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

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?

• There is a 2012 answer by me that offers explanation of gyroscopic precession. This explanation uses symmetry to simplify while still being general. Also: this explanation does not invoke the concept of angular momentum, using less abstract elements instead. What you describe with the word 'resistence' is a responsive effect. There is a self-adjusting aspect to the response of a spinning gyroscope . To read about that effect, follow the link. (If it answers your question, do let me know.) Commented Sep 28, 2021 at 16:28

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.