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Consider the following scenario: https://youtu.be/8H98BgRzpOM?t=27.

How would I calculate the total angular momentum of this system? The spinning wheel is rather easy, and it's $L_{\rm wheel}=I\omega$. However, I am not sure how to account for the angular momentum due to precession. Is it simply $L_{\rm press} = I\Omega$? If so, how would I calculate $I$? Would I simply use inertia of a disk and the parallel axis theorem to find it?

I want to use the total angular momentum to prove the formula $\Omega=\dfrac{mgr}{I\omega}$ by using the fact that $\vec{{\tau }}=\dfrac{d{\vec{L}}}{dt}$.

Lastly, I have a conceptual question: $\Omega$ is said to remain constant assuming $\omega$ is also constant. However when the wheel is held still, $\Omega$ is clearly zero, until it is released, and then $\Omega$ becomes non zero. Why does $\Omega$ rise and then remain constant?

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  • $\begingroup$ For the process of starting gyroscopic precession, see the answer I wrote (2012) about the mechanics of gyroscopic precession $\endgroup$
    – Cleonis
    Nov 17, 2020 at 20:51
  • $\begingroup$ Consider to include an image of the system to make the question self-contained. $\endgroup$
    – Qmechanic
    Nov 17, 2020 at 22:15

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The common approach is to treat the case in approximation only.

When the spin rate is very high the angular momentum associated with the spin is much larger than the angular momentum associated with the precessing motion. (Also, the faster the spin rate the slower the corresponding precession rate.)

Hence the common approach is to neglect the angular momentum associated with the precessing motion. Then the validity of the derived expression is limited to high spin rate, but usually the case of high spin rate is the only one you are interested in.

I'm not sure, but I think that expression $\Omega=\dfrac{mgr}{I\omega}$ is in fact only valid (in appriximation) when neglecting the angular momentum associated with the precessing motion.

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  • $\begingroup$ Ah yes, so the fact that $\Omega$ can suddenly pick up and stop is actually not true, merely the mathematical consequence of the approximation. $\endgroup$
    – user256872
    Nov 18, 2020 at 4:30
  • $\begingroup$ Indeed. Some mechanism has to transfer kinetic energy, to provide the kinetic energy of the precessing motion. In parallel to that, some mechanism must transfer angular momentum to provide the angular momentum of the precessing motion. Those are the juicy parts of the story. For the usual approximation those parts are out of scope. In effect the usual approximationi is a fluke. $\endgroup$
    – Cleonis
    Nov 18, 2020 at 20:32

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