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?

Solution to part (a)

Step 1. Draw a diagram, and take a couple of minutes to set up the problem.

Step 2. We know that the coefficient of friction is given by:

μ_k = f_k/N

Step 3. Bring in torque to tell us something about the force of friction.

τ = rf_k (the angle between r and f_k is 90 degrees).

f_k = τ/r, so the expression for μ_k becomes:

μ_k = τ/rN

Step 4. Use work to tell us about torque:

W = -τ Δθ

The minus sign is because the torque is opposite in direction to the angular displacement.

τ = -W/Δθ

So, μ_k = -W / (rN Δθ)

Step 5. Use the master energy equation to relate the work to the initial and final angular velocities:

W = K_f - K_i = ½ Iω_f² - ½ Iω_i²

So, μ_k = -½ I (ω_f² - ω_i² ) / (rN Δθ)

Step 6. Use a constant acceleration equation to find out something about the angular displacement:

Δθ = θ - θ_o = ½ (ω_f + ω_i) t

The expression for μ_k is now:

So, μ_k = -½ I (ω_f² - ω_i² ) / [½ rNt (ω_f + ω_i) ]

Step 7. Simplify this using the fact that:

ω_f² - ω_i² = (ω_f + ω_i ) * (ω_f - ω_i )

μ_k = -I (ω_f - ω_i) / ( rNt )

Step 8. For a solid disk, I = ½ mr²

μ_k = -½ mr² (ω_f - ω_i) / ( rNt )

μ_k = - mr (ω_f - ω_i) / ( 2Nt )

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 = - mr (ω_f - ω_i) / ( 2Nt )

μ_k = - 1.5 * 0.25 * (120 - 360) / ( 2 * 20 * 10 )

μ_k = 90/400 = 0.225

Solution to part (b)

(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 = (K_f - K_i) / t

For our numerical example:

P_avg = (337.5 - 3037.5) / 10 = -2700/10 = -270 W.