Inverse rotatory water sprinkler

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The discussion centers around the behavior of an S-shaped lawn sprinkler when submerged in water and whether it would rotate clockwise or counterclockwise when water is sucked in instead of squirted out. Participants explore the principles of angular momentum, concluding that the sprinkler would not rotate due to the conservation of angular momentum, as the water entering the tube has zero angular momentum. The conversation also touches on the complexities of deriving equations related to this phenomenon using Bernoulli's principle, with references to existing literature and the Feynman sprinkler problem. Ultimately, the consensus is that while theoretical analysis exists, practical applications show negligible torque during operation. The discussion highlights the intriguing physics behind the sprinkler's mechanics.
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an S-shaped lawn sprinkler (an S-shaped pipe on a pivot) in which water squirts out at right angles to the axis and makes it spin in a certain direction is taken and if you had a lake, or swimming pool (a big supply of water) and you put the sprinkler completely under water, and sucked the water in, instead of squirting it out, which way would it turn? Would it turn the same way as it does when you squirt water out into the air, or would it turn the other way?
 
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What do you think?
I don't want to influence others with my thoughts, just to fix it: 29d64d3f6b992f1ce286086f3855ae8d71956b75 generated from [noparse]http://www.hashgenerator.de/[/noparse] with my reply
 
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clockwise in the s shape looking from above?
because conserving angular momentum, water flows in an anti-clockwise direction the nozzle
(on the whole it moves in an anti-clock wise direction,replacing upper half of s by an upside down L)
so tube must move clockwise.
but looking at it from the point of view of basic forces, force due to change in direction of water on the tube is same in both the cases(water squirting out and flowing into the tube),
so direction of torque must be same?
 
The water leaves with zero angular momentum, and I would not expect water in the lake to rotate opposite to the sprinkler, so based on angular momentum conservation, I expect no rotation.

so tube must move clockwise.
Why?

To extend my first statement: You can generate a force like a rocket, but you cannot generate a (significant) net force by sucking in any medium.
 
Thanks, that confirms my expectation. I did not consider the short period where it gets switched on, as this is negligible for realistic setups - and afterwards, there is no torque.
 
can we obtain the equations in the ideal case?
using bernouilli's principle?
 
It is always possible to obtain equations, but it can be complicated sometimes.
 
i am unable to derive it myself
can you?
i don't care it is complicated.
if the solution is long send a link in pdf please.
in your leisure;-)
 
  • #10
I have no interest in making this.
See the existing literature if you want analytic approaches.
 
  • #11
Try a Google search on "Feynman water sprinkler". There's even a Wikipedia page about it.
 
  • #12
Thanks.
 
  • #14
yes,but it is evaluated in steady state condition in which the external torque is zero.
but it is good enough for me.
thanks
 

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