Solving Potter's Wheel Friction Problem

[check]

So I'm doing an online homework problem:

A potter's wheel - a thick stone disk of radius 0.405 m and mass 112 kg - is freely rotating at 48.0 rev/min. The potter can stop the wheel in 5.67 s by pressing a wet rag against the rim and exerting a radially inward force of 69.1 N. Find the effective coefficient of kinetic friction between the wheel and rag.

Round your answer to three significant figures.



f = \frac {I \omega_i}{r(\Delta t)} = \frac {(112 kg) (0.405 m^2)(48.0 rev/min)} {(0.405 m)(5.67 s)} \left(\frac{2\pi rad}{1 rev}\right) \left( \frac {1 min}{60 s}\right) = 40.2 N

\mu_k = \frac {f}{n} = \frac {40.2}{69.1}=0.582

Trouble is, I keep getting "WRONG".

Am I doing it wrong? I also tried with different # of significant digits and nothin. Any help would be great.

 

[thermodynamicaldude]

Looks like you're using mr^2 for the moment of Inertia...which would work if it were a ring..

..however, the moment of Inertia of a thick solid disk is:

I = 0.5mr^2..

...hope that helps!

 

[check]

Right right right... ok, thanks. I just got it after I posted this. heh heh