Chap 8 - 46, 50, 52, 53, 49, 51, 55, 68

Chap 8 - 46, 50, 52, 53, 49, 51, 55, 68 - 8.52 8.53 The...

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Unformatted text preview: 8.52 8.53 The initial angular velocity of the system is ‘ ml. =[020 rad s 1 rev J: [1407: rad/s The total moment of inertia is given by 1 I = Im +Iqm zmr2 +%MR2 = (80 kg)?”2 +595 kg)(2.0 Inf. Initially, the man is at r = 2.0 m from the axis, and this gives If =3.7>~<102 kg- 1112. At the end, when r=1.0 In, the moment of inertia is If =1.3><102 kg-mz. (a) From conservation of angular momentum, I1F (of =Iia),. , or - I. 3.7x102 kg-mz' = —‘ - —-——-—-— 04012: Is 3 =1.14¢zz' rad s: 3.6 rad s (of [Illa [1.3x102 kg-m2 ( t / ‘2, - . . . . . .‘. - < 1 Cb) The change in kinetic energy 15 AKE=§Iwa§ ~—If (of 2 AKE=%[1.3><102 kg.-In2][1.147: 5‘1] S or AKE = 5.4:(102 J The difference is the work done b the man as he walks _ ' inward. ' 4- The table turns counterclockwise, opposite to the way the woman walks. Its angular momentum cancels that of the woman so the total angular momentum maintains a constant value of [WI = [m +1.We =0. (3) Since the final angular momentum is LW =Iwww+fiwt =O,we have 2 ' I _ mwr _ no: 3 [ It ] r If W :23, = ~[WJELEO m/s) _= 0.360 rad/s. If OI' Hence mm =F].3’60 rad/s counterclockwise—l. 1 1 (b) I’ll,”t =AKE=KEf —O=Emofp +3120? 1 2 1 2 2 WM =§(60.0 kg)(1.50 m/s) +§(5oo kg«m )_(G.360 rad/s) = 8.52 The angular velocity of the system is mi = (0.20 27: rad] = 0.407: rad/s _ s 1 rev - The total moment of inertia is given by I: :m “gm = m2 gm: = (80 kgw +%(25 kg)(2-0 Inf. Initially, the mania at r221} :11 from the axis, and this gives I,- =3.7><102 kg-mz. At the end, when r = 1.0 m, the moment of inertia is If = 1.3)(102 kg - mg. (a) From conservation of angular momentum I f mi = I" (oi , or 1,. 3.7x102 kg- 1112' M _ a)Jr (0.40:: rad/s)—1.14?zr radfs-w 3.6 rad/s “t. (b) The mkmeficne'nfigyis m:%frw§ “%Ifwi2r or 2 2 rad 2 1 2 2 rad 2 AKE=§(1.3><10 1(ng ) 1.147r_ —§(3.7x10 kgm) 0.407:-— , I s s or 3105 = 5.4x 102 J The difference is the work done b the man as he walks ' inwatd. a. 8.53 (a) The table turns counterclockwise, opposite to the way the woman walks. Its angular momentum cancels that of the woman so the total angular momentum maintains a constant value of Liam : Lm + Law 2 0 . Since the final angular momentum is [W] = I” ww +1: (at = O , we have or a) = _[(60.0 kg) (2.00 m) 1 1 2 2 Wm =§(60.U kg)(1.50 m/sf +§(500 kg-m )(0-360 rad/S) = Initially, r =10 m, and 1,:2 [(3.0 kg)(1.0 m)2]+ 3.0 kg- m2 = 9.0 kg - m2 4 Afterward, r = 0.30 m , so If: 2{(3.0 kg)(0.30 m)2]+3.0“kg - m2 :35 kg - m2 (a) From conservation of angular momentum, If m). = If a} , or _ _ IL H 9.0kg-m2 <0. {Who-75 rad/s)- 1 1 (13) KB:- =§Iia)i2 =—2'(9-01_<g-m2)(0.75 rad/s)2 = 1 1 ' Kgf =§Ifwf =5(35 kg-m2)(1.9 rad/5 )2 =m .‘ . .. . . ... . v... . .'. ...._ .. ... __ ....U- .._. 8.51 Th hal 1 fth ck w.=—‘ ——=2.00 ~—. elm angularve outyo epu 15 a 1} _ 0.400111 S fl Since the tension in the string does not exert a torque about the axis of revolution, the angular momentum of the puck is conserved, or If (of = IiQJi . * 2 2 ' Thus, :2)! = m. = 3352- cq. =(0‘ m] (2.00 rad/s)=5.12 rad/s If mrf 0.250 In I_ The net work done on the puck is 1 1 1 ' Wm 2 1031' '35: =50“? “31:- 03 z Eumrflmfi " (NEW ] :23???“ G20? 1 r or Wmt =Q£2:_kgl[(0.250 m)2(5.12 rad/sf—(OAOO mf (2.00 rad/5):] maids w. = 8.55 (a) From conservation of angular momentum, I f Q)! = I . a) so 0f: — mi: we . If 11H“2 . 2 1 1 I I 1 I KE =—I 2=— I +1 I (222.: 1 [~1 of]: 1 (b) ’ 2fwf 2(1 2)[11+I2] " {11+IZJ21 ° 11+}2 KEf I1 01' —— = Since this is less than 1.0, kinetic energy was lost. KEI. I1+ I2 8.68 We treat each astronaut as a point mass,“ m, moving at speed a in a circle of radius 1*. Then the total angular momentum is I L: Iim+Izw=2[(mr2)[Efl:2mm-I r (a) L. 22mg; = 2(75.0 kg) (500 'm/sj-(EDO In) 11: 3.75x103 kg-mz/s 1 1 1 ’ (b) K5: ='2"’117’12: +“2“mzvi- :2[§mvfj K5,. = (75.0 kgx'sm m/s)2 = 1.88 x103 J: (c) Angular momentumis conserved: .1“.JF =Li= 3.75x103 kg-mz/s L 3.75x103 kg-{nE/s d = f = . : ( ) 0’ 2(mrf) 2(75.o kg)(2.50 m) 1 (e) KB}. = 4511;123:050 kg)(10.0 m/sf = .. (1) Wm = KE; “KEI- = 8.46 Using conservation of angular momentum, 'chm = L]: E' H, mm . Thus, (mafiwa = (mr; )wp. Since a) = E at both aphelion and perihelion, this is _ r equivalent to = (111??) Z—p , givmg a r vs: {:1}? =(%$J(M Ian/ss- 8.48 From conservation of angular momentum, If (pi = 12101. . Treating the child as a point mass, this betromes 0),, = (Ii/If) (2),. _ 2 I or (of = - 2250 kg _ 2 (10.0 rev/min) = 7.14 rev/min 250 kg - In +(25.0 kg)(2.00 In) 8.49 The moment of inertia of the cylinder before the putty arrives is Ii=§1—MR2 =%(1o.o kg)(1.00 Inf = 5.00 kg- 1112 -‘~ After the putty sticks to the cylinder, the moment of inertia is If: 13+er 2 5.00 kg- m2 + (0.250 kg)(0.900 m)2 £5.20 kg- m2 Conservation of angular momentum gives If (of 21.0) [U I: 5.00 kg- m2 01' (0 = H (a = 7. 2 . f [If] , [5'20 kgmz 00 rad/s) 6 73 rad/s ...
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This note was uploaded on 04/07/2012 for the course PHYS 111 taught by Professor Dr.jackman during the Fall '07 term at Waterloo.

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Chap 8 - 46, 50, 52, 53, 49, 51, 55, 68 - 8.52 8.53 The...

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