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(B) For a solid sphere rolling without slipping,the moment of inertia is $I = \frac{2}{5}mr^2$ and the condition for rolling is $v = r\omega$,which im

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$\frac{2}{7}$ rotational and $\frac{5}{7}$ translational

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(B) For a solid sphere rolling without slipping,the moment of inertia is $I = \frac{2}{5}mr^2$ and the condition for rolling is $v = r\omega$,which implies $\omega = \frac{v}{r}$.The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} (\frac{2}{5}mr^2) (\frac{v}{r})^2 = \frac{1}{5}mv^2$.The translational kinetic energy is $K_{trans} = \frac{1}{2}mv^2$.The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{5}mv^2 + \frac{1}{2}mv^2 = \frac{7}{10}mv^2$.The fraction of rotational kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.The fraction of translational kinetic energy is $\frac{K_{trans}}{K_{total}} = \frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} 	imes \frac{10}{7} = \frac{5}{7}$.Thus,the sphere has $\frac{2}{7}$ rotational and $\frac{5}{7}$ translational kinetic energy. Hence,option $(b)$ is correct.

$A$ solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

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