Question

A stationary bicycle is raised off the ground, and its front wheel (m = 1.3 kg) is rotating at an angular velocity of 14.2 rad/s (see the drawing). The front brake is then applied for 3.0 s, and the wheel slows down to 3.2 rad/s. Assume that all the mass of the wheel is concentrated in the rim, the radius of which is 0.34 m. The coefficient of kinetic friction between each brake pad and the rim is μ_k = 0.85. What is the magnitude of the normal force that each brake pad applies to the rim?

Answer 1

SOLUTION :

Angular acceleration  

( α ) = Δω / Δt

= (3.2 rad / s - 14. 2 rad / s ) / 3 s

= - 3.67 rad/ s²

Torque

I α = M R²α

Torque

μ N R

Where N is the total normal force applied by the (both) pads

M R α = μ N

Normal force on two pads  

( N ) = 1.3 kg * 0 .34 m * 3.67 / 0 .85

= 1.9084 N

F_N (each pad)

==> 1.9084 N / 2

===> 0.9542 N