The angular momentum of a rotating collection of rocket bits that needn't even touch is very similar to the straight line case 'mv' it's just:
L = sum( r x p)
where r is its position from any origin, p is its linear momentum (i.e. m dr/dt.)
If you do an example L ends up aligned with any rotation axis, and is proportional in length to the rotation speed and the moment of inertia.
You can add angular momentums, they are vectors.
So, if you have two identical disks spinning in opposite directions the L is opposite and equal, they have to be because the ps are opposite and equal. So the angular momentums cancel. That's it.
Incidentally somebody once used this fact to cancel the angular momentum of a bicycle wheel, well they used smaller heavier weights counterrotating, It was still very stable, the stability of bicycles is more to do with the steering geometry 'caster angle'.)
FYI the rotation equivalent for F=d(P)/dt [i.e. F=Ma for constant mass] is just:
T = dL/dt (T = sum(r x f))
That's pretty much everything! You can show that gyroscopes turn the wrong way (it's because of the cross product in L, if you apply a force on one side of a disk, the torque will be at 90 degrees to it, so the L will move at 90 degrees, and as we said earlier, it's aligned with the axis.)
Henry Spencer wrote:
On Thu, 19 Sep 2002, John Carmack wrote:I'm not sure about that. If you have a shaft with two wheels on it, spinning either wheel either way will provide a gyroscopic "stiffness". Two spinning wheels, even going opposite directions, should have twice the stiffness, not none.Nope. Gyroscopic stiffness is simply conservation of angular momentum.When you have a spinning wheel, and you apply a torque to it for a period of time -- that is, you impart some new angular momentum to it -- that new angular momentum adds to the existing angular momentum, using vector addition. If the existing angular-momentum vector is large, adding a small new angular-momentum vector at right angles to it won't move the final angular-momentum vector much. That's gyroscopic stiffness. The net angular momentum of two counter-rotating wheels is zero, so there is no gyroscopic stiffness to be had. (Adding angular-momentum vectors is also a simple way to see how precession of a gyro works.)...I don't think additional rotating masses can't help you do a "plane change" of a rotating mass...There may be large forces induced in the structure that connects the two rotating masses, but the external force is zero. Henry Spencer [EMAIL PROTECTED] _______________________________________________ ERPS-list mailing list [EMAIL PROTECTED] http://lists.erps.org/mailman/listinfo/erps-list
