Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered a uniform disk of mass 4.9 kg and diameter 0.40m. The potter then throws a 2.7 kg chucnk of clay, approximately shaped as a flat disk of radius 8.0 cm, onto the center of the rotating wheel.

What is the frequency of the wheel after the clay sticks to it? Ignore friction/.

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) * 4.9 * 0.20²

                =  0.098

The moment of inertia for clay is,

I_clay = (1/2) M R²

       = (1/2) * 2.7* 0.08²

      =  0.00864

Therefore,

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

                 (0.098) * (2.0 rev/s)   = (0.098+ 0.00864) * ?_final

                            ?_final= 1.84 rev /sec