# 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.

### 3 Answers

- electron1Lv 72 years agoFavorite Answer
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.

- oubaasLv 72 years ago
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

- 2 years ago
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