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Normal Modes - Pendulum on a Moving Block

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A block of mass M can move along a smooth horizontal track. Hanging from the
    block is a mass m on a light rod of length l that is free to move in a vertical plane
    that includes the line of motion of the block. Find the frequency and displacement
    patterns of the normal modes of oscillation of the system firstly by `spotting' the
    normal modes of the system and

    then secondly by writing the Lagrangian L = T - U
    for the system and solving the Euler-Lagrange equations.

    2. Relevant equations

    [tex]\frac{d}{dt}\frac{\delta L}{\delta\dot{q}_i} = \frac{\delta L}{\delta\dot{q}}[/tex]

    3. The attempt at a solution

    Mode 1- Translation
    Mode 2- Pendulum swings, at the same time the Block also oscillates from side to side?

    Kinetic Energy:

    [tex]\frac{1}{2} M\dot{x}^2 + \frac{1}{2}ml^2\dot{\theta}^2[/tex]

    Potential Energy:

    [tex]mg(l - l cos\theta)[/tex]

    Then write down the Lagrangian as L = T-U

    applying the Euler-Lagrange equations for variables [tex]x[/tex] and [tex]\theta[/tex]

    I get [tex]M\dot{x} = 0[/tex]

    and for the [tex]\theta[/tex] coordinate, I just get the trivial pendulum equation....with [tex]\omega = \sqrt{g/l}[/tex]


    Is that it? because this doesnt seem to have yielded me the answer they're looking for I guess...
     
  2. jcsd
  3. Feb 23, 2010 #2

    Gokul43201

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    Is that all? Aren't you missing a term here (for the hanging mass)?
     
  4. Feb 23, 2010 #3
    am I?

    isnt that taken into account by the second term?
     
  5. Feb 23, 2010 #4
    aha! you mean by the translation of the hanging mass as well as the swinging right?
     
  6. Feb 23, 2010 #5

    vela

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    You've omitted the kinetic energy of the pendulum due to translation, for one thing. One approach to obtaining the Lagrangian is to write down the position of each mass, differentiate it with respect to time to get the velocity of the mass, and use that result to express the kinetic energy. For instance, the horizontal position of the mass m is equal to, using your symbols, [itex]x+l \sin\theta[/itex]. The horizontal component of its velocity is then [tex]\dot{x}+(l \cos\theta) \dot{\theta}[/tex]. You can already see the kinetic energy of mass m will have contributions from the pendulum's motion, from the whole system translating, and from a cross term.
     
  7. Feb 23, 2010 #6
    can I write the KE as:

    [tex]\frac{1}{2} m\dot{x}^2 + \frac{1}{2} ml^2\dot{\theta}^2[/tex]

    or do I have to write it as x and y components?
     
  8. Feb 23, 2010 #7

    vela

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    When you include all the components, it'll reduce down to those two components plus, it appears, another cross term. That cross term is important because it corresponds to the interaction between system's translational motion and the oscillatory motion of the pendulum.
     
  9. Feb 23, 2010 #8
    After everything, I got:

    [tex]M \ddot{x} + m \ddot{x} + ml \ddot{\theta}cos \theta - ml\dot{\theta}^2 sin\theta = 0[/tex]

    [tex]m \ddot{x} lcos\theta - m \dot{x} l \dot{\theta} sin \theta + ml^2 \ddot{\theta} - mglsin \theta = 0[/tex]


    how the heck can I solve this?
     
  10. Feb 23, 2010 #9

    vela

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    I haven't actually worked it out, so maybe there's a better way, but this is what I'd try: One of the equations should have been d/dt(something)=0, so the something is a constant. You can use that to eliminate [itex]\dot{x}[/itex] from the other equation, leaving only [itex]\theta[/itex]s. You also want to assume [itex]\theta[/itex] is small, so you can use the small-angle approximations for sine and cosine.
     
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