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Another Bar Pendulum Question

  1. Feb 21, 2007 #1
    1. The problem statement, all variables and given/known data

    How can one determine from a straight line plot, the value of g (acceleration due to gravity) and k (radius of gyration) of a bar pendulum?


    2. Relevant equations

    The time period of a bar pendulum is given by

    [tex]T = 2\pi\sqrt{\frac{I}{Mgd}}[/tex]

    [tex]I = I_{0} + Md^2[/tex]

    where I is the moment of inertia about an axis passing through the pivot, d is the distance between the center of mass and the pivot, M is the mass of the rod.

    In an experimental setup, we are varying d and measuring T for every chosen d.

    3. The attempt at a solution

    Also,

    [tex]I_{0} = Mk^2[/tex]

    So,

    [tex]I = M(k^2 + d^2)[/tex]

    Hence,

    [tex]T = 2\pi\sqrt{\frac{k^2 + d^2}{gd}}[/tex]

    For a fixed value of T, there are two values of d, and

    [tex]d_{1} + d_{2} = \frac{4\pi^2(d_{1} + d_{2})}{T^2}[/tex]

    and

    [tex]d_{1}d_{2} = k^2[/tex]

    But which straight line plot yields k and g? I can see that if k >> d, then we can say that T^2 is proportional to 1/d, but this would be a gross approximation, valid only for points very close to the center of mass.

    Thanks
    Cheers
    vivek
     
  2. jcsd
  3. Feb 22, 2007 #2
    Moment of inertia around a center of mass for perfect rod of lenght 2k is:

    [tex]I_{0}=\frac{1}{12}m(2k)^2[/tex]

    How did you got this equation:

    [tex]d_{1} + d_{2} = \frac{4\pi^2(d_{1} + d_{2})}{T^2}[/tex]
    ?

    Fixed walue of T? What two walues of [itex]d[/itex]?
     
    Last edited: Feb 22, 2007
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