Eric Laithwaite's Talk on Gyroscopes w/ Demos

[tinman]

I always get a little paranoid when people ask me questions that they already know the answers to.

I presume you are initially hanging the gyro rotor straight down, so that an arrow representing gravity would be in a line along the axle but pointing straight down, and that your "pivot" is a u-joint or other system that is free to move in any direction.

What you will see will depend strongly on the length of the axle, from the pivot point to the rotor. If the axle is very long, so that your applied force results in mostly "translation", that is a straight sideways movement with very little tilting of the rotor, then the precession force will be small and you might not notice it. That is, the resulting motion will be mostly in the same direction as the applied force.
If your axle is short enough, so that your applied force produces a good tilt of the rotor out of the horizontal plane, then you will experience stronger precession and the rotor will try to move at right angles to the applied force. What you will actually see when you do the experiment is some combination of motion in the plane of the paper (the direction of applied force) and at right angles to it.
The "center" of the precession motion is in the rotor disc, not at the top of the axle where your pivot is, and if you use a flexible rotor with a rigid axle you will see the rotor itself twist and warp as you try to force the precession.

Believe it or not TK,i didnt know the answer to my question-thus the reason i was asking you,as i know you have had quite some experiance with gyro's-where i have very little.

I may have not stated my question correctly,and the answer you gave leaves me a little lost(didnt tell me what i needed to know). So im guessing i will have to set up something myself,and see if i get the effect im hopeing for. But i will try again asking you before i go to all that trouble. I will try with a clearer diagram (below).

First-if the flywheel is not rotating,and our jet of water is hitting the flywheel side on(as depicted),then we will have x amount of force at the pivot point that i have called force A and force B-a twisting or sheer force on the pivot pin. Now ,what i want to know is--> will force A and force B be the same if the flywheel is spinning at say 10 000 RPM as it is when the flywheel is not spinning-will gyroscopic precession lesson force A and B ?. We will asume that the water jet can rotate with the flywheel during it's precession so as it continues to hit the flywheel in the same spot.

[TinselKoala]

You can think of the response of the rotor like this: When the rotor is spinning, the response to a force applied "from the side" as you have drawn it will simply act as though you are applying the force from a point 90 degrees around the rotor. That's precession in a nutshell.

But in your system there are going to be two forces acting on the rotor, one from your water jet and the other from gravity. As long as the rotor is hanging straight down the gravity force will not cause any precession, but as soon as your water jet (acting as if it were coming from 90 degrees around the rotor) causes the thing to precess out of the straight down (rotor horizontal) position, then gravity also acts to contribute to the precession. So the resultant motion of the rotor won't be purely "as if" the water jet is acting from 90 degrees but will be a combination of the precessing from the water jet and from gravity, so the rotor will move not straight toward or away from the viewer but at some angle between that and the plane of the paper.

If you didn't have any gravity, the rotor would try to move at right angles to the applied water jet force, that is, straight toward or away from the viewer, depending on the direction of rotation of the spinning rotor. Again, the length of the suspension will affect the magnitude of this motion; a very long axle will show less effect than a very short one. This is because the precession happens as a result of the rotor tilting out of the horizontal plane, and with a very long arm it won't tilt much as you push it along with your applied force. If your top suspension is rigged to "slide along" in the direction of your push, so that the rotor always remains horizontal as it is being pushed, then it won't precess since the rotor won't tilt, it will just slide along.

It might be simpler if you just eliminate the water jet and start with the axle horizontal. Then only gravity acting straight down affects your rotor, and the motion will be purely into (or out of) the plane of the drawing. If the rotor is driven at a constant speed by the motor, the whole assembly will just rotate around and around in the same plane (horizontal plane)  as if you are pulling it from the side (90 degrees from the pull of gravity).  If your rotor is coasting (slowing, running out of stored energy), then it will "nod" down as it goes around, until finally it is hanging vertically as in your drawing and then there is no more force acting to precess the thing. This shows that the "antigravity" effect comes from the stored energy of rotation of the rotor.  The overall weight of the apparatus doesn't change, of course. (But if you allow your water jet to push from some other angle than the horizontal (sorry, before I had vertical here) , then it will affect the _measured weight_ to the extent that it is acting in the direction of gravity. Some trigonometry will tell you just how much.)

Yes, by all means construct some apparatus and explore for yourself.

You might also find this page amusing, with the Java applet.

http://butikov.faculty.ifmo.ru/Applets/Gyroscope.html

It can't be adjusted to your precise design but it should help you see what is happening and what to expect. I still haven't found anything that describes the phenomenon that you will encounter if you build and run an apparatus like mine up above, though.