Isai Portfolio

Introduction

I gracefully struggled through this simulation in a controls course that ended up being one of my favorite classes in college. Controls theory and the physical possibilities we are able to achieve are fascinating. I want to give a quick shout out to the following public resource which was the main material we studied in that class, and where a lot of my code comes from. The github repo can be found here. I also want to shout out my professor Dr. Killpack who did a phenomenal job instilling a continous curiosity in engineering and especially in controls.
Throughout this post I will assume you have some knowledge with Euler-Lagrange Equations, Python, Linear Algebra and ODE's.

First Steps

When we use Euler-Lagrange to determine the equations of motion for a given system, we have to start off by defining our generalized coordinates (degrees of freedom) and the number of bodies in our system. Here's a quick little svg of how I'm defining all the coordinates:

The system has two degrees of freedom, one for the angle of the bar and another for the horizontal position of the cart, so I'll define our generalized coordinates as such: $q_1 = \theta$ and $q_2 = z$
There are two bodies, so we'll have 2 equations of motion.

Deriving the Lagrangian

Let's kick things off by writing the position vectors to the center of mass of each body. Modeling the pendulum as a slender arm (1) and the cart as a solid body (2) we get:

\vec{r}_1 = (z + \frac{l}{2}\sin\theta)\hat{i} + \frac{l}{2}\cos\theta\hat{j}

\vec{r}_2 = z\hat{i}

Take the derivative of both to get:

\dot{\vec{r}_1} = (\dot{z} + \frac{l}{2}\dot{\theta}\cos\theta)\hat{i} - \frac{l}{2}\dot{\theta}\sin\theta\hat{j}

\dot{\vec{r}_2} = \dot{z}\hat{i}

Cool, now there's a couple more things we need before writing down the total kinetic energy. Notice that the slender arm will also have an angular velocity defined by: $\vec{\omega}_1 = -\dot{\theta}\hat{k}$ , and an inertia matrix of:

I_{G_1} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & \frac{1}{12} m_1 l^2 & 0 \\ 0 & 0 & \frac{1}{12} m_1 l^2 \end{bmatrix}

Kinetic Energy: $T$

Now we can get the total kinetic energy, which, in our case, is defined by the following:

T = \frac{1}{2} m_1 \dot{\vec{r}}_1 \cdot \dot{\vec{r}}_1 + \frac{1}{2} \vec{\omega}_1^\top I_{G_1} \vec{\omega}_1 + \frac{1}{2} m_2 \dot{\vec{r}}_2 \cdot \dot{\vec{r}}_2

Click the dropdown below if you want to see some good old plugging and chugging. Otherwise, feel free to trust my algebra skills and skip ahead.

In the end, we get the following:

T = \frac{1}{2} (m_1 + m_2) \dot{z}^2 + \frac{l^2}{6} m_1 \dot{\theta}^2 + \frac{l}{2} m_1 \dot{z} \dot{\theta} \cos\theta

Potential Energy: $V$

Defining the initial potential energy at the center of the cart, our total potential energy is given by the gravitational potential energy of the center of mass of the slender rod, which is:

V = \frac{l}{2} m_1 g \cos\theta

Now we can write out the Lagrangian $L = T - V$ :

L = \frac{1}{2} (m_1 + m_2) \dot{z}^2 + \frac{l^2}{6} m_1 \dot{\theta}^2 + \frac{l}{2} m_1 \dot{z} \dot{\theta} \cos\theta - \frac{l}{2} m_1 g \cos\theta

Deriving the Euler-Lagrange Equation

\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right) - \frac{\partial L}{\partial q_k} = Q_k

See the little subscript $k$ in the Euler-Lagrange equation? That represents the number of generalized coordinates, and since we have two ( $q_1 = \theta$ and $q_2 = z$ ) we're going to apply this equation twice.

Once again, there's going to be lots of plugging and chugging, so click the dropdowns below if you want to see that happen. Otherwise, you can trust my partial derivative skills and skip ahead for the final result.

Work in progress

I'm still working on this section. Check back soon for updates!