A stationary bicycle is raised off the ground, and its front wheel (m = 1.3 kg)

Question

A stationary bicycle is raised off the ground, and its front wheel (m = 1.3 kg) is rotating at an angular velocity of 11.6 rad/s (see the drawing). The front brake is then applied for 3.0 s, and the wheel slows down to 3.5 rad/s. Assume that all the mass of the wheel is concentrated in the rim, the radius of which is 0.33 m. The coefficient of kinetic friction between each brake pad and the rim is muk = 0.85. What is the magnitude of the normal force that each brake pad applies to the rim? ___________ N

Explanation / Answer

Given that

Mass of bicycle wheel m = 1.3 kg

Initial angular speed of wheel i = 11.6 rad/s

Final angular speed of the wheel f = 3.5 rad/s

The braking force applied for a time t = 3.0 s

Radius of the wheel r = 0.33 m

Coefficeint of friction k = 0.85

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From rotational dynamics

             f = i + t

then angular acceleration

              = (f - i)/t

                = (3.5 rad/s - 11.6 rad/s)/3.0s

                = -2.7 rad/s2

Now the torque acting on thewheel is

                  = I

                fr = (mr2)

        (k Fn)r = (mr2)

Therefore the normal force

             Fn = mr / k

                 = (1.3 kg)(0.33 m) ( 2.7 rad/s2) / 0.85

                 = 1.36 N

Therefore each brake pad applies

           Fn' = 1.36 N/2 = 0.68 N

           

      

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