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Constrained Lagrangian equetion (barbell)

  1. Jun 16, 2013 #1
    Hi!

    I tried to compute an ideal barbell-shaped object's dynamics, but my results were wrong.
    My Langrangian is:

    ## L = \frac{m}{2} ( \dot{x_1}^2 + \dot{x_2}^2 + \dot{y_1}^2 + \dot{y_2}^2 ) - U( x_1 , y_1 ) - U ( x_2 , y_2 ) ##

    And the constraint is:

    ## f = ( x_1 - x_2 )^2 + ( y_1 - y_2 )^2 - L^2 = 0 ##

    I dervated L + λf by ## x_1, x_2, y_1, y_2 ## and λ:

    ## m \ddot x = - \frac{\partial U}{\partial x_1 } + \lambda ( x_1 - x_2 ) ## (1)
    Four equtions similar to this and the constraint.

    Then I expressed ## \ddot x_2 , \ddot y_1 , \ddot y_2 ## with ## \ddot x_1 , x_1 , x_2 , y_1 , y_2 ## and U's partial derivates' local values:

    ## m \ddot x_2 + \frac{\partial U}{\partial x_2} = - m \ddot x_1 - \frac{\partial U}{\partial x_1} ##
    ## m \ddot y_2 + \frac{\partial U}{\partial y_2} = - m \ddot y_1 - \frac{\partial U}{\partial y_1} ##
    ## ( m \ddot y_1 + \frac{\partial U}{\partial y_1} )(x_1 - x_2) = (- m \ddot x_1 - \frac{\partial U}{\partial x_1})(y_1 - y_2) ## (2)

    After expressing ## (x_1 - x_2) , (y_1 - y_2) ## and ## \lambda ## from equation (1) and (2), substituted it into the constraint eqution I got an equation like this:

    ## (m \ddot x_1 + \frac{\partial U}{\partial x_1} ) * <something> = 0 ##

    I think it's wrong.
    Can you confirm or point on my mistake?
    Thanks :)
     
    Last edited: Jun 16, 2013
  2. jcsd
  3. Jun 16, 2013 #2
    I don't know what a "barbell-shaped dynamics" is, but if in your problem in the plane with two points there is a constraint, the degrees of freedom are 3, not 4, so I would have written the Lagrangian as function of 3 independent generalized coordinates, for example x1, y1 and the angle between the line connecting the two points (P1 = (x1,y1); P2 = (x2,y2)) and the x axis.
     
  4. Jun 16, 2013 #3
    Yes I've tried it now, it looks fine. Thanks!
     
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