Rotational Motion Part 3

Concept:The angular momentum $L$ of a rigid body rotating about an axis is $L = I \omega$, where $I$ is the moment of inertia about that axis and $\omega$ is the angular velocity.Explanation:Let mass $m$ be at one end and mass $2m$ at the other end, with rod length $d$.The centre of mass (CM) is located at distance $x_{\text{CM}}$ from the $m$ end:$$x_{\text{CM}} = \frac{m \cdot 0 + 2m \cdot d}{m + 2m} = \frac{2d}{3}.$$Distance of mass $m$ from CM: $r_1 = \frac{2d}{3}$.Distance of mass $2m$ from CM: $r_2 = d - \frac{2d}{3} = \frac{d}{3}$.Moment of inertia about the CM axis:$$I = m r_1^2 + 2m r_2^2 = m\left(\frac{2d}{3}\right)^2 + 2m\left(\frac{d}{3}\right)^2.$$$$I = m \cdot \frac{4d^2}{9} + 2m \cdot \frac{d^2}{9} = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{6md^2}{9} = \frac{2}{3}md^2.$$Using $L = I \omega$, we get:$$\omega = \frac{L}{I} = \frac{L}{\frac{2}{3}md^2} = \frac{3}{2}\frac{L}{md^2}.$$Answer:The angular velocity is $\frac{3}{2}\frac{L}{md^2}$, which matches option (B).