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Calculating the equations of motion for particle in parabola

  1. May 8, 2017 #1
    I made the problem up myself, so there might very well not be a rational answer that I like!
    1. The problem statement, all variables and given/known data
    A point-particle is released at height h0 is released into a parabola. The position of the particle is given by (x, y) and the acceleration due to gravity is g. All forms of friction etc are ignored.

    2. Relevant equations
    I wanted to see if I could solve this using just the energy equations. So these became:

    Ek = ½mv2
    Eg = mgy
    E = Ek + Eg

    Because the particle rolls down a parabola we also have

    y = x2

    3. The attempt at a solution
    First, I figured out that the total energy must always be mghh. The masses now cancel out. Subsequently, the equation E = Ek + Eg gives you a pretty neat equation for the speed:

    [tex] |v|^2 = 2gh_0 - 2gy [/tex]
    So that
    [tex] v_x^2 + v_y^2 = 2gh_0 - 2gy [/tex]

    Using the chain rule allows you to change the dy/dt term into a dx/dt term. Dividing and taking the square root of both sides gives you:

    [tex] \frac{dx}{dt} = \sqrt{\frac{2gh_0-2gx^2}{4x^2+1}} [/tex]

    I put the formula into desmos and the velocity graph looked pretty reasonable! So I figured I'd try and obtain the equation of motion for the x-axis, but the integral is monstrous:

    [tex]t = \int\sqrt{\frac{4x^2+1}{2gh_0-2gx^2}} dx[/tex]

    I even took about thirty minutes to expand the fraction into something workable. It gives you a square root with four fractions, all four having a linear function in the denominator and a constant in the numerator and an additional constant. I put the functions into a desmos graph at https://www.desmos.com/calculator/6rgsilfnaz. This does not appear very workable. Wolfram Alpha also doesn't give a pretty and regular function as an answer.

    Yet the problem is pretty simple. It's the equations of motion, using simple variables for something rolling down rather simple shape. Did I take a wrong turn somewhere? Am I supposed to end up with an implicit solution? Or is this really a problem that can only be solved numerically?
     
  2. jcsd
  3. May 8, 2017 #2
    Why don't you start out by at least looking at the solution in the limit of short times. Try substituting ##x=-\sqrt{h-\lambda}##, where ##\lambda## is small.
     
  4. May 8, 2017 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Already you are in trouble: from
    $$\left( \frac{dx}{dt}\right)^2 = \frac{2gh_0-2gx^2}{4x^2+1}$$
    you cannot conclude that
    $$\frac{dx}{dt} = \sqrt{\frac{2gh_0-2gx^2}{4x^2+1}}$$
    You need to input some physics in order to select the correct square root.

    Suppose at ##t=0## the particle is to the right of the origin, at ##(x_0,x_0^2)## (where my ##x_0^2## is your ##h_0##). Starting from rest, the particle should slide down the parabola to the left (towards the origin), and so the velocity should become ##< 0##. That is, the correct DE (for small times, at least) is
    $$\frac{dx}{dt} = - \sqrt{\frac{2gh_0-2gx^2}{4x^2+1}}$$
    Perhaps this sign error is the source of your difficulties in obtaining a sensible solution.

    Anyway, that DE persists up until the point in time ##t_1 > 0## where the particle attains the point ##(-x_0,x_0^2)## on the left leg of the parabola. Then the motion reverses, so for some time after ##t_1## your original DE is OK because the velocity should be positive. Then, at the next time ##t_2 > t_1## where point ##(x_0,x_0^2)## is attained, the sign is reversed again, etc.

    Each DE is solvable as an implicit form involving Elliptic functions and other square-root functions, so is an analytical mess. However, numerical solutions should be easy to obtain using modern DE solving packages.

    There is one final difficulty, however: from the DEs above, when ##x = x_0## the solution is ##x(t) = x_0## for all ##t## (that is, ##dx/dt## starts at 0 and remains there). To get started you need to look at a very short initial time interval and use the known force to get ##x(\Delta t)< x_0## (for very small ##\Delta t > 0##), then start the numerical solution from ##t = \Delta t## with known initial condition ##x(\Delta t) < x_0##.
     
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