Question

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

(a) ωM/(M+m)
(b) ωM/(M+2m)
(c) ω(M-2m)/(M+2m)
(d) ω(M+2m)/M

Answer 1

Answer ⇒ Option (b). ''ωM/(M+2m)''

Explanation ⇒

Moment of inertia of the ring will be given by the relation,

I = Mr²

Angular Momentum = I⍵

When the masses are attached, the moment of Inertia will be given by,

I'= Mr²+ 2mr²

= r² (M+2m)

Assuming that the new angular speed be ⍵'. Therefore, the angular momentum = I'⍵'.

By the law of the conservation of the momentum, the angular momentum is always conserved.

∴ I'⍵'=I⍵

⇒ ⍵' = I⍵/I'

⇒ ⍵' =⍵Mr²/(M+2m)r²

∴ ⍵' =⍵M/(M+2m)

Hence, Option (b). is correct.

Hope it helps.

Answer 2

b is the correct answer