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A thin circular ring of mass M and radius r is rotating about its axis with angular speed $ω$ . Two particles of mass m each are now attached at diametrically opposite points. The angular speed of ring will become

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$(\frac{M - 2 m}{M + 2 m}) ω$

$(\frac{M}{M + 2 m}) ω$

$\frac{M ω}{M + m}$

$(\frac{M + 2 m}{M}) ω$

answer is B.

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Detailed Solution

Conservation of angular momentum

$(M r^{2}) ω = (M r^{2} + 2 m r^{2}) ω^{1}$

$\Rightarrow ω^{1} = \frac{M ω}{M + 2 m}$

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