If a solid sphere is rolling,the ratio of its | vedclass

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(A) The rotational kinetic energy of a rolling body is $K_R = \frac{1}{2} I \omega^2$.Since $I = \frac{2}{5} MR^2$ for a solid sphere and $v = R\omega

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$2 : 7$

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(A) The rotational kinetic energy of a rolling body is $K_R = \frac{1}{2} I \omega^2$.Since $I = \frac{2}{5} MR^2$ for a solid sphere and $v = R\omega$,we have $K_R = \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{5} Mv^2$.The total kinetic energy is $K_T = K_{translational} + K_{rotational} = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$.The ratio is $\frac{K_R}{K_T} = \frac{\frac{1}{5} Mv^2}{\frac{7}{10} Mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.

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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