Angular speeds and a basketball

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SUMMARY

The discussion centers on calculating the fraction of total kinetic energy that is rotational for a basketball rolling with a constant linear speed V. The total kinetic energy (KE) is the sum of translational KE, given by the formula 1/2 MV², and rotational KE, expressed as 1/2 Iω². The correct answer for the fraction of rotational kinetic energy is 2/5, derived under the assumption of rolling without slipping, where V equals ωR.

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anil
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I don't know how to solve the following problem:

A basketball rolls along the floor with a constant linear speed V. (a) Find the fraction of its total kinetic energy that is in form of rotational kinetic energy about the center of ball.

I think the answer is 2/5. But i don't know how to do it.
 
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The total KE is the sum of the translational and rotational KE (about center of mass).
The translational KE = 1/2 MV2
The rotational KE = 1/2 Iω2, where I is the rotational inertia of a spherical shell about its center (look it up!)

One can assume that it is rolling without slipping, which means that V = ωR.
 

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