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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

Vedclass · Accepted Answer

$\frac{y^2}{2x}$

Vedclass · Answer

(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 $L$ is given by $L = I \omega$. We can express the kinetic energy in terms of angular momentum as: $K = \frac{1}{2} I \left( \frac{L}{I} \right)^2 = \frac{L^2}{2I}$. Given $K = x$ and $L = y$,we have $x = \frac{y^2}{2I}$. Rearranging for the moment of inertia $I$,we get $I = \frac{y^2}{2x}$. Therefore,the correct option is $B$.

$A$ 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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