Question

.A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.7 rev/s . The wheel can be considered a uniform disk of mass 4.7 kg and diameter 0.34 m . The potter then throws a 3.2-kg chunk 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

This problem can be solved by applying the principle of conservation of angular momentum.

Now =

angular momentum = inertia x rad/s
Now,

Inertia of wheel = 0.5 x 4.7 x 0.17^2 = 0.0679 kg.m^2
Again -

1.7 rev/s = 1.7 x 2pi = 10.68 rad/s

So, initial angular momentum = 0.0679 x 10.68 = 0.725 kg.m^2/s

and, inertia of clay = 0.5 x 3.2 x 0.08^2 = 0.01024 kg.m^2

now we have -

Initial momentum = Final momentum  

0.725 kg.m^2/s = (0.0679 + 0.01024) x omega

=> omega = 0.725 / (0.0679 + 0.01024) = 9.28 rad/s

And the final frequency of the wheel in rev/s = 9.28 / (2*pi) = 1.48 rev/s