Question

6.
A particle of mass 'm' moves on the axis of a ring
of radius 'R' and mass M. If particle at rest is
released from 'P then its kinetic energy at centre
c will be
m
CRP​

Answer 1

Answer:

Mv^2/R + -GM/(x^2 + R^2)^(1/2).

Explanation:

Let the distance from which the particle had flew of the circle be x. So, the distance will be (x^2 + R^2)^(1/2).

So, the potential energy at the point P will be = -GM/(x^2 + R^2)^(1/2) where M is the mass of the particle.

Again we know that the circle is of radius R and then there will be centripetal force of Mv^2/R which will also be in addition to the potential energy when it is at rest.

So, kinetic energy =Mv^2/R + -GM/(x^2 + R^2)^(1/2).