Different results when using generalized coordinates?

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

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

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

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

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

$$$$\Rightarrow \vec{r_1}=0$$$$

Why does this happen?

I will add the diagram of the system:

• 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? Mar 3 '19 at 20:39

If I use the relation

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

• Basis vectors need not to be orthogonal in order to form a basis. Mar 3 '19 at 20:09
• @DanielDuque But you do need it to write the identity as the usual sum of outer products.
– J.G.
Mar 3 '19 at 20:14
• Then how am I supposed to obtain the position in terms of the new basis? Mar 3 '19 at 20:54
• @IchVerloren Switch to an orthonormal one with this first, then switch back by inverting the transformation relating the two bases.
– J.G.
Mar 3 '19 at 21:18
• @J.G. you are right, I should be using the reciprocal vectors Mar 3 '19 at 21:49

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.