Space Rocket Lift Efficiency and the Gyroscope?

Question

I'm no physicist apart from basic 3d web animation, I'm just curious and please feel free to correct my misuse of terms or inadequate speculations.

I've been reading a lot on gyroscopes and aerodynamics and the various concepts of lift, drag, etc. I understand how a rocket utilizes lift and drag and how much it depends on thrust. I also understand that there are several forces acting upon a gyroscope which determines its stability and why the tool is so useful in viscosity navigation, among other things.

It is my speculation that gyroscopes, once spun at the right velocity, can generate equally opposing momentum along it's various directional paths in 2d cross section i.e. up, down, left & right or (+/-) y and (+/-) x, and ideally, these equal, but, opposite forces generate equilibrium by shifting the object's center of gravity (point 0) to a variable location in the object i.e. relative to the direction in which the gyroscope is mounted or positioned. I also believe that these forces are constant relative to the constancy of the object's velocity. e.g. an electric gyroscope peaking at a constant spin velocity.

With that said and in not having any way to test, I ask: If I am correct in saying that all the gyroscope forces given a fixed resistance, are equal and constant up to the given moment of external force (e.g. throwing a spinning electric gyroscope [straight] up in the air), will the upward force of throwing stack up with and amplify the upward (lift) force of the gyroscope? Will that gyroscope rise higher and faster than if it weren't spinning and will it descend much faster in the same manner i.e. a much stronger interaction with gravity upon descent? Also, if this were so, could the design concept of a gyroscope be applied to Space Rocket Designs (i.e. long shaft and a proportionally larger, extruding spinning wheel), where less fuel could be used in launch by piggy backing and amplifying the lift generated by the wheel?

Answer 1

The quick answer is that gyroscopes don't do what you suggest, and indeed don't do anything not expected from the well established laws of mechanics. Because gyrosocopes are so widely used in inertial guidance systems their behaviour has been measured to exquisite precision e.g. in the Gravity Probe B satellite. Any effect of the type you describe would be easily measured.

Having said this, every student is baffled by gyroscopes when they first encounter them, and that certainly includes me. However the equations that describe their motion are very simple. I think the problem is that the gyroscope equivalent of Newton's first law is:

$$ \vec{\tau} = \frac{d\vec{I}}{dt} $$

where $\vec{\tau}$ is the torque and $\vec{I}$ is the angular momentum. The big difference from linear motion is that the torque is given by:

$$ \vec{\tau} = \vec{r} \times \vec{F} $$

where $\vec{F}$ is the force and $\vec{r}$ is the vector from the centre of rotation to the point where the force is applied, and the $\times$ is a cross product. The cross product produces a vector that is at right angles to the two vectors in the product, which means the torque is a vector at right angles to the applied force, and therefore the change in angular momentum is at right angles to the applied force.

It's the fact that gyroscopes respond in a different direction to the force you apply that makes their behaviour seem so counterintuitive. But I must emphasise that it's our intuition that is the problem, not the gyroscope. Once you've mastered the maths, the behaviour of gyroscopes is very simple and entirely predictable.

Response to comment:

The rocket on the left has the gyroscope rotating about it's centre i.e. the gimbal in which it's mounted rotates about the centre. The dotted line shows the thrust provided by the rocket motor. In this case the net torque is zero because the same force is applied to both ends of the gyroscope and the torques cancel out. This rocket would rise normally.

In the right hand drawing the gyroscope is pivoting about it's end. In this case there is a net torque and the rocket would feel a force rotating it out of the plane of the diagram. If the angular momentum of the gyroscope were large enough (it wouldn't be in real rockets) the rocket would indeed rotate round and crash into the ground.