Question

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

• A. $m\omega d^2$
• B. $\frac{1}{2}m\omega d^2$
• C. $2m\omega d^2$
• D. $\frac{1}{4}m\omega d^2$

Answer 1

The correct option is B

$\frac{1}{2}m\omega d^2$

Consider the situation shown in figure. The magnitude of the velocity of the ball $A$ with respect to the centre $O$ is $v$ = $\frac{\omega d}{2}$.

The angular momentum of the ball with respect to the axis is $L_1$ = $mvr$ = $m\left(\frac{\omega d}{2}\right)\left(\frac{d}{2}\right)$ = $\frac{1}{4}m\omega d^2$.

The same is the angular momentum $L_2$ of the second ball. The angular momentum of the system is equal to sum of these two angular momenta i.e., $L$ = $\frac{1}{2}m\omega d^2$.