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(B) Given that the rotational kinetic energy $K_R$ is $50\%$ of the linear kinetic energy $K_T$. $K_R = 0.5 K_T$ Using the formulas $K_R = \frac{1}{2}

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Cylinder

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(B) Given that the rotational kinetic energy $K_R$ is $50\%$ of the linear kinetic energy $K_T$. $K_R = 0.5 K_T$ Using the formulas $K_R = \frac{1}{2} I \omega^2 = \frac{1}{2} (Mk^2) (v/R)^2 = \frac{1}{2} Mv^2 (k^2/R^2)$ and $K_T = \frac{1}{2} Mv^2$,we get: $\frac{1}{2} Mv^2 (k^2/R^2) = 0.5 	imes \frac{1}{2} Mv^2$ $\frac{k^2}{R^2} = 0.5 = \frac{1}{2}$ For a solid cylinder,the moment of inertia is $I = \frac{1}{2} MR^2$,so $k^2 = \frac{1}{2} R^2$,which implies $\frac{k^2}{R^2} = \frac{1}{2}$. Thus,the object is a solid cylinder.

If the rotational kinetic energy of an object is $50\%$ of its linear kinetic energy,then the object is:

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