The Feynman Puzzle

Factual Questions
21

Wait, Scylla We’ve already established that the pressure of the water flowing through the tube would make the tube turn backwards. We were trying to establish the direction the tube would turn if suction was applied to draw water in instead of if pressure was applied to push the water out.

My hose will try to straighten itself out anytime pressure is applied, but you may have a very special hose that doesn’t do that. But anyway, that part has already been established in the original statement.

I still don’t see any difference in the direction the tube will turn based on which direction the water is going. It will still turn backwards in both cases.

22

Jimb:
Sorry if I’m not explaining myself well.

Both our hoses (no puns here) attempt to straighten themselves out as a result of the pressure inside them. They do this because the pressure inside the hose is pushing outward. Like blowing up a ballon.

When you turn on the hose it jumps around as a result of the water colliding against the walls of the hose during turns. It’s trying to keep its momentum just as you described earlier. Once full it just bulges.

23

Basically we are talking about the equivalent of a rotating sprinkler except that the head is underwater, were drawing water through the hose instead of pushing it out, and the outflow is somewhere else that doesn’t come into play. Am I correct?

24

Scylla, your summary of the situation is exactly correct.

25

Good, I was getting worried that I had it all wrong and was about to look very stupid.

The thing to understand is that the whole mass of water is pushing the S curve or sprinkler head equally in all directions, including from the inside out. The head is in fact trapped. It can’t move unless another force is applied to cause it to do so.

26

Scylla, you seem to be arguing against the given portion on the problem. Plus against yourself. Forget the hose itself. Several things will influence whether it will move around, like the weight of the hose with water in it, friction against the ground. But, your sprinkler continues to turn even after it is full of water because the moving water tries to go in a straight line. The sprinkler changes the direction of the water and therefore is forced in the opposite direction. The S tube in the problem does the same thing.

The question is, though, if instead of applying water under pressure, you draw water thru by suction, does the tube (sprinkler) still move backwards?

27

Jimb:

I maintain that the head does not rotate in any direction once it’s under water and suction is applied.

It is not the water changing direction that causes the sprinkler head to rotate when we run water through it. It is the force of the water being expelled from the head.

Like a jet plane.

The water being expelled from the head is pushing it backwards. This is like standing on a sled on a frozen pond and throwing a snowball off. The sled moves in the opposite direction.

The effect of the directional change of the water is nil.

28

As for the hose, forget the other stuff. I’m just talking about the way it jumps or torques when you initially fill it, and then fails to once it’s filled.

Once this occurs as the water is being pushed through the hose, it’s also being pulled by the water ahead of it. Otherwise there would be a vacuum.

29

Bricker: Am I making ANY sense?

30

Bricker:

One of your earlier questions was if the sprinkler would spin if water was being expelled, and the whole apparatus is underwater.

Absolutely.

I liked your pun so much I forgot to answer the question.

31

I can see where that might have confused things, and it’s my fault, sorry.

32

Ok, let’s take your jet engine example. When the plane is setting on the runway and the throttle is pushed forward. The thrust out the back pushes the plane forward. But, if the pilot drops those diverters across the back of the engines, the plane goes backwards. The air still comes in the front of the engine and out the back but since the diverters change the direction of the flow the final result is that the plane goes backwards.

To change the direction of any motion, there must be a force applied. When the direction changes, there is an equal and opposite reaction. So, it shouldn’t matter why or which direction the water is moving, it still has to change directions to follow a S curve.

If there wasn’t a tap in the middle and the flow went from end to end, then the flow and the forces would cancel since they would be making a hard left then a hard right. But with the tap in the middle the flow into both ends are making only hard lefts. They therefore add to each other rather than cancel each other.

33

Jim, I am with Scylla on this.

The jet engine example is good, because I can now tell you the difference between the squirting and the suction cases.

When the tube is squirting, the flow of water is in one direction: right out of the tube. This creates the strong push backwards that we all agree would happen.

BUT - when it’s sucking in water, it’s grabbing the water from right around the nozzle, from all different directions, so to speak. So there wouldn’t be the same force.

How’s this: picture the tube under water… and then we pump out water with green food coloring… we’d SEE the green water shoot directly out. But if we then immediately reversed and started sucking water in, the green water wouldn’t be pulled directly back in. Instead, water from right around the mouth of the tube would be pulled in; the green water that was farthest away wouldn’t be touched.

Wow.

That wasn’t one of the options in the book. Feynman had arguments for forwards, and for backwards (the latter of which we have recreated perfectly, I might add) but never gives any answer.

I think this is it. None of the above. It doesn’t move.

  • Rick
34

Scylla,

I’m going to have to weigh in to agree with others that you are making an incorrect argument. As far as I can tell, you are arguing a problem in dynamics with a statics argument. What you neglect is that, once some of the water is moving, there are dynamic pressure effects that may cause the sprinkler head to rotate (think Bernoulli). You seem to be arguing as if all the pressure forces were entirely static.

I am embarrassed to admit that, despite waaay too much education in physics, I can’t remember the resolution of Feynman’s problem. Frankly, I’m not sure if he ever came up with a satisfactory theoretical prediction, which may be why he resorted to experiments.

Rick (not the OP)

35

The sprinkler head does not turn because it is shaped like an “S”. It turns because water is moving out of it, and the force acting on the water acts also on the sprinkler. It acts in the opposite direction. The sprinkler is also acted upon by the force of the water moving out from its center, but that force remains static, since the center does not move. A straight pipe, capped on the ends, with holes on opposite sides of the horizontal about the axis of rotation would act the same. The static forces of pressure are not what causes the sprinkler to sprinkler to spin, it is the movement of water.

The containing medium would not alter that characteristic, other than to modify the speed with which the rotation took place, because of friction, and momentum from mass. An air “sprinkler” would act the same, in air, or in water. Reversing the direction of the air would reverse the direction of the rotation. It would also involve additional frictional elements because the air drawn in is dynamically involved in the forces opposing rotation, unlike the converse case, where our water appeared at the center from outside of the system. Water propellant in a water environment would do so even more. None the less, it would tend to rotate in the opposite direction, albeit inefficiently.

Now, go out and get your garden hose, and a sprinkler, and go into your bathroom, fill the bathtub, and try it out. You will need a pump, for the reversed flow, unless you live on the top floor of a tall building, and have a very long hose.

<P ALIGN=“CENTER”>Tris</P>

“…it doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are – if it doesn’t agree with experiment, it’s wrong.”
–** R.P. Feynman **

36

There seem to be a few arguments:
[ul][li]JimB: The centrifugal force of going around the curve is the same in the case of squirting or sucking, so it spins the same in either case.[/li][li]hansel: The act of water being sucked in will suck the tube along, in exactly the time-reversed way that squirting makes it spin the other way. It turns out this is correct, though not because of the suction acting like reverse thrust theory.[/li][li]Scylla: It doesn’t go anywhere because the water going into the tube is coming from all directions and is not directed like it is in the case of a squirt.[/ul][/li]
First, lets cover the ‘suction as a time-reversal of squirting’ option. This can’t be because fluid that is moving will willingly change directions when being sucked but not when being squirted. To see the difference, hold your hair dryer up and see how far away you can be and still feel the air blast. Now, hold your vacuum cleaner and see how far away you can feel the suction - it’s a much smaller distance. As Scylla said, the water does go in from all directions. Picture someone exhaling a bunch of cigarette smoke. It shoots out in a fairly straight line, then diffuses. Now imagine seeing that film in reverse - all the smoke shooting straight into the guys mouth like a piece of spaghetti. Not gonna happen. Simpler experiment: Blow out a candle. Now, light it again and try to ‘suck’ it out. (danger :Don’t do this if there is the remotest chance you could hurt yourself and sue me.)

So, the suction option doesn’t fly. What about the centrifugal force? The argument about the jet engine is intriguing, though it is worth noting in passing that it demonstrates that the suction of the engine is not what provides thrust, lending more credence to the above logic.

But what about the change in direction (centrifugal force) vs. reaction thrust (rocket-engine type propulsion)? Without doing any math, let me try to convince you it is the thrust that matters most. If it is the change in direction that is important, a jet engine with reversers deployed and running in reverse would work just as well to stop the plane as with the engine running normally.

Let’s just think about this. A jet engine running normally without the reversers achieves a lot of thrust without any direction changing of exhaust. Thus, the thrust must be important. If we now deploy the reversers and the thrust points in the other direction with 100% efficiency (not the case, but let’s just say) then the engine is working just as well to slow the plane.

But wait! We get the extra bonus of the change of direction of the thrust bouncing off the reversers! That must be ADDED to the engine’s thrust, giving a more efficient engine! So then, jet engines should be mounted backwards and fly with their reversers deployed to take advantage of this.

Nope, if Boeing could eke another 5% efficiency out of a jet engine they’d do it in a heartbeat. So, it must be the thrust that matters most, and since we can’t ‘suck’ thrust, the forces on the inverse sprinkler are going to be smaller.

The final answer is, yes, the change in direction is a component of the forces on the sprinkler (or the jet engine) but it is small compared to the thrust component, and it is exactly balanced by the much smaller suction-anti-thrust component.

In fact, if you have a very good experimental setup, you will get the thing to go in the suction direction, but only slowly. The only reason that an inverse sprinkler goes in the opposite direction of the normal sprinkler is the angular momentum change that Scylla describes above with his jerky water hose. After the flow becomes steady, the sprinkler stays spinning slowly, as opposed to spinning quickly when squirting. And, if you add any friction, the thing will just sit there.

Links:
[ul][li]Not terribly helpful: MIT’s Edgerton Center[/li][li]Much better: University of Maryland Physics[/li]Also good: http://www.wiskit.com/marilyn/sprinkler.html[/ul]

37

Grrr - blew it.
[ul][li]Hansel has the direction correct, but the reason for the direction wrong, and the effect is much smaller than in the normal sprinkler mode.[/li][li]There should be a link at the bottom of my post to This article about Marilyn M.V.S.'s answer[/li]The final result is that it will spin in the opposite direction to the normal sprinkler mode, but much slower.[/ul]

38

The MIT link claims if you do the experiment, it won’t move.

The U of M link claims it will move forward.

Apparently they are both describing their actual experimental results.

NOW, I am well and truly baffled. This is clearly confounding the great minds of our time.

  • Rick
39

Hmm… I still can’t believe I actually registered myself just so I could post this. I guess even my ego can have a field trip some days. Anyway, below is a simple argument as to why, under ideal circumstances, the S-shaped tube will stay still while it sucks in fluids.

Key Concept: Conservation of Linear Momentum
First, let’s examine the case of your everyday lawn sprinkler. Just for sake of reference, let’s place the S-shaped tube on a Cartesian plane with your standard x-y coordinates. Now since the system has symmetry, let’s cut of the lower part of the S-shape tube, so we now have a C-shaped tube, with water entering from it’s lower leg. Furthermore, we’ll also assume the C-shape tube is a half circle.

1): When water enters the C-shaped tube perpendicular to the x-y plane, it has no momentum in the x nor y direction.

2): When water exits the C-shaped tube from the top, it now has momentum in the +x direction.

3): Net change in linear momentum is +x.

4): Since linear momentum of the system has to be conserved, the +x gained by the exiting water must be provided by a corresponding -x of the tube itself. This means the tube will spin counterclockwise.

Now with the above in mind, we can tackle Feynman’s problem with ease.

1): When water is just beginning to get sucked in, it has no momentum in the x nor y direction.

2): When the water leaves the tube, it leaves perpendicular to the x-y plane. It therefore has no momentum in the x or y direction either.

3): Net change in linear momentum is 0.

4): No momentum change necessary for the tube, no movement for the tube.

I’d like to stress that any number of real life factors and imperfections in experimental technique and or equipment can throw the final result off. For example, the entrance and exit for the C-shaped tube must be perfectly parallel. You can probably see that even a slight discrepancy will force the tube to provide momentum in the y direction, thus inducing a net torque that turns the tube. Toss in local variations in pressure and the initial momentum of the fluid, it would be hard to perform this experiment with anything except your mind. I believe the reason why the freshmen physics project found the suction sprinkler to turn in reverse is due to such factors. The fact that they reported the suction sprinkler didn’t work with air tends to confirm this suspicion (not enough momentum change to overcome friction due to the low mass of air).

The principal still stands however :slight_smile:

40

I just gotta throw in my $0.02. After reading these posts, I kept waiting for Bernoulli to be used. He was mentioned, but not used. Here’s my backwards rotation argument: Yes, in the reverse case the water is moving toward the entrance from all directions, but at the entrance to the tube, there will be water moving radially across the tube wall’s face (tube has non-zero wall thickness). This moving water will be at a lower pressure that the water “behind” the tube (sum of vectors on back side of “S”). Because of the dynamic situation, the the will be pushed backwards. Furthermore, this should be a very slight effect. If cajoles, I can produce drawings (only if you promise not to laugh).

To clarifiy what I mean by radially moving water, image you are staring straight into the tube opening. The tube appears to you as a ring of non-zero thickness. This radially moving water moves from outside the ring, across the ring’s thickness, and into the inner circle.