given
m1 = 0.48 kg
m2 = 0.82 kg
M = 0.35 kg
R1 = 0.02 m
R2 = 0.3 m
k = 0.25
vi = 0.82 m/sec
a )
d = 0.7 m away
I = 0.5 M ( R12 + R22 )
= 0.5 x 0.35 x ( 0.022 + 0.032 )
= 0.0002275
K1i = 1/2 m1vi2
= 0.5 x 0.48 x 0.822
= 0.1613 J
U1i = m1 g d
= 0.48 x 9.8 x 0.7
U1i = 3.2928 J
K2i = 1/2 m2 vi2
= 0.5 x 0.82 x 0.822
= 0.275 J
Kroti = 1/2 I wi2
= 0.5 x 0.0002275 x ( vi/R2 )2
= 0.5 x 0.0002275 x ( 0.82/0.03 )2
Kroti = 0.0849 J
the frictional force fk = k m2 g
= 0.25 x 0.82 x 9.8
fk = 2 N
applying law of conservation energy is
( K1i + K2i + U1i + U2i + Krot i ) - fk d = K1f + K2f + Krot f
( 0.1613 + 0.275 + 3.2928 + 0 ) - 2 x 0.7 = 0.5 x 0.48 x vf2 + 0.5 x 0.82 x vf2 + 0.5 x
0.0002275 x ( vf / R2 )2
2.3291 = 0.24 x vf2 + 0.41 x vf2 + 0.126 vf2
0.24 x vf2 + 0.41 x vf2 + 0.126 vf2 = 2.3921
vf2 = 2.3921/0.776
vf2 = 3.0826
vf = 1.755 m/sec
b )
wf = vf / R2
= 1.755 / 0.03
wf = 58.5 rad/sec
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