# Why wouldn't this device work?

1. Sep 3, 2006

### GotMex?

I wasn't sure where to post my question but it seems like it would belong here. I was hoping for an anti-physics category or something hehe. I apologize in advance if I was wrong. Anyway, I've been reading about perpetual motion machines and I understand the physics behind them and why they are impossible.

I found a lot of examples of machines and enjoy figuring out why they won't work. There is just one example I found that I'm still not entirely sure why it won't work. Here's a link to an image describing it.
http://www.kilty.com/graphics/magnet.gif

Wikipedia gave it as an example but didn't give an explanation for why it wouldn't work. The other sites I found gave confusing explanations as well. Anyone have any experience with this example and can tell me why it wouldn't work? Thanks.

2. Sep 3, 2006

### Danger

It's a blatent violation of the 2nd law of thermodynamics. There is no way that the top magnet will be able to pull the ball back up.

3. Sep 3, 2006

### DaveC426913

It's a particualry bad example (despite scoring points for simplicity) because the flaw is so obvious: why would the ball fall down in the first place?

4. Sep 3, 2006

### GotMex?

That's sort of what I was thinking. If the magnet is strong enough to pull it up the ramp, then why wouldn't it be strong enough to keep it from falling.

5. Sep 3, 2006

### Danger

Bingo, buddy! And don't let anyone try to tell you that it's an electromagnet that only comes on at certain times, because it still wouldn't be practical without an external source of energy for the magnet.

6. Sep 4, 2006

### Tido611

If the ball started at the bottom of the ramp, in the little holder and it began to move due to the magnetic force from the magnet then that means that the force of gravity is weaker then the magnetic force moving the ball closer and closer to the magnet, also increaseing the magnetic force so there is no chance the ball would make it down the hole in the top of the ramp.

7. Sep 4, 2006

### DaveC426913

In defense of the poor PM device, I think the rationale is thus:

The ball falls down the sloped ramp. Because the ramp is extremely steep at the top, the ball is virtually in freefall, enough to overcome the magnetism.

Once at the bottom though, the ball is stopped abruptly, losing all its momentum.

Now, since the ball is supported on the diagonal ramp, it doesn't require as much effort for the magnet to lift it.

At least, that's the rationale. It's still flawed of course.

8. Sep 4, 2006

### sdemjanenko

It would work except once the ball falls the lower ramp is steeper so in order to meet up with the straight ramp it must eventually have a slope less than that of the straight ramp. At the point, or even before it reaches the same slope the ball would get caught by the magnet and stuck in place.

9. Sep 4, 2006

### leright

yeah, the ramp is steeper on the downward ramp than the upward ramp.

10. Sep 5, 2006

### sdemjanenko

its only steeper for a portion, then it turns less steep since its a curve, not straight

11. Sep 5, 2006

### DaveC426913

Yes, but by that point the ball has plenty of momentum.

12. Sep 5, 2006

### rcgldr

If the path was a loop, and there was no loss of energy to friction, then you wouldn't need the magnet. The only near frictionless systems are orbits of planets, moons, and stars, since space is almost a pure vacuum. The other scenario is current flowing through a loop of superconducting wire.

13. Sep 5, 2006

### Gokul43201

Staff Emeritus
No experience with this example, but I can tell you why it will not work (only to subsequently find someone else come up with a much simpler and more elegant explanation).

It will not work because you can not build the system without friction. In the absence of friction, it (or some form of it) would work just fine, and as pointed out by Jeff, wouldn't even need the magnet.

However, the magnet does not "offset" the losses due to friction.

Quick reason: the magnetic field is conservative, unlike friction.

Longer explanation:

Magnetic and gravitational fields are conservative, i.e, the work done by these fields is independent of the path. As a result of this, the work done by these fields on a particle, over the course of a loop is zero. (See this by writing the path integral from A to A as the sum of two path integrals from A to B and again from B to A; these integrals are equal in magnitude and opposite in sign, irrespective of the actual paths from A to B and back, and cancel each other off). So, after each loop, while there's a loss of energy from dissipative forces, there is no gain of energy from the magnetic field.

Still longer, and more rigorous proof:

1. Pick any point (call it A) where the velocity of the ball is v(A;n) during the n'th looping of the loop.

2. Watch the ball do a loop (assuming it can; if the ball can not do a loop there's nothing left to prove) and come back to A. During the course of this loop, the total work done by the conservative forces (gravitational + magnetic) is zero. The work done by the non-conservative forces (friction, air resistance), must hence equal the loos of KE of the ball (ref: Work-Energy Theorem). The ball thus has a smaller velocity when next it arrives at A. i.e, $v(A;n+1) \leq v(A;n)$

3. One could still argue that as the ball gets slower with each successive loop, the loss of KE also gets smaller, and the velocity asymptotically approaches a terminal value. However, since air resistance is proportional to velocity, a non-zero velocity will imply a non-zero drag and hence a non-zero loss to the KE. So, the only possible asymptotic value, if one exists, is zero. (For the specific case of the ball on the ramp/hoop, the loss of speed is faster than asymptotic. One can show that there's a non-zero lower bound on the work done by rolling friction, and hence, a finite upper limit on the number of loops before the KE reaches zero at A).

4. Now we make use of a specific flaw in this system. Since the magnetic field has a spatial variation and the gravitation field does not (or it has a much smaller spatial variation), we can safely conclude that that the net field is not zero everywhere on the loop. This in turn implies that the net potential energy is not a constant during the motion, and specifically, there must exist a pair of spots where the net PE are respectively a maximum and a minimum for the loop.

5. If the PE at A is smaller than the maximum PE by some P' > 0 (which we can ensure by our choice of A), we need a minimum KE at A given by: KE(A,min) = PE(max) - P', which then imposes a minimum speed required at point A, v(A,req). Now since v(A) at best approaches zero asymptotically (with the number of loops completed, n), we can always find an n=N when v(A;N) is smaller than any chosen number. Specifically, there exists some N, where v(A;N) < v(A,req). At this N, the ball has insufficient KE to reach the point of maximum PE and hence fails to complete the loop.

Since N is finite the motion is not perpetual.

Last edited: Sep 5, 2006
14. Sep 5, 2006

### DaveC426913

Uh, can you just elaborate on this? You're saying that without friction, this device would operate just fine? How do you figure that? Gravity is still in effect.

15. Sep 6, 2006

### rcgldr

Similar to an orbiting object, if the ball could roll in a loop without friction, and without any extrenal forces it would roll in the loop forever at a constant speed. Adding gravitational or magnetic forces would vary the speed, but without friction losses, the total energy would be conserved, and you'd just have variation between potential and kinetic energy.

I don't the device as pictured would work, even without friction, but if the ramp curves were modified to be more loop like, it could work. Instead of a hole, if there was a top to the ramp to guide the ball downwards, and at the bottom, a guide to loop the ball back upwards, then it could work, if there was no friction. If a hole at the top were to be used, then the hole would need to be far enough away from the magnet that it wouldn't stop the ball. The shape of the bottom ramp would have to loop the ball back around and upwards on the upper ramp.

Last edited: Sep 6, 2006
16. Sep 6, 2006

### J77

If the magnet were powerful enough to pull the ball back up the slope (with enough strength to easily get past the hole)...

and if the hole were big enough for the ball to fall down without skipping over the hole...

and if gravity overcame the magnetic pull and the ball could drop to the start...

Surely, the only way for this motion not to be perpetual is that the magnet would eventually lose strength!

I don't see how friction comes into the argument unless the magnet induces a field such that the ball only just makes it to the hole when the set-up is new - or, likewise, after the magnet loses strength to be at the point where the ball just makes it to the hole.

17. Sep 6, 2006

### bomba923

As explained earlier, magnetism is a conservative force and thus performs zero net work on any mass traveling in a closed path (i.e., where its net displacement is zero). To offset the negative work done unto the mass by the non-conservative force of friction, the magnet must perform positive work unto the mass. Since magnetism is a conservative force, this simply cannot occur within a closed path.

Last edited: Sep 6, 2006
18. Sep 6, 2006

### J77

Is it a closed path?

19. Sep 6, 2006

### bomba923

Isn't it obvious?
The ball starts at the bottom of the ramp. It travels up the ramp and drops through a hole, after which it rolls down a curve back to...the bottom of the ramp.

The starting and ending positions coincide, and hence you have a closed path
(as the displacement between two coinciding points is always zero).

Last edited: Sep 6, 2006
20. Sep 6, 2006

### J77

No, it's not obvious to me that the magnetic field acts in a continuos way along this path.

ie. once the ball drops through the hole - would it not be possible to shield it form the magnetic field before it appears at the bottom of the ramp once more?