When the sprinkler sprays water it will spin counterclockwise. But ...
In 1966, Feynman turned down an offer from the editor of Physics Te...
Physicist Edward C. Creutz, who was in charge of the Princeton cycl...
Note that the hydrostatic pressure is the same in every direction i...
If we analyze the forces acting on the sprinkler and the water we e...
Although this argument is correct it is worth noting that the cross...
Using the same forces analysis we used above but this time for a no...
In the following video you can see someone doing the experiment for...
arXiv:physics/0312087v3 [physics.flu-dyn] 3 May 2004
An elementary treatment of the reverse sprinkler
Alejandro Jenkins
∗
California Institute of Technology, Pasadena, C alifornia 91125
(Dated: CALT-68-2470, Dec. 2003; to appear in the American Journal of Physics)
We discuss the reverse sprinkler problem: How does a sprinkler turn when submerged and made to
suck in water? We propose a solution that requires only a k nowledge of mechanics and fluid dynamics
at the introductory university level. We argue that as the flow of wat er starts, the sprinkler briefly
experiences a torque that would make it turn toward the incoming water, while as the flow of water
ceases it briefly experiences a torque in the opposite direction. No torque is expected when water
is flowing steadily into it unless dissipative effects, such as viscosity, are considered. Dissipative
effects result in a small torqu e that would cause the sprinkler arm to accelerate toward the steadily
incoming water. Our conclusions are discussed in light of an analysis of forces, conservation of
angular momentum, and the experimental results reported by others. We review the conflicting
published treatments of this problem, some of which have bee n incorrect and many of which have
introduced complications that obscure the basic physics involved.
I. INTRODUCTION
In 1985, R. P. Feynman, one of most distinguished
theoretical physicists of his time, published a collection
of autobiographical anecdotes that attracted much at-
tention on account of their humor and outrageousnes s.
1
While describing his time at Princeton as a graduate stu-
dent (1939–1942), Feynman tells the following story:
2
There was a problem in a hydrodynamics
book,
3
that was b e ing discusse d by all the
physics students. The problem is this: You
have an S-shaped lawn sprinkler . . . and the
water squirts out at right angles to the axis
and makes it s pin in a cer tain direction. Ev-
erybody knows which way it goes around; it
backs away from the outgoing water. Now
the question is this: If you . . . put the sprin-
kler completely under water, and sucked the
water in . . . which way would it turn?
Feynman went on to say that many P rinceton physi-
cists, when presented with the problem, judged the solu-
tion to be obvious, only to find that others arrived with
equal confidence at the opposite ans wer, or tha t they
had changed their minds by the following day. Feynman
claims that after a while he finally decided what the an-
swer should be and proceeded to test it experimentally by
using a very large water bottle, a piece of coppe r tubing,
a rubber hose, a cork, and the air pressure supply from
the Princeton cyclotron laboratory. Instead of attaching
a vacuum to suck the water, he applied hig h air pressure
inside of the water bottle to push the water out through
the sprinkler. According to Feynman’s account, the ex-
periment initially went well, but after he cranked up the
setting for the pressure supply, the bottle exploded, and
“. . . the whole thing just blew glass and water in all di-
rections throughout the laboratory . . . ”
4
Feynman
1
did not inform the reader what his answer
to the reverse sprinkler problem was or what the exp e r-
iment revealed before exploding. Over the years, and
particularly after Feynman’s autobiographical recollec-
tions appea red in print, ma ny people have offered their
analyses, both theoretical and experimental, of this re-
verse sprinkler problem.
5
The solutions presented often
have been contradictory and the theoretical treatments,
even when they have been correct, have introduced un-
necessary conceptual complications that have obscur e d
the basic physics involved.
All physicists will probably know the frustration o f be-
ing c onfronted by an elementary question to which they
cannot give a ready answer in spite of all the time dedi-
cated to the study of the subject, often at a much higher
level of sophistication than what the problem at hand
would seem to re quire. Our intention is to offer an ele-
mentary treatment of this problem which should be ac-
cessible to a bright se c ondary school student who has
learned basic mechanics and fluid dynamics. We believe
that our answer is ab out as simple as it can be made,
and we discuss it in light of published theoretica l and
exp erimental treatments.
II. PRESSURE DIFFERENCE AND
MOMENTUM TRANSFER
Feynman spe aks in his memoirs of “an S-shaped lawn
sprinkler.” It should not be difficult, however, to con-
vince yourself that the problem does not depend on the
exact shape of the sprinkler, and for simplicity we shall
refer in our argument to an L-shaped structure. In Fig. 1
the sprinkler is closed: water cannot flow into it or out of
it. Because the water pressure is equal on opposite sides
of the sprinkler, it will not turn: there is no net torque
around the sprinkler pivot.
Let us imagine that we then remove part of the wall
on the right, as pictured in Fig. 2, opening the sprinkler
to the flow of water. If water is flowing in, then the
pressure marked P
2
must be lower than the pressure P
1
,
because water flows from higher to lower pressure. In
both Fig. 1 and Fig. 2, the pressure P
1
acts on the left.
But because a piece of the sprinkler wall is missing in
2
FIG. 1: A sprinkler submerged in a tank of water as seen from
above. The L-shaped sprink ler is closed, and the forces and
torques exerted by the water pressure balance each other.
Fig. 2, the rele va nt pressure on the upper right part of
the open sprinkler will be P
2
. It would seem then that the
reverse sprinkler should turn toward the water, because
if P
2
is less than P
1
, there would be a net force to the
right in the upper part of the sprinkler, and the resulting
torque would make the sprinkler turn clockwise. If A is
the cross section of the sprinkler intake pipe, this torque-
inducing force is A(P
1
− P
2
).
But we have not taken into account that even though
the water hitting the inside wall of the sprinkler in Fig. 2
has lower pr e ssure, it also has left-pointing momentum.
The incoming water transfers that momentum to the
sprinkler as it hits the inner wall. This momentum trans-
fer would tend to make the sprinkler turn c ounterclock-
wise. One of the reasons why the reverse sprinkler is a
confusing problem is that there are two effects in play,
each of which, acting on its own, would make the sprin-
kler turn in opposite directions. The problem is to figure
out the net result of these two effects.
How much momentum is being transferred by the in-
coming water to the inner sprinkler wall in Fig. 2? If
water is moving across a pressure gradient, then over a
differential time dt, a given “chunk” of water will pass
from an area of pressure P to an area of pressure P − dP
as illustrated in Fig. 3. If the water travels down a pipe
of cross-section A, its momentum gain per unit time is
A dP . Therefore , over the entire length of the pipe, the
water picks up momentum at a rate A(P
1
− P
2
), where
P
1
and P
2
are the values of the pressure at the endpoints
of the pip e . (In the language of calculus, A(P
1
− P
2
) is
the to tal force that the pressure gradient across the pip e
exerts on the water. We obtain it by integrating over the
differential force A dP .)
The rate A(P
1
− P
2
) is the same rate at which the
water tra ns fers momentum to the sprinkler wall in Fig. 2,
FIG. 2: The sprinkler is now open. If water is flowing into it,
then the pressures marked P
1
and P
2
must satisfy P
1
> P
2
.
because whatever left-pointing momentum the incoming
water picks up, it will have to transfer to the inner left
wall upon hitting it. Therefore A(P
1
− P
2
) is the force
that the incoming water exerts on the inner sprinkler
wall in Fig. 2 by virtue of the mo mentum it has gained
in traveling down the intake pipe.
Because the pressure difference and the momentum
transfer effects cancel each other, it would seem that the
reverse sprinkler would not move at all. Notice, however,
that we considered the reverse sprinkler only after wa-
ter was already flowing continuously into it. In fact, the
sprinkler will turn toward the water initially, because the
forces will balance only after water has begun to hit the
inner wall o f the sprinkler, a nd by then the sprinkler will
have begun to turn towa rd the incoming water. That is,
initially only the pressure difference effect and not the
momentum transfer effect is relevant. (As the water flow
stops, there will be a brief period during which only the
momentum transfer and not the pressure difference will
be acting on the sprinkler, thus producing a momentary
torque opposite to the one that acted when the water
flow was be ing es tablished.)
Why can’t we similarly “prove” the patently false
statement that a non-sucking sprinkler submerged in wa-
ter will not turn as water flows steadily out of it? In that
case the water is going out and hitting the upper inner
wall, not the left inner wall. I t exerts a force, but that
force produces no torque around the pivot. The pressur e
difference, on the other hand, does exe rt a torque. The
pressure in this case has to be higher inside the sprinkler
than outside it, so the sprinkler turns counterclockwise,
as we expect from experience.
3
FIG. 3: As water flows down a tube with a pressure gradient,
it picks up momentum.
III. CONSERVATION OF ANGULAR
MOMENTUM
We have argued that, if we ignore the transient ef-
fects fr om the switching on and switching off of the fluid
flow, we do not expect the reverse sprinkler to turn at
all. A pertinent questio n is why, for the case of the regu-
lar sprinkler, the sprinkler-water system clearly exhibits
no net angular momentum around the pivot (with the
angular momentum of the o utg oing water cancelling the
angular momentum of the rotating sprinkler), w hile for
the reverse spr inkler the system would appear to have
a net angular momentum given by the incoming water.
The answer lies in the simple observa tion that if the wa-
ter in a tank is flowing, then something must be pushing
it. In the regular sprinkler, there is a hig h pressure zo ne
near the sprinkler wall next to the pivot, so it is this lower
inner wall that is doing the original pushing, as shown in
Fig. 4(a).
For the reverse sprinkler, the highest pres sure is out-
side the spr inkler, so the pushing originally comes from
the right wall of the tank in which the whole system sits,
as shown in Fig. 4(b). The for c e on the regular spr inkler
clearly causes no torque around the pivot, while the force
on the re verse sprinkler does. That the water should ac-
quire a net angular momentum around the sprinkler pivot
in the absence of an external torque might seem a viola-
tion of Newton’s laws, but only because we are neglecting
the movement of the tank itself. Consider a water tank
with a hole in its side, such as the one pictured in Fig. 5.
The water acquires a net angular mo mentum with respect
to any point on the tank’s bottom, but this angular mo-
mentum violates no physical laws because the tank is not
inertial: it recoils as water flows out of it.
6
But there is one further complication: in the reverse
sprinkler shown in Fig. 4, the water that has acquired
left-pointing momentum from the pushing of the tank
wall will transfer that momentum back to the tank when
(a)
(b)
FIG. 4: The force that pushes the water must originally come
from a solid wall. The force that causes the water flow is
shown for both the regular and the reverse sprinklers when
submerged in a tank of water.
it hits the inner sprinkler wall, so that once water is flow-
ing steadily into the reverse sprinkler, the tank will stop
exp eriencing a recoil forc e . The situation is analogous to
that of a ship inside of which a machine gun is fired, as
shown in Fig. 6. As the bullet is fired, the ship recoils ,
but when the bullet hits the ship wall and becomes em-
bedded in it, the bullet’s momentum is transferred to the
ship. (We assume that the collision of the bullets w ith
the wall is completely inelastic.)
If the firing rate is very low, the ship periodically ac-
quires a velocity in a direction opposite to that of the
4
FIG. 5: A tank with an opening on its side will exhibit a
flow such that the water will have an angular momentum with
respect to the tank’s bottom, even though there is no external
source of torque corresponding to the angular momentum.
The apparent paradox is resolved by noting that the tank
bottom offers no inertial point of reference, because the tank
is recoiling due to the motion of the water.
fired bullet, only to stop when that bullet hits the wall.
Thus the ship moves by small steps in a direction op-
posite that of the bullets’ flight. As the firing rate is
increased, eventually one reaches a rate such that the in-
terval between successive bullets being fired is equal to
the time it takes fo r a bullet to travel the length of the
ship. If the machine gun is set for this exa c t rate from
the be ginning, then the s hip will move back with a con-
stant velocity from the moment that the first bullet is
fired (when the ship picks up momentum from the re-
coil) to the moment the last bullet hits the wall (when
the ship comes to a stop). In between those two events
the s hip’s velocity will not change be c ause every firing is
simultaneous to the previous bullet hitting the ship wall.
As the firing rate is made still higher, the ship will
again move in steps, because at the time that a bullet is
FIG. 6: In this thought experiment, a ship floats in the ocean
while a machine gun with variable firing rate is placed at one
end. Bullets fired from the gun will travel the length of the
ship and hit the wall on the other side, where they stop.
being fired, the previous bullet will not have quite made
it to the ship wall. Eventually, when the rate of firing
is twice the inverse of the time it takes for a bullet to
travel the length of the ship, the motion of the ship will
be such that it picks up speed upon the first two shots,
then moves uniformly until the penultimate bullet hits
the wall, whereupon the ship looses half its velocity. The
ship will finally come to a stop when the last bullet has
hit the wall. At this point it should be clear how the
ship’s motion will change as we continue to increase the
firing rate of the gun.
8
For the case of continuous flow of water in a tank
(rather tha n a discrete flow of machine gun bullets in
a ship), there clearly will be no intermediate steps, re-
gardless of the rate of flow. Figur e 7 shows a water tank
connected to a shower head. Water flows (with a co ns e-
quent linear and angular momentum) between the points
marked A and B, before exiting via the shower head.
When the faucet valve is opened, the tank will experi-
ence a recoil from the outgoing water, until the water
reaches B and begins exiting through the s hower head,
at which point the forces on the tank will balance. By
then the tank will have acquired a left-pointing momen-
tum. It will lose that momentum as the valve is closed or
the water tank becomes empty, when there is no lo nger
water flowing away from A but a flow is still impinging
on B.
A. K. Schultz
9
argues that, at each instant, the wa-
ter flowing into the reverse sprinkler’s intake carries a
constant angular momentum around the sprinkler pivot,
and if the sprinkler could turn without any resistance (ei-
ther from the friction of the pivot or the viscosity of the
fluid) this angular momentum would be counterbalanced
by the a ngular momentum that the sprinkler picked up
as the water flow was being switched on. As the fluid
flow is switched off, such an ideal sprinkler would then
lose its angular momentum and come to a halt. At every
instant, the angular momentum o f the sprinkler plus the
incoming water would be zero.
Schultz’s discussion is cor rect: in the absence of any
resistance, the sprinkler arm itse lf move s so as to cancel
the momentum of the incoming water, in the same way
that the ship in Fig. 6 moves to cancel the momentum of
the flying bullets. Resistance, o n the other hand, would
imply that some of that momentum is picked up not just
by the sprinkler, but by the tank as a whole. If we cement
the pivot to prevent the sprinkler from turning at all,
then the tank will pick up all of the momentum that
cancels that of the incoming water.
How does non- ideal fluid behavior affect this analysis ?
Viscosity, turbulence, and other such phenomena all dis-
sipate mechanical energy. Therefore, a non-ideal fluid
rushing into the reverse sprinkler would acquire less mo-
mentum with respect to the pivot, for a given pressure
difference, than predicted by the analysis we carried out
in Sec. II. Thus the pressure difference effect would out-
weigh the momentum transfer effect even in the steady
state, leading to a small torq ue on the s prinkler even
5
FIG. 7: A water tank is connected to a shower head, so that
water flows out. Water in the pipe that connects the points
marked A and B has a right-pointing momentum, but as long
as that pipe is completely filled with water there is no net
horizontal force on the tank.
after the fluid has begun to hit the inside wall of the
sprinkler. Total a ngular momentum is conserved because
the “missing” momentum of the incoming fluid is being
transmitted to the surrounding fluid, and finally to the
tank.
IV. HISTORY OF THE REVERSE SPRINKLER
PROBLEM
The literature on the subject of the reverse sprinkler is
abundant and confusing. Ernst Ma ch speaks of “reaction
wheels” blowing o r sucking air where we have spoken of
regular or reverse sprinklers respectively:
10
It might be supposed that sucking on the re-
action wheels would produce the opposite mo-
tion to that resulting from blowing. Yet this
does not usually take place, and the reason
is obvious . . . Generally, no perceptible rota-
tion takes place on the sucking in of the air
. . . If . . . an e lastic ball, which has one escape-
tube, be a ttached to the reaction- w hee l, in
the manner represented in [Fig. 8(a)], and be
alternately squeezed so that the same quan-
tity of air is by turns blown out and sucked
in, the wheel will continue to revolve rapidly
in the same direction as it did in the case in
which we blew into it. This is partly due to
the fact that the air sucked into the spokes
must participate in the motion of the latter
and ther efore can produce no reactional rota-
tion, but it also r e sults partly from the differ-
ence of the motion which the air outside the
(a)
(b)
FIG. 8: Illustrations from Ernst Mach’s Mechanik
10
: (a). Fig-
ure 153 a in the original. (b). Figure 154 in the original. (Im-
ages in the public domain, copied from the English edition of
1893.)
tube assumes in the two cases. In blowing,
the air flows out in jets, and per fo rms rota-
tions. In sucking, the air comes in from all
sides, and has no distinct rotation. . .
Mach appears to base his tr e atment on the observation
that a “reaction wheel” is not seen to turn when sucked
on. He then sought a theoretical rationale for this obser-
vation without arriving a t one that satisfied him. Thus
the bluster about the explanation being “obvious,” ac-
companied by the tenta tive language ab out how “gen-
erally, no pe rceptible rotation ta kes place” and by the
equivocation about how the lack of turning is “partly
due” to the air “participating in the motion” of the wheel
and partly to the air s ucked “coming in from all sides.”
Mach goes on to say that
11
if we per fo rate the bottom of a hollow cylin-
der . . . and place the cylinder on [a pivot],
after the side has b e en slit and bent in the
manner indicated in [Fig. 8(b)], the [cylinder]
6
will turn in the direction of the long arrow
when blown into and in the direction of the
short arrow when sucked on. The air, here,
on entering the cylinder can continue its r o-
tation unimpeded, and this motion is accord-
ingly compe nsated for by a rotation in the
opposite direction.
This observation is correct and interesting: it shows
that if the incoming water did not give up all its a ngu-
lar momentum upon hitting the inner wall of the reverse
sprinkler, then the device would turn toward the incom-
ing water, as we discus sed at the be ginning of Sec. III.
12
In his introduction to Mach’s Mechanik, mathemati-
cian Karl Menger describes it a s “one of the great sci-
entific achievements of the [nineteenth] century,”
13
but
it seems that the passage we have quoted was not well
known to the twentieth century scientists who com-
mented publicly on the reverse sprinkler. Feynman
1
gave
no answer to the problem and wrote as if he expected and
observed rotation (though, as some have pointed out, the
fact that he cranked up the pressure until the bottle ex-
ploded suggests another explanatio n: he expected rota-
tion and didn’t see it). In Refs. 14 and 15 the authors dis-
cuss the problem and claim that no rotation is observed,
but they pursue the matter no further . In Ref. 16, it is
suggested that students demons trate as an exercis e that
“the direction of rotation is the same whether the flow is
supplied through the hub [of a submerged sprinkler] or
withdrawn from the hub,” a result which is discounted
by almost all the re st of the literature.
Shortly after Feynman’s memoirs appeared, A. T. For-
rester published a paper in which he concluded that if
water is sucked out of a tank by a vacuum attached to a
sprinkler then the sprinkler will not rotate.
17
But he also
made the bizar re claim that Feynman’s original experi-
ment at the Princeton cyclotron, in which he had high
air pres sure in the bottle push the water out, would ac-
tually cause the sprinkler to rotate in the direction of
the incoming water.
17
An exchange on the issue of con-
servation of a ngular momentum between A. K. Shultz
and Forr e ster appeared shortly thereafter.
9,18
The fol-
lowing year L. Hsu, a high school student, published an
exp erimental analysis which found no rotation of the re-
verse sprinkler and questioned (quite sensibly) Forre ster’s
claim that pushing the water out of the bottle was not
equivalent to sucking it out.
19
E. R. L indgren also pub-
lished an experimental result that supported the claim
that the reverse sprinkler did not turn.
20
After Feynman’s death, his graduate research advi-
sor, J. A. Wheeler, published some reminiscences of
Feynman’s Princeton days from which it would appear
that Feynman observed no motion in the sprinkler be-
fore the bottle exploded (“a little tremor as the pres-
sure was first applied . . . but as the flow c ontinued there
was no rea c tion”).
21
In 1992 the journalist James Gle-
ick published a biography of Feynman in which he states
that both Feynman and Wheeler “were scrupulous about
never revealing the answer to the orig inal question” and
then claims that Feynman’s answer all along was that
the sprinkler would not turn.
22
The physical justifica-
tion that Gleick o ffers for this answer is unenlightening
and wrong. (Gleick e choes one of Mach’s comments
10
that the water entering the reverse sprinkler comes in
from many dire c tions, unlike the water le aving a regular
sprinkler, which forms a narrow jet. Although this ob-
servation is correct, it is not particularly relevant to the
question at hand.)
The most detailed and pertinent work on the sub-
ject, b oth theoretical and experimental, was published by
Berg, Collier, and Fe rrell, who claimed tha t the reverse
sprinkler turns toward the incoming wa ter.
24,25
Guided
by Schultz’s arguments about conservation of angular
momentum,
9
the authors offered a somewhat convoluted
statement of the correct observation that the sprinkler
picks up a bit of angular momentum before reaching
a steady state of zero torque once the water is flowing
steadily into the sprinkler. When the water stops flow-
ing, the sprinkler c omes to a halt.
31
The air-sucking reverse sprinkler at the Edgerton Cen-
ter at MIT shows no movement at all.
27
As in the setups
used by Feynman and others, this sprinkler arm is no t
mounted on a true pivot, but rather turns by twisting
or bending a flexible tube. Any tra ns ie nt torque will
therefore cause, at most, a brief shaking of such a device.
The University of Maryland’s Physics Lecture Demon-
stration Facility offer s video evidence of a reverse sprin-
kler, mounted on a true pivot of very low friction, turning
slowly toward the incoming water.
26
According to R. E.
Berg, in this particular setup “while the water is flow-
ing the nozzle rotates at a constant angular sp e e d. This
would be consistent with conservation of angular momen-
tum exce pt for one thing: while the water is flowing into
the nozzle, if you reach and stop the nozzle rotation it
should remain still after you release it. [But, in practice,]
after [the nozzle] is released it starts to r otate again.”
37
This behavior is consistent with non-zero dissipation of
kinetic energy in the fluid flow, as we have discussed. An-
gular momentum is conserved, but only after the motion
of the tank is taken into account.
38
V. CONCLUSIONS
We have offered an ele mentary theoretical treatment of
the behavior of a reverse sprinkler, and concluded that,
under idealized conditions, it should e xperience no torque
while fluid flows steadily into it, but as the flow com-
mences, it will pick up an angular momentum opposite
to that of the incoming fluid, which it will give up as the
flow e nds. However, in the pres e nce of viscosity or turbu-
lence, the reverse sprinkler will expe rience a small torque
even in steady state, which would cause it to accelerate
toward the incoming water. T his torque is balanced by
an opposite torque acting on the surrounding fluid and
finally on the tank itself.
Throughout our discussion, our foremost concern was
7
to emphasize physical intuition and to make our treat-
ment as simple as it could be made (but not sim-
pler). Surely a question about what L. A. Delsasso
called, according to Feynman’s recollection, “a freshman
exp eriment”
4
deserves an answer presented in a langua ge
at the corresponding level of complication. More impor-
tant is the principle, famously put forward by Feynman
himself when discussing the spin statistics theorem, that
if we can’t “reduce it to the freshman le vel,” we don’t
really understand it.
39
We also have commented on the perplexing history of
the reverse sprinkler problem, a history which is interest-
ing not only because physicists of the stature of Mach,
Wheeler, and Feynman enter into it, but also because
it offers a startling illustra tion of the fallibility of great
scientists faced with a question about “a freshman ex-
periment.”
Surely, as the Duchess said to Alice during one of her
adventures in Wonderland, “everything’s g ot a mor al, if
you can only find it.”
40
Acknowledgments
The historical section of this paper owes a great deal to
the list of references on the reverse sprinkler that is given
at the Web site for the University of Maryland’s Physics
Lecture Demonstration Facility.
26
Thanks are due to sev-
eral readers who commented on this paper after it first
appeared in manuscript form, particularly to J. M. Dlu-
gosz, who took it upon himself to clarify the relationship
between this discussion and the account of the experi-
mental results at the University of Maryland. The result
of his inquiries was a useful exchange with R. E. Berg.
∗
Electronic address: jenkins@theory.caltech.edu
1
R. P. Feynman, Surely You’re Joking, Mr. Feynman, (Nor-
ton, New York, NY, 1985), pp. 63–65.
2
R. P. Feynman, Ibid., p. 63.
3
It has not been possible to identify the book to which
Feynman was referring. As we shall discuss, the matter
is treated in Ernst Mach’s Mechanik, first published in
1883.
10
Yet this book is not a “hydrodynamics book” and
the reverse sprinkler is presented as an example, not a
problem. In Ref. 21, J. A. Wheeler suggests that the prob-
lem occurred to them while discussing a different question
in the undergraduate mechanics course t hat Wheeler was
teaching and for which Feynman was the grader.
4
R. P. Feynman, Ibid., p. 65.
5
In the literature it is more usual to see this problem identi-
fied as the “Feynman inverse sprinkler.” Because the prob-
lem did not originate with Feynman and Feynman never
published an answer to the problem, we have preferred
not to attach his name to the sprinkler. Furthermore, even
though it is a pedantic point, a query of the Oxford English
Dictionary suggests that “reverse” (opposite or contrary in
character, order, or succession) is a more appropriate de-
scription than “inverse” (turned up-side down) for a sprin-
kler that sucks water.
6
This observation might seem trivial, but its consequences
can be counterintuitive. The Zapruder film of the 1963
assassination of U.S. president J. F. Kennedy, shows
Kennedy’s head snapp ing backward after the fatal shot,
even though the official theory of the assassination asserts
that the shot was fired from behind Kenned y by gunman L.
H. Oswald. For several decades, conspiracy theorists have
seized on this element of the Zapruder fi lm as evidence that
the fatal shot could not have been fired by Oswald and
must have come instead from in front of the president’s
motorcade. In 1976, L. W. Alvarez published an analysis
of the Zapruder film in which he explained that the jet
of brain tissue that emerged from president’s exit wound
might easily have thrown his head in the direction opposite
to that of the incoming bullet. Alvarez demonstrated this
to his satisfaction both theoretically and experimentally,
the latter by firing at a melon and photographing it as it
moved in the direction opposite to what one would naively
have expected.
7
7
L. W. Alvarez, “A physicist examines the Kennedy assas-
sination film,” Am. J. Phys. 44, 813–827 (1976).
8
Two interesting problems for an introductory un iversity-
level physics course suggest themselves. O ne is to show
that the center of mass of the bullets-and-ship system will
not move in the horizontal direction regardless of the fir-
ing rate, as one expects from momentum conservation. An-
other would be to analyze this problem in the light of Ein-
stein’s relativity of simultaneity.
9
A. K. Schultz, “Comment on the inverse sprinkler prob-
lem,” Am. J. Phys. 55, 488 (1987).
10
E. Mach, Die Mechanik in Ihrer Entwicklung Historisch-
Kritisch Dargerstellt (1883). First published in English in
1893 as The Sci ence of Mechanics: A Critical and Histori-
cal Account of its Development (Open Court, La Salle, IL ,
1960), 6th ed., pp. 388–390.
11
E. Mach, Ibid., p. 390.
12
In Ref. 23, P. Hewitt proposes a physical setup identical
to the one shown in Fig. 8(b), and observes that the de-
vice turns in opposite directions depend ing on whether the
fluid pours out of or into it. Hewitt’s discussion seems to
ignore the important difference between such a setup and
the reverse sprinkler.
13
E. Mach, Op. cit., p. v.
14
P. Kirkpatrick, “A neglected lesson from the Cartesian
diver,” Am. J. Phys. 10, 160 (1942).
15
H. S. Belson, “‘Empty’ hero’s en gine,” Am. J. Phys. 24,
413–414 (1956).
16
Proceedings of the National Science Foundation Confer-
ence on Instruction in Fluid Mechanics, 5–9 September
1960, Exp. 2.2, p. II–20.
17
A. T. Forrester, “Inverse sprink lers: A lesson in the use of a
conservation principle,” Am. J. Phys. 54, 798–799 (1986).
18
A. T. Forrester, “Comments on a letter by A. K. Schultz,”
Am. J. Phys. 55, 488–489 ( 1987).
19
L. Hsu, “Two simple experiments and the resolution of
the Feynman-Forrester conflict,” Am. J. Phys. 56, 307–
8
308 (1988).
20
E. R. Lindgren, “The transport of momentum t heorem,”
Am. J. Phys. 58, 352–357 ( 1990).
21
J. A. Wheeler, “The young Feynman,” Phys. Today 42
(2), 24–28 (1989).
22
J. Gleick, Genius: The Life and Science of Richard Feyn-
man (Pantheon, New York, NY, 1992), pp. 106–108.
23
P. Hewitt, “Figuring physics,” Phys. Teach. 40, 390, 437
(2002).
24
R. E. Berg and M. R. Collier, “The Feynman inverse sprin-
kler problem: A demonstration and quantitative analysis,”
Am. J. Phys. 57, 654–657 ( 1989).
25
R. E. Berg, M. R. Collier, and R. A. Ferrell, “The Feynman
inverse sprinkler problem: A detailed kinematic study,”
Am. J. Phys. 59, 349–355 ( 1991).
26
R. E. Berg et al., University of Maryland Physics Lecture
Demonstration Facility,
<http://www.physics.umd.edu/lecdem/services/demos/demosd3/d3-22.htm>.
27
MIT Edgerton Center Corridor Lab: Feynman Sprinkler,
<http://web.mit.edu/Edgerton/www/FeynmanSprinkler.html>.
28
M. Kuzyk, “Letter,” Phys. Today 42 (11), 129–130 (1989).
29
R. E. Berg and M. R. Collier, “New dev ice lets you un-
water your lawn,” Phys. Today 43 (7), 13 (1990).
30
A. Mironer, “The Feynman sprinkler,” Am. J. Phys. 60,
12 (1992).
31
There are other references in the literature to t he reverse
sprinkler. For a rather humorous exchange, see Refs. 28
and 29. Already in 1990 the American Journal of Physics
had received so many confl icting analyses of the problem
that the editor prop osed “a moratorium on p ublications
on Feynman’s sprinkler.”
30
In one of her 1996 columns for
Parade Magazine, Marilyn vos Savant, who bills herself
as having the highest recorded IQ, offered an account of
Feynman’s experiment which, she claimed, settled that the
reverse sprinkler does not move.
32
Vos Savant’s column em-
phasized the confusion of Feynman and others when faced
with the problem, leading a reader to respond with a letter
to his local newspaper in which he questioned the credi-
bility of physicists who address matters more complicated
than lawn sprinklers, such as the origin of the universe.
33
32
M. vos Savant, “Ask Marilyn,” Parade Magazine, Oct. 6,
1996.
33
A. de Gruyter, “Big Bang Theorists Can’t Simulated Wa-
ter Sprinkler Reversal,” Houston Chronicle, Oct. 26, 1996,
p. A35.
34
J. S. Miller, “Physics in a toy boat,” Am. J. Phys. 26, 199
(1958).
35
R. S. Mackay, “Boat driven by thermal oscillations,” Am.
J. Phys. 26, 583–584 (1958).
36
I. Finnie and R. L. Curl, “Physics in a toy boat,” Am. J.
Phys. 31, 289–293 (1963).
37
R. E. Berg, private communication with J. M. Dlugosz and
A. Jenkins (2004).
38
In the late 1950’s and early 1960’s, there was some inter-
est in the related physics problem of the so-called “putt-
putt” (or “pop-pop”) boat, a fascinating toy boat that pro-
pels itself by heating (usually with a candle) an inner tank
connected to a submerged double exhaust. Steam bubbles
cause water to be alternately blown out of and sucked into
the tank.
34,35,36
The ship moves forward, much like Mach
described the “reaction wheel” turning vigorously in one
direction as air was alternately blown out and sucked in.
39
R. P. Feynman, Six Easy Pieces, (Perseus Books, Cam-
bridge, MA, 1994), p. xxi.
40
L. Carroll and M. Gardner (ed.), The Annotated Alice: The
Definitive Edition, (W. W. Norton, New York, NY, 2000),
p. 91.
In the following video you can see someone doing the experiment for the normal and reversed sprinkler.
[](https://www.youtube.com/watch?v=pkKwSEseJXk&feature=youtu.be&t=470
"The Feynman Sprinkler")
Note that the hydrostatic pressure is the same in every direction in a fluid at a given depth, otherwise the fluid would start to move.
When the sprinkler sprays water it will spin counterclockwise. But what happens when the sprinkler sucks water in?
Although this argument is correct it is worth noting that the cross section (A) of an incompressible fluid flowing along a pressure gradient is not constant. A simple way to see this is using the continuity equation (when a fluid is in motion, it must move in such a way that mass is conserved)
$$
A_1v_1 = A_2 v_2
$$
If $A_1=A_2$ it would mean that $v_1=v_2$ and the fluid would not accelerate.
The author did an oversimplification that it's not completely accurate but doesn't impact the argument. The conservation is enough to prove the magnitude of the momentum transfer but the cross section of the fluid is indeed way more complex.
In 1966, Feynman turned down an offer from the editor of Physics Teacher to discuss the problem in print and objected to it being called "Feynman's problem," pointing instead to the discussion of it in Mach's textbook.
Physicist Edward C. Creutz, who was in charge of the Princeton cyclotron at the time of the incident, and the only person present in the lab besides Feynman, revealed
> "We watched the sprinkler as the pressure built up, forcing water into the top of the sprinkler, and in a few minutes the sprinkler head turned about 5 degrees in the direction a sprinkler usually turns when water is forced into it through its base. There was a little tremor, as Dick called it, and the sprinkler head rapidly moved back to its original position and stayed there. The water flow continued with the sprinkler stationary. We adjusted the pressure to increase the water flow, about five separate times, and the sprinkler did not move, although water was flowing freely through it in the backwards direction, which is consistent with the analysis of Jenkins. The carboy then exploded, due to the internal pressure. A janitor then appeared and helped me clean up the shattered glass and mop up the water. I don’t know what Dick had expected to happen, but my vague thoughts of a time-reversal phenomenon were as shattered as the carboy."
If we analyze the forces acting on the sprinkler and the water we end up with the following diagrams:
where $A\Delta P=A(P_2-P_1)$ is the force on the water due to the pressure difference, $m$ and $v$ are the mass and velocity of the water inside the sprinkler and $F_f$ is the friction force arising from the flow of water inside the sprinkler.
We can write the total force acting on the sprinkler in the horizontal direction as:
$$
F_T = A\Delta P - v \frac{\Delta m}{\Delta t}-F_f
$$
In the case of an ideal fluid $F_f=0$ and assuming that we are already in the phase where the water is flowing continuously inside the sprinkler the condition to get to steady-state motion (a state in which the properties of the system do not change with time) is that $A\Delta P = v \frac{\Delta m}{\Delta t}$ and so $F_T=0$. Since the reaction force $v \frac{\Delta m}{\Delta t}$ when the water hits the sprinkler's inner wall is exactly $A\Delta P$ the total force acting on the sprinkler will be zero.
Note that before getting to the steady state the motion of the sprinkler will be in the direction of $A\Delta P$ since initially there's no water hitting the wall ($v \frac{\Delta m}{\Delta t}=0$) and so $F_T=A\Delta P$.
In the steady state the sprinkler will move in the direction of $A\Delta P$ with constant angular velocity since the total force acting on it is zero.
When we stop sucking in water through the sprinkler the pressure difference ($\Delta P$) is removed and so the total force $F_T=-v \frac{\Delta m}{\Delta t}<0$ acting on the sprinkler in the opposite direction until the fluid and the sprinkler stop moving.
Using the same forces analysis we used above but this time for a non-ideal fluid ($F_f\neq 0$) we end up with:
$$
F_T = A\Delta P - v \frac{\Delta m}{\Delta t}-F_f
$$
where $F_T$ is the total force acting on the sprinkler.
In this case since there's friction between the fluid and the sprinkler we will have momentum and energy transferred from the fluid to the walls of the sprinkler even in the steady state. As a result the pressure difference effect is not completely canceled and $F_T>0$, making the sprinkler move in the direction of the incoming water.