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This source, The Double Pendulum Fractal provides an image of a flip-time fractal for the rods of a double pendulum. The authors state that it is "not energetically possible for either pendulum to flip [within the solid black curve]".

fliptime fractal

Can someone explain to me the proof as to why the double pendulum cannot flip in these regions? Thanks.

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  • $\begingroup$ There are some easy cases. For example, if the initial angles are $\theta_1=\theta_2=0$, then the pendulum has no energy to start with and won't move. This point is directly in the center of the white region. Another easy situation is to hold the upper rod at $\theta_1=0$, then vary the lower rod in the range $-\pi<\theta_2<\pi$. In this case, the total initial potential energy is never enough for the lower mass to reach $\theta_2=\pi$, which is needed for a flip. This corresponds to the vertical line in the center of the white region. $\endgroup$
    – Andrew
    Commented Mar 24 at 20:31

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Preliminaries.

First, let's clarify: the "solid black line" is not very clearly visible in this image, but the article provides a bigger image from which it's clear that it refers to the boundary of the central white region (excluding the two mushroom-shaped protrusions).

Now, getting the equation for this boundary is pretty straightforward:

1. Total energy.

Assume, as the article does, that the rods are identical and have mass $m$ and length $l$. $\theta_1$ is the deviation from the vertical of the top rod, and $\theta_2$ is likewise for the bottom rod. Define, for simplicity, $a=l/2$. Then, when the system is released from rest, its total energy is:

$$U = U_1 + U_2 = mga(1-cos\theta_1) + [mga(1-cos\theta_1) + mga(1-cos\theta_2)]$$ $$U = mga(3-2cos\theta_1-cos\theta_2)$$

2. Minimum energy for at least one pendulum to flip.

To be energetically possible for the bottom pendulum to flip, the total energy must be enough for $\theta_2$ to get to $\pi$. The lowest energy such configuration is when $\theta_1 = 0, \theta_2=\pi$, and the kinetic energy is zero. So the minimum required energy is:

$$U_{barely} = mga(3 - 2cos 0-cos\pi)=2mga$$

For the top pendulum, the minimum energy is higher ($4mga$). So the equation for when the total energy is barely enough for at least one pendulum to flip is:

$$mga(3-2cos\theta_1-cos\theta_2)=U_{barely}=2mga$$

Simplifying:

$$2cos\theta_1+cos\theta_2=1$$,

which gives the shape of the "solid black line".

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  • $\begingroup$ What would be the lowest energy configuration of $U_{barely}$ for different rod-lengths? Is it anything other than 0 and pi? Thanks for the great answer :) $\endgroup$ Commented Mar 25 at 11:35
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    $\begingroup$ You are welcome :) Regardless of the rod lengths, in order for the bottom rod to be able to flip, it must be able to reach the angle $\theta_2=\pi$. The lowest energy configuration where that occurs is when the top rod's center of mass is as low as possible, i.e. when $\theta_1=0$. $\endgroup$ Commented Mar 25 at 18:14

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