Since I can't sketch here I am referring to: Feynmanlectures-Gyroscope

Assume that you have got a gyroscope (as in Figure 20-3 in the link) that pivoted and can turn around any axis. Let $\vec{L}$ be the the angular momentum of the gyroscope spinning around its own axis. There will be a torque $\vec{\tau}$ due to gravity.

Then there is the following relation between angular velocity of precession and the angular momentum and the torque as in this link: $$\vec{\tau}=\vec{\Omega} \times \vec{L}$$

How do I know that $\vec{\Omega}$ is vertical, since there is no vector division?

  • $\begingroup$ Curl your fingers in the orientation of rotation, then the "omegavector" points along your thumb. $\endgroup$ Commented Dec 1, 2014 at 15:40

1 Answer 1


Curl your fingers in the orientation of rotation, then the "omegavector" points along your thumb. In this specific case the gyro has two rotations, one of them got $\vec{\Omega}$ associated to it. This is the rotation about the vertical i.e. the whole of gyroscope precesses, hence the associated "omegavector"* points up along the positive $z$-axis.

The other omegavector $\vec{\omega}$ is the one associated with the spin of they gyro, again, if you curl your fingers along the spin direction, you will get the direction of $\vec{\omega}$. Does this answer your question?

*It's really not called an omegavector.

  • $\begingroup$ So you split the rotation of the gyroscope arbitrarily into two vectors? Is that right? You define $\Omega$ to be vertical. $\endgroup$
    – user50224
    Commented Dec 1, 2014 at 15:56
  • 1
    $\begingroup$ Without going into the equations, since they are described in the link you gave, just from a physical point of view, if you have something rotating about an axis $A$, then there will be an omegavector associated to that rotation pointing along that axis. Do you agree with this? $\endgroup$ Commented Dec 1, 2014 at 16:06

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