- #1
Heatherfield
- 22
- 0
I made the problem up myself, so there might very well not be a rational answer that I like!
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.
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
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?
Homework Statement
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.
Homework 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
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?