Constrained Lagrangian equetion (barbell)


by Jengalex
Tags: barbell, constrained, equetion, lagrangian
Jengalex
Jengalex is offline
#1
Jun16-13, 03:14 AM
P: 2
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 :)
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lightarrow
lightarrow is offline
#2
Jun16-13, 08:14 AM
P: 1,504
Quote Quote by Jengalex View Post
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 :)
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.
Jengalex
Jengalex is offline
#3
Jun16-13, 08:36 AM
P: 2
Quote Quote by lightarrow View Post
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.
Yes I've tried it now, it looks fine. Thanks!


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