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(A) The total kinetic energy $(K_{total})$ is the sum of translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.Given that $K_r =

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Ring

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(A) The total kinetic energy $(K_{total})$ is the sum of translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.Given that $K_r = 50\% 	ext{ of } K_{total}$,it implies $K_r = K_t$.We know $K_r = \frac{1}{2} I \omega^2$ and $K_t = \frac{1}{2} mv^2$.Equating them: $\frac{1}{2} I \omega^2 = \frac{1}{2} mv^2$.Since $v = \omega r$,we have $\frac{1}{2} I \omega^2 = \frac{1}{2} m(\omega r)^2$.Simplifying,$I \omega^2 = m \omega^2 r^2$,which gives $I = mr^2$.The moment of inertia $I = mr^2$ corresponds to a ring.

If the rotational kinetic energy is $50\%$ of the total kinetic energy,then the body is a .......... .

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