Explanation

Let the masses $m$ and $2m$ be at the ends of a rigid rod of length $d$.

First, locate the centre of mass of the system.

Taking origin at mass $m$,

$$ x_{CM} = \frac{m(0) + 2m(d)}{m + 2m} $$

$$ x_{CM} = \frac{2md}{3m} = \frac{2d}{3} $$

So, distance of mass $m$ from centre of mass is

$$ r_1 = \frac{2d}{3} $$

Distance of mass $2m$ from centre of mass is

$$ r_2 = d - \frac{2d}{3} = \frac{d}{3} $$

Moment of inertia about axis passing through centre of mass and perpendicular to rod is

$$ I = m r_1^2 + 2m r_2^2 $$

Substituting values,

$$ I = m\left(\frac{2d}{3}\right)^2 + 2m\left(\frac{d}{3}\right)^2 $$

$$ I = m\frac{4d^2}{9} + 2m\frac{d^2}{9} $$

$$ I = \frac{6md^2}{9} = \frac{2}{3}md^2 $$

Angular momentum is related to angular velocity by

$$ L = I\omega $$

Hence,

$$ \omega = \frac{L}{I} $$

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

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

Therefore, the angular velocity of the system is

$$ \boxed{\frac{3}{2}\frac{L}{md^2}} $$

Related JEE Main Questions