Sorry for the double post. I should have re-read for clarity.
To obviate any misreadings:
I said
More correctly: …if the water is still spinning as it leaves, then the sprinkler felt a torque a little bit ago, namely the “reaction” torque due to the “action” torque it gave to the water.
Maybe it was clear from the context of the paragraph that followed anyway…
-P
Pasta:
Thanks, it kinda makes sense now. Of course, if the water is still spinning, that force will be imparted to the head. duuuh! Why didn’t I think of that? If my little hydraulic fittings don’t spin the water, then they’re okay, right
The compressible water thing sounds fishy to me, but I’ll take your word on it.
So the whole thing depends on how the Fricken experiment is set up? Oh well.
Thanks again Pasta. You got a good noodle there.
Often wrong… NEVER in doubt
This is not a solution to the inverse sprinkler problem, but I wanted to give my take on correcting Scylla’s misconception:
Just because a fluid is (ideally) incompressible doesn’t mean it has no pressure gradients (it means it has no density gradients, which is not the same thing). For a trivial example, on the surface of the Earth, a container of fluid has a static pressure gradient that opposes gravity.
But even in the absence of gravity, pressure gradients are associated with any motion of a fluid. Bernoulli’s Theorem essentially states that energy is conserved along a streamline in an incompressible flow (if we can ignore viscous forces). This energy is generally expressed as an energy density which is the sum of two components (three, if there is gravitational potential energy)–the pressure (which is, statistically, an energy density), and the kinetic energy per unit volume of the fluid. Since the sum of these two quantities is conserved along a streamline, if the velocity (thus kinetic energy) of the fluid increases as it follows its trajectory, the pressure decreases, and vice versa.
This is the source of the lift on airplane wings–the fluid on top of the wing is moving faster than that on the bottom, so the pressure is less and voila, we have net lift. Note that for the overall lift calculations, air may generally be treated as incompressible and inviscid (having zero viscosity). Compressibility and viscosity become essential factors only in the boundary layers around the wing.
Any motion of the fluid in the inverse sprinkler case implies pressure gradients. Whether you treat these as causing the motion or resulting from the motion is a matter of point of view, I think. On the other hand, the only way we have to suck the fluid into the sprinkler is to set up a pressure gradient in the first place (which is what a pump does), so perhaps it is reasonable to consider the pressure gradient as cause of the flow in this case.
As Pasta points out, pressure is a statistical concept, and one could argue the case with Conservation of Angular Momentum or with pressure forces equally well. One implies the other.
Rick (ex-Fluid Dynamicist)
Thanks for the help Rick. I learn more here than in college.
Often wrong… NEVER in doubt
Pasta,
Sorry my ASCII diagram didn’t come out just right - I forgot that HTML is rendered without multiple spaces unless you use . Several of the things in it should have been shifted to the right, including the word “less” should have been near the right side where there is less pressure. You seemed to understand it anyway.
I don’t completely buy that you can’t look at the problem from the standpoint of pressures, and must use conservation of momentum. The pressure that I think is key here is on the container walls - on the left-side wall there is a pressure difference, which equates to a force, that is not balanced by a force from the open end. In the squirting-out case, the pressure is higher on the right side of the wall than on the left, so it and the whole container move to the left. In the opposite case, the pressure is higher outside the container, so the force pushes the left-side wall to the right, turning the sprinkler backwards. The container doesn’t have an up-down force because the forces on the top and bottom walls counter each other.
Just because water is incompressible, that doesn’t mean that there are no differences in pressure.
And if you would rather think of the pressure in terms of statistical events of moving molecules, just think of how the molecules on the left side of the wall hit it with more momentum than on the right, so the conservation of momentum wants to move this wall to the right. But as you said, on the open end there is no pushing or pulling, so the force on the left wall is not balanced.
I still say it moves in the opposite direction when sucking.
I never dreamed, when posting this question, that it would garner so many… um… varied responses.
I have since acquired and read What Do You Care What Other People Think? This is the companion volume to Surely You’re Joking, Mr. Feynman, from which mighty tome I got the puzzle. In What Do You Care, Feynman claims the thing doesn’t move.
For whatever that’s worth.
Does he say why?
Another twist on this experiment is to consider if the same apparatus could be used to pump fluid in both directions if it was spun by an outside force. Many pumps use this principle, and I believe you’ll find that they are far more efficient in the “squirt” mode but will also function in the “scoop” mode.
For those of you who have waited 24 years for a definitive answer:
Well, that is a pretty frustrating article for anyone hoping for definitive proof: in the second video in which the flow is reversed, the “device is prevented from rotating in order to improve the visualization of the flow.” 
How extremely unhelpful!
ETA: I am certainly not faulting you, @PastTense, but rather the authors of the article in question. In any event, thank you for posting the link.