# Dynamic model of an inverted pendulum (part 4)

This page is part of a serie of articles on how to write the model of an inverted pendulum. We strongly recommand to read the previous pages for a better understanding.

## Interactions

We will now solve the system in order to eliminate the interactions between bodies ($$\lambda_i$$). It is proven that:

$$M.\ddot{q} = F - \dfrac{d\phi}{dq}^\top .\Lambda$$

Where :

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

$$\Lambda=\begin{pmatrix} \lambda2 \\ \lambda3 \\ \lambda4 \\ \lambda5 \end{pmatrix}$$

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:

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