Cartesian and polar coordinate in Simple pendulum, Euler-Lagrange

AI Thread Summary
The discussion focuses on the application of the Euler-Lagrange equation to a simple pendulum using both Cartesian and polar coordinates. It highlights that substituting different coordinate systems simultaneously leads to incorrect results, as the pendulum has only one degree of freedom. The Lagrangian must be expressed in terms of a single independent variable, either θ, x, or y, to maintain consistency. The conversation emphasizes the importance of adhering to the constraints of degrees of freedom when applying Lagrangian mechanics. Ultimately, the pendulum's dynamics can only be accurately described using one coordinate system at a time.
Father_Ing
Messages
33
Reaction score
3
Homework Statement
Find the equation of motion of a simple pendulum.
Relevant Equations
L = T-V
Simple-Pendulum-Labeled-Diagram.png

$$L = \frac {mv^2}{2} - mgy$$
It is clear that ##\dot{x}=\dot{\theta}L## and ##y=-Lcos \theta##. After substituting these two equations to Lagrange equation, we will get the answer by simply using this equation: $$\frac {d} {dt} \frac {∂L}{∂\dot{\theta}} - \frac {∂L}{∂\theta }= 0$$

But, What if we only substitute ##\dot{x}=\dot{\theta}L## to the Lagrange?
$$L = \frac {mL^2 \dot{\theta}^2}{2} - mgy$$
Then:
$$\frac {d} {dt} \frac {∂L}{∂\dot{\theta}} - \frac {∂L}{∂\theta }= 0$$ $$\frac {d} {dt} \frac {∂L}{∂\dot{y}} - \frac {∂L}{∂y }= 0$$
So, We get : $$mL^2 \ddot{\theta} = 0$$

But, this answer definitely does not make any sense. Is this caused by using two different types of coordinate system at the same time?
If yes, then why is this prohibited?
 
Physics news on Phys.org
Father_Ing said:
But, this answer definitely does not make any sense. Is this caused by using two different types of coordinate system at the same time?
If yes, then why is this prohibited?
The pendulum has only one degree of freedom (independent variable), which is either ##\theta## or ##x## or ##y##.

What you have described is a different system where ##y## and ##\theta## are separate degrees of freedom. PS Although it may not make much physical sense, as there is no KE associated with movement in the ##y## direction.
 
PeroK said:
The pendulum has only one degree of freedom (independent variable), which is either θ or x or y.
But, all three of them depends on time (because if it is not, I don't think we can differentiate it with respect to time).
 
If your system has ##n## degrees of freedom, then the Euler-Lagrange equations in the form you quote (known as the "second kind") require that the Lagrangian is expressed using ##n## independent generalised coordinates. Your pendulum has only one degree of freedom, so you must express ##L## in terms of either ##\theta##, or ##y##, or ##x##, or some other coordinate, but not any more than one of them!

If the number of coordinates ##\tilde{n}## exceeds the number of degrees of freedom ##n##, then given ##\tilde{n}-n## holonomic constraint equations expressing the dependencies of these coordinates on each other you can:
1. use the constraint equations to eliminate ##\tilde{n} - n## of these coordinates and get an independent set.
2. or use Lagrange multipliers and Lagrange's equations of the "first kind".
 
  • Like
Likes vanhees71 and PeroK
Father_Ing said:
But, all three of them depends on time (because if it is not, I don't think we can differentiate it with respect to time).
I can't see the relevance of that. A simple pendulum has one degree of freedom. Full stop.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top