Cart and pendulum Lagrangian question.

AbigailM

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.

cepheid

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.

AbigailM

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})+mglc os\phi[/itex]

Source: "Problems and Solutions on Mechanics" by Yung-Kuo Lim.