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Physics riddle

  1. Apr 26, 2016 #1

    micromass

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    http://img-9gag-fun.9cache.com/photo/aX9zqD2_460s.jpg [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 26, 2016 #2

    jedishrfu

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  4. Apr 26, 2016 #3
    NOT 2 B
     
  5. Apr 26, 2016 #4

    QuantumQuest

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    Without specific data and judging solely from the diagram I intuitively tend to vote for B.
     
  6. Apr 26, 2016 #5

    Borg

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    If the start and stop height differences are the same, wouldn't the answer be C?
     
  7. Apr 26, 2016 #6

    micromass

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    HINT:

     
  8. Apr 26, 2016 #7

    jedishrfu

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    What was the question again?
     
  9. Apr 26, 2016 #8
    If the lines are frictionless then it shouldn't matter
     
  10. Apr 26, 2016 #9

    Borg

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    Well, when you put it that way... :olduhh:
     
  11. Apr 26, 2016 #10

    PeroK

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    Suppose the blue line had a smaller slope at the beginning and end. The ball could take a long time to get across.

    And, if the red line had a really large dip, it could take a long time to get across.
     
  12. Apr 26, 2016 #11

    PeroK

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    I reckon the red will be behind at the top of the first trough and may just about catch up by the end. But, you need specific data.
     
  13. Apr 26, 2016 #12

    jedishrfu

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  14. Apr 26, 2016 #13

    Borek

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    Brachistochrone is not flat, but everything that goes below is again slower - does it mean trajectories that go "deeper" are slower than brachistochrone, but always faster than the flat/straight one?
     
  15. Apr 26, 2016 #14

    PeroK

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    It's easy enough to compare a straight line down a slope with the limiting case of freefall, followed by horizontal motion. If the height of the slope is ##h## and the horizontal distance is ##d##, then the times are:

    ##t_1^2 = \frac{2(h^2 + d^2)}{gh}## (for the straight line)

    ##t_2^2 = \frac{2(h + d/2)^2}{gh}## (for free fall plus horizontal)

    The straight line takes longer when:

    ##h < \frac{3d}{4}##

    But,of course, the brachisrochrone is always faster than both.
     
  16. Apr 26, 2016 #15
    θOk the answer to this problem was bothering me, so I did some math:

    1) Suppose we have two paths with one ball on each path. The both start at height ho at xo and finish at height h1 at x1. h1 is less than ho. xo, and ho are taken to be the 0 datum. Positive h is downwards.

    2) Suppose further that we know height profile of the paths h(x) between xo and x1.

    How do we calculate which ball will reach x1 first?

    My solution:
    Neglecting friction and all other interesting physics, we have a relationship between the height and the velocity of the ball, v(x) = √[2*g*h(x)]
    We can estimate which ball will get to the finish line first by computing its average velocity in the x direction. Since we know the paths of the ball, we can average over [xo,x1].

    Taking the angle θ to be measured between positive x and tangent to h(x), the average can be computed as: {∫√[2*g*h(x)]*cos(θ(x))dx}/(x1-xo) from xo to x1.
    In this case we would have to show that the product of those expressions is greater than a case where h(x) is constant (as in case 1 for example).

    Now, that integral might be a pain because we have a general function h(x) and a trigonometric function. I thought to myself "How can we relate h(x) to θ?" Ah ha! We know that h(x) = ∫{dh(x)/dx}*dx, and we know that the slope of h(x) can be expressed as the tangent of θ at a point x. So we have an equation in terms of θ, only:

    average(v(x)_x) = { ∫√[2*g* (∫tan(θ(x))*dx)] * cos(θ(x))dx }/(x1-xo)

    And I hate to quit but this is as far as I got before looking up this integral in mathematica and seeing how many terms are in the solution. The only way I know how to do this by hand is through integration by parts, which would have to be done twice since the angle is parameterized by x.

    Anyway, you can see that if h(x) is constant, then the cosine term goes to 1, the tangent term becomes h(x). You need to show that h(x)*cos(θ) on average is greater for the slopey case than h(x) in the flat case.
     
  17. Apr 26, 2016 #16
    Everyone else seems to be seeing a diagram. Not I.
     
  18. Apr 26, 2016 #17

    PeroK

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    A simpler problem, but relevant to this question would be:

    Take two balls with the same initial velocity, to travel a distance ##d## along the track. The first ball goes horizontally. The second ball goes down and up an incline (assume two symmetric straight lines). What's the relationship between the times, the initial velocity and the angle of the incline? What is the optimum angle?
     
  19. Apr 26, 2016 #18

    Borek

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    Actually now that I think about it, it is trivial - let's say we go down, horizontal, up (sides of a rectangle, or more precisely, sides of a right trapezoid). There always exist a rectangle in which going down takes exactly as long as the horizontal leg in the "flat" case. That means going through the sides of the rectangle will take a bit longer than two times the flat case. That in turn means not every path below the flat one is guaranteed to be faster - or, in other words, B is not guaranteed to be faster than A, and to be sure which is faster we need to know exact shape of B and calculate the time it takes to travel both paths.

    Or am I misunderstanding something?
     
  20. Apr 27, 2016 #19

    PeroK

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    Yes, you have to calculate in each case.

    The problem here involves an initial velocity, which complicates everything. If the initial velocity is low, then many paths below the horizontal will be faster. The optimum path will be related to the brachistrochrone, but the initial velocity messes things up. But, steeps paths, curves, straight inclines (down and up) will all be better than going along the flat.

    As the initial velocity increases, it should still be possible to beat the flat path, but only by going down a short distance. I haven't tried the maths (yet) but the equation involves the relative (proportional) increase in average velocity against the proportional increase in distance. The faster the initial velocity, the smaller the proportional increase you get by going down. If the ball is going fast enough, then nothing will be significantly faster than just going along the flat.

    The answer is some paths below the horizontal beat the horizontal path and some do not. The key factors are the initial velocity and the force of gravity (as well as the shape of the path, of course).
     
  21. Apr 27, 2016 #20
    If friction ignored answer A. On the flat section A continues with a steady speed but during the equivelent section of Bs journey the speed changes continually. Imagine just one part of Bs journey for example the first valley it encounters. Now imagine that, compared to the horizontal distance travelled, the valley was incredibly deep, for example horizontal distance travelled = 20m and depth of valley = 20 000 000m. From these numbers it should be easy to see that in general a journey with variable heights will take a longer time to complete.
     
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