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Lagrangian of a system (First pages of L&L)

  1. Apr 16, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    The problem can be found in L&L's book "Mechanics" in the end of the first chapter. (See the last picture of page 12 of http://books.google.com.ar/books?id...v+Davidovich+Landau"&cd=2#v=onepage&q&f=false ). The mass m1 is free to move on the x-axis. While the mass m2 moves like a pendulum. There's the gravitational acceleration [tex]\vec g[/tex]. I must find the Lagrangian of such a system.
    I keep getting [tex]L=\frac{m_1 \dot x ^2}{2}+\frac{m_2 }{2} \left [ \dot x ^2 +2 \dot{\vec x} \dot \theta l \cos (\theta) + \dot \theta ^2 l^2 +2gl \cos (\theta) \right ][/tex]. I see that I have an error: I have a vector [tex]\dot {\vec x }[/tex] which is impossible.
    L&L's answer is [tex]L\frac{1}{2} (m_1+m_2) \dot x^2 +\frac{1}{2}m_2 (l^2 \theta ^2 +2l \dot x \theta \cos \theta)+m_2 g l \cos \theta[/tex].
    I don't understand what I'm doing wrong. Also L&L don't have any [tex]\dot \theta[/tex] term... Mine appeared when I calculated T_2, the kinetic energy of m_2 in polar coordinates. When I derivated the position [tex]\vec r_2 =(\dot {\vec x} + l \sin \theta )\hat i + (l \cos \theta) \hat j[/tex] with respect to time I got some [tex]\dot \theta[/tex] terms.
     
    Last edited: Apr 16, 2010
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  3. Apr 16, 2010 #2

    Gokul43201

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    Looks like a couple of typos in that version of L&L - the [itex]l \theta[/itex] terms in the solution ought to be [tex]l \dot \theta[/tex]. You can tell by inspection that the posted solution is dimensionally off (missing a factor of 1/time).

    As for your concern about having an vector [itex]\mathbf{\dot x}[/itex] ... what about the other vector in that term: [tex]\mathbf{\dot \theta}[/tex]?
     
    Last edited: Apr 16, 2010
  4. Apr 16, 2010 #3

    fluidistic

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    Ok thanks for the clarification.
    If I understand you, [tex]\theta[/tex] is a vector? I'm not really swallowing it. Can you confirm this? It would makes sense anyway, but I don't understand it. Hmm not sure it would make sense since I'd have a [tex]\cos (\text{vector})[/tex] term which doesn't seem OK.
     
  5. Apr 16, 2010 #4

    Gokul43201

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    No, my LaTeX is messed up. For some reason, I can't get inline tex tags to work properly (with \vec and \dot). Fixing it ... give me a minute.

    EDIT: Fixed now.
     
  6. Apr 16, 2010 #5

    fluidistic

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    Ok perfect. I just found the same problem "explained in wikipedia" (see http://en.wikipedia.org/wiki/Lagrangian_mechanics#Pendulum_on_a_movable_support) but I still have my problem with [tex]\dot \vec x[/tex].
    Here is what I do: [tex]\vec r_2 = (\vec x + l \sin \theta)\hat i +(l \cos \theta) \hat j[/tex].
    In order to find the kinetic energy of [tex]m_2[/tex] (or m in wikipedia's article), I need the velocity squared. The velocity is [tex]\dot \vec r_2 =(\dot \vec x + \dot \theta l \cos \theta)\hat i +(- \dot \theta l \sin \theta) \hat j[/tex]. Am I right until now?
    So that [tex]|\dot \vec r_2|^2=(\dot \vec x + \dot \theta l \cos \theta)^2+(- \dot \theta l \sin \theta)^2= \dot x ^2 + 2 \underbrace{\dot \vec x}_{\text{this term}} \dot \theta l \cos( \theta ) + \dot \theta ^2 l^2 \cos ^2 (\theta)[/tex].
    EDIT: I FIND MY MISTAKE! A beginner one. Of course, there is no vector. I expressed the components of r (a vector) as vectors! Problem solved!
    Thanks for your time. Now I go back to play with these Lagrangians. :)
     
    Last edited: Apr 16, 2010
  7. Apr 16, 2010 #6

    fluidistic

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    Out of curiosity, I've plugged the Lagrangian into Euler-Lagrange equation and since I have 2 generalized coordinates (x and theta), I've found 2 equations. Namely the one that is in wikipedia: [tex]\ddot x \cos (\theta) +l \ddot \theta + g \sin (\theta)=0[/tex] if I take [tex]\theta[/tex] as variable and another one if I take x as variable: [tex]m_2 \ddot x +2l \left [ \ddot \theta \cos (\theta) -\dot \theta ^2 \sin \theta \right ]=0[/tex].
    Are they both the equations of motion? Or there's just one of these equations that describe the whole motion of the system?
     
  8. Apr 20, 2010 #7

    Gokul43201

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    How many equations would you require to describe the motion of say a projectile?
     
  9. Apr 20, 2010 #8

    fluidistic

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    I think I made an error in my last post (I've redone the arithmetics and got a different answer), but my question doesn't change.
    In this case I think I'd need 2 equations. One for the horizontal component of the position and one for the vertical component of the position vector; that is if I'm using Cartesian coordinates. If I derivate I get the velocity and if I derivate again I get the acceleration. I'd be done if I'm given the 2 initial conditions [tex](\vec x_0, \vec v_0)[/tex]. I hope I'm right on this.

    But still, in the case of this pendulum I get 2 equations involving theta, x and their derivatives. Both equations are worth 0. Hence I can equate them and write the first minus the second =0 and I've all the information within 1 single equation. Is this right?
     
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