swansont Posted March 17, 2017 Share Posted March 17, 2017 There is always a force on the string whether the radius is increasing, decreasing or the force is balanced and the radius remains constant. (I am aware that I have Omitted certain required provisos, but I am sure we are all intelligent enough to understand that these are implied and we can keep the discussion within the bounds intended). There is always work being done. If anybody has noticed, I have corrected my error and replaced tangential with perpendicular. There cannot be a perpendicular component of the force. Angular velocity can change without there being a force because it changes with the radius, yes, but the component of velocity which affects angular velocity cannot. i.e.: The component of velocity which is perpendicular to the radius cannot be affected without a perpendicular component of force. Balanced? If an object is moving in a circle, there is no balancing of forces. There is only the tension, directed toward the center. And as the path is perpendicular to the force, no work is done. But if the path isn't a circle, there must be another component of force. You agree with an argument that you think supports your position and then disagree when it is pointed out that it does not support your position. This clearly indicates that it is your position which stands and not the validity of the argument. A strong indication that you are subject to confirmation bias. I'm biased in favor of proper physics Link to comment Share on other sites More sharing options...
Bender Posted March 17, 2017 Share Posted March 17, 2017 (edited) Ok, it took me more time than I was hoping, but I had some fun with the mathematical derivation of the example of the rotating professor (or a any isolated system, really) mathematically without using angular momentum, only using Newton's laws of motion and basic geometry.What we have is a central, rotating object with unchanging moment of inertia [math]I[/math] and angular velocity [math]\omega[/math]. There are two equal and small masses ([math]m/2[/math] each) symmetrically attached to this object at radius [math]R[/math]. They can symmetrically move in and out, while their angular velocity is always equal to that of the central object.When the radius changes, the masses have an acceleration with two components, the first is a result of the angular acceleration [math]a_1=\alpha R[/math], the second is the coriolis acceleration caused by the radial speed [math]a_2=2 \omega dR/dt[/math] (a derivation of this can be found here), or combined:[math]a_t=\alpha R+2 \omega dR/dt[/math] (1)We need to find a relation between the linear acceleration of the masses and the angular acceleration of the central object. The force required to accelerate the masses is:[math]F_t = m/2 \cdot a_t[/math]The torque required to accelerate the central object is:[math]T = I \alpha[/math]This torque is a result of the reaction forces from the masses:[math]T=-2 F_t R = m a_t R[/math]Notice that the radial forces do not matter as they cancel out.Now combining the last two equations:[math]I \alpha = - m a_t R[/math]or[math]a_t=-\frac{I \alpha}{m R}[/math]Substituting in (1)[math]-\frac{I \alpha}{m R}=\alpha R+2 \omega dR/dt[/math] or[math]2 \omega R dR/dt=-(\frac{I} {m}+R^2) \alpha[/math] [math]2 \omega R dR/dt=-(\frac{I} {m}+R^2) \frac{d \omega} {dt}[/math][math]2 \omega R dR=-(\frac{I} {m}+R^2) d \omega[/math] rearranging[math]\frac{1}{\omega} d \omega= - \frac{2 R}{(\frac{I} {m}+R^2)} dR[/math] (2)Now I'm going to do a substitution to facilitate the integration:[math]x=(\frac{I} {m}+R^2)[/math][math]dx=2 R dR)[/math]putting this in (2) yields:[math]\frac{1}{\omega} d \omega= - \frac{1}{x} dx[/math]integrating between two radiuses:[math]\int_{\omega_1}^{\omega_2}\frac{1}{\omega} d \omega= - \int_{(\frac{I} {m}+R_1^2)}^{(\frac{I} {m}+R_2^2)}\frac{1}{x} dx[/math]almost there:[math]ln(\frac{\omega_2} {\omega_1})=ln(\frac{\frac{I} {m}+R_1^2} {\frac{I} {m}+R_2^2})[/math]note that the second natural logarithm was inverted because of the minusSo at last, shifting a bit around, we have the answer, which is very generally applicable:[math]\omega_2=\omega_1 \frac{I+m R_1^2} {I+m R_2^2}[/math]Of course, we can do the same calculation using conservation of angular momentum[math]L_1=L_2[/math][math]I_1 \omega_1 + m R_1 v_1=I_2 \omega_2 + m R_2 v_2[/math]replacing v[math]I_1 \omega_1 + m R_1^2 \omega_1=I_2 \omega_2 + m R_2^2 \omega_2[/math][math]\omega_1 (I_1 + m R_1^2)=\omega_2 (I_2 + m R_2^2)[/math][math]\omega_2=\omega_1 \frac{I+m R_1^2} {I+m R_2^2}[/math] That looks surprisingly familiar (not that surprising really ). It sure was a lot easier. You seem very fond of logical proof, here you have an actual logical proof. Edited March 17, 2017 by Bender Link to comment Share on other sites More sharing options...
Mandlbaur Posted March 18, 2017 Author Share Posted March 18, 2017 (edited) Great. We now have three alternative logical proofs predicting one result and one mathematical proof predicting a differing result. When a scientist is provided with two alternative theories predicting differing results, he should conduct an experiment to determine which of them is correct. Fortunately we have arguably the best demonstration available which we can turn into an experiment by measuring it - the one contained in Walter Levin's lecture 20. Measuring this, I have provided significant evidence which shows that the results do not match your mathematical proof. The only question remaining is: Does my theory match the measured results? Edited March 18, 2017 by Mandlbaur Link to comment Share on other sites More sharing options...
Bender Posted March 18, 2017 Share Posted March 18, 2017 (edited) We have two rigorous methods actually making a prediction, one heavily contested idea that has yet to provide any prediction, and no proper experimental results. Perhaps if I'm bored this evening, I'll adjust Levin's estimate to match your measurements exactly. If you do not even have a way to make predictions based on your idea, how can you even hope to confirm them experimentally? Do you really think you can go to a scientific journal with a wild idea, no predictions, no mathematical model and no experimental results? Edited March 18, 2017 by Bender Link to comment Share on other sites More sharing options...
Bender Posted March 18, 2017 Share Posted March 18, 2017 Besides, you really can't dismiss the derivation that easily. Sure, you dismiss the one based on conservation of angular momentum, since that is the subject of this speculative thread. The other derivation, however, is entirely and directly based on Newton's laws of motion. To dismiss it, you first need to dismiss Newton's laws. 1 Link to comment Share on other sites More sharing options...
Mandlbaur Posted March 19, 2017 Author Share Posted March 19, 2017 We have two rigorous methods actually making a prediction, one heavily contested idea that has yet to provide any prediction, and no proper experimental results. Perhaps if I'm bored this evening, I'll adjust Levin's estimate to match your measurements exactly. If you do not even have a way to make predictions based on your idea, how can you even hope to confirm them experimentally? Do you really think you can go to a scientific journal with a wild idea, no predictions, no mathematical model and no experimental results? My prediction is that the magnitude of linear momentum which is perpendicular to the radius is what we need to conserve. Shouldn't be too complicated to make a mathematical model and generate predictions. Taking measurements of a high quality demonstration is "experimental results". My "wild idea" has been proven three times over each with a different approach. Adjusting a prediction to match the results is not mentioned in any description of the scientific method that I have read. I believe it is nothing less than fraud. Besides, you really can't dismiss the derivation that easily. Sure, you dismiss the one based on conservation of angular momentum, since that is the subject of this speculative thread. The other derivation, however, is entirely and directly based on Newton's laws of motion. To dismiss it, you first need to dismiss Newton's laws. I can dismiss a derivation based on the fact that the measured results do not match those predicted by the derived equation (Without having to dismiss Newton's laws or waste my time trying to find the mistake). I have shown this discrepancy for both the ball on a string as well as the rotating professor which pretty much covers all the angles. -2 Link to comment Share on other sites More sharing options...
uncool Posted March 19, 2017 Share Posted March 19, 2017 My prediction is that the magnitude of linear momentum which is perpendicular to the radius is what we need to conserve. Shouldn't be too complicated to make a mathematical model and generate predictions. This seems trivially falsified if we take the case of an object moving with no force affecting it. Link to comment Share on other sites More sharing options...
Bender Posted March 19, 2017 Share Posted March 19, 2017 (edited) My prediction is that the magnitude of linear momentum which is perpendicular to the radius is what we need to conserve. Shouldn't be too complicated to make a mathematical model and generate predictions. Then why don't you? Taking measurements of a high quality demonstration is "experimental results". No it isn't, that's why one is called "demonstration" and the other is called "experiment". My "wild idea" has been proven three times over each with a different approach. After which it has been carefully explained why the "proof" is invalid. Adjusting a prediction to match the results is not mentioned in any description of the scientific method that I have read. I believe it is nothing less than fraud. Thank you for making my point for me. In these descriptions you have read, did they mention "taking the rough estimates of a demonstration out of context"? I can dismiss a derivation based on the fact that the measured results do not match those predicted by the derived equation (Without having to dismiss Newton's laws or waste my time trying to find the mistake). I have shown this discrepancy for both the ball on a string as well as the rotating professor which pretty much covers all the angles. If it wasn't for Poe's law, I could only assume you are trolling at this point. I do your work for you, doing the calculations with your premisses in mind, and you won't even bother looking at them? I is possible you lack the skill, in which case you are welcome to ask for help. If I did make a mistake, it would be very helpfull if you could point it out. Let's try something different: ever heard of helicopters? Don't you think the engineers designing them would have noticed if their calculations completely failed to predict the behaviour of the helicopter? What about the engineers designing satelites and space probes. They should have been in for quite a surpise is their yo-yo de-spin device completely failed to perform or they didn't manage to target their satelites with the reaction wheels they designed. Except, you know, they didn't because it all worked fine. Perhaps you have heard of Kepler's laws of planetary motion? They are entirely consistent with conservation of angular momentum; in fact, the concept of conservation of angular momentum has its origin in Kepler's second law. Don't you think astronomers would, at some point in the last couple of centuries, have noticed something if they completely and utterly fail to predict planetary motion? Edited March 19, 2017 by Bender 2 Link to comment Share on other sites More sharing options...
swansont Posted March 19, 2017 Share Posted March 19, 2017 So there's this demo, where you have a spinning wheel and you tip it over while standing on a platform that can turn. You begin rotating https://m.youtube.com/watch?v=_XgYTP0kB7A Calculating the angular momentum here would be tough, as is the problem with a lot if these demos. But there is this: if angular momentum is not conserved, why does the man stop rotating when he tips the wheel back to its original orientation? Link to comment Share on other sites More sharing options...
Mandlbaur Posted March 20, 2017 Author Share Posted March 20, 2017 After which it has been carefully explained why the "proof" is invalid. As far as I am aware, I have successfully tackled and defeated every reasonable argument presented against any of my logic. Unless you can point out a relevant argument which I have failed to address, please stop making this unfounded claim? So there's this demo, where you have a spinning wheel and you tip it over while standing on a platform that can turn. You begin rotating https://m.youtube.com/watch?v=_XgYTP0kB7A Calculating the angular momentum here would be tough, as is the problem with a lot if these demos. But there is this: if angular momentum is not conserved, why does the man stop rotating when he tips the wheel back to its original orientation? My argument is that since angular momentum is defined by the radius it will change when the radius changes. Angular momentum is effectively conserved in situations where there is no change in radius. (I am obviously referring to the magnitude of the radius) Link to comment Share on other sites More sharing options...
uncool Posted March 20, 2017 Share Posted March 20, 2017 Angular momentum is also conserved when there is a change in radius, as in the example of an object moving past a point with no force acting on the object; the only requirement is that any force be central (that is, in a direction parallel to the radius from the point angular momentum is being measured from). Link to comment Share on other sites More sharing options...
Mandlbaur Posted March 20, 2017 Author Share Posted March 20, 2017 Angular momentum is also conserved when there is a change in radius, as in the example of an object moving past a point with no force acting on the object You are correct, the moment the motion becomes linear, my theory of rotational dynamics falls apart. Link to comment Share on other sites More sharing options...
Bender Posted March 20, 2017 Share Posted March 20, 2017 You are correct, the moment the motion becomes linear, my theory of rotational dynamics falls apart. If it falls apart in that situation, why not in others? Could you also address the other points in my previous post? (I think going back and forth about your so-called proofs is pointless, since you won't accept the arguments against it and I have run out of ways to rephrase them.) Link to comment Share on other sites More sharing options...
Mandlbaur Posted March 20, 2017 Author Share Posted March 20, 2017 If it falls apart in that situation, why not in others? Could you also address the other points in my previous post? (I think going back and forth about your so-called proofs is pointless, since you won't accept the arguments against it and I have run out of ways to rephrase them.) I will not address any of your points until such time that you are prepared to back up your claim that I have ignored arguments presented against my logic. You have now made this accusation three times and this is the third time that I am asking you to point out any argument against my logic that I have failed to address. You are not entitled to make wild claims without providing evidence to back them up just as I am not entitled to do so. I view this as an ad-hominem attack. -1 Link to comment Share on other sites More sharing options...
swansont Posted March 20, 2017 Share Posted March 20, 2017 I view this as an ad-hominem attack. Nobody who understands the meaning considers it to be that, though. Link to comment Share on other sites More sharing options...
imatfaal Posted March 20, 2017 Share Posted March 20, 2017 ! Moderator Note OK this thread has run its course. The OP started with a nonsense claim and has failed to defend it either heuristically, mathematically, or empirically; in fact, the OP's contention has been refuted with great explanations, with mathematical derivations, and through the provision of experimental data/video. The OP has now resorted to a refusal to communicate, logical fallacies, reversal of the burden, and claims of persecution. Thread Locked. The OP does not have permission to reopen a new thread on this topic without first receiving explicit permission from a staff member Link to comment Share on other sites More sharing options...
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