# Period of a ball rolling on a cycloid

1. Dec 17, 2011

### guillefix

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.

Last edited: Dec 17, 2011
2. Dec 17, 2011

### AlephZero

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

3. Dec 17, 2011

### guillefix

4. Dec 17, 2011

### AlephZero

That looks better.

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

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

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

I haven't gone through the rest of your working in detail but the general idea looks right.

5. Dec 18, 2011

### JHamm

You just need to show that the potential energy ($mgy$) is of the form $\frac{1}{2}ks^2$ where $s$ is the arc length and k, m and g are constants.

Last edited: Dec 18, 2011
6. Dec 19, 2011

### guillefix

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

Last edited: Dec 19, 2011