Question

A flat uniform circular disk (radius = 2.30 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A 50.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.70 m/s relative to the ground.

Answer 1

The resulting angular speed = 0.6 rad / s.

Explanation:

Here there is no external torque acting on the system thus we can apply the law of conservation of angular  momentum  

Angular momentum of the man = Iω

Where I = Inertia of the man about the axis of rotation

or         I = M r 2

            I  = 50 * 1.25*1.25 = 78.125

w = Angular velocity of the man, that can be calculated as follows

Tangential velocity of man = v = 2m/s  

So time taken to describe this circle is t = (2*pi* r) / v

Now angle described in 1 revolution θ = 2*pi radians

This angle is subtended in time t = (2*pi* r) / v

Thus angular speed = w = θ/t = 2*pi* ( v/ 2π r) = v/r = 2.70 / 1.25 = 2.16 rad/s

So angular momentum of man = Iw = 78.125 * 2.16 = 168.75.

To conserve the angular momentum before and after,

Angular momentum of disk = angular momentum of the man  

           i.e.             Iw of disk = 168.75

                                disk of I = (disk of M*R^2) / 2

                                              = (1.00 * 102 * 2.30 * 2.30) / 2

                                              = 269.79

                 Thus 269.79 of disk of w = 168.75

      Resulting angular speed of disk = 168.75 / 269.79 = 0.6 ras / s