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Constant solutions to the Lagrange equations of motion

  1. Apr 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle moves on the surface of an inverted cone. The Lagrangian is given by

    t%29%5Cdot%7Br%7D%5E%7B2%7D+r%5E%7B2%7D%5Cdot%7B%5Cphi%7D%5E%7B2%7D%20%5Cright%29-mg%5Calpha%20r.png

    Show that there is a solution of the equations of motion where png.latex?r.png and png.latex?%5Cdot%7B%5Cphi%7D.png take constant values png.latex?r_0.png and png.latex?%5COmega.png respectively

    2. Relevant equations
    The equations of motion png.latex?r.png and png.latex?%5Cphi.png are

    tex?%5Cddot%7Br%7D+%5Calpha%5E%7B2%7D%5Cddot%7Br%7D-r%5Cdot%7B%5Cphi%7D%5E%7B2%7D+g%5Calpha%20=0.png (1)

    png.latex?r%5E2%5Cdot%7B%5Cphi%7D=K.png (2)

    So png.latex?%5Cdot%7B%5Cphi%7D.png can be eliminated from equation (1)

    3. The attempt at a solution
    Would I go about this by setting png.latex?%5Cddot%7Br%7D.png equal to a constant, png.latex?r_0.png , and similarly for png.latex?%5Cdot%7B%5Cphi%7D.png ?
     
    Last edited by a moderator: Apr 17, 2017
  2. jcsd
  3. Apr 24, 2015 #2

    MarcusAgrippa

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    It is not [itex] \ddot{r}, \dot{\phi}[/itex] that you were asked to show provide a solution, but [itex] {r}, \dot{\phi}[/itex]. These requirements define uniform circular motion. The question is thus asking you to use the equations to check that uniform circular motion is a solution.
     
  4. Apr 24, 2015 #3

    Ray Vickson

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    If ##r(t) = r_0## (a constant), what is ##\ddot{r}(t)##?
     
  5. Apr 24, 2015 #4

    MarcusAgrippa

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    What do you get when you differentiate a constant?
     
  6. Apr 24, 2015 #5
    Differentiating a constant is zero.
    I was thinking about the equations in isolation to each other and not as part of a system. So

    [tex]v=r \dot{\phi}=r_0 \Omega[/tex]

    I just need to show that the equations of motion lead to this.
     
    Last edited: Apr 24, 2015
  7. Apr 24, 2015 #6

    Ray Vickson

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    No: they do not "lead to" this; they merely allow this.

    There is a big difference, since (due to the lack of friction) a solution with non-constant ##r## and ##\dot{\phi}## never settles down to a limiting orbit of constant ##r## and ##\dot{\phi}##. However, if you start off with the correct initial conditions, your orbit maintains constant ##r## and ##\dot{\phi}## forever. Essentially, that is what you are being asked to verify.
     
  8. Apr 24, 2015 #7
    So setting the acceleration [itex]\ddot{r}=0[/itex] leaves a constant [itex]r[/itex] and [itex]\dot{\phi}[/itex] which allows the uniform motion.
     
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