Coefficient of friction?

Question

A grindstone in the shape of a solid disk with diameter 0.52 m and a mass of m=50kg is rotating at omega=820 rev/min. You press an ax against the rim with a normal force of F = 230 N, and the grindstone comes to rest in 7.30 s.
http://session.masteringphysics.com/problemAsset/1040261/1/yf_Figure_10_44.jpg

Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

Answer 1

Torque is a vector product of radius R and friction force f applied
T=R (uN) friction force f=uN
also torque is a product of moment of inertia I of a rotating body and its angular acceleration A
T=IA

A=w/t where w - angular belocity
and
I= (1/2) mR^2
We have
(1/2) mR^2w/t=R (uN)

u=(1/2) mR w/(t N)
u=(1/2) 50 (0.52/2) (820 2 pi / 60)/( 7.30 x 230)
u=0.33