Duc-Tri 
In this post, we derive the equations of motion for an inverted pendulum mounted on a cart using the Lagrangian approach. The system consists of a cart with mass \(m_\text{cart}\) that can move horizontally and a pendulum with mass \(m\) attached by a rigid rod of length \(l\). The pendulum’s angle \(\theta\) is measured from the vertical.
1. Kinetic Energy
The total kinetic energy is the sum of the kinetic energy of the cart and that of the pendulum.
1.1. Kinetic Energy of the Cart
The kinetic energy of the cart is given by:
\[\begin{align} K_\text{cart} = \frac{1}{2} m_\text{cart} \dot{x}^2, \label{K_cart} \end{align}\]where \(x\) is the horizontal position of the cart.
1.2. Kinetic Energy of the Pendulum
The position of the pendulum mass \(m\) can be expressed as:
\[\begin{align} x_m = x + l \sin\theta, \quad y_m = l \cos\theta. \end{align}\]Taking time derivatives, we obtain:
\[\begin{align} \dot{x}_m = \dot{x} + l\dot{\theta}\cos\theta, \quad \dot{y}_m = -l\dot{\theta}\sin\theta. \end{align}\]Thus, the kinetic energy of the pendulum is:
\[\begin{align} K_m &= \frac{1}{2} m \left(\dot{x}_m^2 + \dot{y}_m^2\right) \notag \\ &= \frac{1}{2} m \Bigl[\bigl(\dot{x} + l\dot{\theta}\cos\theta\bigr)^2 + \bigl(-l\dot{\theta}\sin\theta\bigr)^2\Bigr] \notag \\ &= \frac{1}{2} m \left[\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2\left(\cos^2\theta + \sin^2\theta\right)\right] \notag \\ &= \frac{1}{2} m \left[\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2\right]. \label{K_m} \end{align}\]Combining \eqref{K_cart} and \eqref{K_m}, the total kinetic energy is:
\[\begin{align} K &= K_\text{cart} + K_m \notag \\ &= \frac{1}{2} m_\text{cart} \dot{x}^2 + \frac{1}{2} m \left[\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2\right]. \end{align}\]2. Potential Energy
For this system, only the pendulum contributes to the gravitational potential energy. Taking the upward vertical direction as positive, the potential energy of the pendulum is:
\[\begin{align} P_m = mgy_m = mgl\cos\theta. \end{align}\]Note: The choice of the reference level for potential energy is arbitrary, and the dynamics remain unaffected by an additive constant.
3. The Lagrangian Function
The Lagrangian is defined as the difference between the kinetic and potential energies:
\[\begin{align} \mathcal{L} &= K - P \notag \\ &= \frac{1}{2} m_\text{cart} \dot{x}^2 + \frac{1}{2} m \left[\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2\right] - mgl\cos\theta. \end{align}\]We choose the generalized coordinates:
\[\begin{align} \mathbf{q} = \begin{bmatrix} x \\ \theta \end{bmatrix}. \end{align}\]4. Derivation of the Equations of Motion
Using the Euler–Lagrange equation for each coordinate \(q_i\):
\[\begin{align} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_i}\right) - \frac{\partial \mathcal{L}}{\partial q_i} = Q_i, \end{align}\]where \(Q_i\) represents the generalized forces. Here, an external force \(F\) is applied to the cart (affecting \(x\)), while there is no external torque on \(\theta\).
4.1. Equation for \(x\)
First, compute the derivative with respect to \(\dot{x}\):
\[\begin{align} \frac{\partial \mathcal{L}}{\partial \dot{x}} = (m_\text{cart}+m)\dot{x} + ml\dot{\theta}\cos\theta. \end{align}\]Taking the time derivative yields:
\[\begin{align} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) &= (m_\text{cart}+m)\ddot{x} + ml\ddot{\theta}\cos\theta - ml\dot{\theta}^2\sin\theta. \end{align}\]Since \(\mathcal{L}\) does not explicitly depend on \(x\):
\[\begin{align} \frac{\partial \mathcal{L}}{\partial x} = 0. \end{align}\]Thus, the Euler–Lagrange equation for \(x\) becomes:
\[\begin{align} (m_\text{cart}+m)\ddot{x} + ml\ddot{\theta}\cos\theta - ml\dot{\theta}^2\sin\theta = F. \label{final_eq_1} \end{align}\]4.2. Equation for \(\theta\)
Next, compute the derivative with respect to \(\dot{\theta}\):
\[\begin{align} \frac{\partial \mathcal{L}}{\partial \dot{\theta}} = ml\dot{x}\cos\theta + ml^2\dot{\theta}. \end{align}\]Taking its time derivative:
\[\begin{align} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) = ml\left(\ddot{x}\cos\theta - \dot{x}\dot{\theta}\sin\theta\right) + ml^2\ddot{\theta}. \end{align}\]The partial derivative with respect to \(\theta\) is:
\[\begin{align} \frac{\partial \mathcal{L}}{\partial \theta} = -ml\dot{x}\dot{\theta}\sin\theta + mgl\sin\theta. \end{align}\]Thus, the Euler–Lagrange equation for \(\theta\) (with no external torque) is:
\[\begin{align} ml\left(\ddot{x}\cos\theta - \dot{x}\dot{\theta}\sin\theta\right) + ml^2\ddot{\theta} + ml\dot{x}\dot{\theta}\sin\theta - mgl\sin\theta = 0. \end{align}\]Notice that the terms \(- ml\dot{x}\dot{\theta}\sin\theta\) and \(+ ml\dot{x}\dot{\theta}\sin\theta\) cancel, allowing simplifying the equation above to:
\[\begin{align} ml\ddot{x}\cos\theta + ml^2\ddot{\theta} - mgl\sin\theta = 0. \label{final_eq_2} \end{align}\]5. Final Dynamical Model
Collecting the results, the dynamical model of the inverted pendulum on a cart is given by:
\[\begin{align} (m_\text{cart}+m)\ddot{x} + ml\ddot{\theta}\cos\theta - ml\dot{\theta}^2\sin\theta &= F, \quad \text{(from \eqref{final_eq_1})}\\ ml\ddot{x}\cos\theta + ml^2\ddot{\theta} - mgl\sin\theta &= 0. \quad \text{(from \eqref{final_eq_2})} \end{align}\]These equations describe the coupled dynamics of the cart and pendulum. They form the basis for further analysis and control design for stabilizing the inverted pendulum.
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