A solid sphere of mass M and radius R is rota | vedclass

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(D) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$. For a solid sphere rotating about its diameter,the moment of inertia is $I

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$1: 5$

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(D) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$. For a solid sphere rotating about its diameter,the moment of inertia is $I_{	ext{sphere}} = \frac{2}{5} M R^2$. Thus,$K_{	ext{sphere}} = \frac{1}{2} \left( \frac{2}{5} M R^2 ight) \omega_{	ext{sphere}}^2 = \frac{1}{5} M R^2 \omega_{	ext{sphere}}^2$. For a solid cylinder rotating about its geometrical axis,the moment of inertia is $I_{	ext{cylinder}} = \frac{1}{2} M R^2$. Given that $\omega_{	ext{cylinder}} = 2 \omega_{	ext{sphere}}$,the kinetic energy is $K_{	ext{cylinder}} = \frac{1}{2} \left( \frac{1}{2} M R^2 ight) (2 \omega_{	ext{sphere}})^2 = \frac{1}{4} M R^2 (4 \omega_{	ext{sphere}}^2) = M R^2 \omega_{	ext{sphere}}^2$. The ratio is $\frac{K_{	ext{sphere}}}{K_{	ext{cylinder}}} = \frac{\frac{1}{5} M R^2 \omega_{	ext{sphere}}^2}{M R^2 \omega_{	ext{sphere}}^2} = \frac{1}{5}$.

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