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.
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)