Lagrange Equations of Motion for a particle in a vessel

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Homework Help Overview

The discussion revolves around the application of Lagrange equations of motion for a particle in a vessel, focusing on the formulation of the Lagrangian and the resulting equations of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants examine the correctness of the Lagrangian and its implications on the equations of motion, questioning the signs of specific terms and the validity of certain assumptions. There is a focus on understanding how certain terms can become negative and the implications of simplifying the motion under specific conditions.

Discussion Status

Participants are actively engaging with the problem, checking each other's reasoning and identifying potential mistakes in the formulation of the equations. Some guidance is offered regarding how to analyze the signs of terms in the equations of motion by considering special cases of motion.

Contextual Notes

There is mention of specific conditions, such as switching off gravity and simplifying the motion to a fixed vertical plane, which may influence the interpretation of the equations being discussed.

Wombat11
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Homework Statement
A particle of mass m moves without friction on the inside wall of an axially symmetric vessel given by Z=(b/2)(x^2+y^2) where b is a constant and z is in the vertical direction. Find the Lagrangian of the system and the Euler-Lagrange equations. (Hint: work in cylindrical polar coordinates with x = r cos θ, y = r sin θ.)
Relevant Equations
I don't know how to put the equations in the computer so i'm just gonna drop a picture.
The final answer should have a negative b^2⋅r(dot)^2⋅r term but I have no idea how that term would become negative. Also I know for a fact that my Lagrangian is correct.
 

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Well, first thing first.
The Lagrangian is a scalar, so I don't understand why did you write ##\hat{\phi}, \hat{z}## etc which are a unit-norm vectors.
As for your solution, from the steps that I checked it seems valid to me.
 
Wombat11 said:
The final answer should have a negative b^2⋅r(dot)^2⋅r term but I have no idea how that term would become negative.
You can usually check the sign of a term in the equation of motion by considering a simple, special case of the motion.

For example, suppose you switch off gravity. In addition, suppose the initial condition is such that the the particle moves only in a fixed vertical plane so that ##\dot \theta## remains zero.

Your equation of motion then simplifies to ##\ddot r = - \frac{b^2r}{1+b^2r^2} \dot r^2##. Does the negative sign on the right side make sense?
 
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OK, my mistake I found a mistake you wrote in the second line ##-2br\dot{r}^2## it should be a plus sign.
 
MathematicalPhysicist said:
OK, my mistake I found a mistake you wrote in the second line ##-2br\dot{r}^2## it should be a plus sign.
I think that mistake is corrected in the next line.
 
Last edited:
TSny said:
I think that mistake is corrected in the next line:
https://www.physicsforums.com/attachments/249157
Indeed.
 

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