Only z component of angular momentum $\vec{L} = I\vec{\omega}$ considered?

Question

The derivation of the angular momentum of a rigid body in terms of the moment of inertia considers only the z component. Why are not the other components considered ?

Isn't the rotation about the z-axis? So the distance along the rotation axis aren't important. — John Alexiou, Oct 18 '16 at 15:29
Possible duplicate of Cancelling internal forces/moments term when deriving inertial matrix — John Alexiou, Oct 18 '16 at 15:33
I think the author is afraid to use cross products which is what you need when considering all the 3D components. — John Alexiou, Oct 18 '16 at 15:35
He begins by saying "consider the body rotating around $Z$ axis. Simple. — Prasad Mani, Oct 18 '16 at 15:40
The thing is: this is not a limitation of any kind. You have a rotating body, thus you have one special axis. You can alway call the axis z. — Bort, Oct 18 '16 at 15:40
@ja72 The relationship between this question and your proposed duplicate is not obvious to me. — rob, Oct 19 '16 at 00:01
@rob In the proposed duplicate the MMOI and angular momentum are formed for a point considering the summation of small masses. The question here ultimately is how different is angular momentum when considering all components. — John Alexiou, Oct 19 '16 at 12:45

Robin Ekman · Accepted Answer · 2016-10-24T01:54:35.690

The $z$ axis is of course not special. However, the angular momentum is a vector quantity, and it is often easiest to compute a vector quantity by calculating each component separately. On the other hand, precisely because the $z$ axis is not special, the same calculation will apply to the $x$ and $y$ components, as long as you change what needs to be changed. This last part means that a rigid body really has three moments of inertia -- one for each axis. The formal way to do it is with an object called the inertia tensor. ¹

You can choose the $z$-axis to be the axis of rotation, and then the $x$ and $y$ components will be zero, so that you only need the $z$ component. This amounts to demanding that $L = L_z\hat z$ where $\hat z$ is a unit vector, but then if $\tau$ is the torque $$\tau = \frac{d}{dt} L = \frac{d}{dt} L_z \hat{z}$$ so $\frac{d}{dt} \hat z$ has to be non-zero in general. Thus the cost is that you have to find how $\hat z$ changes with time. That can be done however, and any mechanics textbook should treat it.

The inertia tensor has six independent components because it's symmetric, but three only encode the transformation between the chosen coordinate system and the principal axes. It's really only the principal moments that are properties of the rigid body itself. Three principal moments + three Euler angles = six components.

Only z component of angular momentum $\vec{L} = I\vec{\omega}$ considered?

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