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Motion Equation

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  1. Oct 31, 2015 #1
    • Missing homework template due to originally being posted in technical forum
    Hello! Nice to meet you. I'm studying a Bachelor of Science and I have a doubt:

    A bead slides without friction on a frictionless wire in the shape of a cycloid with equations

    x = a(θ-sin θ) y = a(1+cos θ)

    where θ's range is 0 to 2π and a>0

    a) Find the equation of motion:

    And I resolved:
    0JSS1Th.png
    I checked in a physics book that the solution is correct

    b) Make the change of variable u= cos (θ/2) and find a linear equation for u(t). Solve this equation and find the solution θ(t) which in the case of a bead at t = 0 falls from rest from the corresponding point θ= π/2

    I don't know how to make the change. Can anybody help me? Thanks and nice to meet you.
     
  2. jcsd
  3. Oct 31, 2015 #2

    Orodruin

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    Exactly what in the change of variables are you having problems with? It is a matter of reexpressing the differential equation in terms of u instead of theta.
     
  4. Oct 31, 2015 #3
    Thanks for answering! I do not understand how I have to make the change of variable, I'm trying to do since Wednesday and not even know where to start the change of variable
     
  5. Oct 31, 2015 #4

    Orodruin

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    It is just a normal change of variables. You insert the new variable in place of the old one using the relation between them.
     
  6. Oct 31, 2015 #5
    I tried doing a simple change of variables. But I obtain a non linear equation..

    JHSWflM.png
     
  7. Oct 31, 2015 #6

    Orodruin

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    Please show your work and how you got there. Your non-linear term should not be there. The other terms are correct.
     
  8. Oct 31, 2015 #7
    0JSS1Th.png

    u= cos (θ/2) then:

    θ = 2 arccos u

    then:

    5GyXjwH.png

    and:
    xew0ky7.png

    Then substitute:

    9J1DJir.png

    Now: 91_11.gif and 91_10.gif and substitute:

    TLtcqXT.png

    We now that cos(arccos (a)) = a and sen(arccos(a) = 6ZAyPdf.png

    7YBhE6B.png

    Using first and second derivates and simplifying:

    WCC5lzs.png

    Then multiplied by TsiQmof.png and dividing by (-4) we obtain:

    proxy.php?image=http%3A%2F%2Fi.imgur.com%2FJHSWflM.png

    Which isn't linear.


    What is your opinion? Thanks!
     
  9. Oct 31, 2015 #8

    Orodruin

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    It is much easier to help you if you stop pasting images and instead use the forum LaTeX to type in your equations (they become quotable).

    Your second derivative is wrong, the second term needs to contain ##(u')^2##. Also, it becomes much much easier if you do not invert the relation ##u = \cos(\theta/2)## but instead differentiate it as it stands.
     
  10. Oct 31, 2015 #9
    ¿##(u')^2##? I don't see this.. I don't understand what you mean. Sorry.
    Thanks Orodruin!
     
  11. Oct 31, 2015 #10
    Now I think that the second derivative is : OsZri7y.png
    But it doesn't contains ##(u')^2##
     
  12. Oct 31, 2015 #11

    Orodruin

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    It does, you did it wrong again. What is the derivative of
    $$
    \frac{1}{\sqrt{1-u^2}}?
    $$
     
  13. Oct 31, 2015 #12
    ##uu' ((1-u^2))^{-3/2}##
    It's that right?
     
  14. Oct 31, 2015 #13

    Orodruin

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    So what is then the derivative of
    $$
    \frac{u'}{\sqrt{1-u^2}}?
    $$
     
  15. Oct 31, 2015 #14
    s3sAL1r.png
    It's ok?

    Thanks!
     
  16. Oct 31, 2015 #15

    Orodruin

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    What happened to the ##u## coming out of the derivative of ##1/\sqrt{1-u^2}##? But again, the best way forward is not to solve for ##\theta## before differentiating, it is to directly differentiate ##u = \cos(\theta/2)##.
     
  17. Nov 1, 2015 #16
    Hello! If you do not mind, I prefer to finish the exercise by this method and later solve by the method you suggest

    Then, the second derivative is:
    7kJBG4v.png

    Replacing first and second derivative:

    zwwYBmm.png

    And simplifying:

    42fM8Lx.png


    Simplifying again:

    64V4Fqp.png


    What do you think? Thanks and have a nice sunday
     
  18. Nov 1, 2015 #17

    Orodruin

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    Looks fine apart from a square disappearing in your middle step (the end result is correct).
     
  19. Nov 1, 2015 #18

    Orodruin

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    Ah yes, and the ##a## should be in the denominator in the end result ...
     
  20. Nov 1, 2015 #19
    Sorry! It was a mistake...
    The linear equation is:
    X22HYPk.png

    The characteristics polynomial is:
    HcfyQCk.png

    With complex roots:

    zhCQBwQ.png

    Ajnzykq.png

    The general solution:

    Tz0YCHI.png

    p1RHjOP.png

    We need to know the particular solution. True?


    Thanks!
     
  21. Nov 1, 2015 #20
    Edit: The equation is homogeneous. No particular solution. But then we can not replace, I can not find the solution
     
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