The two side cycloids are acting as a restrainer. The string needs to wrap aginst one or the other at ALL times. This is why the bob describes a cycloid and not a circular arc. You have been discussing this issue, accused me of trolling, threatened to ban me while all along you didn't even know what a cycloidal pendulum is.
So you might say the point about which the bob oscillates does move and the effective string length does change?
Ok, I think I see the problem. The pendulum doesn't follow a circular trajectory because the length of free string changes. That's one. The length of the string is always 4R, but because the string 'bends' along the inverted cycloidal surfaces, the free length is 4R only when the string is (completely) vertical. But that means the point of contact for the free length of string does change, because it lies along either of the surfaces. That's two. Do we have a starter for the third?
Now you are demonstrating that yo do not to even understand the notion of "length". At least, you are starting to come to grips with the fact that L=4R.
It's getting harder for me to see how you can possibly retract this: Or are you still convinced your equations describe the x' and y' coordinates?
Yes, you are grasping at straws. Give it up, you not only are too lazy to do your own homework, you don't even appreciate when others are doing it for you correctly. Do your TA's know that you have been cheating?
Right, that's enough. Essentially we are all singing from the same hymn sheet so the fact that this thread is now nothing more than a pissing contest is all the more ridiculous. Tach has been given a 3 day ban for trolling.
You realise your equations only describe a cycloid which has a generating circle with radius 4R? Which can't possibly be "the correct starting point" to derive the period of a cycloidal pendulum whose generating circle has a radius of R. Please don't imply that you are capable of lecturing me or anyone else, you make too many mistakes to pull that one.
I see Tach lurking, so for the sake of the facts I recommend he checks out this thread where he got his ass handed to him by AN and Guest over l'Hopital's rule. My memory is fine apparently.
Observation: The ratio between the length of s, when s is a cycloid, and the displacement of the generating circle along x is \( \frac {\pi} {4} \). This is approximately equal to \( \sqrt { \frac {\sqrt 5 + 1} {2}} \). I get a difference of 0.003105. Hope that doesn't introduce any woo to the thread. Just saying, you know? It means that \( dy\; =\; \frac {\pi} {4} dx \), for all x,y, and this is approximately equal to \( dy\; =\; \sqrt {\phi} dx \).
Oops, insufficient carbohydrate intake: I should have written: \( ds\; =\; \frac {\pi} {4} dx \), and \( ds\; \approx\; \sqrt \phi dx \) The ratio between the total y displacement and 1/2 the x displacement is \( \frac {2R} {\pi R}\; =\; \frac {2} {\pi} \).
Damn, scrub those last two posts. The ratios only apply when \( x\; =\; \pi R,\; y\; =\; 2R,\; s\; =\; 4R \) At those values, s is twice the length of y, and the ratio between s and x approximates \( \sqrt \phi \). The ratios vary continuously until the 'halfway' point (which is where a cycloidal pendulum's string is completely straight, with 'unconstrained' length 4R).
Well, this thread has been off track before now. But continuing with the notion of approximation: there are three places that R, the radius between the centre and the point P(x,y) on the perimeter, is perpendicular to the x axis. The second time this happens is the same time the string in the "freely swinging" pendulum is actually free of either cycloid. Then there are two places that R is parallel to x, the first is when y (resp. -y) = R. And the centre of s is where any pendulum will have the greatest velocity, also parallel to x. Why the ratio between the length of s and the 'unrolled' circumference is approximately equal to the golden ratio is anyone's guess, but there it is. It's good enough that you could probably claim that you could build a real one which was a good approximation.
This is demonstrably false, the solution I gave for the case \(\alpha=0\) generalizes to the case \(\alpha > 0\). Here is the solution for the cycloidal pendulum starting at an arbitrary position \(\alpha\). Start with the correct equations of the cycloid: \(x=a( \theta- sin ( \theta) )\) \(y=a(1- cos (\theta))\) Then, the application of the conservation of energy can be written trivially as: \(\frac{mv^2}{2}=mg a[(1-cos(\theta))- (1-cos(\alpha))]\) Therefore: \(v=\sqrt{2ag} \sqrt{ cos(\alpha)- cos(\theta)}\) So: \(dt=\sqrt{ \frac{a}{g}} {\sqrt{\frac{1-cos (\theta)}{ cos(\alpha)- cos(\theta)}} d \theta\) \(T=2 \sqrt{ \frac{a}{g}} \int_\alpha^{2 \pi-\alpha} \sqrt{\frac{1-cos (\theta)}{ cos(\alpha)- cos(\theta)}} d \theta= 4 \sqrt{ \frac{a}{g}} \int_{-1}^1 {\frac{du}{\sqrt{1-u^2}} du=4 \pi \sqrt{ \frac{a}{g}} \) where \(u=\frac{cos(\theta/2)}{cos(\alpha/2)}\) FALSE. See above.
Well lifted from the Wofram site. Your original justification for the extra factor of 4 was (from post #83): The factor of 4 is correct, but the reason is wrong - The integral you give is the time taken for the bob to travel from it's maximum to it's minimum, in other words a quarter of it's period. The factor of 4 has nothing whatsoever to do with the overall length of the string. Also, the OP (remember that?) asked for the equation of motion of the cycloidal pendulum (the bead on a cycloidal track actually, but the EOM would be the same in either case). Your method doesn't provide that. Mine did.
The same exact one that I pointed out to you, twice. So, all along you were contradicting the Wolfram Research solution as I pointed out to you by giving you the links. Actually, I did more than that, I gave you the period of the rolling steel ball.
The method I used and Wolfram's method are not contradictory, except regarding the orientation of the cycloid, which I fail to see how anyone can argue about. Again, unless gravity starts being repulsive there will be no harmonic motion. I refer you to post #91
