Momentus

Cheap gyro, cannibalize a "power ball" and use the gyro inside. Some models also have a built in rev counter. http://www.ebay.co.uk/sch/sis.html?_kw=powerball+gyroscope+exercise+neon+pro

My Hypotheses is that Newtonian Dynamics does not describe behaviour of the offset Gyroscope, as demonstrated by video 5. This has been dismissed as being due to a misunderstanding of the purpose of the experiment resulting in the heavy object being exchanged for the light one. Having done the experiment myself I know that this is not so. The Gyroscope pendulum is another example where the analysis which is correct by the current paradigm does not predict the actual outcome. To start right at the beginning with a spinning gyro suspended from two strings. One string is burnt through leaving the gyro suspended, point of suspension directly above the point of attachment, distance from there to the Centre of mass d. The gyro couple transfers the reaction to support the weight and the gyro remains horizontal. The gyro precesses, which means that it turns about its own centre of gravity. It cannot displace itself. There must be an external force applied in direction of the displacement to accelerate the mass from rest. As the gyro is attached externally by the string, the displacing force must be provided by the string. For the string to provide a horizontal component of force, it must deviate from vertical. This should be visible, initially the gyro stays still then gradually accelerates as the subtended angle increases. It is only after the string has provided this acceleration that the gyro can move in an orbit. There is a way to demonstrate this conclusively. At some point the ratio of d to L will be such that the horizontal force will be too small to displace the mass and the gyro will remain in position with the string circling around it. So, if L then is reduced, would the point of rotation move away from the Gyro?. Would it be a gradual process? Can L be fixed at a length such that the point of rotation would be at d/2? When does K^2's prediction become valid and z become determinate? All of those questions are theoretical. The problem is that it does not happen that way, there is no significant deviation from vertical and the gyro mass moves immediately, it is not accelerated by the string. Some thoughts rather than abuse would be different. Maybe make an effort to understand rather than chant the mantra, Momentus is stupid to think that known science could just possibly be wrong. Momentus .

Oh dear. I talk about Pendulums, you talk about swing ball. Is there no length you wont go to just to make your debating point??

That is an excellent analysis. Your conclusion that the Rotation of the gyro about its own centre of mass which is precession whilst moving with orbital motion of a pendulum means that in any situation, the behaviour must be chaotic - with the exception being when as you say Ω²=ɲ. - this gives the locked behaviour as when the moon orbits the earth. This is a complex interaction, difficult to explain, you did a good job. But and there has to be one I'm sorry to say this not the way a gyro behaves. Indeed one of the fascinations is its smooth motion in a true circle. swinging damps out then moves smoothly. Troll the internet, Google gyro pictures, there are no examples of the chaotic behaviour that should be there. No sarcasm. Your analysis is spot on, the behaviour should be chaotic. But it is not. I was trying to make that point when I said that taking a gyro that is in a stable orbit and shorten the string should result in change.My insistence that the gyro is not pushed, is to ensure that no external force is applied to set the gyro in orbit. Under those conditions, the gyro should remain in the same position rotating (precessing) about its own C of G until some external force is transmitted along the string. For pure precession there is no conflict between the two modes because the C of G cannot move. Yet again this is not how a gyro behaves. No matter how carefully it is released the offset supported gyro will not simply precess about the C of G. Your careful and I hasten to say correct analysis of gyro behaviour is not the real world behaviour. I doubt if any other physicist has taken the trouble to do this. Momentus

If you did know how a pendulum works, then you would know that the size of the orbit radius is a function of the linear momentum initially imparted to the mass. Without that initial momentum the mass will hang directly below the point of suspension, not moving at all. If you had reached that point then the next question would have been where does that linear momentum come from and where does it appear in your formula. And unless and until some other member can take up the burden of explaining the physics of a pendulum? I did a number of experiments with gyros 20 years ago. If you really want to know about the anomaly, then buy a good gyro and invest some time and effort in your own education. Stop thinking that you can learn science by scoring debating points. The experiment is the only arbiter of truth. Momentus. Momentus I can show you the available evidence, I cannot make you think about it. If you think this video was faked, then more of the same is not going to penetrate your deep knowledge of the subject

I have no interest in your belief structure. You can accept the doctrine you have been taught by those whom you think are older and wiser, agree with the view of the masses, name and shame heretics. The experiment does not care what you believe, neither do I feel any obligation to teach you. There is no one on the site who cares enough about what you say to correct the more obvious errors, why should I. Momentus You did not do the simple string experiment, although experiment is rather a grand name for it. The statement that " ̉ۡ varies. With a fixed string length, it increases as the movement radius increases, it has no upper bound."Does not come from your actual experience. It is a canard of magnificent proportions. Perhaps no one reads 'what you wrote'. Momentus I hesitate to taint you by replying to your post. If you read what has been said up to now, there will be no attempt to understand what you have done, only a mad chorus of "experimental error" at best and "charlatan" at least. That there is a flaw in the application of Newton's laws on conservation is the explanation. Momentus Then you should look at video 5.

Whilst I see your point about the use of the word orbit, I am not trying to win a debate, but to explain. circular movement is clumsy don't you think? As you have clearly understood what was meant, I think I will continue to use orbit in this context. A minor point. without gravity, the system would not rotate. I felt that I was being given grief over my typo, unless of course you are addressing Z Man. Do you have an opinion on momentum being a critical factor in calculating deflection? How sad is that. You need a video of a weight on a string? use a mirror.

Great deconstruction of the experiment, which you could not be assed to actualy try. Pity because had you done so you could have corrected the blatant typo (Rad not Diam Doh). We would then have the true and attested behaviour of a mass orbiting a fixed point. This would have included the observation that you, being a seasoned experimenter, ensured that L and ̉ۡ did not vary. "only valid under these conditions: You have a gyro precessing" does that mean it is not an orbiting mass? In the Laithwaite ice tower, did a gyro behave differently from an inert mass? My issue with the K^2 is not one of mathematics but of logic. If the momentum of the mass is unknown, the radius of the orbit cannot be calculated. That is true for any mass be it spinning or inert. Momentus

assuming the equation you have given is z = d Lɲ/(g - Lɲ)Ok the experiment. Do try this at home children. Take a 60 cm Length of string, with a 4oz weight tied on the end(old set of scales). Hold in right hand and swung around, at É rad/sec. gestimate diameter of circle at 20 cm. Increase agitation of right hand, gestimate new radius at 25cm. stop and repeat until boredom sets in or you run out of radii to estimate. During the experiment it will be noted that L, É and M do not vary. yet z does. Which value changes in the right hand site of your equation z = d Lɲ/(g - Lɲ) to reflect the experimental values for z? This experiment is Item 1 K^2's formulae is item 2 I am unable to proceed to item 3 until there is some agreement on item 1 Momentus

Excellent. Ok so assuming that the gyro is orbiting at precession speed at radius of z . Happily turning round and round. The horizontal force component which holds it at this radius is determined by the subtended angle formed by the length of the string and the radius of displacement. Now for a difficult manoeuvre. Increase the speed of the mass and at the same time, move it out to a larger radius orbit, such that É remains the same. This increases the subtended angle, which in turn increases the horizontal force. But the increase in z means that more force is required to hold the mass in orbit. Guess what, do the math and the increase in horizontal force matches the increase required to hold the new orbit. Isn't science wonderful. z = d Lɲ/(g - Lɲ)?? Momentus

Absolutely I am labouring under the impression that any mass, spinning or not, moving in a circle is subjected to centripetal acceleration. There must be a force acting on that mass, which is external to the mass. With a pendulum, be it spinning or not the force is transmitted by the string, and is proportional to the angle subtended by the string from vertical. Pause for reality check Momentus

That you for your contribution. Clearly expressed, but wrong.

Point the first. "point ROUGHLY bellow point of suspension" you may wish to rephrase this. The moving pendulum bob will travel in a regular ellipse, centred EXACTLY below the point of suspension. The bob wants to move in a straight line. The string pulls it constantly to the centre, over a full rotation, all the forces on the point of suspension must be symmetrical, ie equal and opposite. So Again " Point of rotation will not be PERFECTLY under suspension point, resulting in small angle of the string," this should be: "Centre of mass will not be PERFECTLY under suspension point, resulting in small angle of the string," More "O" level than undergrad. And of course none of the above even now that I have corrected it has any bearing on C of M (Conservation of Momentum) You have formulated deflection using L and ̉ۡ. Newtonian dynamics says that cannot be done. Simple logic of a pendulum is that unless the length of the pendulum is changed, the period of oscillation is a constant. No exceptions. Immutable. Fixed. When you try to speed up a pendulum, then all that happens is the deflection increases. Try to slow it down and the radius reduces. The deflection radius therefore depends upon the momentum of the mass (Mv), not solely upon L or ̉ۡ. As v is a function (̉ۡ z ) your formulae needs more boring modification. This is taken in the limit of angle from vertical being not very large. For any pendulum that is in a steady state of rotation, be it Plumb bob or gyro, L ̉ۡ M and v must remain constant to maintain this steady state. If therefore a change is made to L, the length, then there will be a change in the one or more of the other values to a maintain a new state of equilibrium. As Mv must be conserved, then ̉ۡ must be the value that changes. More if you need clarification, Momentus

Running it on a rail is a great idea. You will be able to spin the gyro around the tower by hand and get a direct comparison with the precession motion. Comparing the spin/no spin that way will level the effects of friction. Take care if you approach the Uni. authorities with this. The reaction may well be more extreme than you expect. http://www.gyroscope.com is a good site for gyroscopes. Needless to say This experiment will change your life. May want to give that some thought too. None the less I look forward to finally seeing an honest replication of Laithwaite experiment. The physical symmetries of time, place, rotation. Applies to all matter, not just crystals:confused:Momentus

Did So. still acts like a regular pendulum if you push it. Then do not push it. Then you have a mass orbiting about a point directly below the point of suspension. It is held in orbit by centripetal force imparted by the supporting string. Shorten the string and that force increases tightening the orbit. TO CONSERVE MOMENTUM the rotational speed (RPM) must increase. It does not. Ok. Ok. don't want to spoil your fun, you can push it if you must. Same deal, it still orbits at precession speed not pendulum period. If the claim is that a light tower is not moved by a precessing gyro, surely a heavy tower would budge even less Wow, you are even better at this than K^2. Orwellian in scope. If you are going to lie, lie the big one."There are No Tanks in Bagdad". So in order to prove the claim lets make the tower heavier!! stroke of genius. Heh what about nailing the tower to the bench, then we wont have any of this rubbish about relative masses.