# What is the magnitude of the friction force on the disk?

The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅m2. A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 1.8 s. To bring the wheel to rest, the mechanic gently applies the disk brakes, which squeeze pads against a metal disk connected to the wheel. The pads touch the disk 7.1 cm from the axle, and the wheel slows down and stops in 1.3 s.

Relevance

As the wheel rotates one time, it rotates an angle of 2 π radians.

ω = 5 * 2 π ÷ 1.8 = 10 π ÷ 1.8

This is approximately 17.5 rad/s. Let’s determine the initial angular momentum

L = 0.097 * 10 π ÷ 1.8 = 0.97 * π ÷ 1.8

This is approximately 1.69 kg * m/s. Since the friction force causes the wheel to stop, the final angular momentum is 0 kg * m/s. The friction force causes the torque that stops the wheel from rotating. The distance from the axle to the pads must be in meters.

d = 7.1 ÷ 100 =

Torque = Ff * 0.071

Torque * time = change of angular momentum

Ff * 0.071 * 1.3 = Ff * 0.0923

Ff * 0.0923 = 0.97 * π ÷ 1.8

Ff = (0.97 * π ÷ 1.8) ÷ 0.0923

The friction force is approximately 18.3 N. I hope this is helpful for you.

• ω = 2PI*n/t = 10PI/1.8 = 5.555PI rad sec

torque T = J*Δω/Δt = 0.097*(5.555PI -0)/(1.3-0) = 1.30 N*m

F = T/d = 1.3*100/7.1 = 18.31 N

• The friction force creates a torque on the wheel that stops it, the torque is also the time rate of change of angular momentum.

torque x time = change in angular momentum

t = f(0.0071) = Iw = 0.097(5)(2pi)/1.8

f(0.0071) = 1.693

f = 240 N