Rotational Motion

JEE Main level, single correct, medium

58 of 280

Question 58 of 280

A thin circular ring of mass MM and radius rr is rotating about its axis with an angular speed ω.\omega . Two particles having mass mm each are now attached at diametrically opposite points. The angular speed of the ring will become:

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Answer: (C) ωMM+2m\omega \dfrac{M}{M+2m}

Using conservation of angular momentum
⇒(Mr2)ω=(Mr2+2mr2)ω′\Rightarrow \left(M{r}^{2}\right)\omega =\left(M{r}^{2}+2m{r}^{2}\right){\omega }^{'}
⇒ω′=MωM+2m\Rightarrow {\omega }^{'}=\dfrac{M\omega }{M+2m}.

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