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Unformatted text preview: PH 131 FALL 2005 HOMEWORK # 12 Assigned: 11/14/05 «it» I0: 9,1, Inf1,37); 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, —50“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*OEo (1113).”, (llo)": 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—604662 «(they 2 m” sew) =—
9 ago” 4.02.) = —6o+e(45) : XI The wheel in F13. 1129 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 20mlong 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 0419 ' ’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.1137 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%
0260.0'.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. 1145). 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 slid5 on a horizontal frictionless
surface. If a hoﬁZOIIta] force of magnitude P = 3.0 N is applied to
the block as shown in Fig. 1141, F0? [Ur Bloc}?
“3sz law : ZFK = max
'P—T _—. ”V162.
For no. DOM!
MewInn‘s lam: °<= at/Rb‘ﬁ/B Giﬂwt {m 22.9 “3
{i=1adim 44 wkmkdzau
I :OoOSONaOn'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. 1133). 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—44 200‘6 721’. A‘rlndofndinsommismmwdmnmhod
mmammkwmumm
Wmnmuobjemmﬁdumamm
inclineduanangleol'm‘wi ' .Isstnwninﬁg.
II'JZWIOW .   I  Mga‘ng + T=  M04 ...
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 Fall '08
 wick
 Physics

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