Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


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


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


    [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]


    [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.

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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook