Physlib.ClassicalMechanics.Pendulum.SlidingPendulum

Sliding Pendulum

Description: A simple pendulum of mass $m_2$ attached to a mass $m_1$ as its point of support via a string of length $l$. The mass $m_1$ is free to move horizontally. The Lagrangian of the system is to be found.

Solution: First, the constraints are identified: $$\begin{aligned} y_1 &= 0\\ (x_2 - x_1)^2 + (y_2 - y_1)^2 &= l^2 \end{aligned}$$ And the second constraint gives: $$\begin{aligned} x_2 - x_1 &= l\sin\phi\\ y_2 - y_1 &= y_2 = -\,l\cos\phi \end{aligned}$$ with the generalized coordinate $\phi$ being the angle the string makes with the vertical.

The Lagrangian is obtained as: $$\mathcal{L} = T_1 + T_2 - V_1 - V_2$$ where

$$\begin{aligned} T_1 &= \tfrac{1}{2} m_1 \dot{x}_1^2, & V_1 &= 0,\\[4pt] T_2 &= \tfrac{1}{2} m_2(\dot{x}_2^2 + \dot{y}_2^2) = \tfrac{1}{2} m_2\bigl(l^2\dot{\phi}^2 + \dot{x}_1^2 + 2l\dot{\phi}\dot{x}_1\cos\phi\bigr), & V_2 &= m_2 g y_2 = -m_2 g l \cos\phi \end{aligned}$$

Thus the Lagrangian is $$\mathcal{L} = \tfrac{1}{2}(m_1 + m_2)\dot{x}_1^2 + \tfrac{1}{2} m_2\bigl(l^2\dot{\phi}^2+ 2l\dot{\phi}\dot{x}_1\cos\phi\bigr) + m_2 g l \cos\phi$$