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Question

(B) For a solid sphere of mass $M$ and radius $R$ rolling without slipping on a horizontal plane,the translational kinetic energy $(K_t)$ is given by

Vedclass · Accepted Answer

$5/2$

Vedclass · Answer

(B) For a solid sphere of mass $M$ and radius $R$ rolling without slipping on a horizontal plane,the translational kinetic energy $(K_t)$ is given by $K_t = \frac{1}{2} M v^2$.The rotational kinetic energy $(K_r)$ is given by $K_r = \frac{1}{2} I \omega^2$,where $I = \frac{2}{5} M R^2$ is the moment of inertia about the center of mass and $\omega = v/R$ is the angular velocity.Substituting the values: $K_r = \frac{1}{2} 	imes (\frac{2}{5} M R^2) 	imes (\frac{v}{R})^2 = \frac{1}{5} M v^2$.The ratio of translational kinetic energy to rotational kinetic energy is $\frac{K_t}{K_r} = \frac{\frac{1}{2} M v^2}{\frac{1}{5} M v^2} = \frac{1/2}{1/5} = \frac{5}{2}$.

$A$ solid sphere is moving on a horizontal plane. The ratio of its translational kinetic energy to its rotational kinetic energy is:

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