Two rotating bodies P and Q of masses m with | vedclass

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(D) The rotational kinetic energy $K$ is given by the formula $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.

Vedclass · Accepted Answer

$L_Q > L_P$

Vedclass · Answer

(D) The rotational kinetic energy $K$ is given by the formula $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.Since the kinetic energies are equal,we have $K_P = K_Q$.Therefore,$\frac{L_P^2}{2I_P} = \frac{L_Q^2}{2I_Q}$.Rearranging the terms,we get $\frac{L_Q^2}{L_P^2} = \frac{I_Q}{I_P}$.Taking the square root on both sides,$\frac{L_Q}{L_P} = \sqrt{\frac{I_Q}{I_P}}$.Given that $I_Q > I_P$,it follows that $\frac{I_Q}{I_P} > 1$.Thus,$\sqrt{\frac{I_Q}{I_P}} > 1$,which implies $\frac{L_Q}{L_P} > 1$.Therefore,$L_Q > L_P$.

Two rotating bodies $P$ and $Q$ of masses $m$ with moment of inertia $I_P$ and $I_Q$ $(I_Q > I_P)$ have equal kinetic energy of rotation. If $L_P$ and $L_Q$ are their angular momenta respectively,then:

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