Calculating the equations of motion for particle in parabola

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1. May 8, 2017

Heatherfield

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:

$$|v|^2 = 2gh_0 - 2gy$$
So that
$$v_x^2 + v_y^2 = 2gh_0 - 2gy$$

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:

$$\frac{dx}{dt} = \sqrt{\frac{2gh_0-2gx^2}{4x^2+1}}$$

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:

$$t = \int\sqrt{\frac{4x^2+1}{2gh_0-2gx^2}} dx$$

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. May 8, 2017

Staff: Mentor

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.

3. May 8, 2017

Ray Vickson

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 $x(\Delta t) < x_0$.