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December 18, 2014

December 18, 2014

Posted by **Kristen** on Sunday, January 27, 2008 at 5:18pm.

- AP Physics -
**drwls**, Sunday, January 27, 2008 at 6:19pmAfter falling a distance h, Potential energy m g h is converted to kinetic energy of the rotating shell, the falling mass, and the pulley. Let V be the velcoity of the mass at time t.

The angular velocity of the shell is ws= V/R and its moment of inertia is

Is = (2/3)MR^2

The moment of inertia of the pulley is I, its angular velocity is wp = V/r, and its radius is r.

mgh = (1/2)mV^2 + (1/2)(2/3)MR^2*(V/R)^2 + (1/2) I (V/r)^2

= V^2 [(1/2)m + (1/3)M + (1/2)I/r^2]

V^2 = 2gh/[1 + (2/3)M/m + I/(mr^2)]

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