Period of a ball rolling on a cycloid

In summary, my friend solved this problem already and he explained it to me and gave me some initial hints so I could see if I could solve it. You are trying to show that the period of the cycloid is independent on the point where you leave off the ball, and specifically that it is equal to: T=4pi√(r/g)
  • #1
guillefix
77
0
I have a friend that has solved this problem already and he explained it to me and gave me some initial hints so I could see if I could solve it. You are trying to show that the period of the cycloid is independent on the point where you leave off the ball, and specifically that it is equal to:
T=4pi√(r/g)

I have expanded the integral ∫(dt/dy)*dy to get total T for a quarter of the period so y's bounds are zero and h_0, the maximum height. This is true because the cycloid's minimum goes through the origin and its a refelction over the x-axis of how it is usually represented so it has a bowl shape and we can think of a ball rolling in it. The equations (parametric) for it are:

x=(θ+sinθ)
y=(1-cosθ)

I have also used velocity in terms of the height (y), using the definition of potential energy, so it is:
v=√(2g(h_0-y))
Then I got the vertical component of the velocity by multiplying it by the sin of the angle that the tangent of the cycloid with the horizontal, i.e. by:
1/(√((dx/dy)^2+1))
Then after multiplying it by v, i get the inverse and get dt/dy. Then I put the integral in terms of θ by multiplying by dy/dθ. Afterwards, I have simplified it and got:

T=4√(1/g) * ∫ √((1+cosθ)/(cos(θ_0)-cosθ)) dθ from θ=θ_0 to θ=0 where h_0=1-cos(θ_0)
Note that I have ignored the radius r, but this can be easily added later and just acts as a scalar.

It is this integral that I can't work out how to solve, and I am nearly sure it is right because in page 11 in this document http://tqft.net/papers/cycloid.pdf you can find a very similar integral. However, they don't tell how to solve it, they just show the result. Curiously enough, I don't even get their same result when prompting their same integral into wolframalpha and maple! I am stuck here! please help! My friend is not answering in facebook and anyways i know he did it slaigthly differently, and I am not sure how he did it. Help is appreciated, thanks in advance.
 
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  • #2
guillefix said:
The equations (parametric) for it are:

x=(θ+sinθ)
y=(1-sinθ)

Something is wrong there. Or more likely some things, plural. Also, make sure your cycloid is the correct way up!

See http://mathworld.wolfram.com/Cycloid.html
 
  • #4
guillefix said:
True, I meant y=1-cosθ. But the orientation is correct I think:

That looks better.

To do the integral in the PDF link "by hand", start by substituting [itex] u = \tan (\theta/2)[/itex]

That gives [itex]du = (1 + u^2) d\theta[/itex] and [itex]\cos\theta = (1-u^2)/(1+u^2)[/itex].

That will get rid of the trig functions. If you still can't "see" the answer, substitute [itex]v = u^2[/itex].

I haven't gone through the rest of your working in detail but the general idea looks right.
 
  • #5
You just need to show that the potential energy ([itex]mgy[/itex]) is of the form [itex]\frac{1}{2}ks^2[/itex] where [itex]s[/itex] is the arc length and k, m and g are constants.
 
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  • #6
I finally solved it! and in a much nicer way that they did in the pdf i think (because dealing with infinites is not very nice if avoidable, IMO).

First of all my apologies because my integral wasn't exactly right, it is: ∫√((1+cosθ)/(cos(θ-cosθ_0))) dθ, where θ_0 is a constant and the upper bound is θ_0 and lower bound 0.

Well what I did is first substitute u=cosθ and get:

∫du/√((u-u_0)(1-u)) upper b.=1 lower b.=u_0

Then I substitute v^2=(1-u) and get:

2∫dv/√((v_o)^2-v^2) u.b=V_0, l.b=0

Finally, I substitute v=(v_0)cosθ, and exuberantly rejoice when
2∫ dθ from 0 to π/2 :!)

Which is obviously just pi! I can sleep well tonight certain that the cycloid is isochronic :approve:
 
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Related to Period of a ball rolling on a cycloid

What is a cycloid?

A cycloid is a curve traced by a point on the rim of a circle as it rolls along a straight line. It is a common shape in mathematics and physics, and can be observed in various natural phenomena such as ocean waves and planetary orbits.

How is the period of a ball rolling on a cycloid calculated?

The period of a ball rolling on a cycloid can be calculated using the equation T = 2π√(L/g), where T is the period, L is the length of the cycloid, and g is the acceleration due to gravity. This equation is derived from the principle of conservation of energy.

What factors affect the period of a ball rolling on a cycloid?

The period of a ball rolling on a cycloid is affected by several factors, including the length of the cycloid, the mass of the ball, and the acceleration due to gravity. Additionally, the shape and material of the ball, as well as any external forces acting on it, can also impact the period.

How does the period of a ball rolling on a cycloid compare to other types of curves?

The period of a ball rolling on a cycloid is shorter than that of a ball rolling on a straight line, but longer than that of a ball rolling on a parabola. This is because the cycloid is a brachistochrone curve, meaning it minimizes the time it takes for a ball to travel from one point to another under the influence of gravity.

What real-world applications are there for understanding the period of a ball rolling on a cycloid?

Understanding the period of a ball rolling on a cycloid can have practical applications in fields such as engineering and physics. For example, it can be used to design efficient roller coasters and other amusement park rides, as well as to study the motion of objects in circular or cyclical paths. It can also be applied in the development of clocks and pendulum mechanisms.

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