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I'm solving an imaginary double pendulum (that is, two pendulums whose motion doesn't affect each other). The two pendulums have a "normal" motion, but they are attached.

Taking the point to wich the first pendulum is attached as the origin, the position vectors of $p_1$ and $p_2$ are:

$\vec{r_1}=(l_1sin(\theta_1),-l_1cos(\theta_1))$

$\vec{r_2}=(l_2sin(\theta_2)+l_1sin(\theta_1),-l_2cos(\theta_2)-l_1cos(\theta_1))$

I tried to describe the motion using the coordinates $\theta_1$ and $\theta_2$. I found the new unit vectors from the position of $p_2$.

$\vec{\theta_1}=\frac{1}{h_1}\frac{\partial \vec{r_2}}{\partial \theta_1}=(cos(\theta_1),sin(\theta_1))$

$\vec{\theta_2}=\frac{1}{h_2}\frac{\partial \vec{r_2}}{\partial \theta_2}=(cos(\theta_2),sin(\theta_2))$

If I use the relation

\begin{equation} \vec{A}=\sum_{i=1}^{2}(\vec{A}\cdot\vec{e_i})\vec{e_i} \end{equation} to obtain $\vec{r_1}$ in terms of the new base, I get: \begin{equation} \vec{r_1}=l_1sin(\theta_1-\theta_2)\vec{\theta_2} \end{equation}

But here's the problem. If I use the partial derivatives of $\vec{r_1}$ to find the new base, the result is:

$\vec{\theta_1}=\frac{1}{h_1}\frac{\partial \vec{r_1}}{\partial \theta_1}=(cos(\theta_1),sin(\theta_1))$

$\vec{\theta_2}=\frac{1}{h_2}\frac{\partial \vec{r_1}}{\partial \theta_2}=0$

\begin{equation} \Rightarrow \vec{r_1}=0 \end{equation}

Why does this happen?

I will add the diagram of the system:

enter image description here

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  • $\begingroup$ When you refer to your new basis are you referring to $\hat{e}_{r_1},\hat{e}_{\theta_1}$ for mass 1, and $\hat{e}_{r_2},\hat{e}_{\theta_2}$ for mass 2? Or a single basis: $\hat{e}_{\theta_1}, \hat{e}_{\theta_2}$ for both masses 1 and 2? $\endgroup$
    – Daniel Duque
    Mar 3, 2019 at 20:39

2 Answers 2

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If I use the relation

It's invalid here because your unit vectors aren't orthogonal.

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  • $\begingroup$ Basis vectors need not to be orthogonal in order to form a basis. $\endgroup$
    – Daniel Duque
    Mar 3, 2019 at 20:09
  • $\begingroup$ @DanielDuque But you do need it to write the identity as the usual sum of outer products. $\endgroup$
    – J.G.
    Mar 3, 2019 at 20:14
  • $\begingroup$ Then how am I supposed to obtain the position in terms of the new basis? $\endgroup$
    – IchVerlore
    Mar 3, 2019 at 20:54
  • $\begingroup$ @IchVerloren Switch to an orthonormal one with this first, then switch back by inverting the transformation relating the two bases. $\endgroup$
    – J.G.
    Mar 3, 2019 at 21:18
  • $\begingroup$ @J.G. you are right, I should be using the reciprocal vectors $\endgroup$
    – IchVerlore
    Mar 3, 2019 at 21:49
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If the motion of first pendulum doesn't influence the motion of the second pendulum, then p1 has to be a fixed point.

Since there is no mass or gravity involved, there can be no "normal motion".

If you decide to upgrade to real double pendulum, you'll need a inertial reference frame and 4 position equations.

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