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Q. 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 :

JEE MainJEE Main 2019System of Particles and Rotational Motion

Solution:

Applying angular momentum conservation, about axis of rotation
$L_i = L_f$
$\frac{ML^2}{12} \omega_0 = \left( \frac{ML^2}{12} + m \left( \frac{L}{2} \right)^2 \times 2 \right) \omega $
$\Rightarrow \; \omega = \frac{M \omega_0}{M + 6m}$