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(C) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.For a solid cylinder rotating about its geometrical axis,the moment of iner

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$1: 5$

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(C) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.For a solid cylinder rotating about its geometrical axis,the moment of inertia is $I_c = \frac{1}{2} M R^2$.Let the angular speed of the cylinder be $\omega_c = \omega$.So,$K_c = \frac{1}{2} (\frac{1}{2} M R^2) \omega^2 = \frac{1}{4} M R^2 \omega^2$.For a solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} M R^2$.The angular speed of the sphere is $\omega_s = \frac{\omega}{2}$.So,$K_s = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{\omega}{2})^2 = \frac{1}{5} M R^2 (\frac{\omega^2}{4}) = \frac{1}{20} M R^2 \omega^2$.The ratio of the kinetic energy of the sphere to that of the cylinder is $\frac{K_s}{K_c} = \frac{\frac{1}{20} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} = \frac{1}{20} 	imes \frac{4}{1} = \frac{4}{20} = \frac{1}{5}$.

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