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Maximum angular velocity of disc skidding across surface

  1. May 15, 2017 #1
    1. The problem statement, all variables and given/known data
    coefficient of kinetic friction between the disk and the surface is 0.42

    gravity = 10.6 m/s
    disk mass = 1.75x10^9

    the skid marks are 1280m long, This is due to the fact that uneven friction had set the saucer in very slow rotation around its principal axis. By approximating the saucer by a solid disk, what is the maximum angular velocity attained during the skid?

    2. Relevant equations

    Ke = (.5)(I*w^2)


    3. The attempt at a solution

    fk*d = Ke

    (.42)(1.75x10^9)(10.6)(1280) = (1/4)(mr^2)*w^2)

    masses will cancel

    Sqrt[(4)(.42)(10.6)(1280)(1/r^2)] = w

    i get .926

    but the listed answer is .23

    Not sure If I am approaching this correctly
     
  2. jcsd
  3. May 15, 2017 #2
    got it

    Fk*d = ((.5)mVi^2) + (.5)((Iw^2)

    eventually simplified to

    Sqrt [ (4/r^2)* (.42*10.6*1280 - (1/2)Vi^2) ] = w
     
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