Two thin rings of slightly different radii are joined together to make a wheel (see figure) of radius R. There is a very small smooth gap between the two ring. The wheel has a mass M and its centre of mass is at its geometrical centre. The wheel stands on a smooth surface and a small particle of mass m lies at the top (A) in the gap between the rings. The system is released and the particle begins to slide down along the gap. Assume that the ring does not lose contact with the surface.

(a) As the particle slides down from top point A to the bottom point B, in which direction does the centre of the wheel move?
(b) Find the speed of centre of the wheel when the particle just reaches the bottom point B. How much force the particle is exerting on the wheel at this instant?
(c) Find the speed of the centre of the wheel at the moment the position vector of the particle with respect to the centre of the wheel makes an angle q with the vertical. Do this calculation assuming that the particle is in contact with the inner ring at desired value of `theta`

Answer

(a) First moves to right and then to left
(b) `v_(w)=2m sqrt((gR)/(M(M+m)))`
(c) `v_(w)=sqrt((2m^(2)gRcos^(2)theta(1-cos theta))/((M+m)^(2)+Mmco^(2) theta))`