Register to reply 
Constrained Lagrangian equetion (barbell) 
Share this thread: 
#1
Jun1613, 03:14 AM

P: 2

Hi!
I tried to compute an ideal barbellshaped 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 :) 


#2
Jun1613, 08:14 AM

P: 1,520




#3
Jun1613, 08:36 AM

P: 2




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 