A disc of mass `m` and radius `R` is free to rotate in a horizontal plane about a vertical smooth fixed axis passing through its centre. There is a smooth groove along the diameter of the disc and two small balls of mass `m//2` each are placed in it on either side of the centre of the disc as shown in Fig. The disc is given an initial angular velocity `omega_(0)` and released.

The speed of each ball relative to the ground just after they leave the disc is

Options

A. `(Romega_(0))/(sqrt(3))`
B. `(Romega_(0))/(sqrt(2))`
C. `(2Romega_(0))/3`
D. none of these

Answer

C

Explanation

The angular speed of the disc just after the balls leave the disc is `omega=omega_(0)//3`
Let the speed of each ball just after they leave the disc be `v`. From conservation of energy.
`1/2(1/2mR^(2))omega_(0)^(2)=1/2(1/2mR^(2))omega^(2)+1/2(m/2)v^(2)+1/2(m/2)v^(2)`
Solving we get `v=(2Romega_(0))/3`