Flywheel Momentum Exchange Through Gear

  • Thread starter jimgram
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  • #1
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I've seen several posts with questions regarding transfer of energy and momentum through a gear box. Following is a problem I'm trying to work out but I clearly don't understand the physics involved:

A flywheel with inertia of Ifw initial velocity of ωfwi = 10,000 rpm is connected (assume through a lossless clutch) to a gear reducer with a ratio of n = 0.5. The output of the reducer is connected to a load inertia, Ild which has an initial velocity of ωld = 0. The flywheel will decelerate (giving up momentum) and the load will accelerate (picking up momentum) until the velocity of the load is equal to the velocity of the flywheel times n. At that point, the system will be in equilibrium with both flywheel and load slowing due only to friction.

As a starting point, I assumed that I could determine the ending velocity of the flywheel by using the conservation of momentum law. The flywheel sees the load inertia as Ild * n2. So, the ending flywheel velocity should be the initial total momentum, Lfwi (which is the same as the ending total momentum) divided by the flywheel inertia, Ifw, plus the reflected inertia of the load, Ild * n2.

I can show where energy plus the work done to decelerate the flywheel and accelerate the load is conserved (treating this as an in-eleastic connection), but momentum only is conserved by taking the momentum as the flywheel ending velocity times the sum of the flywheel inertia and the reflected load inertia. I feel certain this is not correct: The flywheel does have an ending momentum of its ending velocity times its inertia and the load does have momentum equal to the ending velocity of the load times its inertia. And momentum must be conserved. Any ideas?

(The worked out math is shown in the attachment)
 

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Answers and Replies

  • #2
K^2
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Momentum is not conserved in a geared system because of the net torque on gearbox' frame. You have to fix the frame of the reducing gearbox for this to work at all.

Strictly speaking, the only thing you can use is that the torques on both ends of the clutch are equal and opposite, and that turns out to be sufficient for this problem.

We know that for both the flywheel and the load the following holds.

[tex]I \frac{d\omega}{dt} = \tau[/tex]

Then we can find final velocity if we happen to know the torque as function of time.

[tex]\omega_f = \omega_i + \frac{1}{I} \int_{t_i}^{t_f} \tau(t) dt[/tex]

We don't know what the torque actually is at every moment of time, but we do know that the torques on flywheel and load are related through clutch and gearbox thus.

[tex]\tau_{ld}(t) = - n\tau_{fw}(t)[/tex]

So let me write out the equation for final velocity of the load.

[tex]\omega_{ldf} = \omega_{ldi} + \frac{1}{I_{ld}} \int_{t_i}^{t_f} \tau_{ld}(t) dt[/tex]
[tex]\omega_{ldf} = \omega_{ldi} - \frac{1}{I_{ld}} \int_{t_i}^{t_f} \frac{1}{n}\tau_{fw}(t) dt[/tex]
[tex]\omega_{ldf} = \omega_{ldi} - \frac{1}{n I_{ld}} \int_{t_i}^{t_f} \tau_{fw}(t) dt[/tex]
[tex]\omega_{ldf} = \omega_{ldi} - \frac{I_{fw}}{n I_{ld}}(\omega_{fwf} - \omega_{fwi})[/tex]

The final line uses the final/initial equation for flywheel, and we end up not needing the actual time dependence of the torque on the clutch.

Finally, we know the relationship between final velocities. They must match at the clutch, so we have.

[tex]\omega_{fwf} = n \omega_{ldf}[/tex]

And the rest is trivial.

[tex]\frac{1}{n} \omega_{fwf} = \omega_{ldi} - \frac{I_{fw}}{n I_{ld}}(\omega_{fwf} - \omega_{fwi})[/tex]

[tex]\omega_{fwf} = n \omega_{ldi} - \frac{I_{fw}}{I_{ld}}(\omega_{fwf} - \omega_{fwi})[/tex]

[tex](I_{ld} + I_{fw})\omega_{fwf} = n I_{ld} \omega_{ldi} + I_{fw} \omega_{fwi}[/tex]

[tex]\omega_{fwf} = \frac{n I_{ld}}{I_{ld} + I_{fw}} \omega_{ldi} + \frac{I_{fw}}{I_{ld} + I_{fw}} \omega_{fwi}[/tex]

Note that if initial velocity of the load is zero, this does look like conservation of angular momentum, and this does have to do with some interesting properties of simple machines in general, and geared systems specifically.
 
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  • #3
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Thank you K2. I did totally neglect the torque on the gear housing. I will attempt your solution and I think I understand the points made. Thanks again
 
  • #4
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K^2: I can't understand the concept that the torque applied to the gear housing can account for the lost momentum since there is no motion transmitted to the housing. in wrestling with this question, I wondered what would be the difference if the gear reducer was replaced with a belt coupling? If the flywheel had a belt with a 2:1 reduction to a load there is no opportunity for torque to applied to a frame.

The equations you provided leads to a plausible result, however, the initial momentum vs. the ending momentum is slightly more than 2X.
 
  • #5
K^2
Science Advisor
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Movement is not necessary. As long as there is torque applied to the housing, there is transfer of angular momentum.

That's the difference between impulse and work. Work is force over spacial displacement. Impulse is force over time displacement. While lack of movement implies no work, there is always time displacement, so as long as there is an applied force, there is momentum transfer.

Same thing applies to rotational movement, except you get angular momentum instead of the linear one, and the work is now torque over angular displacement.
 

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