Two small balls with masses $m$ and 2 m are attached to both ends of a rigid rod of length $d$ and n

Question

Two small balls with masses $m$ and 2 m are attached to both ends of a rigid rod of length $d$ and negligible mass. If angular momentum of this system is $L$ about an axis $(A)$ passing through its centre of mass and perpendicular to the rod then angular velocity of the system about $A$ is :

Professor PC Thomas Classes · Answer

The center of mass divides the rod in the inverse ratio of the masses. Distance of $m$ from CM is $x_1 = \frac{2m}{m+2m}d = \frac{2d}{3}$. Distance of $2m$ from CM is $x_2 = \frac{m}{m+2m}d = \frac{d}{3}$. The moment of inertia about the CM is $I = m(\frac{2d}{3})^2 + 2m(\frac{d}{3})^2 = m(\frac{4d^2}{9}) + 2m(\frac{d^2}{9}) = \frac{6md^2}{9} = \frac{2}{3}md^2$. Angular momentum is $L = I\omega$, so the angular velocity is $\omega = \frac{L}{I} = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2}$.