Lagrange multipliers in a simple pendulum

In summary, the conversation discusses the process of using Lagrange multipliers and constraint equations to find the equation of motion and forces in a simple pendulum system. The modified Lagrangian is written, and the kinetic and potential energies are calculated. The Euler-Lagrange equations yield three generalized coordinates, including the tension force, which is found to be the tension force. The conversation also addresses some uncertainties and clarifications in the process.
  • #1
fluidistic
Gold Member
3,926
262

Homework Statement


Not really a homework question, just want to check out if what I'm doing is right. I challenged myself to find the equation of motion and the forces in the simple pendulum system but with using the Lagrange multipliers and the constraint equation.

Homework Equations


In next part.

The Attempt at a Solution


Let [itex]l[/itex] be the length of the pendulum. I use polar coordinates and set my system of reference in the place where the mass is when at rest. The constraint equation is, I believe, [itex]l=\text{constant}=C[/itex].
I write a modified Lagrangian which takes into account the constraint equation and the Lagrange multiplier. [itex]\tilde L =L-\lambda _ l (l-C)[/itex].
The kinetic energy of the system is [itex]T=\frac{m\dot \theta ^2 l^2}{2}[/itex] while the potential energy is [itex]V=l-l\cos \theta[/itex].
So that [itex]\tilde L=\frac{m\dot \theta ^2 l^2}{2}-(l-l\cos \theta)-\lambda _ l (l-C)[/itex]. I have thus 3 "generalized coordinates", [itex]\lambda _l[/itex] which should be worth the force of constraint, theta and [itex]l[/itex].
Euler-Lagrange equations yield [itex]l=C[/itex] when considering [itex]\lambda_l[/itex] as generalized coordinate, [itex]\theta ^2 +\frac{g}{l}\sin \theta =0[/itex] where I had to consider that [itex]l=C[/itex] so that [itex]\dot l=0[/itex] (that was when considering theta as generalized coordinate) and finally for [itex]l[/itex] they yield [itex]\lambda _l=m \dot \theta ^2 l -mg (1+\cos \theta )[/itex] which has units of Newton. So I am guessing this is the tension force or something like that, I am not really sure. This is a generalized force but I'm not sure with respect to what generalized coordinate. Any explanation is appreciated.
I wonder if all I did is ok. Could someone clarify things up? Thank you!
However I do not know how to get the generalized force, [itex]\lambda _l[/itex].
 
Physics news on Phys.org
  • #2
I tried considering l as the generalized coordinate and setting \theta =\theta _0 with \dot \theta =0 so that it simplifies to \lambda _l = mg (1+\cos \theta _0 ) which is a constant value. Is this alright?
 

Related to Lagrange multipliers in a simple pendulum

What is a simple pendulum?

A simple pendulum is a mass (called a bob) attached to a fixed point by a string or rod. When the bob is pulled to one side and released, it swings back and forth in a regular pattern.

What is the Lagrange multiplier method?

The Lagrange multiplier method is a mathematical technique used to find the maximum or minimum value of a function subject to one or more constraints. It involves introducing a new variable (the Lagrange multiplier) to incorporate the constraints into the function.

How can Lagrange multipliers be applied to a simple pendulum?

In a simple pendulum, the motion of the bob is constrained by the length of the string and the gravitational force acting on it. By using Lagrange multipliers, we can incorporate these constraints into the equation for the motion of the pendulum and solve for the maximum or minimum angle of displacement.

What is the significance of the Lagrange multiplier in a simple pendulum?

The Lagrange multiplier in a simple pendulum represents the tension in the string that keeps the bob in motion. It is a crucial factor in determining the maximum or minimum angle of displacement of the pendulum.

How does the use of Lagrange multipliers affect the motion of a simple pendulum?

By incorporating the constraints of the pendulum into the equation of motion using Lagrange multipliers, we can accurately predict the maximum or minimum angle of displacement of the pendulum, as well as the corresponding tension in the string.

Similar threads

  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
518
  • Advanced Physics Homework Help
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
22
Views
565
  • Advanced Physics Homework Help
Replies
9
Views
3K
  • Advanced Physics Homework Help
Replies
11
Views
1K
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
2K
Back
Top