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 $\omega$. The angular momentum of the system about the axis of rotation is .

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

Answer 1

The correct option is C $\frac{1}{2}m\omega d^2$

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., $L=\frac{1}{2}m\omega d^2$.