Curve for fastest time down a ramp

Question

I came across a physics experiment video showing three balls released from a point A, going down three different kinds of ramps leading to a point B (https://www.youtube.com/watch?v=61S0KW7e-rc)

Assuming no friction and a point mass for simplicity, is there an ideal curve to get it from point A to B in the shortest time (e.g. is it parabolic)? Assuming this significant simplification, can it be proven that for any curve between A and B, the point mass should pass B at the same velocity? I.e. a different curve will only affect the time it takes to get to B, but not the velocity?

Answer 1

Yes, there is a particular curve, called the brachistochrone, which minimizes the time of descent.

https://en.m.wikipedia.org/wiki/Brachistochrone_curve

It happens to be a portion of a cycloid. The mathematics you need to derive this is called the “calculus of variations”. It is not part of most calculus courses.

The fact that the speed (not the velocity) at B is independent of the curve is a consequence of energy conservation. In a uniform gravitational field where the gravitational acceleration is $g$, the potential energy of a mass $m$ at a height $h$ is $mgh$. When you drop by some height, regardless of what path you use to do so, this potential energy gets converted into kinetic energy, which is $\frac{1}{2}mv^2$. This increases the square of the speed by an amount which depends on $h$ but not on the details of the path. For example, if you start from rest at A and B is at a height $h$ lower than A, then the speed at B will be $\sqrt{2gh}$.

Answer 2

Cycloidal is the shape of ramp that minimises the time of descent ... at least for a particle sliding frictionlessly down it - I think for a ball rolling, the optimum curve might well depart from cycloidality (I've often wondered that, but haven't searched into it. I would be glad to get comments about it, actually.). At the very top of the cycloidal ramp, the slope is actually vertical (it's a cusp), so all 'priority', if you like, is given to imparting sheer speed. It then turns towards the horizontal in precisely such a way as best to balance the relative 'priority' of imparting speed & turning the motion horizontal. It's notable that at its ends the cycloid is tangential both to the horizontal axis and to the vertical; so at the bottom all 'priority', now that it has attained its full speed, is now given to directing its motion horizontal. The final speed is fixed by the height of the fall, so the shape of the ramp makes no difference to it.

Actually, I've just realised that I have an issue with this brachistochrone problem. What exactly is the formulation of the problem? Is it that the vertical distance of the fall and the horizontal distance the particle is to move through are both specified? But for a cycloid the horizontal displacement is fixed at π/2 × vertical displacement. So if they are specified independently, is the solution still a cycloid but with relative horizontal & vertical scales adjusted so as to fit?

Because obviously the shortest time for it merely to reach the ground is when the particle simply falls vertically - that way the vertical component of acceleration is always the absolute maximum, without any cosine to diminish it. So if the requirement were merely that the final motion be horizontal, then the optimum ramp would be the limit as the shape tends to vertical drop with infinitesimal bend at the bottom. This is indeed what you would get with a cycloid in the limit of the factor scaling its horizontal extent decreasing without bound. But in that limit, the horizontal distance it has moved through when it reaches the bottom is zero.

So the formulation of the problem must be arbitrary horizontal displacement relative to the vertical.

I've just found a treatment of the brachistochrone for a rolling body (American Journal of Physics 14, 249 (1946); https://doi.org/10.1119/1.1990827): the curve is such that the centre of mass moves on a cycloid!

This means then that for a rolling object of radius ρ, on a standard cycloid of height 2 & horizontal displacement π, the parametric equations for x & y would be (using θ - the angle through which its generating circle has turned - as the parameter)

x = θ - sinθ - ρ.cos(θ/2)

y = 1 + cosθ - ρ.sin(θ/2),

(availing ourselves of the rather convenient fact that for a cycloid, the normal makes an angle θ/2 to the horizontal). If the cycloid is scaled horizontally such that it extends, say, απ instead of π, then the expression is a bit more complicated, since we obviously can't just scale the shape of the rolling body with the cycloid, and the angle that the cycloid's normal now makes to the horizontal is atan(α.tan(θ/2)), so we have to collapse sin(atan()) & cos(atan()) instead of atan(tan()); so it would be, instead

x = θ - sinθ - ρ/√(1+α².tan²(θ/2))

y = 1 + cosθ - ρα.tan(θ/2)/√(1+α².tan²(θ/2)) ...

... a perturbed cycloid, if you will.