Two rotating bodies A and B of masses m and 2 | vedclass

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(C) The rotational kinetic energy $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.Rearranging thi

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$L_B > L_A$

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(C) The rotational kinetic energy $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.Rearranging this for angular momentum,we get $L = \sqrt{2IK}$.Since both bodies have equal kinetic energy $(K_A = K_B = K)$,the angular momentum is directly proportional to the square root of the moment of inertia: $L \propto \sqrt{I}$.Given that $I_B > I_A$,it follows that $\sqrt{I_B} > \sqrt{I_A}$.Therefore,$L_B > L_A$.

Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ $(I_B > I_A)$ have equal kinetic energy of rotation. If $L_A$ and $L_B$ are their angular momenta respectively,then:

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