Question

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

The correct answer is (b)

Explanation: 

Moment of inertia of the ring I = Mr² 

Angular Momentum = I⍵ 

When the masses are attached, the moment of Inertia I'= Mr²+2mr² 

=(M+2m)r² 

Let the new angular speed be ⍵'. So the angular momentum =I'⍵'. 

Since the angular momentum is conserved. I'⍵'=I⍵ 

→⍵' = I⍵/I' =⍵Mr²/(M+2m)r² =⍵M/(M+2m)