Find the Lagrangian of a pendulum plane

In summary, the conversation discusses the movement of a pendulum plane and its position and orientation. The equations for its position, velocity, potential energy, and Lagrangian are also mentioned. The question of whether a certain term should be included is brought up, and it is ultimately decided that it can be disregarded as it does not affect the final equations of motion.
  • #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?
 
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  • #2
A dimension check can disprove one of the two ...
 
  • #3
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}##
 
  • #4
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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  • #5
$$P = (acos(\gamma t) + lsin(\phi(t)), lcos(\phi(t)) )$$
$$v = (-asin(\gamma t)\gamma + lcos(\phi(t))\dot{\phi}, -lsin(\phi(t))\dot{\phi} )$$
$$U = -mg(lcos(\phi(t)) $$
$$L = m(-asin(\gamma t)\gamma + lcos(\phi)\dot{\phi})^2 + (lsin(\phi)\dot{\phi})^2))/2 + mg(lcos(\phi)$$
##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
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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1. What is the Lagrangian of a pendulum plane?

The Lagrangian of a pendulum plane is a mathematical function that describes the dynamics of a pendulum system in terms of its position and velocity. It takes into account the kinetic and potential energy of the pendulum and is used to determine the equations of motion for the system.

2. How is the Lagrangian of a pendulum plane derived?

The Lagrangian of a pendulum plane is derived using the principles of Lagrangian mechanics, which is a mathematical framework for describing the motion of physical systems. It involves setting up the kinetic and potential energies of the pendulum and using the Euler-Lagrange equations to determine the equations of motion.

3. What are the variables in the Lagrangian of a pendulum plane?

The variables in the Lagrangian of a pendulum plane include the position and velocity of the pendulum, as well as the length and mass of the pendulum. Other variables, such as the gravitational constant and the angle of the pendulum, may also be included depending on the specific system being analyzed.

4. How is the Lagrangian of a pendulum plane used in physics?

The Lagrangian of a pendulum plane is used in physics to describe the motion of pendulum systems and to determine the equations of motion. It is also used in the field of classical mechanics to analyze the behavior of mechanical systems and to make predictions about their future motion.

5. Are there any limitations to using the Lagrangian of a pendulum plane?

While the Lagrangian of a pendulum plane is a useful tool for analyzing pendulum systems, it does have some limitations. It assumes that the pendulum is a rigid body and neglects any friction or air resistance, which may not accurately reflect real-world situations. Additionally, it may not be applicable to more complex pendulum systems with multiple degrees of freedom.

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