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Unformatted text preview: 8.22 Choose 135‘gr = O at the level of the base of the ramp. Then, conservation of med1anical
energy gives .(Kgm'JrJ—(EwtwegL 2005,”,m +1610, +1353)“ 152.2 Ram? (3.0 m)2(3.0 rad/s) :__1__.:__L_=—___=2
0” gsine gsins (9.80n1/52)sil120° 8.73 Choose an axis perpendicular to the page and passing through
the center of the cylinder. Then, applying 2': = Ia to the cylinder gives (2T)  R = EMRZJangszgJ, or T = imp _ (1) Now apply = my to the falling objeots to obtain (2m)g—2T=(2m)a,,or a, =g—:%. (2) (a) Substituting equation (2) into (1) yields “3 Mg M ' . Mm
=—— — T, luch T: ' .
4 [4m] w reduces to M + 4m (b) From equation (2) above, a_ __l Ming _ _ Mg _ 4mg
“3 m M+4m g M+4m M+4m 8.66 (a) The center of each wheel moves forward at v = 3.35 m/s and each wheel also turns at angular speed a) = ‘0/ R . The total kinetic energy of the bicycle is
KB = KENS + KB”, , or '  KE IE gmmﬂmwwvz +2£§ImmZJ
1 1 2
3(mﬁmg +211’1m,m,1)'o2 +E(%R2)[%] fl =%[8.44 kg + 3(0.820 1<g)](3.a=sm,/s)2 = (b) In this case, the t0p of each roller moves forward at v = 0.335 m/s . The center of each roller moves forward at 12/2 ; 0.168 m/s and each roller also turns at angular speed aria—225%. The total kinetic energyis KE=KBM +I<Em,or 1 1 '0 2 1
1 1 1 :32
{3mm +mejv2 +EmmR2[4R2] I This gives KE=§£mM +Emmjvz, or 3 1 I I 2
KE =i844 kg +Z[82.0 kg)](0.335m/s) = \ 8.65 Let mp be the mass of the pulley, mI be the mass of the sliding block, and m2 be the
mass of the counterweight. ' I (a) The moment of inertia of the pulley is I = 321—31? R: and its angular velocity at any time is a) = 1 , Where v is the linear Speed of the other masses. The friction force retarding the sliding block is fk 2 than = A (m1 g) . Choose PBS. 2 0 at the level of the counterweight when the sliding mass reaches the
second photogate. Then, from the Workkinetic energy theorem, Wm = [REM +Ksm +PEng — (mm +Ksm +1353) “ft ‘3=%(mi +n12)'0§ FR:J[%]+O gems gg eta—ms, . 1 1 
or %[ml +mz +§mr]vi =§[ml +ne+§mrlvf+ngs“ﬂk("ﬁgl's' 2 — s
Thisreducesto of: vf+WI 7111 +1712 +5171?
and yields _ 0820 m 2+2(0.2081<g)(9.80 m/SZXOJUG m) =
Uf " [ ' ?l 1.45 kg va __ Ill/S H
(b) (Orgm 1.63 m/s . s mononﬁszsaa]
7.1 9=—=% =3.2><io8 .
(a) , m
 {lrevl
9:3.2 18 d =5.0><107 
a» . 7.3 The Earth moves through 21: rad in one year (3.156 X 107 s] . Thus, 27zrad
w=———u= 1.99x10‘7 d .
3156x1075 Alternatively, the Earth moves through 360° in one year (365.242 days). 360°  
Th , =—_= 0.986 d :1
us 4’ 365.2 days eg/ ay \ are" andﬁnd 7.4 we use a: 0.20 rev/ s—0[ 2:: rad] I ‘2
a=__ = 4.2 1 2
305 ha, . 7.10 We will break the motion into two stages: (1) an acceleration period and (2) a
deceleration period. The angular displacement during the acceleration period is 91 :ﬁ=[wf:mi}t=[(5.0 rev/s)(27;rad/1rev)+0
[8.0 s] = 126 rad , and while decelerating, a=[m;:s]f=[EEW%ELWJ(IQS)=188rad. The total displacement is $91 + 6'2 = [(126 +188) rad][ 1 rev ) = 2;: rad 7.13 The pulsar must rotate with angular velocity a) = 15 rev/s (i.e., onehalf revolution per
ﬂash). Thus, from v = r a), the maximum radius is ' om#3.00x103 m/sf 1rev\_ 6
Tm =—a)—_“#l5—re#V/sﬁ—_L2xradJ—n With mass greater than 1.4 times the mass of the Sun and radius comparable to the
Earth, the object is thought to be a neutron star. ...
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 Fall '07
 Dr.Jackman
 Physics

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