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Physical Pendulum least period

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data
    In Fig. 16-41, a stick of length L = 1.6 m oscillates as a physical pendulum. (a) What value of distance x between the stick's center of mass and its pivot point O gives the least period? (b) What is that least period?
    [​IMG]



    2. Relevant equations
    well i'm assuming you have to use the equation for a a physical pendulum which is
    T= 2pi [tex]\sqrt{I/mgh}[/tex]
    where I is the rotational inertia, m is the mass, g is gravity, and h is the distance between the axis of rotation and the centre of the pendulum.

    3. The attempt at a solution
    i tried using that equation but i dont understand the value for I because im assuming if the value for h changes then the other end of the rod changes as well.
     
  2. jcsd
  3. Nov 25, 2007 #2
    Moment of Inertia of the center of mass is

    [tex]I_0=\frac{mL^2}{12}[/tex]

    If it is pivoted distance x from the centre

    [tex]I=I_0+mx^2[/tex]

    [tex]T=2\pi\sqrt{\frac{I}{mgx}}=2\pi\sqrt{\frac{I_0+mx^2}{mgx}}[/tex]

    for T minimum we have
    [tex]\frac{dT}{dx}=0[/tex]
     
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