Question

Two bodies have their moment of inertia $3I$ and $11I$ 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
$\frac{1}{2}{I}_1{\omega }_1^2=\frac{1}{2}\times \frac{1}{2}{I}_2{\omega }_2^2$
$3I\times {\omega }_1^2=\frac{1}{2}\times 11I\times {\omega }_2^2$
$\frac{{\omega }_1}{{\omega }_2}=\sqrt{\frac{11}{6}}$
Ratio of angular momentum is
$\because L=I\omega ,$ then $\frac{L_1}{L_2}=\frac{I_1{\omega }_1}{I_2{\omega }_2}=\frac{3I}{11I}\times \sqrt{\frac{11}{6}}=\frac{3}{\sqrt{66}}$
Hence, $\frac{L_1}{L_2}=\frac{3}{\sqrt{66}}$