Note that the only difference between the cases of s being a flat, inclined plane and a cycloidal curve is the time it takes a rigid body to slide 'frictionlessly' along s. In the case of a plane surface with a slope, \( \Delta s\; =\; \sqrt {\Delta x^2\; +\; \Delta y^2} \), this applies at every instant of s. even if s is a curved path. The velocity of the particle is parallel to s at every instant, so that the Pythagorean relation applies, but instantaneously. The cycloid curve happens to have the "right slope" at each point such that a particle will descend in the same time regardless of the inital position (unlike being released from different points on an inclined plane), as long as the position isn't where the velocity is horizontal (because that's the lowest point of the cycloid).
Your starting point is incorrect. The derivation goes like this: In the frame of the circle generating the cycloid: \(x=R sin(\omega t)\) \(y=R cos (\omega t)\) So, the tangential speed is: \(\frac{dx}{dt}=R \omega cos(\omega t)\) \(\frac{dy}{dt}=-R \omega sin(\omega t)\) In the frame of the ground the speeds are: \(\frac{dx'}{dt}= V- \frac{dx}{dt}=R \omega - R \omega cos(\omega t)\) \(\frac{dy'}{dt}=\frac{dy}{dt}=-R \omega sin(\omega t)\) Integrating wrt time, we get the trajectory in the ground frame: \(x'=R \omega (t-sin(\omega t)+x'_0\) \(y'=-R cos (\omega t)+ y'_0\) But: \(0=x'(0)=x'_0\) \(0=y'(0)=-R+y'_0\) so: \(x'=R ( \omega t-sin(\omega t))\) \(y'=R(1- cos (\omega t))\) With the notation \(\omega t= \theta\): \(x'=R ( \theta-sin(\theta))\) \(y'=R(1- cos (\theta))\) as in here or here and NOT as in your starting point: \(x = R\left(\alpha + \sin \alpha \right), \quad y = R\left(1 - \cos \alpha \right)\) So, your equations for the cycloid are wrong. Now, for the second set of errors, your calculation of the period is also wrong, you knew the right result and you fiddled the proof in order to match the result. Given you starting point, you could not have gotten the right result. Indeed: Here is the correct derivation: \(\omega t= \theta\) \(dx'=R(1-cos (\theta)) d \theta\) \(dy'=-R sin (\theta) d \theta\) \(s^2=dx'^2+dy'^2=2R^2(1-cos(\theta))(d\theta)^2\) Conservation of energy gives: \(\frac{mv^2}{2}=mgy'\) so, \(\frac{ds}{dt}=v=\sqrt{2gy'}\) \(dt=\frac{ds}{\sqrt{2gy'}}=\sqrt{\frac{R}{g}} d \theta\) The period is : \(T=\int_0^{2 \pi} dt=\sqrt{\frac{R}{g}} \int_0^{2 \pi} d \theta=2 \pi \sqrt{\frac{R}{g}} \) Only when one considers that , for the isochronous pendulum, R is actually 4R one obtains: \(T=4 \pi \sqrt{\frac{R}{g}} \) So, there was no way that you would have gotten the answer above given your incorrect starting point. In other words, the correct starting point for obtaining \(T=4 \pi \sqrt{\frac{R}{g}} \) is (look at your own picture) : \(x'=4R ( \omega t-sin(\omega t))\) \(y'=4R(1- cos (\omega t))\) and NOT: \(x = R\left(\alpha + \sin \alpha \right), \quad y = R\left(1 - \cos \alpha \right)\) In other words you started with the wrong premise, used an incorrect derivation and you arrived at the expected result. An interesting case of error cancellation.
Tach, Alphanumeric has beaten you to it, though. And that was in post #2, before you posted anything. . Obviously, t is an angle in radians.
Sure, he used the correct definition of the cycloid, as opposed to prometheus who used the incorrect one and compounded the error with an incorrect derivation of the cycloidal pendulum period. He gave you the correct answer by using the incorrect premise combined with a flawed derivation. An interesting case of multiple error cancellation. That's not the error, the sign of \(sin(\alpha)\) IS. Obviously, you did not understand what the errors are. That goes for your second question as well: You might want to click on the links explaining it. See the thing labelled "L" in the drawing? What is its value, expressed in multiples of "R"?
Ah. L is the length of the string. Then L isn't constant, as prometheus has pointed out to you, but varies as it contacts the left or right cycloidal 'cheek'. Please Register or Log in to view the hidden image! So your equations don't describe the x' and y' coordinates after all. Well dang, you really trying to catch me out here, ain'cha? BTW, how are the two cycloidal cheeks generated for a real pendulum (I bet you don't use a set square)?
I know it isn't nice to keep pointing out someone's mistakes, but Tach is wrong about prometheus using an "incorrect" derivation. Because \(x = R\left(\alpha + \sin \alpha \right), \quad y = R\left(1 - \cos \alpha \right)\) is what you get when you reflect the cycloid curve through the x axis, so the angles are positive rather than negative (i.e. produced anticlockwise instead of clockwise). Which is what you want for a cycloidal track and a bead that moves because of gravity--you want the cycloid to be inverted. Or if you're still catching up: sin(-a) = -sin(a), and cos(-a) = cos(a).
So what are you saying? That the length of the string is constant and that the point about which the bob oscillates does not move? If that was the case the trajectory of the bob would be the arc of a circle. Look at the diagram: Please Register or Log in to view the hidden image! Quite clearly \(O \neq O'\). It is true that the distance from O to the bob is 4R, but it is not true that the distance from O' to the bob (ie L) is 4R. As arfa has already pointed out, the equations you have derived for the cycloid give this: Please Register or Log in to view the hidden image! Firstly, the cycloid has a maximum in the middle, so you clearly aren't going to get harmonic motion unless gravity starts being repulsive. Using my equations for the cycloid: Please Register or Log in to view the hidden image! So now the cycloid is the correct way up to get harmonic motion, but there's also another difference - with my equations the parameter \(\alpha\) goes from \(-\pi\) to \(\pi\) so the position of zero potential energy is at \(\alpha = 0\). Tach's equations have the equavalent of that point (although it is now a maximum, rather than a minimum) at \(\alpha = \pi\). My conventions make more sense for a harmonic motion problem. You are trolling. Up to this point I agree with you. This is wrong. For starters you are assuming that the pendulum / bead is oscillating with the maximum possible amplitude. Secondly, if you allow the general case of a bead reaching a maximum height of a on the one side and \(2 \pi - a\) on the other, the period is \(\int_a^{2 \pi-a} dt\) which when you evaluate it gives you a factor of \(\pi - a\) which is, if you think about it for long enough, the amplitude. You have shown that the period of the cycloidal pendulum depends on the amplitude which as you know, is wrong. This is nothing but a kludge - you know what the answer is and since you haven't gotten it you have to fudge it. Notice with my method there was no kludging of this kind. Your entire post is chock full of sloppy physics, flat out fudge factors and intellectual dishonesty, so you're hardly in a position to label what I posted "interesting error cancellation." If you don't believe me then check out this book which is where the diagram came from. They cover the problem using the same method as I used, which you can see online if you have an amazon account (or £120 to buy the book). Nowhere in your link does the page say that. More dishonesty.
Yep. Look here Nope, try telling this to the Wolfram Research people. You can try sending your stuff to them as an errata to this. I expected you to get all twisted in your knickers, you really can't stand being shown wrong. It has happened before, you know, the piece with the application of the l'Hospital rule, more recently the bit with the vectors representation....
Wrong, see here. You see, the reflection is about the x-axis, so, there is no reason to change the sign of the angle in the description for the x coordinate. You might want to change the sign for y but this will not change the error you and prometheus are now sharing relative to x. Besides, the Wolfram people would sharply disagree with you. If you still disagree, try filing an errata with them.
Actually, in order to find the period it doesn't matter. The whole positive or negative angle thing doesn't make any difference to the length of the period. However, you do want an inverted cycloid curve, and this will change the sign of the angle, but then again, you get the same thing if you reverse the direction of the generating circle. And Tach is still wrong about the length of the cycloidal pendulum: the string is only length 4R when it's vertical.
Did read the Wolfram page? They clearly say that you and prometheus are wrong. Try filing an errata with them, see how far you get.
So how do you explain the fact that the pendulum does not follow a circular trajectory without the string length changing or the point about which the bob oscillates moving? I see what they are doing. They have a cycloid (not an inverted cycloid - I don't quite see how you can disagree with this, since I provided a plot in my previous post) They are calculating the time it takes for the bob to get from a maximum to the minimum and multiplying the answer by 4 to get the period. I think this works because you are taking the derivatives of the x and y equations and squaring them, but in my opinion it's a bit sloppy. I will send them an email about it. You missed out a good deal of their analysis, the stuff about taking the bob to start from a point \(\theta > 0\) was completely missing from your analysis and the reason you multiply the answer by 4 is certainly nothing to do with the length of the string. My comments stand - your post was sloppy and dishonest. The fact you've not replied to the bulk of my posting speaks volumes. Your link and my reply to James R proves that I am willing to accept correction, which is exactly the opposite of what you're accusing me of. I have no memory of discussing l'Hopital's rule (not l'Hospital's rule) with you so maybe you're getting me confused with someone else, or maybe my memory sucks - I'm not going to rule that out. Make no mistake Tach, you're a gnat's pubic hair away from a ban for trolling. As I am involved with this thread I have made an effort not to take any moderating action against you but if there is one more hint of trolling in subsequent posts from you then a ban will be forthcoming.
That is OBVIOUS, the string FOLLOWS the TWO neighboring cycloids at ALL time by WRAPPING around them. See the pictures here. The string length does NOT change. The point about which the bob oscillates does NOT move. You clearly do not even understand what a cycloidal pendulum is.
