Max Velocity on a curved ramp and ideal ramp for highest Velocity

Question

Is it possible to find the max exiting velocity of an object roll down a curved ramp? Since the brachistochrone curve has the path of the fastest time, does this mean it also has the highest exiting velocity? If not which path from point A to B generates the fastest exiting velocity? All accounting for gravity!

Answer 1

Assuming there are no frictional or other non-conservative forces at play, the conservation of energy $E_{pot}+E_{kin}=constant$ will strictly limit velocity to satisfy $v=\sqrt{2gh}$, where $h$ is the amount of vertical drop (displacement in the direction of the force of gravity) and $g$ is the gravitational acceleration constant. This result is independent of path in $(x,y)$. If the object is circular or cylindrical and can roll, and if you add a rolling constraint, which is conservative, then the kinetic energy term will involve angular momentum, which will slow down the center of mass velocity appropriately, but you haven't mentioned that.

The brachistochrone problem is famous because it involves finding the pathway of minimum time between two points in two dimensions with a conservative field in one of those dimensions, and it leads us toward a variational solution. It is a great teaching tool for that. But there is no path to "maximize" velocity as energy conservation specifically limits velocity to a defined value specified solely by the function of the vertical displacement $h$ and $g$. Nothwithstanding, a "straight drop" will deliver the highest velocity soonest.

Answer 2

The quickest path from point A to point B always has the greatest velocity since velocity is inversely proportional to time.

However, I think this question is a bit ambiguous. Assuming air resistance is negligible and the path stays below point A the ball will always have the same velocity at point B.

Is it possible to find the max velocity of an object roll down a curved ramp?

I'm gonna stray more into physics here but:

Using the kinetic energy equation and setting it equal to the gravitational potential energy equation gives you

$$mgh = \frac{mv^2}{2}$$

The masses cancel out so the velocity is $\sqrt{2gh}$.