As I said, it's simply a definition. I can choose zero to be anywhere I like. Sometime. Believe it or not my life does not revolve around this forum. That's their prerogative. As I said before, I got my inspiration from a book that I linked to previously. Why don't you file an errata with the publisher and see how that goes? See what a stupid argument that is now?
I did not dispute that, I am just pointing out that you get the arc length to be a negative entity when \(-\pi <\alpha<0\) with your approach. How do you plan to solve this? Actually, this is not such a stupid idea. Did you file yours?
Quite obviously you say that the distance from \(-\pi\) to 0 is the same as minus the distance from 0 to \(-\pi\), because I have defined zero to be in the middle of the trajectory. It's not complicated at all. Yep. Go for it, and let us know how it goes.
Nope, the change (reversal) in traversing parameter space (\([-\pi,0]\) vs \([0,-\pi]\)) does not warrant changing (reversing) the sign for s since s is an absolute quantity. Arc length is a positive entity, by definition. Your solution to your error is to simply stick a minus sign in front of its expression doesn't cut it. I asked you if you filed your error report, why the deflection? Did you file?
What exactly is your problem? I showed in post 59 that \(ds = 2 R \cos \frac{\alpha}{2} d \alpha\) and hence \(s = 2 R \int_{-\pi}^{\pi} \cos \frac{\alpha}{2} d \alpha = 8R\) in exact agreement with AN's post. There is no issue with negative arc length.
We've been over this, for \(-\pi <\alpha<0\) your \(s\) is negative. Your attempt at reversing the parametrization doesn't fix the problem.
Why is this a problem? For your parameterisation s is going to be something like \(\cos \frac{\alpha}{2}\) which is negative for \(\pi < \alpha <2 \pi\).
\(x = R(\alpha- \sin \alpha), \quad y = R(1+\cos \alpha)=2 R \sin^2 \frac{\alpha}{2}\) Equation of motion: \(\frac{d^2 s}{dt^2} + \frac{g}{4R} s=0\) Solution subject to the boundary conditions: \(s(0) = s_0, \quad \frac{ds}{dt}(0) = 0\) is \(s = s_0 \cos(\omega t)\) where \(\omega^2 = \frac{g}{4R} \) hence the velocity is \(\frac{ds}{dt} = - s_0 \omega \sin (\omega t)\) and the kinetic energy is \(KE = \frac{1}{2} m \omega^2 s_0^2 \sin^2 (\omega t)\) The potential energy is \(PE = mgy\). Using \(y = 2 R \sin^2 \frac{\alpha}{2}\) and \(s = 4 R \sin\frac{\alpha}{2}\) we can show \(y = \frac{s^2}{8R}\). Hence \(PE = \frac{mgs^2}{8R} = \frac{mg s_0^2 \cos^2(\omega t)}{8R} \) Therefore the total energy having subbed for \(\omega^2\) is \(E = KE + PE = \frac{mgs_0^2}{8R}\left(\sin^2 (\omega t) + \cos^2 (\omega t) \right) = \frac{mgs_0^2}{8R} \) and \( \frac{\partial E}{\partial t} = 0\)
By my reckoning for \(x = R(\alpha - \sin \alpha), \quad y = R(1- \cos \alpha)\) we have \(ds = 2 R \sin \frac{\alpha}{2} d\alpha\) and hence \( s = -4 R \cos\frac{\alpha}{2}\) which is negative for \(\alpha\) between 0 and \(\pi\)
Err, you surely meant \(y = R(1+\cos \alpha)=2 R \cos^2 \frac{\alpha}{2}\), right? Kind of oK, so far (modulo the trigonometry flub). How can that be? At \(\alpha=-\pi\) your \(y=0\) and \(PE=0\) which is clearly not the case. At the bottom of the trajectory , i.e. at \(\alpha=0\) your \(PE=2mgR\) , which is clearly not the case.
Yikes! That was a rotten typo. What I actually meant was \(x = R (\alpha + \sin \alpha), \quad y = R(1-\cos \alpha)=2 R \sin^2 \frac{\alpha}{2}\) in line with my original solution. That also clears up the potential energy issue.
You mean \(ds^2=2R^2 (1-cos (\theta))d \theta^2 \) which means \(ds^2=4R^2 \sin^2 \frac{\theta}{2} d \theta^2 \Rightarrow ds=2R \sin \frac{\theta}{2} d \theta\) and s is as I wrote above.
\(ds^2=4R^2 sin^2(\theta/2) d \theta^2 => ds=2R d \theta |sin(\theta/2)|\) ..and this should also fix the problems with your derivations.
No... \(ds^2=4R^2 sin^2(\theta/2) d \theta^2 \Rightarrow ds= \pm 2R d \theta sin(\theta/2)\) which means there is always a region where s < 0. I just did a big lol at this. Right, I think we've established that s can be positive or negative as a function of \(\alpha\)...
