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Homework Statement
Hi everybody! I have a crazy problem about Lagrangian on which I've been working on for two days without being able to figure it out. I have a solution, but I think there is a flaw in it. First here is the problem:
A line rotates with constant angle velocity ##\omega## about a vertical axis, which it intersects at a fixed angle ##\upsilon##. Find the equation of motion of a mass point ##m## sliding without friction on the line under the influence of gravity.
The Attempt at a Solution
I think the whole twist of the problem is about using the right coordinates. I chose cylindrical coordinates as they seemed to help me describe best the system, and here is what I've done:
##q = \rho##
## \varphi = \omega \cdot t##
## z = r \cdot \cos \upsilon##
Now since I use ##\rho## (distance between ##m## and the axis ##z##) as my ##q##, I want to get rid of that ##r## in ##z##:
##\rho = r \cos (\frac{\pi}{2}  \upsilon) = r \sin \upsilon##
##z = r \sin (\frac{\pi}{2}  \upsilon) = r \cos \upsilon##
## \implies \frac{\rho}{z} = \frac{\sin \upsilon}{\cos \upsilon}##
## \implies z = \rho \cot \upsilon##
Here is my main question... To obtain that ##z##, I had to take the moment where the line is perpendicular to the yaxis, but am I allowed to do that? I mean, when the line has rotated out of this position, the ##r## I project on my xaxis (see second graph in attached picture) is not anymore the "real" ##r##, if you see what I mean.
Then to find the Lagrange function and the equation of motion was not too painful:
##L = \frac{1}{2} m \cdot \dot{\rho}^2 + \frac{1}{2} I \cdot \omega^2 + \frac{1}{2} \dot{z}^2 + m \cdot g \cdot z##
## = \frac{1}{2} m (\dot{\rho}^2 + \rho^2 \dot{\varphi}^2 + \dot{\rho}^2 \cot^2 \upsilon) + m \cdot g \cdot \rho \cdot \cot \upsilon##
##\implies \frac{\partial L}{\partial \rho} = m (\rho \dot{\varphi}^2 + g \cot \upsilon)##
## \mbox{and } \frac{d}{dt} \frac{\partial L}{\partial \dot{\rho}} = m\ddot{\rho} (1 + \cot^2 \upsilon)##
##\implies \ddot{\rho}  \frac{\dot{\varphi}^2 + g \cot \varphi}{1 + \cot^2 \varphi} \rho = 0##
Is that correct? I've done two things to check the result:
(1) See what happens when ##\upsilon = \frac{\pi}{2}##, that is when the line is perfectly horizontal. The equation of motion becomes ##\ddot{\rho} = \dot{\varphi}^2 \rho = \frac{v^2}{\rho} =## centripetal acceleration which I guess makes sense since the line is still rotating.
(2) I found the equation of motion with respect to ##\varphi## and found ##\ddot{\varphi} = 0## which again makes sense since ##\varphi = \mbox{ const}##. But I can't explain myself how the equation "knows" that, I don't have the feeling I have really inputed the fact that the angular velocity is constant anywhere! Or did I?
Another student found an equation of motion with respect to ##r##, which I guess could also work though it looks quite a bit more difficult. That's what he got:
##\ddot{r}  \frac{\omega^2}{1 + \cos^2 \upsilon} r + \frac{g \cdot \cos \upsilon}{1 + \cos^2 \upsilon} = 0##
I don't know what to think about it. The verification (1) I did also works for his equation, it might even be the same in another coordinates system! But as far as I know, he also considered ##r## when the line is aligned with the xaxis, which I am still not sure is correct...
You guys always have good advices, hopefully you have an idea. :)
Thanks a lot in advance for your answers.
Julien.
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