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Topic: Sprinkler (Read 28517 times) |
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Franklinstein
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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.
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| « Last Edit: Oct 23rd, 2003, 8:00pm by Icarus » |
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Kozo Morimoto
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Re: New Riddle: Sprinkler
« Reply #1 on: Jul 31st, 2002, 2:56am » |
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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.
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Anonymous Coward
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Re: New Riddle: Sprinkler
« Reply #2 on: Jul 31st, 2002, 7:08am » |
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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.
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anshil
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Re: New Riddle: Sprinkler
« Reply #3 on: Jul 31st, 2002, 8:15am » |
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I just say, typically Feinmann  )
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- chris
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Re: New Riddle: Sprinkler
« Reply #4 on: Jul 31st, 2002, 8:31am » |
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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 
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Franklinstein
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Re: New Riddle: Sprinkler
« Reply #5 on: Jul 31st, 2002, 2:59pm » |
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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!
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bartleby
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I've read about this one before...
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...
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AlexH
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Re: New Riddle: Sprinkler
« Reply #7 on: Aug 3rd, 2002, 6:41pm » |
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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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #8 on: Aug 6th, 2002, 9:30am » |
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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.
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| « Last Edit: Aug 6th, 2002, 9:40am by Archon » |
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Archon
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Re: New Riddle: Sprinkler
« Reply #9 on: Aug 6th, 2002, 9:54am » |
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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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #10 on: Aug 6th, 2002, 10:21am » |
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Hm, allow me to rephrase that counter counter 
Time reversibility doesn't matter, because the force involved at the bend is going to be in the same direction anyway.
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Chronos
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Re: New Riddle: Sprinkler
« Reply #11 on: Aug 6th, 2002, 11:15am » |
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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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #12 on: Aug 6th, 2002, 5:55pm » |
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on Aug 6th, 2002, 11:15am, 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. |
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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.
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Before the pump is turned on, the net angular momentum of the system is zero, so it must also be zero afterwards. |
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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...
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Archon
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Re: New Riddle: Sprinkler
« Reply #13 on: Aug 6th, 2002, 5:58pm » |
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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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #14 on: Aug 6th, 2002, 6:20pm » |
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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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #15 on: Aug 6th, 2002, 8:37pm » |
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Hm, lemme put it this way...
Top down diagram of a system:
Code:
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.
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Chronos
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Re: New Riddle: Sprinkler
« Reply #16 on: Aug 8th, 2002, 1:13pm » |
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For the other force exerted by the outside water: I'll attempt a diagram of the end of the pipe
Code:
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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #17 on: Aug 8th, 2002, 10:32pm » |
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The gun analogy doesn't work, because you have to either treat the guns as external, or take into account the recoil. |
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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.
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Archon
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Re: New Riddle: Sprinkler
« Reply #18 on: Aug 8th, 2002, 10:40pm » |
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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.
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Kozo Morimoto
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Re: New Riddle: Sprinkler
« Reply #19 on: Aug 11th, 2002, 7:22am » |
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"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 
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Archon
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Re: New Riddle: Sprinkler
« Reply #20 on: Aug 13th, 2002, 11:50pm » |
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on Aug 11th, 2002, 7:22am, 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. |
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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. |
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Yes, this is the bernoulli principle I was talking about.
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Archon
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Re: New Riddle: Sprinkler
« Reply #21 on: Aug 14th, 2002, 12:18am » |
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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!
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Kozo Morimoto
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Re: New Riddle: Sprinkler
« Reply #22 on: Aug 14th, 2002, 4:56pm » |
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Quote:| Now, if you can push a medicine ball away from yourself without it pushing back on you, then I might need to reconsider |
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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....
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Archon
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Re: New Riddle: Sprinkler
« Reply #23 on: Aug 14th, 2002, 8:27pm » |
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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? |
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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.
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Jeffrey Daymont
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Re: New Riddle: Sprinkler
« Reply #24 on: Aug 15th, 2002, 2:51pm » |
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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.
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