# Do spinning objects really lose weight?

In summary, the conversation discusses the possibility of gyroscopes becoming lighter when they spin, and whether this claim has any basis in theoretical reasoning. Some suggest that according to Newtonian gravity, a spinning gyroscope should display weight loss, but others argue that there is no mathematical evidence to support this claim. Experimental attempts to measure weight loss have been inconclusive. One participant presents a simplified example of a point mass being whirled in a horizontal circle to illustrate the concept, but others point out flaws in the reasoning. Ultimately, it is concluded that there is no solid evidence to support the idea that spinning objects weigh less.
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?

Might be a myth you know.

... 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.

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.

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

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

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.

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

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).

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.

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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.

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?

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.

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)

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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.

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

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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?

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.

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

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.

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.

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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

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.

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.

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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.

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)

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

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.​

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.

I have done some maths and I need to double check this
T=tension
m=mass
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.

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.

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.

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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.

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.

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

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