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Unformatted text preview: PH 131 FALL 2000 HOMEWORK # 12 Assigned! 1 1/13/00
Due: 11/29/00 CH H 45 4E. The angular position of a point on the rim of a rotating wheel
is given by 6 = 4.0:  3.0:2 + :3. where 9 is in radians and r is
in seconds. What are the angular veloeities at (a) t = 2.0 s and
(b) r = 4.0 5? (c) What is the average angular acceleration for the
time interval that begins at t = 2.0 s and ends at r = 4.0 s? What
are the instantaneous angular accelerations at (d) the beginning
and (e) the end of this time interval? b) :3 4.0 ”6.0.(: +34”)...
co G:2_.a/J) .2 4.0 195 (1,0) +3 {2.02:
5) 60 (f: 40/3) 1 4.0 90 £110) +5 (troy 2‘— 31: —“' JogJet L4v0"2€9’>
I: ll'maﬂbl 6P. The wheel in F  " H _
[g‘ I] 29 has 2.5 revls. You want It) shoot a ZDmlong arrow parallel ‘0 this eight equally spaced spokes and ‘ ' '
a radius of 30 cm. It is mounted axle and through the wheel without hitting my 0f ”“3 SPOkBS’ AS on a ﬁxed axle and is spinning at sume that the arrow and the spokes are very thin. (a) What mini—
mum speed must the arrow have? 03) Does it matter where between
the axie and rim of the wheel you aim? If so. what is the best 5) M 15 MMW 94 to “Mam" ME
7675., NW : Umﬁm ‘10 all55 Bum 0”?
ﬁt TMw5~ m. mew—e» ”*0 12E. Starting from rest, a disk rotates about its central axis with . . _ constant angular acceleration. In 5.0 s, it rotates 25 rad. During that M (1.30  0 time, what are the magnitudes of (a) the angular acceleration and e _.___ (b) the average angular velocity? (0) “mm is the instantaneous 0 angular velocity of the dist: at the end of the 5.0 s? (a) With the (Q :3 2 5 Tar}; ; l : 50/5 ' angular acceleration unchanged. through what additional angle will the disk tum during the next 5.0 s? FlMd ', a) K ’ l3) wau‘a J C)
 5 CL Bram go]. ”13 : U(5on);d) 9006) 6(5) 0
eﬁﬁ/wﬂt via?“
9 =o<+7—
525' mat/3 “ii 0< (“9091 "Be 195 :20 “wt/52"] Fm E‘l ”5) __ —_—. 325W 3 o 44):,
wanna. __ ﬁle/A1," W 5 T‘ 0 Fw'EﬁI. “12
(Q (le' 50b) 2 (1,0 Tad/bi) (5'0 5) C: COO'HXH') @ Fm E9}. ”“15
60H— m») —.= cow—goal 2 0+: (2.900? =— e~<m>~etw = (00 1:2ng = CH LI — "a: P 27F. (21) Whal is the angular speed a: about the polar axis of a point
on Earth‘s surface at a latitude of 40° N? {Earth rotates abOut that ’2'
axis.) (13} What is the linear speed v of the point? What are (c) w
and (d) v for a point at the equator? ssrn g :. \JP : @eosqoo Co : (6.3;HO‘M @0546”) (—7.193ﬂ53d/9
\j :3~§ x10 411
d 1:] (pa 2 SAME A6 EFORE 53E. Calculate the rotational inertia of a wheel that has a kinetic energy of 24400 I when rotating at 602 rovlmin. 55m 565. A communications sateiiiie '
is a solid cylinder with mass 1210
kg, diameter 1.21 m, and length
1.75 m. Prior to launching from
the shuttle cargo bay, it is at
spinning at 1.52 revls about the
cylinder axis (Fig. 1133). Cal
culate the satellite’s (a) rotational
inertia about the rotation axis and
(b) rotational kinetic energy. 42?. The masses and coordiliates of four particles are, as follows:
50g,x = 2.00m,y = 2.0cm: 25g,x = 0,y = 4.0cm; 25 g.;
x=~ —3.0r:m,y = —3.0cm;3{)g,x= ~2.0cm,y =4.Dcm.
What are the rotational inertias of this collection about the (a) x, _
(b) y. and (c) z axes? (d) Syppose the answers to (a) and (b) are A and B. respectively. Then what is the answer to (c) in terms of '
A and B? 5“ §«rxzvlta. (4n (2,. «a
a) I .‘Z 3.9m; Ml +_Z”"‘r'?; T" lXJI“
"(:l 4'1'P. The body in Fig. Il~3? is
pivoted at O, and two forces act
on it as shown. (a) Find an ex—
pression for the net torque on the
body about the pivot. (b) If r] =
1.30 m. r2 = 2.15 m, F, = 420 N, F2 = 4.90 N, a! = ?5.0°, and ; {M}? +mq‘btﬂ—
82 = 60.0°, whal is the net torque about the pivot? ssm Ilw_ . ..__\. [El :“(lﬁé‘erl M alum/d. ‘5
{TL} :YLFzéimgz‘ 54F. A wheel of radius 0.20 m is mounted on a frictionless hori
zontal axis. The rotational inertia of the wheel about the axis is
0.050 kg  n12. A massless cord wrapped around the wheel is at
tached to a 2.0 kg block that slides on a horizontal frictionless
surface. If a horizontal force of magnitude P = 3.0 N is applied to
the block as shown in Fig. 1141, ‘ what is the magnitude of the an 12 _ T L gula: acceleration of the wheel? "T"
Assume the string does not slip on the wheel. G‘NQIUV an :20 K3
R:mdxm a4? wkaeka/m @1 ock’f; acameio'a/L CHM56? 56F. A pulley, with a rotational , ’ inertia of 1.0 X 10‘3 legm2 Gllwm E 2‘ Odom
about its axle and a radius of 10 cm, is acted on by a force applied ’L
tangentially at its rim. The force magnitude varies in time as F = F = 0 ' 50 '6 + 0'30 {— A}  ' 0.50: + 0.30:2, with F in newtons and r in seconds. The pulley is .__,  5 1
initially at rest. At t = 3.0 s whalare (a) is angular acceleration (990 — O 1 I  ['OX [0 ’63 4M and {b} its angular speed? FI'MOli a) o< (5,5) 3—» b a;
E F (X: I: ff: ) (5/5)
I I
'2.
_. o 2 ' C
M,“ ( 51c +06+3(0(o) : 0'54+0»5\L73 @403
I 10'? 1‘ “Race“ —(0 _
pawl 30»): 50(5) +3003") .2 @ :: CL) 2 «59.17 :: 5 (506 +5Oé9ﬂ “(2.705 0 .— 5 3 z
jawed} +jaso+ 94+ : 50E; 5 + gafcja 2 O H
a. 7: ‘5 a
: 25B 0+ (033? CH ll— {6? 669, A uniform spherical shell of mass Maud radius R rotates about €11 m:
a vertical axis on frictionless bearings (Fig. I 145). A massless cord passes around the equator of the shell. over a pulley of rotational I __ I (ﬁrkcﬂiv‘g T)
2 a" inertia I and radius r, and is attached to a small object of mass m. There 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 falls at distance F iM ‘ U it from rest? Use energy considerations. ' __._.—— ° 3 .91 AIlﬁﬂtr‘m‘é
2. 2. T FT“
(an/amt) +é Ema—ﬁt ...
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 Physics, Work

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