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I have been studying Irodov's problem book. There are a lot of good problems that challenge conceptual knowledge, and so I have to ask somebody for a qualitative answer. I can best describe my confusion by example, so let's take problem 1.284:

We have a gyroscope precessing around it's vertical axis (Figure 1.73). The axis of the gyroscope is horizontal.

The gyroscope is hinged at point O

I have managed to find an online solution here. So, since the gyroscope is rotating about it's symmetry axis, I agree that it will have angular momentum in the horizontal plane (it has a radial angular velocity).

However, it seems that this solution neglects the angular velocity of precession, as it relates to the angular momentum of the gyroscope. If the gyroscope is rotating in the horizontal plane, the precession angular velocity vector should point vertically. Shouldn't the gyroscope then have angular momentum in the vertical direction?

Is this maybe due to the referential frame (do we perhaps observe the gyroscope from a frame rotating vertically with angular velocity equal to the precession angular velocity)... ? If this is the case, why is there no centrifugal force?

Something doesn't add up, either way.

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It does have angular momentum in the vertical direction, so that should resolve your issues. There is a horizontal torque, so the upward-tilted angular momentum vector simply traces out a cone, where it's rate of change is always horizontal. The vertical component will usually be very small, but presents no problems, it just stays constant.

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  • $\begingroup$ But take a look at problem 1.286. Here the precession angular velocity is a function of time. So, here the angular momentum should have a vertical component, but the same situation occurs. $\endgroup$ – Knight1805 Nov 13 '16 at 20:58
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No, you are absolutely right, if your origin is the point $O$ in the figure above, then the only torque is in the tangential direction. There is of course an angular velocity $\alpha$ in the $z$ direction. The way it works is the radial component of the angular momentum $\vec{L}$ precesses, and $\frac{d\vec{L}}{dt}=|\vec{L}|\alpha \hat {\theta}$, where $\hat{\theta}$ is the unit vector in the tangential direction. Equating the known torque with $\frac{d\vec{L}}{dt}$ would now give you the precession frequency $\alpha$.

That solution you linked to, probably has notational flaws, because it seems to suggest the radial angular momentum changes through the change $\frac{d\vec{\omega}}{dt}$, which is simply not possible, as there can be no net torque in the radial direction.

The interesting thing is, the gyroscope picks up a small constant angular momentum in the vertical direction as well now, when in the beginning the angular momentum was purely radial. This means the bearings near the point of contact $O$ produce a tangential force and hence a vertical torque, for a very short period initially, to set up the angular velocity $\alpha$ of the gyroscope.

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  • $\begingroup$ But take a look at problem 1.286. Here the precession angular velocity is a function of time. So, here the angular momentum should have a vertical component, but the same situation occurs. $\endgroup$ – Knight1805 Nov 13 '16 at 20:58
  • $\begingroup$ Sorry, I don't have the book with me. I was talking of the simplest possible case of uniform precession, which is what your question seemed to be about. $\endgroup$ – Aritro Pathak Nov 13 '16 at 21:50
  • $\begingroup$ The point of that problem is that the precession angular velocity isn't uniform, but a function of time. My (new) question is why do we neglect the vertical component in that case also... ? $\endgroup$ – Knight1805 Nov 13 '16 at 22:03
  • $\begingroup$ We shouldn't neglect the vertical component of the angular momentum in any situation. $\endgroup$ – Aritro Pathak Nov 13 '16 at 22:46

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