# Coefficient of Friction Between Axe and Grindstone (Torque & MI)

1. Jun 3, 2006

### Luis2101

A grindstone in the shape of a solid disk with diameter 0.510 and a mass of m= 50.0kg is rotating at w = 900rev/min. You press an ax against the rim with a normal force of N= 160N, and the grindstone comes to rest in 7.00sec.

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

----
Simply put, I have no idea how to connect the information here to Coefficient of friction.
I have found Torque using t = I*(Angular Acceleration)
Where Moment of Inertia = 1/2 MR^2 = 1.63kg*m^2
And Angular Acceleration was found using w = w(initial) + angular acceleration*t.
I found the angular acceleration to be -2.14 rev/sec^2 (I converted the angular velocity to rev/sec in finding this).
And torque was equal to -3.5N...

I have all this info so far, but have no idea how to connect it to Coefficient of Friction, I thought maybe there would be some ratio with the Normal Force, but I don't have Friction Force so I'm pretty much stuck...

Any help would be greatly appreciated.

-L.

2. Jun 3, 2006

### arildno

Well, the torque must be the radius times the frictional force..

3. Jun 3, 2006

### Luis2101

That makes perfect sense...

But...
I solved for Friction using torque/radius or (3.5)/(.255) = 13.7N

I then tried to solve for the coefficient of friction using the relationship:
Friction = Mu*Normal Force
So 13.7/160 = 0.0856 for coefficient of friction... which is wrong.

Is that last relationship incorrect in this case?

-L.

4. Jun 3, 2006

### arildno

Well, but you must use angular acceleration measured in radians per second per second in your standard torque equation.
This is where you've gone wrong; multiply your coefficient of friction with $$2\pi$$ to get the right value.

5. Jun 3, 2006

### Luis2101

Ohhh I see...
I have to use radians for angular acceleration otherwise my Torque value is wrong.
Cool, thanks a lot man.

-L.