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First integral of the motion

  1. Feb 3, 2007 #1

    quasar987

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    I have found the following ODE in the context of a mechanics problem and am now asked to find a first integral of this equation.

    [tex](m_1+m_2)\ddot{r}-m_1Cr^{-3}+m_2g=0[/tex]

    I know this means that I'm supposed to find an equation of the form [itex]F(\dot{r},r)=\mbox{const.}[/itex] but I don't know how to achieve that.

    Am I expected to guess a coordinate transformation whose associated constant of the motion (in the sense of Noether's theorem) is of the form [itex]F(\dot{r},r)=\mbox{const.}[/itex]? Or is there a more direct approach? Certainly the equation cannot be integrated directly because what's [itex]\int r^{-3}dt[/itex]??

    Thanks for the help!
     
  2. jcsd
  3. Feb 3, 2007 #2

    AlephZero

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    Multiply the equation by [itex]\dot r[/itex], then integrate.

    That "trick" works for almost all the 2nd order dynamics equations that turn up in textbook and exam questions.
     
  4. Feb 3, 2007 #3

    quasar987

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    Thanks for the tip! :tongue2:
     
  5. Feb 3, 2007 #4

    quasar987

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    I have found the first inetegral of the motion; it is

    [tex]\frac{1}{2}(m_1+m_2)\dot{r}^2+\frac{m_1C}{2}r^{-2}+m_2gr=\mbox{const.}[/tex]

    and I am now asked to give its physical significance.

    My best shot was to remark that the equation above is an equation of energy conservation for a mass [itex]m_1+m_2[/itex] free to move in one dimension and subject to move in a ficticious potential [itex]'V'(r)= \frac{m_1C}{2}r^{-2}+m_2gr[/itex]. But I'm thinking, maybe this is too mathematical and not enough physical?

    Otherwise, how can the equation be interpreted physically?
     
  6. Feb 3, 2007 #5

    quasar987

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    Mmmh, this potential is dubious too because it won't let r go to zero*... but the actual problem is that of a table with a hole in it and two masses linked by a rope passing through the hole in the table, such that one mass slides w/o friction on the table and the other is hanging in the air below it.

    This kinds of baffles physical intuition doesn't it? Instead of the mass spiraling towards the hole, the mass on the table will behave like a kind of planet orbiting the hole... This doesn't make sense; I can't imagine that the mass hanging in the air will ever lift up! :grumpy:


    *unless C is zero, which corresponds to a null initial angular velocity according to my previous calculations, so this makes sense
     
  7. Feb 3, 2007 #6

    AlephZero

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    From the description of the physical problem it should be clear what
    [tex]\frac{1}{2}(m_1+m_2)\dot{r}^2[/tex]
    and
    [tex]m_2gr[/tex]
    represent (they are both energies).

    You didn't give the full question, but if it involves the mass on the table "orbiting" round the hole, shouldn't there be some theta terms in the equations of motion?

    OTOH if C = 0, you don't need to explain what a zero term represents!
     
    Last edited: Feb 3, 2007
  8. Feb 3, 2007 #7

    quasar987

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    Indeed there initially were two coupled equations of motion, but I was able to uncouple them by showing that the second one meant

    [tex]\dot{\phi}^2=Cr^{-4}[/tex]

    Then inject that in the first to get the equation of post #1.
     
  9. Jan 18, 2008 #8
    So the discussion in this thread is really advanced and here I come with a simple question:

    say we have the system

    x' = x
    y' = y

    How do we find a first integral for this?

    I know that f(x) = e^x and f(y) = e^y are solutions for this system...

    But I have no idea where to start with this..
     
  10. Aug 24, 2009 #9
    well, formally one can divide the two equations

    dx/dy=x/y,

    so dx/x=dy/y, yielding ln(abs(x/y))=const.

    HOWEVER, a first integral should be a continuous function, and the above isn't.
    Intuitively, your equations describe a source at x=y=0, and therefore phase volumes grow in time (i.e. there is no conserved quantity that is continuous and is not constant on an open subset of the x-y plane).

    Best, Mathador
     
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