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riddles >> medium >> Sprinkler
(Message started by: Franklinstein on Jul 30th, 2002, 3:16pm)

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Title: Sprinkler
Post by Franklinstein on Jul 30th, 2002, 3:16pm
Here is one that came from the book "Surely your joking, Mr Feinman?" Take one of those sprinklers everyone has seen before, the kind that has two pipes sticking out with the ends slightly bent to the side in opposite directions.

Uncouple the hose from its source and plug it into a pump then submerge the sprinkler in some water, so now when you turn on the pump the sprinkler is drawing water in through the pipes. What happens to the sprinkler?

Feinman didn't reveal the answer in the book but he delighted in giving a convincing argument about what would happen until his colleagues agreed with him, then he would give them a different argument until they agreed that was the correct answer.Then he would remind them of his first argument, etc.

I have an opinion, but have never tested it. It would be interesting to hear from someone who has actually done the experiment.
:-/

Title: Re: New Riddle: Sprinkler
Post by Kozo Morimoto on Jul 31st, 2002, 2:56am
Interesting question...

I would assume that it would go the other way to the normal sprinkler operation.  I sort of imagined it like a vaccuum cleaner trying to 'pull' my hand.  Or trying to imagine a rocket propulsion in reverse...

I can't think of an argument that says it will spin the 'normal' way.

Title: Re: New Riddle: Sprinkler
Post by Anonymous Coward on Jul 31st, 2002, 7:08am
Just a quick guess..  I think it wouldn't spin at all.  A vacuum hose does not 'pull' itself forward.  If it was sucking fast enough I suppose it could cause some turbulence in the water that could cause it to move but otherwise I don't think it would at all.

Title: Re: New Riddle: Sprinkler
Post by anshil on Jul 31st, 2002, 8:15am
I just say, typically Feinmann :o)

Title: Re: New Riddle: Sprinkler
Post by - chris on Jul 31st, 2002, 8:31am
Hmph.

1) Backwards: the pipes suck themselves forward, causing the sprinkler to rotate "backwards" (i.e. the opposite of the direction during normal use)

2) No movement: there is no "actio-reactio" going on, ergo, there can be no cause for movement

3) Forward: the water is sucked in through the nozzle of a pipe, then has go go round the bend in the pipe. The change of impetus of the water in the pipe causes the sprinkler to move in the same direction as during normal use (imagine, say, driving with a car on a big sheet of metal, which is supported by ball-bearings. If you turn left, the sheet of metal you're on will be pushed forward)

I would vote for 2) ... but add that the riddle sucks :)

Title: Re: New Riddle: Sprinkler
Post by Franklinstein on Jul 31st, 2002, 2:59pm
My answer to this puzzle is that the sprinkler would not turn because the fluid forces at the junction would counter balance each other. I did a google search on the problem this morning and found out this was the conclusion that Feynman came up with.

That isn't the end of the story, however. Apparently Feynmans apparatus blew up during the experiment because he was sucking air through it instead of water and the pressure built up. He must not have performed the experiment again (he was chastized for making a mess), because experimenters at the University of Maryland found that the sprinkler turned in the opposite direction! They wrote up a paper in that American Journal of Physics and this was confirmed by Loyola University.

I got this information from a forum on the Marilyn is Wrong! website. Another poster, who made essentially the same argument as me claimed that MIT performed the experiment and got a different conclusion. All of the links on the page were dead so I was never able to determine whether MIT got the sprinkler to move in the same direction. Wouldn't that be something!  :)

Title: Re: New Riddle: Sprinkler
Post by bartleby on Aug 1st, 2002, 7:15am

There have been several experiments done, and no conclusive "answer."

I think the answer is:  It depends.  It depends on the length of the sprinkler arms, the density of the fluid, the friction in the system, etc.  Let's ignore the friction....  Imagine a sprinkler with VERY VERY long arms, sucking molten lead.  I think the lever-effect of the long arms would come into play, multiplying the force of the lead.  Imagine short little arms, sucking oil...

Title: Re: New Riddle: Sprinkler
Post by AlexH on Aug 3rd, 2002, 6:41pm
A friend and I discussed this one a long time ago and I think we decided that the steady state answer is that it wouldn't spin, but that in turning it on you'd get a small kick spinning it backwards which would die out due to the viscosity of the water. This was for a model which used pipes that went out radially and then bent. If you allowed the bent part to start at the center then its possible you'd get some steady state motion. In that case the hose could carry away water which still had angular momentum whereas in a long radial section the angular momentum would bleed away against the sides of the pipe.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 9:30am
This is a very interesting puzzle to me because it is analogous to a problem that I have puzzled over before.
My analogous problem is "why do rocket engines work?"

First let me go over my thinking on my rocket problem, and then apply similar thinking to this one. Let me be clear at the start, though... I do not "know" whether my resoning is correct.

OK, the rocket.
Newton has the obvious answer, action and reaction. The rocket ejects mass downwards (exhaust), therefore the rocket is propelled upwards.
But... (there had to be a but)....
What if we opened the rocket at both ends, such that exhaust could be directed in both directions equally. Then, obviously, the forces counterbalance and the rocket goes nowhere.
In other words, the action of exhaust moving downwards is not critical here. The critical point is that the rocket is designed such that the exhaust can only go downwards. If it goes upwards it collides against the rocket, pushing it up in a simple mechanical collision.
To put it another way....
Assume we have a physical construct in a vacuum. The construct is a metal cylinder of any size, open at both ends. Into this cylinder we inject hydrogen and oxygen, and ignite. Now, pick one of the resulting water molecules. This water molecule will be moving in some random direction. If it collides with the wall of the cylinder, it will "push" the cylinder in the direction it was moving. If, however, its trajectory simply takes it out one of the open ends, then it applies no force to the cylinder. There is no "action at a distance".
Now close one end, making a rocket.
Pick a molecule of water created by the combustion.
If its trajectory takes it straight out the open end of the cylinder without any collisions (with the cylinder walls or any other fuel or exhaust molecules) then this molecule of water has not provided any "thrust" to our rocket. The fact that there are a lot of molecules involved, and the fact that there is only one "way out", however, means that there will, in fact, be collisions with the walls and "top" end of the cylinder, which will impart force, and therefore thrust the rocket in those directions. The "wall" collisions will counterbalance on average, but there is no counterbalance to the closed end. Therefore it is not really the ejection of exhaust that pushes the rocket. Rather it is the collisions of gasses against the "closed end" of the rocket.

OK, that was long.
Now, for the pump.
Again, there is no "action at a distance". The pump creates low pressure within the sprinkler nozzles, but molecules of water moving into those nozzles have not imparted any force unless they actually collide with some part of the sprinkler. However, when they "turn the corner" of the corner of the sprinkler, they must have been acted on by some force (change of direction equals acceleration, acceleration requires force). The direction of that net force must be a line that intersects the "corner" of the sprinkler, and points away from that corner, in order to account for the differnece between the incident and resultant velocities of the water. In other words, if the bend in the sprinkler was shaped like the letter "L", with the open end at the top of the L, then the force will be like a "/", pointing up and to the right. (sum of two forces at right angles, one vertical and upwards to stop the vertical motion of the water, and one horizontally to the right, pushing the water further down the sprinker tube).
The equal and opposite force to this force will be applied to the sprinkler, down and to the left. This pushes the sprinkler away from its "open end". The horizontal forces never matter, because an equal and opposite horizontal force will be happening on the other sprinker nozzle. If the sprinker only had one nozzle, it would also be pushed sideways, and begin some kind of spiral movement I would think.
Interestingly, this means that the sprinker rotates the same way no matter which direction the water is moving. The vector additions are essentially the same if the water is coming out of the sprinker, or being pushed in (yes pushed: theres no such thing as a pull force. It is the higher pressure outside nozzle that pushes water into it)
(edit: i am talking about mechanics here. Let's ignore arguments about the 4 fundamental forces shall we? Yes I know gravity and the strong nuclear force are attractive, and that electromanetism can be depending on polarity. None of these points are really relevant in this context, I think)
In the case of water being ejected, the "resultant" force on the water in order to change its direction at the bend must be up and to the right (again, assuming an L shaped bend with the spout at the top). Therefore the force on the sprinkler is down and to the left. Again, the horizontal force is cancelled out by the force at the opposite nozzle.
Addedum: yes, the two "vertical" forces on either nozzle are also equal and in opposite directions, but they cannot be summed and cancelled because of the different locations on which they act. Both "horizontal" forces (in the "L" diagram) act on the pivot of the nozzle.
Extra complication:
The bernoulli principle may provide extra complication here. In the case of the submerged nozzle, the water at the mouth of each nozzle may well be moving faster than the water at the "bottom" of the "L" bend. The resultant force is vertical and "upwards" on the sprinkler. This would sum negatively with the resultant vertical downwards force of the water changing direction at the bend. Unfortunately, the solution as to whether the magnitude of this force is greather than the force at the L bend probably requires a quantitative analysis, which is beyond my capability to provide. It may depend on the velocity of the water. It may not. Someone with experience with fluid dynamics is required to add to my mechanical analysis.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 9:54am
Here's a counter to what I wrote above, and then a counter to the counter :)

Mechanics is supposedly "time reversible". In other words, it does not depend on the direction of time.
So, lets say that the sprinker is always submerged in order to make the two different situations (water in vs water out) look time reversible.
Then clearly under time reversibility, if the sprinkler goes in one direction in one instance, it must go in the other direction when we reverse time.

I don't think time reversibility applies here, though. Because accelerations are involved. In much the same way that acceleration is the resolution of the "twins paradox". In this pardox, one twin stays on earth while the other flies off at the speed of light and then returns. From each twins frame of reference, the other's clock appears to run slower. But the clocks clearly cannot be both behind and ahead of each other.
See http://zebu.uoregon.edu/~js/glossary/twins_paradox.html. The key is that only one twin undergoes an acceleration.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 10:21am
Hm, allow me to rephrase that counter counter   :P

Time reversibility doesn't matter, because the force involved at the bend is going to be in the same direction anyway.

Title: Re: New Riddle: Sprinkler
Post by Chronos on Aug 6th, 2002, 11:15am
Good analysis, Archon, but you didn't include all of the forces.  Remember, there's also water outside the pipe, pushing on the elbow.  The pressure pushing on the elbow from the outside would be greater than the pressure on the inside of the elbow, which would tend to cause the sprinkler to turn in the reverse of the usual direction.

Another way to look at this problem is with conservation of angular momentum.  The water can leave the pump in any direction, so there's no reason to suppose there's any angular momentum there.  Before the pump is turned on, the net angular momentum of the system is zero, so it must also be zero afterwards.  If the sprinkler is turning in its usual direction, then water drawn into the sprinkler must be rotating in the same direction, at even greater speed.  Thus, we would have angular momentum of the pipe and the water adding, and it's no longer zero.  That can't be.

On the other hand, if the pipe is turning in the direction opposite normal operation, then the water in the tank will be moving in the opposite direction of the pipe, so the angular momenta can cancel.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 5:55pm

on 08/06/02 at 11:15:33, Chronos wrote:
 Good analysis, Archon, but you didn't include all of the forces.  Remember, there's also water outside the pipe, pushing on the elbow.  The pressure pushing on the elbow from the outside would be greater than the pressure on the inside of the elbow, which would tend to cause the sprinkler to turn in the reverse of the usual direction.

Can you explain the origin of this force? I'm not sure what you're talking about here. Yes theres water outside the pipe... it's submerged... but how does that contribute a force? I mean, if it did, then why doesnt the sprinkler move even before you turn it on? As you can see, I haven't grasped the force you are talking about here.

Quote:
 Before the pump is turned on, the net angular momentum of the system is zero, so it must also be zero afterwards.

I admit, I'm not very good with angular momentum. All I can really remember is that it must be conserved in the absence of an external rotational force (ie, torque). But isnt the pump providing a torque force here? I mean, what about in the situation where the pump is ejecting water. Before you turn on the pump the total angular momentum of the system (sprinkler + water) = 0. Surely after you turn the pump on, the total angular momentum of the system is no longer zero...

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 5:58pm
Ah, wait, no, I see what you mean about the conservation of angular momentum. The sprinkler imparts angular momentum to the water when it turns the bend, so the sprinkler must counterbalance that by turning "towards" the nozzle opening. Is that what you mean? Hm. I'll have to think about that.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 6:20pm
OK, I don't agree with the conservation of angular momentum argument.
I say this because the change in direction of the water as it turns the bend is not the result of a torque, because the water is not spinning on an axis, it is just a change in direction. On the other hand, the water is imparting a torque to the sprinkler, so the angular momentum of the sprinkler increases. Angular momentum only applies to the sprinkler, not the water.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 6th, 2002, 8:37pm
Hm, lemme put it this way...
Top down diagram of a system:

Code:
 A.               \___.___D   C   O   \                                              .B

Now, lets say we have a gun at A and B, which fire at the angled walls (at C and D). The bullets are directed towards the pivot at O, and the system spins counterclockwise.
Now reset the system, but move the guns to O, and fire them at C and D. The bullets are directed towards A and B, but the system still spins counterclockwise.

Title: Re: New Riddle: Sprinkler
Post by Chronos on Aug 8th, 2002, 1:13pm
For the other force exerted by the outside water:  I'll attempt a diagram of the end of the pipe

Code:
 A______     B|_____ |    | |

OK, now there's a pressure force pushing up on the wall of the pipe at B.  This is the 3rd law complement of the force that you identified as diverting the water molecules.  There's also a pressure force at A, pushing down.  We know that the pressure inside the pipe is less than the pressure outside, because of the pump.  That means that the force exerted by the water at A is greater than the force exerted by the water at B.  Over most of the pipe, these pressures will cancel out, but the opening at the end of the pipe prevents this here.

As for the angular momentum argument:  If you think of the pump as providing an external torque, then it gets tricky.  So expand the system to include the pump, and the water coming out the end of the pump.  If the water leaves the pump straight down, say, there's no angular momentum in it.

Now, we have water going into the pipe.  That means that the water has a counterclockwise angular momentum.  To balance this, the pipe must have a clockwise angular momentum.

The gun analogy doesn't work, because you have to either treat the guns as external, or take into account the recoil.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 8th, 2002, 10:32pm

Quote:
 The gun analogy doesn't work, because you have to either treat the guns as external, or take into account the recoil.

But surely the guns are external. I mean, I could have the guns braced by a completely independent system. The recoil imparts no force on the spinning system. To put it another way, if I am playing pool and cue the whiteball at a colour, I don't have to take into account the force the whiteball imparts on me when determining what happens to the colour. I know that it will be equal and opposite to the force I impart on it, but that doesn't affect the outcome. Similarly, the pump will be subjected to a force due to the water that it pumps, but the force applies to the pump, not the sprinkler. The pump system imparts a torque to the sprinkler (one way or another), and feels an equal and opposite force, but we dont care about which way the pump (or the pump/delivery hose/bracing system/planet to which it is bolted) spins.

I understand your other force better now though, yes.

I am out of ideas on this one.  :-/

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 8th, 2002, 10:40pm
Actually, I want to say a couple more things about the gun analogy.
If you like, we can make the guns part of the system and take recoil into account. Take my diagram from above, and bolt the guns at O directed at C and D. The gun recoil forces are equal and opposite. The net force is zero. Even if we only had one gun, the recoil still doesn't impart a torque.

My reservation with the gun analogy is that it's not really accurate to think of moving fluid as a "hail of bullets". In my gun I am sure I am certainly correct, but I am not at all sure it is analogous to what we have here.

Title: Re: New Riddle: Sprinkler
Post by Kozo Morimoto on Aug 11th, 2002, 7:22am
"In other words, the action of exhaust moving downwards is not critical here. The critical point is that the rocket is designed such that the exhaust can only go downwards. If it goes upwards it collides against the rocket, pushing it up in a simple mechanical collision.
To put it another way....
Assume we have a physical construct in a vacuum. The construct is a metal cylinder of any size, open at both ends. Into this cylinder we inject hydrogen and oxygen, and ignite. Now, pick one of the resulting water molecules. This water molecule will be moving in some random direction. If it collides with the wall of the cylinder, it will "push" the cylinder in the direction it was moving. If, however, its trajectory simply takes it out one of the open ends, then it applies no force to the cylinder. There is no "action at a distance"."

I think this is incorrect.  You don't need anything to "push" the inside of the cylinder for it to move.  Its the actual force of pushing the water molecule that moves the rocket/cylinder.  The rocket doesn't just ignite the H2 and O2, it has a nozzle to direct the ignition.

If you sit in an office chair with wheels on a flat smooth surface, then flap your arms about, your chair wouldn't move.  However, using the same action, you throw a medicine ball or a shot put or a bag of cement, your chair would move in the opposite direction.  The bag of cement doesn't have to hit you in the face for the chair to move.

I think there is some kind of vaccuum action going on with the sprinkler in water.  For the water to be sucked into the 2 nozzles, it would create a vaccuum just outside of the nozzle for a brief amount of time.  This will cause the nozzle to move towards the vaccuum (from the pressure behind) and the surrounding water to move towards the vaccuum as well.  So the overall process would rotate the sprinkler in the opposite direction to its normal use.  It'll probably depend all on viscosity and the vaccuum pressure etc etc.  Just like if there is not enough water pressure, the sprinkler won't rotate - the water will just dribble out of the 2 nozzles.

So can someone do this experiment in their back yard swimming pool and tape it and put the divx on the net so that this can be settled once and for all!  If only I could attach the pool filter to a sprinkler - pool filter inlet is too wide to be attached to a normal hose sprinkler :-(

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 13th, 2002, 11:50pm

on 08/11/02 at 07:22:32, Kozo Morimoto wrote:
 However, using the same action, you throw a medicine ball or a shot put or a bag of cement, your chair would move in the opposite direction.  The bag of cement doesn't have to hit you in the face for the chair to move.

That's true, but it doesn't change the argument. If you are in physical contact and exerting a force on the ball, then it is exerting an equal and opposite force on you. Now, if you can push a medicine ball away from yourself without it pushing back on you, then I might need to reconsider :) I do not understand how exhaust going "down" can exert an upwards force on the rocket unless the rocket has also exerted a downwards force on the exhaust. Where does this downwards force come from if not from the fact that the "top" of the combustion chamber is closed off?

Quote:
 I think there is some kind of vaccuum action going on with the sprinkler in water.

Yes, this is the bernoulli principle I was talking about.

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 14th, 2002, 12:18am
I'm going to change my answer to the sprinker question, although I maintain my reasoning...

The conservation of momentum/bernoulli argument and the mechanical collision argument seem to lead in opposite directions, but on further consideration, I think I have a holistic answer.

The mechanical argument is certainly justified and suggests it will spin in the same direction.
The bernoulli/momentum argument suggests it will spin in the opposite direction.

I think the deciding factor is the initial state of the system.

Clearly, the force with which the water "hits" the elbow joint and is directed around the bend is the same as the force accelerating the water into the sprinkler in the first place. Therefore, these two forces should cancel out completely. Essentially, the sprinker should not move at all.

But if there is any turbulence inside the sprinkler hose, or any air trapped, then there will be fractionally less force applied in "going around the bend" at some times.

Therefore, in a frictionless environment, the sprinkler would probably accelerate slightly in the backwards direction until the "suction" was counterbalanced by the "collision". At this point the forces cancel and in a frictionless environment the sprinkler would continue to rotate "backwards". However it is clearly not a frictionless environment, and I believe it would soon come to a stop.
If you were to cover the holes for a moment and then let the water come in again, it would, in my opinion, probably begin to spin "backwards" again for a brief period, before once again the forces balanced and we return to the same situation.

So my position is this: the kinetic force and the conservation of momentum / bernoulli pressure work equally and in opposite directions, but the momentum argument applies very slightly before the kinetic force takes effect. Therefore the forces are slightly unbalanced and the sprinkler rotates backwards until friction brings it to a stop.

I'm still hoping someone can actually confirm this though. I can't get this damn riddle out of my head!

Title: Re: New Riddle: Sprinkler
Post by Kozo Morimoto on Aug 14th, 2002, 4:56pm

Quote:
 Now, if you can push a medicine ball away from yourself without it pushing back on you, then I might need to reconsider

This may be getting a bit overboard and silly, but what would happen if you have a electro magnetic ballistic device whereby you can project a heavy mass (like an iron ball) by 'pulling' it along a rail/barrel?  (e-magnet at the tip of the barrel, put the iron ball at the other end, switch on the magnet, ball moves towards the tip and switch off magnet once the ball is a the tip.  the ball would continue on its way past the tip)  Since the mass isn't pushing against you, wouldn't this mean that you won't move?  The tip would move towards the ball and the ball would move towards the tip....

Title: Re: New Riddle: Sprinkler
Post by Archon on Aug 14th, 2002, 8:27pm

Quote:
 This may be getting a bit overboard and silly, but what would happen if you have a electro magnetic ballistic device whereby you can project a heavy mass (like an iron ball) by 'pulling' it along a rail/barrel?

You still cant get past Newton's third law: every action has an equal and opposite reaction. The electromagnet exerts a force on the ball, and the ball exerts an equal and opposite force on the electromagnet. The only difference is that the force is no longer due to collision (gas pressure) but is now due to electromagnetism instead (yes yes ok fundamentally they are the same but that's beyond the scope)

This is something you can actually play with pretty easily... get 2 little bits of iron. Coil a wire around one and clip the ends to a battery, making an electromagnet. Put the two near each other on a table or something. Assuming the field is strong enough to overcome friction, both of them will move, each towards the other one.

I didn't want to bring this up because I thought it would compilicate things, but this is how an ion rocket works. Instead of directing the exhaust downwards by collision, you instead used plasma (ionized gas) as exhaust and direct it backwards with an electric field. It is a much more efficient method of propulsion because every atom/molecule of exhaust contributes to the thrust of the rocket.

Title: Re: New Riddle: Sprinkler
Post by Jeffrey Daymont on Aug 15th, 2002, 2:51pm
I've been enjoying the discussion. Here's a couple of thoughts:

If you disconnected the pump leaving the base of the sprinkler open to the water from underneath, then turned the sprinkler head by hand, would the water travel from the center out to the ends or from the outside in? if the ends are not bent I think we could agree that the water would travel from the center to the outside. But if you spun it in the direction opposite from how it spins normally there would be a force pushing the surrounding water back into the opening. The directon that the water flows would depend on the size of the openings, inside diameter of the tubes, and the length of the tubes. A careful balance of these dimensions might even produce a result where the water inside the sprinkler does not move at all. This might support the answer stated earlier 'it just depends'. I don't know exactly how this helps with the original problem though.

The other imagined experiment is to attach the sprinkler to the inside of the space shuttle and poke a hole through the wall so that the air in the shuttle is sucked out into space through the sprinkler. Now you have a very low viscosity substance being pulled through at a very high force. My hunch is that it would spin in the same direction as it would when sprinking water -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. NASA has yet to approve my experiment.

Title: Re: New Riddle: Sprinkler
Post by Chronos on Aug 17th, 2002, 3:00pm

Quote:
 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.

Title: Re: New Riddle: Sprinkler
Post by Jeffrey Daymont on Aug 19th, 2002, 5:37pm

on 08/17/02 at 15:00:41, 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.

Title: Re: New Riddle: Sprinkler
Post by Chronos on Aug 20th, 2002, 4:07pm

Quote:
 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.

Title: Re: New Riddle: Sprinkler
Post by James Fingas on Sep 3rd, 2002, 12:37pm
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.

Title: Re: New Riddle: Sprinkler
Post by Pietro K.C. on Sep 9th, 2002, 2:21pm
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. ;D

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. :)

Title: Re: New Riddle: Sprinkler
Post by James Fingas on Sep 10th, 2002, 11:54am
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.

Title: Re: New Riddle: Sprinkler
Post by Pietro K.C. on Sep 10th, 2002, 2:53pm
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. :P

Title: Re: New Riddle: Sprinkler
Post by James Fingas on Sep 11th, 2002, 9:28am
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.

Title: Re: New Riddle: Sprinkler
Post by Orius Khan on Sep 16th, 2002, 9:57pm
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...  ;D

-----
[There are no 'Z's in my email address.]

Title: Re: New Riddle: Sprinkler
Post by francis montagnese on Oct 12th, 2002, 10:50am
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. ;D  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.

Title: Re: New Riddle: Sprinkler
Post by Rommel on Nov 11th, 2002, 9:03pm
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.

Title: Re: New Riddle: Sprinkler
Post by James Fingas on Nov 12th, 2002, 9:44am
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.

Title: Re: New Riddle: Sprinkler
Post by GL ChienFou on Nov 12th, 2002, 2:29pm
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?

Title: Re: New Riddle: Sprinkler
Post by James Fingas on Nov 13th, 2002, 6:03am
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 ;)

Title: Re: New Riddle: Sprinkler
Post by Chronos on Nov 13th, 2002, 12:38pm

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.

Title: Re: New Riddle: Sprinkler
Post by Kozo Morimoto on Nov 28th, 2002, 3:56am
http://www.physics.umd.edu/lecdem/outreach/QOTW/arch4/q061.htm

Title: Re: New Riddle: Sprinkler
Post by tdent on Dec 3rd, 2002, 11:57am
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.

Title: Re: New Riddle: Sprinkler
Post by James Fingas on Dec 3rd, 2002, 1:17pm
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.

Title: Re: New Riddle: Sprinkler
Post by jon_G on Dec 3rd, 2002, 5:09pm
the horse is officially dead, we can stop beating it. Good job fingas.

Title: Re: New Riddle: Sprinkler
Post by mike1102 on May 29th, 2003, 1:35pm
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!

Title: Re: Sprinkler
Post by Max M Rosentreter on May 16th, 2004, 6:19pm
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... :-/

Title: Re: Sprinkler
Post by Icarus on May 16th, 2004, 6:48pm

on 05/16/04 at 18:19:23, 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.

Title: Re: Sprinkler
Post by Lozboz on Jun 19th, 2006, 8:15pm
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.

Title: Re: Sprinkler
Post by Whiskey Tango Foxtrot on Jun 19th, 2006, 9:58pm
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.

Title: Re: Sprinkler
Post by Icarus on Jun 20th, 2006, 3:08pm
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.

Title: Re: Sprinkler
Post by SWF on Jul 27th, 2006, 10:37pm
Most of the proposed explanations are not accounting for the relevant factors in this problem. This is a challenging problem, and not just a Physics 101 exercise (if it was so simple Feynman and the other great physicists at Los Alamos would not have been confused by it).

The submerged sprinkler is acted upon by all the fluid that touches it. To evaluate the forces and moments, the flow field in the fluid must be found, which is governed by the Navier-Stokes equations. Solving these equations is not easy and simplfying assumptions are often made. If it is assumed that viscous effects are negligible compared to intertia of the fluid ("ideal fluid"), the Navier-Stokes equations are simplfied, and the forces acting on the sprinkler are unchanged if the direction of the fluid flow is reversed. If instead it is assumed that inertia effects are neglible compared to inertia effects ("Stokes flow") then all forces are reversed when direction of fluid flow is reversed.  The above comments include some other assumptions, such as the fluid is incompressible, steady state flow conditions, and the effects of external body forces such as gravity are not included.

In the case of an ideal fluid (i.e. viscous effects negligible) the only forces acting on the sprinkler are pressure forces, which may be obtained from the flow field by use of the Bernoulli equation. With ideal flow in two dimensions the surface of the sprinkler is traced by streamlines, and from the Bernoulli equation pressure is lowest where velocity is highest- such as where the fluid exits the nozzle. After sketching streamlines for sprinklers of a couple of shapes, I am confident that with ideal flow the shape can be chosen to make the sprinkler rotate either toward or away from an exiting flow . However, the shape that most resembles the common sprinkler rotates in a direction opposite from what most people probably expect (i.e. opposite the way the sprinkler turns when it ejects water when surrounded by air). And being that this is ideal flow, the direction of rotation for a sprinkler of a given shape is the same whether fluid is being forrced out or drawn in.

Of course, real fluid is not the same as an ideal fluid. An important effect not included in the above simplfication is flow separation. For outward flow from the nozzles there is flow separation from the nozzle edges that results in a jet or column of fluid being fired into the surrounding fluid. Separation occurs down stream of the flow direction, and when flow is drawn into the sprinkler any flow separation occurs inside the sprinkler- there is not a column of fluid sucked out of the surrounding fluid. The flow outside a sprinkler during suction resembles the flow field during ideal flow. It is not like the inverse of when flow goes outward, and this is a big differecne between the two situations. During outward flow, the turbulence in the jet of fluid causes energy losses- Bernoulli's equation can't be used to find pressure on the entire surface of the sprinkler, With these losses the outward flowing case could have higher pressure near nozzle outlet, and rotate in the opposite direction from when fluid is drawn into the sprinkler.

Title: Re: Sprinkler
Post by Pyotr on Jan 25th, 2012, 10:14am
Just on a side note, Port-a-Loo operators seem to be quite calm about just sticking in a hose to suck out toilet filth, while fire-fighters are usually quite concerned about a rampant nozzle.

From this, I surmised that however the inverse sprinkler behaved, it would depend on pressure differentials – sucking only gives one bar, while pumping can go a lot higher.

And yes, I did watch the video linked to above demonstrating the immersed sprinkler.

Title: Re: Sprinkler
Post by Grimbal on Jan 27th, 2012, 10:24am

on 07/30/02 at 15:16:00, Franklinstein wrote:
 Feinman didn't reveal the answer in the book but he delighted in giving a convincing argument about what would happen until his colleagues agreed with him, then he would give them a different argument until they agreed that was the correct answer.Then he would remind them of his first argument, etc.

And yet, isn't it obvious what result he observed?
The result was that the device exploded under the pressure.  This means that he didn't see any clear motion in either direction and decided to increase the pressure until he saw something happen.