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Rotational Kinetic Energy and Conservation of Momentum

  1. Sep 1, 2011 #1
    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.
  2. jcsd
  3. Sep 2, 2011 #2
    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.
  4. Sep 2, 2011 #3
    Another try - the attachment didn't seem to come along

    Attached Files:

  5. Sep 2, 2011 #4


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    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.
  6. Sep 2, 2011 #5


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    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.
  7. Sep 2, 2011 #6


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    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.

    Yes. That was my point.
  8. Sep 2, 2011 #7
    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?
  9. Sep 2, 2011 #8
    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.
  10. Sep 2, 2011 #9

    Ken G

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    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.
  11. Sep 2, 2011 #10

    Ken G

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    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).
  12. Sep 3, 2011 #11


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    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.
  13. Sep 3, 2011 #12
    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.
  14. Sep 3, 2011 #13


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    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.
  15. Sep 3, 2011 #14


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    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:blushing:

  16. Sep 3, 2011 #15
    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" [Broken]
    Last edited by a moderator: May 5, 2017
  17. Sep 4, 2011 #16


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    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.
  18. Sep 4, 2011 #17

    Ken G

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    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.
  19. Sep 4, 2011 #18


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    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.

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

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

    What losses are that?
  20. Sep 4, 2011 #19

    Ken G

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    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.
    Loss-less transmission cannot transfer momentum across a velocity difference, period. Not without some way to store the energy.
  21. Sep 4, 2011 #20


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    Perfectly elastic deformation is a way to store energy.
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