Lagrangian Mechanics for Double Pendulum

Double Pendulum Equations of Motion

Consider a double pendulum with lengths , l_2and masses , m_2.

1. Kinetic Energy

The kinetic energy $is: 2474140T = \frac{1}{2}(m_1 + m_2) l_1^2 \dot{\theta}_1^2 + \frac{1}{2} m_2 l_2^2 \dot{\theta}_2^2 + m_2 l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos(\theta_1 - \theta_2)2474140

2. Potential Energy

The potential energy$ is: 2474140V = -(m_1 + m_2) g l_1 \cos\theta_1 - m_2 g l_2 \cos\theta_22474140

From Lagrange's equation: 2474140\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 02474140