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Homework Help: Classical Mechanics: Lagrangian for pendulum with oscillating support

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Greetings! This is an example problem at the end of Chapter 1 in Mechanics (Landau):
    A simple pendulum of mass m whose point of support oscillates horizontally in the plane of motion of the pendulum according to the law [tex] x=acos(\gamma t) [/tex].
    Find the Lagrangian.

    2. Relevant equations

    3. The attempt at a solution
    [tex] x = a cos(\gamma t) + l sin(\phi) ; y = l cos(\phi) [/tex]

    Where, [tex] \phi [/tex] is the angle the pendulum makes with the vertical. Then:

    \dot{x}= -a\gamma sin(\gamma t)+l cos(\phi) \dot{\phi} ; \dot{y} = -l sin(\phi) \dot{\phi}[/tex]

    L = T - U = \frac{1}{2} m ( \dot{x}^2 + \dot{y}^2) - U

    Which gives (ignoring full time derivative terms):
    L = \frac{1}{2} ml^2 \dot{\phi}^2 - m l a \gamma sin (\gamma t) cos(\phi)\dot{\phi} + m g l cos(\phi)

    Answer given in the book:
    L = \frac{1}{2} ml^2 \dot{\phi}^2 + m l a \gamma^2 cos (\gamma t) sin(\phi) + m g l cos(\phi)

    My solution is different from the one given in the book. Could someone please tell me where I'm going wrong?
  2. jcsd
  3. May 12, 2015 #2
    for sure, there is a total derivated connecting both.
  4. May 12, 2015 #3


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    Funny, the internet pdf (2nd edition) has a problem 2 and a problem 3 that show similarities. If I look at problem 2 (and remove m1) that's just what you find:
    in your case x(t) is given, and ##(-a\gamma \sin\gamma t)^2 ## isn't interesting (can you tell why not?), so that leaves your expression !

    Note that ##a sin\gamma t## is not as clear as ##a\sin\gamma t## ( a \sin \gamma t ) because some folks might interpret the ##asin## as ##\arcsin## ( \arcsin ).
    In addition, TeX recognizes the \sin as a function and takes care of proper spacing !

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