Foucault Pendulum - Lagrangian

[linm]

Homework Statement

I should find the Lagrangian of a Foucault Pendulum in a coordinate system on the earth. That means L = T - V where T is the cinetic energy of the pendulum and V the potential energy.

Homework Equations

$$v' = v + [\omega, r]$$
[,] denotes the cross product

The Attempt at a Solution

I write the potential energy as
$$T = \frac{1}{2}mv'^2 = \frac{1}{2}m( (\dot{x}+ \dot{y} + \dot{z}) + \omega*sqrt(x^2+ y^2 + z^2)*sin(\phi))^2$$

where $$\phi$$ denotes the latitude.
The potential energy is given by
$$V = mgz$$
But when I solve this I get strange results. Does anyone see my problem. Thanks a lot for your help!

 

[Jhero]

Hello,

First of all, it is important to note that the Lagrangian of a system is defined as the difference between the kinetic and potential energies, not the sum. So the correct form of the Lagrangian for the Foucault Pendulum would be L = T - V.

Secondly, your expression for the kinetic energy seems to be incorrect. The kinetic energy of a pendulum is given by T = 1/2 * m * v^2, where v is the velocity of the pendulum. In this case, the velocity v would be the tangential velocity of the pendulum's bob.

So, the correct form of the kinetic energy would be T = 1/2 * m * (v + \omega x r)^2, where \omega is the angular velocity of the Earth and r is the distance of the pendulum's bob from the axis of rotation.

Next, the potential energy for a Foucault Pendulum is given by V = m * g * h, where h is the height of the pendulum's bob above the ground. In this case, h would be equal to z, the vertical displacement of the pendulum's bob.

Therefore, the complete expression for the Lagrangian would be:

L = 1/2 * m * (v + \omega x r)^2 - m * g * z

I hope this helps to clarify your doubts. Let me know if you have any further questions. Good luck with your calculations!