# B Weighted wheel 2nd question

1. Aug 30, 2016

### cameron1

Good afternoon and thank you all that helped with the other wheel. It did in fact loose energy from the slide as the position once the movement happened, the energy was lost as the leverage was lower once the movemnt happened and therefore not recoverable.

Which leads me to this wheel and question.
As the wheel rotates, very slowly as to have a minimal centrifugal centripetal force affecting it, the weight in the 10 through 2 o clock position will compress the spring, as the wheel rotates further the spring un-compresses and pushes the weight back to the outside of the wheel.
Question - Does this wheel require more energy to turn when compared to the identical wheel moving at the same speed (rpm) with the weights fixed to the outside so they can not move?

My thoughts, no, since the energy (leverage) is lost at the top but stored in the spring and recovered on the other side as it goes back down. I am wondering if there is energy loss due to the compression of the spring, loss of leverage temporarily.
Thanks for any input. I understand these are pretty mundane questions and all revolve around newtons laws of thermodynamics but some clarity or guidance and confirmation would be great.

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2. Aug 30, 2016

### jbriggs444

You do not lose energy to "leverage".

As I wrote in the previous thread, the only discrepancy between the effort required to turn the wheel with a simple mechanism and the effort required to turn the wheel with a complex mechanism must be due to irreversible effects. Friction free slides, ideal springs and rigid levers on ideal bearings do not suffer from irreversible effects. Accordingly, there can be no increase in effort required.

As others wrote in the previous thread, in the real world, no mechanisms are perfect. You will suffer from irreversible losses. In general, the more complicated the mechanism, the worse the losses. But do not look for those losses in leverage. Look for them in friction, vibration, impact and air resistance.

3. Aug 30, 2016

### Staff: Mentor

Due to conservation of energy, any quality built wheel will require approximately zero energy to turn.

4. Aug 30, 2016

### cameron1

regardless to what happens inside the wheel? My understanding is the same, almost zero energy to "turn" the wheel once it is up to speed, minus bearings, wind and friction. My focus is on what happens to a weighted wheel, is that still the answer, because it is a wheel and whatever happens on one side must counter act on the other side...
The energy to turn the wheel stays the same regardless to what is happening inside the wheel.
That is my understanding.

5. Aug 30, 2016

### jbriggs444

It depends on what happens inside the wheel certainly. The hopper on a rotary cement mixer takes energy to turn because the blades and paddles turn the mixture and viscous friction dissipates the input energy. Let the cement harden and it takes no energy to turn (though will be badly unbalanced).

If there are no dissipative losses then mechanical energy is conserved. One need not even consider what happens on one side of the other. Mechanical energy is (except for such losses) conserved regardless.

This is very close to a rules violation on the grounds that you are describing a perpetual motion machine. The fact that you are running it in reverse is the only thing saving you.

6. Aug 30, 2016

### cameron1

That's interesting that you point that out. Besslers wheels and these "devices" is what I was evaluating. Not that they could ever work, but do they actually require more work to turn because of what was happening inside them.

7. Aug 30, 2016

### Staff: Mentor

Correct*.

*Ignoring friction and other dissipation losses, which seems appropriate here.

8. Aug 30, 2016

### skatche

Strictly speaking, what you've designed here is a wheel with a lower moment of inertia than if the weights were fixed to the outer edge. So you'll actually have an easier time accelerating and decelerating the wheel -- at low speeds. At high speeds, the weights will be pushed out to the edge anyway by centrifugal effects. Interestingly, the same can be said of your previous wheel design: that one would actually dissipate energy, so it would require a constant input to keep moving, but again, that effect disappears at high speeds due to centrifugal forces.

How does the lower moment of inertia affect the total energy input to bring the wheel up to a certain (large) speed? At low speeds the moment of inertia is lower, and at high speeds it's the same as in the fixed-weight case; since it's never greater, it seems as though the total energy input might be lower. But my mathematical intuition is that it should turn out the same either way; after all, the end product looks the same. Still not sure how to justify or refute that without some calculations.

Last edited: Aug 30, 2016
9. Aug 30, 2016

### jbriggs444

One of the useful bits about energy conservation is that it allows you to skip right past the messy calculations to the result that you already know must be true.

10. Aug 30, 2016

### A.T.

What is "energy to turn the wheel"? Even if the inner stuff is lossless, kinetic energy of the weights might build up over time, and that energy needs to be provided from outside.

11. Aug 30, 2016

### cameron1

yes, when I say energy to turn the wheel, it could be generic, a motor, wind, rolling down a hill... some sort of energy is needed to turn it.
skatche- regarding the centrifugal effects. yes once rpm is up a bit, no movement would occur as the weights or anything that can move will move to the outer edge.
But I am observing the effects of what happens at the specific rpm where cf is not high enough to keep the weights stuck to the outside, slow enough to have gravity have a moment to move the weights, if a little. I am trying to evaluate if at that specific rpm when compared to the wheel where the weights would not move, is the energy required to move it different at all.
I make the same evaluation, when we keep the rpm at that level and the weights have a moment to move, the wheel may in fact turn slightly easier that compared to if the weights were fixed. I am guessing that it would be somewhere between the two values if one measured with the weights fixed to the outside and fixed the weights to the inner most position. But I don't want to confuse acceleration here, just the input to keep it moving.

12. Aug 30, 2016

### skatche

Right, but I mean, already I'm having second thoughts. It seems at low angular speeds the system has a lower moment of inertia -- I didn't check it carefully, it was a bit of fudging on geometric observations. However, for the energy input to be the same in both cases, there would need to be speeds at which the system has greater moment of inertia than the fixed-weights case (it might even depend on the configuration of the springs at each moment). I wonder if that might have to do with the fact that the springs will resonate at a certain frequency, or if my initial assumption is even correct.

Last edited: Aug 30, 2016
13. Aug 30, 2016

### skatche

Cameron, with the exception of negligible effects due to small differences in aerodynamics, I think it would be just as easy to keep your system running at a given speed as it would to keep the fixed-weight wheel running. The question is, "how much energy is lost to air resistance?" and that depends only on the shape of the object. There's also friction at the axle, etc., but that'll be the same in both cases. The only real difference would be due to friction in the weight-shafts, so lube those babies up.

Last edited: Aug 30, 2016
14. Aug 30, 2016

### skatche

Careful -- kinetic energy is measured relative to a stationary reference frame, not the spinning frame of the wheel. At low speeds, the weights with springs will have much lower kinetic energy than the fixed weights, which you can see by the fact that they're traversing a tighter circle.

(I hope multiple-posting isn't frowned on here.)

15. Aug 30, 2016

### Staff: Mentor

No. Energy is required to accelerate it, but not to turn it at constant speed. Newton's first law tells us that the force/energy required to keep an object in a state of constant motion (including rotation) is zero.

Again: if there are no friction or other such losses, the energy required to turn the wheel is zero.

16. Aug 31, 2016

### A.T.

The moving weights aren't traversing circles relative to the stationary reference frame you insist on.

17. Aug 31, 2016

### A.T.

If the moment of inertia changes, due to the weight movement, maintaining a constant speed might require energy input / or release energy.

18. Aug 31, 2016

### skatche

You may be right -- it may be an ellipse or some complicated curve. But the basic point is the loop they traverse is shorter than the loop they'd traverse if they were fixed to the inner edge of the wheel.

19. Aug 31, 2016

### A.T.

There is no reason to assume that the complicated curve will always be shorter than the circle which fixed weights would traverse.

20. Aug 31, 2016

### skatche

I can see how this may explain the unusually high moment of inertia at certain speeds -- mainly at resonance, I'd wager -- but at low speeds it's plainly obvious how the weights will travel.

21. Aug 31, 2016

### A.T.

Your intuition is based on the lossy real world. But if one assumes no friction and perfectly elastic collisions, the kinetic energy of the weights can build up quite quickly, even at low rotation speeds.

22. Aug 31, 2016

### skatche

That's simply not the case -- in the idealised, frictionless case the wheel can spin in perpetuity without any energy input whatsoever, hence no increase in the kinetic energy of the weights. At low enough speeds -- lower than resonance -- the weights will travel a nice gentle circle within the larger circle of the wheel.

Last edited: Aug 31, 2016
23. Sep 17, 2016

### physicsBabyMetal

Unless the wheel is unbalanced from the very beginning and at all time, if not any balanced wheel will eventually came to a stop by friction.

24. Sep 18, 2016

### CWatters

This thread has correctly concluded that conservation of energy is alive and well. I think we can arrive at the same answer another way.. After one revolution the system is back where it started even if the masses have moved on some complicated path in the meantime. However gravity is a conservative field so work done is independent of the path.

So no matter how complicated the path if it ever returns to the starting position the work done will be zero.

The same approach can be applied understand similar magnet based machines (yes I know magnetic fields aren't really conservative or non-conservative) that try to use "shields".

Last edited: Sep 18, 2016
25. Sep 18, 2016

### A.T.

If that's the case, the answer is indeed trivial. But it's not obvious why the above should be true, when you spin a lossless mass-spring system.