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(B) The total kinetic energy of a rolling body is the sum of its rotational kinetic energy and translational kinetic energy.$K_{	ext{total}} = K_{	e

Vedclass · Accepted Answer

$2/7$

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(B) The total kinetic energy of a rolling body is the sum of its rotational kinetic energy and translational kinetic energy.$K_{	ext{total}} = K_{	ext{rotational}} + K_{	ext{translational}}$For a solid sphere,the moment of inertia about its center of mass is $I_{	ext{cm}} = \frac{2}{5} mR^2$.Rotational kinetic energy is given by $K_{	ext{rot}} = \frac{1}{2} I_{	ext{cm}} \omega^2$.Substituting $I_{	ext{cm}}$ and using the pure rolling condition $v = R\omega$ (or $\omega = v/R$):$K_{	ext{rot}} = \frac{1}{2} \left( \frac{2}{5} mR^2 ight) \left( \frac{v^2}{R^2} ight) = \frac{1}{5} mv^2$.Translational kinetic energy is $K_{	ext{trans}} = \frac{1}{2} mv^2$.Total kinetic energy is $K_{	ext{total}} = \frac{1}{5} mv^2 + \frac{1}{2} mv^2 = \left( \frac{2+5}{10} ight) mv^2 = \frac{7}{10} mv^2$.The ratio of rotational kinetic energy to total kinetic energy is:$\frac{K_{	ext{rot}}}{K_{	ext{total}}} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.

$A$ solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is ................

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