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Kinetic Energy in Rotational Motion Problem

  1. Nov 15, 2006 #1
    Ok the problem is:

    Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of 1.30 m and a mass of 72.0 kg. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600 m/s^2.

    What I did was solve for the angular velocity through the radial acceleration:

    3600m/s^2 = rw^2

    w = 52.6 rad/s

    Then I solved for the moment of inertia:

    I = mr^2 = 72.0kg(1.30m)^2 = 122 kg*m^2

    Finally I plugged it all into the rotational kinetic energy equation:

    K = (1/2)(122m*m^2)(52.6rad/s)^2 = 1.68*10^5 J

    The actual answer is 8.42*10^4, exactly half of what I got. I don't suppose someone could explain where the *(1/2) is coming from?
     
  2. jcsd
  3. Nov 15, 2006 #2

    OlderDan

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    The moment of inertia of a disk is??????
     
  4. Nov 15, 2006 #3
    hehe (1/2)mr^2
     
  5. Apr 7, 2010 #4
    Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of = 1.16m and a mass of 73.0 kg. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3510 m/s^2.

    CAN SOMEONE PLEASE SOLVE THIS FFS IVE TRIED SO MANY OPTIONS AND ITS NOT WORKING!@@!_#!@#!@!@#)!@(#*)!@(#)((!@*
     
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