Why can't we make a perpetual motion machine by using a magnet to pull up a piece of metal, then letting it...
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Obviously, a perpetuum mobile isn't possible by any law in physics, because energy can't be "created" or "destroyed", only transformed.
This said, I've had an idea for a perpetuum mobile and can't seem to find my mistake (I'm actually very close to building and trying it).
Here's the plan:
Take a piece of wood and attach its upper end to a screw so it can swing like a pendulum. Then, attach a magnetic metal at the lower end of the wood. Now, place two magnets at each side of the pendulum, so that at the maximum amplitude, the metal will barely touch the magnets and will swing back, due to its weight.
Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit. Thus, on the "way back", it will have a slightly higher amplitude. So it swings to the other side, closer to the magnet, which will pull the pendulum a bit more upwards, thus increasing the amplitude furthermore.
This could theoretically go on and the pendulum will never stop, it will actually gain more momentum at the start.
So the conditions are:
- The Metal must be heavy enough so it doesn't stick to the magnets
- The Metal must be magnetic enough so that we gain amplitude instead of losing it each swing
And that's basically it. I am aware that the construction couldn't work, but struggle to find where I made my mistake.
Anyways, if it does work and you guys build it before I do: I want 50% of all profits and want you to name it Perpenduluum Mobile :D
newtonian-mechanics electromagnetism energy-conservation dissipation perpetual-motion
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Florian Claaßen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Obviously, a perpetuum mobile isn't possible by any law in physics, because energy can't be "created" or "destroyed", only transformed.
This said, I've had an idea for a perpetuum mobile and can't seem to find my mistake (I'm actually very close to building and trying it).
Here's the plan:
Take a piece of wood and attach its upper end to a screw so it can swing like a pendulum. Then, attach a magnetic metal at the lower end of the wood. Now, place two magnets at each side of the pendulum, so that at the maximum amplitude, the metal will barely touch the magnets and will swing back, due to its weight.
Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit. Thus, on the "way back", it will have a slightly higher amplitude. So it swings to the other side, closer to the magnet, which will pull the pendulum a bit more upwards, thus increasing the amplitude furthermore.
This could theoretically go on and the pendulum will never stop, it will actually gain more momentum at the start.
So the conditions are:
- The Metal must be heavy enough so it doesn't stick to the magnets
- The Metal must be magnetic enough so that we gain amplitude instead of losing it each swing
And that's basically it. I am aware that the construction couldn't work, but struggle to find where I made my mistake.
Anyways, if it does work and you guys build it before I do: I want 50% of all profits and want you to name it Perpenduluum Mobile :D
newtonian-mechanics electromagnetism energy-conservation dissipation perpetual-motion
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Florian Claaßen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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– David Z♦
1 hour ago
add a comment |
$begingroup$
Obviously, a perpetuum mobile isn't possible by any law in physics, because energy can't be "created" or "destroyed", only transformed.
This said, I've had an idea for a perpetuum mobile and can't seem to find my mistake (I'm actually very close to building and trying it).
Here's the plan:
Take a piece of wood and attach its upper end to a screw so it can swing like a pendulum. Then, attach a magnetic metal at the lower end of the wood. Now, place two magnets at each side of the pendulum, so that at the maximum amplitude, the metal will barely touch the magnets and will swing back, due to its weight.
Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit. Thus, on the "way back", it will have a slightly higher amplitude. So it swings to the other side, closer to the magnet, which will pull the pendulum a bit more upwards, thus increasing the amplitude furthermore.
This could theoretically go on and the pendulum will never stop, it will actually gain more momentum at the start.
So the conditions are:
- The Metal must be heavy enough so it doesn't stick to the magnets
- The Metal must be magnetic enough so that we gain amplitude instead of losing it each swing
And that's basically it. I am aware that the construction couldn't work, but struggle to find where I made my mistake.
Anyways, if it does work and you guys build it before I do: I want 50% of all profits and want you to name it Perpenduluum Mobile :D
newtonian-mechanics electromagnetism energy-conservation dissipation perpetual-motion
New contributor
Florian Claaßen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Obviously, a perpetuum mobile isn't possible by any law in physics, because energy can't be "created" or "destroyed", only transformed.
This said, I've had an idea for a perpetuum mobile and can't seem to find my mistake (I'm actually very close to building and trying it).
Here's the plan:
Take a piece of wood and attach its upper end to a screw so it can swing like a pendulum. Then, attach a magnetic metal at the lower end of the wood. Now, place two magnets at each side of the pendulum, so that at the maximum amplitude, the metal will barely touch the magnets and will swing back, due to its weight.
Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit. Thus, on the "way back", it will have a slightly higher amplitude. So it swings to the other side, closer to the magnet, which will pull the pendulum a bit more upwards, thus increasing the amplitude furthermore.
This could theoretically go on and the pendulum will never stop, it will actually gain more momentum at the start.
So the conditions are:
- The Metal must be heavy enough so it doesn't stick to the magnets
- The Metal must be magnetic enough so that we gain amplitude instead of losing it each swing
And that's basically it. I am aware that the construction couldn't work, but struggle to find where I made my mistake.
Anyways, if it does work and you guys build it before I do: I want 50% of all profits and want you to name it Perpenduluum Mobile :D
newtonian-mechanics electromagnetism energy-conservation dissipation perpetual-motion
newtonian-mechanics electromagnetism energy-conservation dissipation perpetual-motion
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edited 8 hours ago
knzhou
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44.7k11122216
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asked 14 hours ago
Florian ClaaßenFlorian Claaßen
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3712
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Florian Claaßen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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– David Z♦
1 hour ago
add a comment |
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– David Z♦
1 hour ago
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– David Z♦
1 hour ago
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– David Z♦
1 hour ago
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6 Answers
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Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit.
You've neglected to account for the magnetic attraction as the pendulum bob goes back to its central position.
On the outwards leg, you are correct that the magnet's attraction will pull on the bob and give it more energy than it would have in the absence of the magnet. However, in the return leg, the pendulum bob is trying to get away from the magnet's attractive force, and this will claim back all of the additional energy.
(... if the system is perfect, that is. Real-world magnetic materials will show some amount of hysteresis, so the bob will lose slightly more energy on the way back than it gained on the way out.)
This type of mistake is quite common when you have a core dynamics which is known to be conservative, and still seems to be producing energy - you're just conveniently neglecting to take into account the parts of the cycle where that force performs work against your system. For a similar example in action, see What prevents this magnetic perpetuum mobile from working?.
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Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
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– Florian Claaßen
12 hours ago
11
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Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
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– Emilio Pisanty
12 hours ago
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Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
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– Florian Claaßen
12 hours ago
2
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No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
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– Emilio Pisanty
12 hours ago
add a comment |
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Perpetual motion is impossible due to dissipation, or if you prefer second principle of thermodynamics and not to energy conservation.
If your analysis of the suggested setup was correct, you could create mechanical energy for free !
In first analysis, neglecting the (unavoidable) losses in the ferromagnetic medium, your system is conservative : what you have is a modified pendulum, where the confinement potential contains not only the gravitational part but also a magnetic component. Actually the magnetic force slightly decrease the recall torque that you would have with gravity alone, and the amplitude of motion will indeed be larger. But you nevertheless will have a turning point where the kinetic energy vanishes, and when going back, you will reach exactly the same angle for the turning point on the other side.
This correspond to an oscillator with constant amplitude, because losses have been neglected.
The sources of losses are at least : friction in air, friction on axe, ferromagnetic hysteresis, Foucault currents. So the amplitude will decrease and the perpetual motion be reduced to perpetual immobility...
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The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
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– tfb
13 hours ago
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@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
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– Jhor
13 hours ago
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Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
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– tfb
12 hours ago
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A magnet strong enough to pull the metal up will be too strong to let it fall again.
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Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
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– JMac
7 hours ago
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If you would use static magnets for that, than while swinging back, the metal is closer to the magnet attracting it in opposite direction relative to its movement and thus takes more momentum away as it gains from the other magnet. Thus the next amplitude is lower. The result is, that your amplitude overall would even be damped relative to the free pendulum.
The accelerating force by the magnet on the far side is proportional to $cfrac{1}{(d+sin phi l)^2}$ and the deccelerating force by the magnet on the close side is proportional to $cfrac{1}{(d-sin phi l)^2}$.
Where d is the distance between pendulum mass and either magnet in rest state, $phi$ is the deflection angle and l is the pendulum length. Since the Magnets are identical, they carry the same prefactor. Thus resulting force adding to the positively accelerating gravitational force is proportional to
$$ cfrac{1}{(d+sin phi l)^2} - cfrac{1}{(d-sin phi l)^2} = cfrac{(d-sin phi l)^2 - (d+sin phi l)^2}{(d-sin phi l)^2 (d+sin phi l)^2} $$
As the denominator is now again the same for both contributing forces we can neglect it for the ongoing computation and result in the total magnetic force proportional to:
$$ (d-sin phi l)^2 - (d+sin phi l)^2 = d^2 - 2dsin phi l + sin^2 phi l^2 - d^2 - 2dsin phi l - sin^2 phi l^2 = -4dsin phi l $$
thus the oscillation given by the gravitational force is damped by the magnetic forces.
Edit: My original train of thoughts was wrong on this one. At least the calculations show that the oscillator is driven by a smaller force compared to the free oscillator (just gravity). Thus without any energy loss the 'magnetic oscillator' would just oscillate slower without any change in amplitude.
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I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
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– Jhor
13 hours ago
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I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
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– Patrik Puchert
11 hours ago
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No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
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– Jhor
11 hours ago
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I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
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– Patrik Puchert
11 hours ago
2
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I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
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– JMac
10 hours ago
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show 5 more comments
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Clearly, the device with magnets is not a perpetual motion machine, as explained by others. However why not consider the Universe, either one or an endless series with no beginning and no end, a perpetual motion machine?
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Replace the magnets with a lock that catches the pendulum and then releases a separate pendulum next to it. That pendulum then swings over and gets locked on the opposite side, releasing the other pendulum and so the process repeats. No need for magnets at all. The hard part is extracting work from the contraption. You aren't gonna save society by having some balls swinging back and forth. Even if you could extract work it wouldn't be enough to light a single home. You need to devise a system that scales to such massive levels as that it would replace a dam or power plant if built large enough.
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"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
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– azurefrog
4 hours ago
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@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
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– immibis
2 hours ago
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Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit.
You've neglected to account for the magnetic attraction as the pendulum bob goes back to its central position.
On the outwards leg, you are correct that the magnet's attraction will pull on the bob and give it more energy than it would have in the absence of the magnet. However, in the return leg, the pendulum bob is trying to get away from the magnet's attractive force, and this will claim back all of the additional energy.
(... if the system is perfect, that is. Real-world magnetic materials will show some amount of hysteresis, so the bob will lose slightly more energy on the way back than it gained on the way out.)
This type of mistake is quite common when you have a core dynamics which is known to be conservative, and still seems to be producing energy - you're just conveniently neglecting to take into account the parts of the cycle where that force performs work against your system. For a similar example in action, see What prevents this magnetic perpetuum mobile from working?.
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1
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Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
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– Florian Claaßen
12 hours ago
11
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Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
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– Emilio Pisanty
12 hours ago
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Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
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– Florian Claaßen
12 hours ago
2
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No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
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– Emilio Pisanty
12 hours ago
add a comment |
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Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit.
You've neglected to account for the magnetic attraction as the pendulum bob goes back to its central position.
On the outwards leg, you are correct that the magnet's attraction will pull on the bob and give it more energy than it would have in the absence of the magnet. However, in the return leg, the pendulum bob is trying to get away from the magnet's attractive force, and this will claim back all of the additional energy.
(... if the system is perfect, that is. Real-world magnetic materials will show some amount of hysteresis, so the bob will lose slightly more energy on the way back than it gained on the way out.)
This type of mistake is quite common when you have a core dynamics which is known to be conservative, and still seems to be producing energy - you're just conveniently neglecting to take into account the parts of the cycle where that force performs work against your system. For a similar example in action, see What prevents this magnetic perpetuum mobile from working?.
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1
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Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
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– Florian Claaßen
12 hours ago
11
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Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
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– Emilio Pisanty
12 hours ago
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Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
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– Florian Claaßen
12 hours ago
2
$begingroup$
No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
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– Emilio Pisanty
12 hours ago
add a comment |
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Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit.
You've neglected to account for the magnetic attraction as the pendulum bob goes back to its central position.
On the outwards leg, you are correct that the magnet's attraction will pull on the bob and give it more energy than it would have in the absence of the magnet. However, in the return leg, the pendulum bob is trying to get away from the magnet's attractive force, and this will claim back all of the additional energy.
(... if the system is perfect, that is. Real-world magnetic materials will show some amount of hysteresis, so the bob will lose slightly more energy on the way back than it gained on the way out.)
This type of mistake is quite common when you have a core dynamics which is known to be conservative, and still seems to be producing energy - you're just conveniently neglecting to take into account the parts of the cycle where that force performs work against your system. For a similar example in action, see What prevents this magnetic perpetuum mobile from working?.
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Now, in my head, if you give the pendulum a little impulse, it will swing up in one direction and get attracted by the magnet just a tiny bit.
You've neglected to account for the magnetic attraction as the pendulum bob goes back to its central position.
On the outwards leg, you are correct that the magnet's attraction will pull on the bob and give it more energy than it would have in the absence of the magnet. However, in the return leg, the pendulum bob is trying to get away from the magnet's attractive force, and this will claim back all of the additional energy.
(... if the system is perfect, that is. Real-world magnetic materials will show some amount of hysteresis, so the bob will lose slightly more energy on the way back than it gained on the way out.)
This type of mistake is quite common when you have a core dynamics which is known to be conservative, and still seems to be producing energy - you're just conveniently neglecting to take into account the parts of the cycle where that force performs work against your system. For a similar example in action, see What prevents this magnetic perpetuum mobile from working?.
answered 13 hours ago
Emilio PisantyEmilio Pisanty
84.2k22208423
84.2k22208423
1
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Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
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– Florian Claaßen
12 hours ago
11
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Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
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– Emilio Pisanty
12 hours ago
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Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
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– Florian Claaßen
12 hours ago
2
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No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
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– Emilio Pisanty
12 hours ago
add a comment |
1
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Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
$endgroup$
– Florian Claaßen
12 hours ago
11
$begingroup$
Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
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– Emilio Pisanty
12 hours ago
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Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
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– Florian Claaßen
12 hours ago
2
$begingroup$
No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
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– Emilio Pisanty
12 hours ago
1
1
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Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
$endgroup$
– Florian Claaßen
12 hours ago
$begingroup$
Thanks, I didn't think of that. Your absolutely right, but let me propose a solution to the problem: What if I replace the magnets with electromagnets which get powered by the dynamo that gets charged by the pendulum swinging. Now, what if the electromagnet gets powered when the pendulum is swinging towards it, but not powered once the pendulum swings away?
$endgroup$
– Florian Claaßen
12 hours ago
11
11
$begingroup$
Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
$endgroup$
– Emilio Pisanty
12 hours ago
$begingroup$
Then you will need to supply power to the electromagnet when you turn it on, and you will not be able to harvest equal amounts of power from the electromagnet when you turn it off. Let's get this straight: electromagnetism conserves energy. That's a theorem within the theory, and the only way you can get around it is by stepping away from electromagnetism. The fact that perpetual-motion machines don't work is not a "problem" in need of a "solution", and if that's what you're looking for, then this site is not the venue for it.
$endgroup$
– Emilio Pisanty
12 hours ago
$begingroup$
Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
$endgroup$
– Florian Claaßen
12 hours ago
$begingroup$
Right, right, thanks for the quick response. I was not looking for an actual perpetuum mobile but rather the mistake I made while thinking up the model. You have been of great help, thanks :)
$endgroup$
– Florian Claaßen
12 hours ago
2
2
$begingroup$
No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
$endgroup$
– Emilio Pisanty
12 hours ago
$begingroup$
No worries. When asking about perpetual-motion machines, you should keep firmly in mind the type of population that asks the majority of those questions. If your text reads the same way as the comments by that population (which your first comment here definitely satisfies), then you should be prepared for others' view of those questions to come through such a lens. For an example of how to step away from that tone, see the question I linked to.
$endgroup$
– Emilio Pisanty
12 hours ago
add a comment |
$begingroup$
Perpetual motion is impossible due to dissipation, or if you prefer second principle of thermodynamics and not to energy conservation.
If your analysis of the suggested setup was correct, you could create mechanical energy for free !
In first analysis, neglecting the (unavoidable) losses in the ferromagnetic medium, your system is conservative : what you have is a modified pendulum, where the confinement potential contains not only the gravitational part but also a magnetic component. Actually the magnetic force slightly decrease the recall torque that you would have with gravity alone, and the amplitude of motion will indeed be larger. But you nevertheless will have a turning point where the kinetic energy vanishes, and when going back, you will reach exactly the same angle for the turning point on the other side.
This correspond to an oscillator with constant amplitude, because losses have been neglected.
The sources of losses are at least : friction in air, friction on axe, ferromagnetic hysteresis, Foucault currents. So the amplitude will decrease and the perpetual motion be reduced to perpetual immobility...
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Jhor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
$endgroup$
– tfb
13 hours ago
$begingroup$
@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
$endgroup$
– Jhor
13 hours ago
$begingroup$
Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
$endgroup$
– tfb
12 hours ago
add a comment |
$begingroup$
Perpetual motion is impossible due to dissipation, or if you prefer second principle of thermodynamics and not to energy conservation.
If your analysis of the suggested setup was correct, you could create mechanical energy for free !
In first analysis, neglecting the (unavoidable) losses in the ferromagnetic medium, your system is conservative : what you have is a modified pendulum, where the confinement potential contains not only the gravitational part but also a magnetic component. Actually the magnetic force slightly decrease the recall torque that you would have with gravity alone, and the amplitude of motion will indeed be larger. But you nevertheless will have a turning point where the kinetic energy vanishes, and when going back, you will reach exactly the same angle for the turning point on the other side.
This correspond to an oscillator with constant amplitude, because losses have been neglected.
The sources of losses are at least : friction in air, friction on axe, ferromagnetic hysteresis, Foucault currents. So the amplitude will decrease and the perpetual motion be reduced to perpetual immobility...
New contributor
Jhor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
$endgroup$
– tfb
13 hours ago
$begingroup$
@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
$endgroup$
– Jhor
13 hours ago
$begingroup$
Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
$endgroup$
– tfb
12 hours ago
add a comment |
$begingroup$
Perpetual motion is impossible due to dissipation, or if you prefer second principle of thermodynamics and not to energy conservation.
If your analysis of the suggested setup was correct, you could create mechanical energy for free !
In first analysis, neglecting the (unavoidable) losses in the ferromagnetic medium, your system is conservative : what you have is a modified pendulum, where the confinement potential contains not only the gravitational part but also a magnetic component. Actually the magnetic force slightly decrease the recall torque that you would have with gravity alone, and the amplitude of motion will indeed be larger. But you nevertheless will have a turning point where the kinetic energy vanishes, and when going back, you will reach exactly the same angle for the turning point on the other side.
This correspond to an oscillator with constant amplitude, because losses have been neglected.
The sources of losses are at least : friction in air, friction on axe, ferromagnetic hysteresis, Foucault currents. So the amplitude will decrease and the perpetual motion be reduced to perpetual immobility...
New contributor
Jhor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Perpetual motion is impossible due to dissipation, or if you prefer second principle of thermodynamics and not to energy conservation.
If your analysis of the suggested setup was correct, you could create mechanical energy for free !
In first analysis, neglecting the (unavoidable) losses in the ferromagnetic medium, your system is conservative : what you have is a modified pendulum, where the confinement potential contains not only the gravitational part but also a magnetic component. Actually the magnetic force slightly decrease the recall torque that you would have with gravity alone, and the amplitude of motion will indeed be larger. But you nevertheless will have a turning point where the kinetic energy vanishes, and when going back, you will reach exactly the same angle for the turning point on the other side.
This correspond to an oscillator with constant amplitude, because losses have been neglected.
The sources of losses are at least : friction in air, friction on axe, ferromagnetic hysteresis, Foucault currents. So the amplitude will decrease and the perpetual motion be reduced to perpetual immobility...
New contributor
Jhor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 13 hours ago
New contributor
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answered 13 hours ago
JhorJhor
2045
2045
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Jhor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Jhor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
$endgroup$
– tfb
13 hours ago
$begingroup$
@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
$endgroup$
– Jhor
13 hours ago
$begingroup$
Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
$endgroup$
– tfb
12 hours ago
add a comment |
$begingroup$
The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
$endgroup$
– tfb
13 hours ago
$begingroup$
@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
$endgroup$
– Jhor
13 hours ago
$begingroup$
Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
$endgroup$
– tfb
12 hours ago
$begingroup$
The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
$endgroup$
– tfb
13 hours ago
$begingroup$
The key thing is that the forces are conservative, as you say. If the forces are conservative there is a well-defined potential energy at any angle $theta$ and that's basically all you need.
$endgroup$
– tfb
13 hours ago
$begingroup$
@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
$endgroup$
– Jhor
13 hours ago
$begingroup$
@TV I agree of course but it is worth to discuss in more detail the misconceptions involved in the OP. And magnets remains so mysterious, if not magical, for many people...
$endgroup$
– Jhor
13 hours ago
$begingroup$
Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
$endgroup$
– tfb
12 hours ago
$begingroup$
Oh yes, sorry, I only really added the comment as I was half-way through an answer which said that (now abandoned as yours is better).
$endgroup$
– tfb
12 hours ago
add a comment |
$begingroup$
A magnet strong enough to pull the metal up will be too strong to let it fall again.
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$endgroup$
1
$begingroup$
Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
$endgroup$
– JMac
7 hours ago
add a comment |
$begingroup$
A magnet strong enough to pull the metal up will be too strong to let it fall again.
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$endgroup$
1
$begingroup$
Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
$endgroup$
– JMac
7 hours ago
add a comment |
$begingroup$
A magnet strong enough to pull the metal up will be too strong to let it fall again.
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$endgroup$
A magnet strong enough to pull the metal up will be too strong to let it fall again.
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answered 8 hours ago
rhwhw4j645rhwhw4j645
1
1
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rhwhw4j645 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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rhwhw4j645 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$begingroup$
Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
$endgroup$
– JMac
7 hours ago
add a comment |
1
$begingroup$
Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
$endgroup$
– JMac
7 hours ago
1
1
$begingroup$
Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
$endgroup$
– JMac
7 hours ago
$begingroup$
Only if it were strong enough to pull up more than gravity pulls down. The question mentions that the magnet can't stick, which is what I believe that was getting at.
$endgroup$
– JMac
7 hours ago
add a comment |
$begingroup$
If you would use static magnets for that, than while swinging back, the metal is closer to the magnet attracting it in opposite direction relative to its movement and thus takes more momentum away as it gains from the other magnet. Thus the next amplitude is lower. The result is, that your amplitude overall would even be damped relative to the free pendulum.
The accelerating force by the magnet on the far side is proportional to $cfrac{1}{(d+sin phi l)^2}$ and the deccelerating force by the magnet on the close side is proportional to $cfrac{1}{(d-sin phi l)^2}$.
Where d is the distance between pendulum mass and either magnet in rest state, $phi$ is the deflection angle and l is the pendulum length. Since the Magnets are identical, they carry the same prefactor. Thus resulting force adding to the positively accelerating gravitational force is proportional to
$$ cfrac{1}{(d+sin phi l)^2} - cfrac{1}{(d-sin phi l)^2} = cfrac{(d-sin phi l)^2 - (d+sin phi l)^2}{(d-sin phi l)^2 (d+sin phi l)^2} $$
As the denominator is now again the same for both contributing forces we can neglect it for the ongoing computation and result in the total magnetic force proportional to:
$$ (d-sin phi l)^2 - (d+sin phi l)^2 = d^2 - 2dsin phi l + sin^2 phi l^2 - d^2 - 2dsin phi l - sin^2 phi l^2 = -4dsin phi l $$
thus the oscillation given by the gravitational force is damped by the magnetic forces.
Edit: My original train of thoughts was wrong on this one. At least the calculations show that the oscillator is driven by a smaller force compared to the free oscillator (just gravity). Thus without any energy loss the 'magnetic oscillator' would just oscillate slower without any change in amplitude.
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$endgroup$
$begingroup$
I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
$endgroup$
– Jhor
13 hours ago
$begingroup$
I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
$endgroup$
– Patrik Puchert
11 hours ago
$begingroup$
No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
$endgroup$
– Jhor
11 hours ago
$begingroup$
I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
$endgroup$
– Patrik Puchert
11 hours ago
2
$begingroup$
I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
$endgroup$
– JMac
10 hours ago
|
show 5 more comments
$begingroup$
If you would use static magnets for that, than while swinging back, the metal is closer to the magnet attracting it in opposite direction relative to its movement and thus takes more momentum away as it gains from the other magnet. Thus the next amplitude is lower. The result is, that your amplitude overall would even be damped relative to the free pendulum.
The accelerating force by the magnet on the far side is proportional to $cfrac{1}{(d+sin phi l)^2}$ and the deccelerating force by the magnet on the close side is proportional to $cfrac{1}{(d-sin phi l)^2}$.
Where d is the distance between pendulum mass and either magnet in rest state, $phi$ is the deflection angle and l is the pendulum length. Since the Magnets are identical, they carry the same prefactor. Thus resulting force adding to the positively accelerating gravitational force is proportional to
$$ cfrac{1}{(d+sin phi l)^2} - cfrac{1}{(d-sin phi l)^2} = cfrac{(d-sin phi l)^2 - (d+sin phi l)^2}{(d-sin phi l)^2 (d+sin phi l)^2} $$
As the denominator is now again the same for both contributing forces we can neglect it for the ongoing computation and result in the total magnetic force proportional to:
$$ (d-sin phi l)^2 - (d+sin phi l)^2 = d^2 - 2dsin phi l + sin^2 phi l^2 - d^2 - 2dsin phi l - sin^2 phi l^2 = -4dsin phi l $$
thus the oscillation given by the gravitational force is damped by the magnetic forces.
Edit: My original train of thoughts was wrong on this one. At least the calculations show that the oscillator is driven by a smaller force compared to the free oscillator (just gravity). Thus without any energy loss the 'magnetic oscillator' would just oscillate slower without any change in amplitude.
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$endgroup$
$begingroup$
I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
$endgroup$
– Jhor
13 hours ago
$begingroup$
I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
$endgroup$
– Patrik Puchert
11 hours ago
$begingroup$
No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
$endgroup$
– Jhor
11 hours ago
$begingroup$
I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
$endgroup$
– Patrik Puchert
11 hours ago
2
$begingroup$
I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
$endgroup$
– JMac
10 hours ago
|
show 5 more comments
$begingroup$
If you would use static magnets for that, than while swinging back, the metal is closer to the magnet attracting it in opposite direction relative to its movement and thus takes more momentum away as it gains from the other magnet. Thus the next amplitude is lower. The result is, that your amplitude overall would even be damped relative to the free pendulum.
The accelerating force by the magnet on the far side is proportional to $cfrac{1}{(d+sin phi l)^2}$ and the deccelerating force by the magnet on the close side is proportional to $cfrac{1}{(d-sin phi l)^2}$.
Where d is the distance between pendulum mass and either magnet in rest state, $phi$ is the deflection angle and l is the pendulum length. Since the Magnets are identical, they carry the same prefactor. Thus resulting force adding to the positively accelerating gravitational force is proportional to
$$ cfrac{1}{(d+sin phi l)^2} - cfrac{1}{(d-sin phi l)^2} = cfrac{(d-sin phi l)^2 - (d+sin phi l)^2}{(d-sin phi l)^2 (d+sin phi l)^2} $$
As the denominator is now again the same for both contributing forces we can neglect it for the ongoing computation and result in the total magnetic force proportional to:
$$ (d-sin phi l)^2 - (d+sin phi l)^2 = d^2 - 2dsin phi l + sin^2 phi l^2 - d^2 - 2dsin phi l - sin^2 phi l^2 = -4dsin phi l $$
thus the oscillation given by the gravitational force is damped by the magnetic forces.
Edit: My original train of thoughts was wrong on this one. At least the calculations show that the oscillator is driven by a smaller force compared to the free oscillator (just gravity). Thus without any energy loss the 'magnetic oscillator' would just oscillate slower without any change in amplitude.
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$endgroup$
If you would use static magnets for that, than while swinging back, the metal is closer to the magnet attracting it in opposite direction relative to its movement and thus takes more momentum away as it gains from the other magnet. Thus the next amplitude is lower. The result is, that your amplitude overall would even be damped relative to the free pendulum.
The accelerating force by the magnet on the far side is proportional to $cfrac{1}{(d+sin phi l)^2}$ and the deccelerating force by the magnet on the close side is proportional to $cfrac{1}{(d-sin phi l)^2}$.
Where d is the distance between pendulum mass and either magnet in rest state, $phi$ is the deflection angle and l is the pendulum length. Since the Magnets are identical, they carry the same prefactor. Thus resulting force adding to the positively accelerating gravitational force is proportional to
$$ cfrac{1}{(d+sin phi l)^2} - cfrac{1}{(d-sin phi l)^2} = cfrac{(d-sin phi l)^2 - (d+sin phi l)^2}{(d-sin phi l)^2 (d+sin phi l)^2} $$
As the denominator is now again the same for both contributing forces we can neglect it for the ongoing computation and result in the total magnetic force proportional to:
$$ (d-sin phi l)^2 - (d+sin phi l)^2 = d^2 - 2dsin phi l + sin^2 phi l^2 - d^2 - 2dsin phi l - sin^2 phi l^2 = -4dsin phi l $$
thus the oscillation given by the gravitational force is damped by the magnetic forces.
Edit: My original train of thoughts was wrong on this one. At least the calculations show that the oscillator is driven by a smaller force compared to the free oscillator (just gravity). Thus without any energy loss the 'magnetic oscillator' would just oscillate slower without any change in amplitude.
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Patrik Puchert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 5 hours ago
a CVn
642517
642517
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answered 14 hours ago
Patrik PuchertPatrik Puchert
663
663
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$begingroup$
I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
$endgroup$
– Jhor
13 hours ago
$begingroup$
I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
$endgroup$
– Patrik Puchert
11 hours ago
$begingroup$
No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
$endgroup$
– Jhor
11 hours ago
$begingroup$
I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
$endgroup$
– Patrik Puchert
11 hours ago
2
$begingroup$
I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
$endgroup$
– JMac
10 hours ago
|
show 5 more comments
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I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
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– Jhor
13 hours ago
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I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
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– Patrik Puchert
11 hours ago
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No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
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– Jhor
11 hours ago
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I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
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– Patrik Puchert
11 hours ago
2
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I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
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– JMac
10 hours ago
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I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
$endgroup$
– Jhor
13 hours ago
$begingroup$
I firmly disagree with your answer. The amplitude must neitger decrease nor increase when friction and other losses are neglected. See my answer.
$endgroup$
– Jhor
13 hours ago
$begingroup$
I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
$endgroup$
– Patrik Puchert
11 hours ago
$begingroup$
I don't think so. I don't know the exact term for the lets call it magneto attractive force, but it should scale as 1/r². with that in mind and for simplicity using small angle approximation we result in the magnetic force ~(1/(d+sin $phi$ l)²-1/(d-sin $phi$ l)²) again ~ (d-sin $phi$ l)²-(d+sin $phi$ l)² resulting in a net magnetic force proportional to -4dsin $phi$ l. Where d is the distance to either magnet for the undeflected pendulum, $phi$ is the defelction angle and l is the length of the pendulum. Thus the net force accelerating the pendulum is reduced by the magnets.
$endgroup$
– Patrik Puchert
11 hours ago
$begingroup$
No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
$endgroup$
– Jhor
11 hours ago
$begingroup$
No problem with that. Except that this expression goes not has the dimension of a force.Any wayyou ould have to add to the gravitational potential - mglcosphi a magnetic term in dlcosphi and you have a conservative potential giving rise to a periodic oscillation of constant amplitude. To understand your mistake in evaluating the work of this magnetic force, have a look to @EmilioPisanty's answer.
$endgroup$
– Jhor
11 hours ago
$begingroup$
I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
$endgroup$
– Patrik Puchert
11 hours ago
$begingroup$
I only said that the force contributed by the magnets if proportional to what I wrote. Of course gravity is there, but gravity is also there without magnets, so its not interesting when determining the difference. Of course there is some prefactor, which is the same for both magnets (given that they are identical) and thus again not of interest when determining the (scaling of) difference.
$endgroup$
– Patrik Puchert
11 hours ago
2
2
$begingroup$
I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
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– JMac
10 hours ago
$begingroup$
I suggest using Mathjax to format your equations, it is very hard to read the math here. Also, I don't really understand how adding another conservative force would change the amplitude for an ideal pendulum. The amplitude for such a pendulum depends completely on the starting conditions. If you started the magnetic pendulum at the same angle as a only-gravity one, they should both still end up in the same place on each period when you assume no energy losses. I don't see how you would expect a change in amplitude given that both are conservative systems.
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– JMac
10 hours ago
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show 5 more comments
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Clearly, the device with magnets is not a perpetual motion machine, as explained by others. However why not consider the Universe, either one or an endless series with no beginning and no end, a perpetual motion machine?
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add a comment |
$begingroup$
Clearly, the device with magnets is not a perpetual motion machine, as explained by others. However why not consider the Universe, either one or an endless series with no beginning and no end, a perpetual motion machine?
$endgroup$
add a comment |
$begingroup$
Clearly, the device with magnets is not a perpetual motion machine, as explained by others. However why not consider the Universe, either one or an endless series with no beginning and no end, a perpetual motion machine?
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Clearly, the device with magnets is not a perpetual motion machine, as explained by others. However why not consider the Universe, either one or an endless series with no beginning and no end, a perpetual motion machine?
answered 1 hour ago
Steve d'ApolloniaSteve d'Apollonia
6314
6314
add a comment |
add a comment |
$begingroup$
Replace the magnets with a lock that catches the pendulum and then releases a separate pendulum next to it. That pendulum then swings over and gets locked on the opposite side, releasing the other pendulum and so the process repeats. No need for magnets at all. The hard part is extracting work from the contraption. You aren't gonna save society by having some balls swinging back and forth. Even if you could extract work it wouldn't be enough to light a single home. You need to devise a system that scales to such massive levels as that it would replace a dam or power plant if built large enough.
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4
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"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
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– azurefrog
4 hours ago
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@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
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– immibis
2 hours ago
add a comment |
$begingroup$
Replace the magnets with a lock that catches the pendulum and then releases a separate pendulum next to it. That pendulum then swings over and gets locked on the opposite side, releasing the other pendulum and so the process repeats. No need for magnets at all. The hard part is extracting work from the contraption. You aren't gonna save society by having some balls swinging back and forth. Even if you could extract work it wouldn't be enough to light a single home. You need to devise a system that scales to such massive levels as that it would replace a dam or power plant if built large enough.
New contributor
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
4
$begingroup$
"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
$endgroup$
– azurefrog
4 hours ago
$begingroup$
@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
$endgroup$
– immibis
2 hours ago
add a comment |
$begingroup$
Replace the magnets with a lock that catches the pendulum and then releases a separate pendulum next to it. That pendulum then swings over and gets locked on the opposite side, releasing the other pendulum and so the process repeats. No need for magnets at all. The hard part is extracting work from the contraption. You aren't gonna save society by having some balls swinging back and forth. Even if you could extract work it wouldn't be enough to light a single home. You need to devise a system that scales to such massive levels as that it would replace a dam or power plant if built large enough.
New contributor
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Replace the magnets with a lock that catches the pendulum and then releases a separate pendulum next to it. That pendulum then swings over and gets locked on the opposite side, releasing the other pendulum and so the process repeats. No need for magnets at all. The hard part is extracting work from the contraption. You aren't gonna save society by having some balls swinging back and forth. Even if you could extract work it wouldn't be enough to light a single home. You need to devise a system that scales to such massive levels as that it would replace a dam or power plant if built large enough.
New contributor
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
user224619user224619
1
1
New contributor
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user224619 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
$begingroup$
"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
$endgroup$
– azurefrog
4 hours ago
$begingroup$
@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
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– immibis
2 hours ago
add a comment |
4
$begingroup$
"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
$endgroup$
– azurefrog
4 hours ago
$begingroup$
@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
$endgroup$
– immibis
2 hours ago
4
4
$begingroup$
"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
$endgroup$
– azurefrog
4 hours ago
$begingroup$
"The hard part is extracting work from the contraption" -- The harder part is rewriting physics.
$endgroup$
– azurefrog
4 hours ago
$begingroup$
@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
$endgroup$
– immibis
2 hours ago
$begingroup$
@azurefrog A (approximately) perpetual motion machine is relatively trivial if you don't have to extract work from it. Any object orbiting the sun, for example, is one.
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– immibis
2 hours ago
add a comment |
protected by David Z♦ 1 hour ago
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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Comments are not for extended discussion; this conversation has been moved to chat.
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– David Z♦
1 hour ago