The point is that I you were arguing against the Wolfram method all along, calling it a "kludge", something that does "not work for arbitrary \(\alpha\)", remember? What are you talking about now? I am talking about showing you how to calculate the period of the steel ball rolling on the cycloidal track. The Wolfram page is very consistent in the definition of the cycloidal track as \(x=a( \theta - sin (\theta))\) \(y=a(1- cos (\theta))\) Are you still arguing that the Wolfram entry is incorrect?
I acknowledge and have acknowledged that the method you took from Wolfram does give the correct period for any amplitude. You didn't do a very good job of explaining it in the first instance. Yes. Taking the cycloid to be \(x=a( \theta - \sin (\theta))\), \(y=a(1- \cos (\theta))\) does not give a confining potential. Taking the cycloid to be \(x=a( \theta + \sin (\theta))\), \(y=a(1- \cos (\theta))\) does, and the angle conventions make more sense for a periodic problem. At this point you should probably look it up.
That is false, you simply refused to listen because I was contradicting you and you proceeded to hurl abuse instead of trying to understand my point. But a "confining potential" is exactly what I have used in the equations of conservation of energy. Besides, the "confining potential" is a function of y, not x. We are agreeing on the definition of y, we are disagreeing on the definition of x. If one uses \(x=a( \theta + \sin (\theta))\), you cannot get the simplifications in the subsequent calculations of the integral defining the period. This is how I knew that you started with the wrong equation. Look, we can go about this all day long, the point is that if you use the Wolfram approach, you need to use \(x=a( \theta - \sin (\theta))\). Otherwise, the integral does not come out right.
Projection much? Energy is conserved regardless of whether the bob is confined to undergo SHM or not, so you can always make the statement that KE = PE. There are two things that you are overlooking: Firstly, this is not a simple definition of V(x) and V(y). These are parametric equations so changing x(t) will certainly change the behaviour in the y direction as well. Secondly, in addition to the change in x(t) I have changed the range of the parameter t. I honestly do not see how you can argue with this, since I provided plots of both versions of the cycloid in post #91. Seriously. Go back and look at them. I haven't checked, but I would be very surprised if the integral changed at all with either method, as long as you are careful enough to change the limits of the integral. Granted, if you perform the integral with the wrong limits it's certainly not going to work in either case.
Hey, who called the method a "kludge"? Who claimed that it was "intellectually dishonest" and that it is "wrong" and it would not work for "arbitrary \(\alpha\)? Sure they are, see how the parametric equation of the cycloid is derived in first place: 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 Good idea, why don't you do the calculations?
Considering you haven't actually responded to much of the substance in my posts, I would say that's intellectually dishonest of you. The factor of 4 was something you clearly didn't understand as you made up a (wrong) explanation for it, and you haven't noticed the error I slipped in about it. I have clearly shown in the post #91 that the equations you are using for the cycloid cannot possibly result in harmonic motion because a particle on that trajectory would simply fall to on of the minima. It's quite telling that you have computed the time taken for a particle to go from one side of the cycloid to the other and then multiplied the answer by 2 to get the "period." This is more reminiscent of a tunnelling problem in quantum mechanics than SHM. This has got to be about the cheekiest thing you've written in a while. Cast your eyes back to #59 where I actually showed the setup I was using undergoes SHM, as the equation of motion is \(\frac{d^2s}{dt^2} = - \omega^2 s\).
Yet, by your own recognition, this is exactly what the Wolfram entry is also doing. Besides, As to "falling of the minima", the functions are periodic, you only need to change the parametrization from \([0,2 \pi]\) to \([-\pi, \pi]\) and you can see y going from 2R to 0 and back to 2R. So, you are clearly wrong on this subject. Which is quite correct, since the period means that the particle needs to make a complete oscillation. So, you are wrong on this issue just the same. I am asking you to calculate the integral I used , not to repeat the equation of motion you formed. You claimed that the integral was going to come out the same using your expression for x, that was what we were discussing.
I'm sure Wolfram has been wrong in the past too. It's not a crime you know. The parametric equations defining the cycloid are periodic, yes, but that doesn't automatically lead to periodic motion of a bead on that trajectory. This is not right. Changing the limits of the parameter without changing the equations will shift you along by, in this case half a rotation. Instead of a cycloid you'll have two half cycloids stuck together at a cusp. Time taken to go from one side to the other is not half the period if there's no harmonic motion. Remember I have actually proved that there is harmonic motion. You have not.
Your edits are very annoying... I speculated that it would be the same. It's not exactly the same as a claim, and it's pointless since to can read the angular frequency from the equation of motion anyway.
True, this is why I suggested that you file an error report with them, see how they respond. I never made any connection between the periodic nature of the cycloid and the periodic nature of the motion. Yes, you are correct on this. Let's try a different way: both you and I (Wolfram) use the same equation for y: \(y=R(1-cos( \theta))\). So, for \(0< \theta < 2 \pi\). in BOTH cases, y varies from 0 to 2R and back to 0. So, your track doesn't "hold water" either because it is as inverted as mine. It is very easy to prove that the motion is harmonic, You only need to give up on the idea that what Wolfram is describing requires "repulsive gravity", to use your exact terms.
I will post it here when I get it. Would you care to clarify what you meant by "the functions are periodic, you only need to change the parametrization from \([0,2 \pi]\) to \([-\pi, \pi]\) and you can see y going from 2R to 0 and back to 2R." In wolfram's case, having \(0< \alpha < 2 \pi\) gives you a cycloid. In my case the same range gives you two halves of an inverted cycloid joined at a cusp. With my equations the range to use to get an inverted cycloid is \(-\pi< \alpha < \pi\). Please please please go back and look at post #91 where I plotted both of these out. My response to this is "do it then."
Precisely what I said, the equations describing x and y are periodic, nothing to imply that the motion in itself is periodic. This can only be established from writing the equations of motion, as I already explained in post 2. ...which is the cycloid I am using I saw your plots, there is nothing to imply that you will not get a periodic motion using the Wolfram formula for the cycloid. Your fixation with the fact that the cycloid is inverted can be fixed quite easily, by mirroring about the x axis, as in \(y=-R(1-cos (\theta))\). You can tell Wolfram that, if there is "any" error in what they did, it is easily fixable by changing the sign for y. Now , for \(0 < \theta < 2 \pi\) y goes from 0 to -2R back to 0. Happy now?
Well why have you not done this then? Not really. I'd be a lot happier if I knew exactly what that extra minus sign was doing. Simply flipping the sign of a parametric equation when you want to is not exactly a physical argument.
That minus sign will be squared away when you work out \(ds^2\). This is certainly a case of pythagoras hiding a multitude of sins.
For a simple reason: I explained to arfa brane that I wasn't going to do his homework. This website is not for sloving students homeworks. Nothing, it satisfies your insistence that the track "holds water".
Riiight! The same is not true about your \(x=R(\theta +cos (\theta))\). Have you tried the integral with your sign being the opposite to the one used by Wolfram? At the very first step you get \(ds^2=R^2 (d \theta)^2 (1+ cos (\theta))\) instead of \(ds^2=R^2 (d \theta)^2 (1- cos (\theta))\).
'Phew'. I guess another question about this is "why does changing the sign for the sine of t not change the value for the period?" And the answer appears to be: "because of the relation between the squares of the sides of a right triangle, since squaring a negative is the same as squaring a positive displacement". We note that in the case of a pendulum the y displacement is negative by convention (unless, as noted, gravity becomes repulsive). It's the same reason you expect a plumb bob to hang vertically with the bob at the lowest point, rather than at the highest point. (Here, we insert the relevant Homer Simpson reaction).
Now I have a question: how do you get ? What motivates this choice, given the physical axioms (gravity acts in the -y direction, etc)?
Seeing as I've already posted a solution it hardly seems like much of an objection. Yes, and as it is it gives a weird result which if I'm completely honest I don't understand, because it should work just fine.
And yet, here you are, doing my homework. Thanks, btw. I mean that, no really; even if you only believe you're doing "my homework" because it gives you something to bitch about.
