Chapter 9: Problem 6

A particle moves on the surface of a sphere of radius \(a\) under the action of the earth's gravitational field. Find the \(\theta, \phi\) equations of motion. (Comment: This is called a spherical pendulum.)

Short Answer

Expert verified

The equations of motion are \( \theta'' = \text{sin}(\theta)\text{cos}(\theta)\phi'^2 - \frac{g}{a}\text{sin}(\theta) \text{ and }\theta \phi' = \frac{k}{a^2\text{sin}^2(\theta)} \.

Step by step solution

Understand the coordinates and forces involved

Define the coordinates in terms of spherical coordinates \(r, \theta, \text{ and } \phi\), where \(r = a\) is constant. The forces involved are the tension in the string and gravitational force.

Express the kinetic energy

Using spherical coordinates: \[ T = \frac{1}{2}mv^2 = \frac{1}{2}m\big(a^2\big(\theta'^2 + \big(\text{sin}^2(\theta) \big)\big)\big) \phi'^2\big) \], where \( \theta' \) and \( \phi' \) are the time derivatives of \( \theta \) and \( \phi \), respectively.

Express the potential energy

Potential energy is due to gravity: \[ V = mga(1 - \text{cos}(\theta)) \]

Write the Lagrangian

The Lagrangian is given by: \[ L = T - V = \frac{1}{2}ma^2\big(\theta'^2 + \text{sin}^2(\theta) \phi'^2\big) - mga(1 - \text{cos}(\theta)) \]

Derive the Euler-Lagrange equations

For \( \theta \): \[ \frac{d}{dt}\big(\frac{\text{d}L}{\text{d}\theta'}\big) - \frac{\text{d}L}{\text{d}\theta} = 0 \] For \( \phi \): \[ \frac{d}{dt}\big(\frac{\text{d}L}{\text{d}\phi'}\big) - \frac{\text{d}L}{\text{d}\phi} = 0 \]

Solve the Euler-Lagrange equations for \( \theta \)

Substitute the Lagrangian into the Euler-Lagrange equations. For \( \theta \): \[ ma^2\theta'' = ma^2\text{sin}(\theta)\text{cos}(\theta)\phi'^2 - mga\text{sin}(\theta) \]

Solve the Euler-Lagrange equations for \( \phi \)

Substitute the Lagrangian into the Euler-Lagrange equations. For \( \phi \): \[ ma^2\big(\text{sin}^2(\theta) \phi'\big)' = 0 \] which implies \( \phi' = \frac{k}{a^2\text{sin}^2(\theta)} \), where \( k \) is a constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

spherical coordinates

Spherical coordinates provide a way to describe a location in three dimensions using three variables: radial distance (r), polar angle (θ), and azimuthal angle (φ). This system is particularly useful for problems involving spheres, such as our spherical pendulum. Here, the radial distance 'r' is a constant equal to the sphere's radius, a. The angles θ and φ describe the position of the particle on the sphere's surface.
Spherical coordinates simplify the mathematical treatment of systems with symmetry about a point, making them a natural choice for our problem.

Lagrangian mechanics

Lagrangian mechanics is a reformulation of classical mechanics that uses the Lagrangian function to describe the dynamics of a system. It allows for more straightforward handling of complex physical situations involving constraints.
The Lagrangian, L, is the difference between the kinetic energy, T, and the potential energy, V, of a system: \[ L = T - V \]
In our spherical pendulum problem, the kinetic and potential energies are expressed in terms of spherical coordinates, simplifying their calculation.

Euler-Lagrange equations

The Euler-Lagrange equations provide the equations of motion in Lagrangian mechanics. These are derived by applying the principle of stationary action to the Lagrangian function.
For each generalized coordinate, q, the Euler-Lagrange equation is: \[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \]
In our problem, we have two coordinates, θ and φ. By substituting the Lagrangian into these equations, we derive the equations of motion for θ and φ, which describe how the spherical pendulum moves over time.

kinetic energy

Kinetic energy is the energy associated with the motion of an object. For our spherical pendulum, it depends on the particle’s velocity as it swings on the sphere’s surface.
In spherical coordinates, the kinetic energy, T, is given by: \[ T = \frac{1}{2}m\left(a^2(\theta'^2 + \sin^2(\theta) \phi'^2)\right) \]
where \( \theta' \) and \( \phi' \) are the time derivatives of θ and φ, respectively, representing the particle's angular velocities. This expression captures all the energy components due to movement along the angles θ and φ.

potential energy

Potential energy in the context of our spherical pendulum comes from gravity. As the particle moves under the Earth's gravitational field, its height changes relative to a reference point, influencing its potential energy.
The potential energy, V, for our system is: \[ V = mga(1 - \cos(\theta)) \]
Here, 'm' is the mass of the particle, 'g' is the gravitational acceleration, 'a' is the radius of the sphere, and θ is the polar angle. This formula indicates that potential energy depends solely on the angle θ, which measures the particle's elevation compared to the lowest point on the sphere.