# Lagrangian, weird pendulum

1. Feb 13, 2012

### fluidistic

1. The problem statement, all variables and given/known data
A mass m is attached to one end of a light rod of length $l$. The other end of the rod is pivoted so that the rod can swing in a plane. The pivot rotates in the same plane at angular velocity $\omega$ in a circle of radius R. Show that this "pendulum" behaves like a simple pendulum in a gravitational field $g=\omega ^2 R$ for all values of l and amplitudes of oscillations.

2. Relevant equations

$L=T-V$.
I used the trigonometric formula $\sin \theta _1 \sin \theta _2 + \cos \theta _1 \cos \theta _2 =\cos (\theta _1 - \theta _2)$ for a part.
3. The attempt at a solution
I place my coordinate system (Cartesian) in the center of the circle that the pivot describes. $\vec r = x \hat i +y \hat j=(R \cos \theta _1 +l \cos \theta _2 ) \hat i +(R \sin \theta _1 + l \sin \theta _2 ) \hat j$.
Since $T=\frac{m}{2} ( \dot { \vec r} )^2$, I calculated it to be worth $\frac{m}{2}[\omega ^2 R^2+ \dot \theta _2 ^2 l^2+2Rl\omega \dot \theta _2 \cos (\theta _1 - \theta _2)]=L$.
I've got the Euler-Lagrange equation to be (for $\theta _2$) $l^2 \ddot \theta _2 -Rl\omega \sin (\theta _1 - \theta _2 )(\omega - \dot \theta _2 )-Rl \omega \dot \theta _2 \sin (\theta _1 - \theta _2 )=0$. I don't think I can simplify this anymore.
However the answer should be $ml^2 \ddot \theta =-mR\omega ^2 l \sin \theta$ where their $\theta$ is worth my $\theta _2$.
I don't know what I did wrong.
The given answer starts with $x=R\cos ( \omega t )+l \cos (\omega t + \theta )$. I don't really understand why the argument of the second cos term is this way though. Also I don't see why my "method" failed.

2. Feb 15, 2012

### BruceW

You've done everything correctly so far. You just need to go a little further. You can simplify your equation (hint - expand the brackets). And then you have the answer. You might not see instantly that you have the answer, but you really do. Their $\theta$ is not the same as your $\theta_2$, because their $\theta$ is the angle of the pendulum as seen from a coordinate system which is going round with the pivot. Also, they give the argument of the second cos term in that way they did, which is the explicit way to write it, but your way of writing it is exactly the same (since your $\theta_1$ is a function of time, the same as their explicit form).

3. Feb 15, 2012

### fluidistic

Thank you very much, your words are really encouraging.
I reach $\ddot \theta _2 -\frac{g\sin (\theta _1 - \theta _2 )}{l}=0$. (I used the fact that $g= \omega ^2 R$.)
According to wikipedia the DE of a simple pendulum is $\ddot \theta +\frac{g\sin (\theta )}{l}=0$.
So to get this result, I'd have to call $\phi =\theta _1 - \theta _2$ and then show that $\ddot \theta _2 = -\ddot \phi$.
So let $\phi =\theta _1 -\theta _2$. I have that $\dot \phi = \omega - \dot \theta _2$ and thus $\ddot \phi = - \ddot \theta _2$.
This implies that $\ddot \phi + \frac{g \sin \phi }{l}=0$ as required.

4. Feb 15, 2012

### BruceW

I probably would have left it as $\ddot \theta _2 -\frac{g\sin (\theta _1 - \theta _2 )}{l}=0$. But you did better to show that $\ddot \theta _2 = -\ddot \phi$ and get the equation into exactly the required form, nice!