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Typo in Landau mechanics pendulum problem?

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Attempting a mechanics problem from Landau's mechanics (3rd edition) I get a different answer, as shown below. Error by me, or typo in the textbook? I can't find any errata page for the text, but since it's an older book, perhaps no such page is maintained.

    Chapter 1 problem 3a is to calculate the Lagrangian of a pendulum where the point of support is moving in a circle (figure in this google books url).

    2. Relevant equations

    See below.

    3. The attempt at a solution


    The coordinates of the mass are

    [tex]p = a e^{i \gamma t} + i l e^{i\phi},[/tex]

    or in coordinates

    [tex]p = (a \cos\gamma t + l \sin\phi, -a \sin\gamma t + l \cos\phi).[/tex]

    The velocity is

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

    and in the square
    [tex]\dot{p}^2 = a^2 \gamma^2 + l^2 \dot{\phi}^2 - 2 a \gamma \dot{\phi} \sin\gamma t \cos\phi + 2 a \gamma l \dot{\phi} \cos \gamma t \sin\phi=a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi).[/tex]

    For the potential our height above the minimum is

    [tex]h = 2a + l - a (1 -\cos\gamma t) - l \cos\phi = a ( 1 + \cos\gamma t) + l (1 - \cos\phi).[/tex]

    In the potential the total derivative [itex]\cos\gamma t[/itex] can be dropped, as can all the constant terms, leaving

    [tex]U = - m g l \cos\phi, [/tex]

    so by the above the Lagrangian should be (after also dropping the constant term [itex]m a^2 \gamma^2/2[/itex]
    [tex]\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi) \right) + m g l \cos\phi.[/tex]

    This is almost the stated value in the text
    [tex]\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \sin (\gamma t - \phi) \right) + m g l \cos\phi.[/tex]

    It looks like an innocent enough typo (text putting in a [itex]\gamma[/itex] instead of a [itex]\dot{\phi}[/itex]). Also oddly, there's a second reference after that point that also doesn't make sense where they refer to the omission of the total derivative [itex]m l a \gamma \cos( \phi - \gamma t)[/itex] ... a term that I didn't have when multiplying out my velocity?

    Is there consensus that there are a pair of typos here, and if not, can somebody spot the error in my calculation?
     
    Last edited: Jul 13, 2012
  2. jcsd
  3. Jul 13, 2012 #2
    Your calculation looks correct.

    However, notice that the second term in the kinetic energy is:
    [tex]
    \begin{align}
    & m a l \, \gamma \, \dot{\phi} \, \sin(\gamma t - \phi) \\
    = & -m a l \, \gamma \, (\gamma - \dot{\phi}) \, \sin(\gamma t - \phi) + m a l \, \gamma^2 \, \sin(\gamma t - \phi) \\

    = & \frac{d}{dt} \,\left(m a l \, \gamma \, \cos(\gamma t - \phi) \right) + m a l \, \gamma^2 \, \sin(\gamma t - \phi)
    \end{align}
    [/tex]
    After dropping the total time derivative, you get the master's result.
     
  4. Jul 14, 2012 #3
    Very clever (of both you and the master;). Thanks!
     
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