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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 kgmz. (a) From conservation of angular momentum, I1F (of =Iia),. , or
 I. 3.7x102 kgmz'
= —‘  ————— 04012: Is 3 =1.14¢zz' rad s: 3.6 rad s
(of [Illa [1.3x102 kgm2 ( 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)(20 Inf. Initially, the mania at r221} :11 from the axis, and this gives I, =3.7><102 kgmz. 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 radfsw 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 kgm )(0360 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 9.0kgm2
<0. {Who75 rad/s) 1 1
(13) KB: =§Iia)i2 =—2'(901_<gm2)(0.75 rad/s)2 = 1 1 '
Kgf =§Ifwf =5(35 kgm2)(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:2mmI r (a) L. 22mg; = 2(75.0 kg) (500 'm/sj(EDO In) 11: 3.75x103 kgmz/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 kgmz/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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 Fall '07
 Dr.Jackman
 Physics

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