Constrained Lagrangian equetion (barbell)

1. Jun 16, 2013

Jengalex

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. Jun 16, 2013

lightarrow

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.

3. Jun 16, 2013

Jengalex

Yes I've tried it now, it looks fine. Thanks!