Exemple: Simple Pendulum (Cartesian parameterization)

Let's suppose that we have a pendulum of mass m, which describes a circular route of radius a that starts at an angle theta0 with speed 0.

This same problem is already described in the link Exemple: Simple Pendulum (Angular parameterization). But instead of using the angle theta as parameter, we are going to use the variables x and y which describes the cartesian position of the mass.

Initial conditions

We start at the position (x0, y0) and without any velocity, so, we have that

$\begin{matrix} x(t=0) = x_0 \\ y(t=0) = y_0 \end{matrix} \ \ \ \ \ \ \ \ \begin{matrix} \dot{x}(t=0) = 0 \\ \dot{y}(t=0) = 0 \end{matrix}$

System's energy

The kinetic energy is given by the speed how the mass is moving in space. So it's only:

$\bg_white E_{k} = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2\right)$

The potential energy is given only by the position of y and so we have

$\bg_white E_{p} = mgy$

So, the total energy is

$E = \dfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right ) + mgy$

In a matricial form we have that

$E = \dfrac{1}{2} \begin{bmatrix} \dot{x} & \dot{y} \end{bmatrix} \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} + \begin{bmatrix} 0 & mg \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix}$

The lagrangian

$\begin{align*} L_{x}(E) & = \dfrac{d}{dt}\left(\dfrac{\partial E}{\partial \dot{x}}\right) - \dfrac{\partial E}{\partial x} \\ & = 2m\ddot{x} \end{align*}$

$\begin{align*} L_{y}(E) & = \dfrac{d}{dt}\left(\dfrac{\partial E}{\partial \dot{y}}\right) - \dfrac{\partial E}{\partial y} \\ & = 2m\ddot{y} - mg \end{align*}$

As we don't have external force:

$\begin{cases} L_{x} = 0 \\ L_{y} = 0 \end{cases} \Rightarrow \begin{cases} m\ddot{x} = 0 \\ m\ddot{y} - mg = 0 \end{cases}$

In a matrix notation, we have

$L = \begin{bmatrix} L_{x} \\ L_{y} \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \begin{bmatrix} \ddot{x} \\ \ddot{y} \end{bmatrix} - \begin{bmatrix} 0 \\ mg \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Constraint function

As before we had only one degree of freedom for theta, the real problem has only one degree of freedom. Once we have two variables, and one degree of freedom, we need one contraint function g.

$\text{Degrees Of Freedom}\ + \ \text{Constraint Function} = \text{Number Variables}$

So, as the point (x, y) is in a circle, we add the constraint function g that keeps the point in the circle:

But instead of having g=0 we will use the equation

$\alpha^2 g + 2\alpha \dot{g} + \ddot{g} = 0$

Which means

$\alpha^2(x^2+y^2-L^2) + 4\alpha (x\dot{x} + y\dot{y}) + 2(x\ddot{x} + y\ddot{y} + \dot{x}^2 + \dot{y}^2) = 0$

We write it in a matrix notation to get

$\begin{bmatrix} 2x & 2y \end{bmatrix} \begin{bmatrix} \ddot{x} \\ \ddot{y} \end{bmatrix} = - \alpha^2 \cdot g - 2\alpha \dot{g} - 2(\dot{x}^2 + \dot{y}^2)$

So, our real system equation is

$\begin{bmatrix} m & 0 & 2x \\ 0 & m & 2y \\ 2x & 2y & 0 \end{bmatrix} \begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \lambda \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ -\alpha^2 \cdot g - 2\alpha \cdot \dot{g} - 2(\dot{x}^2 + \dot{y}^2) \end{bmatrix}$

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Exemple: Simple Pendulum (Cartesian parameterization)

Initial conditions

System's energy

The lagrangian

Constraint function

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