Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.6 rev/s . The wheel can be considered a uniform disk of mass 6.0 kg and diameter 0.50 m . The potter then throws a 2.2-kg chunk of clay, approximately shaped as a flat disk of radius 7.1 cm , onto the center of the rotating wheel.

Part A What is the frequency of the wheel after the clay sticks to it?

Answer 1

By conservation of angular momentum:

   I ?_initial = I?_final

The moment of inertia for wheel is,

     I_wheel   = (1/2) M R²

                = (1/2) *6.00 * 0.25²

                =  0.1875

The moment of inertia for clay is,

I_clay = (1/2) M R²

       = (1/2) * 2.2* 0.071²

      =  0.00554

Therefore,

         I_wheel * ?_initial   =   (I_wheel + I_clay) * ?_final

                 (0.1875) * (1.6 rev/s)   = (0.1875+ 0.00554) * ?_final

                            ?_final= 1.55 rev /sec