A thin smooth rod of length L and mass M is rotating freely with… | JEE Main 2019 (9 Apr, Shift 2)

Q60JEE Main 2019 · 9 APR · SHIFT 2Medium

Angular Momentum & its Conservation

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 :-

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Answer: (c)

No external torque: $I_1\omega_0=I_2\omega$ .
$I_1=\frac{ML^2}{12}$ (beads at centre), $I_2=\frac{ML^2}{12}+2m\left(\frac L2\right)^2=\frac{(M+6m)L^2}{12}$
$\omega=\frac{M\omega_0}{M+6m}$
So the answer is (c).

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