A thin smooth rod of length $L$ and mass $M$ is rotating freely with angular speed $\omega_0$ about an axis perpendicular to the rod and passing through its center. Two beads of mass $m$ and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be
Answer: (C) $\dfrac{M\omega_0}{M + 6m}$
No external torque acts about the axis, so angular momentum is conserved. The beads at the centre add nothing to the moment of inertia at first; at the ends each adds $m(L/2)^2$:
$$\frac{ML^2}{12}\omega_0 = \left(\frac{ML^2}{12} + 2m\frac{L^2}{4}\right)\omega$$
$$\omega = \frac{M\omega_0}{M + 6m}$$
Solution by Sreeraj P, M.Sc Physics