# Lagrangian for a Spherical Pendulum (Goldstein 1.19)

• Yosty22
In summary: The left hand side is the total time derivative of ∂L/∂Ω. It is zero because ##\frac{\partial L}{\partial \Phi} = 0##.
Yosty22

## Homework Statement

Find the Lagrangian and equations of motion for a spherical pendulum

## Homework Equations

L=T-U and Lagrange's Equation

## The Attempt at a Solution

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I found the Lagrangian to be L = 0.5*m*l222sin2(θ)) - mgl*cos(θ) where l is the length of the rod, ω is (theta dot) and Ω is (phi dot). Here, the angle θ is measured vertically down from the z-axis and Φ is measured in the xy-plane.

My question comes when solving the Euler-Lagrange equation for Φ, namely the term: (d/dt)(∂L/∂Ω).
The inner term, ∂L/∂Ω is easy enough: ∂L/∂Ω = ml2Ωsin2(θ). The trick for me is coming when finding the total time derivative of that. I've seen two sources online that give different values, but what I did was:

d/dt(∂L/∂Ω) = d/dt(ml2Ωsin2(θ)) = ∅*ml2*sin2(theta) + 2ml2Ωsin(θ)cos(θ)ω

Here, ∅ = (phi double dot). Is this right? A lot of things I have seen online leave out the ω = (theta dot) factor in the second term. This has to be there for a total time derivative, right?

If you think about the value of ##\frac{\partial L}{\partial \Phi}##, you should see that there is no need to worry about carrying out the total time derivative ##\frac{d}{dt} \frac{\partial L}{\partial \Omega}##.

Also, I'm not sure how you are defining ##\theta##. Does ##\theta = 0## correspond to the pendulum hanging vertically? That is, does the potential energy increase if ##\theta## increases? If so, I think you should check the sign of your potential energy function in the Lagrangian.

TSny said:
If you think about the value of ##\frac{\partial L}{\partial \Phi}##, you should see that there is no need to worry about carrying out the total time derivative ##\frac{d}{dt} \frac{\partial L}{\partial \Omega}##.

Also, I'm not sure how you are defining ##\theta##. Does ##\theta = 0## correspond to the pendulum hanging vertically? That is, does the potential energy increase if ##\theta## increases? If so, I think you should check the sign of your potential energy function in the Lagrangian.

Yeah, I just caught the potential energy problem myself and fixed it. However, I'm not quite sure what you mean about not needing to carry out the total time derivative. In the equation for Φ, ∂L/∂Φ = 0, so this would mean that the physical quantity described by (d/dt)(∂L/∂Ω) is conserved. Am I misunderstanding this? That is, would it just be the value corresponding to ∂L/∂Ω that is conserved rather than (d/dt)(∂L/∂Ω)?

Yosty22 said:
That is, would it just be the value corresponding to ∂L/∂Ω that is conserved rather than (d/dt)(∂L/∂Ω)?
Yes. The equation (d/dt)(∂L/∂Ω) = 0 implies that ∂L/∂Ω is conserved.

Yosty22

## What is the Lagrangian for a spherical pendulum?

The Lagrangian for a spherical pendulum is a mathematical expression that describes the energy of the system in terms of the position and velocity of the pendulum. It takes into account the gravitational potential of the Earth, the kinetic energy of the pendulum, and the tension in the string.

## How is the Lagrangian derived for a spherical pendulum?

The Lagrangian for a spherical pendulum is derived using a mathematical framework called Lagrangian mechanics, which is based on the principle of least action. This involves writing down the kinetic and potential energy of the system and using the Euler-Lagrange equations to find the equations of motion.

## What are the advantages of using the Lagrangian in studying a spherical pendulum?

The Lagrangian approach allows for a more elegant and concise description of the dynamics of a spherical pendulum compared to using Newton's laws of motion. It also allows for the incorporation of constraints, such as the length of the pendulum string, without the need for additional equations or calculations.

## What is the significance of the Lagrangian's equations of motion for a spherical pendulum?

The equations of motion derived from the Lagrangian for a spherical pendulum provide a complete and accurate description of the system's behavior. They can be solved to determine the position, velocity, and acceleration of the pendulum at any given time, and can also reveal interesting properties of the system, such as conservation of energy or periodic motion.

## Are there any limitations to using the Lagrangian for a spherical pendulum?

The Lagrangian approach may become more complicated for more complex systems with multiple pendulums or non-spherical shapes. Additionally, it may not be the most practical method for solving certain problems, as it requires a good understanding of calculus and mechanics. In these cases, other mathematical techniques may be more suitable.

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