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