Find the Lagrangian of a pendulum plane

  • #1
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?
 

Answers and Replies

  • #2
BvU
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A dimension check can disprove one of the two ...
 
  • #3
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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}##
 
  • #4
BvU
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Ah, sorrry, you are correct. My bad. So we have to check the
executing all that is necessary, i found
Can you please post it ?

And what is
after eliminates explicit dependence (and only dependence) of time
 
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  • #5
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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
 
  • #6
TSny
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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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