Gyroscopic force - measuring it

OK, I been kicking around a couple ideas that in my tiny little fantasy world will win me the Nobel Prize for Physics before I turn 40. Of course they’re probably not gonna work because I’m no good at physics whatsoever, but it’s fun to think about.

With that in mind, I remember the ‘left-hand’ rule from high school, which demonstrates the principle of applied force moving at a right angle to the direction of a spinning wheel - the principle that allows a gyrosope to stand on its end and remain upright while its wheel is spinning. My question is - what’s the formula for determining how much force is being applied?

Your question is a bit confusing to me, so I’m going to answer the question I think you are asking, “What are the equations that determine the motion of a gyroscope?”

In my notation here, underlining indicates a vector, ‘x’ indicates a vector cross-product.

Angular momentum of a particle about a point:
[ul][li]r is the vector from the point to the particle[/li][li]p is the momentum vector, equal to mv (mass * velocity).[/li][li]Angular momentum L = r x p.[/li][/ul]

So, for a given gyroscope, you can calculate it’s angular momentum by integrating all the little dLs for all the little chunks defining a dr and a dp. The simplest case of a gyroscope would be a perfectly thin ring of radius r and mass m, rotating at an angular velocity of [sym]w[/sym] in the x-y plane, in a counterclockwise direction as viewed from above (z > 0.) Such a gyroscope would have L = [sym]w[/sym]r[sup]2[/sup]mk, where k is a unit vector in the positive z direction.

Note that the direction of L is perpendicular to the plane of the disk of the gyroscope - remember that a vector cross product uses the “right-hand rule” - r x p means point the fingers of your right hand in the direction of r, curl your right fingers in the direction of p, and your extended thumb points in the direction of the result. This page at Montana State gives a slightly different definition which gives the exact same result, complete with a picture.

Now, given the angular momentum of the gyroscope, you can calculate how torques effect the motion of the gyroscope.

Torque about a point:
[ul][li]r is the vector from the point to the location where a force is being applied[]F is the Force vector being applied[]Torque N = r x F[/li][/ul]

Now, in linear motion we have F = ma, where a is acceleration. Using calculus notation of [sup]dx[/sup]/[sub]dt[/sub] to mean “the time derivative of x”, this equation can be written as
[ul][li]F = m[sup]dv[/sup]/[sub]dt[/sub] OR,[/li][li]F = [sup]dp[/sup]/[sub]dt[/sub][/li][/ul]

Similarly, in rotational motion we have N = [sup]dL[/sup]/[sub]dt[/sub].

So, when you apply a torque to the gyroscope, the angular momentum changes in the direction of the torque vector. The weirdness of a gyroscope comes from the right-hand rule of what the torque vector actually is - it’s at right angles to the force you are applying. For my intuitive explanation, see my post in the Basic Science you don’t understand thread. For a diagram including vectors (but with slightly different notation than I use) see they hyperphysics page on precession. Be sure to scroll down to see all the diagrams. The hyperphysics site is great for exploring this kind of question, BTW.

Next, you want to know what determines the speed at which th gyroscope precesses, and that would be a matter of applying the above to a notional gyroscope. I’m out of time for today, but perhaps another person will be along to discuss this further, if anyone cares. Also, stopping now gives you a chance to tell me how far off I am from your actual question. :slight_smile:

Well, I picked up some hints from the pages you linked to, douglips, and I think I see where my problem lies. The concept I was wondering about is torque, and it seems on the surface of it that it is not the same thing as force.

Or is it? Since t=Ia is Newton’s 2nd law for rotation (in other words, the equivalent of F=ma for motion), is torque a force exerted in a given direction?

Torque is to force as rotation is to motion. That is, torque is the rotational version of force. Force is always applied in a direction. Torque is like force, but applied with an axis and a rotational direction.

OK, but which way is that force traveling? In the direction of rotation or along the axis of rotation? The diagrams in douglips’ link seems to indicate it’s the axis, but your definition “axis and rotational direction” seems to indicate otherwise.

If you apply a force on a lever at some distance from the fulcrum, you are exerting a torque on the lever with a magnitude equal to the force times the distance from the fulcrum to your hand (assuming you are pushing at right angles to the lever.) The trick is that the torque vector is oriented at right angles to both the force vector and the lever.

So, think of the torque vector as being the axis around which you are trying to push the thing, and the magnitude of it is proportional to the force you are exerting and to the distance from the axis.

This hyperphysics page has diagrams that may be particularly relevant. Be sure to scroll up and down to see everything there is to see.