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(B) The rotational kinetic energy $K$ of a body rotating with angular velocity $\omega$ and moment of inertia $I$ is given by $K = \frac{1}{2} I \omeg

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$y^2/2x$

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(B) The rotational kinetic energy $K$ of a body rotating with angular velocity $\omega$ and moment of inertia $I$ is given by $K = \frac{1}{2} I \omega^2$.The angular momentum $L$ of the body is given by $L = I \omega$.From the angular momentum equation, we have $\omega = \frac{L}{I}$.Substituting this into the kinetic energy equation:$K = \frac{1}{2} I \left( \frac{L}{I} \right)^2 = \frac{1}{2} I \left( \frac{L^2}{I^2} \right) = \frac{L^2}{2I}$.Given $K = x$ and $L = y$, we have $x = \frac{y^2}{2I}$.Rearranging for the moment of inertia $I$, we get $I = \frac{y^2}{2x}$.

$A$ body is rotating about its own axis. Its rotational kinetic energy is $x$ and its angular momentum is $y$. Hence, its moment of inertia about its own axis is

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