Statement#

Consider a double pendulum consisting of:

• First pendulum: mass $m_1$, length $L_1$, angle $\theta_1$ from vertical
• Second pendulum: mass $m_2$, length $L_2$, angle $\theta_2$ from vertical (attached to $m_1$)

Derive the equations of motion using the Lagrangian formalism.

Required Topics#

• Lagrangian mechanics
• Euler-Lagrange equations
• Generalized coordinates
• Kinetic and potential energy
• Coupled differential equations
• Trigonometry

Setup#

Generalized coordinates: $\theta_1$ and $\theta_2$

Positions:

• Mass 1: $x_1 = L_1\sin\theta_1$, $y_1 = -L_1\cos\theta_1$
• Mass 2: $x_2 = L_1\sin\theta_1 + L_2\sin\theta_2$, $y_2 = -L_1\cos\theta_1 - L_2\cos\theta_2$

Lagrangian: $L = T - U$ (kinetic minus potential energy)

What to Derive#

1. Express $T$ (kinetic energy) in terms of $\theta_1$, $\theta_2$, $\dot{\theta}_1$, $\dot{\theta}_2$
2. Express $U$ (potential energy) in terms of $\theta_1$ and $\theta_2$
3. Form the Lagrangian $L = T - U$
4. Apply Euler-Lagrange equations for both coordinates:
   • $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}_1}\right) - \frac{\partial L}{\partial \theta_1} = 0$
   • $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}_2}\right) - \frac{\partial L}{\partial \theta_2} = 0$
5. Write out the two coupled differential equations

Solution#

Solution coming soon.