physics

20 kg , 30 cm diameter disk is spinning at 300 rpm. How much friction force must be applied to rim to bring the disk to a halt in 3.0s?

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asked by Justin
  1. The angular deceleration rate must be
    alpha = (300*2 pi/60 rad/s)/3.0 s
    = (10/3) pi = 10.47 rad/s^2

    The required torque L is given by
    L = I*alpha,
    where I is the moment of inertia of the disc, (1/2) M R^2

    Therefore

    L = (1/2) M R^2 *10.47 rad/s^2
    The units will be Newton-meters

    Once you have determined L, use
    F = L/R for the required friction force, in Newtons.

    R is, of course, the radius, 0.15 m.

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    posted by drwls
  2. Thanks

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    posted by Justin
  3. Thanks for the help!

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    posted by Amy

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