Physics
posted by José .
A mass m=400g hangs from the rim of a wheel of radius r=15cm. When released from rest, the mass falls 2.0m in 6.5sec. Find the moment of inertia of the wheel

In 6.5 seconds, potential energy loss = 0.4 kg*2.0m *g = 7.84 J
Final velocity of mass:
Vf = 2*2m/6.5s = 0.6154 m/s
(twice the average velocity)
Final angular velocity of wheel:
Wf = Vf/R = 4.10 rad/s
K.E. gain = P.E loss
(M/2)Vf^2 + (I/2)Wf^2 = 7.84 J
Solve for moment of inertia, I.
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