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Chronos
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 Re: New Riddle: Sprinkler   « Reply #25 on: Aug 17th, 2002, 3:00pm » Quote Modify

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 the force of air hitting the inside of the tube behind the openings would be much higher than any other force of the air surrounding it.
Not true.  The pressure outside the tube will be approximately one atmosphere, at least initially.  The pressure inside the tube will be somewhere between 1 atm and vacuum.  So the force exerted by the air surrounding the tube will be higher than the force exerted by the air inside.
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Jeffrey Daymont
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 Re: New Riddle: Sprinkler   « Reply #26 on: Aug 19th, 2002, 5:37pm » Quote Modify Remove

on Aug 17th, 2002, 3:00pm, Chronos wrote:
 Not true.  The pressure outside the tube will be approximately one atmosphere, at least initially.  The pressure inside the tube will be somewhere between 1 atm and vacuum.  So the force exerted by the air surrounding the tube will be higher than the force exerted by the air inside.

So focusing on the end of the sprinkler tube, there is one side that is open (designed to spout water) and one side that is closed. On the side that is closed you have 1atm of pressure and on the open side there is less than 1atm of pressure suggesting that the tube should be pushed towards the open end. But what about the force of the air rushing to fill the vacuum? If you blew a stream of air into the opening the sprinkler arm would be pushed away, but that means applying a stream of air higher than 1atm. with the air being sucked through from a supply of air that is already 1atm can the pressure ever exceed 1atm? That doesn't seem possible.

I'm still haunted by the rocket analogy where the exhaust is pushing the rocket from behind. Perhaps that analogy isn't correct since the fuel exhaust is being fired in all directions while the sprinkler head is not simultaneously pulling water in AND pushing water out.

So here's the model as I picture it now. As the open side of the sprinkler drains water from in front of it, water would flow from all directions (including from behind it) to replace it. The pressure from behind the nozzle is what would push it in the direction of the opening (opposite the directon it would normally turn). The amount of pressure on the sprinkler arm would be proportionate to the surface area of the closed end of the sprinkler arm and the surface area of a sphere with a radius equal to the distance from the closed end and the opening of the sprinkler arm. So the force to pull water through the sprinkler would translate to a much smaller force being applied to the arm itself. A high volume of water would need to flow to make a noticeable movement in the sprinkler arm.

Unless, of course, I'm overlooking any important dynamics of fluids in motion.
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Chronos
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 Re: New Riddle: Sprinkler   « Reply #27 on: Aug 20th, 2002, 4:07pm » Quote Modify

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 But what about the force of the air rushing to fill the vacuum?
That force is the pressure force.  As for the blowing, don't just blow into the open end of the tube.  Blow on the other side, too.

In retrospect, blowing into the end is the same situation as the gun firing into the end, and I think this is how I should have answered that, too.
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James Fingas
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 Re: New Riddle: Sprinkler   « Reply #28 on: Sep 3rd, 2002, 12:37pm » Quote Modify

This is a tricky question--why is it in medium?

Here is an interesting experiment that is REALLY EASY to do. Take a garden hose and a bucket. Turn the garden hose on and start filling the bucket. Holding the garden hose over the bucket you can feel a thrust generated by the garden hose. When the bucket gets half full, immerse the garden hose in the bucket. The thrust you felt earlier goes away! Pull it back out, and you can feel the thrust again.

I'm not sure why this happens.

My feeling is that the sprinkler turns in the opposite direction. Reversability is an excellent thought--from what I remember from fluid mechanics, the key to reversability is that the fluid is thick and moves very slowly. If you imagine a pipe slowly sucking up honey, then it seems to me that it would go in the opposite direction to usual.
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Pietro K.C.
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 Re: New Riddle: Sprinkler   « Reply #29 on: Sep 9th, 2002, 2:21pm » Quote Modify

I think that, before we start applying the principle of reversibility, it needs to be established whether the inflow of water into the sprinkler, in the case of the pump establishing a smaller pressure inside it, is time-symmetrical to the outflow of water in the opposite case. I see no reason to believe that it is, but cannot disprove it either. Yet.

I would like to add that reversibility is quite independent of acceleration, and would be very uninteresting indeed if it could only be applied to uniform rectilinear motion!

Also, I think the reversibility you're referring to, James, is a different kind, the thermodynamical kind. In this sense, it has to do with practical reversibility (would the time-inverted version of this event occur spontaneously?), while we are talking about theoretical reversibility (does the time-inverted version of this event violate any laws of physics?).

Hope this gives someone the good idea I've still not had.
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James Fingas
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 Re: New Riddle: Sprinkler   « Reply #30 on: Sep 10th, 2002, 11:54am » Quote Modify

Pietro,
I don't know whether reversible fluid flows are thermodynamically reversible, but I certainly mean that it is time-symmetric.

Our professor showed us this cool movie, showing how some fluid behaviour is time-reversible. The experiment consisted of a cylindrical container filled with glycerin, with a cylindrical rod mounted in the middle. A handle on the rod stuck into the air so you could turn the rod. As you turned the rod, it caused a corresponding rotating motion of the glycerin.

The experiment was to draw a square in the glycerin using some dye, and then rotate the rod in the middle. As you rotated the rod, you could see the square deforming. As you kept rotating, the square was stretched around the rod, and eventually, after a number of turns, it looked like a blurry circle.

However, when you rotated the rod back to its original position, the square came back, exactly like when it started. It was just a little blurrier due to dye diffusion.

The criteria for reversible fluid flow is that the Reynold's number is sufficiently small. The Reynold's number is a measure of the relative importance of viscosity and fluid inertia. Basically, when you can neglect inertia, then the flow is reversible (in time). How this relates to thermodynamic reversibility is a mystery to me.

What I propose is that we analyze the problem at low Reynold's numbers (neglect inertia, reversible flow), and also analyze at high Reynold's numbers (neglect viscosity). If the answers are the same, I'd bet they're the same for medium Reynold's numbers too, and if they're different, then obviously there's a critical point in between where the sprinkler doesn't turn.
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Pietro K.C.
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 Re: New Riddle: Sprinkler   « Reply #31 on: Sep 10th, 2002, 2:53pm » Quote Modify

I guessed at the thermodynamical thing exactly because of the high viscosity-reversibility relationship you pointed out. I'm not really sure, as I've had only a half-semester on fluid dynamics in university, as part of a classical mechanics course (which included stuff like Lagrangian dynamics and vibrations). What it seems to me is that, at high viscosities, the "fluid elements" move around a lot less, so the behavior would resemble more closely that of a solid, which is thermodynamically (i.e. in a practical sense) reversible. This is purely qualitative, but imagine stirring honey. The honey doesn't move around a lot, as compared to water.

What you described in your professor's experiment is precisely what I mean by thermodynamical reversibility. After stirring the dye, there was a practical way of putting it back almost the way it was. One way to explain this is to say the system was "thermodynamically stable" (I'm actually kind of making up the terms as I go along, I've forgotten most of them). I think of it like a gas chamber (which is an extreme case) - if you blow into the gas, you'll change the position of the individual molecules, but altogether the system will look very similar to what it was before. The other extreme case is if you exchange the glycerin by water and the dye by common kitchen salt, NaCl. You can stir backwards all you like, the salt isn't coming down again.

My conjecture was that that high-viscosity fluid flow is (almost) thermodynamically reversible, because of the qualitative argument above. The professor's experiment, plus trying to imagine something diluting thoroughly in honey, seem to agree. I did not mean to imply you were wrong in saying the flow was time symmetric! I do believe that it is, as all physics today.

My point was the following: can anyone prove that, in setting up a contrary pressure scheme in the sprinkler, we also create the EXACT time inverse of normal water flow? Or, conversely, that this inverse water flow implies the EXACT alternate pressure field? Because that would be necessary for a complete reversibility argument, I think.

Just that. Gotta go have dinner, more classes await.
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James Fingas
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 Re: New Riddle: Sprinkler   « Reply #32 on: Sep 11th, 2002, 9:28am » Quote Modify

Pietro,

My answer is yes, we can. All we have to do is make the water flow slow enough, and the Reynold's number will be sufficiently small, giving us a sufficiently reversible water flow.

Of course, we also need to consider faster water flows to solve the problem completely.
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Orius Khan
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 Re: New Riddle: Sprinkler   « Reply #33 on: Sep 16th, 2002, 9:57pm » Quote Modify Remove

I think the answer is that Archon keeps posting counter arguments to himself until the skies open up and a voice from above tells us what really happens...

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[There are no 'Z's in my email address.]
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francis montagnese
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 Re: New Riddle: Sprinkler   « Reply #34 on: Oct 12th, 2002, 10:50am » Quote Modify Remove

it seems to me that when the sprinkler "sucks" in the water it wants to move in the direction that the open end of the tube is facing but, when the water enters the tube it hits the inside wall of the tube and propels it in the opposite direction.  because of these two forces pulling against each other they cancel each other out.  You have to also factor in the viscosity of the water, this also has a "breaking affect on the sprinkler.

Therefore I think that the sprinkler would remain stationnairy.
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Rommel
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 Re: New Riddle: Sprinkler   « Reply #35 on: Nov 11th, 2002, 9:03pm » Quote Modify Remove

I think that the sprinklers would create relative negative pressure, causing them to turn in the opposite direction. The water hitting the bend would not be enough to compensate, as the bend would not be a 90 degree bend.
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James Fingas
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 Re: New Riddle: Sprinkler   « Reply #36 on: Nov 12th, 2002, 9:44am » Quote Modify

I believe that, ignoring fluid viscosity, the sprinkler would remain stationary. Here is my argument, from the point of view of conservation of angular momentum:

Consider a sprinkler that regularly turns clockwise:

\
|
O
|
\

When the water enters the sprinkler head normally, it has no angular momentum (it just flows straight up into the head). As the water exits the head, supposing the head were not turning, it would have a net angular momentum in the counterclockwise direction.

If the head rotates clockwise, it does two things. One, it decreases the counterclockwise angular momentum of the water exiting, and it also provides a reaction force, through friction in the seal between head and body of the sprinkler. This is quite clear.

Now we consider what happens when the sprinkler is submerged. Ignoring fluid viscosity, we consider the fluid as a gas at very low pressures-each gas molecule travels independently of all other gas molecules. Before the sprinkler starts sucking, all water in the test bucket has a net angular momentum of zero.

As water is sucked into the sprinkler head, it temporarily gains angular momentum in the clockwise direction. However, by the time that water reaches the middle of the head, it has lost all or almost all of that momentum. If it has lost all of that momentum, then there is no net angular force imparted on the sprinkler head. However, if a small amount of momentum remains, the water is in effect swirling around in a small whirlpool inside the middle of the head, providing a force on the non-rotating part of the sprinkler, in the clockwise direction. The reaction force is, of course, a counterclockwise force on the rotating part of the sprinkler head.

Therefore, I propose that, at high Reynold's numbers, we can expect either no rotation, or a slight counterclockwise rotation of the sprinkler head.

Since both the inertial and viscous properties of water point to a counterclockwise rotation, the I must conclude that, for all flows of water (ie. all Reynold's numbers), there is either no rotation or a slow counterclockwise rotation (against the usual rotation direction). There will be no rotation if the friction in the bearing between the sprinkler head and body is too high.
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GL ChienFou
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 Re: New Riddle: Sprinkler   « Reply #37 on: Nov 12th, 2002, 2:29pm » Quote Modify Remove

gah! More plumbing!  The sprinkler is a red herring. The only consideration is "Does the system impart angular momentum to the water?". If it does then the sprinkler will turn, if it doesn't then it won't.

At very low speeds we will get laminar flow and the acceleration of the water near to the nozzle will give angular momentum to the water which will move towards the nozzles, which consequently will move "backwards" until steady state is obtained. Whether this overcomes friction is another question. At higher speeds we get turbulent flow and cavitation effects, but my gut feel is still that the partial vacuum at the inlet will still "suck" the sprinkler backwards.

The engineering way of testing this (ie do we gain angular momentum) is to take a length of rubber hose, a large cone full of water. Arrange the hose so that it is parallel to the edge and close to it, just slightly submerged on a track that leads directly to the bottom, using a float and fishing weight arrangement and then run the hose as a syphon. As the water level drops the hose will travel down the cone, near the edge and always just submerged. As the cone becomes nearly empty we will have a large magnification of any imparted angular momentum. My guess is that the water will be swirling towards the pipe inlet. If it is then the sprinkler would turn backwards.

Now who's got a 10,000 gallon conical water tower?
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James Fingas
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 Re: New Riddle: Sprinkler   « Reply #38 on: Nov 13th, 2002, 6:03am » Quote Modify

It occurs to me that a simpler way to test it would just be to submerge a sprinkler in water, suck water backwards into it, and see if it turns backwards. This seems like it would be equivalent (in some sense) to the stated problem. Go ahead--call me crazy
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Chronos
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 Re: New Riddle: Sprinkler   « Reply #39 on: Nov 13th, 2002, 12:38pm » Quote Modify

Quote:
 Since both the inertial and viscous properties of water point to a counterclockwise rotation, the I must conclude that, for all flows of water (ie. all Reynold's numbers), there is either no rotation or a slow counterclockwise rotation (against the usual rotation direction). There will be no rotation if the friction in the bearing between the sprinkler head and body is too high.
I wasn't considering friction, but I agree that in a real physical system it'd probably be enough to prevent movement.  But yes, if the various frictional forces are small enough, it'll rotate opposite its usual direction.
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Kozo Morimoto
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 Re: New Riddle: Sprinkler   « Reply #40 on: Nov 28th, 2002, 3:56am » Quote Modify

http://www.physics.umd.edu/lecdem/outreach/QOTW/arch4/q061.htm
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tdent
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 Re: New Riddle: Sprinkler   « Reply #41 on: Dec 3rd, 2002, 11:57am » Quote Modify

Surely it does come down to total angular momentum, which we can take to be zero at all times. Any given bit of water that flows through the exit pipe starts off with zero ang mom and ends up with zero ang mom (assuming the tubes are set radially on the sprinkler head). Hence the net force in steady operation is zero.

Then the question is what is the ang mom of the water while it is flowing, which must have been created by a transient torque when the flow started, and which must be equal and opposite to the ang mom of the sprinkler. We don't know what the water is doing outside the tube, but it seems reasonable to imagine that its ang mom is either negligibly small, or in the same direction as the water in the tubes (i.e. in the reverse direction to the ang mom of water in 'normal' operation).

So the sprinkler's ang mom and velocity is also in the opposite direction to 'normal' operation.

However, if the sprinkler is not allowed to turn during the initial transient, it should not turn when released.

Of course, this answer is based on an infinitely large tank of water and on the assumption that there really is a 'steady state' in which the ang. mom. of the water is a constant. The same discussion also results in the surprising answer that there is _no net force on the sprinkler_ in steady _normal_ operation, if the sprinkler is underwater in a large tank. The resolution to this paradox must be in the definition of 'transients': for an infinitely large tank, the ang. mom. of water in the tank during 'normal' operation might become indefinitely large, so one might never reach the 'steady state' regime. I don't know if the question of transients also obeys reversibility at sufficiently small Reynolds number.
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James Fingas
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 Re: New Riddle: Sprinkler   « Reply #42 on: Dec 3rd, 2002, 1:17pm » Quote Modify

tdent,

I don't really agree that the angular momentum is zero at all times. Certainly if we consider a bit of water that has just entered the sprinkler nozzle, it has some net angular momentum. From the angular momentum point of view, as the fluid enters the nozzle, it gains a lot of angular momentum (because it speeds up), and then when it goes around the bend, it loses it again.

The key to the angular momentum question is whether or not the fluid has any angular momentum immediately before it enters the nozzle. If it doesn't, then the sprinkler won't spin. The force exerted by the sprinkler on the water to get it to accelerate into the nozzle will balance out with the force of deceleration as it goes around the bend.

However, if the water does have some amount of angular momentum immediately before it enters the nozzle, then the sprinkler will spin. I argue that it does have some, because the sucking motion will tend to suck more water from in front of the nozzle than from behind the nozzle--the water has farther to go if it comes from behind the nozzle. Water will tend to come more from point A than from point B in the diagram below.

AB
\
|
o
|
\

Now picture holding the nozzle so it can't rotate. Because the water is mostly being sucked clockwise, over time the water in the bucket will start to spin clockwise. We know from this that the sprinkler must be exerting a net clockwise force on the water, and by Newton's theory, therefore the water must be exerting a net counterclockwise force on the sprinkler. If we had let the sprinkler head rotate, this counterclockwise force would make the sprinkler rotate counterclockwise.
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jon_G
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 Re: New Riddle: Sprinkler   « Reply #43 on: Dec 3rd, 2002, 5:09pm » Quote Modify

the horse is officially dead, we can stop beating it. Good job fingas.
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mike1102
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 Re: New Riddle: Sprinkler   « Reply #44 on: May 29th, 2003, 1:35pm » Quote Modify Remove

physics 101 people......
Q: Why does the sprinkler turn?
A: Because there is a force on it.
Q: What is that force?
A: force = d/dt(momentum) = d/dt(mv) = v*dm/dt + m*dv/dt. v*dm/dt is "thrust", i.e., the rate at which mass (of the water) exits a nozzel. Because of the geometry of the sprinkler, the thrust produces a torque and the sprinkler rotates. We could submerge the sprinkler in a big tub of water (instead of running it in air) turn on the water hose and the force (thrust) would still be there as long as water flows out the nozzel - and it would still turn clockwise - allbiet slower (more drag on the sprinkler).
Q: What happens to the sprinkler if we suck in water instead of exhaust it?
A: We still have a non-zero v*dm/dt (mass is still flowing) so the thrust is still present, although the direction has changed - we still have a lever-arm to produce torque, so the sprinkler will rotate in the opposite direction because the thrust has changed direction.

Jees..... I hope this right!
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Max M Rosentreter
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 Re: Sprinkler   « Reply #45 on: May 16th, 2004, 6:19pm » Quote Modify Remove

U guys are so gd smart.  It's insane.

I am in AP Physics in high school, and I'm lost.

...I wonder if anyone will read this...
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Icarus
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 Re: Sprinkler   « Reply #46 on: May 16th, 2004, 6:48pm » Quote Modify

on May 16th, 2004, 6:19pm, Max M Rosentreter wrote:
 U guys are so gd smart.  It's insane.   I am in AP Physics in high school, and I'm lost.

Not so much smart (though they are that too), as better educated and more experienced than you are. Get some college physics in line, and a lot of experiential knowedge on forces and reactions, and you will find that some of it is easier to understand. Not all of it: some of it is false, and some of it is ill-explained. I have a degree in physics are have struggled to make sense of some posts. I suggest looking at James Fingas' post (the 3rd above yours) for a simple, easy-to-understand description as to what goes on.

Quote:
 ...I wonder if anyone will read this...

Wonder no more! Forums are sorted by the date & time of the last post, so when you posted, this thread got moved to the top of the list (other than the stickies), and marked as "new" for forum members, so many people will be looking in to see what the new posts were.
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Lozboz
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 Re: Sprinkler   « Reply #47 on: Jun 19th, 2006, 8:15pm » Quote Modify

Well, what causs the sprinkler to go clockwise normally? Is it the force of the water coming out at an angle? Perhaps the sprinkler can only turn clockwise because there is a mechanism in it's stand that stops it from turning counterclockwse? Maybe it depends on the wind?
If it's one of these then the result of the experiment will be different.
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Whiskey Tango Foxtrot
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 Re: Sprinkler   « Reply #48 on: Jun 19th, 2006, 9:58pm » Quote Modify

Hey there Lozboz.  Just to let you know, this topic has been inactive for a bit over two years, which usually means one of two things; either the solution has already been posted, or people much smarter than you or I might ever be have decided that this problem is too difficult.

In this case the answer has been posted, if my memory is not failing me.

The easiest way to explain what happens is that a sprinkler's normal motion is related to the flux of water through the spigot on the sprinkler.  When the sprinkler is attached to a vacuum or something similar, its function is reversed, as well as its flux.  This creates the counter-motion, so that the sprinkler is essentially sucking itself into the empty space it has created.

And as to the wind, it has no effect whatsoever.  It does not drive the sprinkler and in any kind of scientific experiment, a variable like wind would be removed as completely as possible.
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 Re: Sprinkler   « Reply #49 on: Jun 20th, 2006, 3:08pm » Quote Modify

The sprinkler turns normally because of the water coming out at an angle. There is no mechanism in an ordinary sprinkler of this type. This is Newton's Third Law in action. The angle in the pipe pushes the water so that it changes direction from straight out. The water pushes back against the pipe by Newton's Third Law, which causes the pipe to rotate.
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