Do spinning objects really lose weight?

[Dadface]

Several years ago there were widely reported claims that gyroscopes and the like became lighter.Although these claims have generally become discredited could there be any truth to them and could proof come from theoretical reasoning?

 

[dE_logics]

Might be a myth you know.

 

[turin]

... could there be any truth to them and could proof come from theoretical reasoning?

Of course. However, science does not pose speculation as theory. I know of no theory (in the scientific sense) that suggests that gyroscopes become lighter when they spin. However, I am not confident enough in my knowledge of GR to be certain of the lack of theoretical support. There are some weird things hiding in the seemingly simple theory of GR.

 

[Dadface]

Perhaps any proof could come more readily from Newtonian gravity which must conform to GR within its own domain and perhaps the proof could be very simple.What if I posed a seemingly speculative statement which is that according to Newtonian gravity a spinning gyroscope should display a weight loss.Whether this is measurable or not depends on several factor which include the sensitivity of the weighing instrument and the structure and spin speed of the gyroscope.

 

[gunslingor]

Only if it has wings such that it acts, slightly, like a helocopter.

 

[jambaugh]

As a matter of SR and GR, a spinning object having more energy=mass than its stationary counterpart would weigh slightly more.

 

[A.T.]

Rotating objects have a lot of momentum stored, that can be used to lift the object, creating the illusion that the object is very light.

 

[Andy Resnick]

Is the OP's question in regards to Podletkov's "results"?

 

[Vanadium 50]

I think this is the "Tohoku Top" experiment, Hayasaka et al. Phys. Rev. Lett. 63, 2701 - 2704 (1989). It was repeated with a null result by Nitschke et al. Phys. Rev. Lett. 64 2115 (1990), and Faller et al. Phys. Rev. Lett. 64 824 (1990).

 

[Dadface]

Even disregarding relativistic and aerodynamical effects Newtonian mechanics may predict that a weighing instrument can,in principle, detect a weight loss and the proof may be quite simple.In terms of SR when the gyroscope gains mass/energy something else loses it . There have been experimental attempts to measure weight loss but none have been conclusive.

 

[turin]

What if I posed a seemingly speculative statement which is that according to Newtonian gravity a spinning gyroscope should display a weight loss.

The amount of speculation is irrelevant. This is pointless unless you can show it, either by simulation, or analytical proof, or something. That's one of the beauties of science: it uses math, so predictions of a given theory can be made very precise. When you use math to make a claim based on a theory, we don't call it a speculation, we call it a testable prediction. Please show us the mathematical evidence that Newtonian gravity predicts weight loss of a spinning gyro. I make this request sincerely, because, if it is true, I find it very interesting, but I don't see how to extract this prediction from Newtonian gravity.

 

[Vanadium 50]

There have been experimental attempts to measure weight loss but none have been conclusive.

I just posted two references showing it isn't there and that the Tohoku experiment was wrong. Did you read them?

 

[winstonsmith]

In a spinning object angular momentum is acting 90 degrees from the base. if the objects mass is spinning and even a part accounted for through spinning then perhaps the object would weigh less although its mass remains the same.
This seems an obvious result and the fact that it is not part of established physics with its own formula suggests to me that it mus be very difficult to model both mathematically and empirically or that I have looked at it too simplistically. Like Turin I would like to see the maths.

 

[Dadface]

Sorry I don't know how to present diagrams or mathematical equations.I will illustrate this with the simplest example I can think of-a point mass being whirled in a horizontal circle by a light inextensible string.Normally, when a free body force diagram is drawn and if air resistance is ignored ,there will be two forces acting on the mass
1.the tension in the string which acts along the line of the string where it joins the mass.
2.The weight of the mass which acts vertically downwards
If we carry out the conventional analysis we would describe that the vertical component of the tension supports the weight of the mass and that the horizontal component of the tension provides the centripetal force and no surprises are revealed.
Now we need to be a little bit more fussy and look at our free body force diagram again.The weight,in fact,does not act vertically downwards it points towards the Earth's centre.Because of this the weight provides some of the centripetal force,the tension will be smaller and there will be an apparent but not a real weight loss.Obvious but if there is a mistake in my reasoning or if this has been done before I apologise.A couple of extra points
1.The above analysis can be extended for a gyroscope as it is usually imagined.
2.The biggest fractional weight loss(everything else being equal) should be recorded by the largest diameter gyroscope with most of its mass being concentrated in a thin ring on its circumference.
3.The spinning object somewhat resembles a satellite and if its plane of orbit coincides with a central plane of the Earth and if its spin speed has the right value for its height above the Earth then the apparent weight loss will be total.You could cut the string and watch it go.I wouldn't try this because you will need a very long piece of string and a very long ladder.
(If I haven't made my presentation clear sketch a free body force diagram with a round earth)

 

[Phrak]

Now we need to be a little bit more fussy and look at our free body force diagram again.The weight,in fact,does not act vertically downwards it points towards the Earth's centre.

The weight is a vector, pointing vertically down.

 

[Dadface]

Weight is a force,it is a vector and its direction is towards the Earth's centre of gravity.

 

[Phrak]

Weight is a force,it is a vector and its direction is towards the Earth's centre of gravity.

How does that differ from vertically down?

 

[Dadface]

Hello Phrak,it was your previous comment that made me realize that I could have made my presentation clearer and so I edited it by adding an extra sentence in brackets.Have a look try it out and that should answer your question.

 

[granpa]

lol. the point mass is partially orbiting the center of the earth. I never thought of that.

 

[turin]

The weight,in fact,does not act vertically downwards it points towards the Earth's centre.

The difference being? Is this untrue for an object that is not spinning?

Because of this the weight provides some of the centripetal force,the tension will be smaller and there will be an apparent but not a real weight loss.

Why is there an apparent weight loss? I see no apparent weight loss here.

1.The above analysis can be extended for a gyroscope as it is usually imagined.

I guess by "extended" you mean "changed completely". I see no connection between lessening of tension in the string and lessening of weight of a spinning gyro.

 

[Phrak]

Hello Phrak,it was your previous comment that made me realize that I could have made my presentation clearer and so I edited it by adding an extra sentence in brackets.Have a look try it out and that should answer your question.

Technically you are probably wrong though partially correct, even adhering to Newton, and ignoring Einstein.

Most gyroscopes a-spinning, that I know of, are located in commercial aircraft. Nine in each, usually. Of these aging dinosaurs, not yet replaced by laser gyros, one is mounted with it's axis vertically oriented, while the others point in other directions. The ones that point in other directions will not experience your 'orbital effect', as I might call it. They won't weigh less. Of the remaining, some are on the ground, while the aircraft is doing it's airport thing, spinning up and spinning down, and often this means in proximity to some mountain range. These will weigh more, due to local gravitational distortions, so they're out. The remaining have a negligable, and unmeasurable increase in weight, if we are so inclined to ignore Einstein. I am just way too amused by this problem.

Did I overlook any viable population of gyros spinning with axiis more or less vertical? I did leave out the vast population of discarded and forgotten toy gyroscopes. These are sitting idle, and weigh about the same.

 

[Dadface]

Thank you for replying and let me answer the points that were raised.
turin raised three points.

1.There is no difference of course but it was relevant of me to mention it so that the free body force diagram could be drawn with the correct lines of action.

2.With the situation I described the weight(apparent and not real) is felt through the tension and if the tension changes the apparent weight changes also.If the mass was moving through a vertical circle the effects are really easy to measure, even with a pair of bathroom scales.The tension varies from a maximum at the bottom to a minimum at the top these changes becoming less pronounced as the plane of rotation approaches the horizontal(by horizontal I mean in a plane at 90 degrees to an Earth radius).Stick an electric motor on your scales and watch the weight reading fluctuate.In fact you don't even need a motor just stand on your scales and lift an arm rapidly up and down(then do the hokey cokey or a moon dance)

3.I agree that extended could be described as changed completely but when we carry out the analysis we could start by considering a point object and then summing the total effect by integration.Hence my closing comment number 2 in post number 14

If this is still difficult to see just do the free body force diagram with a round Earth and an imaginary extremely long length of string and the mass orbiting above the equator.

Phrak I agree that with commercially built gyroscopes the orbital effect is probably immeasurably small but because of current practical and instrumentational limitations rather than theoretical limitations.I am not sure ,however,. that the Newtonian analysis is at odds with any Einsteinium analysis

 

[Vanadium 50]

It seems we've lost sight of the original question: do spinning objects really lose weight?

The answer is "no". People have looked for such an effect and not seen it.

 

[Dadface]

But theoretically there is an apparent,but not real, weight loss somewhat akin to that experienced in free fall.The fact that such an effect has not been seen yet is irrelevant and due to the effect ,at present, being immeasurably small.You can see the effect for yourself by reading point 2 in post number 22 the vertical movement resulting in apparent,alternate and measurable weight losses and gains.With the analysis I presented in post number 14 I described motion in a horizontal circle where the apparent change will be a weight loss only and which,because of the geometry of the system,will be extremely small.I think there are still people out there who believe there is a real weight loss.Although one must keep an open mind my personal belief is that a real weight loss is most unlikely and that the claims put forward about anti gravity devices should,at present, remain in the realms of science fiction.

 

[Vanadium 50]

You can see the effect for yourself by reading point 2 in post number 22 the vertical movement resulting in apparent,alternate and measurable weight losses and gains.

I don't see anything beyond the fact that shaking a scale will make the needle rock back and forth.

 

[Dadface]

Thats right,it rocks back and forth.Put it another way-you are standing on scales in a lift.when the lift is at rest or moving with constant velocity the scales will measure your true weight.when the lift is accelerating down the scales will give a lower reading and when it is accelerating up the scales will give a higher reading.The principle is the same.(for lift read elevator)

 

[Vanadium 50]

Yes, but shaking a scale and watching the needle move doesn't mean that a spinning object loses weight.

 

[Phrak]

It seems we've lost sight of the original question: do spinning objects really lose weight?

Nope. No, mention of spinning. Not any object, just gyroscopes.

Oh, wait. I see the title and question are at odds.​

 

[Dadface]

Vanadium 50 I do not understand what you mean by your last post.If you mean that there is not a real loss of weight then yes,this is a point I have been stressing.The weight loss is not real it is apparent.An astronaut in orbit experiences an apparent weight loss.If the spaceship were to suddenly dematerialise the astronaut would remain in orbit his weight providing the centripetal force needed.A gyroscope or any object spinning in a horizontal plane is in a sort of partial orbit and there is a corresponding apparent partial loss of weight.I think I have explained this in more detail in my previous posts.

 

[Dadface]

I have done some maths and I need to double check this
T=tension
m=mass
r=radius
v=speed
g= acceleration due to gravity

1.If the normal assumption is taken which is that the weight acts at 90 degrees to the plane of rotation then the tension is given by:
T squared=m^2*v^4/r^2 plus m^2*g^2
2.If now we take it that the weight acts towards the Earth's centre we get that:
T squared=m^2*v^4/r^2 plus m^2*g^2 minus(2 * m^2*v^2 g cos theta)/r
Here theta is the angle that the line of action of the weight makes with the horizontal. If theta becomes zero we can calculate the speed needed for T to become zero(total apparent weight loss) and the speed we calculate is exactly that needed for a satellite.The mass or ring will itself be a satellite moving around the Earth in a central plane.
Excuse my poor attempt at presenting equations.

 

[turin]

Dadface, why do you keep talking about this ball on the end of a string idea? This is not a spinning object; it is an "orbitting" object.

 

[Dadface]

1. spin-vb to rotate or cause to rotate rapidly,as on an axis.
2. rotate-vb to turn or cause to turn about an axis revolve or spin
3 revolve-vb-to move or continue to move around a centre or axis;rotate
(definitions taken from Collins concise dictionary)
I did not want to get involved in semantics but I think that words such as spinning are more appropriate than the word orbiting when applied to real gyroscopes and the like.In post number 19 granpa summarised it wonderfully when he wrote that the point mass is partially orbiting the centre of the earth.It is a phraseology that I have borrowed from(thank you granpa)
If we are to be fussy about the terminology used I have never referred to a ball on a string but I have referred to a point mass(post 14) later being more economical with words and referring to it as a mass.To reiterate on a previous comment,by considering a point object we can ,by a process of integration,sum the effect on the whole object.
In the anticipation of any criticism of my reference to a light inextensible string I want to point out in advance that this itself is nothing more than a theoretical construct that enables the above calculation to be carried out.
p.s.I haven't carried out the full calculation for an extended object because my maths is probably not up to it .Because of lack of use I have forgotten most of my maths and I was never particularly good at it in the first place(it was the 1960s).If you are able or inclined to do a fuller analysis I would be interested by any findings you make but whatever these be there will be an apparent weight loss. If I ever embarked on the task I would calculate the fractional apparent weight loss for a gyroscope where most of the mass is contained within a thin but finite ring on the circumference of the gyroscope this structure being the one that will display a loss which is greater than that displayed by a gyroscope where the mass is more evenly spread out.

 

[turin]

If you are able or inclined to do a fuller analysis I would be interested by any findings you make but whatever these be there will be an apparent weight loss.

I'm still waiting for the justification for this claim. I suspect that it is untrue.

 

[Dadface]

Hello turin.You may be conviced if you do a free body force diagram with a round earth.I am new to forums and a dinosaur when it comes to using computers (despite having attended several courses)and I don't know ,as yet, how to present diagrams.

 

[Vanadium 50]

I think you need to better explain what your question is. It's jumped from spinning objects to shaking scales to orbiting objects.

 

[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.