Do spinning objects really lose weight?

[Dadface]

I agree that I could have I could have explained my original question more clearly but then we can all improve upon the parlance we use.My original intention in starting this thread was to promote some sort of discussion in the hope that some people may work it out for themselves.In retrospect I realize now that this thread would have been better introduced in the brain teasers forum.My excuse is that I am new to forums and I am still getting used to the procedures one should follow.
Several weeks ago I posted a copy of my findings to a well known magazine.It was written in a tongue in cheek style and I am hoping that they will publish it as an amusing snippet in the letters page.If I hear nothing I will try elsewhere,but written more formally.Thanks to a comment made by granpa I am going to call it the satellite effect.

 

[Doc Al]

Dadface, it sounds to me that your question (at least the latest version) is about the change in apparent weight of an object on the Earth due to the Earth's rotation. Is that what this is all about? (Of course the Earth's rotation has no effect on the actual weight of an object--the strength of the Earth's gravitational attraction.)

 

[Dadface]

It is an apparent weight loss but not due to the Earth's rotation.Although there are effects due to the rotation of the Earth I did not bring this into my analysis.What I considered was the rotation of the object itself.

 

[turin]

You may be conviced if you do a free body force diagram with a round earth.

I don't need to draw the diagram; the situation is too simple. I see only two forces and one acceleration. One of those forces is weight. The other force is tension. The acceleration is centripetal. To determine weight, I use Newton's law of universal gravitation, regardless of whether or not the ball is orbitting, rotating, or whatever you want to call it, so I get the same weight as if the ball were just sitting still at that same height on a table top. AFAIK, Newton's law of universal gravitation does not have a velocity-dependent term or factor in it.

 

[Dadface]

AFAIK Newtons law does not have a velocity dependant term in it etc and I agree with everything else you have written.To reiterate I am not claiming a real weight loss it is an apparent weight loss of the type an astronaut in orbit experiences.Since you did not draw the diagram can I enquire which ,if any, of the two forces you identified provides the centripetal force which is needed for the centripetal acceleration?The conventional view is that for an object moving in a horizontal circle, it is the horizontal component of the tension and that the vertical component of the tension supports the weight.In your mind did you see the weight vector pointing vertically downwards?That direction is arguably correct for a flat Earth but not for a round earth. The weight of an object acts towards the Earth's centre and because it has a horizontal component it provides some of the centripetal force needed thereby reducing the tension and resulting in an apparent weight loss. In short we are in agreement about the forces themselves but it is the directions of those forces that I am asking you to consider.Please just scribble it out on a scrap of paper.

 

[Dadface]

The fractional loss of weight is given by:

f^2=(2v^2.g.r.cos phi)/(v^4+g^2.r^2)
v= speed.
g=acceleration due to Earth's gravity at the particular height of the object/ring.
phi = the angle that the weight vector makes with the orbital plane.
r =radius of circle.
Again I need to check my maths.
Please note the following:
1.The fractional loss is independant of the mass.
2.When phi becomes equal to zero and v^2=rg the fractional loss becomes unity and the mass/ring becomes a satellite.

 

[turin]

The weight of an object acts towards the Earth's centre and because it has a horizontal component it provides some of the centripetal force needed thereby reducing the tension and resulting in an apparent weight loss.

Why does reducing the tension result in an apparent weight loss? Tension and weight are two completely different things. The directions of the forces, and in fact the entire fbd, are irrelevent. We don't need to separate force vectors into components. Weight is an input into the fbd, not something that you calculate from it. We already know that weight can contribute to centripetal force (i.e. anything that is in orbit around the Earth gets its centripetal force from its weight, as you say). I don't understand what is the big deal. And I am still waiting for you to relate this to gyros ...

 

[Dadface]

Resolving the two forces into components is the easiest way to solve the problem.How else can you determine what provides the centripetal force and what supports the weight?The centre of the circle in which the mass moves is not along the line of of the string-it is below it and the centripetal force is directed towards the centre of the circle.
I will put it another way:
Imagine the mass was moving in a vertical circle by means of a light rod driven by a motor which is supported from a stand which is sitting on scales.Let the speed be constant and equal to v and let the radius of the circle be equal to r.The centripetal force needed to maintain this circular motion is given by mv^2/r.Now consider the mass when it is at different positions
Topmost point...At this point the two forces are in the same line ,the direction of both is towards the centre and together they provide the centripetal force(C.F.).We can write:
Mg+T =C.F...(Newton second law)
Bottommost point...At this point the two forces are in the same line but act in opposite directions,T towards the centre and Mg away from the centre.We can write:
T-Mg=C.F.
At other points we have to calculate the component of the weight towards the centre.The measured weight of the mass is felt through the tension and the tension varies from a maximum at the bottom ,showing an apparent weight increase to a minimum at the top showing an apparent weight decrease.You are probably familiar with systems of the type I have just described but my analysis of motion in a horizontal circle works on exactly the same principle the weight itself having a horizontal component and the apparent weight change being a loss at all parts of the circle.It does not matter if a method of support other than a string is used or if it is a gyroscope a spinning top or anything else,the principle is the same.

 

[turin]

This will likely be my last post.

Resolving the two forces into components is the easiest way to solve the problem.

I see no problem to solve.

... the apparent weight change ...

In order to continue this discussion, I need you to define "apparent weight", and to identify the object that has this apparent weight.

It does not matter if a method of support other than a string is used or if it is a gyroscope a spinning top or anything else,the principle is the same.

What principle? I do not care about the centripetal force acting on the different parts of a spinning gyroscope. In order to continue this discussion, I want to know about the support force provided by the ground or table-top that supports the spindle of a spinning gyroscope.

BTW, as a suggestion, I would be happy to identify the time average of such a support force as the "apparent weight" of a spinning gyroscope.

 

[nuby]

Here's a thought experiment. If a spherical object on a string was spinning around so fast .. that if released it would fly into outer space. Would it experience any weird effects wrt to its weight or Earth's rotation?

 

[turin]

Here's a thought experiment. If a spherical object on a string was spinning around so fast .. that if released it would fly into outer space. Would it experience any weird effects wrt to its weight or Earth's rotation?

According to GR? I do believe so. (you can look up GP-B as an example of why I believe so). According to Newtonian gravity? I don't believe so.

 

[Dadface]

Hello turin
1.The problem I was referring to was the problem of calculating the apparent weight or apparent weight loss.
2.The apparent weight is the force that would be measured by a weighing insrument.You asked me to identify the object that has this apparent weight and I thought I had already done this in earlier parts of the thread but there are numerous other examples most of them involving accelerated motion.Heres a couple more.
a. Hold a spring balance in your hand and weigh a bag of potatoes.Accelerate your hand up and the balance will give a higher reading,accelerate it down and it will give a lower reading,drop the balance and everything is in free fall.
b.Drive over a hump backed bridge too fast and leave the road and any weighing instrument that may be on the road..
c.Build a gigantic ring that surrounds but is above the equator.
Specifically I was talking about objects spinning or rotating in horizontal circles.The apparent weight changes(losses and gains)happening in vertical circles are already well documented and accepted and very easily measurable eg by a slightly unbalanced flywheel on a set of scales.By considering a gyroscope where most of the mass is concentrated in a very thin ring on its outer edge I have derived an equation giving the apparent weight and by plugging numbers into this the apparent weight can be predicted.It is the support force ie the apparent weight that I have calculated.
4.If predictions such as this can be made within the domain of Newtonian mechanics then they should be inherent in the more encompassing Einsteinium theory.It would be interesting to see what GR predicts and whether or not there are any deviations.

 

[HallsofIvy]

What is the difference between "vertically down" and "toward the Earth's center"?

I would say that "down" is defined as "the direction in which gravity (weight) acts". Generally, that will be toward the Earth's center but if you are standing near a large mountain, say, the direction will be very slightly toward the mountain.

 

[Doc Al]

Hello turin
1.The problem I was referring to was the problem of calculating the apparent weight or apparent weight loss.
2.The apparent weight is the force that would be measured by a weighing insrument.You asked me to identify the object that has this apparent weight and I thought I had already done this in earlier parts of the thread but there are numerous other examples most of them involving accelerated motion.

Note that the question "Do spinning objects really lose weight?" is much less interesting when all you are talking about is "apparent" weight (the support force). The simple answer to the question in the thread title is: No, of course not. If I drop a ball (remove the support force of my hand) would anyone say that the weight of the ball really changed?

By considering a gyroscope where most of the mass is concentrated in a very thin ring on its outer edge I have derived an equation giving the apparent weight and by plugging numbers into this the apparent weight can be predicted.It is the support force ie the apparent weight that I have calculated.

Are you claiming that the force required to support a gyroscope is less when it's spinning? (Let's assume its center of mass is not accelerating.) Just to be clear: Rest the unspinning gyroscope on a scale--the support force is mg. Set it spinning and rest it on the same scale--the support force is less than mg? Is this what you are claiming?

 

[Dadface]

I agree with you HallsofIvy and there are other systematic effects such as those due to the Earth's spin,density variations within the Earth and the other deviations the Earth has from being exactly spherical.An even more detailed analysis would take into account the location of the spinning/rotating object.

 

[Dale]

In Newtonian physics a spinning object does not lose weight, either real or apparent. The example provided in post 14 is not relevant. A spinning object rotates about an axis through its own center of mass, i.e. spinning does not cause the center of mass to accelerate. This is not the case in the analysis of post 14. There are no net external forces on an object whose center of mass is not accelerating, whether or not the object is spinning about an axis through that center of mass is irrelevant. Spinning will add internal forces, but will not change the external forces.

Obviously (from Newton's 2nd law) an object whose center of mass is accelerating is undergoing different external forces than one whose center of mass is not accelerating.

 

[Doc Al]

Exactly!

 

[turin]

You asked me to identify the object that has this apparent weight and I thought I had already done this in earlier parts of the thread
...
Specifically I was talking about objects spinning or rotating in horizontal circles.

Fair enough.

By considering a gyroscope where most of the mass is concentrated in a very thin ring on its outer edge I have derived an equation giving the apparent weight and by plugging numbers into this the apparent weight can be predicted. It is the support force ie the apparent weight that I have calculated.

Again, fair enough. However, if you are referring to this equation ...

The fractional loss of weight is given by:

f^2=(2v^2.g.r.cos phi)/(v^4+g^2.r^2)
v= speed.
g=acceleration due to Earth's gravity at the particular height of the object/ring.
phi = the angle that the weight vector makes with the orbital plane.
r =radius of circle.

then I disagree with your equation. (BTW, when you say v=speed, I'm assuming that you mean the tangential speed of the ring of the gyroscope.)

Since you politely took the time to clarify these issues, then I will humor your request for a fbd. I model a gyroscope as an infinitesimally thin ring of uniform mass density attached to the outer edge of a massless disk of radius, r, attached to the middle of a massless spindle of length, 2h. Neverminding the issue of stability (which I think is anyway irrelevant to the issue of apparent weight loss), I suppose that the spindle is vertical, and supported on a weight scale, of negligible thickness, on the surface of the Earth. I model the Earth as a perfect sphere of uniform mass density, mass M and radius R.

The following involves some basic notions of calculus. I would begin with an fbd that is symmetric about the vertical, with two differential masses, dm, on the opposite ends of a horizontal massless rod of length, 2r, and this horizontal massless rod would be fastened perpendicularly in the middle to the top end of a vertical massless rod, of length, h (so that I have a T-shaped arrangement of massless rods). The bottom end of the vertical massless rod would be supported by a scale. I assume that the horizontal rod with the two differential masses on its ends is rotating at such a rate that the masses are traveling at speed v into or out of the paper. From this diagram, I could calculate the tension in the left half of the horizontal rod to be:

$$d\vec{T}=+\hat{x}\left(v^2r-\frac{GMr}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}\right)dm$$

and the tension in the right half of the horizontal rod to be:

$$d\vec{T}=-\hat{x}\left(v^2r-\frac{GMr}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}\right)dm$$

Of course you recognize the first term in parenthesis as the centripetal acceleration. The second term that is subtracted is the horizontal component of weight. If you want to compare to your results, assuming that I have understood your definitions correctly, you can replace:

$$\frac{r}{\sqrt{\left(R+h\right)^2+r^2}}\rightarrow\cos\phi$$

Anyway, when I add all horizontal forces on all parts of the ring, I determine that the total horizontal force on the ring vanishes, regardless of v.

$$\hat{x}\cdot\vec{F}_{tot}=0$$

Now, all of that was actually irrelevant, because I don't care what is the horizontal force on the ring anyway. I want to know the force that the scale must apply vertically upward in order to prevent the spindle from accelerating vertically downward. I calculate the vertical component of the weight on each of the two differential masses to be the same:

$$dF_{gy}=-\frac{GM\left(R+h\right)}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}dm$$

In order to integrate this, I will call the linear mass density of the ring, $\lambda$. After doing the (trivial) integral, I find that the vertical weight component on the entire ring is

$$F_{gy}=-\frac{2\pi{}GM\lambda{}\left(R+h\right)r}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}$$

Since the only other force acting on the system is the support force from the scale, I find that the apparent weight is

$$F_{aw}=\frac{2\pi{}GM\lambda{}\left(R+h\right)r}{\sqrt{\left(\left(R+h\right)^2+r^2\right)^3}}$$

Notice that this is independent of v. This means that I will measure the same apparent weight regardless of how fast the gyroscope is spinning. In particular, I will measure the same apparent weight regardless of whether or not the gyroscope is spinning.

In terms of your notation, my equation says:

$$f=0$$

regardless of v. So, I guess we're stuck.

 

[Dadface]

First I would like to reply to Doc Al
1.When I opened the thread I simply posed a question and never once have I suggested that it was a real weight loss having stressed that it is an apparent weight loss on several occasions.
2.Yes I am making that claim-a spinning or rotating object or indeed any accelerating object at all can display apparent weight changes(gains and or losses)and in several cases these are easily measurable.I have given many examples of this throughout this thread and I can give many more a lot of them coming from everyday life such as driving your car with an unbalanced tyre.If there is anything new here it is that I am suggesting that rotation in a horizontal plane results in apparent weight losses.The changes may be small but in principle they are measurable.

Dalespan imagine a spinning disc with angular velocity w at its outer edge.Now proceed from the outer edge towards the centre of the disc.Are you telling me that we reach a point where the disc is no longer spinning and has zero angular velocity?


turin thank you very much indeed for presenting your maths.I will go through your method and check it against mine(which seems much simpler)but unfortunately an assumption you made does not fit with the more detailed analysis needed .I refer to the "horizontal massless rod of length 2r".The rod cannot be horizontal otherwise there is no component to support the weight .This may seem,at first sight, to be a trivial point but it is not and is at the crux of what I have been trying to explain.As you know I have been trying to make it clear that marking in the force directions correctly is all important here.
As an example of the above, imagine a mass on a string being whirled in a horizontal circle at ever increasing speeds .As the speed increases the mass rises and the angle to the horizontal decreases.It can never reach the horizontal exactly because this would need an infinite speed.

 

[Dale]

Dalespan imagine a spinning disc with angular velocity w at its outer edge.Now proceed from the outer edge towards the centre of the disc.Are you telling me that we reach a point where the disc is no longer spinning and has zero angular velocity?

Certainly not, how did you ever come to that conclusion?

Look, my point is simple. If the centers of mass of two identical objects are undergoing identical accelerations then, by Newton's 2nd law (f=ma), the external forces (e.g. weight) on each object are necessarily identical.

 

[Doc Al]

2.Yes I am making that claim-a spinning or rotating object or indeed any accelerating object at all can display apparent weight changes(gains and or losses)and in several cases these are easily measurable.I have given many examples of this throughout this thread and I can give many more a lot of them coming from everyday life such as driving your car with an unbalanced tyre.If there is anything new here it is that I am suggesting that rotation in a horizontal plane results in apparent weight losses.The changes may be small but in principle they are measurable.

I still have no idea why you think that just because something is spinning, less support force is required.

DaleSpam's post #51 sums things up rather nicely. If you claim otherwise, please explain why Newton's laws do not apply.

 

[Dale]

I DaleSpam's post #51 sums things up rather nicely.

Thank you!

Given the rather strange misunderstanding he posted in response to my explanation, I expect that communicating will be difficult.

 

[Dadface]

DocAl I am not claiming that Newtons laws do not apply,on the contrary I am using Newtons laws.Precisely Dalespam if the spinning object has a perfectly symmetrical structure and if it rotates in an exactly horizontal plane then the two objects ,if they are symmetrically situated and structured will undergo identical accelerations the only point of similarity being that the magnitudes of those accelerations are the same and not their directions.Please note that my analysisis primarily focussed on non symmetric structures

 

[Dale]

Precisely Dalespam if the spinning object has a perfectly symmetrical structure and if it rotates in an exactly horizontal plane then the two objects ,if they are symmetrically situated and structured will undergo identical accelerations the only point of similarity being that the magnitudes of those accelerations are the same and not their directions.Please note that my analysisis primarily focussed on non symmetric structures

Where are you getting all of this? I said nothing about symmetry, I said nothing about horizontal planes, and I said nothing about magnitude vs. direction of acceleration. The symmetry of the object is irrelevant as is whether or not the object is spinning about a vertical axis.

 

[Dadface]

Dalespam" I am getting all this from "your comment in post 55 when you referred to the "centres of mass of two identical objects undergoing identical accelerations".Perhaps I misunderstood what you meant but your comment was referring to my reference to a single disc so what am I to understand about your "two identical objects"?I thought that you were referring to objects on the disc itself as evidenced by my comments in post 58.Even though there was a mutual misunderstanding here my reference to force direction is most relevant as is the axis of spin.Please spend a few moments and take part in the following thought experiment:

1.There is an extremely tall structure built vertically at one of the Earth's poles and you are standing on top of this structure with an extremely long length of light inextensible string attached to a point mass.

2.You start to whirl the mass in a horizontal circle playing out the string so that it lengthens and you adjust the speed so that the mass does not collide with the earth.

3.You carry on doing this until the mass is moving parallel to and above the equator-it is now in orbitand can become a satellite
Of course you need the right orbital speed (approximately 8km/s for a close orbit)

When the mass becomes a satellite the weight of the object provides all of the centripetal force,the tension becomes zero and any device at the pole which is weighing the mass through the tension will record an apparent but not real total loss of weight.The mass can break loose of the string and it will still remain in orbit.
Does the apparent weight suddenly disappear in one go when the mass becomes a satellite around the equator?The answer is of course not the change is gradual; increasing as the string lengthens and it is Newtons laws that make these predictions there being an apparent weight loss no matter how small the radius of the circle.Any object spinning or rotating in a horizontal plane can be described as being in partial orbit.

 

[Dale]

2.You start to whirl the mass in a horizontal circle playing out the string so that it lengthens and you adjust the speed so that the mass does not collide with the earth.

I already told you that this is not relevant. The motion you describe here is not spinning since it is rotation about an axis that does not pass through the object's center of mass.

Let A and B be two arbitrary but identical rigid objects of mass m. A is not spinning, but B is spinning (rotating about an axis through its center of mass). A is experiencing a net external force f and therefore A's center of mass is accelerating at a rate a=f/m. The center of mass of B is also accelerating at the same rate a. From Newton's 2nd law the force acting on B is also f. Therefore, since the forces must be equal a spinning object must weigh the same as a non-spinning one.

 

[Dadface]

DaleSpam you seem to have dismissed a system I chose to analyse on the basis of the word "spinning".I have already referred to this word and its definition in post 32,please take a look.May I also point out that the analysis,in principle,also applies to an extended system on a spindle.
As for your second point I am assuming you mean that the "arbitary objects" are identical
in terms of the predictions made by Newtons laws.They are not by virtue of the fact that one is spinning and the other is not spinning.If the spinning object was balanced on a very thin spindle it can remain spinning until It slows down enough so that it falls over.In principle you may be able to balance the non spinning object but it will be in a state of unstable equilibrium,the slightest wobble and over it goes.Newtons laws,therefore, make different predictions for the two objects,predicting that whilst the spinning object is going at a fast enough rate it will not fall over even when there is a slight wobble.These predictions,of course, are confirmed by observations.
There are several other differences and even the elastic properties of the spinning object are relevant to the analysis when this is carried out in even greater detail.Early in this thread jambaugh introduced S.R into the analysis and pointed out that the spinning object becomes more massive.The mass changes may be extremely and immeasurably small but nevertheless we can feel fairly confident that there are changes.
If we get back to classical physics Newtons laws require that each point object,be it a single particle or one just one component part of a macroscopic object,requires a centripetal force for it to remain in its curved path.The main difference between my analysis and those that are conventionally carried out is that I show that the weight of the object itself contributes to the centripetal force needed,even when the plane of rotation is horizontal.
Can we agree that a full detailed analysis of rotating/spinning objects is very complicated?

 

[Doc Al]

As for your second point I am assuming you mean that the "arbitary objects" are identical
in terms of the predictions made by Newtons laws.They are not by virtue of the fact that one is spinning and the other is not spinning.If the spinning object was balanced on a very thin spindle it can remain spinning until It slows down enough so that it falls over.In principle you may be able to balance the non spinning object but it will be in a state of unstable equilibrium,the slightest wobble and over it goes.Newtons laws,therefore, make different predictions for the two objects,predicting that whilst the spinning object is going at a fast enough rate it will not fall over even when there is a slight wobble.These predictions,of course, are confirmed by observations.

But the prediction that you are making, that the "weight" is different for a spinning object, is not confirmed by experiment.

There are several other differences and even the elastic properties of the spinning object are relevant to the analysis when this is carried out in even greater detail.Early in this thread jambaugh introduced S.R into the analysis and pointed out that the spinning object becomes more massive.The mass changes may be extremely and immeasurably small but nevertheless we can feel fairly confident that there are changes.

None of which has anything to do with what you are claiming about spinning objects.

If we get back to classical physics Newtons laws require that each point object,be it a single particle or one just one component part of a macroscopic object,requires a centripetal force for it to remain in its curved path.The main difference between my analysis and those that are conventionally carried out is that I show that the weight of the object itself contributes to the centripetal force needed,even when the plane of rotation is horizontal.

Again you equivocate between "an object being twirled on a string" and "a spinning object". Sure, if you twirl an object on a giant string, then the direction of the gravitational force will have a radial component. That's a far cry from making a claim about the apparent weight of a spinning object.

 

[Dadface]

Doc Al spinning, turning,rotating, orbiting choose whatever word you prefer, the principle is the same.Sketch a giant disc ideally with most of its mass concentrated on its outer edge,being supported above a round earth,mark in the gravitational forces, particularly at the outer edges,and see the radial component.Now set the disc spinning and watch the disc support system changing ,stretching etc so as to accommodate the resulting centripetal force.Now ask yourself the question- "Is it changes in the support system only that provides the centripetal force"?.The answer is no ,the weight provides some of the centripetal force the amount of which depends on the geometry and structure of the system, the speed of rotation and the radius of the circle.

 

[Dale]

DaleSpam you seem to have dismissed a system I chose to analyse on the basis of the word "spinning".I have already referred to this word and its definition in post 32,please take a look.

Yes, I am dismissing the system you chose to analyse. It is irrelevant to your own OP. In your OP you specifically referred to gyroscopes. Gyroscopes spin, so an analysis of orbital motion is simply not relevant as both turin and I have pointed out.

Your use of a dictionary definition of the word "spin" is also not relevant. Word useage in physics is much more precise (and mathematical) than in general useage. In physics, spinning is rotation about an axis that passes through the object's center of mass. Gyroscopes spin. Satellites orbit. Analysis of orbital motion is not relevant to the OP about gyroscopes.

Everybody here knows that orbiting objects have less apparent weight than non-orbiting objects and nobody is disputing that. Everybody (besides you) also knows that the fact that orbiting objects have less apparent weight is irrelevant to the question about whether or not spinning objects have less apparent weight.

May I also point out that the analysis,in principle,also applies to an extended system on a spindle?

I don't think so. If you think it does then please demonstrate that.

 

[Dadface]

In my OP I referred to "gyroscopes and the like" and when I presented an analysis I did so by referring to a point object and I have stated that the effect on a whole object can be found by integration,so it does apply to gyroscopes as well as to orbiting objects.
With your second point I have been dismissed on a point of semantics.
With your third point I am told that everybody here besides me knows that the apparent weight loss of orbiting objects is irrelevant to the question about spinning objects this being despite the fact that I have been pointing out that with a spinning object there is a horizontal weight component and that the spinning object can be described as being in partial orbit.
With your fourth point you seemed to suggest that my analysis cannot,in fact,be applied
to an extended system. If you think this is true then please demonstrate that.May I suggest that you read my previous post.

 

[Dale]

I have stated that the effect on a whole object can be found by integration,so it does apply to gyroscopes as well as to orbiting objects.

Then please perform the integration and demonstrate that your analysis also applies to spinning objects rather than just orbiting objects.

you seemed to suggest that my analysis cannot,in fact,be applied to an extended system. If you think this is true then please demonstrate that.

I asked you first , besides the burden of proof is always on the person making the unorthodox claim.

 

[Dadface]

Thank you Dalespam, Doc Al and everybody else, especially turin. I found all of your feedback useful and it forced me to think through things more carefully.I think I have already done the maths correctly but I am going to get this checked out.If I knew how to post a smilie face I would do so here.

 

[turin]

turin thank you very much indeed for presenting your maths.

You are quite welcome.

I refer to the "horizontal massless rod of length 2r".The rod cannot be horizontal otherwise there is no component to support the weight .

This is simply untrue in my model. I clarify that the horizontal rod in my model is perfectly rigid, and it isn't even supposed to represent a real rod, but an infinitesimal slice of a disk that supports the massive ring of the gyroscope. There is nothing wrong with the rod remaining horizontal and supporting the weight of the gyroscope. I suppose I should have also calculated the balance of torques, but I was too lazy. In short, the symmetry of my model leads to a balance of torques.

You can always point to something in a model that fails to perfectly represent nature. So, of course, if you insist on finding something in the model that does not perfectly describe nature, then no one on this forum, or anywhere else in the physics community of whom I am aware, can offer you a satisfactory contention. Why not object to the infinitesimal thickness of the ring, or the use of massless rods (or strings)? And what of the Ehrenfest Paradox?

If you want to start introducing the effect of material properties (as it seems that you do from one of your comments to DaleSpam), then count me out.

 

[Dadface]

Hello turin.In principle your model differs from mine in one respect.I see that the horizontal rod bends, although very slightly for a "rigid" rod and when I take into account all of the angles involved,including those for the weight vectors, I obtain a different answer.Basically I am just being fussy ,overly so perhaps for a normal gyroscope where we can assume that any bending etc is negligibly small in that any changes become difficult,if not impossible to measure.Thank you again,I really appreciate your feedback and perhaps we will meet in other threads.