I suppose one cannot draw a little hart on its side, one cannot stand on it nor out any pendulum there, it is so slippery that one cannot observe Magnus effect in a stream of a liquid or gas, it is made of stuff that is so ideal that there is no compression, it would not change shape because of rotation. One cannot stand on it so one cannot see how the sun and other stars are moving around.

I suppose that it is so ideal that any rotation does not have sense at all.

Stam is right. It is connected to the Coriolis force.

I suppose one cannot draw a little hart on its side, one cannot stand on it nor out any pendulum there, it is so slippery that one cannot observe Magnus effect in a stream of a liquid or gas, it is made of stuff that is so ideal that there is no compression, it would not change shape because of rotation. One cannot stand on it so one cannot see how the sun and other stars are moving around.

I suppose that it is so ideal that any rotation does not have sense at all.

I'm wondering how on earth you are going to see the ball, save measure its movements, if you forfeit electrodynamics. Quantum or classical, measurement is interaction with what you measure.

If we assume you may somehow see the ball, the answer is simple. Since you designed you ball to be rotationally symmetric, may only use it to detect a rotation of which the centre does not coincide with that of the ball.

Maybe the ball is so slippery that the proposed wave solutions do not work either (one would have to think hard about what perfectly slippery means in terms of boundary conditions). What if we prepare some region of space with a considerable variation in the gravity force. Then, by using only radial directed forces on the ball, we should be able to maneuver it into that region (say close to a neutron star). There the center of mass and center of gravity is detectably shifted with respect to each other, which I believe imply observable effects on ball motion.

Simpler: What is the motion of a rotating ball in a varying gravity field, all calculated classically and non-relativistically? In GR it is known that spinning particles do not follow geodesics exactly, due to the coupling of spin to spacetime curvature.

Notes added: I think one can make a model for seeing the ball, in terms of photons bouncing back in such a way that only a radial directed force (read momentum transfer) is exerted on the ball.

I think the solution by Remi, in terms of (gravitational) spin-orbit forces, is in the same category as mine.

More additions:

Most other proposed solutions, although involving interesting and relevant phenomena for realistic cases, I find to be outside the spirit and intent of the question. If the sphere was distorted, it would not be an ideal sphere. Sound or light should be treated as phonons or photons; i.e. particles scattering off a perfectly slippery surface. Temperature is not mentioned explicitly, but that should be considered an oversight, so that it should be considered zero. Actually, perfectly slippery must be interpreted to mean zero absorption coefficient (otherwise another loophole is opened) hence zero emission. Measured Sagnac effect requires instruments on the surface; they will slip off.

GR induced frame dragging due to rotating mass is a valid answer, in my opinion, although a bit challenging to measure.

I can add the reference to the Sagnac effect which is also sensitive to the rotation of the ball.

One can detect the rotarion of a reference frame, the other explanation for the movement observed would be if all the universe was rotating in the opposite direction.

I did a little check of how classical mechanics of rigid bodies has been taught at NTNU (by me). That ruled out my proposed solution; an inhomogeneous gravity field can lead to a torque, which changes angular momentum. But that is, according to premises of the question, not observable. Therefore, for a perfect sphere, the center-of-mass motion is unchanged. For the same reason Remi's proposal does not work either. There is no spin-orbit coupling (between perfect spheres) in Newton gravity (relativistic generalizations are explicitly ruled out in the question). Electromagnetic waves are also explicitly ruled out. Sound waves is just an effective description of particle mechanics, which cannot work for individual particles; hence they cannot work either. GR effects are explicitly ruled out in the question.

My conclusion that only the answers by Pavlo himself and Igor are correct, according to how the question is formulated.

I found the premises of the question to be quite clear -- on second reading. But it is for Pavlo to decide if I have misunderstood him (and likewise with regard to the other posters). It is obviously not a very practical or realistic question, but thought-provoking. I liked it!

Note added: As I said (implicitly) in my previous post, the answer is in my judgement No. No answers in the opposite direction, including several attempts I have given, satisfies the premises of the question.

Stam> The second is the probe. if the probe isn't specified, the question is meaningless.

But Stam, the question is to find a probe, if there is any -- within the constraints of the question. There seems to be none.

Fortunately, in research the best approach is to change the question, if the original one proves impossible :-)

My take on that one: how do you know the sphere is rotating? If not only it is infinitely slippery, but has no marks of any sort linked to it, then it is impossible to state that it rotates, and it is then indeed impossible to find any effect of the rotation.

But if there are, say, four red dots, non-coplanar, marked on your sphere, then you can make a reference frame fixed to those four dots, and then see whether a freely moving particle goes in a straight line in that reference frame. If it does, the sphere is not rotating.

Actually, I would agree the question not to be metaphysical at all: If an inertial system is defined, say, by directions to distant radio sources, we ask whether this particular construct rotates or not, and we are dealing with an object quite nearly as impalpable as that delightful sphere.

Remi> will cause the sphere to precess and lock to the other similar-sized perfect body.

Since we cannot see the rotation, we cannot see the precession either. And, since the sphere is perfect by assumption, it will not generate Newtonian gravity forces which depends on rotation, precession or nutation. One would have to go to relativistic (gravity-magnetic) effects for that (ruled out in the question). It does not matter how many asymmetric gravitating bodies you arrange around. At first I thought it would be possible too. Not so, not with a perfect sphere.

The lock-in you mention is probably a more interesting physical effect for real bodies, but that is another question.

If your ball is so slippery, that you are unable to create moment of force, then you are right.

Thank you all for the answers. As has been mentioned by Kåre Olaussen spin-orbit coupling absent in classical mechanics, and really, this question appeared in my brain when I was a student and learned spin-orbit coupling.

As for Stam Nicolis's notice about incorrect question and the absence of a probe: I thought that "absolute slippery" condition is enough. This is a just "ideal imaginary experiment" and you can imagine any ideal probe (may be instead of "ideally rough substance" which could stop motion of ideal slippery substance).

F. Leyvraz, of cause, any mark could not be put on the absolute slippery surface.

Simple picture: imagine that you are in 18 century and take this ball in your hands and you can not decide (due to it's ideal slippery) is it rotating or not.

Slippery refers to the tangential, not the normal components. Normal incidence on a rotating perfect conductor leads to different reflection than from a non-rotating conductor, precisely because the former defines a non-inertial frame. Classical mechanics laws means solving the equations of motion, nothing more and nothing less. While the term ``perfect conductor'' simply is one way of realizing the perfect slip condition, the non-inertial character of the rotating frame is the only relevant issue.

The reason for the confusion, of course, is the refusal to translate ambiguous words to equations. If this is done, things become clear. If the goal is just metaphysical speculation and increasing the number of answers, this doesn't have anything to do with physics and simply abuses the forum's counters.

On the issue of marks: I am afraid it is a bit deeper than just the question of making some paint stick. If you can in any way identify a constituent particle of your sphere, then you have a mark and you can trivially decide whether the system is turning. If there is no way to identify any constituent particle, then I am back to my original remark: what do you mean by saying that it is turning? Is this any different from wondering whether a structureless point is turning on its axis or not?

Francois> If there is no way to identify any constituent particle, then I am back to my original remark: what do you mean by saying that it is turning?

GR would introduce frame-dragging in the vicinity. That is in principle detectable (only ruled out in the question). More generally, I think almost any relativistic theory of gravity will introduce velocity dependent forces which can be detected, by probes which are not spherical or slippery.

Pavlo> spin-orbit coupling absent in classical mechanics

I don't think "spin-orbit coupling" is generally absent in rigid-body Newtonian mechanics. The center-of mass will move as if the total force worked on it. In addition there will be a torque which changes the state of rotation. Dependent on the force distribution the total force and torque will in general change with orientation of the body at fixed center-of-mass (but not for a perfect sphere); that leads to a possible coupling between rotational and translational motion. It seems wrong/misleading to call this effect spin-orbit coupling though.

Stam Nicolis The reason for the confusion, of course, is the refusal to translate ambiguous words to equations. If this is done, things become clear. If the goal is just metaphysical speculation and increasing the number of answers, this doesn't have anything to do with physics and simply abuses the forum's counters.

Equations here is a classical Newton's laws. What we able to do in range of this setting? We can throw this ball in no-air gravity field F = Mg and look at trajectory (it will be parabolic - it is not depend on rotating). OK. We can let the ball on an inclined surface and measure time and speed, math law again F=Mg and again rotating doesn't make influence to rolling. ОК. We can put this ball into the bag and try to rotate this bag in different ways and measure energy which we must apply, does this energy depends on rotating axis - no, again. The math equations here is E = Jw2/2, J is moment of inertia, w - rotating frequency. E.t.c. I thought that my conditions are clear and intuitively understandable what methods of classical mechanics I meant. There is no big choice: throw, rolling, rotating...

An extended body, such as a sphere, isn't, only, characterized by its mass, but by how the mass is distributed in space-by its inertia tensor. Apparently the corresponding part of the classical mechanics course, where this is treated, was skipped,  else the nonsensical  statements given above couldn't have been made (of course the question of this thread wouldn't have been posed, either).  It might, therefore, be useful, to consult a textbook on classical mechanics, that presents this topic.

http://www.damtp.cam.ac.uk/user/reh10/lectures/ia-dyn-chapter10.pdf

for instance. 

And, once that material has been understood, to  try to calculate how a particle is reflected off a non-rotating vs. a rotating sphere-and, therefore, from the properties of such a particle, to try and learn how to  deduce the properties of the motion of the sphere. 

@ Kare: Indeed, I understand about GR. But even there, in the absence of any marks allowing to describe a point on the sphere, how do you arrive at a description of the sphere involving, say, a rotating energy-momentum tensor?

In other words, given a sphere defined as a pure continuum (no atoms) how do we infer whether a motion exists which carries the form into itself? In my mind, once you have given the sphere this level of abstraction, the concept of rotation ceases to be meaningful.

On the other hand, as soon as some kind of marks are allowed, then my solution, albeit trivial, is immediate. So I agree I am a  bit of spoiler.

To some extent I would differ from Pavlo in saying that rotation in those circumstances is not a deep property, but a mere issue of convention with no physical consequences whatever, in any theory whatever.

I thought of the sphere as a fat neutrino; you are not going to describe it's spin in terms of a classical energy-momentum tensor (Pauli once tried for the electron). Then, quantum mechanics aside, how do we know that such particles have spin? One way might be through their spin-orbit coupling, but that requires some kind of charge. Maybe gravity, being universal, always implies the existence of a-kind-of charge. We are, by some means, able to "see" the sphere; that must imply that it interacts with something which we can interact with.

In particle reactions one can verify the neutrino spin indirectly, through conservation of total angular momentum. Here weak interactions are involved.

But, what if there is a dark matter particle with spin, only interacting via a scalar field (an axion field, for example) except gravitation (which provides an in-principle objective existence of the spin, and a way to measure it). How could we be able to measure its spin? Is it possible without involving gravity? Even with quantum mechanics? That is a modified, more realistic, version of Pavlo's question.

There is a more scary question along this line: What if cold dark matter consists of particles which only act with normal matter via gravitation. How can we know if it even exists, instead of being a modification of the gravity equations?

The problem of rotation will remain unsolved so long as the luminiferous medium remains banished from orthodox physics. For something to be rotating, it has to be rotating relative to something, such as to cause inertial effects. That something was wrongly dropped from physics in the 20th century. See "The Double Helix Theory of the Magnetic Field".

One hopefully learns that, when describing particles with spin, the total angular momentum, orbital + spin, is the relevant quantity. How this is done in practice is by means of the spin connection. This is discussed in general relativity courses when writing the Dirac equation in curved spacetime, in which the Dirac equation, of course, is a classical field equation. 

If dark matter particles have only gravitational interactions with ordinary matter, it will not be possible to detect them, in practice, only their collective effects, that are, already, measured, can be probed-that's well known. It also is irrelevant for the question that's a homework problem of classical mechanics. 

It's not useful to discuss more complicated issues, when the simpler issues of classical mechanics are, apparently, not assimilated. One should start by working out what happens to two particles, that collide, elastically, with velocities that aren't collinear and go on from there, to the case of a rigid body colliding, elastically, with a point particle. And then, try to understand what happens, when one tries to step off a moving bus or train, first when the later is moving on a straight line, then when it's turning. 

You need to study the Newton's Cradle. See this link here http://gsjournal.net/Science-Journals/Essays-Mechanics%20/%20Electrodynamics/Download/5737 It's all about rotation on a smaller scale.

http://gsjournal.net/Science-Journals/Essays-Mechanics%20/%20Electrodynamics/Download/5737

Even if it isn't ``gimballed'', an extended object reacts to a torque, because it has an inertia tensor:, since all its points don't coincide.  A sphere has an inertia tensor, that can be calculated. And can be found, by computing the result of the elastic collisions of particles off it. This is ghostwriting undergraduate mechanics homework assignments. 

Because it's got a frictionless surface without any markings on it, then the only way that rotation could be observed would be by virtue of an expansion in the equatorial plane due to centrifugal force. This expansion will always happen if rotation is occurring, but it may not be to a sufficient degree of magnitude to be easily observable.

Stam> And can be found, by computing the result of the elastic collisions of particles off it.  This is ghostwriting undergraduate mechanics homework assignments. 

I think you have to show us how, by a calculation, if you cannot apply tangential forces to the surface.

Stam> but whose tangential component will be equal to the linear velocity of the point on the sphere it strck; so it will fly off at an angle

Your particle enters with zero momentum in the direction tangential to the sphere (at the point of impact), and leaves with a non-zero momentum in the tangential direction. Where can this change of momentum come from?

Stam> Tangential forces aren't relevant

It can only come from a (very strong) tangential force acting over a (very short) time. Tangential forces are important!

The solution you give corresponds to a "sticky" (as long as the contact lasts) surface, exactly the opposite of a perfectly slippery one -- which in my vocabulary means exactly zero tangential forces.

This class of problems may belong introductory university mechanics, but that does not mean that they are completely uninteresting or trivial. They are very nice to illustrate how the conservation laws of momentum, angular momentum, and energy can be applied and checked. There are two cases where no energy is converted to heat, the case of a perfectly slippery sphere and the case of a perfectly "sticky" sphere. Everything in between will lead to generation of some heat due to friction.

Note added: The links provided by Stam are not very enlightening! I see no focus on the essential principles -- the conservation laws. A lot of technical details. What is the point of calculating the moment of inertia if you don't understand when, and how, to use it?

You're all missing the point. The only way the rotation will be observable will be via the inertial forces, and in particular, the centrifugal force. When an object rotates, relative to the background stars, it experiences centrifugal stress within its structure. This will translate into expansion in the equatorial plane, although it may be such a minimal effect as to be virtually non-observable.

Dear Paul Pushkin You have asked a very funny question. 

1 If you mean absolutely perfect body with zero friction. Such a body can only be created by God. Speculate about the measurement of its rotation is equivalent to reasoning about how many angels can fit on the end of the pen.

 2 You can create a new branch of science "mechanical metamaterials" for example: study the properties of an ideal spherical "matryoshka dolls», in which each sphere is rotated in its direction.

I liked your question. I wish you all the best. Ivan I. Dolgov 20160929

Дорогой Pavlo Pyshkin Вы задали очень весёлый вопрос.

1 Если Вы имеете в виду абсолютно идеальное тело с нулевым коэффициентом трения. Такое тело может быть создано только Богом. Рассуждать об измерении его вращения равносильно рассуждениям о том, сколько ангелов может поместиться на конце пера.

2 Вы можете создать новое направление науки «механические метаматериалы» например: исследование свойств идеальной сферической «матрёшки», в которой каждая сфера вращается в своём направлении.

Вопрос мне понравился. Желаю Вам всего хорошего.

И.И. Долгов, 20160929

Similar problem wrangled by Newton (https://en.wikipedia.org/wiki/Rotating_spheres). And what about frame dragging in the case of Einstein's two rotating globes? (http://science.jrank.org/pages/11027/Relativity-General-Relativity.html) Good luck, undergraduates!

We can't realize the rotation of your slippery ball because our senses can't detect it.

If we measure the static of your ball by any instrument ( f.i. another touching ball with signs on it's surface) we can fix the rotation - very simple!

Franz> - very simple!

Please tell us more about the construction of your instruments (any or all of them). Keep in mind that the ball is too slippery to f.i. transfer angular momentum to your touching ball with signs on.

Incidentally: each point on the sphere defines a frame. If the sphere isn't rotating, all these are inertial frames, due to spherical symmetry;  if it is, they aren't-and they aren't all equivalent, due to spherical symmetry being broken.  So it's possible to deduce what each frame on the sphere would measure, regarding  a given point moving outside the sphere and not even intersecting its path-and the results are correspondingly different, if the sphere is rotating or not.  And of course the other way  around, too. In particular, how the velocity of a particle is transformed from the frame in which the sphere is rotating to the frame on any given point on its surface. That's why tangential forces aren't relevant, once one has specified that it's a sphere and that it rotates in a certain way. Everything else follows from there and vice versa.

The bottom line: a non-rotating sphere has spherical symmetry: any and all physical quantities describing its interaction with any sort of probes are invariant under SO(3), the isometry group of the sphere. A uniformly rotating sphere has axial symmetry: the physical quantities are invariant under that particular SO(2) subgroup that leaves the the axis of rotation invariant. These two groups are not isomorphic. So whatever means observers far away use, they can remark that in one case they're measuring quantities invariant under SO(3); in the other case they can only agree on quantities invariant under an SO(2) subgroup. If the rotation isn't uniform, the subgroup of SO(3) will be different.  

The surface of the sphere implies that there must be a boundary condition imposed there, that's all-that should be compatible with the symmetries: SO(3) for the non-rotating sphere and some subgroup of it for the rotating sphere.

That's what gives content to the statement ``measuring the rotation of a sphere''. It means measuring certain invariants of subgroups of SO(3). And to anticipate: the same procedure holds for any object.  The physical quantities are invariant under the symmetry group of the object-so the interaction is, too.

Any other details don't matter. 

Pavlo Pyshkin, your question is easily solved.

The answer is YES!

Just push the sphere upwards with your finger, next to the contact point with the table. If it doesn't immediately fall according to the classical laws for non-spinning spheres, it is spinning. If you don't know the spinning direction, you can test this at several places around the slippery contact point (different longitudes next to the contact point).

All the best!

@Thierry,

nice try, but in the spirit of the Pavlo's question gravity should be excluded. 

This is an interesting tricky question. Such a classical macroscopic ball cannot exist without electromagnetism. If it is not macroscopic then we have elementary particles, but they are not solid spheres and obey the laws of quantum mechanics. To have a solid macroscopic sphere there must be some interaction between its constituents (EM). But then you can immerse it in some fluid and observe the motion of molecules of the fluid due to adhesive (electric) forces. In fact, there are always such a liquid or gas, we can make it of the same material ball was made. One could object that  friction is zero, but there is no way to avoid adhesive forces, friction is never exactly zero.  Point is that if we exclude electromagnetism then there is no ball, so,  there is no any "deep internal property" in "it".

@Miroslav: Nice try, but if gravity would be excluded, acceleration would not.

If one accelerates the sphere exactly as I explained (a direction outside the diameter axis), still then, the outcome of spinning and non-spinning spheres immediately appears. A non-spinning sphere will immediately and strongly be deviated by the acceleration, while a spinning would not.

Your guess about "a classical macroscopic ball cannot exist without electromagnetism" is infantile and beside the point. Infinite slippery doesn't exist either.

Thierry> If one accelerates the sphere exactly as I explained (a direction outside the diameter axis)

Why didn't you read the question? You cannot apply a force in a "direction outside the diameter axis" (i.e., outside a radial direction), because that implies a tangential force --  which the sphere is too slippery to allow.

@Thierry,

well, describe me ordinary macroscopic ball without the EM field, you can surely do that because you are "mature", I presume. My point was about the "deep internal property", far gone conclusion from the simple non.existent problem. But why bother to read the others comments carefully, there are nice words like "infantile" that cover all.

Kåre, maybe do you need a little lesson of elementary undergraduate physics. What will happen if I push with my finger the sphere just outside the diagonal axis, even if the sphere is infinitely slippery?

You have a mass, either spinning or not, and you have the vector of acceleration pointing upon the mass in a given direction. You will agree that if the mass is not spinning, I can apply this acceleration. Only inertia of the mass and the acceleration vector are implied.

If you are still not convinced, you can integrate a number of impacts of small bullets in the explained direction, which will result in the force (and acceleration) anyway.

When it is spinning and infinitely slippery, of course I can apply the same steady acceleration vector, either directly, or by the integration of a number of impacts of small bullets in the same direction!

When the mass is not spinning, the acceleration vector will cause a momentum, which only depends upon its inertia and the acceleration (or applied force).

When the mass is spinning, the inertia is way different, and the gyroscopic effect will result in a totally different deviation, due to the high angular momentum in addition to (the non-spinning) inertia.

Voilà. Without any mathematical nonsense...

Hi Thierry,

                 I took it from the way the question was worded that the ball had no markings and so we could neither see it spinning nor precessing. I understood your point that if it is spinning and that if we punch it, it will precess. But what if it has got no markings? Presumably we are not allowed to attach it to a pivot.

                 The question in many respects is a tautology, in that it reduces to "can we detect that a  ball is spinning if we are denied the use of all the standard manners whereby we could detect it?".

@Thierry,

FIrst:

"Just push the sphere upwards with your finger, next to the contact point with the table. If it doesn't immediately fall according to the classical laws for non-spinning spheres, it is spinning."

Then:

"number of impacts of small bullets in the same direction"

It seems that not only the ball is slippery. 

"Kåre, maybe do you need a little lesson of elementary undergraduate physics."

Such sentences a far from courteous.

Now about "elementary undergraduate physics":

Here we see that there is the "high angular momentum in addition to (the non-spinning) inertia". How one can tell what is the ball inertia? You can produce the pure acceleration with the force vector passing through the center, but then one will get just overall inertia. If we are not sure about the rotation then how to tell what is the inertia of non-spinning ball?

The bullet will not produce any rotation. It is enough to resolve the momentum vector of bullet  into components, one through the center of the ball and one tangential. The first one will just accelerate the center off mass of the ball during the impact, but that doesn't help. The tangential component cannot rotate the ball because there is no a friction. A little of "mathematical nonsense" helps sometimes. 

Well, dear Miroslav, first you mix up two different solutions I gave, which is not so smart, and then you want to insult me on that. Not smart either.

Then: if you would read the question, the mass and the radius of the sphere are given. Please rely on Newton for the non-spinning case, and to Euler for the spinning case.

"The bullet will not produce any rotation." has absolutely nothing to do with the issue. Neither "one through the center of the ball and one tangential".

Just the inertia in the two cases matters, and the steady directional force a bit outside the sphere's diameter, preferably in absolute local coordinates.

If the bullets follow that direction, the sphere will be deviated differently in both cases because the rotation inside the sphere differs in both cases. One can try out different directions for the force to find out the spinning orientation.

If you want to make the sphere so non-physical that even the impact of the bullets would not alter the direction of the physical components of stresses inside the sphere, you can prove nothing at all of course and there is no solution. That would mean an infinity speed of stresses inside the sphere. And such a case would only prove that the question itself is nonsense, only valid for mathematicians' free time occupation.

However, the point is that the impact of the force-component will not immediately reach the center of the sphere when the sphere is spinning, since the sphere is physical. The force-field will become deviated inside the sphere.

Hence, even in the case of an absence of gravity, the answer is YES.

This is what I meant by "if mathematicians are unqualified to solve simple physical questions, how could they possibly solve important ones?"

@Thierry,

I "mixed" nothing, I commented your solution, then you offer another one, quite different, but suggesting that it is the same as first. It is hard to follow all that. I am glad that the word "slippery" is noticed by you because I am really mad because of your offensive comment to Kare.

"If you want to make the sphere so non-physical that even the impact of the bullets would not alter the direction of the physical components of stresses inside the sphere, you can prove nothing you want of course and there is no solution."

I "make" nothing, Pavlo defined conditions, if there is no friction there is no rotation by bullets. Now you are to admit that, so where is an apology to Kare? But yes, we can prove nothing that way. 

"The force-field will become deviated inside the sphere."

Well, that changes nothing. If the initial bullet's momentum vector was directed to the center (hard to achieve) there is no change in the direction of the momentum vector of the center of mass due to internal forces (momentum preservation, possible internal forces obey the third Newton's law). Angular momentum is also preserved (there is no torque). We cannot "see" inside the ball, so, what we might detect? 

Deviation of the sphere's trajectory of course,  differently in both cases,  due to the *physical* sphere!

Besides, Miroslav, "I commented your solution, then you offer another one, quite different, but suggesting that it is the same as first." 

Well, do you accept your beloved equivalence principle,  or don't you, just because *I* wrote this? 

I can only quote Dr. Watson: "What ineffable twaddle!"  "I never read such rubbish in my life."

This is what I meant by "if mathematicians are unqualified to solve simple physical questions, how could they possibly solve important ones?"

These individuals downvote or critic without content, and exhibit not only a great cowardice, but also an immense incompetence to a sensible, scientific reply.

Dear mathematicians! A nice example to understand this, is the set of 5 colliding balls. On the picture, you see the first ball going to hit the others. For a physical sphere, there are indeed many more little balls inside, but let's start with this setup. The last ball will leave the group a few milliseconds later as you know, because the speed of propagation is not infinite and it is not "glued" to the other balls like in the physical sphere, so, we now disregard that the last ball would leave the group in the example.

Let us now put the set on a rotating disk. The first ball will still hit the group radially, but the last ball will not anymore be positioned diametrically and opposite to the original position where the first ball has hit the group, because of the time delay. Hence, the last ball will leave the group on another path than when the set were not rotating.

Now, consider that the 4 balls would be glued together. The total set would be displaced a little, according to the equity of impulsion between the impacting ball and the other balls.

So, we clearly see that, spites the infinitely slippery surface, still it is clearly possible to detect very easily when the physical sphere is spinning.

Remember that applying a force is the same as applying a number of succeeding impacts.

So, the path of motion of the spinning sphere will be altered compared with the path of the non spinning physical sphere.

Christian,

Indeed, at the moment of impact in a certain starting point at the sphere's surface, of which the coordinates wrt the room are known, the impulse is following a radial path inside.

That impulse is transmitted internally to the rest of the sphere, at a certain transmission velocity, of which the average direction remains radial wrt the sphere.

However, in coordinates expressed wrt the room, the impulse localization will change with an angular velocity during that internal transmission, if the sphere is spinning. That angular velocity is related to the velocity of transmission, the diameter of the sphere, and the velocity at the surface of the sphere at the impact point.

When the impulse reached the opposite surface of the initial position wrt the sphere's coordinates, it will, according to the law of conservation of impulse momentum, make the sphere move a bit forward.

That place will differ from the radial line of initial impulsion if the sphere is spinning.

Since a force is a succession of impulses, the sphere's path will be different from the expected one (radially wrt the coordinates of the sphere *and* the room), if the sphere is spinning.

One can perform the test at several places in order to find the precise spin axis direction and its angular velocity.

As you see, the physical solution has been clarified without the use of maths.

Another challenge would be to find out what is the minimal number of impact tests that are necessary in order to find both the precise spin axis direction and its angular velocity.

That might be easier solved by using maths. Could you find it?

Zero friction doesn't exist anywhere, but if it did the rotation could still be detected by classical optics. The object has to have either a reflective surface or a radiating surface, or some proportional combination of the two which makes the rotation measurable except maybe in the axial direction.

In terms of GR there is a frame dragging of local space with very small magnitude, which in principle could be measured since idealities are allowed

So,  still relativists and mathematicians are unable to defend themselves,  like chicken.. They just prefer to down vote... 

Jerry! Read the question! "The question is: can you (!!! without references to quantum mechanics, electrodynamics, general relativity, ...!!!) detect rotation of this ball by using only classical mechanics laws?"

So, no optics, no GRT.

Christian, "elastic properties of the ball that have not been specified in the question" What an idiotic remark! If we talk about mechanics, we talk of a real physical sphere, not a mathematical one!!!

Okay, Christian, you don't like the elasticity explanation? Let's do it the "Donald Trump way":

Take a rifle and shoot straight on a spinning sphere, radially, so that it enters the sphere. Make the balance inside the sphere with the mass of the bullet, its speed, the mass inside the sphere, and the speed of that mass of the sphere, in contact with the bullet, radially and transversally.

See what happens....

Maybe you can add up vectors, at least?

Anyway, I will stop this thread since there is no decent reply from anyone here.

Thierry> we talk of a real physical sphere

No, we were talking about an idealised situation, as formulated in the original question. Most people tried to find solutions which respected the spirit of the question/intention of the questioner. You curiously seem to find satisfaction in insulting them for not being willing to cheat.

The Popular upvote of Igor's answer is deserved, because it was the first correct answer. Contrary to your proposals.

Well, Kåre and Christian, predending a null result because of an inpropriate formulation of the question would be an enormous insult at the address of Pavlo Pyshkin, who intended to speak about real mechanics, not mathematics, and a cowardly way out for those who are not able (n)or willing to think about real physics. It confirms what mathematicians are really worth in the world of physics.

Running away with the statement that the question itself is a null situation, like Igor Goliney does, seems to be the standard way of pretending to do physics for mathematicians...

However, with the "Trump way" I explained above, they cannot run away anymore. The result is: YES, rotation will be detected by mechanics.

Christian, "You like it the Trump way? Why am I not surprised?" 

Just like for the MMX null result, the expression ""You like it the Trump way?" is of no value wrt to what I wrote. As usual.

And as to "continuation of insults", you just insulted  but 50% of the U.S. population.

Dear all, I am going "off the aether"  ;-) for a moment...

Merry Xmas and happy new year!

Best wishes to all !

Another, slightly seasonal, comment, if I may:

The stream of answers to this question reminds me (just a little) of the Bohr-Einstein debate, though I suspect it's become a little more about persistence than physics... 

A Merry Xmas and a physical New Year to all!

If we just consider the question alone in the absence of all the qualifiers, the answer is simple. Rotation relative to the background stars will induce a centrifugal force in the equatorial plane. However, the add on qualifiers to the original question are more or less asking whether we can detect rotation in some unlikely ideal context which has been specially designed so that we can't detect it.

If we just consider the question alone in the absence of all the qualifiers, the answer is simple. Rotation relative to the background stars will induce a centrifugal force in the equatorial plane. However, the add on qualifiers to the original question are more or less asking whether we can detect rotation in some unlikely ideal context which has been specially designed so that we can't detect it.

Mr. Srinivasarao, You've just repeated verbatim what I said yesterday

Frederick@

Remember Oscar Wilde: “Imitation is the sincerest form of flattery that mediocrity can pay to greatness.”