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 individual balls and of the system about the axis of rotation.

• A. $\frac{1}{4}mw{d}^2,\frac{3}{4}mw{d}^2$
• B. $\frac{1}{4}mw{d}^2,\frac{1}{2}mw{d}^2$
• C. $\frac{1}{4}mw{d}^2,mw{d}^2$
• D. $\frac{1}{8}mw{d}^2,\frac{1}{4}mw{d}^2$

Answer 1

The correct option is B

$\frac{1}{4}mw{d}^2,\frac{1}{2}mw{d}^2$

Consider the situation shown in figure. The velocity of the ball A with respect to the centre O is $\left(\frac{\omega d}{2}\right)$ Themagnitude of angular momentum of the ball with respect to the axis is $L_1$= mvr =

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 because both mass are rotating anticlockwise i.e., L=$\left(\frac{1}{2}\right)m\omega d^2$.