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 ………..

Question

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 ………..

Infinity Learn · Accepted Answer

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

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