Is a Spinning Gyroscope Weightless?

This video is about gyroscopic motion

[berntolovhellstrom]

He applied a rotation force that was converted to the lift force he referred to. You did not apply that force and therefore your result won’t be comparable.

[AB-dbpz]

One critical thing you missed in your experiment is how Laithwaite didn't just allow the gyro to precess at it's natural rate, but (you can see it in his Christmas Lecture to the Royal Society) he accelerated the gyro in it's precession direction and that is when it got raised up by a spring in the axle. Can you add a feature to speed up the precession of your gyro (after burning the string) and report back? Very interesting work!

[Danell]

The mystery is not "what does it weigh?"
The mystery is "how could he lift it with such relative ease?"
Mystery not explained.

[CA-jxby]

All you had to do was repeat the original experiment standing on a bathroom scale.

[joelg]

This is way off dude. It wasn't a question of how much it weighed. You didn't address the original dilemma.

[markcampbell]

He “lifted” 40lbs like a feather over his head, something he couldn’t have done without the gyroscope in action. Seems like you strayed from the point he was making.

[dunkelec]

Where is the part of the graph of the sensors for the NON spinning gyro?

[dwalden]

I’m not a physicist, but my take is pretty simple: The rotational momentum is transferred into upward force, given a little help. Also: Notice that he could only push it up and over in one direction, which follows the spinning direction. If spun in the opposite direction, he would be able to lift it up in the opposite way.

[bricology]

I think that the phenomenon that Laithwaite experienced has to do with the "conservation of angular momentum". He wasn't so much *lifting* the weight of the rotor, as much as the energy from the rotor spinning around its own axis was transferred to Laithwaite as he pirouetted in place. As Laithwaite began to tilt the spinning rotor from the horizontal plane, it needed to transfer energy, and as he turned, the energy of the rotor was transferred through the shaft to him, in the form of torque and "pulling" on the horizontal plane. His body's mass largely negated the horizontal "pull", and his grip on the handle negated the torque. I think it would be *much* easier for a 175 lb.(or whatever) man to resist a 40 lb. horizontal pull than to do a 40 lb. lift.

It seems to me that this could be empirically tested in at least three ways:
1. Have a person replicate Laithwaite's experiment, but not directly grip the handle; instead have the interface between the shaft and the experimenter's hands be a torque meter.
2. Also, if my hypothesis is correct, the speed at which the experimenter turns in place should be critical, and would be set by the rotor (its mass and speed); the experimenter turning too fast or too slow should affect the ability to successfully replicate it.
3. Measure the loss of speed in the rotor both without being moved from the horizontal plane, and from Laithwaite's experiment. My guess is that in the latter, it will lose considerable speed because of transfer of energy to the experimenter, due to the conservation of angular momentum.

[personx]

I used to deal with aircraft gyros when I was in the navy and happen to know that what I saw done was real.
You miss the point dude.
The question is whether doing what he did results in energy savings after the energy is spent spinning up the gyro. If it does then the method has practical value.

[repairitdontreplaceit]

Prof. Laithwaite wasn't  lifting the gyro, the gyro was lifting itself
because the professor was accelerating its precessional motion . as others have said . please re run the experiment under the same conditions the prof did

[brianwilson]

I was going to call out your entire experiment for being completely off track from what you were supposed to test....Then after looking at comments I see EVERYONE called you out already. So yeah.

[asdasdasd]

"while the experiment was a success but it didn't support Laithwaite claims" 
Prof. Laithwaite wasn't  lifting the gyro, the gyro was lifting itself because the professor was accelerating its precessional motion.
The setup of your experiment including position of load cells is quiet wrong. 
Only one load cell is needed and must be positioned on the gyros arm, between swivelling point and the gyro.
Gyro not spinning = full weight 
Gyro spinning and naturally precessing = zero weight
forcing precession speed = negative weight

[timellis]

The correct way to execute this would have been to stand on a scale and weigh yourself. then weigh a 40lb gyro on a shaft. Then stand on the scale and pick up the gyro above your head and see if you gained any weight. I went to community college.

[wwtapsable]

You missed one part I'd the experiment he also pushed the gyroscope in the direction of precession

[Neil-Aspinall]

You did not answer how he lifted the spinning gyroscope with ease. Your data set misses the point junior scientist.

[BritishBeachcomber]

As an undergraduate, I was fortunate to attend a lecture by Prof Laithwaite. A great showman, he later acknowledged that gyroscopes behave fully in accord with Newtonian mechanics.

[gregc]

When I saw this video in the side bar, my first reaction was: "Why would anyone ask such a stupid question in the first place???"
After watching the video in the video: I note that he didn't hold the bar and gyro stationary over his head but kept it in motion. He was letting the gyro precess. He was effectively using some of the stored angular momentum to help support the weight of the gyro, but the trick only works if you follow the gyro's lead, so to speak.

[carmelpule]

The spinning gyro is not weightless but all the vertical gravity weight and accelerating forces are transferred to the central support pivot.
Recently I took a deeper interest in gyros. Having met Professor Eric Laithwaite many years ago and recently following what was done over the years, including the famous lifting machine by Sandy Kidd, and many others trying to use the gyro in an inertial propulsion system, I felt that many do not fully understand the engineering dynamics of the gyro and so I decided to look at it from just an engineering point of view.
I treated the gyro as a set of mass particles m(x, y, z, t) distributed over the face of a rotating flat disc being subjected to the distributed accelerations, a(x, y, z, t) of a "cycloidal trajectory" decided upon by the active geometry of the gyro. This is basically a spinning mass in the shape of a flat thick circle of large diameter going around in a circle about a central pivot point. Each mass particle in the spinning gyro flat circular wheel is also being translated as shown in the video and it will in fact be subjected to an acceleration force of the form F= m.a. where a is a vector. The final behavior of the gyro is the massive integral of all those accelerations working on all those particles existing in space and time. If one draws the three-dimensional voluminous cycloidal trajectory moved by one particle, then one has a good ideal of what one would expect of it.
Basically, as an engineer, what I did was to use the past experience of mass particles going around curves as in the case of a mass particle going around a circle. This is a case where the speed stays constant and it is the direction that changes in the velocity component of the mass particle. With the mass particles existing in the gyro, then when the thick circle spins, the mass particle in an un-torqued spinning circle will go through the curve of a flat circle with its circumferential constant speed but changing direction all the time, hence generating the conventional centrifugal force acting outwards on a mass particle traversing that particular curve/trajectory of a flat circle. If as shown in the video, the gyro happens to be operating with a horizontal spinning axis then, then as a plan view, the flat disc of a gyro will be no longer flat, but it will take a three-dimensional dept. If the gyro axle is operating at a higher angle with the horizontal, such that when stationary but spinning, the plan view would be an ellipse, and when the gyro is free to precess then the trajectory would be an elliptical cycloid going around the central support pivot.
Any curve in the plan view of the trajectory of the cycloid followed by a particle on the spinning wheel of a gyro can easily be drawn using a pair of compasses and a pencil and some paper and the answer would be very illuminating. This is the remarkable conclusions. The conclusion is rather very the extended axle of a spinning flat circular disc is acting as an extended radial arm on a central pivot and each radius of the gyro wheel extends in quadrature ( 90 degrees) with the spinning axis, then a voluminous room is required to define the volume through which the gyro fits. The gyro can no longer be treated as a spinning flat disc. It becomes a VOLUMINOUS AFFAIR in which the new volume will permit sideways/lateral accelerations to the particles in the disc of a gyro in addition to the ones in a flat circle. It is these lateral accelerations due to a three-dimensional cycloidal trajectory that gives the gyro all its characteristics. If one draws merely the PLAN view of what the gyro is doing in the video above one will find that the shape of the " disc" is no longer elliptical as when the gyro spins and does not traverse but does not move as a unit with respect to the ground. The shape becomes a distorted ellipse.
Now one may consider in the video that the plan view of diameter and flatness of a gyro disc can be fitted in an ellipse when the gyro is torqued at the top and bottom of an instantaneous vertical diameter while the horizontal diameter remains dormant. One may think of this torque action as subjecting the circumference of the top half of the gyro circle to an elliptical lateral acceleration distributed pattern while the lower half would be subjected to the opposite acceleration pattern. There, with that alone one may predict where one mass particle will be when the gyro is torqued and accelerated laterally on the disc, as it spins through traversing the circumference with the angular spinning Ws.
Let us assume that the circumferential lateral acceleration in a torqued gyro is not elliptical but simply a constant through half of each circumference then taking the top half of the circumference,
lateral acceleration to gyro disc a= k
lateral velocity to gyro disc = integral of acceleration a=kdt= k.t
lateral distance covered to gyro disc = integral of velocity v= k.t.dt= S= (k.t^2)/2
Just seeing that integral over the top circumferential half a spin will see the mass particle move out and that is PRECESSION. One may proceed to integrate over the lower side where the acceleration is reversed in the other direction.
These two simple integrals on the upper and lower circumferential halves of mass particles in a laterally torqued flat spinning disc will indicate a concave curve and a convex curve.
All this means is that THE UPPER HALF OF A TORQUED SPINNING GYRO GOING AROUND A CENTRAL PIVOT ON AN EXTENDED ARM ( or without one) IS SUBJECTED TO AN INBOUND CENTRIFUGAL FORCE WHILE THE LOWER HALF IS SUBJECTED TO AN OUTBOUND CENTRIFUGAL FORCE.
In the video, this is what holds the weight of the gyro seemingly floating, as it has an inbound centrifugal force action on the top horizontal half of the gyro wheel and another outbound centrifugal force acting out on the lower horizontal half of the spinning wheel. If one had to draw the actual " ellipses" in the plan view of the cycloidal paths one will find that the " translating and spinning ellipses " are distorted on the upper and lower curve with the upper curve having a small radius of curvature while the lower curve have a larger radius of curvature than the normal ellipse expected in an inclined circle. And here is the beauty and elegance of the gyro. Each spinning mass particle will have to traverse a tighter curve when navigating the top half of the spinning gyro wheel and a shallower, less tight curve when going through the lower half of the spinning wheel. This means that the inbound centrifugal force generated by a mass particle when traversing the upper half-circle is higher than when it is traversing the lower half AND WHAT IS MORE THE UPPER ONE HAS AN INBOUND CENTRIFUGAL FORCE WHILE THE LOWER ONE HAS AN OUTER CENTRIFUGAL FORCE. This all happens in the voluminous cycloidal path around the central pivot, and it also happens when any gyro is torqued.
The vertical and horizontal torques generated by these two opposite acting centrifugal forces, moving about on each quarter face of the gyro disc depends on Spin Ws, Mass m, Diameter d, extended radius Ra, Wp, etc, and it is what balances the gravitational weight ( m.g.R arm) and the dynamic angular torque ( I, Ws, Wp.)
if the gyro is spinning very fast the inbound cancels out the outbound centrifugal force and a gyro operating as shown in the video shows no resultant centrifugal force. If one digs deeper one would find that there is no lateral reaction on the central pivot except the vertical weight of the gyro and any vertical acceleration of the mass of the gyro.
It is remarkable how a massive integral of the product F= ma in the form of a distributed force due to a million distributed mass particle operated upon through a double integral on the acceleration only as the mass is constant, or say ( m(x, y, z, t) * a(x, y, z, t)) gives all those beautiful elegant artistic engineering results.

[Sammyjankis]

The gyro effect allowed him to increase the distance travelled during the lift as he circled it around him he increased the mechanical advatage similar to rolling it up a mild incline over several feet instead of lifting directly opposite gravitation force