Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.8rev/s . The wheel can be considered a uniform disk of mass 4.7kg and diameter 0.34m . The potter then throws a 3.1kg chunk of clay, approximately shaped as a flat disk of radius 13cm , onto the center of the rotating wheel.
What is the frequency of the wheel after the clay sticks to it?
I got 1.64 rev/s and it is not correct. Can anyone explain what I might be doing wrong?

Answer 1

P = angular momentum = I omega

I1 = (1/2) m r^2 = (1/2)(4.7)(.17)^2
= .0679

P1 = (.0679) 2 pi (1.8)

I2 = I1 + (1/2)(3.1) (.065)^2
= .0679 + .00655 = .0744

P2 = P1
so
.0744 (2 pi)(rps) = .0679 * 2 pi *1.8

rps = 1.64 revs/s so I agree with you wholeheartedly

Answer 2

In I2 you halfed the radius of the flat disk as if it were a diameter like pottery wheel. it's not 0.065 it's 0.13