# Torque and angular acceleration of a grindstone

A grindstone in the shape of a solid disk with diameter 0.550 m and a mass of m = 50.0 kg is rotating at omega = 840 rev/min. You press an ax against the rim with a normal force of F = 160 N , and the grindstone comes to rest in 7.60 s. Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

To be honest, I want to draw a force diagram, but the thing is that the force then pushes into the grindstone. And then I was thinking about using Wtot-Wfriction=Change in Kinetic energy. But the thing is that I don't know how to use the Fnormal. If anyone can point me in the right direction, that would be great.

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Doc Al
Mentor
And then I was thinking about using Wtot-Wfriction=Change in Kinetic energy.
Better to use Newton's 2nd law for rotation.
But the thing is that I don't know how to use the Fnormal.
Hint: How does Fnormal relate to the friction force?

second law for rotations? torque=rXF? and Oh, fk=fnmewk. but I still don't know where to fit it into the equation/ or is that my F?

Doc Al
Mentor
Yes, the friction force is the force that produces the torque.

so then torque=r X (fn*mewk)=I*alpha
but then how do I relate omega to alpha? is it alpha=domega/dt?

Doc Al
Mentor
so then torque=r X (fn*mewk)=I*alpha
Good.
but then how do I relate omega to alpha? is it alpha=domega/dt?
Yes. α = Δω/Δt. But convert ω to units of radians/second.

So i did that using my numbers and I get that the answer is 29.83 which makes absolutely no sense because mew cannot be bigger than 1. Did I do somethign wrong in the calculations: I got that my equations were:
r*Fnmewk=1/2mr^2*(w*2pi/t)
so
mewk=(1/2mr^2*(w*2pi/t))/(r*F)
so mew k=29 which is wrong
so did I just plug it into my calculator wrong or what?

Doc Al
Mentor
Did I do somethign wrong in the calculations: I got that my equations were:
r*Fnmewk=1/2mr^2*(w*2pi/t)
Note that you are given ω in units of rev/minute. How can you convert to radians/second?