A ring of mass M = 90 gram and radius R = 3 meter is kept on a frictionless horizontal surface such that its plane is parallel to horizontal plane. A particle of mass m = 10 gram is placed in contact with the inner surface of ring as shown figure. An initial velocity v=2 m/s is given to the particle along the tangent of the ring.  The magnitude of the force of interaction in milli-newton between them after 1 sec from the start is ………..

There is no external force acting on the system 'ring + particle' , it means the COM will move with constant velocity and center of ring and particle will rotate about COM. Velocity of COM is Vcm=mVm+M Radius of Rotation of particle is x= M.R M+m The normal reaction between the particle and ring provides required centripetal force  for circular motion of the particle about COM. ∴N=mv−vcm2x=mv2M(m+M)R=(0.01)(2)2(0.09)(0.1)(3)=6×10−3 newton