A body is rolling down an inclined plane. If | vedclass

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(D) Rotational kinetic energy is $K_{R} = \frac{1}{2} I \omega^{2} = \frac{1}{2} Mk^{2} \left(\frac{v}{R}\right)^{2}$ (since $I = Mk^{2}$ and $v = R \

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Solid ball

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(D) Rotational kinetic energy is $K_{R} = \frac{1}{2} I \omega^{2} = \frac{1}{2} Mk^{2} \left(\frac{v}{R}ight)^{2}$ (since $I = Mk^{2}$ and $v = R \omega$).This simplifies to $K_{R} = \frac{1}{2} Mv^{2} \left(\frac{k^{2}}{R^{2}}ight)$.Translational kinetic energy is $K_{T} = \frac{1}{2} Mv^{2}$.According to the problem,$K_{R} = 40\% K_{T}$.Therefore,$\frac{1}{2} Mv^{2} \left(\frac{k^{2}}{R^{2}}ight) = \frac{40}{100} 	imes \frac{1}{2} Mv^{2}$.This gives $\frac{k^{2}}{R^{2}} = \frac{40}{100} = \frac{2}{5}$.For a solid sphere (solid ball),the moment of inertia is $I = \frac{2}{5} MR^{2}$,so $k^{2} = \frac{2}{5} R^{2}$,which means $\frac{k^{2}}{R^{2}} = \frac{2}{5}$.Hence,the body is a solid ball.

$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

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