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Angular frequency of a pendulum

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data

    A physical pendulum consists of a small 1.5kg mass at the bottom end of a uniform 1.00m long 1.5kg stick swinging about its upper end. The moment of inertia of the pendulum about its upper end is 2.00kg*m^2. What is the angular frequency


    2. Relevant equations

    sqrt(mgd/I)=w

    I = I(cm)+md^2 = I = (1/3)(1.5)(1^2)+ (1.5)(d^2) = 2



    3. The attempt at a solution

    m = 1.5+1.5
    g = 9.81
    d = 1
    I = 2

    w = sqrt([3*9.81*1]/2) = 3.83 rad/s

    According to my solutions manual this is wrong and the correct answer is 3.32 rad/s and somehow d = .75. I don't know how they calculated that, can anyone help?
     
  2. jcsd
  3. Apr 1, 2012 #2

    ehild

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    Do you know what d means in the formula for angular frequency?

    ehild
     
  4. Apr 1, 2012 #3
    I thought d was distance, but I am guessing that it isn't?
     
  5. Apr 1, 2012 #4

    ehild

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    Distance of what?

    ehild
     
  6. Apr 1, 2012 #5
    Wait, is supposed to be distance to center mass? The lecture notes don't say this, but I think this is right.

    [1.5(.5)+1.5(1.0)]/3 = .75
     
  7. Apr 1, 2012 #6

    ehild

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    Yes, it is the distance of the CM from the pivot.:smile:

    ehild
     
  8. Apr 1, 2012 #7
    Thanks
     
  9. Apr 2, 2012 #8
    Can someone explain to me how this was done? Is that the equation for the center of mass? Where did the number 0.5 come from? Thanks!
     
  10. Apr 2, 2012 #9
    Yes that is the center of mass equation.

    The 0.5 comes from the uniform rod of 1m.
     
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