Question

Two particles balls A and B, each of mass m, are attached rigidly to the ends of a light rod of length l. The system rotates about the perpendicular bisector of the rod at an angular speed $\omega$. Calculate the angular momentum of the individual particles and of the system about the axis of rotation.

Answer 1

Angular Momentum $\overline{L}=r\times mv$,

For the Ball 'A' , $r=-\frac{l}{2}\hat{i}$, $mv=-\frac{m\omega l}{2}\hat{j}$,

$\therefore \overline{L}=\frac{m\omega l^2}{4}(\hat{i}\times \hat{j})$ $\Rightarrow \overline{L}=\frac{m\omega l^2}{4}\hat{k}$ .......1

For the Ball 'B' , $r=\frac{l}{2}\hat{i}$, $mv=\frac{m\omega l}{2}\hat{j}$,

$\therefore \overline{L}=\frac{m\omega l^2}{4}(\hat{i}\times \hat{j})$ $\Rightarrow \overline{L}=\frac{m\omega l^2}{4}\hat{k}$ .......2

Adding equation 1 and 2, We get Angular Momentum of the system

$\overline{L}=\frac{m\omega l^2}{2}$