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.