Different Shape for Fliptime Fractal

Question

This is related to my previous question. I am generating images of how long it takes for a double pendulum to flip in different configurations. I was trying to find the shape of where it isn't energetically possible to flip where $3l_1 = l_2$ and $m_1 = m_2$, where $l$ is the lengths of the pendulums and m is the mass of the pendulums.

Here is my attempt (as directed by the answer to my question): The length of the first rod is $l_1$ and the length of the second rod is $3_l1$. Let $a = \frac{l_1}{2}$

Total energy:
$mga(1-cos\theta_1) + [mga(1-cos\theta_1) + 3mga(1-cos\theta_2)]$
$mga(1-cos\theta_1 + 1 - cos\theta_1 + 3 - 3cos\theta_2)$
$mga(5 - 3cos\theta_2 - 2cos\theta_1)$

Minimum energy configuration: $\theta_1 = 0$ and $\theta_2 = \pi$
Plugging...
$mga(5 - 3cos\pi - 2cos0)$
$mga(5 - 3(-1) - 2(1)$
$6mga$

Setting them equal and simplifying:
$mga(5 - 3cos\theta_2 - 2cos\theta_1) = 6mga$
$ -1(3cos\theta_2 + 2cos\theta_1) = 1$

Plotting in desmos, this does not form a square. When I try this method with $2l_1 = l_2$, it produces a shape that is a square. These results are incorrect (see the source in my previous question).

What am I doing wrong (did the authors of the pdf, in the previous post, mislabel their graphs?) Thanks :)

Answer 1

There is a mistake at the first line, more specifically in the second bracket term. I'll set $l_1=1$. The height for the first pendulum's center of mass is: $$ h_1 = -\frac12\cos\theta_1 $$ For the second pendulum's, it's: $$ h_2 = -\cos\theta_1-\frac{l_2}2\cos\theta_2 $$ The total potential energy (setting $mg=1$) is therefore: $$ E_p = h_1+h_2 = -\frac12(3\cos\theta_1+l_2\cos\theta_2) $$ In your case, $l_2 = 3$: $$ E_p = -\frac32(\cos\theta_1+\cos\theta_2) $$ The minimal energy flip is $E_p = 0$ (reached both for $\theta_1=0,\theta_2=\pi$ and $\theta_1=\pi,\theta_2=0$) so the curve is indeed: $$ \cos\theta_1+\cos\theta_2 = 0 $$

For $l_2=2$: $$ E_p = -\frac12(3\cos\theta_1+2\cos\theta_2) $$ the minimal energy flip is $E_p = -\frac12$ (reached for $\theta_1=0,\theta_2=\pi$), so the curve is indeed: $$ 3\cos\theta_1+2\cos\theta_2 = 1 $$ which is not a square (but it is not considered in the slides nor in the article).

Trying out for $l_2=4$ is also consistent with the slides.

Hope this helps.