Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Multiple Flywheel System

  1. Sep 24, 2009 #1
    This is a real-world problem - NOT AN EXAM OR HOMEWORK (although it may sound as if it was produced by a deranged physics professor).

    A three-flywheel system has two identical variable inertia flywheels, each connected to the inputs of a differential. The common output of the differential is connected to the 3rd flywheel. The initial conditions are:
    1. Angular velocity of variable-inertia flywheel A, wa = 200 rad/sec
    2. Angular velocity of variable-inertia flywheel B, wb = -200 rad/sec
    3. Angular velocity of fixed-inertia flywheel C is from the standard equation for differentials:
    wc = (wa+wb)/2, so initially, wc = 0
    4. The inertia range for both flyweels A & B is: Ia, Ib = 0.4 m2*kg to 0.6 m2*kg
    5. The initial setting for the inertia of flywheel A & B is: Ia, Ib = 0.5 m2*kg
    6. The inertia of flywheel C is fixed: Ic = 5 m2*kg

    The momentum for each flywheel is: L = I*w, so: La = 200*m2*kg/sec; Lb = -200*m2*kg/sec; Lc = 0

    Ignoring friction of any sort, if the inertia of flywheel A is decreased from 0.5 m2*kg to 0.4 m2*kg, at the same time the inertia of flywheel B is increased from 0.5 m2*kg to 0.6 m2*kg, and this change occurs linearly over a period of 10 seconds, find:

    1. The ending angular velocities for all three flywheels
    2. The momentums of each flywheel

    Owing to the work done in changing the inertia’s of wheels A & B, kinetic energy is not conserved in the flywheels.
    The initial momentum can’t be zero, so I assume that the initial momentum is 400*m2*kg/s (the velocity relationship is wc = (wa-wb)/2). Therefore, the ending momentum must be 400*m2*kg/s divided into the three flywheels.

    I am into my 2nd year working on this problem in one way or another. It certainly has a solution because it's actually being done. But I'm at a loss as how to model it. Anyone out there needing your brain cells stimulated? (Please don't come back with a simple solution - my ego couldn't stand it).
  2. jcsd
  3. Sep 25, 2009 #2
    I'm interested in this, but for me the equation 'wc = (wa+wb)/2, so initially, wc = 0' does not make sense in the first place - I think this is because I don't understand the
    actual mechanical linkage when you use the term 'differential.

    If you don't mind, can you clarify that? i.e what does the coupling 'do' in terms of energy linkage (assuming zero losses due to friction).

    Also, your statements :
    'Owing to the work done in changing the inertia’s of wheels A & B, kinetic energy is not conserved in the flywheels.'
    - why do you say this?

    I think that when these questions are clear, then an analysis of the changes in kinetic energy will give you the answer (if you assume no losses due to friction).
  4. Sep 25, 2009 #3


    User Avatar
    Gold Member

    I think it's worthwile to notice that the initial angular velocities of the two variable inertia flywheels run counter, so that the the inital total angular momentum is zero.

    I will refer to the shaft from the fixed inertia flywheel to the diff as the 'prop shaft', and to the other two shafts as the 'left shaft' and the 'right shaft'

    The easy version of the problem, of course, is when the prop shaft would be locked in a fixed position. Then the two variable inertia flywheels will simply exchange angular momentum, as if they are connected by a fixed rod. (The diff only reversing the rotation direction of the shafts.)

    In this setup the prop shaft is connected to a flywheel too. That is a nice problem. (The inner gearing of a diff has its own inertia, making the inner gearings act as a flywheel. We either neglect that inertia, or lump it in with the total moment of inertia of flywheel C.)

    I'm unsure whether all three angular momenta can be just trown on a heap, disregarding the actual inner working of the diff. Nice problem, I'll get back if I can make some progress on it.

  5. Sep 25, 2009 #4
    haward: A good way to understand the working of the differential is to reference the comments by Cleonis: If the "prop shaft", that is, the shaft coupling flywheel C to the differential is locked, then the gear set in the differential acts exactly like a reversing connection between flywheel A and B. This reversing gear set, however, is normally free to rotate about an axis, that axis being the output to flywheel C - thus, a differential. Also, check out the Antikythera mechanism (however, Wikipedia mistakingly reports that it is not a differential - see attachment). Therefore, if FW A and FW B counter-rotate at equal speed, the output FW C will be zero. Your 2nd question regarding the conservation of kinetic energy- once the inertia's are changed and you calculate the resulting momentum and the initial energy and ending energy, you'll find that these values are not equal, again because of the work required to shift the inertia.
    Cleonis: Yes, the inertia of the inner gearing and body of the diff should be assumed to be included in FW C. I did not assume that the momentum of FW A and B would cancel since they are essentially coupled by a reversing gear. They both have momentum and the momentum from each can be available to the output shaft in the same direction, thus adding.

    Attached Files:

    Last edited: Sep 25, 2009
  6. Sep 25, 2009 #5
    With the following assumptions:
    a) the two variable inertia flywheels geared to act in an opposite way on the fixed inertia flywheel so that angular momentum Lc = La - Lb = 0 as initial state
    b) Ic = 5 m2*kg (given) and Lc=Ic * wc = 0 as initial state
    c) Ia = 0.5 m2*kg and Ib = 0.5 m2*kg initial state and
    wa = 200 rad/sec
    wb = 200 rad/sec
    => La= 100 and Lb = 100 initial state, with Lc = 0

    d) Ia = 0.4 m2*kg and Ib = 0.6 m2*kg final state

    e) ASSUMING no energy is put into either flywheels A or B and that angular momentum is conserved during the process of changing the MOMENTS of inertia of flywheels a and B, and that the rate of change is the same in moments of inertia is the same in each flywheel, and no friction losses etc we can say that

    i) La = 100 will remain constant because this is a variable inertia flywheel, and it is the moment of inertia that is varied; angular momentum remains the same at 100.
    Lb = 100 also

    ii) The rotational rates will change however due to the changes in moments of inertia, so that:

    La=100= 0.4 * wa(final state) => wa(final sate) = 250 rads/sec
    Lb=100= 0.6 * wb(final state) => wb(final sate) = 166.666. rads/sec

    Lc = La -Lb = 0 i.e does not change, hence wc = 0

    Different assumptions:

    IF energy were put into the system, or taken out of the system, this would change the calculations.

    IF the flywheels were geared so that Lc = La + Lb so that Lc = 100 + 100 = 200 initial state then Lc (final state) woudl remain = 200, and wc initial = wc final =

    Lc = 200 = 5 * wc => wc = 40 rads/sec

    IF work were done this would result in a change of angular momentum in each flywheel A and B.
    From the details given, there is no way to know or calculate if this is the case (by definition, a variable inertia flywheel changes the MOMENT of inertia, and this results in a change of angular velocity so that momentum is conserved).
    Any additional energy added or taken out would need to be known, or calculated by knowing the final angular velocities of both A nd B.

    Since the scenario here is to calculate final angular rates and momentum without any indication that energy is put in/out, then I think this is scenario suggests no changes in momentum.
  7. Sep 26, 2009 #6


    User Avatar
    Gold Member

    This assumption is incorrect.

    Flywheels A and B are described as 'variable inertia flywheels'. That means a mechanism is in place to alter the moment of inertia; a mechanism that alters the distance of the masses to the central axis of rotation.

    To sustain circumnavigating motion a centripetal force must be exerted. To contract a rotating system the centripetal force must exceed the required amount. During the contraction phase the centripetal force is doing work. That is, during the contraction phase the centripetal force is pumping energy in the system, increasing the angular velocity.

    During contraction of a rotating system angular momentum is conserved, but the rotational kinetic energy changes; work is being done.

    I posted in this thread because I'm especially interested in rotation. I recently manufactured a Java simulation that illustrates that http://www.cleonis.nl/physics/ejs/angular_acceleration_simulation.php" [Broken]

    In the case of a single variable inertia flywheel all of the work that is done goes into accelerating that flywheel. In the three flywheel setup the kinetic energy that is pumped in is free to transfer to other components of the system.

    The final state of the three flywheel system must be some equilibrium state, but its hard to say what equilibrium.

    Last edited by a moderator: May 4, 2017
  8. Sep 26, 2009 #7
    The basis for analysis is that no external energy is input. It has been found that the increasing centripetal force (FW A as w increases) can be made to balance the decreasing centripetal force (FW B as wb decreases). In haward's analysis (i,ii) that since L may remain constant, Lc = 0 can only be true if wc is zero. But if wa and wb are different, then owing to the differential, wc = (wa-wb)/2, therfore Lc could not be zero and also, La + Lc could not equal 200 (the total angular momentum available).

    Cleonis is correct that there is an equilibrium state and that is exactly my problem

    Cleonis: In your Java simulation, have you found that as the center mass (the 'control mass') drops there is a state in which that mass = centripetal force and therefore equibrium is achieved and the mass drops no further?
  9. Sep 26, 2009 #8


    User Avatar
    Gold Member

    My plan is to develop a version of that simulation that focusus on the process of developing to an equilibrium state.

    The equilibrium is an energy equilibrium (not a force equilibrium). The mechanism that provides the centripetal force constitutes a reservoir of potential energy. The rotational motion constitutes a reservoir of kinetic energy.

    In the final state the energy levels in those two reservoirs are in equilibrium. Its a general theorem of physics that when a complex system is allowed to develop freely then in the end state the system's energy will have become distributed evenly over the available degrees of freedom; the equipartition theorem.
    The equipartition theorem is best known in statistical mechanics, but I believe there are applications in macroscopic mechanics too.

    The tricky part is to identify what the energy reservoirs are.

  10. Sep 26, 2009 #9


    User Avatar
    Gold Member

    Applying the principles to the three flywheel setup.

    Flywheel C is a fixed-inertia flywheel, connected to the prop shaft that leads to the diff.
    Flywheels A and B are variable-inertia flywheels, connected to the left and right shaft.
    In the initial state, A and B have the same amount of angular velocity, in counterrotation relative to each other.

    For exploration take the following version:
    All three flywheels are initally non-rotating. A and B are set to the same moment of inertia.

    Then when there is energy input to flywheel C the energy will spread through the system.
    [STRIKE]Turning of the prop shaft sets up counterrotating motion in the flywheel pair A-B. For the flywheel pair A-B the amount of counterrotation is a reservoir of kinetic energy in itself. So flywheel C on one hand, and the flywheel pair A-B on the other hand, can be counted as two connected reservoirs of kinetic energy. Those reservoirs will develop towards equilibrium relative to each other. Equilibrium state will be when they store the same amount of kinetic energy. [/STRIKE]

    Turning the prop shaft causes co-rotating motion of the flywheel pair A-B.
    The total motion of flywheel pair A-B can be decomposed in a co-rotating component and a counterrotating component.

    How will kinetic energy redistribute itself between A and B? I think A and B will be in equilibrium when they store the same amount of kinetic energy. Hence if A and B have different moments of inertia, they will end up with different angular velocities.

    [STRIKE]That's my theory - I haven't even started on calculations.[/STRIKE]
    Well, I posted too early. I think I'm on the right track, but I'm not there yet.

    Last edited: Sep 26, 2009
  11. Sep 27, 2009 #10


    User Avatar
    Gold Member

    We're dealing with angular momentum and angular kinetic energy. Angular momentum is directional, it's a vector quantity, and angular kinetic energy is not directional, it's a scalar quantity.

    The flywheels A and B are a system with two degrees of freedom: their respective angular velocities.

    If A and be are purely counterrotating then the co-rotating component of their motion is zero.

    Transforming to a representation in terms of co-rotating component and counterrotating component entails a shift in what is taken as zero point of angular velocity. I will use the following names:
    Vco = co-rotating angular velocity
    Vcounter = counterrotating angular velocity

    Let the starting angular velocities of A and B be 200 and -200 respectively. When the angular velocity of the prop shaft changes from 0 to 10, and the amount of counterrotation is to remain the same, then the angular velocities of A and B will shift to 210 and -190 respectively.

    Let's try some things.
    For simplicity we assume the moment of inertia is 1 unit.
    Rotational kinetic energy is 1/2*I*w2.
    Calculating the total angular kinetic energy for A and B, in the case of Va = 210 and Vb = -190
    [tex]\frac{1}{2}*210^2 + \frac{1}{2}*(-190)^2 = 40100 [/tex]

    Now let's rewrite that as follows:
    [tex]\frac{1}{2} (V_{counter} + V_{co})^2 + \frac{1}{2} (-V_{counter} + V_{co})^2 [/tex]

    Writing out the above expression several terms drop away against each other, resulting in:
    [tex] 2 * \frac{1}{2} {V_{counter}}^2 + 2 * \frac{1}{2} {V_{co}}^2 [/tex]

    This indicates that indeed we can express the kinetic energy of A and B in terms of the sum of a co-rotating component and a counterrotating component. (The more general case where A and B do not have the same moment of inertia still wants proof.)

    In the three flywheel setup flywheel C only interacts with the co-rotating component of motion. I expect that in the final state there is an equal amount of kinetic energy in flywheel C and in the co-rotating component of flywheel pair A-B.

    Remaining question: must yet another interaction be attributed, directly between flywheel A and flywheel B? I'm inclined to say no, I think the only conduit of energy transfer is via C, but I'm not sure.

    Last edited: Sep 27, 2009
  12. Sep 27, 2009 #11
    You may be on the right track. I'm not sure I understand the term "co-rotating ang. vel." (unless it is simply referring to the ang. vel. of FW C - I.E. the prop shaft).

    My efforts have so far involved the assumption that FW's A & B are connected (via the differential) by a counter-rotating gear train, thus the use of negative notation for the counter-rotation would not be proper.

    However, as previously stated, that led me nowhere. Essentially, any terms (e.g ke, momentum, torque, ang. vel.) that are taken as an initial assumption, are then used to determine the transfer to the output (FW C), which in turn then changes the value of the initial assumption. This leads me to think that at each moment in time, there is an asymtotic solution to the state of the output. My experience leaves me short at that point.
  13. Sep 27, 2009 #12


    User Avatar
    Gold Member

    I use the term 'co-rotating angular velocity' for a velocity that I attribute to the flywheel pair A-B.

    Quoting myself:
    I will use the following names:
    Vco = co-rotating angular velocity
    Vcounter = counterrotating angular velocity

    Let the starting angular velocities of A and B be 200 and -200 respectively. When the angular velocity of the prop shaft changes from 0 to 10, and the amount of counterrotation is to remain the same, then the angular velocities of A and B will shift to 210 and -190 respectively.

    In other words, the status of angular velocities 210 and -190 for A and B respectively can in abstraction also be seen as a combination of A and B rotating at +200 and -200 respectively, and A and B rotating at +10 and + 10 respectively.

    In calculations you are free to represent the rotational status of A and B in either way, as in both representations the total kinetic energy of A and B evaluates to the same quantity.
    Mechanically only the co-rotating component interacts with flywheel C. I think that without the decomposition in counterrotating component and co-rotating component the problem cannot be computed.

    Last edited: Sep 27, 2009
  14. Sep 27, 2009 #13
    Setting beginning and ending conditions and knowing that total begining momentum must equal total ending momentum:

    Initial: w for FW A = 200; w for FW B = 200; Ia = .5 m^2*kg; Ib = .5 m^2*kg
    then La = 100 and Lb = 100 (my mistake in stating the original problem), so total momentum is 200*m^2*kg/s

    Also given is the ouput FW C inertia is 5*m^2*kg

    Using your ending conditions: w for FW A = 210; w for FW B = 190; w for FW C = 10
    Now, the momentum for FW C is 50 m^2*kg/s, so FW A and FW B must have a combined momentum of 150*m^2*kg/s.

    If it could be assumed that both wheels have equal momentum (I.E. 75 each) then the the change in inertia can be computed: Ending inertial for FW A is .357143 and the ending inertia for FW B is .394737; Both inertia's have to decrease.

    Of course there are multiple soultions - as long as La and LB together equal 150*m^2*kg/s. But to maintain an equal velocity change for FW A and FW B (equal but opposite), the final inertia for each wheel will not represent an equal change (as per the example above, both had to decrease). So, beginning with known inertia variation, the ending angular velocities will not change equally, but at some point, the difference will be 20 and the output will be 10 and at his point, the output momentum will be 50 and La and Lb together will be 150. Can we determine the respective angular velocities based on inertia values for FW A and B which have varied equally but inversely?
  15. Sep 27, 2009 #14


    User Avatar
    Gold Member

    About the total momentum:
    The relevant constraint is that during the entire process there is no net change of total angular momentum.

    The reason I emphasize this is that there is ambiguity in the attribution of total angular momentum. Since FW A and B are counterrotating, shouldn't the total angular momentum be counted as zero? Is there a right way and a wrong way to do the momentum bookkeeping?

    I don't think there is a "right way" and a "wrong way", the important thing is to be consistent with the bookkeeping.

    In the initial state I prefer to count the momentum of A and the momentum of B as opposite, so I attribute zero angular momentum to the initial state. All my next moves should remain consistent with that, including what I count as increase of angular momentum and decrease of angular momentum. That's very tricky, I find myself making sign errors all the time.

    It'll be several days until I can get back to this problem. In the meantime, good luck.

  16. Sep 27, 2009 #15
    Thank you for your valuable insights. My reason for counting both momentums as postive is that I can extract both to the output as useable momentum. This is because of the use of the differential. If I wanted to set up a situation where all momentum came from either FW A or FW B, I use this momentum in the same rotational direction for FW C.
    I can't say that this is correct. Could it be (as I think you may be suggesting) that there is no momentum available for FW C until there is a delta-w from the input FW's?
  17. Sep 28, 2009 #16
    Using the parameters given in the original premise (but corrected):
    wa=200; wb=200; Ia=.5 to .4; Ib=.5 to .6;
    Initial momentum: La=100; Lb=100; Ltotal=200

    Given ending output condition: wc=10, Ic=5, therefore Lc=50; therefore La+Lb=150
    Using an iterative method, find La and Lb that satisfy wc=10 and Lc=50;
    Answer: La(end)=64.8; Lb(end)=85.2; therefore wa=162; wb=142

    I don't think this is actually a solution because there is no garuntee that these conditions can or will actually happen. It seems this would need to depend on the rate-of-change (given a 10 second period) which would determine the amount of torque developed on the output shaft and therefore the final momentum imparted to FW C. But it seems to be a step in the right direction??
  18. Sep 29, 2009 #17
    I believe I have a solution, though not a very elegant one. A Mathcad .pdf is attached showing the steps. There may be a next step possible to avoid the need to manually enter the ending FW C ang. vel. but for now I think this is adequate.

    Essentially, I am starting with an assumed ending velocity for the output, then altering it to cause the torque on both input flywheels to be equal.

    The kinetic energy equations leave me confused but I can't see an error in the rest of the data. The kinetic energy seems to indicate a high input of work to vary the inertia's but I think more or less one negates the other.

    Thank you for your help. You incouraged me to think along a different direction. If you see glaring errors I would definitely appreciate feedback

    Attached Files:

    Last edited: Sep 30, 2009
  19. Sep 29, 2009 #18


    User Avatar
    Gold Member

    I haven't looked into the attached document yet, but I want to make some remarks.

    I expect that the final outcome does not depend on how fast the inertia of flywheels A and B is changed. 10 seconds, 10 minutes, 10 hours; in a thought experiment, with a perfectly frictionless gear system, I expect no dependency on a time factor.

    I will discuss a number of cases, each simpler than the three flywheel case, to focus on aspects. I will abbreviate 'moment of inertia' to 'inertia.

    - The case of two flywheels connected by a solid rod. If one of them is a variable inertia flywheel, and it contracts, then the final state will be uniquely determined. The underlying assumption is zero elasticity. If there is any elasticity then response can lag behind impressed force ever so slightly. In a thought experiment perfect rigidity can be assumed, and with perfect rigidity response to change is immediate, so every state of motion is an equilibrium state. (The case that is illustrated with the Java simulation I mentioned earlier in the thread is one where there can be a state of dis-equilibrium (which over time resolves). So it's a very different case, and unfortunately unhelpful to understand the three flywheel case.)

    Because of the above considerations I have changed my mind: now I don't think kinetic energy can play a helpful role in the calculation.

    - The case of two flywheels, A and B, connected via gears, in a gear ratio of 1:1 If A has an inertia that is twice as large as the inertia of B, then whenever some force pumps energy in the system the velocity change of B will be twice as large as the velocity change of A. (Compare the case of collision with linear velocity; the lightest particle undergoes the largest change in velocity.) The two flywheels will exert the same torque on each other (action and reaction equally strong, and opposite in direction).

    - The case of two flywheels, A and B, connected via gears, in a gear ratio of 2:1 . The gear ratio influences how much torque each flywheel can exert on the other one. The amount of change of velocity that flywheel A will impart to flywheel B is now determined by two factors:
    - the inertia ratio of A to B
    - the gear ratio of A to B

    - The case of the three flywheels, connected by a diff; the original problem. What will happen in the case where only one of the flywheels, say flywheel A, is a variable inertia flywheel. What if flywheel A is the only one to have energy pumped in it? The torque that is instantly exerted upon the other parts of the system will propagate through the gear system instantaneously. How the change of angular velocity will distribute throughout the system will be subject to inertia ratios, and gear ratios.

    - The case of both flywheels A and B changing angular velocity is a superposition of the two cases.

    Applying these ideas in an actual calculation will take some serious concentration. Earliest opportunity to give it a try is next weekend.

  20. Sep 30, 2009 #19


    User Avatar
    Gold Member

    Today the consequences of the above dawned on me. Flywheel A and flywheel B are variable inertia ratio flywheels, so the inertia ratio relative to the other flywheels changes over time.

    That leads me to the conclusion that the outcome will be affected by the specific order of events. That is, first changing flywheel A, and after that flywheel B, will give a different outcome than first B, then A. A and B simultaneously will give yet another outcome.

    In retrospect I think I was badly off track when I tried to do something with the energy. Right now I think that the only way to get an answer is to apply numerical analysis, using moment of inertia and torque.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Multiple Flywheel System
  1. Hollow flywheel (Replies: 5)

  2. Momentum of Flywheel (Replies: 10)

  3. Flywheel & Energy (Replies: 6)

  4. RPM of a Flywheel (Replies: 4)