How? Calculate torque from twisting a gyroscope

Question

Please be patient, I'm not a physics guy.

I'm building a robot that flips itself over using a gyroscope. The gyroscope spins horizontaly at first, then a servo changes the gyro's rotating axis, making the robot flip in the opposite direction.

How can I calculate the torque of the flip, given:

• gyro's rotational speed (rpm or equivalent)
• gyro's diameter
• gyro's mass
• speed at which rotating axis is changed

Is there a formula for this?

Answer 1

For a spinning flywheel, the external torque vector determines the rate of change of the angular momentum vector , L, which equals Iw. For a uniform disk, I, (The rotational inertia) = (½) m R^2, and the angular velocity vector, w, (in radians/sec) is defined as being along the axis of rotation. (You may want to look up the associated “right hand rules”.) The problem is, when you push on the axle of a spinning flywheel, it does not move in the direction of your push, so that the direction of applied torque and the reaction torque change with time.

Answer 2

Before giving the formula, note that the response of the gyroscope wheel will not be in opposite direction.

For names of direction of rotation let me use the standard names of aircraft principle axes: roll, pitch and yaw.

Take the spin axis of the gyro wheel as rolling, and apply a large torque that will pitch the gyro wheel fast. Then in response the gyro wheel will yaw.
But then: if that yawing motion is fast too it will itself give rise to a response.
So: to construct a robot that can flip itself is not straightforward. I'm guessing it can be done, but be prepared to see first attempts move in a way that is counter-intuitive to you.

As to a formula:

The usual letter for rotation rate is the greek letter 'omega' $\omega$
I will use a subscript to distinguish between spin rate and pitch rate

The usual letter for torque is the greek letter 'tau' $\tau$

I will use an expression for a highly simplified configuration, a wheel with all of its mass located at the outer rim. Of course an actual wheel is a disk. Still: usually a gyro wheel has as large proportion of its mass close to the rim. You will need to guess the error that is introduced by this simplification so you can compensate for it.

spin rate: $\omega_s$
pitch rate: $\omega_p$
wheel mass: $M$
wheel radius: $R$

formula for the magnitude of the resulting torque when a spinning gyro wheel is pitched:

$$ \tau = \omega_s \omega_p M R^2 $$

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Responding to a comment:

There is the dimensionality check:
dimensionality of torque: $Nm$
Hence: dimensionality of torque: $M L^2 T^{-2}$
(kilogram, meter, meter, per second, per second)

There is the consistency with the following approximate formula of rate of precession (L for angular momentum of the spinning gyro wheel):

$$ \omega_p = \frac{\tau}{L} $$

(The above formula is a good approximation only when the rate of precession is slow compared to the spin rate. When the rate of precession is fast then the angular momentum of the precession is a significant contribution.)

(My apologies for not giving the derivation I used. I concur that I really should. I integrated around the rim of the gyro wheel. That is an unusual derivation method, so if I were to give the derivation here I would have to explain many of the steps.)