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A smooth uniform rod length $L$ and mass $M$ has two identical beads of negligible size, each of mass $m,$ which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity $ω_{0}$ about an axis perpendicular to the rod and passing through the mid-point of the rod. There are no external forces. When the beads reach the end of the rod, the angular velocity of the system is :

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$(\frac{M}{M + m}) ω_{0}$

$(\frac{M}{M + 4 m}) ω_{0}$

$(\frac{M}{M + 6 m}) ω_{0}$

$(\frac{M}{M + 2 m}) ω_{0}$

answer is C.

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Detailed Solution

Conservation of angular momentum,

$I_{1} ω_{1} = I_{2} ω_{2}$

$ω_{2} = \frac{I_{1}}{I_{2}} . ω_{1} = \frac{(M L^{2} / 12)}{[(M L^{2} / 12) + 2 m {(L / 2)}^{2}]} ω_{0} = (\frac{M L^{2}}{M L^{2} + 6 m L^{2}}) ω_{0} = (\frac{M}{M + 6 m}) ω_{0}$

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