A stationary bicycle is raised off the ground, and its front wheel (m = 1.20 kg)

Question

A stationary bicycle is raised off the ground, and its front wheel (m = 1.20 kg) is rotating at an angular velocity of 13.6 rad/s (see the figure). The front brake is then applied for 4.73 s, and the wheel slows down to 2.47 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 ?k = 0.948. What is the magnitude of the normal force that each brake pad applies to the rim?

Explanation / Answer

Inertia of the rim = 1.20*(0.33)^2 = 0.131and alpha = (13.6-2.7)/4.73 = 2.304 Torque = I*alpha = 0.30749 = 2*k *N ==> N = 0.1621

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