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Bead sliding on a wire - calculus of variations

  1. May 6, 2016 #1
    I am asked to find the shape of a wire that will maximize the speed a sliding bead when it reaches the end point(Similar to the brachistochrone problem expect that the speed is to be maximized and not time minimized).

    But shouldn't the speed at the end be independent of the shape of the wire? The potential energy of the bead at the start is ##U = mgh## and the final kinetic energy is ##\frac{1}{2} m v^2##

    equating these you get, ##v = \sqrt{2gh}##. and that simply depends on ##h## and nothing else.

    Is this wrong?
     
  2. jcsd
  3. May 6, 2016 #2

    TSny

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    Sounds right. Have you stated the problem completely and exactly as given?
     
  4. May 6, 2016 #3
    This is the problem statement. Am I interpreting this right?

    A bead slides fritionlessly on a wire, starting at rest from the origin (0,0) to some point (X,-Y) where X and Y are positive constants.
    (Here the +y direction is vertically upward) Aerodynamic drag is negligible. Find the shape of the wire between the two
    points which maximizes the speed of the bead when it reaches the fixed end point (X,-Y).
    Assume the length of the wire between the initial and final points is some fixed value L, greater than the distance between the points.
     
  5. May 6, 2016 #4

    TSny

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    I think you're right. As long as Y is fixed, all shapes lead to the same final speed.
     
    Last edited: May 6, 2016
  6. May 7, 2016 #5

    nrqed

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    Is this a problem from a textbook or from your teacher? I wonder if the teacher simply did not realize that the question was actually trivial (I know that I myself have been guilty of sometimes not thinking through a question completely before asking it to students)
     
  7. May 7, 2016 #6
    It is from a professor and maybe he made a mistake...
     
  8. May 7, 2016 #7

    nrqed

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    Ok, then either he did not realize that the answer was trivial or it is a trick question to test whether the students think before trying to do something complicated. In either case, you are right that the answer is that the shape does not matter.
     
    Last edited: May 7, 2016
  9. May 8, 2016 #8

    vela

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    The only complication I see is that you have to require y<0 everywhere (except at the start), otherwise the bead can't make it to the end of the wire because it'll come to rest and possibly turn around.
     
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