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Angular Velocity Dynamics (Easy?)

  1. Dec 22, 2006 #1
    A 73.9 kg shaft with a radius of 8.9 cm is spinning at a rate of 364.5 rpm. If a board leans against the outside providing a frictional force of 59.92 N, how long will it take for the shaft to stop rotating?

    Answer says 2.094 s.

    Problem I have is that my Inertia table doesn't list shaft inertia or anything. Unsure how to go about starting this?
     
    Last edited: Dec 22, 2006
  2. jcsd
  3. Dec 22, 2006 #2

    Doc Al

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    Staff: Mentor

    Treat the shaft as a cylinder--see if that works.
     
  4. Dec 22, 2006 #3
    So Inertia = Solid cylinder = 1/2mr squared.
    I = .5 * 73.9 * .089 sq
    I = .293

    Do I use the Net Torque = 0 from here? I'm still pretty lost for some reason from here.
     
  5. Dec 22, 2006 #4
    Alright I figured it out surprisingly by conceptual thinking.

    Force: I/r
    1) 1/2mr = 3.28855
    2) * angular velocity = 3.28855 * 38.17 radians per second

    3) 125.52 / F2 = time
    = 2.094
     
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