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(D) The moment of inertia of a solid sphere about its diameter is $I_{	ext{sphere}} = \frac{2}{5} M R^2$. The rotational kinetic energy is $K_{	ext{

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$1: 5$

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(D) The moment of inertia of a solid sphere about its diameter is $I_{	ext{sphere}} = \frac{2}{5} M R^2$. The rotational kinetic energy is $K_{	ext{sphere}} = \frac{1}{2} I_{	ext{sphere}} \omega_{	ext{sphere}}^2 = \frac{1}{2} 	imes \frac{2}{5} M R^2 \omega_{	ext{sphere}}^2 = \frac{1}{5} M R^2 \omega_{	ext{sphere}}^2$.The moment of inertia of a solid cylinder about its geometrical axis is $I_{	ext{cylinder}} = \frac{1}{2} M R^2$. Given the angular speed $\omega_{	ext{cylinder}} = 2 \omega_{	ext{sphere}}$,the rotational kinetic energy is $K_{	ext{cylinder}} = \frac{1}{2} I_{	ext{cylinder}} \omega_{	ext{cylinder}}^2 = \frac{1}{2} 	imes \frac{1}{2} M R^2 (2 \omega_{	ext{sphere}})^2 = \frac{1}{4} M R^2 (4 \omega_{	ext{sphere}}^2) = M R^2 \omega_{	ext{sphere}}^2$.The ratio of their kinetic energies 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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