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6.A particle of mass 'm' moves on the axis of a ringof radius 'R' and mass M. If particle at rest isreleased from 'P then its kinetic energy at centrec will bemCRP |
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Answer» 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). |
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