A horizontal uniform circular disc of mass M and radius R is rotating freely about a frictionless vertical axis passing through its centre. A person of mass m is standing at the edge of the disc. The initial rotational kinetic energy of the system is E . The person slowly walks towards the centre of the disc and stops. If M = 4m , the work done by the person in walking to the centre is E x . The value of x is

Step-by-step solution

The initial moment of inertia of the system (disc + person at the edge) is: I_i = 1 2 MR^2 + mR^2 Given M = 4m , we have: I_i = 1 2 (4m)R^2 + mR^2 = 2mR^2 + mR^2 = 3mR^2 When the person reaches the centre, their distance from the axis of rotation is zero. The final moment of inertia of the system is: I_f = 1 2 MR^2 + 0 = 2mR^2 Since no external torque acts on the system, the angular momentum L is conserved. The final kinetic energy is related to the initial kinetic energy by: K_f = L^2 2I_f = K_i I_i I_f K_f = E (