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Homework Help: Problem 4, Landau/Lifshitz, page 12

  1. Apr 25, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    The problem can be found there, page 12, problem 4: http://books.google.com.ar/books?id...ntcover&dq=mechanics&cd=1#v=onepage&q&f=false.



    2. Relevant equations
    [tex]\vec v = \vec \omega \wedge \vec r[/tex].
    [tex]L=L_1+L_2+L_3[/tex].

    3. The attempt at a solution
    I've made an attempt and then saw the solution given in the book and I don't get it. First, there is no mention of a gratitational field, hence why is there a "g" term in the solution?
    My attempt: None of the 3 particles have potential energy, hence the Lagrangian is simply the sum of the kinetic energies of the particles.
    For each one of the 2 particles [tex]m_1[/tex], I've found that [tex]T=\frac{m_1}{2} \Omega ^2 l^2 \sin ^2 (\phi)[/tex] since they describe a circular motion of radius [tex]l \sin \phi[/tex]. So it only remains to find the kinetic energy of [tex]m_2[/tex].
    Choosing the origin at point A, [tex]x=2l \cos \phi[/tex]. Now to calculate [tex]\dot x[/tex], I think that only [tex]\phi[/tex] may vary thus [tex]\dot x = -2l \dot \phi \sin \phi[/tex]. So I get [tex]T_2=2m_2 l^2 \sin ^2 (\phi )[/tex].
    Which gives me [tex]L=l^2 \sin ^2 (\phi) (m_1 \Omega ^2 +2 \dot \phi ^2 m_2)[/tex].
    I wonder if my answer is correct if I assume no external gravitational field. L&L didn't specify the field but in the answer there's a "g"...
    Thanks for any kind of help.
     
  2. jcsd
  3. Apr 26, 2010 #2

    gabbagabbahey

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    Usually whenever the word "vertical" is used in a mechanics problem, it's safe to assume there is an earth-like gravitational field [itex]g[/itex] present.

    Even without a gravitational field present, wouldn't there be an additional contribution to the kinetic energy of each [itex]m_1[/itex] when [itex]\theta[/itex] is changing?

    And your kinetic energy for [itex]m_2[/itex] makes no sense to me...did you mean [tex]T_2=2m_2 l^2\dot{\theta}^2\sin^2\theta[/tex]?

    Edit: Maybe your version of the text uses different labels for the angles...in my version, [tex]\Omega=\dot{\phi}[/tex] and [itex]\theta[/itex] represents the angle the massless rods make with the axis of rotation.
     
  4. Apr 26, 2010 #3

    fluidistic

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    Ok thanks a lot. I will remember this.
    Hmm. Ok a (or 2?) term is missing. I think I've calculated the KE the masses have with their circular motion. It would remain the term of the vertical (and horizontal?) KE. Unless I'm misunderstanding something.
    Yes I meant this, I made a typo error.

    I see. So when you said "theta" in the upper question, which notation did you use?
     
  5. Apr 26, 2010 #4

    gabbagabbahey

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    Not a factor of 2, but a term involving [itex]\dot{\theta}[/itex], corresponding to the motion of each [itex]m_1[/itex] when the rotating frame is stretched or squished.

    The one where theta is the angle the upper supports make with the axis of rotaion/
     
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