Lagrange Equations of Motion for a particle in a vessel

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SUMMARY

The discussion focuses on the Lagrange Equations of Motion for a particle in a vessel, specifically addressing the presence of a negative term, -b^2⋅r(dot)^2⋅r, in the final equation. Participants confirm the validity of the Lagrangian and identify a correction needed in the equation of motion, where a negative sign should be changed to a positive sign in the term -2br⋅dot{r}^2. The simplification of the equation of motion under specific conditions, such as switching off gravity and fixing the motion in a vertical plane, is also highlighted as a method to check the sign of terms.

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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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