So it bothers me that I cannot come up with a function of x that when plotted would give me the cycloid. I can easily make the graph of the cycloid by graphing parametric equations, but from there I am completely at a loss. I really don't want to give in and just look on the internet for the answers, but I am also unsure of whether or not this is just plain beyond my abilities right now. I'll put up some of my musings and let's see where we can get from there.
First, let's define some constants: w = angular momentum of circle r = radius of circle
Now, to define height(y) as a function of time(t) we can write the following equation:
$ y=2r\sin^2\frac{wt}{2} $
I get this equation because I know the angle formed at the base of the circle by the x axis (tangent to the circle at all times) and the secant line between the dot and the intersection of x axis and circle is half the angle of rotation at time t. I can use this fact, the fact that I know the length of the secant line (by using the central angle), and the fact that the angle between the x axis and the height is 90 degrees to define the height(y) as a function of time, radius, and angular momentum. (Since we're keeping radius and angular momentum a constant, time is all that matters right now)
In the same fashion I found the x component of the height by subtracting the displacement of the intersection of the height and the x axis and the intersection of the circle and the x axis from the intersection of the x axis and the circle. That displacement can be found by using the same facts used to find the height, only taking the cosine rather than the sine. We then subtract that from the circle intersection (which is a function of time, angular momentum, and radius) and define the x component.
$ x=wrt-2r\sin\frac{wt}{2}\cos\frac{wt}{2} $
From here I did a lot of messing around with trig identities and algebra and even some integrals and derivatives, but all to no avail. For now I am unable to define y as a function of x to get a cycloid shape. Any suggestions? Jhunsber
Hmmm. Well, Josh, you could solve that top equation for $ t $ in terms of $ y $ and get
$ t=\frac{2}{w}\sin^{-1}\left(\sqrt{\frac{y}{2r}}\right). $
Then you could plug that into the second equation and get
$ x=2r\sin^{-1}\left(\sqrt{\frac{y}{2r}}\right)-\sqrt{2ry}\sqrt{1-(y/2r)}, $
but that is pretty ugly. I think you have found the reason that you see the cycloid described parametrically in all the books! (See page 694 of the book.) --Bell 13:55, 13 October 2008 (UTC)
- Thanks Professor Bell. Now I can sleep at night once again. His Awesomeness, Josh Hunsberger
