Physics related derivation of equation of cycloid

Question

So I was trying to find the path traversed by a point on the rim of a rolling disc. I eventually landed up at an equation but when I went to check it out in the internet, I couldn't find any similar ones. Most of the derivations required knowledge of higher mathematics that I don't have. I am not sure if this could be classified as a check my work type of question, but this is not a homework question, I did this purely by self interest, so I would really appreciate it if someone answered, as I am almost certain it is wrong, given the complexity of other derivations and this seems to be a way too simple approach.

We know that the $\omega$ vector makes an angle with the horizontal $\omega t$. So the net velocity of the point resolved in the horizontal direction is $$R\omega \cos(\omega t) + v$$ where v is the velocity of the center of mass and $t$ is time taken from when the point is at the topmost level. Similarly, the velocity resolved in the vertical direction is $$R\omega \sin(\omega t)$$

Now the coordinates can be represented as $$x=\int_{}^{}(R\omega \cos(\omega t) + v) dt$$ and $$y=\int_{}^{}(R\omega \sin(\omega t)) dt$$

Then I brought t to one side and equated the equations to get

$$y^2=R^2-x^2-v^2t^2+2xvt$$

We assume the disc isn't slipping so $v=R\omega$ and we substitute the value for t from the second integral as $ t=1/\omega[\arcsin(y/R)]$ into this equation to get

$$y^2=R^2-x^2-R^2[\arccos(y/R)]^2+2xR \arccos(y/r)$$

This seems to agree with the fact that it is independent of the angular speed of the disc but other than that I am not sure of how to verify it. It would be nice if I could get this in the form of $y$ as a function of $x$ so that I could check in a graph plotter.

Answer 1

Your derivation is correct, although your assumption about $v$ (it's constant) must be made before evaluating the relevant integral. Physically speaking, make the transformation to the moving frame:

$y' = y$
$x' = x - vt$,

and the implicit form becomes $y'^2 + x'^2 = R^2$. So, this is indeed a cycloid, because we see a circular path in the moving frame.

In your implicit form, you can't get $y = f(x)$ with standard functions, as you've discovered for yourself. You can, if you'd like, solve for $x = f(y)$ instead, and plot result with the domain restricted to $y \in [-R,R]$.