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(D) For rolling without slipping,the velocity is $v = r \omega$. Total kinetic energy $E = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2}

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Thin circular ring

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(D) For rolling without slipping,the velocity is $v = r \omega$. Total kinetic energy $E = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$. Substituting $v = r \omega$,we get $E = \frac{1}{2} I \omega^2 + \frac{1}{2} m r^2 \omega^2 = \frac{1}{2} (I + m r^2) \omega^2$. Rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2$. Given that $K_{rot} = 50 \% 	ext{ of } E$,so $K_{rot} = \frac{1}{2} E$. Substituting the expressions: $\frac{1}{2} I \omega^2 = \frac{1}{2} [\frac{1}{2} (I + m r^2) \omega^2]$. $I = \frac{1}{2} I + \frac{1}{2} m r^2$. $\frac{1}{2} I = \frac{1}{2} m r^2$,which implies $I = m r^2$. The moment of inertia $I = m r^2$ corresponds to a thin circular ring.

$A$ body is rolling without slipping on a horizontal plane. If the rotational kinetic energy of the body is $50 \%$ of its total kinetic energy,then the body is

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