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Shortest period of a simple pendulum

  1. Aug 28, 2009 #1
    A uniform disk of radius 0.2m and 5.6 kg mass has a small hole distance d from the disk's center that can serve as a pivot point. What should be the distance d so that this physical pendulum will have the shortest possible period? What will be the period at this distance?

    So, I know that the period is defined as: T=2[tex]\Pi[/tex][tex]\sqrt{\frac{I}{mgd}}[/tex]
    And I know that the moment of Inertia is I=[tex]\frac{1}{2}[/tex]mr[tex]^{2}[/tex]

    The first part of the question, which I omitted, asks to find the distance d for a specified period and I got that portion with no problem. I figured that I needed to take the derivative of the equation and set it equal to zero to find the minimum period but when I do that I simple come up with zero... Not quite sure where I am making the error here!!!
    Last edited: Aug 28, 2009
  2. jcsd
  3. Aug 29, 2009 #2


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    All I can deduce from your equation is that as d→0, mgd→0 so the fraction of I/mgd→0 and T→0 as well.
  4. Aug 29, 2009 #3


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    I is the pendulum's moment of inertia around the pivot, not around the center of mass.
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