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Lagrangian Mechanics

  • #1
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Homework Statement


We have a mas m attached to a vertical spring of length (l+x) where l is the natural length.

Homework Equations


Find the Lagrangian and the hamiltonian of the system if it moves like a pendulum

The Attempt at a Solution


we know that the lagrangian of a system is defined as:
L=T-V

T=mv2/2
V=mgy-[kr2/2]

where
x1=(l+x)sinθ
y1=(l+x)cosθ

would you chose your variables the same way?
 

Answers and Replies

  • #2
Orodruin
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The potential energy in your spring gets lower and lower as the spring extends if you use your expression for the potential energy. This is unphysical.
 
  • #3
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May I add that the strength of Lagrangian/Hamiltonian approach is that it takes advantages of constraints. You want to use generalized coordinates instead of familiar Cartesian coordinates.

Possible choices

[itex]l[/itex] - for the length of the spring (I don't think that resting length is important for this problem, which is why I chose [itex]l[/itex] )

$$\theta$$ - for angular displacement of the pendulum that this spring is attached to

I believe this system has 2 degrees of freedom and thus 2 generalized coordinates. I recall doing a problem similar to this in my Classical Mechanics lecture where the length was given as a function of t.....in that case [itex]l[/itex] is already predetermined and this system has only one degree of freedom $$\theta$$. I hastily thought this through and hastily wrote this response, if time permits I'll give it a more careful examination later.
 
  • #4
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I would use polar coordinates with the origin at the fixing point and theta the angle between the spring and the vertical. That way you have orthogonal coordinates so you don't get cross-terms in the kinetic energy, and the potential energy of the spring is expressed easily in terms of r. V is just mgr(1-cos(theta)) then.

If you define theta as the angle between the spring and the horizontal then the potential energy will be slightly neater, but then your equilibrium position will be at theta=pi/2 and that will only make life difficult if you want to find the frequency of small oscillations about the equilibrium.

If you go ahead with your coordinates, you'll find a bit of algebra will get you to polar coordinates in the end, though as has been mentioned you need to reverse the sign in your elastic potential energy. I think the sign on your gravitational potential energy needs to be reversed too, but I'm unsure of exactly how you've defined your coordinates. Just remember the gravitational potential increases as the mass moves up!
 

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