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(B) For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5}MR^2$. Since the sphere is rolling without slipping,the velocity of t

Vedclass · Accepted Answer

$2: 7$

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(B) For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5}MR^2$. Since the sphere is rolling without slipping,the velocity of the center of mass is $v = R\omega$,where $\omega$ is the angular velocity. The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} (\frac{2}{5}MR^2) \omega^2 = \frac{1}{5}M(R\omega)^2 = \frac{1}{5}Mv^2$. The translational kinetic energy is $K_{trans} = \frac{1}{2}Mv^2$. The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{2+5}{10}Mv^2 = \frac{7}{10}Mv^2$. The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$. Thus,the ratio is $2: 7$.

If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is

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