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Two insects P and Q are firmly sitting at the ends of a massless semicircular wire of radius R and two more insects A and B are firmly sitting at the bottom of the wire. The wire is given an angular velocity $ω_{0}$ about a vertical axis through its centre
as shown in the figure. Mass of each insect is M. Now A and B crawl to the opposite ends to meet P and Q. Final angular velocity attained by the rod is equal to

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$\frac{ω_{0}}{4}$

$\frac{ω_{0}}{2}$

$ω_{0}$

$2 ω_{0}$

answer is B.

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Detailed Solution

$\begin{array}{l} {\vec{L}}_{i} = {\vec{L}}_{f} \\ (2 {mR}^{2}) ω_{0} = (2 {mR}^{2} + 2 {mR}^{2}) ω^{'} \\ ω^{'} = \frac{ω_{0}}{2} \end{array}$

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