You need to ask prometheus, the derivation from base principles shows the opposite sign: \(x=R(\theta -cos (\theta))\) , in line with what Wolfram says.
Actually, I think the unresolved problem is why you get a correct result using this: Where the formula for x is not \( x = R\left(\alpha - \sin \alpha \right) \). Despite Tach's objections, things appear to work out. This does not augur well for other discussions with Tach about rolling wheels and the Doppler effect. Much less a formal debate (how's that going by the way, should I check again in another week or so?)
So, you worked out the integral resulting from the equation of energy conservation? I don't see you doing any calculations , so how did you divine the above? Has nothing to do with the debate on the Doppler effect. Or, did you do some calculations that no one knows about? Since you aren't contributing anything, all you can do is wait and see.
I'm more interested in how you found a problem with prometheus' calculations, which don't use equations of energy conservation, but probably do assume that gravity is conservative (at least, when its represented as a force):
Prometheus understands. The fact that you don't understand, is inconsequential. One thing that I see wrong with the solution that he wrote is that he arrives to: \(s=4R sin (\frac{\alpha}{2}) \) \(s\) needs to increase monotonically when \(\alpha\) varies from 0 to \(2 \pi\) whereas the above gives \(s=0\) at both ends of the interval. Not only that, it does not produce the correct total arc length. On the other hand, if you look at AN's solution for the first part of the problem, using \(x=R(\theta- cos(\theta))\) produces the right answer. What does this tell you? If you turned in your homework as done by AN and prometheus, an astute TA will notice the contradiction between the two halves of the homework since they contain different equations for the cycloid (of course they do, since they are done by two different people). Besides, AN has a slight mistake in his derivation as well. So, did you turn it in?
Yes, I agree that prometheus understands. Which is why he's demonstrated that most, if not all, of your "objections" have been made because you don't understand the problem. It's why you keep accusing me of the same thing. Which doesn't bother me in the least, coming from someone who appears to be as incapable as you. That you've managed to drop several clangers (do I really need to go back and drag them out again?) while accusing prometheus and myself of the same "crime" only speaks to your somewhat unconventional approach, shall we say. Or shall we say "fucked up attitude"?
Still haven't managed to explain why a steel ball rolling along a cycloidal track isn't a cycloidal pendulum, have we? Still can't explain why it's only a pendulum for small angles? I wouldn't call that a "slight mistake", more of a big clanger.
Do you want me to do all your homeworks while you sit and do absolutely nothing? I gave you the period in an earlier post, do you think you could derive it all by yourself? Can you insert the friction in the equations all by yourself? No? That's too bad.
I don't see any equation in this: So you haven't "gave me the period in an earlier post", at all. You have gave a lot of unsolicited crapola though. Your claim that a rolling ball will have a different equation of motion is hard to understand: the steel ball rolling along a cycloidal track moves in exactly the same path as a ball sliding along the same track ("frictionlessly"). The only difference if it's rolling, I assert, is that it will have a smaller velocity and so the period will be longer; the equations of motion however, will be the same equations in either case (how can they possibly describe two different paths, when the paths are physically constrained to be identical??).
Riiight, did you see "you need to derive"? as in all by yourself Please Register or Log in to view the hidden image!
Did you miss the part where I don't "need" to do what you say I do? Which is, I assume, the same rationale you're using for your own somewhat specious claims. Or alternatively, that you need to use because you know you can't support your claims. You won't post any equations (ever) that support your claim that you need a different set of e.o.m for a rolling ball and a sliding ball, because you don't know what they are, or where to start looking. What you'll do instead is keep deflecting from your own inadequacies by accusing others of having them.
Actually , the above applies perfectly to you. And no, you won't manage to get me to post the e.o.m for you. You need to learn how to do this all by yourself and stop waiting for others to do your homework for you.
Ah. I see what you're saying: since you have no idea how the equations are different, although you're quite "sure" they must be, you also believe that noone else including me knows either. These little chats are just so damn ...informative.
I know because I solved them, this is how I got the period formula. And no, you won't be able to bait me into showing how they are derived. I don't know about others but I definitely know that you are unable to derive them. This is why you troll the thread, in the hope that someone will show you.
To which I respond thusly: "Bullshit!". You've got nada, 'omz. I know because the period of a cycloidal pendulum is \(4\pi \sqrt {\frac {R} {g}} \), this will only change by a constant factor if the pendulum happens to be a rolling body. This has to be true because the kinetic energy of rotation doesn't "go into" another set of equations; the rolling body follows the same path (i.e. it obeys the same e.o.m. for its position as a freely swinging or sliding body). The only difference is that the body "converts" some of its potential into rotation, so it will only experience a longer period. There are no "different" equations, only a constant factor that affects the period's length. You can't explain this and perhaps will never be able to.
The math and the equation of the period proves that you don't know what you are talking about. Nice try, though. Not only longer, also "different", as in "angle dependent". Time to roll up your sleeves and start doing the calculations all by yourself. You had about 10 days already and , nada, zero, niente, zilch. So how about you stopped shouting and you started calculating? Bzzt, wrong. Try finding the period all by yourself for a change.
What "angle" is the motion of the pendulum "dependent" on? Note: the track is curved into a cycloid; "the angle" between the rolling or sliding ball and anything stationary will depend on the shape of the track. The shape of the track is described by the same equations in both cases. The equations will successfully predict the position of the rolling or sliding ball and will not be "different", except for a constant factor.
