# Inverted pendulum : model without friction

## Introduction

This post presents the model of an inverted pendulum. This device is composed of an activated trolley and a pendulum which has its center of mass above its pivot point. In this post, we will assume the following hypotheses:

• $m_1$ is the mass of the trolley.

• $m_2$ is the mass of the pendulum.

• $I_1$ is the moment of inertia of the trolley.

• $I_2$ is the moment of inertia of the pendulum.

• $L$ is the distance between the pivot and the center of mass of the pendulum.

• $\Theta$ is the angle of inclination of the pendulum.

• $x_1$ and $y_1$ are the coordinates of the center of mass $C_1$ of the trolley .

• $\Theta_1$ is the orientations of the trolley.

• $x_2$ and $y_2$ are the coordinates of the center of mass $C_2$ of the pendulum.

• $\Theta_2$ is the orientations of the pendulum.

The following notation will be used in this article to avoid too complex equations for the derivatives: $\frac{d\Theta}{dt} = \dot{\Theta}$

And for the second derivative: $\frac{d^2\Theta}{dt^2} = \ddot{\Theta}$

## Constraints

As the trolley and the pendulum are linked by a pivot, the following equations can be written. This equations expresses the coordinate and orientation of each body of our system. For the trolley (body 1): $\begin{array}{r c l} x_1 &=& x_1 \\ y_1 &=& 0 \\ \Theta_1 &=& 0 \\\end{array}$

For the pendulum (body 2): $\begin{array}{r c l} x_2 &=& x_1 - L.sin(\Theta)\\ y_2 &=& y_1 + L.cos(\Theta) &=& L.cos(\Theta)\\ \Theta_2 &=& \Theta\end{array}$

Let’s define the vector $\vec{q}$, the position of our system : $\vec{q}=\begin{pmatrix} x_1 \\ \Theta_2 \end{pmatrix}$

Let’s define the constraints on position: $\phi = \begin{pmatrix} y_1 \\ \Theta_1 \\ x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ x_1 - L.sin(\Theta) \\ L.cos(\Theta) \end{pmatrix}$

Let’s calculate the constraints on velocity: $\dot{\phi} = \begin{pmatrix} \dot{y_1} \\ \dot{\Theta_1} \\ \dot{x_2} \\ \dot{y_2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \dot{x_1} - L.\dot{\Theta_2}.cos(\Theta_2) \\ -L.\dot{\Theta_2}.sin(\Theta_2) \end{pmatrix}$

Let’s calculate the constraints on acceleration: $\ddot{\phi} = \begin{pmatrix} \dot{y_1} \\ \ddot{\Theta_1} \\ \ddot{x_2} \\ \ddot{y_2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \ddot{x_1} - L.\ddot{\Theta_2}.cos(\Theta_2) + L.\dot{\Theta_2}^2.sin(\Theta_2) \\ -L.\ddot{\Theta_2}.sin(\Theta_2) - L.\dot{\Theta_2}^2.cos(\Theta_2) \end{pmatrix}$

## Dynamics

The fundamental principle of dynamics can be apply to our system. For the trolley (body 1): $\begin{array}{r c l} m_1.{a_1}^x = m_1.\ddot{x_1}&=& \| \vec{F_t} \| + \lambda_1 \\ m_1.{a_1}^y = m_1.\ddot{y_1}&=& 0 + \lambda_2 \\ I_1.\ddot{\Theta_1} &=& 0 + \lambda_3 \\\end{array}$

For the pendulum (body 2): $\begin{array}{r c l} m_2.{a_2}^x = m_2.\ddot{x_2} &=& 0 + \lambda_4 \\ m_2.{a_2}^y = m_2.\ddot{y_2} &=& \| \vec{F_g} \| + \lambda_5 = -g.m_2 + \lambda_5\\ I_2.\ddot{\Theta_2} &=& 0 + \lambda_6 \\\end{array}$

Where $\lambda_1$, $\lambda_2$, $\lambda_3$, $\lambda_4$, $\lambda_5$ and $\lambda_6$ are the Lagrange’s coefficients. These coefficients represent the interaction between bodies. In the present system, this is the interaction between the pendulum and the trolley.

## Interactions

We will now solve the system in order to eliminate the interactions between bodies ( $\lambda_i$). It is proven that: $M.\ddot{q} = F - \frac{d\phi}{dq}^\top .\Lambda$

Where : $M=\begin{pmatrix} m_1 & 0 \\ 0 & I_2 \end{pmatrix}$ $\ddot{q}=\begin{pmatrix} \ddot{x_1} \\ \ddot{\Theta_2} \end{pmatrix}$ $F=\begin{pmatrix} \| \vec{F_t} \| \\ 0 \end{pmatrix}$ $\frac{d\phi}{dq} = \begin{pmatrix} \frac{dy_1}{dx1} & \frac{dy_1}{d\Theta_2} \\\frac{d\Theta_1}{dx1} & \frac{d\Theta_1}{d\Theta_2} \\\frac{dx_2}{dx1} & \frac{dx_2}{d\Theta_2} \\\frac{dy_2}{dx1} & \frac{dy_2}{d\Theta_2} \end{pmatrix} =\begin{pmatrix}0 & 0 \\0 & 0 \\1 & -L.cos(\Theta_2) \\0 & -L.sin(\Theta_2)\end{pmatrix}$ $\Lambda=\begin{pmatrix} \lambda2 \\ \lambda3 \\ \lambda4 \\ \lambda5 \end{pmatrix}$

Applied to our system, this gives: $\begin{pmatrix} m_1 & 0 \\ 0 & I_2 \end{pmatrix} .\begin{pmatrix} \ddot{x_1} \\ \ddot{\Theta_2} \end{pmatrix} =\begin{pmatrix} \| \vec{F_t} \| \\ 0 \end{pmatrix} -\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & -L.cos(\Theta_2) & -L.sin(\Theta_2) \end{pmatrix} .\begin{pmatrix} \lambda2 \\ \lambda3 \\ \lambda4 \\ \lambda5 \end{pmatrix}$

and: $\begin{array}{r c l}m_1.\ddot{x_1} &=& \| \vec{F_t} \| -\lambda4 \\I_2.\ddot{\Theta_2} &=& \lambda4.L.cos(\Theta_2) + \lambda5.L.sin(\Theta_2)\end{array}$

## Solving the system

Based on the previous equations, we can formulate $\lambda_4$ and $\lambda_5$: $\begin{array}{r c l}\lambda_4 &=& m_2.\ddot{x_2} \\\lambda_5 &=& m_2.\ddot{y_2} + g.m_2 \\\end{array}$

The previous equations become: $\begin{array}{r c l}m_1.\ddot{x_1} &=& \| \vec{F_t} \| -m_2.\ddot{x_2} \\I_2.\ddot{\Theta_2} &=& m_2.\ddot{x_2}.L.cos(\Theta_2) + (m_2.\ddot{y_2} + g.m_2).L.sin(\Theta_2)\end{array}$

Thanks to the previous constraints on acceleration: $\begin{array}{r c l}m_1.\ddot{x_1} &=& \| \vec{F_t} \| -m_2.(\ddot{x_1} - L.\ddot{\Theta_2}.cos(\Theta_2) - L.\dot{\Theta_2}^2.sin(\Theta_2) ) \\I_2.\ddot{\Theta_2} &=& m_2.(\ddot{x_1} - L.\ddot{\Theta_2}.cos(\Theta_2) + L.\dot{\Theta_2}^2.sin(\Theta_2) ).L.cos(\Theta_2) + (m_2.(-L.\ddot{\Theta_2}.sin(\Theta_2) - L.\dot{\Theta_2}^2.cos(\Theta_2)) + g.m_2).L.sin(\Theta_2)\end{array}$

Rewriting the previous equation gives us: $\begin{array}{r c l}(m_1+m_2).\ddot{x_1} &-& m_2.L.cos(\Theta_2).\ddot{\Theta_2} &=& \| \vec{F_t} \| - m_2.L.\dot{\Theta_2}^2.sin(\Theta_2) \\-m_2.L.cos(\Theta_2).\ddot{x_1} &+& (I_2+ m_2.L^2).\ddot{\Theta_2} &=& m_2.g.L.sin(\Theta_2)\end{array}$

And: $\begin{pmatrix} m_1+m_2 & - m_2.L.cos(\Theta_2) \\-m_2.L.cos(\Theta_2) & (I_2+ m_2.L^2)\end{pmatrix} . \ddot{q} =\begin{pmatrix} \| \vec{F_t} \| - m_2.L.\dot{\Theta_2}^2.sin(\Theta_2) \\ m_2.g.L.sin(\Theta_2)\end{pmatrix}$

The general equation of the system is: $\ddot{q}=A^{-1}.B$

Where: $A=\begin{pmatrix} m_1+m_2 & - m_2.L.cos(\Theta_2) \\-m_2.L.cos(\Theta_2) & (I_2+ m_2.L^2)\end{pmatrix}$ $B= \begin{pmatrix} \| \vec{F_t} \| - m_2.L.\dot{\Theta_2}^2.sin(\Theta_2) \\ m_2.g.L.sin(\Theta_2)\end{pmatrix}$

## State space representation

Based on the previous result, it is now possible to write the state space model of the system. Let’s define $X$ the state of the system: $X=\begin{pmatrix} x_1 \\ \Theta_2 \\ \dot{x_1} \\ \dot{\Theta_2} \end{pmatrix}$

The state space representation is: $\dot{X}=\begin{pmatrix} \dot{x_1} \\ \dot{\Theta_2} \\ \left[A^{-1}.B\right] \end{pmatrix}$

## Acknowledgements

I want to thank Sébastien Lagrange from the University of Angers for his help and explanations.