A thin circular ring of mass M and radius r is rotating about its axis

1.	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» 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.*

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»

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.

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