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Find period of circular disk physical pendulum

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform circular disk whose radius R is 40.0 cm is suspended as a physical pendulum from a point on its rim.

    (a) What is its period of oscillation?

    (b) At what radial distance r < R is there a point of suspension that gives the same period?

    R = .40 m
    g = 9.81 m/s^2
    h = .40m

    2. Relevant equations

    T = 2pi(I/(mgh))^.5
    I = .25mR^2

    3. The attempt at a solution

    I don't understand why R is .40 m but h isn't. I arrived at a solution of .63 s, but the actual solution is 1.55 s.
     
  2. jcsd
  3. Jun 18, 2009 #2

    cepheid

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    Your moment of inertia is wrong.

    EDIT: Have you heard of the parallel axis theorem?
     
    Last edited: Jun 18, 2009
  4. Jun 19, 2009 #3
    Read a little on it once you mentioned it. The axis of rotation Rz, and is perpendicular to the disk and goes through the attachment point, right?
     
  5. Jun 19, 2009 #4

    cepheid

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    Yes, the axis of rotation is perpendicular to the disk surface and does pass through the pivot point. I am not sure what you mean by, "The axis of rotation Rz" though.

    Anyway, basically you know the moment of inertia of the disk around its centre of mass, and can use this result plus the parallel axis theorem to determine the moment of inertia for rotation around the pivot point.
     
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