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Cycloid Lagrangian

  1. May 15, 2017 #1
    1. The problem statement, all variables and given/known data
    A point like particle of mass m moves under gravity along a cycloid given in parametric form by
    $$x=R(\phi+\sin\phi),$$
    $$y=R(1-cos\phi),$$
    where R is the radius of the circle generating the cycloid and ##\phi## is the parameter (angle). The particle is released at the point (##x=\pi R, y=2R##) from rest.

    1) What types of constraint apply to this system and how many degrees of freedom are needed to describe the motion?

    2) Show that the lagrangian of the system is
    $$L=2mR^2\dot{\phi}\cos^2\frac{\phi}{2}-2mgR\sin^2\frac{\phi}{2}$$

    3) Introduce a new generalised coordinate ##s=4Rsin\frac{\phi}{2}## and express the Lagrangian in terms of s.
    2. Relevant equations


    3. The attempt at a solution
    1) I am not sure what the constraints are. It's just that the particle is forced to move in a cycloid. Could someone help me here?
    The degrees of freedom is 2 since x and y are dependent on R and ##\phi##. So R and ##\phi## are the variables required.

    2) The lagrangian is L=T-V.
    Am I correct in thinking ##T=\frac{1}{2}m(\dot x^2+\dot y^2)##? I have tried taking the time derivative of the right hand side of the given equations of x and y. I got $$\dot x=R(\dot \phi+\dot \phi \cos\phi),$$
    $$\dot y=R(\dot \phi \sin\phi).$$
    Substituting into T, $$T=R^2m\dot \phi ^2 (1+cos\phi).$$
    Similarly, $$V=mgy=mgR(1-\cos\phi)$$
    Then $$L=R^2m\dot \phi ^2 (1+cos\phi)-mgR(1-\cos\phi)$$
    I'm looking through the double angle formulas and can't really determine how to get to the given lagrangian.

    3) What to do?

    Thanks in advance
     
  2. jcsd
  3. May 15, 2017 #2

    Orodruin

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    1) Is R a degree of freedom? If it is, where are the derivatives of R in 2?

    2) Just apply the double angle formula and the trigonometric one.

    3) It is a variable substitution. What is your problem with it?
     
  4. May 16, 2017 #3
    1) Is R a degree of freedom? If it is, where are the derivatives of R in 2?
    Sorry, I realised I initially had R as a degree of freedom but after taking the derivative, it became too complex I figured R is not a degree of freedom. This makes sense because it's just the radius of a circle. So one degree of freedom, ##\psi##

    2) Just apply the double angle formula and the trigonometric one.
    Okay

    3) It is a variable substitution. What is your problem with it?
    I don't know where to substitute? Should I solve for R, so ##R=\frac{s}{4\sin\frac{\phi}{2}}## and plug it in the x and y equation?
     
  5. May 16, 2017 #4

    TSny

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    I'm not sure exactly what the question is asking here. But standard texts often categorize constraints using scary words. For example, have you run into the terms "holonomic", "nonholonomic", "scleronomous", and "rheonomous"?

    (I studied Lagrangian mechanics in 1970 and this is the first chance that I've ever had to use any of those words. Thanks for the opportunity. :oldsmile: )
     
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