Posted by **José** on Wednesday, March 21, 2012 at 10:32pm.

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

- Physics -
**drwls**, Wednesday, March 21, 2012 at 11:10pm
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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