Rotational Motion

JEE Main level, single correct, medium

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Question 223 of 280

A thin smooth rod of length LL and mass MM is rotating freely with angular speed ω0{\omega }_{0} about an axis perpendicular to the rod and passing through center. Two beads of mass mm and negligible size are at the center of the rod initially. The beads of mass mm 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) Mω0M+6m\dfrac{M {\omega }_{0}}{M+6m}

From conservation of angular momentum.
ML212ω0=[ML212+m(L2)2+m(L2)2]ω′\dfrac{M{L}^{2}}{12}{\omega }_{0}=\left[\dfrac{M{L}^{2}}{12}+m{\left(\dfrac{L}{2}\right)}^{2}+m{\left(\dfrac{L}{2}\right)}^{2}\right]\omega '
ML212ω0=(ML212+mL22)ω′\dfrac{M{L}^{2}}{12}{\omega }_{0}=\left(\dfrac{M{L}^{2}}{12}+\dfrac{m{L}^{2}}{2}\right)\omega '
ω′=Mω0M+6m\omega '=\dfrac{M{\omega }_{0}}{M+6m}

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