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Rotational inertia and torque

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 120g Frisbee is 28cm in diameter and has about half its mass spread uniformly in a disk, and the other half concentrated in the rim. With a quarter-turn flick of the wrist, a student sets the Frisbee rotating at 560rpm.

    (a) What is the rotational inertia of the Frisbee?

    (b) What is the magnitude of the torque, assumed constant, that the student applies?

    2. Relevant equations
    I=1/2*m*r2

    I=m*r2

    [tex]\tau=I\alpha[/tex]

    3. The attempt at a solution

    I got the first part of this by using I=1/2*.60kg*.14m2 for the disk then adding I=.60kg*.14m2 for the rim which gives me 0.001764kg/m2
    The part that is messing me up is finding the torque, I think the best way to find the torque is to find the angular acceleration of the frisbee, but I havn't been able to get it. I am pretty sure I need to use the rotational/kinematic equations... Any help would be great.
     
  2. jcsd
  3. Mar 11, 2009 #2

    LowlyPion

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    Welcome to PF.

    First of all maybe recalculate your I ?
     
  4. Mar 11, 2009 #3

    LowlyPion

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    So long as you convert angles to radians and ω to radians/sec then you can use the kinematic analogs to motion:

    http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

    You are accelerating the frisbee from rest to 560 rpm over 1/4 a turn (according to the problem anyway).
     
  5. Mar 11, 2009 #4
    Thanks LowlyPion, I didn't realize it was only a quarter turn.
     
  6. Mar 11, 2009 #5

    LowlyPion

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    Be sure and correct your moment of Inertia calculation.

    3/2(.60)(.14)2 is not .00176

    Edit: Wait. I see it should have been .06 kg not as you wrote it. Your answer for I is correct making that change.
     
    Last edited: Mar 11, 2009
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