hw12 - PH 131 FALL 2005 HOMEWORK 12 Assigned «it I0 9,1...

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Unformatted text preview: PH 131 FALL 2005 HOMEWORK # 12 Assigned: 11/14/05 «it» I0: 9,1, Inf-1,37); Hanan, Mimi/20 (No? chm, wan? if) Theangularposition ofapoimtmmefim ofarotatingwheel is given by 9 = 4.0: — 3.0:2 + :3, where 0 is in radians and r is in seconds. What are the angular velocities at (a) t = 2.0 s and (b) t = 4.0 s? (c) What is the average angular accelcmtion for the time interval that begins at: = 2.0 s and ends at r = 4.0 5? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval? a) LO: 0'9 : 0.3- (4.91, —5-0“7’2’-H73) w a 4.0 +01:- +31?" a. CDC-£22,.OA) :: 4.0 —g.5 (1,0) +3 (1.0-)1: 1’) ”0:400: 4*O-E-o (1113).”, (l-l-o)": 29 Ms C) Ava. 0M6. “CCEQQ’Ht'M : I .42 _ 032-“)! 2 ‘23 “'4'0) I'M/6 4": {11"4“ L4.0~2-€D"> al) °<__ 093 .2 $00 Goé+34921—6-04662 «(they 2 m” sew) =— 9 ago” 4.02.) = —6-o+e(4-5) :- XI The wheel in F13. 11-29 has " eight equally spaced spokes and a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5 owls. You want to shoal a 20-m-long arrow parallel to this axle and through the wheel without hining any of the spokes. As- sume dull the arrow and the spokes are very Ihin. (a) What mini- mum speed must the snow have? (13) Does it matter where between the axle md rim of the wheel you aim? If so. what is [he best location? cl. [0— n ® SmfingfmmmLadiskmmtitscflumluiswim . , _ mummgularlccelerufiomlnS.Os,itrotate525md.Duringlhat M C00 -0 time.whatmthemagufimdsof{a)fl1emgulumlemionand 8 =_ O (b) It: average. angular velocity? (c) What is the instantaneous O angular velocity ofthc disk at the end of the 5.0 s? (d) With the 9 =25 7.152. J J: : 50/) - angular acceleration unchanged. through what additional angle will thedisktumduringflwnemSfls? [31‘de a)o( ’ L) wavfi,c) a O Cl.» [0- '2} @(3)Whuismemguluspeeduaboutthepolaraxisofapoim [fit f “MS 041-9 ' ’ssulfaceaxalatitudeofwmmmthmmesahomthat axis.)(h)Whalisfl1elinequeed vofthepoim? Wham(c)w g: 400 19— AMA/Mud and(d)v fora pointanhcequawr? ssm \J = "i “ l - UP SECOSLIO ‘0 ~ G’shmém 630549.9(7'1W’054A P r: - x10 4M5 —/ E] (Pa = SAME A6 EFOQE E} v: - 12a) (5 3;:an any? axle) raw/6) E ck \0—33 @ Calculate the rotational inertia of a wheel that has a kinetic energy of244m I when routing at 602 rcvfmiJL ssm ' cu, 10—3—3 @mmandooordinmoffourpufidesmns follows: Glf‘lw: SOg.x=2.0cm.y=2.0m;253.x=0.y=4.0m;25g,I x=-3.0m.y=-3.0m;30g,x-=-2.0an,y=4.0cm. ml: 508 Whatmmemufionalimrfiasafdfisooflecfionabomdw(n)x._ (b)y.and(c)zucs?(d)Supposclheanswersm(a)and(b)mA _ . _ ”V12. " 2‘5"} and B. respectively. Than what Isthc mwcrm (c) 1n terms of . AnndB? L “(=1 - : [go (1.0 ‘4 {mfl(o’)"+ 131313)";L '30 ("“9113 (44?" : 957 1013. (m ckla— [{1— @fl: body in Fig.11-37 is pivolndato,andtwofmm unitasshown.(a) Fmdancx- pl'cssionforlhenettorqueondae . s . __ l mmtmepim (bunt: mnsrnobmn- qufl. "T- 1.7: l.30m,r2=2.15m.F.=4.20N.F,=4.90N.Bl=75.0°,and ; m 4.091% 02-60.0'.whuismemtlmqneabwlthcpivot? gm Ilw_ Ii) :YLFlgimgz _ F? =“E’4m .A pulley. with a national Inertia of 1.0 x 10" kg om: about its axle and a radius of 10 cm. is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F s ' 0.50! + 0.309. with F in newth and r in seconds. he pulley is initiaily at rest. A: r = 3.0 a what an: (a) its angular acceleration and (b) its angular speed? ckio 451- @A uniform spherical shell of mass Maud radius R rotates about a vertical axis on frictionless hearings (Fig. 11-45). A massiessaxd passes around the equator of the shell. over a pulley of rotational inertia I and radius r. and is attached to a small object of mass m. Then: is no friction on the pulley‘s axle: the cord does not slip on the pulley. What is the speed of the object after it fails a distance i: from mast? Use energy considerations. Ck 10*‘33 33 AwheelofndiusOJOm is matedmafrictioufiesshm‘i- zonal axis. Th: rotational inenia of {he wheel about the axis is 0.050 113- m2. A masslcss cord wrapped around the wheel is at— tached to a 2.0 kg block than slid-5 on a horizontal frictionless surface. If a hofiZOIIta] force of magnitude P = 3.0 N is applied to the block as shown in Fig. 11-41, F0? [Ur Bloc}? “-3sz law : ZFK = max 'P—T -_—. ”V162. For no. DOM! Mew-Inn‘s lam: °<-= at/Rb‘fi/B Giflwt {m 22.9 “3 {i=1-adim 44 wkmkdzau I :OoOSONa-On'z— [Y week's «enigma—£5001 , x 2 Q )- P = 3.0” _ F't'rnot: cg 0 A communications satellite - is a solid cylinder with mass 1210 kg, diameter 1.21 m, and length 1.75 m. Pn'or to launching from the shulxle cargo buy, it is set spinning at 1.52 rev}: about the cylinder axis (Fig. 11-33). Ca]- culate the satellite-’5 (a) rotational inertia about the rotation axis and (b) rotational kinetic energy. Z. '1. '2. ad- L) K194, 3‘ I00 #51}le E30 7;“:- > 5 You! lGEThflywhadofmenfineismmingazsnmm 1...), :— 25“9 ”wt/5 MafinkmfimflMMuumm LO __ 0 mdmmmafluflflamflflhcwm- 1- " fimfinndffiofdnflyflmLtb)memglefinnd)dzmgh .é _£ :- Q0 .0A . mmnmmmmmmumnnmba ,2. I (3)41; ”0.9:; «ev' 0") of: Am Mafia” ,_ "ZSOTM/I’L 4t- {71—4-4 200‘6 721’. A‘rlndofndinsommismmwdmnmhod- mmammkwmumm Wmnmuobjemmfidumamm inclineduanangleol'm‘wi ' .Isstnwninfig. II'JZWIOW .- - - I - -Mga‘ng + T= --- M04 ...
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