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 International Skeptics Forum Split Thread: Pendular Motion (And Relativity)

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 30th July 2018, 04:07 AM #81 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers If your analysis within one reference frame yields two answers, then one of them is wrong. I'll try and say it again. The work done by a force is the vector product of the force and the displacement through which it moves. The latter is frame dependent. The work done is therefore frame dependent. You are assuming that a force acting on a body for a fixed amount of time produces a frame-independent change in kinetic energy. This is simply wrong. As long as you're making that mistake you'll get erroneous results, which will give the false impression that the absolute motion of the frame of reference can be detected; in fact, what you are detecting is an arithmetical error arising from the fact that you're calculating the energy change in the wrong frame of reference. Dave I do not assume the frame independent change in kinetic energy. I am trying to establish a conservation of energy analysis within one reference frame.
 30th July 2018, 04:19 AM #82 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG 2. We attach the wheel to a concrete block on the ground. The ball is on the edge r=1m. We accelerate the ball to 1m/s tangential velocity. It means \omega = 1rad/s then we will stop acceleration we will keep constant rotation. What energy is in the ball? 1/2m_ball v^2 + 1/2I_ball \omega^2 Wrong. You're double counting the energy of the ball. You can either express it as rotational energy or as kinetic energy, but you can't calculate both separately then add them together. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 30th July 2018, 05:39 AM #83 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers Wrong. You're double counting the energy of the ball. You can either express it as rotational energy or as kinetic energy, but you can't calculate both separately then add them together. Dave Dave, with all due respect what you know, you are wrong in this case. Imagine the ball is cut from the rod when tangential velocity is 1m/s. The ball will fly away with velocity v and it will continue to rotate with \omega. When the ball is cut from the train car and the train starts to decelerate the ball will fly away with v but there is no rotation. I hope you are going to start to see the issue. Please, see the posts 1 and 7 from the other forum. The rotational kinetic energy is 1/1000 from the translational kinetic energy in the post 7 example.
 30th July 2018, 06:13 AM #84 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by SDG Dave, with all due respect what you know, you are wrong in this case. Imagine the ball is cut from the rod when tangential velocity is 1m/s. The ball will fly away with velocity v and it will continue to rotate with \omega. When the ball is cut from the train car and the train starts to decelerate the ball will fly away with v but there is no rotation. I hope you are going to start to see the issue. Please, see the posts 1 and 7 from the other forum. The rotational kinetic energy is 1/1000 from the translational kinetic energy in the post 7 example. Your scenario 2 is not clear in that regard. You say "We accelerate the ball to 1m/s tangential velocity." however it would be the wheel with r= 1m that would be accelerated thus. Now you can also accelerate the ball on the edge of the wheel with a rotational velocity of 1 rad /s (like the wheel) but you don't have to in order to accelerate the wheel and that additional rotational momentum (of just the ball not the wheel itself) is not explicitly noted in the scenario description. Heck, you could accelerate just the ball to 1000 rad/s and still have the wheel turning at 1 rad /s. If they can rotate independently of each other then their rotations can be, well, independent of each other. If they can't rotate independently of each other then the ball has to rotate with the wheel which means that it doesn't rotate in a frame co-moving (rotating with) the wheel. __________________ BRAINZZZZZZZZ Last edited by The Man; 30th July 2018 at 06:15 AM.
 30th July 2018, 06:26 AM #85 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG Dave, with all due respect what you know, you are wrong in this case. Sorry, yes, you're right; the ball has rotational energy because it isn't a point mass. OK, that's established. What's the next step? Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 30th July 2018, 07:08 AM #86 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Again, as I mentioned before part of the problem is the focus on energy not momentum. Since distance is frame dependent kinetic energy is too. Again since speeds aren't relativistic time remains basically the same and thus momentum remains the same. In being released from the wheel the ball can only carry away whatever momentum it has prior to being released. __________________ BRAINZZZZZZZZ
 30th July 2018, 07:37 AM #87 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by Dave Rogers Sorry, yes, you're right; the ball has rotational energy because it isn't a point mass. OK, that's established. What's the next step? Dave Also not being a point mass means it has physical extents and thus a moment of inertia, the rotational equivalent of mass. For solid sphere that would be I = (2/5)MR2 http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi __________________ BRAINZZZZZZZZ
 30th July 2018, 08:38 AM #88 MRC_Hans Penultimate Amazing     Join Date: Aug 2002 Posts: 21,575 Originally Posted by SDG The reason why I introduced the train acceleration is to point out the difference between straight line acceleration and curvilinear acceleration from energy point of view in one reference frame. I'll try again. Let us simplify the scenarios with considering rigid rods beneath the balls. 1. We attach the ball to the train car floor. We accelerate the train in a straight line to 1m/s velocity then we stop acceleration and we will maintain constant straight line motion. What energy is in the ball? 1/2m_ball v^2 2. We attach the wheel to a concrete block on the ground. The ball is on the edge r=1m. We accelerate the ball to 1m/s tangential velocity. It means \omega = 1rad/s then we will stop acceleration we will keep constant rotation. What energy is in the ball? 1/2m_ball v^2 + 1/2I_ball \omega^2 Do we have an agreement? If the ball has a velocity of 1m/s, and a weight of 1 kg, then it has a kinetic energy of 0.5 N regardless of whether it is in a round or straight path. Hans __________________ If you love life, you must accept the traces it leaves.
 30th July 2018, 08:43 AM #89 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by MRC_Hans If the ball has a velocity of 1m/s, and a weight of 1 kg, then it has a kinetic energy of 0.5 N regardless of whether it is in a round or straight path. Hans No, he's right; there's an additional component of energy arising from the fact that the ball is spinning about its own centre. Or, mathematically equivalently, because the part of the ball furthest from the centre of the wheel is moving at a greater velocity than that part nearest to the centre. The dependence of velocity on radius is linear, but the kinetic energy goes as the square of the velocity, so the energy of a ball of finite radius is greater than that of a point mass. However, it's a small difference, and I suspect the claimed relativity violation will come down to a small proportion of that small difference, so unless there's a mathematically exact formula posted then the case will not be proven. That's rather a high bar to clear. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 30th July 2018, 12:02 PM #90 MRC_Hans Penultimate Amazing     Join Date: Aug 2002 Posts: 21,575 Originally Posted by Dave Rogers No, he's right; there's an additional component of energy arising from the fact that the ball is spinning about its own centre. Or, mathematically equivalently, because the part of the ball furthest from the centre of the wheel is moving at a greater velocity than that part nearest to the centre. The dependence of velocity on radius is linear, but the kinetic energy goes as the square of the velocity, so the energy of a ball of finite radius is greater than that of a point mass. However, it's a small difference, and I suspect the claimed relativity violation will come down to a small proportion of that small difference, so unless there's a mathematically exact formula posted then the case will not be proven. That's rather a high bar to clear. Dave Yes, sure, there is a rotational energy. There is also a kinetic energy in the spoke, or connecting rod. And a rotational energy. And in the wheel hub. And axle. Excuse me, but apart from detailing all the energies and vectors in a hypothetical wheel, what are we discussing here? Is there a purpose to all this? Hans __________________ If you love life, you must accept the traces it leaves.
 31st July 2018, 05:08 AM #91 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers Sorry, yes, you're right; the ball has rotational energy because it isn't a point mass. OK, that's established. What's the next step? Dave Now we need to recognize that when the same amount of energy is applied in both cases then the velocities of the balls are not going to be the same. The second scenario is such that some energy goes into the rotation of the ball and therefore less energy goes to translational velocity. Does this make sense?
 31st July 2018, 05:15 AM #92 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by The Man Again, as I mentioned before part of the problem is the focus on energy not momentum. Since distance is frame dependent kinetic energy is too. Again since speeds aren't relativistic time remains basically the same and thus momentum remains the same. In being released from the wheel the ball can only carry away whatever momentum it has prior to being released. The energy analysis and momentum analysis should yield the same result for the velocities. If there would be a discrepancy I would say something is wrong. We can try to discuss both.
 31st July 2018, 05:16 AM #93 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by The Man Also not being a point mass means it has physical extents and thus a moment of inertia, the rotational equivalent of mass. For solid sphere that would be I = (2/5)MR2 http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi Absolutely correct. This was calculated in the other forum as well.
 31st July 2018, 05:17 AM #94 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG Now we need to recognize that when the same amount of energy is applied in both cases then the velocities of the balls are not going to be the same. The second scenario is such that some energy goes into the rotation of the ball and therefore less energy goes to translational velocity. Does this make sense? That's one way of expressing it. Another is that the linear variation of velocity from the inner to the outer edge of the finite width ball in the second case leads to a nonlinear variation of elements of kinetic energy, so the velocity of the centre of the ball is less than in the first case. As MRC_Hans says, we seem simply to be analysing some Newtonian mechanics here; where is this actually going? (I'd also point out that Newtonian mechanics embodies the principle of relativity no less than Einsteinian - the big achievement of SR is the unification of the PoR with the frame invariance of the speed of light in a vacuum - so any result that appears to violate SR should violate Newtonian mechanics no less.) Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 31st July 2018, 05:19 AM #95 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by MRC_Hans If the ball has a velocity of 1m/s, and a weight of 1 kg, then it has a kinetic energy of 0.5 N regardless of whether it is in a round or straight path. Hans Hans, It was already shown as a false statement and it appears to me Dave already understands why this is the false statement. SDG
 31st July 2018, 05:30 AM #96 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers That's one way of expressing it. Another is that the linear variation of velocity from the inner to the outer edge of the finite width ball in the second case leads to a nonlinear variation of elements of kinetic energy, so the velocity of the centre of the ball is less than in the first case. As MRC_Hans says, we seem simply to be analysing some Newtonian mechanics here; where is this actually going? (I'd also point out that Newtonian mechanics embodies the principle of relativity no less than Einsteinian - the big achievement of SR is the unification of the PoR with the frame invariance of the speed of light in a vacuum - so any result that appears to violate SR should violate Newtonian mechanics no less.) Dave This is omega for the cycloid: https://theelectromagneticnatureofth...01r1v0.1_w.png In all the points on the cycloid during an acceleration of the wheel different amounts of energy go to rotational kinetic energy, therefore different velocities of the ball in B, L, T points are the issues for the relativity. The relativity is calculated with a point mass where the rotational kinetic energy is ignored. This appears to me as a problem.
 31st July 2018, 05:40 AM #97 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by SDG Now we need to recognize that when the same amount of energy is applied in both cases then the velocities of the balls are not going to be the same. The second scenario is such that some energy goes into the rotation of the ball and therefore less energy goes to translational velocity. Does this make sense? Which brings us right back to the rolling wheel or ball scenario. Where the translational velocity is only a direct result of the rotation itself. Difference being in that case the translational velocity of the wheel or ball is considered the lateral motion of the rotational center, not particularly the edges or extremities. The driving point of contract has to be moving laterally slower than the center does as the center overtakes that point. Similarly the non-contact point opposite the driving point moves faster laterally than the center as it overtakes the center. Heck, just on the rotating wheel or ball (not moving laterally) the input energy result in different tangential velocities at different distances from the riotational center, this case the velocity of the center is zero. So yes you can get different translational velocities just from simple rotation. Even in a co-rotating frame with the ball or wheel if I toss an object from one point on that radius to another further or closer to the center it will take a curved path. The Coriolis force or effect, an inertial force resulting from the rotation. So that differing of velocity is even detectable in a co-moving (with the rotation) frame. TLDR: Yes, different velocities of just different points on a rotating body is just a simple fact of rotation. https://en.wikipedia.org/wiki/Coriolis_force ETA: Angular momentum http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#am __________________ BRAINZZZZZZZZ Last edited by The Man; 31st July 2018 at 05:57 AM.
 31st July 2018, 06:06 AM #98 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by SDG The energy analysis and momentum analysis should yield the same result for the velocities. If there would be a discrepancy I would say something is wrong. We can try to discuss both. Sure they "should" but it is just easier to get things mixed up with the energy analysis. "If there would be a discrepancy I would say something is wrong." with the analysis. ETA: In discussing both it should be noted that for the rolling wheel scenario the angular momentum of the wheel is orthogonal to the direction of motion and the linear momentum. __________________ BRAINZZZZZZZZ Last edited by The Man; 31st July 2018 at 07:12 AM. Reason: typo
 31st July 2018, 06:30 AM #99 MRC_Hans Penultimate Amazing     Join Date: Aug 2002 Posts: 21,575 Originally Posted by SDG The energy analysis and momentum analysis should yield the same result for the velocities. If there would be a discrepancy I would say something is wrong. We can try to discuss both. The reason we have a rotational energy is that we have arbitrarily decided to work with a portion of the wheel. Since the hypothetical wheel is constructed as balls on spokes, this makes logical sense, but it is still arbitrary. When you calculate the velocity of the balls on the wheel, you are presumably using the center of the ball. But this means that some of the ball travels faster, and some slower, and this gives a rotational energy if you detach the ball. Again, the different analysis differ if we do not take all vectors into account. This is a problem for our method of calculation (if we need it to be that precise), but it is not a problem for the theory. I am still wondering what we are supposed to find, here. Hans __________________ If you love life, you must accept the traces it leaves.
 31st July 2018, 06:30 AM #100 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG This is omega for the cycloid: https://theelectromagneticnatureofth...01r1v0.1_w.png Please define what you mean by "omega for the cycloid" rigorously. I have a suspicion its meaning is not what you take it to be, because... Originally Posted by SDG In all the points on the cycloid during an acceleration of the wheel different amounts of energy go to rotational kinetic energy, therefore different velocities of the ball in B, L, T points are the issues for the relativity. ...you can't be talking about the angular velocity, and hence rotational kinetic energy, of each ball about its centre here. As these are attached to the wheel, all are rotating at the same angular velocity as the wheel and each other. Originally Posted by SDG The relativity is calculated with a point mass where the rotational kinetic energy is ignored. This appears to me as a problem. I think the problem is that you're mixing up different elements of rotational kinetic energy in different frames of reference. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 31st July 2018, 07:22 AM #101 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by Dave Rogers Please define what you mean by "omega for the cycloid" rigorously. I have a suspicion its meaning is not what you take it to be, because... ...you can't be talking about the angular velocity, and hence rotational kinetic energy, of each ball about its centre here. As these are attached to the wheel, all are rotating at the same angular velocity as the wheel and each other. I think the problem is that you're mixing up different elements of rotational kinetic energy in different frames of reference. Dave Also remember that the cycloid motion requires both a lateral and a rotational component. So what has happened is explicitly a combination of lateral and rotational motions. So just trying to identify it as purely rotational iongores that included and required lateral component. Again as I noted upthread just to perceive a cycloid motion means you have to be able to perceive the lateral motion already. The rotational motion becomes just superfluous in detecticting a lateral motion that you can already just perceive directly. __________________ BRAINZZZZZZZZ
 31st July 2018, 10:23 AM #102 SDG Thinker   Join Date: Jul 2018 Posts: 240 Dave and Hans, What I have in mind is the post #8 from the other forum. There are elastic rods that hold the balls. The bottom parts of the elastic rods are attached to the wheel. At the moment I will talk about the ground reference frame and let us consider the bottom and top balls B and T. There are points where the bottom of the rods are attached to the wheel. Let's call these points BA for the bottom ball and TA for the top ball. These points have the velocities after the first second of the acceleration. BA velocity is 0.09m/s and TA velocity is 1.01m/s These points have the \omega_ground in the ground reference frame for positions 0 and \pi*100 from this plot: https://theelectromagneticnatureofth...01r1v0.1_w.png The ground \omega_ground = v_ground/r_ground The r_ground plot: https://theelectromagneticnatureofth...01r1v0.1_r.png The v_ground plot: https://theelectromagneticnatureofth...01r1v0.1_v.png The elastic rods transfer the acceleration force to the balls. The energy transferred through this force goes to the rotational kinetic energy and the translational kinetic energy. The question of the day: Are the B and T velocities going to have the same delta to the BA and TA velocities if the B and T balls absorb different rotational kinetic energy due to the fact that they are going through different ground \omega_ground?
 31st July 2018, 10:32 AM #103 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by The Man Also remember that the cycloid motion requires both a lateral and a rotational component. So what has happened is explicitly a combination of lateral and rotational motions. So just trying to identify it as purely rotational iongores that included and required lateral component. Again as I noted upthread just to perceive a cycloid motion means you have to be able to perceive the lateral motion already. The rotational motion becomes just superfluous in detecticting a lateral motion that you can already just perceive directly. This is the point, the normal component of the normal and tangential coordinate system analysis defines how much of the energy goes to the rotational kinetic energy. It was demonstrated in the post #80 http://www.internationalskeptics.com...5&postcount=80 that the tangential component does not contribute the rotational kinetic energy.
 31st July 2018, 10:49 AM #104 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by SDG Dave and Hans, What I have in mind is the post #8 from the other forum. There are elastic rods that hold the balls. The bottom parts of the elastic rods are attached to the wheel. At the moment I will talk about the ground reference frame and let us consider the bottom and top balls B and T. There are points where the bottom of the rods are attached to the wheel. Let's call these points BA for the bottom ball and TA for the top ball. These points have the velocities after the first second of the acceleration. BA velocity is 0.09m/s and TA velocity is 1.01m/s These points have the \omega_ground in the ground reference frame for positions 0 and \pi*100 from this plot: https://theelectromagneticnatureofth...01r1v0.1_w.png The ground \omega_ground = v_ground/r_ground The r_ground plot: https://theelectromagneticnatureofth...01r1v0.1_r.png The v_ground plot: https://theelectromagneticnatureofth...01r1v0.1_v.png The elastic rods transfer the acceleration force to the balls. The energy transferred through this force goes to the rotational kinetic energy and the translational kinetic energy. The question of the day: Are the B and T velocities going to have the same delta to the BA and TA velocities if the B and T balls absorb different rotational kinetic energy due to the fact that they are going through different ground \omega_ground? No, the question of the day was "Please define what you mean by "omega for the cycloid" rigorously." Which of course would mean "omega_ground" here. Angular velocity represented by "omega" dose not itself have a lateral component only rotational. The graphical representations again include lateral components where labeled as purely rotational (omega). Angular velocity is measured from the axis of rotation, which is not the ground unless the center of the wheel is on the ground. As the tangential velocity is different for the balls that would indicate, if the balls are the same mass, that they are different distances from the center of rotation. Yes kinetic energy will be different for similar masses at different distances from the center of rotation. Not because of some "omega_ground" and inorder for there to even be an "omega_ground" the ground would have to be at the center of rotation and hence the center of the wheel. __________________ BRAINZZZZZZZZ
 31st July 2018, 11:01 AM #105 The Man Scourge, of the supernatural     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 13,371 Originally Posted by SDG This is the point, the normal component of the normal and tangential coordinate system analysis defines how much of the energy goes to the rotational kinetic energy. It was demonstrated in the post #80 http://www.internationalskeptics.com...5&postcount=80 that the tangential component does not contribute the rotational kinetic energy. Which has noting to do with the post you quoted or the rotational motion being superfluous to the lateral motion you would already have to be able to directly perceive in order to even just perceive the cycloid motion of some point or points on the wheel. Oh and if the wheel is initially driven by a tangential force (or forces) then that's where all the rotational kinetic energy comes from. Also the motion of the center of the wheel (the lateral motion referenced for the cycloid motion) isn't "tangential". That's why I have been referring to it as just lateral motion. __________________ BRAINZZZZZZZZ
 31st July 2018, 12:39 PM #106 MRC_Hans Penultimate Amazing     Join Date: Aug 2002 Posts: 21,575 Originally Posted by SDG Dave and Hans, What I have in mind is the post #8 from the other forum. There are elastic rods that hold the balls. The bottom parts of the elastic rods are attached to the wheel. At the moment I will talk about the ground reference frame and let us consider the bottom and top balls B and T. There are points where the bottom of the rods are attached to the wheel. Let's call these points BA for the bottom ball and TA for the top ball. These points have the velocities after the first second of the acceleration. BA velocity is 0.09m/s and TA velocity is 1.01m/s These points have the \omega_ground in the ground reference frame for positions 0 and \pi*100 from this plot: https://theelectromagneticnatureofth...01r1v0.1_w.png The ground \omega_ground = v_ground/r_ground The r_ground plot: https://theelectromagneticnatureofth...01r1v0.1_r.png The v_ground plot: https://theelectromagneticnatureofth...01r1v0.1_v.png The elastic rods transfer the acceleration force to the balls. The energy transferred through this force goes to the rotational kinetic energy and the translational kinetic energy. The question of the day: Are the B and T velocities going to have the same delta to the BA and TA velocities if the B and T balls absorb different rotational kinetic energy due to the fact that they are going through different ground \omega_ground? I see you making very complex descriptions of simple things. So I will give you simple answers: There will be a difference in the velocities and forces acting on the balls through the cycle because they are under acceleration: Gravity. There will be no difference due to any linear speed (speed of train). It will be entirely irrelevant what their trajectory looks like from a stationary reference. Hans __________________ If you love life, you must accept the traces it leaves.
 31st July 2018, 01:06 PM #107 JeanTate Master Poster   Join Date: Nov 2014 Posts: 2,016 Originally Posted by MRC_Hans The reason we have a rotational energy is that we have arbitrarily decided to work with a portion of the wheel. Since the hypothetical wheel is constructed as balls on spokes, this makes logical sense, but it is still arbitrary. When you calculate the velocity of the balls on the wheel, you are presumably using the center of the ball. But this means that some of the ball travels faster, and some slower, and this gives a rotational energy if you detach the ball. Again, the different analysis differ if we do not take all vectors into account. This is a problem for our method of calculation (if we need it to be that precise), but it is not a problem for the theory. I am still wondering what we are supposed to find, here. Hans (my bold) At one time, it was that a detailed analysis of some idealization of the physical system described would be clear evidence that Special Relativity is inconsistent (or wrong). I do not know if that's still the (or an) unstated goal.
 31st July 2018, 02:04 PM #108 MRC_Hans Penultimate Amazing     Join Date: Aug 2002 Posts: 21,575 Originally Posted by JeanTate (my bold) At one time, it was that a detailed analysis of some idealization of the physical system described would be clear evidence that Special Relativity is inconsistent (or wrong). I do not know if that's still the (or an) unstated goal. Yeah, I gathered that, but we don't seem to find that. We are rummaging around in some details of very trivial physics, bickering about how deep we should delve into details. We are not going anywhere. Hans __________________ If you love life, you must accept the traces it leaves.
 1st August 2018, 01:07 AM #109 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG The question of the day: Are the B and T velocities going to have the same delta to the BA and TA velocities if the B and T balls absorb different rotational kinetic energy due to the fact that they are going through different ground \omega_ground? They are not "absorbing different rotational kinetic energies"; their rates of rotation are, by the definition of the problem, identical. Now, listen. You're the one trying to prove something here. You're the one who bears the burden of proof. So you're the one who should be answering questions, not the people you're trying to convince. So, as MRC_Hans said, the question of the day is still "Please define what you mean by "omega for the cycloid" rigorously." Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 1st August 2018, 02:39 AM #110 SDG Thinker   Join Date: Jul 2018 Posts: 240 \omega_ground = v_ground/r_ground Any curvilinear motion has a curvature K. The radius is 1/K. \omega=v/r What is wrong with this definition? Sent from my iPhone using Tapatalk
 1st August 2018, 02:57 AM #111 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG \omega_ground = v_ground/r_ground Any curvilinear motion has a curvature K. The radius is 1/K. \omega=v/r What is wrong with this definition? What's wrong with it is that it's the source of your error. You're taking the rate of curvature of the cycloid, multiplying it by the ground frame velocity, and equating this to the angular velocity of the ball. This is incorrect, because the ball is not traversing the cycloid at constant dx/dt; it slows to a dead stop at the discontinuity of the cycloid, which it must do for its acceleration not to be infinite, and has a maximum dx/dt at the peak of each cycle. In fact, if you calculate the dx/dt value properly, rather than incorrectly assuming it's constant, you'll find that in fact the angular velocity of the ball is equal to that of the wheel at all times - which it must be, because it's stated as such implicitly in the definition of the problem. The instantaneous angular velocity of the ball is not, therefore, v/r; it's 1/r dx/dt, where dx/dt is a variable, not a constant. That's your error, and that's the source of your supposed violation of relativity. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 1st August 2018, 03:06 AM #112 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers What's wrong with it is that it's the source of your error. You're taking the rate of curvature of the cycloid, multiplying it by the ground frame velocity, and equating this to the angular velocity of the ball. This is incorrect, because the ball is not traversing the cycloid at constant dx/dt; it slows to a dead stop at the discontinuity of the cycloid, which it must do for its acceleration not to be infinite, and has a maximum dx/dt at the peak of each cycle. In fact, if you calculate the dx/dt value properly, rather than incorrectly assuming it's constant, you'll find that in fact the angular velocity of the ball is equal to that of the wheel at all times - which it must be, because it's stated as such implicitly in the definition of the problem. The instantaneous angular velocity of the ball is not, therefore, v/r; it's 1/r dx/dt, where dx/dt is a variable, not a constant. That's your error, and that's the source of your supposed violation of relativity. Dave Where do you see the discontinuity of the cycloid in the discussed example? The example shows a curtate cycloid. It has inflection points where r goes to infinity. Sent from my iPhone using Tapatalk
 1st August 2018, 03:14 AM #113 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG Where do you see the discontinuity of the cycloid in the discussed example? The example shows a curtate cycloid. It has inflection points where r goes to infinity. You're discussing irrelevancies. The rate of rotation of the ball may be expressed as 1/r(x) dx/dt, where 1/r(x) is the curvature of the cycloid as a function of x and dx/dt is the rate of change of the x co-ordinate of the centre of the ball. However, dx/dt is not constant and equal to v, the velocity of the train relative to the ground frame of reference, so the rate of rotation of the ball is not in general equal to v/r(x). That's your mistake, and if and when you recognise that, you'll recognise that mistake is the source of the frame dependence you think you've identified. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 2nd August 2018, 09:04 AM #114 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers You're discussing irrelevancies. The rate of rotation of the ball may be expressed as 1/r(x) dx/dt, where 1/r(x) is the curvature of the cycloid as a function of x and dx/dt is the rate of change of the x co-ordinate of the centre of the ball. However, dx/dt is not constant and equal to v, the velocity of the train relative to the ground frame of reference, so the rate of rotation of the ball is not in general equal to v/r(x). That's your mistake, and if and when you recognise that, you'll recognise that mistake is the source of the frame dependence you think you've identified. Dave Where do you see the curvature of the cycloid is a function of x only? The curvature is an x,y function. https://en.wikipedia.org/wiki/Curvature The velocity is the x,y function as well. The linked plot is v_mag not v_x. I do not understand what you are talking about. Why are you fighting \omega_ground? Are you saying that \omega_ground is not the correct \omega? Is the \omega of the wheel axle preferred one? Is the moving reference frame preferred one?
 2nd August 2018, 09:17 AM #115 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG I do not understand what you are talking about. Yes, you're making that clear. Choosing the X axis to be in the direction of motion of the train, you have taken the rate of rotation of the ball on the perimeter of the wheel to be the rate of curvature of the cycloid multiplied by the velocity of the train. This would be only valid if the ball were moving in such a way that the X component of its velocity in the direction of the train were constant, which it is not; there is a sinusoidal term in the X component of its velocity that you are neglecting. As a result, your calculation of its rate of rotation is wrong. If you include it, you'll find that it cancels out with the X component of the cycloid omega to give a constant rate of rotation of the ball. That's your error, and the source of the effect you believe you've identified. I don't really have any more to say on this. I've now explained your mistake to you three times. It's now incumbent on you to try to understand it for yourself. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 2nd August 2018, 09:54 AM #116 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers Yes, you're making that clear. Choosing the X axis to be in the direction of motion of the train, you have taken the rate of rotation of the ball on the perimeter of the wheel to be the rate of curvature of the cycloid multiplied by the velocity of the train. This would be only valid if the ball were moving in such a way that the X component of its velocity in the direction of the train were constant, which it is not; there is a sinusoidal term in the X component of its velocity that you are neglecting. As a result, your calculation of its rate of rotation is wrong. If you include it, you'll find that it cancels out with the X component of the cycloid omega to give a constant rate of rotation of the ball. That's your error, and the source of the effect you believe you've identified. I don't really have any more to say on this. I've now explained your mistake to you three times. It's now incumbent on you to try to understand it for yourself. Dave What you are talking about is true only for the ideal cycloid where 'top' of the wheel has velocity 2v. That's the only case where the \omega_ground is constant and on top of that it is 1/2 of the moving reference frame \omega. This is not true for curtate and prolate cycloids! Check it out. I did the math, I did the work, I know what I am talking about.
 2nd August 2018, 01:32 PM #117 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG I did the math, I did the work, I know what I am talking about. You did the math wrong, and no, you don't know what you're talking about. The centre of the ball can only have a constant dx/dt when it's at the centre of the wheel, in which case it's describing a straight line rather than a cycloid. In any other position it has a varying dx/dt, so you can't simply multiply the curvature of the cycloid by the magnitude of the train velocity to get the rate of rotation. That's four times. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right
 3rd August 2018, 12:05 AM #118 SDG Thinker   Join Date: Jul 2018 Posts: 240 Originally Posted by Dave Rogers You did the math wrong, and no, you don't know what you're talking about. The centre of the ball can only have a constant dx/dt when it's at the centre of the wheel, in which case it's describing a straight line rather than a cycloid. In any other position it has a varying dx/dt, so you can't simply multiply the curvature of the cycloid by the magnitude of the train velocity to get the rate of rotation. That's four times. Dave Dave, your calculation is OK in relation to the center of the wheel, but this is mixing reference frames. You take the moving reference frame \omega and claim it to be the ground reference frame \omega. The curtate cycloid has an inflection point. The radius of the curvilinear motion in the inflection point is infinity in the ground reference frame. \omega is 0 in the inflection point in the ground reference frame. This is what is necessary for the conservation of energy analysis. The reference frames cannot be mixed. Your approach is against the definition of the \omega=v/r in one reference frame. This is where the disconnect is between us. So unless you recognize that \omega = v/r and v, r have to be from the same reference frame, we are not going to get anywhere. I leave it with you, SDG
 3rd August 2018, 12:33 AM #119 Kid Eager Philosopher     Join Date: Nov 2010 Posts: 6,696 More grist for the mill, cutting out the (alleged) middle man, as it were: http://theelectromagneticnatureofthings.com Mind you, it might be complete bollocks, but I leave it to y'all with the mad physics skills to interpret all the squiggly symbols. __________________ What do Narwhals, Magnets and Apollo 13 have in common? Think about it.... Last edited by Kid Eager; 3rd August 2018 at 12:34 AM.
 3rd August 2018, 12:55 AM #120 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 28,166 Originally Posted by SDG Dave, your calculation is OK in relation to the center of the wheel, but this is mixing reference frames. You take the moving reference frame \omega and claim it to be the ground reference frame \omega. The curtate cycloid has an inflection point. The radius of the curvilinear motion in the inflection point is infinity in the ground reference frame. \omega is 0 in the inflection point in the ground reference frame. This is what is necessary for the conservation of energy analysis. The reference frames cannot be mixed. Your approach is against the definition of the \omega=v/r in one reference frame. This is where the disconnect is between us. So unless you recognize that \omega = v/r and v, r have to be from the same reference frame, we are not going to get anywhere. I leave it with you, SDG No. The above is either irrelevant or incorrect. Your definition of the problem gives a funtional form for the angular velocity of the wheel, and the angular velocity of the balls attached to its perimeter must follow the same functional form. Your mathematical analysis gives a different functional form for the angular velocity of the ball to the one you put in the definition of the problem. No further analysis is needed to conclude that your mathematical analysis is incorrect. Dave __________________ Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right? Tony Szamboti: That is right

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