12
$\begingroup$

I know that this machine does not work, via thermodynamics. I am asking for an analysis in terms of mechanics and magnetism.

Anyway, so here is the machine:

A magnetic brings a ball up a hole, and then drops it through a hole. http://cabinetmagazine.org/issues/51/pertpetual2.jpg

The magnet (the red ball) pull the ball up the ramp, and then it drops the ball through the hole, which then rolls down, and go ups the ramp again. Thermodynamics shows that this can not work.

From a mechanics and magnetism perspective, what happens when you do this, and why can't it happen?

I have a source saying that it would work if it were frictionless and we didn't try to extract energy from it (here), so in a certain sense, it is very close to possible (they didn't provide a full analysis though.)

Another Image:

$\endgroup$
  • 3
    $\begingroup$ I like the sketch before your edit a bit better actually. $\endgroup$ – Bernhard Jun 7 '15 at 18:04
  • 5
    $\begingroup$ Why should it work, i.e. why do you think that the ball doesn't simply get stuck to the magnet? $\endgroup$ – ACuriousMind Jun 7 '15 at 18:08
  • 1
    $\begingroup$ If you assume frictionless and do not extract energy, there are a lot of ways to make a perpetual machine: the curved piece put horizontal, and the ball going forward and back again will work. But if you are into it, a simple pendulum will do just fine, no need for a magnet at all. $\endgroup$ – rodrigo Jun 7 '15 at 22:07
  • 6
    $\begingroup$ IIRC, "would work if it were frictionless and we didn't try to extract energy from it" is true of almost every proposed perpetual motion machine (which are often dressed up pendulums or spinning wheels, which would also keep going forever under those conditions) $\endgroup$ – Random832 Jun 8 '15 at 2:29
  • 5
    $\begingroup$ Real magnets are very strong when touching the object and quite weak at a distance. If the magnet was powerful enough to pull the small ball at that distance by the time it got to the hole at the top of the ramp it would be accelerating so fast that it would jump off the ramp and stick to the magnet. The other possibility is that the magnet is not strong enough to pull the ball in the first place. Either way you get equilibrium very quickly! $\endgroup$ – CJ Dennis Jun 8 '15 at 11:38
23
$\begingroup$

There is no problem assuming that the ball will fall trough the hole. Even is the magnetic force is large, it only needs to be larger than the gravity component along the inclined surface. This component is $mg \cos \theta$. Once the ball gets to the hole the gravity felt by the ball increases to mg, so it can happens that the ball that initially went up now goes goes down the hole.

However, notice that as the ball moves downward the gravity force starts to diminish again (is becoming more horizontal also the magnetic force decreases as the ball moves away from the magnet. If you make a graph about both forces you will find that there is always a point on the ball's path when both forces have the same magnitude but opposite sign. That will be the equilibrium position of the ball, where it will stay at rest if you extract all its initial kinetic energy. Thus it is not a perpertum mobile after all.

$\endgroup$
  • 1
    $\begingroup$ Actually you have to address the friction of the ball on the ramps, because that is the main reason it is not a perpetum mobile. The fact it have a equilibrium position is not the reason. $\endgroup$ – Mindwin Jun 8 '15 at 13:18
  • 1
    $\begingroup$ @Mindwin You are not correct, using your criteria a frictionless pendulum would be a perpetum mobile. $\endgroup$ – user66432 Jun 8 '15 at 14:34
  • 1
    $\begingroup$ Wouldn't it be PM if it is completely frictionless? Is this answer wrong? physics.stackexchange.com/a/69014/48721 $\endgroup$ – Mindwin Jun 8 '15 at 15:33
  • 6
    $\begingroup$ yes, a perpetual motion machine is not one that keeps moving forever in absence of friction, but one that still keeps moving when you extract work (an unlimited amount) from it. $\endgroup$ – user66432 Jun 8 '15 at 15:37
  • $\begingroup$ @Mindwin Nothing is frictionless....except nothing. $\endgroup$ – J... Jun 9 '15 at 2:13
37
$\begingroup$

If the magnet is powerful enough to pull the ball up from the bottom of the ramp, the force on it will be quite strong at the top of the ramp.

If so, why would the ball drop through the hole? The pull from the magnet will overwhelm gravity.

Even if you constructed one where the ball could fall through the upper hole, I don't see any reason why it should go through the lower hole. If the magnet can pull it up the ramp, the pull from the magnet should prevent it from reaching the hole on the lower ramp.

$\endgroup$
  • 1
    $\begingroup$ To extend your first paragraph: While going up the ramp in addition to the growing magnetic force the ball will gain linear momentum pointing towards the magnet. $\endgroup$ – pabouk Jun 9 '15 at 9:23
12
$\begingroup$

If I were designing the experiment

  1. I would make the inside surfaces of the ramp out of of mu metal to shield the ball once it falls in the hole, otherwise a strong magnet will be pulling it back up the lower incline.

  2. I would use an iron ball with a smooth glass coating to reduce friction

  3. I would use a glass upper ramp , again to reduce friction

  4. A judicious use of mu metal could shadow the magnetic field as the ball reaches the upper hole so as to be sure to drop.

It might work for some time if correctly designed. Suppose it does, certainly no energy can be extracted from the system . Even if the friction were zero there would be loss of the magnetization of the original magnet over time: to pull the ball energy must be supplied to the iron ball's magnetic domains and this over time will demagnetize the red ball. Of course all the other considerations of friction and impact energy loss as it falls (maybe etc, just these two come to mind) will lead the ball to finally stop in the lower ramp.

Here is a different kind of perpetual motion, where thermodynamics is more evident.

$\endgroup$
  • 3
    $\begingroup$ Mu metal isn't cavorite - the magnetic field would go through the hole and around the sides. $\endgroup$ – Random832 Jun 8 '15 at 2:26
  • 1
    $\begingroup$ @Random832 yes it would but the strength from the hole would be much/enough diminished. without mu metal the field will be very strong in the beginning of the curved ramp and will pull the ball up against gravity, to similar strength as pulling it up the top ramp: en.wikipedia.org/wiki/Magnet#Force_between_two_bar_magnets . The force is roughly 1/r^2 for dipoles, but the spill over field from the hole will not be a dipole . $\endgroup$ – anna v Jun 8 '15 at 3:00
2
$\begingroup$

Newtonian Physics certainly precludes the existence of any 'truly' perpetual motion. Under certain conditions, we can achieve near perpetual motion by judicious application of principles and then form questions like "Although no work can be removed from the system, is the lifetime of the magnetic force in a magnet sufficient that running for several years is an acceptable answer?"

Regarding the device device shown in the question, which is a modification of John Wilkins' device similar to this one, the problems inherent in Wilkin's machine are compounded with the design shown here. Gravity is the overwhelming force to overcome in this experiment, friction plays only a small part, but we also can't overlook the current induced in the movement of the ball, which does eventually become the principle force degrading the perpetuation of motion. Here however, if the magnet was strong enough to overcome the inertia and the gravitational force of the weight on an incline, it would have sufficient energy to attach to the magnet at the top. Even without friction, and let's say that we started the ball at the top to give it an initial kinetic force, when it reached the 2nd hole, the magnet would not only have to overcome gravity, but the overall motion away from the magnet at the top, as it passed through the 2nd hole it would behave more like a skier doing a ski jump and fly away from the machine. So the magnet would also have to over come this motion, and reverse the balls direction as it is now also moving away with a velocity equal to the kinetic energy from dropping the ball minus friction.

Wilkins overcomes this with an additional ramp at the bottom, which by converting the linear motion into angular motion which reverses the direction of the ball and conserves that kinetic motion to move the ball partway up the ramp. Even so, this is not enough to overcome gravity, and friction, and the counter force of an induced current to enable it to pull the ball all the way up, while being weak enough to allow the ball to drop through the hole. I believe, but have not tested or proven, that perhaps a reduction of gravity would make this machine feasible, like on the moon. It appears also, that because the ball is made of metal that is attracted to the magnet, the forces in the ball can set up a induced current counter field, due to the material the ball is made of and its in the field. A reverse lenz effect is present that would also contribute a force the magnet would have to overcome. This would require a superfine balance between a completely attractive force, and the opposing forces. The machine looks very promising; but, on a very subtle level has counter forces from gravity, friction, and a reverse lenz effect which counter the kinetic energy built up in the balls motion which pretty much stop the action from perpetuating fairly quickly.

In the video I link, the original author of the video overcomes this effect through the pulsing of the strength of an electromagnet hidden in the base holding the permanent magnet. By turning the electromagnet off at the right moment either manually or by inductive sensing, the ball drops. So, I agree, it is very close to possible; but, it just isn't possible. I have a sense that being able to tweak gravity might be enough to make the device function under the right gravitational force, but have a real concern that eddy currents in the metal ball would provide the eventual force that would counter any possibility of it working.

As a magnetic field problem, due to Maxwell's laws and Lenz' law, the problem is a lot more complex mathematically than it looks, we tend to overlook the ball's effect on the magnetic field as it would be moving and shifting the flux density which would induce a current which would affect the movement and flux density in opposition to the movement.

$\endgroup$
  • $\begingroup$ Did you mean precludes or allows in the first sentence? $\endgroup$ – user1717828 Jun 8 '15 at 0:26
  • $\begingroup$ Precludes, as in impossible. Truly, as in totally perpetual. Even planetary orbits decay eventually, Picking nits over a general concept that something that will run for a lifetime is considered perpetual enough. Not even the universe, as a closed system, is truly perpetual. The same holds for the concept of "free" energy, in two ways. One, free as in costs nothing. And free as in able to do work so cheaply to almost cost nothing. 1st: improbable, 2nd exists in magnets and charged batteries. So the concept is maligned. $\endgroup$ – Cyberchipz Jun 9 '15 at 3:08
  • $\begingroup$ youtube.com/watch?v=V70w3cxDJIM $\endgroup$ – mcodesmart Jun 10 '15 at 1:16
1
$\begingroup$

The very reason for the ball to move up is the presence of the magnet. So, as long as the magnet is present in its place, it will not allow the ball to move in the downward direction. Conversely, if the magnet can allow the ball to go in the downward direction, it would be incapable of pulling it upwards. Therefore the cyclic motion is impossible.

Many have answered the question earlier correctly, but in parts. Perpetual motion of the first kind implies extraction of work from a body which keeps moving on its own, in a cycle . Since the body keeps on moving on its own, the energy we extract comes free with no input (leading to violation of the law of conservation of energy). For the sake of completeness - perpetual motion of the second kind says : when the input energy is heat, we cannot extract an equal quantity of energy in any other form as out put, using even an ideal device (no friction etc) that works in cycles. The design details of the device don't alter the conclusion.

Even in the diagram shown, the essentials are just the vertical component of the force due to the magnet and that due to gravity . Gravitational force can be assumed to be constant irrespective of the position of the ball on its motion in the cycle.

We may add, in the absence of friction, simple harmonic motion rules, and, a pendulum oscillates for ever, an ideal spring loaded mass oscillates for ever and so on. The moment we try to extract work, the oscillations decrease in amplitude and finally cease, when the initially stored energy got extracted.

Consider the lowest point, A, on the cyclic path. Consider a point B to the left, and a point B' vertically below B on the curved path. Th ball will move from A to B only if the vertical component of the magnetic force is greater than the gravitational force 'mg'. Similarly, it will move from A to B' if the vertical component of the magnetic force is greater than mg. The vertical component of the magnetic force at B and B' is greater than the vertical component of magnetic force at A. Consequently, if the ball moves from A to B it necessarily moves from A to B'. On the other hand, if the ball moves from B to A then, and only then, it would move from B' to A.

Alternately, you can consider the highest point, H, on the cyclic path. If the vertical component of the magnetic force is greater than the gravitational force 'mg' at B, the ball will move from B to H, if not, the ball will move from H to B. So is the case with motion between B' and H. Consequently, if the ball moves from H to B it necessarily moves from H to B'. On the other hand, if the ball moves from B to H then it would necessarily move from B' to H.

$\endgroup$
  • $\begingroup$ as the answer which is marked as correct suggest, your first paragraph is wrong. The effective force of gravity changes because of the angle of the two ramps. $\endgroup$ – Noldig Jun 10 '15 at 12:15

protected by Qmechanic Jun 7 '15 at 23:07

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).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.