Question

A simple pendulum (Problem 4) is suspended from a mass \(M\) which is free to move without friction along the \(x\) axis. The pendulum swings in the \(x z\) plane and gravity acts in the negative \(z\) direction. Find the Lagrangian and Lagrange's equations for the system.

Short answer

The Lagrangian \( L = \frac{1}{2}M(\dot{X})^2 + \frac{1}{2}m\left[ (\dot{X} + L \dot{\theta} \cos \theta)^2 + (L \dot{\theta} \sin \theta)^2 \right] + mgL \cos \theta \). The Euler-Lagrange equations are derived from this Lagrangian.

Step-by-step solution

- Define the coordinates and kinetic energy

Consider the mass of the pendulum bob is m and it is connected to the mass M which moves along the x-axis. Let’s denote the length of the pendulum as L, and the angle it makes with the vertical axis as θ. The x-coordinate of pendulum bob is given by M's position plus L sin θ, and its z-coordinate is L cos θ. The kinetic energy of the system includes both translational and rotational parts.

- Write expressions for velocities

The velocity of mass M is \(\frac{dX}{dt}\). The velocity components of the pendulum bob are \(\frac{d}{dt}(X + L \sin \theta)\) in the x-direction and \(\frac{d}{dt}(-L \cos \theta)\) in the z-direction. Use these to write the kinetic energy components.

- Kinetic energy expression

Combine the velocity components to find the total kinetic energy: \[ T = \frac{1}{2}M(\dot{X})^2 + \frac{1}{2}m\left[ (\dot{X} + L \dot{\theta} \cos \theta)^2 + (L \dot{\theta} \sin \theta)^2 \right] \].

- Express the potential energy

The potential energy of the pendulum system is purely gravitational, given by \[ V = -mgL \cos \theta \].

- Construct the Lagrangian

The Lagrangian \( L \) is defined by \ L = T - V \. Substitute in the kinetic and potential energy expressions from the previous steps.

- Euler-Lagrange equations

Use the Lagrangian to write down the Euler-Lagrange equations with respect to the generalized coordinates X and θ. These equations will be \(\frac{d}{dt} \frac{\partial L}{\partial \dot{X}} - \frac{\partial L}{\partial X} = 0 \) and \(\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} = 0 \).

- Simplify and solve

Simplify these equations to get the final form of the equations of motion for the system.

Key concepts

simple pendulum

A simple pendulum consists of a mass (called the bob) attached to a string or rod of fixed length. It is one of the most common physical systems studied in mechanics because it is relatively simple yet rich in dynamics.
In our problem, the simple pendulum is connected to a movable mass, making the system more complex. The bob swings in a plane, and the pendulum's motion can be described using angles and lengths.

• The length of the pendulum is fixed.
• The bob swings in a vertical plane due to gravity.
• The movable mass adds translational dynamics to the system.

Understanding the motion of a simple pendulum helps in analyzing various physical systems in real life, such as clocks and sensors.

Euler-Lagrange equations

The Euler-Lagrange equations form the backbone of Lagrangian mechanics. They provide a systematic method to derive the equations of motion for a system.
The Lagrangian, represented by \( L= T - V \), simplifies the complexity involved in mechanical systems by focusing on kinetic (T) and potential (V) energies.
The general form of the Euler-Lagrange equations is: \[ \frac{d}{dt} \frac{\textrm{d}L}{\textrm{d} \theta} - \frac{\textrm{d}L}{\textrm{d} \theta} = 0 \]

• First, write the Lagrangian by determining T and V.
• Next, identify the generalized coordinates and velocities.
• Finally, apply the Euler-Lagrange equations to find the equations of motion.

In our problem, these equations help determine how the pendulum bob and the movable mass interact dynamically.

kinetic energy

Kinetic energy is the energy of motion. For our system, kinetic energy comes from both the translational motion of the movable mass and the rotational motion of the pendulum bob.
The kinetic energy, T, is given by the sum of translational and rotational components:
\[ T = \frac{1}{2}M(\frac{dX}{dt})^2 + \frac{1}{2}m\bigg[ (\frac{d}{dt}(X + L \theta \theta)cos(\theta))^2 + (L \theta \theta(sin(\theta))^2 \bigg] \]
Here, M is the mass of the moving mass, X is its position, m is the mass of the bob, L is the length of the pendulum, and \( \theta \) is the angle.

• Translational kinetic energy (\frac{1}{2} M(\frac{dX}{dt})^2)
• Rotational kinetic energy (\frac{1}{2}m\bigg[ (\frac{d}{dt}(X + L \theta (cos\theta)))^2 + (L \theta (sin(\theta)))^2 \bigg])

Kinetic energy helps in understanding how fast or in what manner the masses are moving.

potential energy

Potential energy in a pendulum system is primarily gravitational, arising due to the height from a reference point.
In our problem, the potential energy, V, is only due to the height of the pendulum bob.
\[ V = - mgL cos(\theta) \]

• m: mass of the pendulum bob
• g: acceleration due to gravity
• L: length of the pendulum
• \(\theta\): angle of the pendulum with the vertical

The negative sign indicates that potential energy decreases as the pendulum swings downward.
Potential energy helps in analyzing how the position of the bob affects its overall energy within the system.

generalized coordinates

Generalized coordinates are the set of parameters used to describe the configuration of a mechanical system, specifically tailored to exploit the system's constraints and symmetries.
For our problem, we use:

• \( X \) - Position of the moving mass along the X-axis
• \( \theta \) - Angle of the pendulum with the vertical axis

These coordinates simplify our analysis by focusing on just the essential motions, allowing for efficient computation of the system’s dynamics.
They help reduce the complexity of the problem by transforming it into a principally equivalent but simpler form.