A stationary bicycle is raised off the ground, and its front wheel (m = 1.70 kg) is rotating at an angular velocity of 21.9 rad/s (see the figure). The front brake is then applied for 1.35 s, and the wheel slows down to 4.47 rad/s. Assume that all the mass of the wheel is concentrated in the rim, the radius of which is 0.330 m. The coefficient of kinetic friction between each brake pad and the rim is μk = 0.501. What is the magnitude of the normal force that each brake pad applies to the rim?

To solve this problem, we can use the concept of torque and angular acceleration.

First, let's calculate the angular acceleration of the wheel when the brake is applied. We can use the equation:
Δω = α * Δt, where Δω is the change in angular velocity, α is the angular acceleration, and Δt is the change in time.

Δω = 21.9 rad/s - 4.47 rad/s = 17.43 rad/s

Δt = 1.35 s

So, the angular acceleration, α = Δω / Δt = 17.43 rad/s / 1.35 s = 12.91 rad/s².

Next, we need to calculate the torque exerted by the friction force on the wheel. Torque is calculated using the equation:
τ = I * α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

The moment of inertia of a solid disk rotating around its axis is given by:
I = (1/2) * m * r², where m is the mass of the wheel and r is the radius of the wheel.

m = 1.70 kg
r = 0.330 m

I = (1/2) * 1.70 kg * (0.330 m)² = 0.92 kg.m²

Now, we can calculate the torque:
τ = 0.92 kg.m² * 12.91 rad/s² = 11.90 N.m

The friction force can be calculated using the equation:
τ = F_friction * r, where F_friction is the friction force and r is the radius of the wheel.

11.90 N.m = F_friction * 0.330 m

F_friction = 11.90 N.m / 0.330 m = 36.06 N

Since there are two brake pads, each brake pad applies half of the total normal force. Therefore, the magnitude of the normal force that each brake pad applies to the rim is 36.06 N / 2 = 18.03 N.