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Homework Help: Normal modes / Rod on a string

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform rod of length a hangs vertically on the end of an inelastic string of
    length a, the string being attached to the upper end of the rod. What are the
    frequencies of the normal modes of oscillation in a vertical plane?

    Answer: [tex]\omega^2 = (5 \pm \sqrt{19})g/a[/tex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I have tried a few attempts, all failed, like trying to take 2 pivots, one at the top of the string, one at the string-rod interface, or the rod COM, and trying to use torque and such like...no successes, so if someone could push me in the right direction, I would be grateful.
     
  2. jcsd
  3. Feb 17, 2010 #2

    vela

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    I wrote down a Lagrangian in terms of two coordinates, the angle the string makes with the vertical and the angle the rod makes with the vertical. Using small angle approximations, I was able to derive those normal mode frequencies.

    What level course is this for?
     
  4. Feb 17, 2010 #3
    2nd year undergrad, we havent done Lagragian mechanics yet though...

    I will try it anyway...How can I express the translational Kinetic energy of the rod?
     
    Last edited: Feb 17, 2010
  5. Feb 17, 2010 #4
    ok, is it:

    [tex](1/2)I\dot{\phi}^2 + (m/2) (d/dt((a/2)sin\phi + asin\theta))^2[/tex]

    I is the moment of inertia of rod about 1 end, (1/3)ma^2

    phi is the angle the rod is to vertical

    theta is angle of string to vertical

    and once i havent the lagragian in terms of the angles, do I minimize the action or?
     
    Last edited: Feb 17, 2010
  6. Feb 18, 2010 #5

    vela

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    You've only included its motion in the horizontal direction; you need to account for its motion vertically too. Also, you have to subtract the potential energy to get the Lagrangian.

    But as you haven't learned about the Lagrangian formulation of mechanics, I don't think this is going to be a very wise use of your time. (Besides, I just kind of hand-waved my way to the answer from piecing together stuff from my old classical mechanics homework. I wouldn't be able to explain why my guess worked.)
     
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