A grindstone in the shape of a solid disk with diameter 0.520 and a mass of = 50.0 is rotating at = 860 . You press an ax against the rim with a normal force of = 250 (Figure 1) , and the grindstone comes to rest in 7.40 .

Question

A grindstone in the shape of a solid disk with diameter 0.520 and a mass of  = 50.0 is rotating at  = 860 . You press an ax against the rim with a normal force of  = 250 (Figure 1) , and the grindstone comes to rest in 7.40 .
Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

Damon · Answer

angular momentum initial = I omega
I = (1/2) m r^2 = (1/2)(50).26^2
860 what? rpm?
860 rev/min * 1/60 min/sec * 2 pi radians/rev = 90 rad/sec
so Iw= 152

torque * time = change of angular momentum

250 N * mu * 7.4 sec = 152

I am just guessing at your units for everything