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Interesting Puzzle

  1. Mar 28, 2009 #1
    Some of you may have heard this before, but its one of those classic problem that I love debating, so please weigh in if you think you have the answer...

    You have an S-shaped sprinkler, when water is run out of it the sprinkler rotates about its center axis in the opposite direction to the flow of the water.

    Now imagine submerging this sprinkler in a tank of water an sucking in water. What direction does the sprinkler rotate in?
  2. jcsd
  3. Mar 29, 2009 #2
  4. Mar 29, 2009 #3
    I'm actually not quite satisfied with the wikipedia entry. I suppose that's because I didn't really understand it. The sprinkler rotates because not all of the fluid enters the spout. Maybe that's what the wiki authors meant when they were talking about viscosity, but their language seemed unnecessarily obscure.

    A rocket will expel all of its propellant. But imagine if you could run a rocket backwards. Most of the propellant would get sucked back into the nozzle. But some of the particles would miss and slide by the nozzle rim on the outside. In the case of the sprinkler, it's these uncaptured particles that impart a net torque on the sprinkler.

    The wikipedia article suggested that it's difficult to demonstrate, because the friction of the sprinkler bearings usually overwhelms the very small torque created. That might be true in an incompressible fluid. In his book, Feynman mentioned that he was doing it in a liquid. But it's actually very easy to set up a Feynman sprinkler in air, and I recall with fondness seeing this done in an undergrad lecture many years ago.

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  5. Mar 29, 2009 #4


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    So a sucking sprinkler, rotates in the opposite direction then a spraying one. This seems quite intuitive to me. Why is this interesting or what is here to debate. I guess I miss something.
  6. Mar 29, 2009 #5
    I also do not understand why the sprinkler doesn't turn. It seems there may be two opposing forces at work. The first is that when there is no suction the open end of the sprinkler tube has equal pressure on all sides. When we start the suction, the pressure of the open end of the tube disappears so it would seem that sprinkler would turn in the direction of the open end.

    On the hand, the water flowing through the tube would have centrifugal effect as it rounds the curve in sprinkler and this force would tend to oppose the previous force. The first force I believe would be in proportion to the velocity of liquid in the tube whereas the centrifugal force would increase as the square of the velocity.

    Perhaps this is why Feynman kept increasing the pressure, expecting to see one or other of the forces dominate.
  7. Mar 29, 2009 #6
    Me too
  8. Mar 29, 2009 #7
    It's interesting because you aren't exactly right. It's an interesting problem of viscosity and angular momentum.
  9. Apr 3, 2009 #8


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    Last edited by a moderator: Apr 24, 2017
  10. Apr 3, 2009 #9
    Last edited by a moderator: Apr 24, 2017
  11. Apr 3, 2009 #10
    Yes, you are missing something, but it's hard to see without a picture. Even though the website quoted below draws the wrong conclusion, it does have a few handy pictures. The point is that all the forces and torques on a reverse sprinkler (appear to) cancel each other out.

    A normal sprinkler will rotate very fast, because there are large net torques on it. A reverse sprinkler will rotate very very slowly, and for a different reason. The two situations are not inverses of each other.

    Nice pictures, but they make the same mistake that (according to wikipedia) the great Ernst Mach made when he tackled the problem a hundred years ago.

    The issue here is that the Feynman sprinkler appears easy to model, but it's not. A simplistic model of it would allow us to conclude that all the torques cancel each other out, and that there's no rotation. But that model neglects the fact that in the real world, not all of the water molecules given an impulse by the reverse sprinkler wind up getting sucked into it.

    It's reminiscent of the physicists of yesteryear who argued fiercely that it's impossible to throw a curve ball. Their model of baseballs in flight allowed them to erroneously conclude that a curve ball was just an optical illusion. But the real phenomenon required a more sophisticated model, because real baseballs do curve, no matter what an idealized baseball does.
    Last edited by a moderator: Apr 24, 2017
  12. Apr 4, 2009 #11
    Great post Cantab!!!!!!!
  13. Apr 4, 2009 #12


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    Because of viscosity in the water, any flow into the hole of the reverse sprinkler will also cause the neighboring water to flow against the sprinkler itself, and this external flow exerts an additional opposing torque to the sprinkler. The opposing momentum related torques are both internal and external to the sprinkler.

    Another approach to this would be to look at the angular momentum of the sprinkler and water.
  14. Apr 4, 2009 #13
    Jeff, that is a beautiful sentence! Very illustrative, thanks!
  15. Apr 4, 2009 #14
    I was reading this yesterday afternoon and I had a thought. I'm not sure if anyone else had this thought, but I drew this little diagram quickly to show what I meant.

    It doesn't matter which way the way the water is flowing through the tube, the change in momentum is the same. This creates a force opposite of it on the tube, which always points in the same direction. This could definitely contribute to the net torque being different in my opinion.

    This is what I drew real quick yesterday:

    http://photos-h.ll.facebook.com/photos-ll-snc1/v2618/80/112/13753843/n13753843_46024463_7827909.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  16. Apr 4, 2009 #15


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    I see, I considered only the pressures and missed the inelastic collision in the inverse sprinkler. Is this correct?

    http://img15.imageshack.us/img15/5135/feynmansprinklerweb.png [Broken]
    Last edited by a moderator: May 4, 2017
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