A ball rools without slipping. The radius of gyration of the ball about – its own axis is K. If the radius of the ball is R then the fraction of total energy associated with its rotational energy will be

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$\frac{K^{2}}{R^{2}}$

$\frac{R^{2}}{K^{2} + R^{2}}$

$\frac{K^{2} + R^{2}}{R^{2}}$

$\frac{K^{2}}{K^{2} + R^{2}}$

answer is B.

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Detailed Solution

$\begin{array}{r} K . E_{R} = \frac{1}{2} {mν}^{2} (\frac{K^{2}}{R^{2}}) \\ K . E_{Toal} = \frac{1}{2} {mν}^{2} (1 + \frac{K^{2}}{R^{2}}) \end{array}\} \frac{{KE}_{R}}{K . E_{total}} = \frac{K^{2}}{K^{2} + R^{2}}$

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