A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω.
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :
(1) \(\omega\frac{M}{M+m}\)
(2) \(\omega\frac{M+2m}{M}\)
(3) \(\omega\frac{M}{M+2m}\)
(4) \(\omega\frac{M-2m}{M+2m}\)
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Correct option is (3) \(\omega\frac{M}{M+2m}\)
Using conservation of angular momentum
(Mr2)ω = (Mr2 + 2mr2)ω'
ω' = \(\frac{M\omega}{M+2m}\)
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