Cart and pendulum Lagrangian question.

In summary, the author finds the small angle frequency by rewriting the equation of motion as the amplitude of the pendulum's angle with the normal, and then using a trig identity to find the frequencies.
  • #1
AbigailM
46
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After finding the equations of motion of a pendulum in an accelerating cart:

[itex]\ddot{\phi} + \frac{acos\phi +gsin\phi}{l}=0[/itex]

,the method that Taylor uses in Prob 7.30 for finding the small angle frequency, is to rewrite [itex]\phi[/itex] as [itex]\phi_{0}+\delta \phi[/itex]. Then you can use a trig identity in the equation of motion to get the frequencies.

My question is what is the reasoning for rewriting [itex]\phi[/itex] the way we do?
To me it looks like just the amplitude since [itex]\phi_{0}[/itex] is just the angle that the pendulum makes with the normal when it is at rest relative to the cart.

Thanks.
 
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  • #2
AbigailM said:
After finding the equations of motion of a pendulum in an accelerating cart:

[itex]\ddot{\phi} + \frac{acos\phi +gsin\phi}{l}=0[/itex]

,the method that Taylor uses in Prob 7.30 for finding the small angle frequency, is to rewrite [itex]\phi[/itex] as [itex]\phi_{0}+\delta \phi[/itex]. Then you can use a trig identity in the equation of motion to get the frequencies.

My question is what is the reasoning for rewriting [itex]\phi[/itex] the way we do?
To me it looks like just the amplitude since [itex]\phi_{0}[/itex] is just the angle that the pendulum makes with the normal when it is at rest relative to the cart.

Thanks.

I've been mulling over your problem.

If you just do a vertical force balance, and a horizontal force balance, between the tension and the weight, you'll find that there is an equilibrium angle at which the pendulum will hang, given by ##\phi_0 = \arctan(a/g)##. You can also obtain this by setting ##\ddot{\phi} = 0## in your equation of motion.

Now, I think what your author is investigating is, what happens if we perturb the system around this equilibrium point? With what frequency will the pendulum oscillate around it? So in other words he's saying, let's add a small perturbation ##\delta \phi## to the system so that the new angular position is ##\phi = \phi_0 + \delta \phi##. That's the motivation for doing this.

By the way, it's been a while for me, and I can't figure how to set this up using Lagrangian methods. Maybe you can enlighten me? I know that L = T - V, but I'm not sure what do as far as choosing the generalized coordinates, or accounting for the constant acceleration. The way I did it was to first transform into the non-inertial frame of the cart, and then use Newton's laws to arrive at the equation of motion. Specifically, I said that ##\sum \tau = I \alpha##, where ##\alpha = \ddot{\phi}## and then solved. I had to add in a horizontal pseudo-force of magnitude ma in the direction opposite the acceleration of the cart in order to explain what was countering the gravitational torque at equilibrium. I got the same equation of motion as you, except that my ##g\sin\phi## and ##a\cos\phi## terms were of opposite sign.
 
  • #3
Thanks for the help cepheid! That makes sense. ugh a botched sign I hate those. Hrmmm I've never tried your method. The only other method I've seen is where this problem is solved in Taylor Ch9 using [itex]m\boldsymbol{\ddot{r}} = \boldsymbol{F}-m\boldsymbol{A}[/itex], for solving non-inertial problems.

So for this problem using the lagrangian I started in cartesian coordinates finding positions and velocities for x and y, then converting to [itex]\phi \hspace{2 mm}and\hspace{2 mm} \dot{\phi}[/itex] for my generalized coordinates.

[itex]x=lsin\phi +\frac{1}{2}at^{2}[/itex]

[itex]y=-lcos\phi[/itex]

Calculate the derivates and then [itex]v^{2}=\dot{x}^{2}+\dot{y}^{2}[/itex]

The potential energy is [itex]-mglcos\phi[/itex]

The lagrangian is:
[itex]\frac{1}{2}m(\dot{\phi^{2}}l^{2}+2at\dot{\phi}lcos\phi+a^{2}t^{2})+mglcos\phi[/itex]

Source: "Problems and Solutions on Mechanics" by Yung-Kuo Lim.
 
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1. What is the Cart and pendulum Lagrangian question?

The Cart and pendulum Lagrangian question is a classic problem in physics that involves finding the equations of motion for a cart connected to a pendulum. The goal is to determine the motion of the cart and pendulum system based on the forces acting on it.

2. What is the Lagrangian approach to solving this problem?

The Lagrangian approach is a mathematical method for solving problems in classical mechanics that involves finding the Lagrangian function, which is a single expression that describes the system's energy. This function is then used to derive the equations of motion for the system.

3. How is the Lagrangian function derived for the Cart and pendulum system?

The Lagrangian function for the Cart and pendulum system is derived by considering the kinetic and potential energies of the system. The kinetic energy is determined by the motion of the cart and pendulum, while the potential energy is determined by the height of the pendulum and the position of the cart. These energies are then combined to form the Lagrangian function.

4. What are the advantages of using the Lagrangian approach for this problem?

One advantage of using the Lagrangian approach is that it provides a more elegant and concise way of solving problems in classical mechanics. It also allows for a more systematic and general approach to solving problems, making it easier to apply to more complex systems.

5. Are there any limitations to using the Lagrangian approach for this problem?

One limitation of using the Lagrangian approach for the Cart and pendulum problem is that it can be more difficult to apply for systems with non-conservative forces. It also requires a good understanding of calculus and physics principles to effectively use this method.

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