# Energy by Impluse Momentum Through a Gear Reducer

1. Jan 17, 2012

### jimgram

I have an inertia torque source (e.g. a flywheel) which is the input to a simple gear reducer, say with a ratio of 3:1, and the output is coupled to a purely inertial load (e.g. no friction or external torque).

Let's say the flywheel has an initial ωfw of 120, so then the output of the gear reducer and velocity of the load is 40. By inertia laod, I mean to accelerate it and the simplest way is to decrease the moment of inertia of the flywheel.

The initial kinetic energy is 1/2*Ifwofwo2 + 1/2*Ild*(ωfwo/3)2. If I reduce the moment of inertia of the flywheel by 1/2, then the ending kinetic energy is: 1/2*1/2*Ifwo*(ωfwo*Ifwo/1/2*Ifwo)2.

The impulse momentum is the integral of torque wrt time.

I can show that the work done to change the moment of inertia due to mpulse momentum on the gear housing is (-Le2/2*Ie)-(-Li2/2*Ii).

The momentum balance is correct when the impulse momentum is added and the energy balance is correct for for a ratio of 1 (I.E. no gear reducer) but is incorrect for any ratio not equal to 1.

I think this must be because there is energy associated with the impulse momentum on the gear housing. Can anyone confirm and if so explain how it's determined? If not, any other ideas as to the energy imbalance. This couldn't get much simpler but I can't get the energy balance straight.

2. Jan 19, 2012

### Andrew Mason

It would help to clarify the problem if you could explain exactly what you are doing. How do you change the moment of inertia of the flywheel? If the flywheel is spinning at the time you reduce its moment of inertia, this will require energy. Where does the energy come from?

With respect to the load connected to the gear box, it is not clear when do you connect it to the flywheel. Is the flywheel spinning freely and then you connect it to the gear box/load? Or is it always connected?

AM

3. Jan 19, 2012

### rcgldr

In a two flywheel system, where the flywheels are somehow coupled, you need something to store and extract energy during transitions in flywheel speed in order to conserve energy and momentum. For example if the flywheels were coupled with a lossless coiled spring, the spring could store and return energy during transitions.

Last edited: Jan 19, 2012
4. Jan 19, 2012

### jimgram

We have two flywheels coupled together through a 3:1 gear reducer. Initially, they both are spinning - fwa is at 120 rad/s, therefore fwb must be spinning at 1/3 times that or 40 rad/s.

I then change the moment of inertia for fwa - it has a mechanism that allows the input of a force to re-position the primary mass changing it's radius from the axis of rotation (I = ms*r2). If I reduce the moment of inertia for fwa to 1/2 of its initial value, the angular velocity of both flywheels must increase to satisfy the conservation of momentum. I mis-stated the first requirement: the work done to re-position the mass of fwa is: (-Lend2/2*Iend)-(Linit2/2*Iinit). This is an input of energy from an external source and must be added to the energy balance value in order to achieve conservation of energy.

In order to achieve conservation of momentum when using a rear reduction the impluse momentum on the gear reducer housing must be added in with the final momentum of both flywheels. The impluse momentum on the gear reducer is$\int$ τgear housing dt. No problems so far.

But the final sum of energies still will not show conservation of energy. My question is: Is there some way that the torque on the gear reducer, which must be accounted for for conservation of momentum, also contributes to the energy balance?

5. Jan 19, 2012

### rcgldr

Shouldn't the work done be equal to the change in energy of the entire system? You could also calculate the work done as the integral of force times Δr for the masses moved in fwa. The force = m v^2 / r and with v and r both varying over time, the integral could become difficult. You would need to find a relationship between v and r on fwa in this system.

6. Jan 20, 2012

### jimgram

As long ago as 2008 I posed a similar question minus the gear reducer. I received a response from mikelizzi with the exact observation that the work done is the integral of force Δr, as follows:

W = ∫ F dr ; cf = F = mass*ω2*r ; But ω varies as the mass is re-positioned

Angular momentum = L =Iω = mr2*ω or ω = L/mr2

Combining these equations:

W = ∫ mω2*r dr = ∫ m(L/mr2)2*r dr = ∫ L2/mr3 dr = L2/m * ∫ 1/r3 dr = -L2/2mr2

So, W = -Lf2/2*If - -Lo2/2*Io

This works out perfectly and for a system using only a variable inertia flywheel; both momentum and enegy are conserved, regardless of any additional non-variable inetia. But when you introduce a gear reduction between the variable inertia flywheel and any other inertial load, and calculate the torque and impulse momentum on the gear housing, you will arrive at: momentum is conserved - energy is not.

7. Jan 20, 2012

### rcgldr

The issue here is that you need to include the angular momentum of the entire system, not just a single flywheel. Calling the source flywheel "a" and the inertial load flywheel "b", and assuming a massless gear system:

Angular momentum = Iaωa + Ibωb

8. Jan 20, 2012

### jimgram

Correct. L in the work equation is meant to include the total system momentum

9. Jan 20, 2012

### rcgldr

If a flywheel is a solid uniform disk, then angular inertia = 1/2 m r2. A hoop or a hollow cylinder has angular inertia of m r2.

However part of flywheel "a" mass is radially movable, so the angular inertia is something other than 1/2 m r2 depending on the distribution of the movable mass versus the non-movable mass on flywheel "a".

10. Jan 20, 2012

### jimgram

That's a very good point. But even though, this equation maintains the conservation of energy even when there is non-variable inertia specified. For example, I can set a maximum and minimum inertia range. The minimum inertia is the portion that is non-variable. As you pointed out earlier, the equation uses the total momentum and the total inertia.

It seems that only when a gear reduction is introduced between a portion of the non-variable inertia the energy balance goes away

11. Jan 20, 2012

### Andrew Mason

You still have not explained the problem very clearly.

So initially you have this first flywheel spinning at angular speed ω and the second flywheel coupled to it at a 3:1 gear ratio so it is spinning at ω/3? Then you change the moment of inertia of the first flywheel by decreasing the radius at which the flywheel's mass is located? And you want to calculate the change in energy of the system? Is that the problem?

How is the system mounted? Is it just spinning in space? If it is spinning without constraint in space then how does the gear work? What prevents the whole thing from spinning together (ie all at the same angular speed)? If it is not constrained in space, why would angular momentum be conserved?

Since you have to add energy to the system to change the moment of inertia, why would energy be conserved here?

AM

Last edited: Jan 20, 2012
12. Jan 21, 2012

### rcgldr

That issue could be eliminated by having both flywheels share the same axis.

geared roller - update - This wouldn't work since the roller could rotate at any speed unless an external torque was applied to hold it in place.

The energy and force to move the masses could be an internal potential source, such as a capacitor and idealized motor / generator that is lossless and consumes no energy if not moving. In such a system, it would seem that total energy (potential + kinetic) would have to be preserved, as well as angular momentum. The issue is figuring out how the radius of the movable masses affects the angular velocity of the system, which then lets you determine the centripetal force versus radius of those masses, then the integral of force times Δr, from r0 to r1 of those masses would equal the energy exhanged between the potential source and angular kinetic energy of the system.

Last edited: Jan 21, 2012
13. Jan 21, 2012

### jimgram

AM - The gear reducer housing is firmly attached to earth. The torque on the housing is the sum of the input torque and the output torque (with proper directions applied I.E. if it's a non-reversing gear reducer then both input and output are positive, etc.)

The impulse momentum on the housing is: ∫τgh(t) dt. Using this value for Lgh(gh=gear housing) plus Lfwa plus Lfwb we will show the momentum is conserved (L(total original) = L(total final).

Regarding your question "why would energy be conserved": If the only energy put into the system or taken out of the system is the work done to re-positon the mass of fwa (Ekms) then Ek(fwa orig.) + Ek(fwb orig.) must equal Ek (fwa final) plus Ek (fwb final) minus Ekms.

rcgldr - I can't really visualize the mechanism you are describing but I know that any mechanism that changes angular velocity will alo change torque and will therefore have a reaction torque applied to it. I think the consequence otherwise would be law-breaking.

14. Jan 21, 2012

### rcgldr

You're correct, the mechanism could simply rotate at the same speed at both flywheels, in which case they have the same angular velocity. As you mentioned, if the geared mechanism is "held" in place, it transfers a torque to whatever is holding the gear in place, but that gives me an idea. What if you coupled two sets of of equally spaced and oppositely rotating flywheels together such as:

fwb0---fwa0-+-fwa1---fwb1

where ωb0 = 1/3 ωa0 = - 1/3 ωa1 = - ωb1

with the gearing and flywheel axis attached to a frame that links all 4 flywheels together. Any torques on the frame and gearing should cancel and the equal and opposing torques would effectively hold the gearing mechanism in place since there is no net torque. This would allow the gearing system to be considered massless.

The angular momentum of the system would always be zero:

Ib0 ωb0 + Ib1 ωb1 = 0
Ia0 ωa0 + Ia1 ωa1 = 0

Note that if the inertias of Ia0 and Ia1 are equally changed, the angular momentum of each set of flywheels {fwa0, fwb0} and {fwa1, fwb1} would not have to be conserved, only canceled by an equal an opposing change in angular momentum of the other set. I don't know if this changes the situation too much from the one you originally wanted.

The energy and force to move the masses radially in fwa0 + fwa1 could be an internal potential source, such as a capacitor and idealized motor / generator that is lossless and consumes no energy if the masses are not moving radially. The total energy of this closed system, potential and angular kinetic, should remain constant.

Last edited: Jan 21, 2012
15. Jan 21, 2012

### jimgram

rcgldr - I not particularly interesting in removing the effect of torque on the gear housing. This is an extremely common application. Cars, machinery of all kinds include transmissions and they are normally bolted to a frame. I think that impulse momentum handles the conservation of momentum equations quite well. I'm just having trouble getting my mind around the problem of conserving energy.

You gave me a good idea regarding the equations for determining the work done due to re-positioning the flywheel mass. I now think that's where the error is.

16. Jan 22, 2012

### rcgldr

I'm not sure what that is. I did a web search for "impulse momentum" and didn't get any hits. In the case of cars or machinery, eventually most of these end up attached to the earth (or at least resting on it), in which case when there are equal and opposing torques between the object and the earth, then angular momentum is conserved only if you include the earth as part of the system.

17. Jan 22, 2012

### jimgram

Think of a flywheel spinning 300 rad/s. It's coupled to another flywheel by a fixed 20:1 gear reducer, so the second flywheel is spinning 15 rad/s. Both flywheels have an inertia of 5 m2 kg, the momentum L of the first flywheel is 5*300=1500 Nms and the second flywheel has L=5*15=75 Nms, so the total momentum is 1575 Nms.

Now increase speed of the first flywheel by reducing its inertia to 3 m2 kg. It's new velocity will be the total sytem momentum seen by the first flyweel, which is ωfw1o*Ifw1o divided by the total new system inertia, which is Ifw1f + (Ifw2/n2) = 498 rad/s

The new momentum for the first flywheel is 3*498=1494 Nms and for the second flywheel it's 5*(498/20)=124.5 making the total of the two flywheels 1618.5 Nms so we picked up 118.5 Nms of momentum, which we know isn't possible. But in order for the first flywheel to accelerate the second flywheel torque was applied through the gear reducer. Impulse momentum is F*t or in this case torque times time which is the same as I*Δω which is the same as ΔL. This can be calculated thus: Lgr = ∫ τgr dt which will be found to be equal to 118.5 Nms regardless of the time period chosen.

18. Jan 22, 2012

### rcgldr

So impulse momentum just accounts for the momentum lost in the two flywheel system? If the entire system is in space, then angular momentum would be conserved, resulting in the entire system rotating. If the system is attached to the earth, then the 'lost" angular momentum was put into the earth.

19. Jan 22, 2012

### jimgram

In space (in an orbiting satellite for example) torque applied to a gear housing transfers to the satellite body and will follow the Δmomentum=torque equation. Knowing the moment of inertia of the satellite with respect to where the gear housing is anchored to it allows the calculation of ω for the satellite about that point.

This can't be done for anything anchored to Earth causing a force or torque. Momentum must be accounted for as torque*t = impulse = ΔL in order to achieve a conservation of momentum.

My original question was: Is there an energy component because of the gear housing torque that must be accounted for to achieve a conservation of energy?

20. Jan 22, 2012

### rcgldr

It can, the change in momentum of the earth will be just as significant as any change in momentum in any other component of a device plus earth closed system, but due to the massive size of the earth, the change in velocity of the earth will be extremely tiny. There have been a few threads that dealt with issues like the acceleration of a car that included the energy and momentum imparted to the earth, in order to show energy and momentum are conserved as viewed from any inertial frame of reference.

There's no reason why you can't consider an idealized situation where there are no losses in the gearing mechanism, so energy of the system would be conserved. Momentum of the entire system is only conserved if the entire system is free to rotate (perhaps mounted at it's center of rotation to a frictionless axle), or if you include the effects on any object that the system is attached to, such as the earth.