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Rotational KE problem

  1. Mar 17, 2009 #1
    The problem statement, all variables and given/known data

    Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of = 1.30m and a mass of 66.0kg . To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3600m/s² .
    What is the maximum kinetic energy that can be stored in the flywheel?


    The attempt at a solution

    I took I = 3/2MR² (at rim)
    where R = 2.6m
    Then K = 1/2 * I * ω², where ω² = a/R

    so K = (1/2)(3/2)(66kg)(2.6m)²(3600/2.6) = 463320J
    which is wrong..
    The ans is 7.72x10^4 J


    can someone pls tell me where i gone wrong?
    Thanks!!
     
  2. jcsd
  3. Mar 17, 2009 #2

    rl.bhat

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    Moment of inertia of flywheel is 1/2*M*R^2 and radius is 1.3 m.
     
  4. Mar 17, 2009 #3
    Radial acceleration = [tex]m r^2 \omega[/tex]
    Find out [tex]\omega[/tex]
    Calc. K.E using the equation [tex]\frac{1}{2} I \omega^2[/tex]..
     
  5. Mar 17, 2009 #4

    LowlyPion

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    Not quite. The equation is correct, but ω = v/r = 3600/1.3

    As noted by rl.bhat a uniform disk has a moment of inertia of of 1/2mr².
     
  6. Mar 17, 2009 #5
    The Moment of Inertia isn't always 1/2 mr^2, is it?
    It does depend upon the axis chosen, with M.I. of the new axis found out using parallel and perpendicular axis theorems.
     
  7. Mar 17, 2009 #6

    LowlyPion

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    Rotating about its central axis it is.
    http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html#icyl2

    The || and ⊥ axis theorem are useful of course for other axes of rotation.
     
  8. Mar 20, 2009 #7
    i got it. thanks!
     
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