A grinding wheel in the form of a uniform solid disk is spun up to an angular velocity of ω_{i}. The wheel has a mass m and a radius r. A tool to be sharpened is pressed against the wheel with a normal force of N.
(a) If, after t seconds, the wheel's angular velocity is ω_{f}, what is the coefficient of friction between the tool and the grinding wheel? Assume no other torques act on the wheel.
(b) What is the average rate of kinetic energy loss over this time period?
Step 1. Draw a diagram, and take a couple of minutes to set up the problem. The freebody diagram of the wheel shows the normal force and the force of kinetic friction. The wheel is rotating clockwise  we'll take that to be the positive direction.
Step 2. Apply Newton's Second Law for rotation. The force of friction is the only force giving a torque, and that torque is counterclockwise:
Στ = Iα
rf_{k} = Iα
Step 3. Substitute in for the force of kinetic friction:
f_{k} = μ_{k}N
Therefore rμ_{k}N = Iα
μ_{k}  = 

Step 4. Use a constant acceleration equation to find α in terms of the angular velocities:
ω_{f} = ω_{i} + αt
This gives  α  = 

So, μ_{k}  = 

Step 5. For a solid disk, I = ½ mr^{2}
μ_{k}  = 

= 

Let's take an example where the disk has a mass of 1.5 kg, a radius of 25 cm, and an initial angular speed of 360 rad/s. The tool to be sharpened is pushed against the wheel with a normal force of 20 N, and after t = 10 seconds the wheel's angular speed is 120 rad/s.
μ_{k}  = 

= 

=  0.225 
(b) What is the average rate of kinetic energy loss over the 10 second interval?
The rate of energy loss is the power loss. One way to get the average power is to use:
P_{avg} = τ ω_{avg}
Another method is to calculate the initial and final kinetic energies, subtract to find the energy loss, and divide by the time interval.
P_{avg}  = 

= 

For our numerical example:
P_{avg}  = 

= 

=  270 W 