What force am I feeling when I poke a gyroscope to spin off axis of rotation?

Question

So this question is very simple however there is so much material on gyroscopes that I am getting my wires crossed.

What I understand, if you poke a gyroscope it will react 90 degrees in the direction of spin to your poke due to precession. This makes it so your poke "experiences" a resistance to motion from the gyroscope. This resistance is what I am after. Is it torque? I've seen formulas to get resistance torque, is that it? There is also a lot of mention of rigidity of gyroscopes but I can't find formulas for that. Is the resistance angular momentum?

Answer 1

A gyroscope has an angular momentum:

$$ \vec L=I\vec\omega$$

If you poke it so that it's spinning in a different direction, $\omega'$, the angular momentum is now:

$$ \vec L'=I\vec\omega'$$

Suppose the poke lasted $\Delta t$, then you applied a torque:

$$\vec\tau = \frac{\vec L'-\vec L}{\Delta t}\rightarrow \frac{d\vec L}{dt}=I\frac{\Delta{\omega}}{\Delta t}$$

Of course your poking finger just applies at force $\vec f$ at some distance from $\vec r$ from the pivot:

$$ \tau = \vec r \times \vec f = rf\sin{\theta}=I\Delta{\omega}/\Delta t$$

So the resistance to your force is:

$$ \frac{I}{r\sin\theta}\frac{\Delta\omega}{\Delta t}= \frac{I\alpha}{r\sin\theta}$$

where $\alpha$ is the angular acceleration.