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

• A. $L_B>L_A$
• B. $L_A>L_B$
• C. $L_A=\frac{L_B}{2}$
• D. $L_A=2L_B$

Answer 1

The correct option is A $L_B>L_A$

The kinetic energy of rotation is given by $\frac{1}{2}I{\omega }^2$.
Since the kinetic energy of rotation is same,
$I_A{\omega }_A^2=I_B{\omega }_B^2$
$⟹\frac{{\omega }_A}{{\omega }_B}=\sqrt{\frac{I_B}{I_A}}$
The ratio of the angular momentum becomes,
$\frac{L_A}{L_B}=\frac{I_A{\omega }_A}{I_B{\omega }_B}$
$=\sqrt{\frac{I_A}{I_B}}<1$
$⟹L_B>L_A$