m/sreaches the other side.Bv0.2yA1=0.125=1.6 rad>sv=220.125=8y2125 mmu=sin-1 a15125b=6.21°30 mmh=125-125 cos 6.21°=0.90326 mm T1+V1=T2+V21 122 (8)(0.2)2+2 c5 (8)(0.125)2d(1.6)2+0 =-(0.90326)(10-3)8(9.81)+1 (8)v2(0.125)2+12 c25 (8)(0.125)2d(v)22v=1.836 rad>s (HB)2=(HB)3 c25 (8)(0.125)2d(1.836)+8(1.836)(0.125) cos 6.2°(0.125 cos 6.2°) - 8(0.22948 sin 6.2°)(0.125 sin 6.2°) =c25(8)(0.125)2dv3+8(0.125)v3 (0.125)v3=1.7980 rad>s T3+V3=T4+V41 2 c25 (8)(0.125)2d(1.7980)2+12(8)(1.7980)2(0.125)2+0 =8(9.81)(0.90326(10-3))+12 c25 (8)(0.125)2d(v4)2 +1 (8)(v4)2(0.125)22v4=1.56 rad>sSo thaty2=1.56(0.125)=0.195 m>sAns.81391962_09_s19_p0779-0826 6/8/09 4:59 PM Page 814© 2010 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.*19–44.The 15-kg thin ring strikes the 20-mm-high step.V1Determine the smallest angular velocity v1the ring can haveso that it will just roll over the step at Awithout slipping180 mmA20 mmThe weight is non-impulsive.(HA)1=(HA)215(v1)(0.18)(0.18-0.02)+C15(0.18)2D(v1)=C15(0.18)2+15(0.18)2Dv2v2=0.9444v1+R©Fn=m(aG)n ; (15)(9.81) cos u-NA=15v22 (0.18)When hoop is about to rebound,N-0.Also,cos u=160A180,and sov2=6.9602 rad>sv6.96021=0.9444=7.37 rad>sAns.81491962_09_s19_p0779-0826 6/8/09 5:00 PM Page 815© 2010 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.•19–45.The uniform pole has a mass of 15 kg and fallsfrom rest when u=90°.It strikes the edge at Awhenu=60°.If the pole then begins to pivot about this pointCafter contact,determine the pole’s angular velocity justafter the impact.Assume that the pole does not slip at Basit falls until it strikes A.3 mConservation of Energy:Datum is set at point B.When the pole is at its initial andAfinal position,its center of gravity is located 1.5 m and 1.5 sin 60° m=1.299 mabove the datum.Its initial and final potential energy are 15(9.81)(1.5)=220.725 N#mB0.5 mand 15(9.81)(1.299)=191.15 N#m.The mass moment of inertia about point BisI1uB=12 (15)A32B+15A1.52B=45.0 kg#m2.The kinetic energy of the pole before the impact is 12 I1B v21=2 (45.0)v21=22.5v21.Applying Eq.18–18,we have T1+V1=T2+V2 0+220.725=22.5v21+191.15 v1=1.146 rad>sConservation of Angular Momentum:Since the weight of the pole is nonimpulsiveforce,the angular momentum is conserved about point A.The velocity of its masscenter before impact is yG=v1 rGB=1.146(1.5)=1.720 m>s.The mass momentof inertia of the pole about its mass center and point Aare I1G= (15)A32B=11.25 kg#m212and IA=1 (15)A32B+15a1.5-0.5b2=24.02 kg#sin 60°m212Applying Eq.19–17,we have (HA)1=(HA)2 (myG)(rGA)+IG v1=IA v2 [15(1.720)]a1.5-0.5sin 60°b+11.25(1.146)=24.02v2 v2=1.53 rad>sAns.81591962_09_s19_p0779-0826 6/8/09 5:00 PM Page 816© 2010 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.19–46.The 10-lb block slides on the smooth surface whenvthe corner Dhits a stop block S.Determine the minimumvelocity vthe block should have which would allow it to tipB1 ftCABover on its side and land in the position shown.Neglect thesize of S.Hint:During impact consider the weight of the1 ftblock to be nonimpulsive.ADSDCConservation of Energy:If the block tips over about point D,it must at least achievethe dash position shown.Datum is set at point D.When the block is at its initial andfinal position,its center of gravity is located 0.5 ft and 0.7071 ft abovethe datum.#Itsinitial and final potential energy are 10(0.5)=5.00 ftlband10(0.7071)=7.071 ft#lb.The mass moment of inertia of the block about point DisID=1a1032.2bA12+12B+a1032.2bA20.52+0.52B2=0.2070 slug#ft212The initial kinetic energy of the block (after the impact) is 1 I1D v22= (0.2070) v22Applying Eq.18–18,we have22. T2+V2=T3+V3 12 (0.2070) v22+5.00=0+7.071 v2=4.472 rad>sConservation of Angular Momentum:Since the weight of the block and the normalreaction Nare nonimpulsiveforces,the angular momentum is conserves about point D.Applying Eq.19–17,we have (HD)1=(HD)2 (myG)(r¿)=ID v2 ca1032.2byd(0.5)=0.2070(4.472) y=5.96 ft>sAns.81691962_09_s19_p0779-0826 6/8/09 5:00 PM Page 817© 2010 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.19–47.The target is a thin 5-kg circular disk that canzrotate freely about the zaxis.A 25-g bullet,traveling at600 m>s,strikes the target at Aand becomes embedded init.Determine the angular velocity of the target after the200 mmimpact.Initially,it is at rest.300 mm600 m/sA100 mmConservation of Angular Momentum:Referring to Fig.a,the sum of the angularimpulse of the system about the zaxis is zero.Thus,the angular impulse of thesystem is conserved about the zaxis.The mass moment of inertia of the target about the zaxis is I1mr2=1z= (5)A0.32about the zaxis when the bullet is embedded in the target,44B=0.1125 kg#m2.Since the target rotates the bullet’s velocity is(vb)2=v(0.2).Then,(Hz)1=(Hz)20.025(600)(0.2)=0.1125v+0.025Cv(0.2)D(0.2)v=26.4 rad>sAns.81791962_09_s19_p0779-0826 6/8/09 5:00 PM Page 818© 2010 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.*19–48.A 2-kg mass of putty Dstrikes the uniform 10-kgplank ABCwith a velocity of 10 m>s.If the putty remainsattached to the plank,determine the maximum angle uof300 mmDswing before the plank momentarily stops.Neglect the sizeAuBCof the putty.800 mm400 mmConservation of Angular Momentum:Referring to Fig.a,the sum of the angularimpulses about point Bis zero.Thus,angular impulse of the system is conservedabout this point.Since rod ACrotates about point B,(vGAC)2=v2rGAC=v2(0.2)and (vD)2=v2rGD=v2(0.3).The mass moment of inertia of rod ACabout its mass center is I11GAC= ml2= (10)A1.221212B=1.2 kg#m2.Then, (HB)1=(HB)2 2(10)(0.3)=1.2v2+10Cv2(0.2)D(0.2)+2Cv2(0.3)D(0.3)v2=3.371 rad>sAns.Conservation of Energy:With reference to the datum in Fig.a,V2=AVgB2=WAC (yGAC)2+WD(yGD)2=0and V 3=AVgB3=WAC (yGAC)3-WD(yGD)3 =10(9.81)(0.2 sin u)-2(9.81)(0.3 sin u)=13.734 sin uThe initial kinetic energy ofthe system is T 2=12 I v11GAC2 2+2 mAC (vGAC)2 2+2 mD(vGD)2 2 =12 (1.2)A3.3712B+ 12 (10)C3.371(0.2)D2+12 (2)C3.371(0.3)D2=10.11 JSince the system is required to be at rest in the final position,T3=0.Then,T2+V2=T3+V310.11+0=0+13.734 sin uu=47.4°Ans.811962_09_s19_p0779-0826 6/8/09 5:00 PM Page 819© 2010 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.•19–49.The uniform 6-kg slender rod ABis given a slightAhorizontal disturbance when it is in the vertical position androtates about Bwithout slipping.Subsequently,it strikes thestep at C.The impact is perfectly plastic and so the rodrotates about Cwithout slipping after the impact.Determine the angular velocity of the rod when it is in thehorizontal position shown.1 mCConservation of Energy:From the geometry of Fig.a,u=tan-1 a0.2250.3b=36.87°and .BC=20.32+0.2252=0.375 mThus,rCG=0.5-0.375=0.125 m.With 0.225 mreference to the datum,V1=W(yG)1=6(9.81)(0.5)=29.43 J,V2=V3=W(yG)3= 6(9.81)(0.5 sin 36.87°)=17.658 J,and .VB4=W(yG)4=6(9.81)(0.225)= 13.2435 JSince the rod is initially at rest,T1=0.The rod rotates about point Bbefore impact.Thus,(v)0.3 mG2=v2rBG=v2 (0.5).The mass moment of inertia of the rod about its masscenter is I=1 ml2=112 (6)A12B=0.5 kg#m2.Then,T11G2=2 m(vG)2 2+2 IGv212= 12 (6)Cv12(0.5)D2+2 (0.5)v2 2=1v2 2.Therefore,T1+V1=T2+V20+29.43=1v2 2+17.658v2=3.431 rad>sThe rod rotates about point Cafter impact.Thus,vG=vrCG=v(0.125).Then,T=1 m(v+12 I11G)2Gv2=2(6)Cv(0.125)D2+2 (0.5)v2=0.296875v22so thatT3=0.296875v3 2and T4=0.296875v4 2T3+V3=T4+V40.296875v3 2+17.658=0.296875v4 2+13.2435v4 2-v3 2=14.87(1)Conservation of Angular Momentum:Referring to Fig.b,the sum of the angularimpulses about point Cis zero.Thus,angular momentum of the rod is conservedabout this point during the impact.Then,(HC)1=(HC)26C3.431(0.5)D(0.125)+0.5(3.431)=6Cv3(0.125)D(0.125)+0.5v3v3=5.056 rad>sSubstituting this result into Eq.(1),we obtainv4 2-(5.056)2=14.87v4=6.36 rad>sAns.81991962_09_s19_p0779-0826 6/8/09 5:01 PM Page 820© 2010 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.19–50.The rigid 30-lb plank is struck by the 15-lb hammerhead H.Just before the impact the hammer is grippedloosely and has a vertical velocity of 75 ft>s.If the coefficientof restitution between the hammer head and the plank ise=0.5,determine the maximum height attained by the 50-lbblock D.The block can slide freely along the two verticalguide rods.The plank is initially in a horizontal position.DCConservation of Angular Momentum:Referring to Fig.a,the sum of the angularBimpulses about point Bis zero.Thus,angular momentum of the system is conserved1 ftHabout this point during the impact.Since the plank rotates about point B,(vD)2=v2(1)and (vG)2=v2(1.25).The mass moment of inertia of the plank about 3 ftits mass center is I1 ml2=1 a30G=bA4.52B=1.572 slug#121232.2ft2.Thus,0.5 ftA(HB)1=(HB)21532.2 (75)(3)=5032.2 Cv30152(1)D(1)+32.2Cv2(1.25)D(1.25)+1.572v2-32.2 (vH)2(3)4.581 v2-1398(vH)2=104.81(1)Coefficient of Restitution:Here,(vA)2=v2(3) T.Thus, A+cBe=(vA)2-(vH)2(vH)1-(vA)10.5=-v2(3)-(vH)2-75-03 v2+(vH)2=37.5(2)Solving Eqs.(1) and (2),v2=17.92 rad>s (vH)2=-16.26 ft>s=16.26 ft>s TConservation of Energy:With reference to the datum in Fig.b,V2=AVg= WD(yG)2=0and .gB3=WD(yG)3=50hB2V3=AVAvDB2=v2(1)=17.92(1)=17.92 ft>sand (vD)3=0Thus,T12=2 m)150D(vD2 2=2 a32.2bA17.922B=249.33 ft#lband T3=0ThenT2+V2=T3+V3249.33+0=0+50hh=4.99 ftAns.82091962_09_s19_p0779-0826 6/8/09 5:01 PM Page 821© 2010 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.19–51.The disk has a mass of 15 kg.If it is released from150 mmrest when u=30°,determine the maximum angle uofrebound after it collides with the wall.The coefficient ofrestitution between the disk and the wall is e=0.6.WhenCu=0°,the disk hangs such that it just touches the wall.Neglect friction at the pin C.150 mmuDatum at lower position of G.T1+V1=T2+V20+(15)(9.81)(0.15)(1-cos 30°)=1v 2 c32 (15)(0.15)2dv2+0=3.418 rad>s a:+be=0.6=0-(-0.15v¿)3.418(0.15)-0v ¿=2.0508 rad>sT2+V2=T3+V31 c3 (15)(0.15)2d(2.0508)222+0=0+15(9.81)(0.15)(1-cos u)u=17.9°Ans.82191962_09_s19_p0779-0826 6/8/09 5:01 PM Page 822© 2010 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.*19–52.The mass center of the 3-lb ball has a velocity ofz(vG)1=6 ft>swhen it strikes the end of the smooth 5-lbslender bar which is at rest.Determine the angular velocity2 ftAof the bar about the zaxis just after impact if e=0.8.2 ftBO0.5 ft(vG)1 ϭ 6 ft/sGConservation of Angular Momentum:Since force Fdue to the impact is internaltothe system consisting of the slender bar and the ball,it will cancel out.Thus,angularmomentum is conserved about the zaxis.The mass moment of inertia of the slender r ϭ 0.5 ftbar about the zaxis is I15(yB)2z=12 a32.2b19–17,we haveA42B=0.2070 slug#ft2.Here,v2=2.Applying Eq. (Hz)1=(Hz)2 Cmb (yG)1D(rb)=Iz v2+Cmb (yG)2D(rb) a332.2b(6)(2)=0.2070c(yB)22d+a332.2b(yG)2(2)(1)Coefficient of Restitution:Applying Eq.19–20,we havee=(yB)2-(yG)2(yG)1-(yB)10.8=(yB)2-(yG)26-0(2)Solving Eqs.(1) and (2) yields(yG)2=2.143 ft>s (yB)2=6.943 ft>sThus,the angular velocity of the slender rod is given byv(yB)26.9432=2=2=3.47 rad>sAns.82291962_09_s19_p0779-0826 6/8/09 5:01 PM Page 823© 2010 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.•19–53.The 300-lb bell is at rest in the vertical positionbefore it is struck by a 75-lb wooden post suspended fromtwo equal-length ropes.If the post is released from rest atu=45°,determine the angular velocity of the bell and thevelocity of the post immediately after the impact.Theuu3 ftcoefficient of restitution between the bell and the post is4.5 fte=0.6.The center of gravity of the bell is located at pointGand its radius of gyration about Gis .kG=1.5 ftGConservation of Energy:With reference to the datum in Fig.a,V1=AVg= -W(yB1G)1=-75(3 cos 45°)=-159.10 ft#lband V2=AVgB2=-W(yG)2= -75(3)=-225 ft#lb.Since the post is initially at rest,T1=0.The post undergoes curvilinear translation,T11752=2 m(vP)2 2=2c32.2d(vP)2 2.Thus,T1+V1=T2+V20+(-159.10)=1752 c32.2d(vG)2 2+(-225)(vP)2=7.522 ft>sConservation of Angular Momentum:The sum of the angular impulses about pointOis zero.Thus,angular momentum of the system is conserved about this pointduring the impact.Since the bell rotates about point O,(vG)3=v3rOG=vThe mass moment of inertia of the bell about its mass center is IG=13(4.5). mkG 2= 3001232.2 A1.52B=20.96 slug#ft2.Thus, (HO)2=(HO)3 7532.2 (7.522)(3)=3007532.2 Cv3(4.5)D(4.5)+20.96v3-32.2(vP)3(3)209.63v3-6.988(vP)3=52.56(1)Coefficient of Restitution:The impact point Aon the bell along the line of impact (xaxis) is C(vA)3C(Dx=v3(3).Thus, =vA)3(vDx-(vP)3P)2-C(vA)2Dx A:+B e0.6=-v3(3)-(vP)3-7.522-03 v3+(vP)3=4.513(2)Solving Eqs.(1) and (2),v3=0.365 rad>s (vP)3=3.42 ft>sAns.82391962_09_s19_p0779-0826 6/8/09 5:01 PM Page 824© 2010 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.19–54.The 4-lb rod ABhangs in the vertical position.A 2-lb block,sliding on a smooth horizontal surface with avelocity of 12 ft>s,strikes the rod at its end B.DetermineAthe velocity of the block immediately after the collision.Thecoefficient of restitution between the block and the rod at Bis .e=0.83 ft12 ft/sBConservation of Angular Momentum:Since force Fdue to the impact is internaltothe system consisting of the slender rod and the block,it will cancel out.Thus,angular momentum is conserved about point A.The mass moment of inertia of the slender rod about point Ais .I1 a4bA32B+4A= A1.52B=0.3727 slug#1232.232.2ft2Here,v(yB)22=3.Applying Eq.19–17,we have (HA)1=(HA)2 Cmb (yb)1D(rb)=IA v2+Cmb (yb)2D(rb) a232.2b(12)(3)=0.3727c(yB)223d+a32.2b(yb)2(3)(1)Coefficient of Restitution:Applying Eq.19–20,we havee =(yB)2-(yb)2(yb)1-(yB)1 A:+B0.8=(yB)2-(yb)212-0(2)Solving Eqs.(1) and (2) yields (yb)2=3.36 ft>s :Ans.(yB)2=12.96 ft>s :82491962_09_s19_p0779-0826 6/8/09 5:02 PM Page 825© 2010 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.19–55.The pendulum consists of a 10-lb sphere and 4-lbrod.If it is released from rest when u=90°,determine theangle uof rebound after the sphere strikes the floor.Takee=0.8.0.3 ft2 ftOu0.3 ftI1 a32.2b(2)2+21010A=45a32.2b(0.3)2+a32.2b(2.3)2=1.8197 slug#ft23Just before impact:Datum through O.T1+V1=T2+V20+4(1)+10(2.3)=1 (1.8197)v22+0v2=5.4475 rad>sSince the floor does not move, A+cB v=2.3(5.4475)=12.529 ft>se=0.8=(vP)-00-(-12.529)( vP)3=10.023 ft>sv3=10.0232.3=4.358 rad>sT3+V3=T4+V412 (1.8197)(4.358)2+0=4(1 sin u1)+10(2.3 sin u1)u1=39.8°Ans.82591962_09_s19_p0779-0826 6/8/09 5:02 PM Page 826© 2010 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.*19–56.The solid ball of mass mis dropped with avelocity v1onto the edge of the rough step.If it reboundshorizontally off the step with a velocity v2,determine theangle uat which contact occurs.Assume no slipping whenthe ball strikes the step.The coefficient of restitution is e.uvr1v2Conservation of Angular Momentum:Since the weight of the solid ball is anonimpulsive force,then angular momentum is conserved about point A.The mass moment of inertia of the solid ball about its mass center is I2G= mr2.Here,v=y2 cos u52r.Applying Eq.19–17,we have (HA)1=(HA)2 Cmb (yb)1D(r¿)=IG v2+Cmb (yb)2D(r–) (my2y2 cos u1)(r sin u)=a5 mr2barb+(my2)(r cos u) y2y=57 tan u(1)1Coefficient of Restitution:Applying Eq.19–20,we havee=0-(yb)2(yb)1-0e=-(y2 sin u)-y1 cos uy2e cos y=u(2)1sin uEquating Eqs.(1) and (2) yields 5e cos u7 tan u=sin u tan2 u=75 eu=tan-1 ¢7A5e≤Ans.826
