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Mass point sliding on a rotating line (Lagrangian)

  1. May 20, 2016 #1
    1. The problem statement, all variables and given/known data

    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.

    3. 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 y-axis, but am I allowed to do that? I mean, when the line has rotated out of this position, the ##r## I project on my x-axis (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 x-axis, 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.
     

    Attached Files:

  2. jcsd
  3. May 20, 2016 #2
    Actually in the case of the equation of motion of my friend, the verification (1) provides a slightly different result as mine:

    ##\ddot{r} = \omega^2 \cdot r = \frac{v^2}{r}##

    That also looks like the centripetal acceleration, but with ##r## instead of ##\rho## (which is the distance between ##m## and the z-axis). I think mine is correct, because the distance between ##m## and the rotation axis is ##\rho##, but I would not put my hand in the fire for that... What do you guys think?


    Julien.

    EDIT: I just realized that none of our equations produce ##\ddot{\rho}/\ddot{r} = -g## for ##\upsilon = 0##... Are we both wrong, or is there something I am missing here?
     
    Last edited: May 20, 2016
  4. May 20, 2016 #3

    TSny

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    ##\varphi## is not a degree of freedom in this problem since ##\dot{\varphi} = \omega = ## constant. There is only one degree of freedom. For me, it seemed more natural to let ##r## be the variable. It works out nicely with ##r## , but you can just as well use ##\rho##. ##r## and ##\rho## are proportional to each other. A nice thing about the Lagrangian approach is that you do not need to adopt any particular coordinate system, such as Cartesian, cylindrical, spherical, etc. You can use any convenient "generalized coordinates."

    Looks like you forgot the negative sign for the potential energy when setting up ##L = T - V##.

    You must have made an algebra error in solving for the equation of motion in terms of ##\ddot{\rho}##. Notice that your equation contains ##\dot{\varphi}^2+g \cos \varphi## which is dimensionally inconsistent.

    Also, you can let ##\frac{1}{1+\cot \varphi ^2} = \sin ^2 \varphi##.
     
  5. May 20, 2016 #4

    TSny

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    The last term on the left is incorrect.
     
  6. May 20, 2016 #5

    TSny

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    For ##\upsilon = 0##, ##\rho = 0 = ## a constant. So, ##\dot{\rho} = \ddot{\rho} = 0##. The equation of motion for ##\rho## will just reduce to ##0=0##. However, the equation of motion for ##r## should reduce to ##\ddot{r} = -g##.
     
  7. May 20, 2016 #6
    @TSny Hi and thank you for your answer!

    Taking in consideration all your remarks, I come up now with:

    ##\ddot{\rho} - \frac{\rho \dot{\varphi}^2 - g \cot \upsilon}{\sin^2 \upsilon} = 0##

    I'm also gonna try with r, though I am unsure how to treat the rotational energy then. I'm at a party, I sat in a corner to do the equation again so I'm only writing a short message now. Thanks for your answer! :P

    Julien.
     
  8. May 20, 2016 #7

    TSny

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    The ##\sin^2 \upsilon## factor is in the wrong place.

    Your expression for ##T## that you had in terms of ##\rho## is correct. You can easily rewrite it in terms of ##r## since ##\rho## is proportional to ##r##.
    Enjoy!
     
  9. May 20, 2016 #8
    Really? It was a factor of ##\ddot{\rho}## but then I divided the whole expression by ##\sin^2 \upsilon##.
     
  10. May 20, 2016 #9

    TSny

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    The factor of ##\ddot{\rho}## was ##1+\cot^2 \upsilon##, which does not equal ##\sin^2 \upsilon##.
     
  11. May 20, 2016 #10
    Yes my mistake, I've realized that: ##\ddot{\rho} - \sin^2 \upsilon (\rho \dot{\varphi}^2 - g \cot\upsilon) = 0##
     
  12. May 20, 2016 #11

    TSny

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    Looks good.
     
  13. May 20, 2016 #12
    @TSny Thanks a lot, that's great. ^^ And with ##r## I get:

    ##\ddot{r} - \frac{\sin^2 \upsilon \dot{\varphi}^2}{1 + cos^2 \varphi} r + \frac{g \cos \upsilon}{1 + \cos^2 \upsilon} = 0##

    To get that I wrote ##\rho = r \cos (\frac{\pi}{2} - \upsilon) = r \sin \upsilon##. There might be a way to simplify that, I haven't looked into it yet.


    Julien.
     
  14. May 20, 2016 #13

    TSny

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    I don't think those denominators should be there.

    Yes.
     
  15. May 20, 2016 #14
    Yeah I just did it again starting from x, y and z and I don't get those denominators anymore. Instead just:

    ##\ddot{r} - r \dot{\varphi}^2 \sin^2 \upsilon + g cos \upsilon = 0##
     
  16. May 20, 2016 #15

    TSny

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    OK. Good work.
     
  17. May 21, 2016 #16
    @TSny Thanks a lot. At the end it was a good exercise since I solved the problem using cartesian, cylindrical and spherical coordinates. Using spherical coordinates was by far the easiest (the Lagrangian comes after two lines of calculation).


    Julien.
     
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