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If a spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotation is

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$\frac{3}{5}$

$\frac{3}{7}$

$\frac{2}{7}$

$\frac{2}{5}$

answer is B.

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Detailed Solution

$\begin{array}{l} f_{r} = \frac{\frac{1}{2} I ω^{2}}{\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}}, where I = \frac{2}{5} m r^{2}, \\ ω = \frac{v}{r} \\ Solve to get f_{r} = \frac{2}{7} \end{array}$

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