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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:
A
$L_B > L_A$
B
$L_A > L_B$
C
$L_A = \frac{L_B}{2}$
D
$L_A = 2L_B$
Explanation
$K = L^2/2I$; for equal K, larger I means larger L.
Detailed Solution
$K_A = K_B \Rightarrow \frac{L_A^2}{2I_A} = \frac{L_B^2}{2I_B}$
$\because I_B > I_A$, $\therefore L_A^2 < L_B^2 \Rightarrow L_A < L_B$
$\because I_B > I_A$, $\therefore L_A^2 < L_B^2 \Rightarrow L_A < L_B$