Question

The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅ m 2 . A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 2 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 1.4 s . What is the magnitude of the friction force on the disk?

Answer 1

Here,

I = 0.097 Kg.m^2

inital angular speed, w = 5 * 2pi/2 rad/s

w = 15.71 rad/s

let the frictional force is f

NOw, Using second law of motion

f * 0.071 * 1.4 = 0.097 * (15.71)

f = 15.3 N

the magnitude of frictional force is 15.3 N