Two small balls with masses m and 2m 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 :

We need to find the angular velocity of a system of two balls (masses $$m$$ and $$2m$$) attached to a rod of length $$d$$, given angular momentum $$L$$ about the axis through the center of mass.

Find the center of mass. Taking the position of mass $$m$$ as origin:

$$x_{cm} = \frac{m \cdot 0 + 2m \cdot d}{m + 2m} = \frac{2d}{3}$$

So mass $$m$$ is at distance $$\frac{2d}{3}$$ from the CM, and mass $$2m$$ is at distance $$d - \frac{2d}{3} = \frac{d}{3}$$ from the CM.

Calculate the moment of inertia about the CM:

$$I = m\left(\frac{2d}{3}\right)^2 + 2m\left(\frac{d}{3}\right)^2 = m \cdot \frac{4d^2}{9} + 2m \cdot \frac{d^2}{9} = \frac{4md^2 + 2md^2}{9} = \frac{6md^2}{9} = \frac{2md^2}{3}$$

Use $$L = I\omega$$ to find $$\omega$$:

$$\omega = \frac{L}{I} = \frac{L}{2md^2/3} = \frac{3L}{2md^2}$$

The correct answer is Option (2): $$\frac{3}{2}\frac{L}{md^2}$$.