# Homework Help: Grindstone and kinetic friction problem

1. Jul 24, 2011

### armolinasf

1. The problem statement, all variables and given/known data

A grindstone in the shape of a solid disk with diameter .52m and mass 50 kg is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N and the grindstone comes to a rest in 7.5 seconds. Find the coefficient of kinetic friction of the ax on the grindstone.

3. The attempt at a solution

Since the friction causes the grindstone to come to a stop, then the initial kinetic energy of the grindstone must equal the negative of the work done by friction. And the force of friction is given by the coefficient of friction, k , times the normal force, n, so the work done will be that force times the distance x.

Using the work energy theorem for the grindstone we would get W=-.5Iv^2, where I is the moment of inertia and v is its initial speed.

I=MR^2/2=(50*.26^2)/2=1.69 kgm^2
v=850 rev/min=89rad/sec
x=89(7.5)(2pi.26)=1090 m

knx=-.5Iv^2
k=(.5Iv^2)/nx
k=(.5)(1.69)(89^2)/(160)(1090)=.038

This is obviously wrong and it doesn't match the answer of.482 given by my text, but where am I going wrong? Thanks for the help.

2. Jul 24, 2011

### Staff: Mentor

Velocity will not be constant over the 7.5 seconds that the axe is in contact with the grindstone. So your "distance" figure, based upon total revolutions of the circumference, will not be accurate.

3. Jul 24, 2011

### HallsofIvy

A shortcut: With a constant force, the deceleration will be constant and so the angular velocity decreases linearly. That means that the average angular velocity will be the simple average of the beginning and ending angular velocities: $(850+ 0)/2= 425$.
revolutions per second.

(I can see no reason to divide by $2\pi$ to convert "revolutions per second" to "radians per second" when you then multiply by [itex]2\pi[/tex] to find the circumference of the wheel.)

4. Jul 24, 2011

### armolinasf

Ok so I could still use the work energy relationship to find k. The distance would then be the average angular velocity times r, or x=110.5. Using this i get a more reasonable answer of .380 but still not the .482 in my book...

5. Jul 24, 2011

### Staff: Mentor

Your value for x looks suspect. Can you show your calculation?

6. Jul 24, 2011

### armolinasf

It was suspect...so average angular velocity is 425 rev/min/60sec=(7.08 rev/sec).26=1.84 m/sec. 1.84*7.5sec=13.81 meters.

Using this value for x i get k=3.03

7. Jul 24, 2011

### Staff: Mentor

(425 rev/min) * (1 min / 60 sec) * (2pi rad/rev) * (0.26 m) = 11.57 m/sec

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