Question

Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with momenta 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. $L_A = \frac{L_B}{2}$
• B. $L_A = 2 L_B$
• C. $L_B > L_A$
• D. $L_A > L_B$

Answer 1

Answer: (C)

$L_B > L_A$

Answer 2

$K.E_{A} = K.E_{B}$
$\frac{1}{2} I_{A} \omega^{2}_{A} = \frac{1}{2} I_{B} \omega^{2}_{B}$
$\frac{\omega_{A}}{\omega_{B} } = \sqrt{\frac{I_{B}}{I_{A}}}$ .....(i)
$L_{A} =I_{A}\omega_{A} L_{B} = I_{B} \omega_{B}$
$\frac{L_{A}}{L_{B}} = \frac{I_{A}}{I_{B}} \times\frac{\omega_{A}}{\omega_{B}} = \frac{I_{A}}{I_{B}} \times \sqrt{\frac{I_{A}}{I_{B}}}$
$= \sqrt{\frac{I_{A}}{I_{B}}} < 1$
$\left[L_{A} < L_{B}\right]$