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Simple Harmonic Motion, rigid pendulum

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A small hole is drilled in a meter stick is to act as a pivot. The meter stick swings in a short arc as a physical pendulum. How far from the center of mass should the pivot point be for a period of 1.65 seconds?

    2. Relevant equations

    Moment of inertia as given in book: I=(1/3)ml^2
    T=2[tex]\pi[/tex][tex]\sqrt{}[/tex]I/mgh
    (where t is period, I is moment of inertia, m is mass g is gravity of course, and h is the distance from the pivot oint to the center of mass)

    3. The attempt at a solution
    I substituted the I in the second equation. The masses cancelled each other out in the square root, l is 1 since it is a meter stick and I used the given T so I end up with the equation...

    1.65=2[tex]\pi[/tex][tex]\sqrt{}[/tex]((1/3)/gh)

    And then I solved for h getting .493 meters but my book says this is incorrect.
     
  2. jcsd
  3. Apr 6, 2010 #2

    vela

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    The formula you used for the moment of inertia is wrong. The expression you used is for a long rod pivoted at the end, but that's not the case here. You'll want to use the parallel-axis theorem to get the correct expression.
     
  4. Apr 6, 2010 #3
    That's really weird because the book has an example with a meter stick and uses that formula for I.
     
  5. Apr 6, 2010 #4

    vela

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    In the example, the meter stick is probably pivoted at its end. That's not what's happening in this problem, so you can't use that expression for I.
     
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