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Barbell in gravitational field (Lagrange)

  1. May 26, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! Here is a new Lagrange problem I am trying to solve, and I would like to have your opinion about my solution so far!

    A barbell composed of two masses ##m_1## and ##m_2##, idealised as particles and separated by a distance ##a## from each other, moves in the Earth's gravitational field.
    a) Write the Lagrange function using spherical coordinates for the relative vector and cartesian coordinates for the center of mass.
    b) Derive the equations of motion and give minimum one special solution.
    c) Which quantities are conserved in this system?

    2. Relevant equations

    Lagrange function, Lagrange equations of motion

    3. The attempt at a solution

    So first I drew the "situation" (see attached picture) and forgot everything about a barbell. :biggrin: I assumed ##m_2## was bigger than ##m_1## to have an idea of what's going on, but that' not so important. I called ##\vec{R}## the vector going from the origin to the center of mass and ##\vec{r}## the vector going from ##m_2## to ##m_1##. ##\vec{r_1}## and ##\vec{r_2}## are the vectors going from the origin to respectively ##m_1## and ##m_2##.

    a)
    So first I wrote ##\vec{R}## and ##\vec{r}## in terms of ##\vec{r_1}## and ##\vec{r_2}## :

    ##\vec{R} = \frac{\vec{r_1} \cdot m_1 + \vec{r_2} \cdot m_2}{m_1 + m_2}##
    ##\vec{r} = \vec{r_1} - \vec{r_2}##

    I define the reduced mass as ##m := \frac{m_1 \cdot m_2}{m_1 + m_2}## and I rewrite those to get new expressions for ##\vec{r_1}## and ##\vec{r_2}##:

    ##\vec{r_1} = \frac{m}{m_1} \vec{r} + \vec{R}##
    ##\vec{r_2} = \frac{m}{m_2} \vec{r} + \vec{R}##

    Preliminary, I wrote ##T## and ##V##:

    ##T = \frac{1}{2} m_1 \cdot \dot{r}_1^2 + \frac{1}{2} m_1 \cdot \dot{r}_2^2##
    ##V = m \cdot g \cdot z##

    Of course ##m## is still the reduced mass and so ##z## refers to the z-position of the center of mass. After substituting ##\vec{r_1}## and ##\vec{r_2}##, I get an expression for ##L##:

    ##L = \frac{1}{2} m \cdot \dot{\vec{r}}^2 + \frac{1}{2} (m_1 + m_2) \dot{\vec{R}}^2 - m \cdot g \cdot z##

    That's nice. But is it correct? Unfortunately the problem asks for ##\vec{R}## to be expressed in cartesian coordinates and ##\vec{r}## to be expressed in spherical coordinates... Therefore:

    ##\vec{R}^2 = x^2 + y^2 + z^2 \implies \dot{R}^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2##
    ##\vec{r} = r \cdot (\cos \theta, \sin \theta) \implies \dot{\vec{r}} = \dot{r} \cdot (\cos \theta, \sin \theta) + r \cdot \dot{\theta} \cdot (-\sin \theta, \cos \theta)## since ##\varphi## is constant (I would think) because both masses accelerate at the same rate ##g##.

    And that brings me to that Lagrange function:

    ##L = \frac{1}{2} (m_1 + m_2) (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) + \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - mgz##

    Uuf...Is that correct? I'm not completely sure about the transformation from vectors to scalars, especially in the case of ##\vec{r}##; I've squared "inside" because I think the kinetic energy should square the velocity in direction of ##\hat{r}## and also in direction of ##\hat{\theta}##. Is that right?

    b)
    So hopefully a) is correct to begin with. I'm a bit confused by the next question because if it was up to me I would have written the Lagrange function as well as the equations of motion in terms of ##R## and ##r##, but because of the choice of coordinates in the end I had to write them in terms of ##z## and of ##r## (because the other equations give a ##0## acceleration, I will write them anyway). I get those:

    ##\ddot{x} = 0##
    ##\ddot{y} = 0##
    ##\ddot{\theta} = 0##
    ##\ddot{\varphi} = 0##

    ##\frac{\partial L}{\partial z} = -m \cdot g## and ##\frac{d}{dt} \frac{\partial L}{\partial \dot{z}} = (m_1 + m_2) \cdot \ddot{z}##
    ##\implies \ddot{z} = \frac{-m}{m_1 + m_2} g = \frac{- m_1 \cdot m_2}{(m_1 + m_2)^2} g##

    ##\frac{\partial L}{\partial r} = m \cdot r \cdot \dot{\theta}^2## and ##\frac{d}{dt} \frac{\partial L}{\partial \dot{r}} = m \cdot \ddot{r}##
    ##\implies \ddot{r} = r \dot{\theta}^2##

    Mm... I don't know what to think of those. The square on ##m_1 + m_2## in the ##\ddot{z}## equation looks strange, and I have nothing to say really about the second one. At least the zeros in the other equations are coherent! :-p

    I didn't go further yet as I am too unsure about my solution. Can anybody take a look and tell me if that's correct/wrong?


    Thanks a lot in advance for your answers!


    Julien.
     

    Attached Files:

  2. jcsd
  3. May 26, 2016 #2

    TSny

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    Should you use the reduced mass in the expression for V?
    Looks good except for the m in the last term and there should be a term involving ##\dot{\varphi}##
    [EDIT: For a barbell, what is the value of ##\dot{r}##?]

    No, I don't think this is a correct statement. ##\varphi## is free to change
     
    Last edited: May 26, 2016
  4. May 26, 2016 #3
    Shouldn't I? :biggrin: I thought it would be the same to treat the system as a one-object system with the center of mass and the reduced mass. Otherwise I guess I would write it that way:

    ##V = g \cdot (m_1 \cdot z_1 + m_2 \cdot z_2)##

    There must be a way to write it in term of angles... I need to think a bit about it.

    I guess the missing term would be ##r \cdot \dot{\varphi}^2 \sin^2 \theta##?

    No, I don't think this is a correct statement. ##\varphi## is free to change[/QUOTE]

    I don't get why. Since the force of gravity is the only one acting on the system, shouldn't both masses fall at the same acceleration?


    Julien.
     
  5. May 26, 2016 #4
    I forgot to thank you for your answer, sorry for that, I really appreciate your help. :)
     
  6. May 26, 2016 #5

    TSny

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    See if you can write the expression in parentheses in terms of the z coordinate of the center of mass (##z##).
    Yes

    Each mass feels two forces. The acceleration of the two masses will not generally equal each other.
     
  7. May 26, 2016 #6
    Now I'm really puzzled: what is the second force? The tension in the rod between the masses? How does it show in the equations? Maybe through ##\vec{r}##..

    By the way I missed one of your questions in the last post: I guess ##\ddot{r} = 0## since the distance between the two masses doesn't change...

    Julien.
     
  8. May 26, 2016 #7

    TSny

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    Yes. If the barbell were spinning rapidly, this tension force could be much greater than the force of gravity on the masses.
    The tension force does not enter the Lagrangian. The tension force is a constraint force keeping ##r## constant. You are taking account of the effect of the tension force by restricting ##r## to always equal ##a##.
    Yes, ##r = a = ## constant. So, ##\dot{r} = \ddot{r} = 0##. ##r## is not one of the "generalized coordinates" in setting up the Lagrangian. It is just a constant.

    If you replaced the rod by a stiff spring so that ##r## could vary, then you would need to include ##r## as a generalized coordinate in the kinetic energy. You would also need to include the potential energy of the spring in ##L##. Then you could use the Lagrange equations to determine the equation of motion for ##r##.
     
  9. May 26, 2016 #8
    About rewriting ##V##, that's what I get now:

    ##z = \frac{m_1 \cdot z_1 + m_2 \cdot z_2}{m_1 + m_2}##
    ##\implies V = z \cdot g \cdot (m_1 + m_2)##


    Julien.
     
  10. May 26, 2016 #9

    TSny

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    Yes, that's it.
     
  11. May 26, 2016 #10
    @TSny Super. And thanks for your explanation, there's a lot for me to learn in that post.

    So does that all mean that at the end I should get:

    ##L = \frac{1}{2} (m_1 + m_2) (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) + \frac{1}{2} m (r^2 \dot{\theta}^2 + r \dot{\varphi}^2 \sin^2 \theta) - (m_1 + m_2) g z##?

    Though I understood what you meant with ##\dot{r} = \ddot{r} = 0##, it looks a bit strange in my process since I've substituted ##\dot{\vec{r}}##. But I guess then that the vector may rotate so its acceleration is not necessarily zero. Is that a correct assumption?


    Thanks a lot for all those answers.

    Julien.
     
  12. May 26, 2016 #11

    TSny

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    Yes. I would go ahead and let ##r = a## to emphasize that ##r## is just a constant.

    Yes. ##\vec{r}## is not a constant due to changing orientation. But ## r= |\vec{r}| = a## is a consant.
     
  13. May 26, 2016 #12
    Waoh I'm somehow fascinated by that. I've rewritten this step to visualize it better:

    ##\vec{r} = r \cdot (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta)##
    ##\implies \ddot{\vec{r}} = \dot{r} (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta) + r \cdot \dot{\theta} (\cos \theta \sin \varphi, \cos \theta \sin \varphi, -\sin \theta) + r \cdot \dot{\varphi} (-\sin \theta \sin \varphi, \sin \theta \cos \varphi, 0)##
    ##= 0 + r \cdot \dot{\theta} \cdot \hat{\theta} + r \cdot \dot{\varphi} \cdot \sin \theta \cdot \hat{\varphi}##

    That's great! But then isn't there a power two missing in my post #3 on the ##r##?


    Thanks a lot again, your answers are very clear as always.


    Julien.
     
  14. May 26, 2016 #13

    TSny

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    Typos: ##\dot{\vec{r}}## instead of ##\ddot{\vec{r}}## on the left. Middle term on right should be ##r \cdot \dot{\theta} (\cos \theta \cos\varphi, \cos \theta \sin \varphi, -\sin \theta) ##
    Yes! Sorry I overlooked that in your previous posts.
     
  15. May 26, 2016 #14
    @TSny Thanks for your answer, that was very nice. I'm gonna go back to the equations of motion and post it again later. Are you sure concerning the ##\sin\varphi##? Why would the ##\cos\varphi## be differentiated in the second expression?
     
  16. May 26, 2016 #15

    TSny

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    Pretty sure.
    It shouldn't. I think you were the one who differentiated the ##\cos\varphi## in the second expression. But, I might not be seeing it. I think I need a little break.
     
  17. May 26, 2016 #16
    @TSny No you are right I typed ##\sin## instead of ##\cos## though I had it right on my sheet. Sorry for that. For the equations of motion, I found:

    ##\ddot{z} = -g##
    and
    ##\ddot{\theta} = 2 \dot{\varphi}^2 \cos \theta \sin \theta##

    The detail of all the derivatives are in the attached picture. But take a break please, I might also need one. :)

    Thanks a lot again for all this help.


    Julien.
     

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    Last edited: May 26, 2016
  18. May 26, 2016 #17

    TSny

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    That looks pretty good. However, check the factor of 2 in the equation for ##\ddot{\theta}##.
     
  19. May 27, 2016 #18
    @TSny Yes the 2 disappears because of the ##\frac{1}{2}## before the ##m##... Well, getting there was sure not easy, and now I've got that strange looking differential equation:

    ##\ddot{\theta} - \dot{\varphi}^2 \cos \theta \sin \theta = 0##

    which looks very difficult to solve to me because ##\theta## is inside the ##\cos## and the ##\sin## so I can't get a proper factor :/. I'm also still pretty bad at differential equations... I suppose I could rewrite the equation like that:

    ##d\dot{\theta} = \dot{\varphi}^2 \cos \theta \sin \theta dt##

    But that doesn't help really since each factor of the right side should be integrated with respect to time... Any clue?

    For ##z## I simply get ##z = -\frac{1}{2} g t^2 + \dot{z}_0 t + z_0##. Not sure if I can substitute the ##\dot{z}_0## and ##z_0## with anything since I was not given any indication about an initial velocity/position. I guess I could arbitrarily define ##z_0 = 0## but for the velocity I don't think I can do anything so far.

    PS: back to the spherical coordinates shortly, cause I would like to check I understand this right: for ##\vec{r}## we use a different origin (namely ##m_2##) as for ##\vec{R}##, right? And ##\theta## is the angle between ##\vec{r}## and ##z'## while ##\varphi## is the angle between ##\vec{r}## and ##x'##?

    Thanks a lot again, I'm getting there. Slowly. :-p

    Julien.
     
    Last edited: May 27, 2016
  20. May 27, 2016 #19

    TSny

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    Yes, this looks complicated since it contains the two variables ##\theta## and ##\varphi##.
    Your result for the equation of motion for ##\varphi## is not correct. (See attached picture below). You should find that the corrected differential equation for ##\varphi## can be easily integrated to get an expression for ##\dot{\varphi}## in terms of ##\theta##. You can then use this to eliminate the ##\dot{\varphi}## in the above equation for ##\ddot{\theta}##.
    Note that you are only asked to come up with a specific solution to the equations of motion, rather than having to find the general solution.
    This is good. You have the general solution for the equation of motion for ##z##. When solving a second order differential equation, you will have two arbitrary constants that are determined by the initial conditions. Since you are asked to come up with just one specific solution to the equations of motion, you are free to pick any initial conditions that you want.
    Yes. You defined ##\vec{r}## to be the position of m1 relative to m2. (You could have chosen ##\vec{r}## to be the position of m2 relative to m1. Either way is fine.)
    What is the meaning of the primes here? In spherical coordinates, ##\varphi## is the angle between the x -axis and the projection of ##\vec{r}## onto the x-y plane.
     

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  21. May 27, 2016 #20
    Thanks again for your answer. I'm pulling my hair on the equation of motion for ##\varphi##. I keep on getting

    ##\ddot{\varphi} \sin^2 \theta + 2 \cos \theta \sin \theta \dot{\varphi}^2 = 0##

    but that's not what you seem to suggest (or is that easily integrable? Maybe I just don't know how to solve that). Note that I didn't simplify the sin so that you can still see the process through. I'm not sure how to integrate with two variables depending on time. :/

    That's great, thanks!

    The primes refer to a cartesian coordinate system having ##m_2## for origin and with its axes parallel to the corresponding x,y,z-axes. A bit farfetched but it kind of help me to visualise which angles we were talking about. :)

    Julien.
     
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