Double pendulum

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An example of a double pendulum.

A double pendulum with 1 second shutterspeed.

In horology, a double pendulum is a system of two simple pendulums on a common mounting which move in anti-phase.

In mathematics, in the area of dynamical systems, a double pendulum is a pendulum with another pendulum attached to its end, and is a simple physical system that exhibits rich dynamic behavior. The motion of a double pendulum is governed by a set of coupled ordinary differential equations. Above a certain energy its motion is chaotic. Also see pendulum (mathematics).

The double pendulum consists of two thin rods (moment of inertia, $I=\frac{1}{12} M \ell^2$ ) connected by a pivot and the end of one rod suspended from a pivot. It is natural to define the coordinates to be the angle between each rod and the vertical. These are denoted by θ₁ and θ₂. The position of the centre of mass of the two rods may be written in terms of these coordinates. If the origin of the Cartesian coordinate system is assumed to be at the point of contact of the wall and the first pendulum, then the centre of mass is located at:

$x_1 = \frac{\ell}{2} \sin \theta_1,$

$x_2 = \ell \left ( \sin \theta_1 + \frac{1}{2} \sin \theta_2 \right ),$

$y_1 = -\frac{\ell}{2} \cos \theta_1$

and

$y_2 = -\ell \left ( \cos \theta_1 + \frac{1}{2} \cos \theta_2 \right ).$

This is enough information to write out the Lagrangian.

1 Lagrangian
2 Chaotic motion
3 References
4 See Also
5 External links

[edit] Lagrangian

The Lagrangian is given by

$L = \frac{1}{2} m \left ( v_1^2 + v_2^2 \right ) + \frac{1}{2} I \left ( {\dot \theta_1}^2 + {\dot \theta_2}^2 \right ) - m g \left ( y_1 + y_2 \right )$

where the first term is the kinetic energy of the bodies, the second term is the kinetic energy of the center of mass of each rod, and the last term is the potential energy of the bodies in a uniform gravitational field.

Substituting the coordinates above and rearranging the equation gives

$L = \frac{1}{6} m \ell^2 \left [ {\dot \theta_2}^2 + 4 {\dot \theta_1}^2 + 3 {\dot \theta_1} {\dot \theta_2} \cos (\theta_1-\theta_2) \right ] + \frac{1}{2} m g \ell \left ( 3 \cos \theta_1 + \cos \theta_2 \right ).$

There is only one conserved quantity (the energy), and no conserved momenta. The two momenta may be written as

$p_{\theta_1} = \frac{\partial L}{\partial {\dot \theta_1}} = \frac{1}{6} m \ell^2 \left [ 8 {\dot \theta_1} + 3 {\dot \theta_2} \cos (\theta_1-\theta_2) \right ]$

and

$p_{\theta_2} = \frac{\partial L}{\partial {\dot \theta_2}} = \frac{1}{6} m \ell^2 \left [ 2 {\dot \theta_2} + 3 {\dot \theta_1} \cos (\theta_1-\theta_2) \right ].$

These expressions may be inverted to get

${\dot \theta_1} = \frac{6}{m\ell^2} \frac{ 2 p_{\theta_1} - 3 \cos(\theta_1-\theta_2) p_{\theta_2}}{16 - 9 \cos^2(\theta_1-\theta_2)}$

and

${\dot \theta_2} = \frac{6}{m\ell^2} \frac{ 8 p_{\theta_2} - 3 \cos(\theta_1-\theta_2) p_{\theta_1}}{16 - 9 \cos^2(\theta_1-\theta_2)}.$

The remaining equations of motion are written as

${\dot p_{\theta_1}} = \frac{\partial L}{\partial \theta_1} = -\frac{1}{2} m \ell^2 \left [ {\dot \theta_1} {\dot \theta_2} \sin (\theta_1-\theta_2) + 3 \frac{g}{\ell} \sin \theta_1 \right ]$

and

${\dot p_{\theta_2}} = \frac{\partial L}{\partial \theta_2} = -\frac{1}{2} m \ell^2 \left [ -{\dot \theta_1} {\dot \theta_2} \sin (\theta_1-\theta_2) + \frac{g}{\ell} \sin \theta_2 \right ].$

[edit] Chaotic motion

Graph of the time for the pendulum to flip over as a function of initial conditions

The double pendulum undergoes chaotic motion, and shows a sensitive dependence on initial conditions. The image to the right shows the amount of elapsed time before the pendulum "flips over", as a function of initial conditions. Here, the initial value of θ₁ ranges along the x-direction, from −3 to 3. The initial value θ₂ ranges along the y-direction, from −3 to 3. The colour of each pixel indicates whether either pendulum flips within $10\sqrt{g/\ell\ }$ (green), within 100 (red), 1000 (purple) or 10000 (blue). Initial conditions that don't lead to a flip within $10000\sqrt{g/\ell\ }$ are plotted white.

The boundary of the central white region is defined in part by energy conservation with the following curve:

$3 \cos \theta_1 + \cos \theta_2 = 2. \,$

Within the region defined by this curve, that is if

$3 \cos \theta_1 + \cos \theta_2 > 2, \,$

then it is energetically impossible for either pendulum to flip. Outside this region, the pendulum can flip, but it is a complex question to determine when it will flip.

[edit] References

Eric W. Weisstein, Double pendulum (2005), ScienceWorld. (Contains details of the complicated equations involved.)
Peter Lynch, Double Pendulum, (2001). (Java applet simulation.)
Northwestern University, Double Pendulum, (Java applet simulation.)
Theoretical High-Energy Astrophysics Group at UBC, Double pendulum, (2005).
Meirovitch, Leonard (1986). Elements of Vibration Analysis, 2nd edition, McGraw-Hill Science/Engineering/Math. ISBN 0-07-041342-8.