Euler-Lagrange equation for paraboloid plane

In summary, the conversation discusses a classical mechanics question involving a paraboloid shaped plane of mass M and a point mass m placed on it. The Euler-Lagrange equations are used to solve for the motion of the system, but there may be errors in the calculations due to inconsistencies in the use of masses.
  • #1
wavemaster
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I have a classical mechanics question I couldn't conclude. The reason seems to be mathematical. It's this:
There's a paraboloid shaped plane of mass M, which is standing on a frictionless surface and can slide freely. It's surface is [tex]y=ax^2[/tex]. A point mass m is place on the plane. Solve the Euler-Lagrange equations of the system for little mass.

My (unsuccessful) solution is as follows:I chose [tex]x_m[/tex] as the x (horizontal) coordiante of point mass, and for sliding, plane it's [tex]x_M[/tex]. I defined another coordinate for point mass: [tex]x[/tex] is the horizontal distance of point mass from the center (or bottom) of the parabol. So, coordinates of m are
[tex]x_m = x_M + x[/tex]
[tex]y_m = ax^2[/tex]
so velocities are
[tex]\dot{x_m} = \dot{x_M} + \dot{x}[/tex]
[tex]y_m = a2x\dot{x}[/tex]

So kinetic energy of the system is:
[tex]T = \frac{m\dot{x_m}^2}{2} + \frac{M\dot{x_M}^2}{2}[/tex]
[tex]T = \frac{m ( \dot{x_M}^2 + \dot{x}^2 + 2\dot{x}\dot{x_M} + 4a^2x^2\dot{x}^2 )}{2} + \frac{M\dot{x_M}^2}{2}[/tex]
and potential is
[tex]V = mgy = mgax^2[/tex]

and
[tex]{\cal L} = T-V[/tex]

Now, Euler-Lagrange equations are:
(1) [tex]\frac{d}{dt} \frac{\partial {\cal L}}{\partial \dot{x_M}} - \frac{\partial {\cal L}}{\partial x_M}[/tex]
(2) [tex]\frac{d}{dt} \frac{\partial {\cal L}}{\partial \dot{x}} - \frac{\partial {\cal L}}{\partial x}[/tex]

The solution of the first one yields (of course, if I done the math correctly):
[tex]\ddot{x_M}(m+M) + \ddot{x}m = 0[/tex]
and second is
[tex]\ddot{x} + \ddot{x_M} + 4a^2x\dot{x}^2 4a^2x^2\ddot{x} + 2gax = 0[/tex]

I tried combining two solutions and got this non-linear differential equation:

[tex]\ddot{x}(1- \frac{m}{m+M} + 4a^2x^2) + 4a^2x\dot{x}^2 + 2gax = 0[/tex]

Maybe there was a mathematical mistake in my solution, maybe this differential equation could be solved with some tricks, or maybe I'm downright wrong by choosing such coordinates. I don't know.

Is there someone who can solve this question?
 
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  • #2


Hello,

Thank you for posting your question. I can see that you have put a lot of effort into solving this problem and have made some good progress. However, there are a few things that could be causing the issue you are facing.

Firstly, in your kinetic energy equation, you have a term that is 4a^2x^2\dot{x}^2. This term is not present in the potential energy equation and it does not have a corresponding term in the Euler-Lagrange equations. This could be causing the non-linearity in your final differential equation.

Secondly, when you derived your Euler-Lagrange equations, you used the Lagrangian \cal L = T - V, which is correct. However, in your kinetic energy equation, you have used the mass m for both the point mass and the paraboloid, whereas in the potential energy equation, you have used different masses (m for the point mass and M for the paraboloid). This could also be causing some errors in your final equation.

I would suggest going back and checking your calculations, making sure that all of the terms in your kinetic and potential energy equations have corresponding terms in the Euler-Lagrange equations. Also, double check your use of masses in the equations.

I hope this helps and good luck with your solution!
 

What is the Euler-Lagrange equation for a paraboloid plane?

The Euler-Lagrange equation for a paraboloid plane is a mathematical formula used to find the stationary points of a functional. It is derived from the calculus of variations and is used to solve optimization problems.

What is the purpose of the Euler-Lagrange equation?

The Euler-Lagrange equation is used to find the points where a functional attains its maximum or minimum value. This is useful in solving optimization problems in various fields such as physics, engineering, and economics.

What is the difference between a paraboloid plane and a paraboloid surface?

A paraboloid plane is a two-dimensional surface that is shaped like a bowl or a saddle. On the other hand, a paraboloid surface is a three-dimensional surface that can have a curved or flat shape. The Euler-Lagrange equation can be used to find the stationary points of both paraboloid planes and surfaces.

How is the Euler-Lagrange equation derived for a paraboloid plane?

The Euler-Lagrange equation is derived by applying the calculus of variations to a functional that represents the energy of the system. This functional is then minimized or maximized using the Euler-Lagrange equation, which involves taking the derivative of the functional with respect to the dependent variable and setting it equal to zero.

What are some real-life applications of the Euler-Lagrange equation for a paraboloid plane?

The Euler-Lagrange equation has various applications in different fields. In physics, it can be used to determine the path of a particle under the influence of a potential field. In engineering, it can be used to optimize the shape of structures such as bridges and buildings. In economics, it can be used to find the optimal production levels for a given cost function.

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