Question

Two balls of masses m and $2m$ are attached to the ends of a light rod of length L. The rod rotates with an angular speed $\omega$ about an axis passing through the centre of mass of system and perpendicular to the place. Find the angular momentum of the system about the axis of rotation.

• A. $\frac{2}{3}m\omega L^2$
• B. $\frac{1}{3}{\omega }^{2}Lm$
• C. $\frac{2}{3}{\omega }^{2}Lm$
• D. $\frac{1}{3}Lm$

Answer 1

The correct option is A $\frac{2}{3}m\omega L^2$

Location of CM,

$x_{CM}=\frac{2m\times 0+m\times L}{2m+m}=\frac{L}{3}$

The velocity of ball A w.r.t. the centre of mass C is

$v_A=\omega x_{CM}=\frac{\omega L}{3}$

$v_B=\omega (L-x_{CM})=\frac{2\omega L}{3}$

Thus, angular momentum of a system about C,

$L=2m{v}_A{x}_{CM}+m{v}_a(L-x_{CM})$

$=2m\times \frac{\omega L}{3}\times \frac{L}{3}+m\times \frac{2\omega L}{3}\times \frac{2L}{3}$

$=\frac{2m\omega L^2}{3}$.