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(B) For a solid sphere,the moment of inertia is $I = \frac{2}{5}MR^2$. The total kinetic energy $K_{total}$ is the sum of translational kinetic energy

Vedclass · Accepted Answer

$2/7$

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(B) For a solid sphere,the moment of inertia is $I = \frac{2}{5}MR^2$. The total kinetic energy $K_{total}$ is the sum of translational kinetic energy $K_T$ and rotational kinetic energy $K_R$. $K_T = \frac{1}{2}Mv^2$ and $K_R = \frac{1}{2}I\omega^2$. Since the ball rolls without slipping,$v = R\omega$,so $\omega = v/R$. Substituting $I$ and $\omega$: $K_R = \frac{1}{2} (\frac{2}{5}MR^2) (\frac{v}{R})^2 = \frac{1}{5}Mv^2$. The total kinetic energy is $K_{total} = K_T + K_R = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$. The fraction of energy associated with rotational motion is $\frac{K_R}{K_{total}} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.

$A$ spherical ball is rolling on a table without slipping. What fraction of its total kinetic energy is associated with its rotational motion?

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