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    Quickest Path Down a Slide

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    See the attached file for full problem description, as there are a number of equations that cannot be expressed in plain text.

    The Solution uses a different constant of convenience:
    A) The b and the C of the provided question are related as b = 1/2C^2
    B) It takes the negative y below zero, unlike the reversed direction of y in the attached file.

    These details do not affect the essence of the answer.

    The Solution is provided in plain TEX source as well as in a PDF.

    This is the famous brachistochrone variational problem of determining the path of quickest descent along a frictionless slide, under the influence of a constant gravitational acceleration, g.
    Assume a two dimensional motion, where x is the horizontal spatial coordinate (perpendicular to the gravitational field), and y is the spatial coordinate which increases down (in the direction of the gravitational acceleration vector, g), where the path to be determined is y(x), which extends from the origin (o,o), to the destination point (x_0 ,y_0).
    A) Write down the time functional (see the attachment for the equation)
    Where y(x) is the path of the brachistochrone, for the minimum time of transport, which is to be determined from the variational calculation. The integrand, f(y,y_x,x), can be determined by expressing the time differential as a ratio of the spatial path differential, ds, divided by the velocity, v, along the path, or dt =ds/v.
    Note that, in general, the spatial path differential can be expressed in terms of the two-dimensional, x - y, differentials, as ds = sqrt(dx^2+dy^2).
    Also, note that the velocity, v, along the path, can be determined by energy conservation (kinetic plus potential energy), E = KE + PE, where the initial gravitational potential energy, PE = -mgy, is continually converted into kinetic energy, KE=mv^2/2, as the object moves in the downward vertical direction, or positive y direction.
    Since the velocity can be expressed in terms of y, and the spatial differential, ds , can be expressed in terms of a function of the derivative of the path, dy(x)/dx = y_x, times the x differential, dx, the time differential, dt , can be expressed in the form given above, dt = dxf (y,y_x,x), specifically, write down this time functional, t, as an integral over the x coordinate, from 0 to x0. Also, write down the specific integrand, f (y,y_x,x).
    B) Note that the extremum of the time functional can be determined by the zero of the time variation, (attached), which is solved by the Euler equation, (attached), and for the case that the integrand, f , does not explicitly depend on x, where f = f(y, y_x), then the variational problem is solved by the alternative form of the euler equation, (attached), where K is a constant.
    Given this simplified approach to the solution of the variational problem for the minimum time of descent, you should be able to use the integrand, f (y, y_x), that you found in a), above, in this alternative Euler equation to rewrite the equation to be solved (attached), where C is a constant that is related to the K constant.
    C) This equation can be solved by the variable substitution (attached) where dy I dx can be expressed in terms of d(phi) | dx. You should be able to transform this differential equation, in y and x, into a differential equation in (phi) and x.
    Ultimately the differential equation can be solved by integration, where the integral over x, on one side of the equation can be performed, and the integral over (phi), on the other side of the equation can be performed. Given the boundary condition that at x = 0, then (phi) = 0, since y = 0, the specific solution for x (phi), can be found. The point is that you will have the solution of x and y, which is a cycloid, expressed in a parameterizatron with an angle variable, (phi).

    Write down the explicit final brachistochrone path, x(phi) and y(phi), where the path domain is (attached) so that the end point is (attached). Sketch the path.
    D) Calculate the time of flight along the brachistochrone path, (attached) expressed using the given parameters, g, C . The point is that you should be able to transform the integral of the time along x, into an integral along (phi), using the calculation that you went through in c), above, where the integral will be easy to solve.
    E) Also, calculate the time of flight along the straight line path, (attached). Note that this can be calculated with this straight line function, using the original time functional, (attached). Again, this integral is not too bad, and can be done, expressed in the given parameters, g,c.
    F) Finally, show that the ratio of the straight line path time to the minimum time for the brachistochrone, is (attached), which is fairly significant, since the straight line path is the shortest distance between the two points; however, the brachistochrone path turns out to be optimal, giving a 20% decrease in time. Describe physically why the brachistochrone path might exhibit a shorter slide time than the straight line path.

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    Solution Preview

    The derivation is in the attached pdf file.

    I chose a different constant of convenience because I think it makes things simpler:
    my b and the C of your texts are related as b = 1/2C^2
    Also, I took negative y below zero, unlike the reversed direction of y in your texts.
    These details do not affect the essence, and I think my choice is better than that suggested in the text.

    Here is the plain TEX source

    centerline{bf Brachistochrone }

    The motion starts from $(x,y)=(0,0)$ and the mass moves down to positive x.
    The potential energy transformed into kinetic energy is $-mgy$ and the kinetic energy is $mv^2/2$,
    therefore the speed is
    The time spent on travelling infinitesimal displacement is
    dt = {sqrt{dx^2+dy^2}over v} = {dxsqrt{1+y'^2}over sqrt{-2gy}},
    where $'$ means the derivative with respect to x ($y'=dy/dx$).
    From equation (2) we find
    f(y.y',x) = {1over sqrt{2g}} sqrt{1+y'^2over -y} ...

    Solution Summary

    The famous Brachistochrone Variation problem is worked out in an attached PDF.