Rotational Kinetic Energy and Conservation of Momentum

[jimgram]

There are several references to “lost” kinetic energy when trying to analyze motion based on the Conservation of Momentum and the Conservation of Energy laws. Generally, the answer to apparent discrepancies include references to ‘elastic’ or ‘in-elastic’ collisions and whether or not they are ‘perfect’. There are always answers that somewhere, energy is used that wasn’t accounted for (e.g. ‘sound’, ‘heat’, ‘deformation’, etc.). All of this is understandable. When it comes to rotational kinetic energy, however, there seems little explanation is available. I’m sad to admit that I’ve struggled with this question for a few years now and I’ve received lots of insightful help from this forums members.

I’ve set up a problem involving flywheels: A first flywheel (fwa)is pre-charged with given angular velocity, and thus an initial value for momentum and for kinetic energy. This flywheel can drive a second flywheel (fwb) through an infinitely-variable-transmission (IVT). It’s important to understand that these devices do exist and, though generally limited in torque, are quite efficient. See Torotrak or simply Google IVT. This is used in the Formula 1 race circuit to enable flywheel-based regenerative braking. Many CVT’s and IVT’s use hydraulics or motor generators and do have significant losses, but the purely mechanical ball-disc or cone-disc versions, which rely on friction drive (bear in mind they are not designed to slip, hence the limited-torque comment), have very low losses. Furthermore, when combined with a differential to form a split-path transmission, they can operate to zero (or infinity) to some value, n, which allows a clutchless neutral.

Initially, I neglected the torque on the transmission in my calculations and was corrected by more than one forum member. I don’t feel too bad about missing that since reading in my Halliday Resnick Walker physics book that in 1986, the spacecraft Voyager 2, on its 1986 flyby of the planet Uranus, was set in unwanted rotation by this flywheel effect every time its tape recorder was turned on at high speed. The ground staff at the Jet Propulsion Laboratory had to program the onboard computer to turn on counteracting thruster jets every time the tape recorder was turned on or off.

So, here’s where I am with this problem first posted last January. The give parameters are:

A. Flywheel_a inertia = 23.4*m2*kg
B. Flywheel_b inertia = 169.8*m2*kg
C. Initial angular velocity of FW1 = 10,000*rpm
D. Initial angular velocity of FW2 = 0
E. The transmission ratio (n) = 0...0.075 (I.E. 1/n = inf…..13.33)
F. The period change ratio: t = 14 sec

You can set n (ratio) as a function of t (time) this way: n(t)=t*ne/te
Where ne=ending ratio (.075) and te=ending time (14*s).

I decided to mount the IVT in a space satellite having a moment of inertia about the IVT of Isa. This way I can make the moment small enough that the transmission torque will cause significant rotation of the satellite or so large (e.g. Earth) that any motion is minor.

Step one, I believe, is to determine the angular velocity of the first flywheel as the IVT loads it with the second flywheel by changing the IVT ratio from 0 to 13.33 over a period (t) of 0 to 14 seconds. I received much help in deriving this initial equation, and it turns out to be fairly simple:

The angular velocity of the flywheel as a function of time (and
thus a function of n (the ivt ratio from zero to ne) is equal to
the initial momentum of the flywheel (Ifwa*ω fwai) divided by the
total inertia as seen by the flywheel (Ifwa+(n2*Ifwb):

ω fwa(t) = (Ifwa*ω fwai)/(Ifwa+(n2*Ifwb)

Then, the velocity of the second flywheel follows:

ω fwb(t) = n(t)*ω fwa(t)

From here, torques can be calculated: tor = I*(dw/dt)

and as well as momentum and energy as a function of time. However, since we know momentum is conserved, we need to take the initial momentum of flywheel A and subtract the ending momentum of flywheel A and B in order to determine the momentum of the satellite (this is the momentum imparted by the torque on the IVT). We now have satisfied the mechanical (torque) constraints and also the Conservation of Momentum Law.

Now, since we have one and only one momentum of the satellite, and we know its moment of inertia (which we can pretend is any value), we can easily determine at what rate it is spinning as a function of time. I chose to use a very high value so that its angular velocity is very low since this velocity would then need to be subtracted from that of flywheel B and thus I would have needed to come up with a solution for all three masses simultaneously (over my pay grade).

Regardless, it can easily be seen that the total kinetic energy can be almost whatever you want it to be – it increases by the square of the angular velocity of the satellite, or decreases likewise. It can increase beyond the initial energy level of flywheel A alone but I would guess that’s because I haven’t subtracted the satellites’ angular velocity from that of flywheel B. But, when the inertia of the satellite is on the order of a small planet, and therefore its angular velocity is negligible, I have ‘lost’ a non-negligible amount of energy. Obviously, I need help.

 

[jimgram]

I hoped someone would be curious enough to do the basic math and offer a plausible explanation of the mis-match of ending kinetic energy to beginning kinetic energy. Maybe this is just too simple.

In any case I've attached a pdf file showing my work and results. Please don't tell me I have to re-write the laws of physics.

 

[jimgram]

Another try - the attachment didn't seem to come along

 

[A.T.]

There are always answers that somewhere, energy is used that wasn’t accounted for (e.g. ‘sound’, ‘heat’, ‘deformation’, etc.). All of this is understandable. When it comes to rotational kinetic energy, however, there seems little explanation is available.

There is no difference between rotational kinetic energy and kinetic energy. It is the same physical quantity. So I don't quite see why you need specific information.

As for your flywheel example, why not make it simpler:

Two identical flywheels are counter rotating at the same speed: Rotational kinetic energy is not zero.

Then the transmission connects them so they brake each other and stand still: Rotational kinetic energy is zero.

It is the same as an inelastic collision.

 

[cmb]

There is no difference between rotational kinetic energy and kinetic energy. It is the same physical quantity.

I don't follow this statement. There are many things that are the same physical quantity but that are different!?

I would say there is a fundamental difference: Any given object in linear motion has zero linear kinetic energy in its inertial frame, whereas no frame exists in which a rotating object has zero kinetic energy.

Or further; identify the minimum kinetic energy of two objects [or an ensemble of objects] and you have then identified their 'collision frame'.

For a rotating object, there would appear to be no inertial frame for which the kinetic energy can be zero. It would seem to me that you'd have to consider rotating objects as an ensemble of particles moving wrt each other, and presumably you would end up at the conclusion that the frame in which the kinetic energy is a minimum is one where the axis of rotation is stationary (this seems inevitable, but I would not presume it is obvious).

That being said, I think your end point [that it is like the collision of any other objects] is generally right. In the case of a CVT connection, this would act like a sticky collision. Any work done which is not accounted for by the work done from the reaction of the gearbox on its mounts and the mass on which it is fixed would have to be accounted for within the workings of the gearbox. If anything (if I understood what was said correctly) this looks like it could rather be a demonstration that a CVT is always lossy during a period of continuously changing ratio, howsoever it is carefully designed.

 

[A.T.]

I don't follow this statement. There are many things that are the same physical quantity but that are different!?

What I meant here: You can represent any rotating object as a bunch of point masses that have some linear velocity, and thus linear kinetic energy. The rotational kinetic energy is just the sum of a bunch of linear kinetic energies.

That being said, I think your end point [that it is like the collision of any other objects] is generally right. In the case of a CVT connection, this would act like a sticky collision. Any work done which is not accounted for by the work done from the reaction of the gearbox on its mounts and the mass on which it is fixed would have to be accounted for within the workings of the gearbox. If anything (if I understood what was said correctly) this looks like it could rather be a demonstration that a CVT is always lossy during a period of continuously changing ratio, howsoever it is carefully designed.

Yes. That was my point.

 

[jimgram]

That being said, I think your end point [that it is like the collision of any other objects] is generally right. In the case of a CVT connection, this would act like a sticky collision. Any work done which is not accounted for by the work done from the reaction of the gearbox on its mounts and the mass on which it is fixed would have to be accounted for within the workings of the gearbox. If anything (if I understood what was said correctly) this looks like it could rather be a demonstration that a CVT is always lossy during a period of continuously changing ratio, howsoever it is carefully designed.

This is the conclusion that I feared. But I cannot understand how the Law of Conservation of Energy alone can make my finely-tuned, well designed, and nearly frictionless IVT dissipate energy, which would require work, and hence power and then I suppose it would get hot! But why? Is this the same as saying "whenever there is a gear train - even with just two gears - you will see a loss of energy in the transfer of momentum?

 

[Adam_L]

I am not physics master, but I know just a little.

Like you said, "nearly frictionless IVT dissipate energy".

Friction(kenetic or static) during its course of action results in a loss of energy. There is not getting around this factoid. You many have heard it before, the main engineering marvel would be to design/create a system in which does not generate ANY heat, thus no loss of energy and forward momentum.

In 99.9% of mechanical applications, there is going to be at least one point of friction(or several). Magnatism may be the only real way to get beyond this fact of life.

 

[Ken G]

The answer here doesn't have to do with whether or not frictionlessness is truly possible. It is a thought experiment, and we can do whatever we want-- we can say it is free of any dissipation or friction. But what then happens? The flywheel has no way to communicate its angular momentum to the other flywheel! These two issues are tightly coupled-- this system has to dissipate energy in order to transfer angular momentum, even if it is perfect. It's the same with linear momentum and energy-- there is nothing fundamentally different in your problem with having two long metal planks sliding atop each other. I think that's what A.T. was saying-- rotational and linear kinetic energy that are internal to a system are the same thing. Neither depends on reference frame of an observer.

 

[Ken G]

In fact, it might be useful to imagine a potential energy source that can store the missing kinetic energy. Use a spring of some kind to couple the flywheels (or planks, it makes no difference). Now you get the same loss of kinetic energy when the two flywheels are moving together, but here it isn't dissipated-- it's locked in the springs. If you then lock down the flywheels, the energy will stay in the springs, but if you release them, the springs will release their energy and the flywheels will oscillate, like a pendulum, with the angular momentum sloshing back and forth periodically. The bottom line is, if you want one flywheel to transfer momentum to the other, you need a shared force between them, and that force must result either in the storage of potential energy (if it is a conservative force), or the dissipation of that energy (if it isn't).

 

[cmb]

This is the conclusion that I feared. But I cannot understand how the Law of Conservation of Energy alone can make my finely-tuned, well designed, and nearly frictionless IVT dissipate energy, which would require work, and hence power and then I suppose it would get hot! But why? Is this the same as saying "whenever there is a gear train - even with just two gears - you will see a loss of energy in the transfer of momentum?

In theory, you could engineer a fixed gear-mesh of evolutes on each cog that is entirely non-sliding and therefore friction free. You might even conceive of it such that the surfaces meet always at a precise zero speed and a lubricant that causes no loss of energy into the structure of the gears. This is a practical thought experiment.

But, no, there are some thought experiments that are 'not permitted', e.g. why not just conjoure up a time-machine and go back to collect some energy and bring it back to the future. A thought experiment that results in an impracticality serves only to show the axioms of that thought experiment are wrong.

So let's consider the 'perfect' CVT: Let's say our perfect CVT is running at some given speed. If you are able to change the gear ratio at will from this ratio, then it means that there is some gear ratio above that, and some gear ratio below, to which you can cause the mechanism to move to. This means that there must be some form of 'incipient' gear change available. This is quite true in practice; for example, in the torotrak there is a disc running on the inside of a toroidal shell. That disc contacts the shell at some point, which traces out a circular route around the shell. The reason it is possible to change ratio with that arrangement is that if a torque is applied to the rotating disc it will begin to step sideways from that circular track. In doing so, you have a part of that contact area at the 'old' ratio and another part of that disc at the 'new' ratio. The disc can be progressively forced across different contact paths by these means. But it means that during a ratio change, there will always be at least two parts of the disc running at a higher and a lower ratio. You cannot have a higher and lower ratio running simultaneously onto the same output shaft, there has to be some slippage. As there has to be a finite frictional reaction between the drive disc and the toroidal shell, so there is always some heating due to frictional slippage.

Now, you can tune this to be very efficient but not completely non-lossy. By slowing the rate at which the mechanism can (and does) change, you should theoretically reduce this slippage loss, but to change a ratio means there is always some small increment of ratio leading to two 'gear ratios' being driven simultaneously, hence there is always slip.

The only alternative is to have an infinite number of fixed ratios through which you change. Then we enter some 'infinitesimal' type calculations - how does the calculation above (transfer between two inertial wheels) stack up if we have some change of ratio, delta[ratio] through which each instantaneous coupling at that new ratio leads to some lossy process? Then we decrease that ratio ad infinitum until we have integrated that delta[ratio] between the higher [one wheel rotating] and lower [both wheels locked together] operating speeds. I suspect you will end up with losses consistent with satisfying the numbers you've posted.

Other cvt mechanisms, for example the DAF style with a belt gripped between two pulleys, is similar; within the region where the metal chain is gripped, there is a tiny deformation of surfaces and as the pulleys are adjusted, so the chain on one side of the pulley is rotating slightly faster than the chain on the other side, and so they must be slipping somehow with respect to each other (IIRC the DAF system used a belt in compression, rather than tension. I guess this was a solution to avoiding the chain digging into the pulley so hard that it 'self-tightened' itself destructively, rather than relieve the reaction force on the pulley thereby encouraging a little slippage during ratio changing. I think with later material improvements the preference tends to be to have the chain in tension these days, e.g. the Audi multi-tronic system.)

There are also magnetic gears that are capable of a slip funtion, thus a cvt effect, and of course there is the Prius hybrid which uses an ICE motor on one side of a differential and an electric one on the other. Theoretically the differential may run 'perfectly', but the 'slippage' is then implemented within the electric motor and there would be associated losses there.

 

[jimgram]

cmb: You have explained in significant detail how complex and VARIABLE the probable frictional losses are in various types of transmissions. But, as a thought experiment, doesn't it seem most unlikely that one simple formula -E= 1/2*ms*v^2 - can predict with a degree of accuracy just what those frictional losses will be - and how does this equation account for all of those minute variable losses.

That being said, the energy "loss" per the Mathcad worksheet I attached earlier is about twice what the ending kinetic energy is for both flywheels. In other words, I've lost over half the energy that I started with. That can't be hypothetical friction loss.

It's been suggested that the work done to decelerate flywheel A (minus work) and accelerate flywheel B (positive work) must be subtracted from the total ending energy. When this is done, the ending energy equals the beginning energy exactly. If this is the solution, I do not understand the logic.

 

[cmb]

cmb: You have explained in significant detail how complex and VARIABLE the probable frictional losses are in various types of transmissions. But, as a thought experiment, doesn't it seem most unlikely that one simple formula -E= 1/2*ms*v^2 - can predict with a degree of accuracy just what those frictional losses will be - and how does this equation account for all of those minute variable losses.

I'd not be surprised if it were to all boil down to something as simple as that. This is because there must be some maximum efficiency, as per the example I gave (integrate the losses whilst tending your theoretical gearbox to an infinite set of cogs).

Once you have established your 'perfect' gearbox with an infinite set of cogs and infinite set of gear changes (each change with an 'approaching zero' amount of energy loss), I think you can see that it would tend towards some given, non-variable result.

 

[RonL]

Hi Jim,
Without going back to your other threads to refresh my memory, is there a reason the transmission has to be mechanical gears ?
If design is flexible, and my not knowing anything beyond the flywheels, my thoughts as in the last thread, go to the counter rotating flywheels with a narrow change in speed increase and decrease between the two.
Electric energy transfer might (should) be more efficient than mechanical.

Each flywheel can be 1/2 of the motor/generator unit, and as was mentioned, spring storage represents a Conservative method between the flywheels (I would make it a gas spring) consider the nitrogen bladder in most hydraulic breakers and many other applications. A sealed compressed air design might be much less expensive.

The slight over speed and draw down of each flywheel can be very tightly controlled with the electric portion, while the flywheels bounce between the high and low pressures of the gas springs.

If this fits in your design (or can) the thermal changes in the gas portion can help compensate for some loss from frictional areas.

Maybe there is some part of all this that might help

Ron

 

[jimgram]

Thanks for your thoughts RonL. While this is related to earlier posts I'm looking for something here fundamentally more basic. If the simplest 3-mass situation cannot be solved without maintaining the basic laws of physics and without resorting to all kinds of ambient friction, I then need to question some much more far-reaching assumptions.

cmb: As I previously indicated, I'm not looking for a small amount of kinetic energy mis-match. I'm certain that the small amount of friction in the Torotrac IVT will not eat up 50% of the initial flywheel energy.

The fundamental question is: With rotational dynamics in its simplest form, how can you conserve momentum at L=I*v AND kinetic energy at Ek=1/2*I*ω²?

Here’s a blog where the thread starter thinks that the answer is the creation of ‘free’ energy:
(please don't hold this against me - I use it only as an example that this subject can be a little confusing)
http://www.energeticforum.com/renewable-energy/4493-create-energy-conservation-momentum-law.html"

 

[A.T.]

The fundamental question is: With rotational dynamics in its simplest form, how can you conserve momentum at L=I*v AND kinetic energy at Ek=1/2*I*ω²?

Same as with linear dynamics: You need perfectly elastic mechanisms, which turn all deformation energy into work, and none of it into heat.

Such mechanisms do not violate energy conservation. But it gets tricky with the second law of thermodynamics. It seems it would be OK too, because the second law doesn't demand that the entropy increases, just that it doesn't decrease.

 

[Ken G]

I still think a key point is being overlooked here. It makes no difference how nearly frictionless the process is, the whole idea of getting one flywheel to spin up another one is a definitively lossy process. Same for one plank sliding across the other. This means that the more frictionless you make the process, the longer it takes. That's it, that's all you need-- in the limit of a totally frictionless apparatus, it would take forever. Friction is not actually the enemy here, it is your friend-- a perfectly frictional coupling would transfer the rotation instantly, a very frictionless one would take a real long time. That's the point behind a transmission-- it's not the dissipation you want to avoid, you accept the dissipation as part of your goal. What you want to avoid is for the dissipation to happen suddenly and jerkily.

Now, there can also be losses over and above simply what is required to get one flywheel to share its angular momentum. Those are bad losses, not the good one you want. So design should focus on minimizing that.

 

[A.T.]

the whole idea of getting one flywheel to spin up another one is a definitively lossy process.

What do you mean by "definitively"? Is there a law in physics which says that there must be loses here? We might not be able to build a loss-less transmission, but it would not violate any fundamental laws of physics.

Same for one plank sliding across the other.

A loss-less transmission would have no sliding, just perfectly elastic deformation.

That's it, that's all you need-- in the limit of a totally frictionless apparatus, it would take forever.

I don't want a totally frictionless apparatus. I want just static friction, which doesn't cause any losses. Losses come from sliding friction.

Now, there can also be losses over and above simply what is required to get one flywheel to share its angular momentum..

What losses are that?

 

[Ken G]

What do you mean by "definitively"? Is there a law in physics which says that there must be loses here?

Yes. There is always a definite kinetic energy loss whenever momentum is transferred across a velocity difference, it's in the definitions of the quantities involved. When there is no other place to put that energy, as in this case, it must go into heat, again by definition. Ergo, there is a definitive loss of bulk kinetic energy whenever momentum is transferred across a velocity difference, it's right in the equations, and is not avoidable by any reduction of friction-- that only makes the transfer take longer.

A loss-less transmission would have no sliding, just perfectly elastic deformation.

Loss-less transmission cannot transfer momentum across a velocity difference, period. Not without some way to store the energy.

 

[A.T.]

A loss-less transmission would have no sliding, just perfectly elastic deformation.

Loss-less transmission cannot transfer momentum across a velocity difference, period. Not without some way to store the energy.

Perfectly elastic deformation is a way to store energy.

 

[Ken G]

Fine, then it's like the springs I mentioned a few posts back. The point I'm making is, it doesn't make the least bit of difference what kind of force is used to transfer the angular momentum, whether dissipative or elastic-- the very process of doing that is going to require the removal of kinetic energy. So any effort made to prevent removal of kinetic energy just slows the process down, that removal should not be viewed as an odd side effect of the process, but rather it is the process. The OP seemed to say that in a thought experiment, some way to transfer the momentum could happen without losing energy, and then the discussion turned to the issue of degree of dissipation, but that's a red herring. If the two flywheels are going to share angular momentum, then we want them to lose kinetic energy, and we must find a way to get them to lose kinetic energy, no matter what way we choose.

 

[A.T.]

The OP seemed to say that in a thought experiment, some way to transfer the momentum could happen without losing energy,

It is possible. In the linear case you need perfectly elastic collision. In the angular case you need perfectly elastic transmission.

If the two flywheels are going to share angular momentum,then we want them to lose kinetic energy,

I rather think what the OP wants is to transfer angular momentum without loosing kinetic energy. Just like perfectly elastic collisions do with linear momentum.

 

[Ken G]

It is possible. In the linear case you need perfectly elastic collision. In the angular case you need perfectly elastic transmission.

I mean without losing any of the kinetic energy of the flywheels. That cannot happen, by definition of the quantities. Nothing to do with energy conservation, that's all I'm saying. It's all conservation of angular momentum-- if you want to share the angular momentum of the flywheel, you must find a way to remove kinetic energy. The two tasks are the same, so efforts to do the first without the second are misguided, regardless of how ideal the system is or what you plan to do with the removed energy.

 

[jimgram]

I think Ken G is saying in effect that in order to transfer momentum (linear or rotational) you must relay on friction (I.E. one plank on top of another), while AT is saying you must have an in-elastic collision. Both of you are saying that a way to store energy si required in order to vary the speed of one mass by transferring momentum from another mass (either by the deformation of a mass or a spring -same thing).

Can there be a way to apply a force to a body that is not a collision? For example with an IVT?

In any case, coupling a spinning flywheel with a small inertia to a stationary flywheel with a large inertia results in a major reduction in velocity of the small flywheel (and thus a major reduction in kinetic energy (ω²) and a small increase in velocity of the big flywheel. The sum of the resultant kinetic energies can be far less than the kinetic energy of the small, fast flywheel initially. I do not believe this major difference can be expalined by in-eleastic collisions or friction. If the coupling was a friction clutch instead of an IVT then its easy to see that the clutch should dissipate considerable energy. BUT, if it works out mathematically the same energy loss for a clutch and for an IVT, then why would anyone ever use an expensive IVT?

 

[Ken G]

I think Ken G is saying in effect that in order to transfer momentum (linear or rotational) you must relay on friction (I.E. one plank on top of another),

Not quite, I'm saying it makes no difference what type of force you use, the issue at hand depends only on Newton's third law (forces come in action/reaction pairs, whether elastic or inelastic). Let me put it this way:

1) to change the spin of the flywheels, you will need a force on each.
2) that force will come in an action/reaction pair (or a large number of them)-- equal and opposite on the two flywheels.
3) treating the two flywheels as a single system, the equal and opposite forces must yield no net torque, so angular momentum is conserved (or linear momentum for the two planks), so it can only be transferred, not created.
4) transfer of angular momentum across a difference in rotation rate will require a drop in total rotational kinetic energy.
Thus, a desire to transfer angular momentum implies a desire to reduce the rotational kinetic energy of the system. Any effort to do the former is an effort to do the latter, and it doesn't make any difference if one is going to try and reduce dissipation. Reducing dissipation in the absence of a conservative potential energy source is counterproductive because it just slows down the transfer. Replacing dissipation by conservative forces also offers no particular advantage, if the goal is to achieve two equal rotation rates-- it would only be an advantage if you wish to call upon the energy storage to return the flywheels to their original rotation rates. That latter desire seems irrelevant to the basic question that is being asked.

As for why pay for an IVT, that would have to do with avoiding additional energy dissipation above and beyond what we have been talking about, but that would just mean losing no rotational angular momentum in the net. So the goal of the design should be to isolate the full system from external torques, not to reduce dissipation in the angular momentum transfer. Note that an infinitely frictional connection would also accomplish the minimum energy losses-- but would create a very jerky transition and might not even be possible without destroying something. Maybe heat is an issue that an IVT avoids, though it doesn't really seem to store the energy anywhere else that I can see, but that depends on the details that I don't know.

 

[A.T.]

I mean without losing any of the kinetic energy of the flywheels. That cannot happen, by definition of the quantities.

It can happen theoretically if you have a prefectly elastic transmission that can temporally store energy.

Think of a perfectly elastic linear collision of two identical balls (one moving, one at rest):

O--> O

The left ball will transfer all momentum to the right ball and stop. Kinetic energy is conserved.Now make it two flywheels, made of two connected elastic balls each (one spins, one is at rest):

   O-->
   |
   +  
   |
<--O 
 
   O
   |
   +  
   |
   O

If you bring them together, so that two balls collide, the spinning top one will transfer all angular momentum to the bottom one. Kinetic energy is conserved.

if you want to share the angular momentum of the flywheel, you must find a way to remove kinetic energy.

If you want to transfer a specific amount of momentum, then you might have to remove kinetic energy. Just like in the linear case, you are not free to choose how much momentum is transferred if kinetic energy is to be conserved. That's because your system of equations has now only one solution (it is not undetermined anymore):

- 2 variables: final (angular) velocities of the two objects after transfer
- 2 equations: (angular) momentum conservation & (rotational) kinetic energy conservation

If you demand that the final velocities are different from those that this system of equations yields, then in fact you are demanding that energy is removed.

There is one slight difference between the linear and the angular case though: The masses of the colliding balls don't change during the collision. But you could build flywheels with a variable moment of inertia. That makes your system under-determined again, so you might be able to conserve kinetic energy for multiple different combinations of final velocities.

 

[cmb]

I just wrote out a reply, then it dawned on me where this question should go to clarify it.

You cannot have an infinitely smooth change of gear without 'stickiness', for the reasons I gave above, so I tried to get us looking at the situation of how you could decelerate one by the other through an infinite number of gears.

But that itself is not particularly clear, so let me rephrase the question a little, see if it is where you want to go with it.

Let us imagine a rotating inertial wheel connected to a 2:1 gear ratio. The gears of this gear-set are of negligible mass and all is friction-free. Right next to the rotating output gear is another inertial wheel of the same size and mass as the first, which is stationary. At the flick of a mental switch, studs pop out of the stationary inertial wheel which the rotating gear strikes in a lossless elastic collision, then the studs promptly pull back into the inertial wheel.

In the moment that the gear and inertial wheel struck each other, there is a lossless exchange of momentum. The inertial wheel accelerates by, let's say, w (omega). The gear, being attached by a 2:1 gear and running faster, has an effectively lower moment of inertia. It will bounce back at 2w, but we find that at the inertial wheel it is connected to it has only slowed by w.

So here we have what I think is the proposed curiosity. The question is, if you have successfully spread out the momentum, the kinetic energy of both wheels is now lower. {Because W^2>(W-w)^2 + w^2 (where W was the original rotational velocity, and w is the change)}.

Where did the energy go!?

The answer is that it must have been reacted into the gearbox. In a stationary satellite, the whole thing would have started rotating when this elastic impact happened. The best way to visualise this is to imagine that the 2:1 ratio is an epicyclic gear box. Imagine the outer gear is the inertial wheel and it is rotating. Now imagine the centre gear is the other inertial wheel, of the same moment of inertia. What would be happening here is the planetary gears would be rotating to keep this scenario moving (all friction free, of course!). To effect a momentum transfer in this arrangement, you would have to momentarily, and elastically, alter the speed of the planetary gears. But this is impossible, unless you have a 'base' against which to react with.

So, picutre the eplicyclic freewheeling on the outer, stationary gear on the inner. Whether we apply friction to the moving planetaries to change their speed (and thus transfer momentum between outer and inner), or try to arrest them momentarily in an elastic event, whatever holds those planetaries has to react the momentum exchange. However you construct your gearbox, it is topologically exactly the same as this, and the kinetic energy is transferred into the reaction structure of the gearbox, if it is a sudden elastic transfer of momentum.

 

[Ken G]

It can happen theoretically if you have a prefectly elastic transmission that can temporally store energy.

Sure, but we covered that-- only by returning to the original configuration can it do that. I have no problem with perfect elasticity or absence of dissipation, it is just the definition of the quantities that matters here, and they give that to get the two flywheels moving together always requires removing kinetic energy. It is the goal to do that, not an avoidable consequence. To see where the energy went, all you will need to do is see how the angular momentum came across, they will always be the same thing.

Think of a perfectly elastic linear collision of two identical balls (one moving, one at rest):

O--> O

The left ball will transfer all momentum to the right ball and stop. Kinetic energy is conserved.

Yes, it can happen in situations where you are not moving the momentum over slowly, or where you are not reaching the same motion between the two bodies, but rather "jumping across" to a different situation on the "far side" of the scenario of interest here. You are right there are more general ways to conserve energy, but not in the case of trying to get the two objects moving together, which is the situation being discussed here. I should have specified that.

If you bring them together, so that two balls collide, the spinning top one will transfer all angular momentum to the bottom one. Kinetic energy is conserved.

True, but again it is not relevant to the OP scenario of either gradual momentum transfer, or reaching a combined movement.

If you demand that the final velocities are different from those that this system of equations yields, then in fact you are demanding that energy is removed.

Precisely my point-- removal of kinetic energy is the goal of the flywheel scenario, not some kind of side effect that we should be trying to minimize, or should wonder about how we could do away with with a more ideal system.

There is one slight difference between the linear and the angular case though: The masses of the colliding balls don't change during the collision. But you could build flywheels with a variable moment of inertia. That makes your system under-determined again, so you might be able to conserve kinetic energy for multiple different combinations of final velocities.

Changing the moment of inertia would provide a way to do work on the system. If there is some other way to do work, you can get any kinetic energy you want-- which is what people do who claim to run cars on water (I'm not saying you are claiming anything like that).

 

[Delta Kilo]

The fundamental question is: With rotational dynamics in its simplest form, how can you conserve momentum at L=I*v AND kinetic energy at Ek=1/2*I*ω²?

I don't see a problem. You have 3 bodies: flywheel A, flywheel B and spacecraft C and two conservation laws: $E=\frac{I_a \omega_a^2}{2}+\frac{I_b \omega_b^2}{2}+\frac{I_c \omega_c^2}{2}$, $L = I_a \omega_a + I_b \omega_b + I_c \omega_c$ You have 1 extra degree of freedom so you have a range of solutions. Eg. you can start with $\omega_a = \Omega, \omega_b = \omega_c = 0$ (flywheel A spins, flywheel B ands spacecraft are at rest), find E and L and then solve it for $\omega_a = \omega_c$ (flywheel A is at rest relative to spacecraft ).

The upshot of it: the gearbox will exert torques on its mounts. If you are concerned with angular momentum conservation, these would have to be accounted for.

 

[rcgldr]

you can start with ... flywheel A spins, flywheel B and spacecraft are at rest ... then solve for ... flywheel A is at rest relative to spacecraft .

I think the end goal was flywheel A and B rotating at the same speed. If you just want flywheel A to stop and flywheel B to spin at the same speed that flywheel A was spinning at, then a lossless (elastic) coiled spring connecting the two flywheels could accomplish that, storing energy during the transition.

Even if you had a middle flywheel, connnected to each outer flywheel with a lossless motor / generator and capacitor (in order to store energy). Then flywheel A could be slowed down transferring angular momentum and energy into the middle flywheel and capacitor. Then the process reversed to speed up flywheel B. The problem here is that since angular momentum is conserved, when flywheel A and B are spinning at the same speed, and the middle flywheel returned to it's initial rest state, you end up with energy stored in the capacitor. To end up with zero energy in the capacitor, the middle flywheel will end up rotating in the opposite direction of flywheels A and B when they're both spinning at the same speed.

 

[Dadface]

Some of the energy transfer systems being suggested here suffer from their own inherent losses.An idealised spring system can store only fifty percent of the energy needed to distort it and an idealised capacitor charging system can store only fifty percent of the energy needed to charge it.
The idea of elastic transmissions by collision is interesting and although it can work for linear motion(perfectly for some collisions with microscopic objects eg identical gas atoms under certain conditions) it would be interesting to see how it can be achieved,in practise,with rotational motion and macroscopic structures like flywheels.
If the energy is transferred by some sort of gear/clutch linkage then,assuming an idealised system,the maximum energy of the whole assembley,as a result of linkage is given by:

Final Energy=Initial Energy*I/(I+Ix)

I= moment of inertia of original flywheel.

Ix equals effective moment of inertia of everything else( second wheel plus clutch/gears etc)

Although the equation applies to a system which has been massively simplified I think it illustrates that considerable energy losses are inevitable.

 

[cmb]

Some of the energy transfer systems being suggested here suffer from their own inherent losses.An idealised spring system can store only fifty percent of the energy needed to distort it and an idealised capacitor charging system can store only fifty percent of the energy needed to charge it.

AFAIK - that's not right. You're saying that a capacitor being repeatedly charged then discharged would heat up by the total power that the capacitor is handling? (because it is causing as much loss as power that it passes?!)

Or you are saying that if I have a 10 kg mass I attach on a spring, and after letting go the spring stretches 1 m with the mass on the end of it, that the spring will heat up by 50J!?

By these means you can calculate how many times a spring can bounce - each bounce would reduce the energy stored in the spring by 50%. Not many bounces, then? So are you saying a spring in simple harmonic motion would never have a displacement in excess of 70% of the previous displacement!?

I think you are confusing the issue of this thread, as well as purveying some errors.

 

[Ken G]

Yes, I think a factor of 1/2 slipped in on you there Dadface-- springs and capacitors can in principle store 100% of the work done on them.

 

[Dadface]

AFAIK - that's not right. You're saying that a capacitor being repeatedly charged then discharged would heat up by the total power that the capacitor is handling? (because it is causing as much loss as power that it passes?!)

Or you are saying that if I have a 10 kg mass I attach on a spring, and after letting go the spring stretches 1 m with the mass on the end of it, that the spring will heat up by 50J!?

By these means you can calculate how many times a spring can bounce - each bounce would reduce the energy stored in the spring by 50%. Not many bounces, then? So are you saying a spring in simple harmonic motion would never have a displacement in excess of 70% of the previous displacement!?

I think you are confusing the issue of this thread, as well as purveying some errors.

1.Consider ,for example,a helical spring with a mass M added resulting in a final static extension of x:

PE lost =Mgx

Energy stored=0.5Mgx (average force times extension)

2.Similarly the energy stored in a charged capacitor=0.5QV(Q= charge V=volts)

In the case of the spring if a mass is added and released the system overshoots the equilibrium position and vibrates with damped harmonic motion before coming to rest.In the case of the capacitor energy is lost by Joule heating as the charging current flows through the connecting wires.Mainly heat losses in both cases.

 

[cmb]

1.Consider ,for example,a helical spring with a mass M added resulting in a final static extension of x:

PE lost =Mgx

Energy stored=0.5Mgx (average force times extension)

eh? Why is the 'average force' not Mg!? Did its mass change somewhere during the fall?