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August 30, 2014

August 30, 2014

Posted by **alicia** on Tuesday, July 10, 2012 at 5:23pm.

- physics -
**drwls**, Wednesday, July 11, 2012 at 12:03amThe total kinetic energy is the sum of

Translational KE = (1/2)MV^2 = 39.5 V^2, and

Rotational KE = 2*(1/2)*I*w^2

= I*V^2/R^2 = 0.978 V^2

I is the moment of inertia of a single wheel.

The fraction of the energy that is rotational is

0.978V^2/(0.978+39.5)V^2 = 0.0243

Note that the V cancels out.

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