# The new chaotic pendulum

###### content
1. Plan of the study
2. Simple pendulum
3. Spherical pendulum
4. Inclined pendulum
5. Chaotic pendulum

## Preliminary studycase of the spherical pendulum

We will now study the case of the spherical pendulum. We add another degree of freedom to the case of the simple pendulum : the mass can now rotate around the $z$ axis, doing an angle $\varphi$ with the $x$ axis. The movement will now be tri-dimensionnal.

### Positions and speeds

We begin by seeking the mass' coordinates and corresponding speeds, with respect to the angles $\theta$ and $\varphi$ : $$\left \{ \begin{array}{r c l} x & = & l \sin \theta \cos \varphi \\ y & = & l \sin \theta \sin \varphi \\ z & = & -l \cos \theta \end{array} \right . \; \Rightarrow \; \left \{ \begin{array}{r c l} \dot{x} & = & l (\dot{\theta} \cos \theta \cos \varphi - \dot{\varphi} \sin \theta \sin \varphi ) \\ \dot{y} & = & l (\dot{\theta} \cos \theta \sin \varphi + \dot{\varphi} \sin \theta \cos \varphi )\\ \dot{z} & = & l \dot{\theta} \sin \theta \end{array} \right .$$

### energies, Lagrangian, movement equations

We may then find the potential and kinetic energies, and thus the corresponding Lagrangian $\mathcal{L}$ :
$$\begin{eqnarray*} E_k & = & \frac{1}{2}m v^2=\frac{m}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^2)=\frac{m l^2}{2}(\dot{\theta}^2+\dot{\varphi}^2\sin^2\theta) \\ E_p & = & -\vec{P}\cdot\vec{r}=mgz=-mgl\cos\theta \\ \\ \mathcal{L} & = & E_k - E_p = \frac{m l^2}{2}(\dot{\theta}^2+\dot{\varphi}^2\sin^2\theta) + mgl\cos\theta \end{eqnarray*}\\$$ From this we deduce the Euler-Lagrange equations : $$\begin{eqnarray} \frac{\partial \mathcal{L}}{\partial \theta} = \frac{\text{d}}{\text{d} t}\frac{\partial \mathcal{L}}{\partial\dot{\theta}} \; & \Rightarrow \; & \ddot{\theta}=\dot{\varphi}^2\sin\theta\cos\theta-\frac{g}{l}\sin\theta \\ % \frac{\partial \mathcal{L}}{\partial \varphi} = \frac{\text{d}}{\text{d} t}\frac{\partial \mathcal{L}}{\partial\dot{\varphi}} \; & \Rightarrow \; & \ddot{\varphi}=-2\dot{\theta}\dot{\varphi}\cot\theta \end{eqnarray}$$

### Integrating the equations

We can now integrate numerically the new-found coupled equations (we will use here Wolfram Mathematica 8 ).
Finally, we get the following trajectories, which of course depend of the initial conditions we chose :

This case will be encountered for $\alpha=0$ and $\dot{\varphi} \neq 0$

We can notice that the central area is never reached by the pendulum, and the "bowl"-shaped trajectories : the pendulum now describes a part of a sphere.
Like for the case of the simple pendulum, the mechanic energy is conserved, and the kinetic energy $E_k$ becomes zero for a same value of the potential enery $E_p$ – the pendulum then reaches a maximal height and "falls" again.

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