I'm calculating a hypothetical flywheel/IVT/reducer/vehicle scenerio where a flywheel with a moment of inertia of 312 ft(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}*lb and ang. velocity of 10,000 rpm is connected to 30-in dia. driving tire of a vehicle. The vehicle is 2000 lbs. The flywheel will accelerate the vehicle 0-60 mph in 10 seconds.

I need to solve for the ending angular velocity of the flywheel. This seems simple enough based on the conservation of momentum. Initially, the vehicle has zero momentum and, in 10 seconds, it has gained momentum: p=mass*v or 176,000 ft*lb/s. This translates to angular momentum of the tire: L=p*r, so L=220,000ft^{2}*lb/s.

The initial momentum of the flywheel is L=I*w which is 327,100 ft^{2}*lb/s, so the ending momentum of the flywheel is equal to the initial momentum minus the momentum of the tire: L_{e}=107,100 ft^{2}*lb/s.

Based on this I originally assumed that I could find the ending velocity of the flywheel by using this known ending momentum: w_{e}=L_{e}/I This gives an ending flywheel omega of 3274 rpm.

At 60 mph the tire is turning 672 rpm, so there needs to be a gear reducer between the flywheel and the tire of 4.872:1 (this is assuming that at maximum car speed, the IVT is at 1:1).

If the torque is calculated based on delta*momentum it is found that the torque to decelerate the flywheel is equal to the torque on the tire to accelerate the vehicle. But this can not be correct since there is a 4.872:1 gear box between the two.

I then found as error in my approach in that the moment of inertia changes by the inverse of the square of the ratio. Applying this, however, throws things in much more confusing results. It seems impossible to base an inertia value of the output of a gear reducer on the square of the ratio since immediately the conservation of momentum law is violated.

Any help will be greatly appreciated.

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# Flywheel vehicle drive

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