Eduprobe
Question:
A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing through its centre O with two point masses each of mass M/8 at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is (8/9)ω and one of the masses is at a distance of (3/5)R from O. At this instant, what is the distance of the other mass from O?
23R
13R
35R
45R
Solution:
From conservation of angular momentum, Li = Lf since there is no external torque acting on the system.
Li = MR²ω
Lf = MR²(8/9)ω + (M/8)((3/5)R)²(8/9)ω + (M/8)x²(8/9)ω
where x is the distance of the 2nd mass from the center.
Solving we get
x = (4/5)R