Angular speeds and a basketball

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To solve the problem of a basketball rolling with a constant linear speed V, the total kinetic energy consists of translational and rotational components. The translational kinetic energy is calculated as 1/2 MV², while the rotational kinetic energy is 1/2 Iω², with I being the rotational inertia of the sphere. Assuming the basketball rolls without slipping, the relationship V = ωR can be applied. The fraction of total kinetic energy that is rotational can be determined, and the answer is suggested to be 2/5. Understanding these concepts is crucial for calculating the distribution of kinetic energy in rolling objects.
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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