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

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 ?

• This is from Kleppner - pg. 253 Oct 18 '16 at 15:16
• Don't use  in title.
– user36790
Oct 18 '16 at 15:24
• Isn't the rotation about the z-axis? So the distance along the rotation axis aren't important. Oct 18 '16 at 15:29
• Possible duplicate of Cancelling internal forces/moments term when deriving inertial matrix 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. Oct 18 '16 at 15:35

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.