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 5.5 kg and diameter 0.44 m . The potter then throws a 2.5-kg chunk of clay, approximately shaped as a flat disk of radius 7.4 cm , onto the center of the rotating wheel.

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

Answer 1

Solve this problem using conservation of angular momentum.

Now-

angular momentum = I x w (moment of inertia x angular velocity)
Initially -
Moment of inertia of wheel, I1 = 0.5 x 5.5 x 0.22^2 = 0.1331 kg.m^2

w1 = 1.7 rev/s x 2pi = 10.68 rad/s

Again, inertia of clay = 0.5 x 2.5 x 0.074^2 = 0.006845 kg.m^2

So, final moment of inertia, I2 = 0.1331 + 0.006845 = 0.139945 kg.m^2

w2 = ?

Initial angular momentum = final angular momentum

=> 0.1331 x 10.68 = 0.139945 x w2

=> w2 = (0.1331 x 10.68) / 0.139945 = 10.16 rad/s

convert this in rev/s -

w2 = 10.16 / (2*pi) = 1.62 rev/s