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Old Nov 15, 2012, 07:32 AM
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The below examples help me understand the (fairly subtle) difference between a system in which the net momentum is changing and a system in which the net momentum stays the same…

Imagine you are standing next to a long, straight, level stretch of railroad track (frictionless of course) with a string of boxcars on it extending for a mile in either direction. If you were to exert a force of magnitude F on one of the boxcars, and the direction of the force was to the right as you faced the track, then Newton’s Second Law says the string of boxcars would accelerate to your right at a rate of F/m (where m is the total mass of the string of boxcars). You would experience a “reaction” force to your left of magnitude F. Just basic Dynamics.

Now imagine that you took the track directly underneath the string of boxcars and bent it into a circle with a circumference of 2 miles (so you’d be able to connect the first and last boxcars). If you were to again exert a force of magnitude F (to your right) on a boxcar, the result would appear very much the same. The boxcars in front of you would accelerate to the right at a rate of F/m, and you would experience a reaction force to your left of magnitude F. There's a difference though...

In the case of the straight track, the center of mass of the string of boxcars accelerates to your right at a rate of F/m (the string experiences net momentum change). In the case of the circular track, the boxcars in front of you accelerate to your right at a rate of F/m, but their counterparts on the opposite side of the circle accelerate to your left at a rate of F/m. The center of mass of the circular string of boxcars remains at the center of the circle (there’s no net momentum change). The harder you push to the right, the harder the track pushes to the left to keep the train on the track. The circular track illustrates that it’s possible to achieve a reaction force by pushing against an object (and putting that object in motion), but without transferring net momentum to that object. All that is required is that you cause something else push on that object with equal magnitude in the opposite direction.

When you exert a force on the circular string of boxcars, you are changing the momentum of the individual boxcars, but when you sum up the momentum change of all the boxcars that are in motion, the net momentum change is zero (momentum is a vector quantity). In this case it is not accurate to say that the force you exert is accompanied by a change in the momentum of the string of boxcars. Instead, the force you exert is accompanied by a reaction force between the string of boxcars and the railroad track.

If you don’t see a fundamental difference between the above examples, then there’s not much point in trying to connect this analogy to a wing pushing against the air. On the other hand, if you see that the straight track example represents momentum exchange and the circular track represents balanced reaction forces read on…

Which of the above examples is a better analogy for a lifting wing in steady flight? If you look at the transverse flow pattern in the wake of a lifting wing, it’s pretty apparent that the downwash travels not straight down, but in closed (or nearly closed) paths. While this is in no way a quantitative proof that there is or is not net momentum left in the wake, it certainly suggests the possibility that there is more at work here than pure vertical momentum exchange.

A quantitative analysis of the momentum in the wake of a lifting wing can be done, and it shows that there is not always equality between the lift and the air’s rate of momentum change.
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