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(B) For a solid sphere,the moment of inertia is $I = \frac{2}{5}MR^2$. The rotational kinetic energy is $K_R = \frac{1}{2}I\omega^2 = \frac{1}{2}(\fra

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$2:7$

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(B) For a solid sphere,the moment of inertia is $I = \frac{2}{5}MR^2$. The rotational kinetic energy is $K_R = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{5}MR^2\omega^2$. Since the sphere is rolling without slipping,$v = R\omega$,so $K_R = \frac{1}{5}Mv^2$. The total kinetic energy is $K_{Total} = K_{Translational} + K_{Rotational} = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$. The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_R}{K_{Total}} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.

If a solid sphere is rolling,the ratio of its rotational kinetic energy to the total kinetic energy is given by

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