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Physical Pendulum Problem

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform disk of radius R = 1.40 m and a 6.0 kg mass has a small hole a distance d from the disk's center that serves as a pivot point.

    What should be the distance d so that this physical pendulum will have the shortest possible period?

    2. Relevant equations

    T = 2π[tex]\sqrt{\frac{I}{MgL}}[/tex]

    3. The attempt at a solution

    Using the parallel axis theorem and moment of inertia of a disk I found the period as:

    T = 2π[tex]\sqrt{\frac{\frac{1.4^{2}}{2}+d^{2}}{9.8d}}[/tex]

    When I find the minimum of this function I get d = 0.990 which is the right answer.

    What I don't understand is why the d isn't zero. Won't T approach zero as d approaches zero?
     
    Last edited: Nov 14, 2009
  2. jcsd
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