? asked in Science & MathematicsPhysics · 2 years ago

What is the magnitude of the friction force on the disk?

The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅m2. A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 1.8 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 1.3 s.

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  • 2 years ago
    Favorite Answer

    As the wheel rotates one time, it rotates an angle of 2 π radians.

    ω = 5 * 2 π ÷ 1.8 = 10 π ÷ 1.8

    This is approximately 17.5 rad/s. Let’s determine the initial angular momentum

    L = 0.097 * 10 π ÷ 1.8 = 0.97 * π ÷ 1.8

    This is approximately 1.69 kg * m/s. Since the friction force causes the wheel to stop, the final angular momentum is 0 kg * m/s. The friction force causes the torque that stops the wheel from rotating. The distance from the axle to the pads must be in meters.

    d = 7.1 ÷ 100 =

    Torque = Ff * 0.071

    Torque * time = change of angular momentum

    Ff * 0.071 * 1.3 = Ff * 0.0923

    Ff * 0.0923 = 0.97 * π ÷ 1.8

    Ff = (0.97 * π ÷ 1.8) ÷ 0.0923

    The friction force is approximately 18.3 N. I hope this is helpful for you.

  • oubaas
    Lv 7
    2 years ago

    ω = 2PI*n/t = 10PI/1.8 = 5.555PI rad sec

    torque T = J*Δω/Δt = 0.097*(5.555PI -0)/(1.3-0) = 1.30 N*m

    F = T/d = 1.3*100/7.1 = 18.31 N

  • 2 years ago

    The friction force creates a torque on the wheel that stops it, the torque is also the time rate of change of angular momentum.

    torque x time = change in angular momentum

    t = f(0.0071) = Iw = 0.097(5)(2pi)/1.8

    f(0.0071) = 1.693

    f = 240 N

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