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...How is this relevant to the topic here? You posted contradictory stuff. It takes 38 posts to figure out what you actually did, it was corrected, apparently to your dismay. So instead of learning, you go off and propose a new thing, this 'my calculation' which is strangely (but not unexpectedly) unspecified. Let me know how it works out.
...QuoteSo a simplification, disc spinning in an intergalactic space as per the op but there is no rod, no centripetal acceleration.What are the A points accelerations when we consider two sets of input parameters:1. Ω=(0,0,1), ω=(0,2,0)2. Ω=(0,0,0), ω=(0,2,1)Spins cannot have two values, so this makes no sense....
So a simplification, disc spinning in an intergalactic space as per the op but there is no rod, no centripetal acceleration.What are the A points accelerations when we consider two sets of input parameters:1. Ω=(0,0,1), ω=(0,2,0)2. Ω=(0,0,0), ω=(0,2,1)
The point of contention is the relationship between orbit and spin angular velocities.
In both cases the flywheel has the same rotation around K,k.
Are the A points accelerations going to be the same in both instances?
Accelerometers in ISS show near zero values, although it's accelerated toward the earth center significantly.
At time t=0 we cut the B from the rod and we are left with the spinning wheel.Is ω=(0,2,1) the correct spinning wheel initial angular velocity after the cut?
Quote from: Jaaanosik on 17/04/2023 22:03:55At time t=0 we cut the B from the rod and we are left with the spinning wheel.Is ω=(0,2,1) the correct spinning wheel initial angular velocity after the cut?It can't speed up (to 2.236) like that, but it did very much have a k component to its motion due to a fairly high precession rate (significant torque being applied).
That would result in unstable motion, a disk spinning on a different axis than the one perpendicular to its face. It would wobble and not have a defined spin rate, but it would have a defined (constant) angular momentum axis perhaps something in the neighborhood of 10° up from the y axis or something. That's just a guess. I'm not sure if it would be accurate to specify an angular velocity to your wobbling disk since it continuously changes and the acceleration of a given point on it does not point perpendicular to the axis. None of the usual trivial formulas would apply, and only some fairly involved calculus would be able to compute the acceleration of a given A point.
Consider the solar system. It has definite constant angular momentum on an axis more or less corresponding to the orbital plane of the planets. The majority of that angular momentum is accounted for by Jupiter (more than everything else combined). But the solar system has no angular velocity. There's no rate at which the solar system revolves due to it not being a rigid object. But neither do even unstable rigid objects like the wingnut I mentioned. A disk spinning on a skewed axis would exhibit predictable but not regular angular motion.
...What you describe is a pseudo-rotation radius based on the momentary frame-dependent curvature of the trajectory of a point on a cycloid. That's a valid (albeit pretty complicated) way to go about it, but assign it a new symbol. Don't overload Ω like you've done since Ω is fixed at 1 (and the instruments at B can measure it directly). So call it δ or something. Then the equations will work out and you get the correct answer for any frame....
over 90% percent of what I said is textbook stuff.
Are you going to help the truth to come out?
Quote from: Jaaanosik on 26/04/2023 12:33:30over 90% percent of what I said is textbook stuff.I disagree with that since I see no mention of your funny trochoid analysis in the one page of book you show.If you're going to use trochoid mathematics for the acceleration of point B, you also need to use it for the A points, but you're not doing that. None of that is in the tiny bit of textbook you mention although I imagine it's covered somewhere.Anyway, I stopped contributing since all the answers are already above and you purpose seems to be to demonstrate physics to be wrong and not to figure out where the error is in your calculations. QuoteAre you going to help the truth to come out?I did, and done correctly there are no contradictions. The error has been pointed out. Stop defending something the textbook doesn't say and understand the corrections.
Hi. I've had several attempts at writing a reply BUT I wasn't sure it would make any difference. What sort of reply or response would you like?Quote from: Jaaanosik on 26/04/2023 12:33:30Are you going to help the truth to come out? I do not have Sean Carroll (a physicist) in my friends list or anything like that. There are limits on what I could do.Let's not consider your idea for a moment, just consider an arbitrary idea presented by some person in the world. Is it efficient to get the world's leading physicists to apply their time to look at all these people's ideas? Isn't it more efficient for them to concentrate on peer-reviewed research papers in the journals? Now, sometimes a good idea that appeared on a forum is going to be missed. That's a shame, don't get me wrong, that is a real shame - but if all the physicists were reading forums instead of journals then hardly any new research gets done by them and that seems to be a much greater loss. This forum is quite small, there aren't many users, who do you think is here and what do you think they can or should do?Best Wishes.
What do you think about the presented calculations and the conclusion there is a preferred frame?
Quote from: Jaaanosik on 27/04/2023 16:38:45What do you think about the presented calculations and the conclusion there is a preferred frame?I think you need to keep an open mind and listen to the responses you got to your question. I'm sure it is very exciting to think you have discovered something that has been missed by millions of physicists. Isn't it much more likely that you have just made an error in your analysis?
Quote from: Jaaanosik on 26/04/2023 12:33:30over 90% percent of what I said is textbook stuff.I disagree with that since I see no mention of your funny trochoid analysis in the one page of book you show.If you're going to use trochoid mathematics for the acceleration of point B, you also need to use it for the A points, but you're not doing that. None of that is in the tiny bit of textbook you mention although I imagine it's covered somewhere....
The textbooks are clear, there are only Ω and ω.
The generic motion equations are used for an instantaneous acceleration calculations and they are supposed to work for any inertial observer if the relativity is true.
Generic curved trajectory has two accelerations, normal and tangential.
The point B is at the top of the cycloid in K'1 frame and there is only pure normal/centripetal acceleration, there is no tangential acceleration.
The pursuit of the truth is the most noble inclination of the human nature.
Interesting though, Halc stopped posting when I asked him if Ω=1 makes the rest frame the preferred frame, isn't it?
Why are K'1, K'2 curvature radius' dismissed in their respective frames?
Quote from: Halc on 26/04/2023 14:08:57If you're going to use trochoid mathematics for the acceleration of point B, you also need to use it for the A points, but you're not doing that.I did in my post #32.
If you're going to use trochoid mathematics for the acceleration of point B, you also need to use it for the A points, but you're not doing that.
Quote from: Jaaanosik on 21/04/2023 17:23:28The textbooks are clear, there are only Ω and ω.The textbook (or at least the one page you show) discusses a formulation using the proper frame of the motion, the one frame where everything is back in the same place after a rotation. It does not discuss the trochoid method at all.The trochoid method works, but it doesn't have 4 lines (components) like you're showing. It needs to be done separately based on the path taken by the point in question (in this case, each of the A points, each of which has a different curvature. You've not computed that, but are instead attempting to mix the trochoid method for a different point with what the textbook does.So find a textbook that discusses the trochoid method. You have a point taking a sort of helical path through space, and at any given point it indeed has a curvature and speed from which a momentary radius can be determined.You haven't done that for any of the A points, and you did it for B only in the one frame with only X motion. Do it (compute the acceleration of B) in the frame were V=(0,1,0) instead to illustrate what is involved. Having done that, try to generalize to the curve of say A4.This has nothing to do with relativity theory. The book discusses Newtonian mechanics....
Only because you've chosen a frame (1,0,0) where there's no tangential acceleration. Again, try (0,1,0) where there very much is tangential acceleration. Does your method work (for point B) in that frame? Do you even have a method? What is the curvature and speed in that frame?
Halc, thanks for your post!
Let us settle the relativity.
So selection of moving point B is arbitrary and acceleration at A is acceleration of B and relative acceleration between A/B.
QuoteOnly because you've chosen a frame (-1,0,0) where there's no tangential acceleration. Again, try (0,1,0) where there very much is tangential acceleration. Does your method work (for point B) in that frame? Do you even have a method? What is the curvature and speed in that frame?So in K'1, K'2 the point B follows cycloid motion trajectory and the acceleration analysis reflects curvature radius of those respective frames.K'1 B acceleration is pure normal with r=4m, K'2 B acceleration is pure tangential.My calculations account for that.
Only because you've chosen a frame (-1,0,0) where there's no tangential acceleration. Again, try (0,1,0) where there very much is tangential acceleration. Does your method work (for point B) in that frame? Do you even have a method? What is the curvature and speed in that frame?
Quote from: Jaaanosik on 01/05/2023 00:05:41Halc, thanks for your post!There's an option to do it in actions, top of the list.QuoteLet us settle the relativity.By relativity, I mean relativity theory, which has totally different equations and properties than does Newtonian physics. So for instance, if I put an accelerometer at the end of a 2770 km string and spin it at a proper rate of 1000 RPM, under Newtonian physics it will read about 2.9e8 m/s² but relativity theory says it will read about 1.14e9 m/s². Under Newtonian physics, the spin is 1000 RPM relative to any inertial frame, but the angular speed is frame dependent in relativity theory. Those differences are what I'm talking about, not the fact that speed is relative to something, which is true for both theories.Your examples have not in any way touched on relativity theory or the equations involved with it, so any assertions about it being wrong are baseless since it has never been discussed in this topic.Your book page at the bottom correctly says "it is important to realize that the acceleration of a particle as observed in a translating system x-y is the same as that observed in a fixed system X-Y if the moving system has a constant velocity".You contradict this with your calculations showing non-identical accelerations as observed in different moving systems.
QuoteSo selection of moving point B is arbitrary and acceleration at A is acceleration of B and relative acceleration between A/B.Yes, but you are not correctly computing the latter since they are not based on the motion of A relative to B.As I said in my prior post, I'm mostly repeating myself, as are you. This is unproductive.QuoteQuoteOnly because you've chosen a frame (-1,0,0) where there's no tangential acceleration. Again, try (0,1,0) where there very much is tangential acceleration. Does your method work (for point B) in that frame? Do you even have a method? What is the curvature and speed in that frame?So in K'1, K'2 the point B follows cycloid motion trajectory and the acceleration analysis reflects curvature radius of those respective frames.K'1 B acceleration is pure normal with r=4m, K'2 B acceleration is pure tangential.My calculations account for that.I still don't see you doing it in K'3 using frame (0,1,0). I'm challenging the validity of your method, and you seem to evade it rather than demonstrate it. What is the radius of curvature of the path of point B in that frame? What is the velocity of B in that frame? Most importantly, what is the acceleration of B in that frame, and how do you arrive at that using your method?I don't think you have a method.