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Feynman Sprinkler

  1. Nov 19, 2013 #1
    Was reading my old copy of Surely You're Joking Mr. Feynman again; ran into the anecdote about the Feynman sprinkler.

    In the book, one guy argues that the water going into the sprinkler will cause the same centripetal force as the water going out of the sprinkler, causing a torque in the same directions for both cases. This results in the sprinkler accelerating in the same direction for both spraying and sucking.

    However, Wikipedia says that an ideal Feynman sprinkler will accelerate "backwards;" can anyone tell me why the above argument was false?

    PS: Apologies for misspelling Feynman. (Edit: corrected that for you. Friendly Admin.)
    Last edited by a moderator: Nov 21, 2013
  2. jcsd
  3. Nov 21, 2013 #2
    Well let's see ... Let's say a submarine has gotten a leak, a strong water stream is hitting the wall opposite to the hole, the the centripetal force in this case is the force that the wall exerts on the water.

    The wall where the hole is has a reduced area, because of the hole. So this wall experiences a reduced force.

    I guess wikipedia would say that the missing force on the wall with the hole is larger that the force caused by the water hitting the other wall.
  4. Nov 21, 2013 #3


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    It wouldn't work with a straight pipe with holes on the sides of the end of the pipe. Assume the sprinkler is "S" shaped. The water flowing along the opposite wall effectively applies an averaged force at a smaller radius than the holes at the end.

    There are still the issues that the water surrounding the sprinkler will resist any sprinker movement, and that there is induced flow of water surrounding the intake flow of water into the holes due to viscosity, some of which interacts with the outer surface of the tubing of the sprinkler.
  5. Nov 21, 2013 #4
  6. Nov 22, 2013 #5

    Well ok. My perhaps too simple theory was:

    A sprinkler is a rocket.
    A reverse sprinkler is a reverse rocket.
    A submarine with a leak is also a reverse rocket.
    So therefore a reverse sprinkler is equivalent to a leaking submarine.
  7. Nov 22, 2013 #6


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    I should have stated that I wouldn't expect it to work with a straight pipe. However after re-thinking this, even with straight pipe, it's possible that a stagnation zone forms near the end of the pipe, which could change the effective radius of the opposing force when the inflow is redirected towards the center of the sprinkler. Still, I seem to recall that the sprinkler was supposed to be S shaped.
  8. Nov 23, 2013 #7
    In case of the submarine I think the forces would all cancel out so the sub would not move.
    This can be shown by looking at momentum. Let's say the hole has an area A and the outside water pressure is p. Because of it's reduced area the water outside will exert a smaller force on the wall with the hole. The difference in force is p*A. So the water outside the sub exerts an effective force of p*A on the sub. That force should theoretically increase the momentum of the sub by p*A*1s every second.
    But the water that rushes into the sub got accelerated by the same force p*A. So that water will also receive a momentum of p*A*1s every second. When it hits the other wall it transfers that momentum to the sub.
    So in short everything cancels out.
    And btw. I also think that a reverse sprinkler is equivalent to a leaking submarine, so the sprinkler won't move except for dissipative effects as explained in the article Gordianus mentioned.
  9. Nov 23, 2013 #8


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    I understand your description regarding how the momentum of stream as it hits the opposite wall would just cancel the difference in pressure, and the submarine would initially have a slight sideways movement cancelled as the stream hits the wall.

    Question is, though, does a reverse rocket accelerate forward or backwards?

    That is why this question about the sprinkler has stumped physicists, beginning with Mach, Feynman, and others.
  10. Nov 23, 2013 #9


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    Why would a centripetal force turn the sprinkler? It acts radially and creates no torque.

    What does "ideal" mean here?

    A simplistic view, that neglects the viscosity of the fluid, leads to the conclusion that the sucking sprinkler will not turn at all. Because there is no net momentum change of the fluid in the plane of rotation: fluid starts at rest, and ends up moving perpendicular to the plane, so momentum in the plane is zero again.

    However, in real life not all fluid set into motion ends up sucked into the pipe. So the fluid gains momentum in the plane of rotation, and the sucking sprinkler rotates opposite to the normal one, but much slower given the same flow rate in the pipe.


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    Last edited: Nov 23, 2013
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