Find the Lagrangian of a pendulum plane

AI Thread Summary
The discussion focuses on deriving the Lagrangian for a pendulum subjected to horizontal harmonic motion. The position of the pendulum is defined in terms of its angle and the harmonic motion parameters, leading to a specific expression for the Lagrangian. Participants debate the correctness of a middle term in the Lagrangian, with one noting that dimensional analysis can clarify discrepancies. It is established that two Lagrangians can yield the same equations of motion if they differ by a total time derivative. The conversation references Landau's work, emphasizing the importance of understanding how certain terms can be omitted without affecting the resulting equations of motion.
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Homework Statement
Pendulum plane, which suspension executes a harmonic motion.
Relevant Equations
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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:

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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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