Solving KE for a Bead Rolling Along an Ellipse Using Lagrange's Method

  • Thread starter Thread starter elegysix
  • Start date Start date
  • Tags Tags
    ellipses
AI Thread Summary
The discussion revolves around finding the equation of motion for a bead rolling along the bottom half of a frictionless ellipse using Lagrange's method. The user is uncertain whether to use kinetic energy expressed in polar coordinates or Cartesian coordinates. They initially consider two forms of kinetic energy but lean towards using the Cartesian form, despite concerns about its appropriateness. The conversation emphasizes the importance of expressing kinetic energy in terms of generalized coordinates for clarity. Ultimately, the suggestion is to start with the Cartesian coordinates and substitute the ellipse's equations to derive the necessary expressions.
elegysix
Messages
404
Reaction score
15
I'm trying to find the equation of motion of a bead, which is constrained to roll along the bottom half of this frictionless ellipse, by using langrange's method - L(\phi,\dot{\phi}).
Here's the setup:
given the bottom half of an ellipse:
\mathbf{r}(\phi) = acos(\phi)\mathbf{\hat{i}}-bsin(\phi)\mathbf{\hat{j}}
where 0<\phi<\pi, and \mathbf{\hat{i}} , \mathbf{\hat{j}} are the unit vectors for x and y, and \phi is measured from the positive x-axis counterclockwise.

my question is should I use

1) KE=\frac{m}{2}(\dot{r}^{2}+r^{2}\dot{\phi}^{2})
or
2) KE=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})

where \dot{x} \:\&\: \dot{y} come from x=acos(\phi) , and y=-bsin(\phi)

which one?

I'm confused because I have polar type coordinates, r(\phi) in terms of Cartesian components.


Assuming that the second equation for KE is not appropriate to use here,

If I use the first the first equation for KE, then is it 'Legal' to find and use \mathbf{\dot{r}} like this? or is this wrong?

\mathbf{\dot{r}}=-a\dot{\phi}sin(\phi)\mathbf{\hat{i}}-b\dot{\phi}cos(\phi)\mathbf{\hat{j}}

--------------------------------------------------------------------------------------------------------------
As my only other option I'm aware of, I know that the definitely correct, brute force approach looks like this:
\mathbf{r}(\phi)_{polar}=|\mathbf{r}(\phi)_{cartesian}|\mathbf{\hat{r}}=\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{r}}

which means we're in polar coordinates, so

\mathbf{\dot{r}}(\phi)=\dot{r}\mathbf{\hat{r}}+r\dot{\phi}\mathbf{\hat{\phi}}

\mathbf{\dot{r}}(\phi)=\frac{d(\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)})}{dt}\mathbf{\hat{r}}+\dot{\phi}\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{\phi}}

and L=KE-PE... and then I have to solve \frac{dL}{d\phi}-\frac{d(\frac{dL}{d\dot{\phi}})}{dt}=0
which, that route is becoming a nightmare very quickly, and I don't want any trouble with the algebra police... any advice?
 
Physics news on Phys.org
The idea in these types of problems is to find an expression for the KE purely in terms of your generalized coordinates. Sometimes you may know the KE immediately in those terms, but other times you may have to start from Cartesian coordinates where you know the expression for KE and then you can transform to your generalized coordinates.

That is a long-winded way of saying that I would start with 2) and substitute in the a cos and b sin terms.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top