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

Answer 1

Answer: (D)

$L_B>L_A$

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,

$\frac{L_A}{L_B}=\frac{I_A{\omega }_A}{I_B{\omega }_B}$

$=\sqrt{\frac{I_A}{I_B}}<1$

$⟹L_B>L_A$