Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ ($I_B I_A$) h

Question

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 $L_B$ be their angular momenta respectively, then:

Accepted Answer

The rotational kinetic energy ($K$) is related to angular momentum ($L$) and moment of inertia ($I$) by the relation $K = \frac{L^2}{2I}$.
Given that the kinetic energies of rotation are equal: $K_A = K_B$.
$\frac{L_A^2}{2I_A} = \frac{L_B^2}{2I_B}$
Rearranging the terms, we get:
$\frac{L_B^2}{L_A^2} = \frac{I_B}{I_A}$
It is given that $I_B > I_A$, so the ratio $\frac{I_B}{I_A} > 1$.
Therefore, $\frac{L_B^2}{L_A^2} > 1 \implies L_B^2 > L_A^2 \implies L_B > L_A$.

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