Register to reply

Constrained Lagrangian equetion (barbell)

by Jengalex
Tags: barbell, constrained, equetion, lagrangian
Share this thread:
Jengalex
#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 :)
Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
lightarrow
#2
Jun16-13, 08:14 AM
P: 1,521
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
#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!


Register to reply

Related Discussions
Raychaudhuri equetion Special & General Relativity 2
Rotating barbell? Introductory Physics Homework 1
Derive lagrangian: finding potential energy of a particle constrained to a surface Advanced Physics Homework 4
Barbell vs dumbell General Physics 3
Barbell Underwater Introductory Physics Homework 6