Say I got a so massive rod and a relatively light spinning flywheel. Fixing an end on a pivot, and the disk on the other endpoint, as expected, the rod would undergo precession.

If we call the plane defined by the precessing rod x-y plane, then once the precession start, a z-direction angular momentum is formed concurrently (which seemed to be neglected usually). But the cause of precession is a torque prependicular to existed angular momentum due to gravity. There's no torque on z-direction.

And we have: $\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.$ This is a vector equation, so direction really counts.

More clearly, there's a obvious contradiction that a torque has made angular momentum change not only in its own direction i.e. what torque, at first, made the massive rod precess? Anyone to point out where I went wrong?

Diagram: Brief Diagram: Why upward angular momentum exist?

  • $\begingroup$ A picture would help. $\endgroup$
    – mmesser314
    Commented Jan 28, 2016 at 14:22

2 Answers 2


The left hand diagram is in the plane of the couple produced by the weight of the rod and disc and the reaction force at the pivot point.

enter image description here

The right hand diagram is my attempt at a 3D schematic.

The key to understanding what is going on is first to realise that the direction of the external torque $\tau$ is at right angles to the direction of the angular momentum of the rod and the disc, $\vec L_{\text{old}}$.

The change in angular momentum $\Delta \vec L$ must be in the same direction as that of the torque and so is at right angles $\vec L_{\text{old}}$.
This means that the magnitude of the angular momentum $|\vec L_{\text{old}}|$ cannot change and so $|\vec L_{\text{old}}|$ = $|\vec L_{\text{new}}|$.

So the angular momentum of the system changes direction, the precession, but there is no change in magnitude.

Now you cannot have it all ways. Up to now all that has been mentioned is the total angular momentum of the rod and the disc.

Your statement about the angular momentum resulting from the precession can be dealt with in the following way.
Resolve the angular momentum into a $z$ component and a component in the $x-y$ plane as shown below:

enter image description here

Apologies for the different annotation but it saved me time using an old diagram.

What do you notice. The angular momentum in the $z$ direction does not change. This is correct because as you pointed out there is no torque in that direction.

Later after seeing your diagram

Does the precession angular momentum, upward angular momentum as you call it, ever change? No!

  • $\begingroup$ So which torque, at first, created the precession angular momentum? Is it a instaneous one exerted the time when I put it on the rod? $\endgroup$
    – YiFei
    Commented Jan 28, 2016 at 15:08

This is something that greatly confused me too when I first learned about this!

If the disk was not rotating, but the rod and disk system was rotating about the z-axis, then you are right, we would have a non-zero $L_z$. But the disk is spinning. When precession begins the net torque on the system still has no z-component. If you were to add up all of the angular momenta of each particle in the system, you would find that $L_z$ is still $0$.

In other words, there is no torque in the z-direction that is required to get the system to precess. You can't think of it as a stationary disk rotating about a rod. The net torque is in the x-y plane due to gravity, and this determines the change in $\vec L$ over time.


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