Equation of motions for simple pendulum in cartesian coordinates instead of generalized coordinates [closed]

Question

I am trying to write the equation of motions for a simple pendulum but instead of writing them in generalized coordinates ($\theta$), I want to write them in cartesian coordinates (x, y), as I will need them later for further research.

I am writing this matrix in python and it's driving an openGL simulator, this is how I am testing it.

I think my constraint equations are wrong. This is the problem definition I have:

I have a simple pendulum, hinged at (0,0), with an arm of length L with no mass, and on the other end at position (x,y) there is a mass m. The two forces applied on m are g (pointing downards) (0, -g) and $F_{pend}$ (pointing to 0,0) ($F_{pend}^x, F_{pend}^y$)

The 4 unknowns are $\ddot x, \ddot y, F_{pend}^x, F_{pend}^y$.

Newton's Law gives us the first two equations:

$m \ddot x - F_{\text{pend}}^x = 0$

$m \ddot y - F_{\text{pend}}^y = -mg$

The constraint on the length of the arm gives us the third equation, after taking the second derivative:

$x^2 + y^2 - l^2 = 0$

$2 x \dot x + 2 y \dot y = 0$

$2 \dot x^2 + 2 x \ddot x + 2 \dot y^2 + 2 y \ddot y = 0$

$x \ddot x + y \ddot y = - \dot x^2 - \dot y^2$

At this point I need the 4th equation, which I have no idea what it could be. I currently think is:

$F_{pend} · \langle-y, x\rangle = 0$

$F_{pend}^x (- y) + F_{pend}^y (x) = 0$

This makes the matrix Ax=b for which we can run it to compute x:

$A = \begin{bmatrix} m & 0 & -1 & 0\\ 0 & m & 0 & -1\\ x & y & 0 & 0\\ 0 & 0 & -y & x \end{bmatrix}$

$ x^T = \begin{bmatrix} \ddot x & \ddot y & F_{pend}^x & F_{pend}^y \end{bmatrix}$

$ b^T = \begin{bmatrix} 0 & -mg & - \dot x^2 - \dot y^2 & 0 \end{bmatrix}$

The way I run the simulator is the following:

Initialize: $(x, y) = (-L, 0)$, $(\dot x, \dot y) = (0, 0)$, $dt = 0.001$

Loop forever:

• Run matrix to compute $(\ddot x, \ddot y)$
• Update position $(x, y) = (x, y) + (\dot x, \dot y) * dt$
• Update velocity $(\dot x, \dot y) = (\dot x, \dot y) + (\ddot x, \ddot y) * dt$

I know this is wrong because when I run this in the simulator I get a very weird motion that looks nothing like a pendulum and the mass drops forever.

Could anyone help me in finding the 4th equation, or even tell me if there are errors in my reasoning?

Answer 1

If the goal is to obtain the movement equations in cartesian coordinates a good procedure is to use the lagrarangian formulation. Thus calling $p=(x,y)$ we have

$$ L = \frac 12 m \dot p\cdot\dot p - m g (y-l_0)+\lambda(x^2+y^2-l_0^2) $$

The movement equations give

$$ \cases{ m\ddot x - 2\lambda x = 0\\ m\ddot y -2\lambda y + m g = 0\\ x^2+y^2-l_0^2=0 } $$

now deriving the last equation twice regarding $t$ and solving for $\ddot x,\ddot y,\lambda$ we have

$$ \left\{ \begin{array}{rcl} l_0^2x''(t) & = & x(t) \left(g y(t)-x'(t)^2-y'(t)^2\right) \\ l_0^2y''(t) & = & y(t) \left(x'(t)^2+y'(t)^2\right) - g x(t)^2\\ 2l_0^2\lambda & = & m \left(g y(t)-x'(t)^2-y'(t)^2\right) \\ \end{array} \right. $$

here $\lambda$ represents the tension on the rope.