Sliding Pendulum

Tag: LnL_1.5.2

Source:

Textbook: Landau and Lifshitz, Mechanics, 3rd Edition

Chapter: 1 The Equations of motion

Section: 5 The Lagrangian for a system of particles

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

A. Configuration space

The configuration space of the sliding pendulum system.

The generalized coordinates are the horizontal position of the support mass and the angle that the string makes with the vertical.

The horizontal position x₁ of the support mass.

The angle φ that the string makes with the vertical.