given
m_{1} = 0.48 kg
m_{2} = 0.82 kg
M = 0.35 kg
R_{1} = 0.02 m
R_{2} = 0.3 m
_{k} = 0.25
v_{i} = 0.82 m/sec
a )
d = 0.7 m away
I = 0.5 M ( R_{1}^{2} + R_{2}^{2} )
= 0.5 x 0.35 x ( 0.02^{2} + 0.03^{2} )
= 0.0002275
K_{1i} = 1/2 m_{1}v_{i}^{2}
= 0.5 x 0.48 x 0.82^{2}
= 0.1613 J
U_{1}_{i} = m_{1} g d
= 0.48 x 9.8 x 0.7
U_{1}_{i} = 3.2928 J
K_{2i} = 1/2 m_{2} v_{i}^{2}
= 0.5 x 0.82 x 0.82^{2}
= 0.275 J
K_{rot}_{i} = 1/2 I w_{i}^{2}
= 0.5 x 0.0002275 x ( v_{i}/R_{2} )^{2}
= 0.5 x 0.0002275 x ( 0.82/0.03 )^{2}
K_{rot}_{i} = 0.0849 J
the frictional force f_{k} = _{k} m_{2} g
= 0.25 x 0.82 x 9.8
f_{k} = 2 N
applying law of conservation energy is
( K_{1i} + K_{2i} + U_{1i} + U_{2}_{i} + K_{rot} i ) - f_{k} d = K_{1}_{f} + K_{2f} + K_{rot f}
( 0.1613 + 0.275 + 3.2928 + 0 ) - 2 x 0.7 = 0.5 x 0.48 x v_{f}^{2} + 0.5 x 0.82 x v_{f}^{2} + 0.5 x
0.0002275 x ( v_{f} / R_{2} )^{2}
2.3291 = 0.24 x v_{f}^{2} + 0.41 x v_{f}^{2} + 0.126 v_{f}^{2}
0.24 x v_{f}^{2} + 0.41 x v_{f}^{2} + 0.126 v_{f}^{2} = 2.3921
v_{f}^{2} = 2.3921/0.776
v_{f}^{2} = 3.0826
v_{f} = 1.755 m/sec
b )
w_{f} = v_{f} / R_{2}
= 1.755 / 0.03
w_{f} = 58.5 rad/sec
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