How to Handle Constraints in Lagrangian Mechanics?

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SUMMARY

The discussion centers on formulating a constrained Lagrangian in polar coordinates for a system described by the functional J = ∫ L(ψ, r, r') dψ, where r' = dr/dψ. Participants confirm that the constrained Lagrangian can be expressed as Lc = L(ψ, r, r') - λ(r - r(ψ)), introducing a generalized force to maintain the motion along the curve. A further inquiry involves whether to express the parameter "a" as a function of r when applying the constraint, leading to the formulation Lc = L(a(r), ψ, r, r') - λ(r - f(ψ)). The consensus is that both approaches are valid, but the implications of each should be carefully considered.

PREREQUISITES
  • Understanding of Lagrangian mechanics and the Euler-Lagrange equation.
  • Familiarity with polar coordinates and their application in mechanics.
  • Knowledge of generalized forces and constraints in dynamical systems.
  • Basic proficiency in calculus, particularly in dealing with integrals and derivatives.
NEXT STEPS
  • Study the derivation and application of the Euler-Lagrange equation in constrained systems.
  • Explore the concept of generalized coordinates and their role in Lagrangian mechanics.
  • Investigate the implications of introducing parameters as functions of other variables in Lagrangian formulations.
  • Learn about the method of Lagrange multipliers for handling constraints in optimization problems.
USEFUL FOR

This discussion is beneficial for physicists, mechanical engineers, and students studying classical mechanics, particularly those interested in advanced topics related to constraints in Lagrangian systems.

Sunfire
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Hello,

I have the functional

J = ∫ L(ψ, r, r') dψ, where r'=dr/dψ. L is written in polar coordinates (r,ψ).

Now I want to constrain the motion to take place on the polar curve r = r(ψ). Can I write the constrained lagrangian as

Lc=L(ψ, r, r') - λ(r - r(ψ)) and then solve the Euler-Lagrange equation

Does this make sense?

Thanks!
 
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You should probably call the curve something else, like ##\rho(\psi)##. But yes, you should be able to do that, the extra ##\lambda(r - \rho)## term will give you a generalized force constraining the object to the curve.
 
CompuChip said:
... yes, you should be able to do that, ...

Thanks CompuChip...
Along these lines - say, the Lagrangian is, as agreed, Lc = L(a, ψ, r, r') - λ(r - f(ψ))
where r=f(ψ) is the constraint and "a" is a parameter.

The question is - I know that when imposing the constraint, it so happens that a = a(r). I mean, "a" becomes a function of r under the imposed constraint.

This is a challenge... Do I write

Lc = L(a(r), ψ, r, r') - λ(r - f(ψ)) and then solve the Euler-Largange eq.; or do I solve the E-L eq. and not worry about a(r)? It's a puzzle. Would you or someone know?
 
Last edited:

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