The Brachistochrone Problem: cycloid curve

In summary, The bead on the cycloid curve will either slide down or slide down and back up, depending on the ratio of x to y. If x/y is less than π/2, the bead will continue to slide down the curve. If it is greater than π/2, the bead will start to slide back up the curve. The point (x,y) represents the right-hand endpoint of the bead's path, while the origin is the left-hand endpoint. This can be observed kinematically, as the bead's position on the curve is determined by the ratio of x to y. The param
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physicsbeginnerss
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Homework Statement
Show, in Figure 4.4, that for a point like P3, x3/y3 > π/2 and for P2, x2/y2 = π/2.
Relevant Equations
$$\frac{y1}{x1}>\frac{P2A}{AO}=\frac{2a}{a\pi}=\frac{2}{\pi}$$
This is 'Boas mathematical Methods in the Physical Sciences' homework p484.(Calculus of Variations)

problem2 section4 number 2

calculus1.jpg

The bead is rolling on the cycloid curve.(Figure 4.4)
And the book explain that
'Then if the right-hand endpoint is (x, y) and the origin is
the left-hand endpoint, we can say that the bead just slides down, or slides down and
back up, depending on whether x/y is less than or greater than π/2 (Problem 2).'

My question is how can be it's just slides down or slides down and back up depending on
the magnitude of x/y? I mean it's depending on potential energy that it back up with origin height.(That contains origin potential energy)

The book explains when it comes to P2 the circle has rolled halfway around so OA=$$\frac{1}{2}\cdot 2a\pi=a\pi$$.
For any point P1 on arc
OP2, P1 is below the line OP2, and the coordinates (x1, y1) of P1 have
$$\frac{y_1}{x_1}>\frac{P2A}{AO}=\frac{2a}{a\pi}=\frac{2}{\pi}$$ or $$x_1/y_1 < π/2$$.

And p3 is applied by same solution that x3/y3 would be greater than $$frac{\pi}{2}$$.

Here is my understanding.

KakaoTalk_20200628_210902569.jpg

This is what papers says and I drewed it and understand how it works.

But if $$\frac{x_3}{y_3}$$ is less than $$\frac{\pi}{2}$$
diffcycloid.jpg

In this curve it's hard for the bead back up to p3, But!, Is this can satified with the parametric equation of cycloid?
## $$x=a(\theta-sin\theta)$$
## $$y=a(1-cos\theta)$$

My question is can the last picture be satisfied with the parametric equation of a cycloid or not? And if not, what is the alternative solution that $$\frac{x_3}{y_3}$$ less than $$\frac{\pi}{2}$$ that is still cycloid curve.
 

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Can you fix your Latex? Use double dollar signs.
 
  • #3
PeroK said:
Can you fix your Latex? Use double dollar signs.
I just did it.
 
  • #4
physicsbeginnerss said:
Then if the right-hand endpoint is (x, y) and the origin is
the left-hand endpoint, we can say that the bead just slides down, or slides down and
back up, depending on whether x/y is less than or greater than π/2
Perhaps it is intended as a purely kinematic observation. If the point of contact at some instant is at (x, y) (y measured positive down from the origin) then:
if x/y < π/2 the bead is still on the downward slope
if x/y > π/2 the bead has started back up
 
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Likes physicsbeginnerss

1. What is the Brachistochrone Problem?

The Brachistochrone Problem is a mathematical problem that asks for the shape of a curve that will allow a ball to travel from one point to another in the shortest amount of time, under the influence of gravity.

2. What is a cycloid curve?

A cycloid curve is a specific type of curve that is formed by tracing a point on the circumference of a circle as it rolls along a straight line. It is the solution to the Brachistochrone Problem.

3. Who first solved the Brachistochrone Problem?

The Brachistochrone Problem was first solved by the mathematician Johann Bernoulli in 1696. He challenged other mathematicians to solve the problem and offered a prize to the first person who could do so.

4. What is the significance of the Brachistochrone Problem?

The Brachistochrone Problem is significant because it is one of the first examples of the calculus of variations, which is a branch of mathematics that deals with finding the optimal solution to a problem. It also has applications in physics and engineering, such as in designing roller coasters and finding the optimal path for a spacecraft to travel.

5. Is the cycloid curve the only solution to the Brachistochrone Problem?

No, the cycloid curve is not the only solution to the Brachistochrone Problem. There are infinitely many curves that can solve the problem, but the cycloid is the curve that minimizes the travel time. Other solutions include the tautochrone curve, which minimizes the travel time for a pendulum, and the catenary curve, which minimizes the travel time for a cable under its own weight.

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