A ring of mass M and radius R is rotating with angular speed `omega` 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) omega` and one fo the masses is at a distance of `(3)/(5) R` from O. At this instant the distance of the other mass from O is

Options

A. `(2)/(3)R`
B. `(1)/(3)R`
C. `(3)/(5)R`
D. `(4)/(5)R`

Answer

d

Explanation

Let the distance of other mass from the centre `O` be x Applying Law of conservation of angular momentum Initial Angular momentum = final angular momentum
`MR^(2) omega = [MR^(2) + (M)/(g) (3/(5) R)^(2) + (M)/(8) x^(2)](8)/(9)omega`
`R^(2) = [(200R^(2) + 9R^(2) + 25x^(2))/(200) ](8)/(9)`
on solving we get `x = (4)/(5) R` .