Find the Lagrangian of a pendulum plane

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Homework Help Overview

The discussion revolves around finding the Lagrangian for a pendulum that is subjected to horizontal harmonic motion. The original poster presents a mathematical expression for the position of the pendulum and its Lagrangian, while questioning the correctness of a specific term in their derivation.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the Lagrangian and expresses uncertainty about a middle term in their expression. Some participants suggest performing a dimensional analysis to verify the correctness of the terms involved. Others inquire about the elimination of explicit time dependence and seek clarification on the derivation process.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the Lagrangian formulation. There is a recognition of the need to check assumptions and terms, and some guidance is offered regarding the relationship between different forms of the Lagrangian. However, no consensus has been reached on the correctness of the original poster's expression.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of assistance provided. The discussion includes references to established texts, such as Landau, which may influence the interpretation of the problem.

LCSphysicist
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Homework Statement
Pendulum plane, which suspension executes a harmonic motion.
Relevant Equations
\n
Pendulum plane, which suspension executes a horizontal harmonic motion $$x = acos(\gamma t)$$

Position P, orientation x to right and y points below, phi is the pendulum's angle wrt y.

$$P = (acos(\gamma t) + lsin(\phi(t)), lcos(\phi(t)) )$$

So executing all that is necessary, i found it, after eliminates explicit dependence (and only dependence) of time

$$\frac{m(l^2\dot{\phi }^{2} -2la\gamma \dot{\phi }cos(\phi )sin(\gamma t)) }{2} + mglcos(\phi )$$

BUt the answer is:

1603636131542.png


Not sure about this middle term, is it right?
 
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A dimension check can disprove one of the two ...
 
BvU said:
A dimension check can disprove one of the two ...
I don't see it.
##mla \gamma^2## has the same unit as ##mla \gamma \dot{\phi}##
##\gamma## and ##\dot{\phi}## has unit ##s^{-1}##
 
Ah, sorrry, you are correct. My bad. So we have to check the
LCSphysicist said:
executing all that is necessary, i found
Can you please post it ?

And what is
LCSphysicist said:
after eliminates explicit dependence (and only dependence) of time
 
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$$P = (acos(\gamma t) + lsin(\phi(t)), lcos(\phi(t)) )$$
$$v = (-asin(\gamma t)\gamma + lcos(\phi(t))\dot{\phi}, -lsin(\phi(t))\dot{\phi} )$$
$$U = -mg(lcos(\phi(t)) $$
$$L = m(-asin(\gamma t)\gamma + lcos(\phi)\dot{\phi})^2 + (lsin(\phi)\dot{\phi})^2))/2 + mg(lcos(\phi)$$
##L = m((asin( \gamma t) \gamma)^2 + (l \dot{\phi})^2 - 2asin(\gamma t) \gamma lcos(\phi) \dot{\phi})/2 + mgl(cos(\phi))##

Taking ## m((asin( \gamma t) \gamma)^2## off we get my answer.

I know it is strange, but Landau did it too, as he said in the solution, he takes off this terms. I don't know why, but it makes sense so i follow it.
Makes sense because when we will need to calc the $$\frac{\partial L}{\partial q}-\frac{d}{dt} \frac{\partial L}{\partial \dot q}=0$$ we wil not need this terms at all
 
Two Lagrangians yield the same equations of motion if they differ by a total time derivative of some function of the coordinates and time. See the 4th paragraph of page 4 in Landau. Show that your middle term differs from Landau's middle term by a total time derivative.
 
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