Question

Two bodies have their moment of inertia $3 I$ and $11 I$ respectively about their axis of rotation. If the kinetic energy of first body is half of the second then their angular momentum will be in the ratio of

• A. $1$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{4}{\sqrt{11}}$
• D. $\frac{3}{\sqrt{66}}$

Answer 1

Answer: (D)

According to the question
$\frac{1}{2} I_{1} \omega_{1}^{2}=\frac{1}{2} \times \frac{1}{2} I_{2} \omega_{2}^{2}$
$3 I \times \omega_{1}^{2}=\frac{1}{2} \times 11 I \times \omega_{2}^{2}$
$\frac{\omega_{1}}{\omega_{2}}=\sqrt{\frac{11}{6}}$
Ratio of angular momentum is
$\because L=I \omega,$ then $\frac{L_{1}}{L_{2}}=\frac{I_{1} \omega_{1}}{I_{2} \omega_{2}}=\frac{3 I}{11 I} \times \sqrt{\frac{11}{6}}=\frac{3}{\sqrt{66}}$
Hence, $\frac{L_{1}}{L_{2}}=\frac{3}{\sqrt{66}}$