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Consider a non-planar rigid body rotating about a fixed axis (say, the Z-axis, chosen vertically). Let the origin $O$ is chosen somewhere on the Z-axis. Let $\textbf{r}_i$ represent the position vector of the $i^{th}$ particle of the rigid body. Then, by definition, the angular momentum of the body about $O$ is given by $$\textbf{L}=\sum\limits_{i}m_i\textbf{r}_i\times(\vec{\omega}\times\textbf{r}_i)=\sum\limits_{i}m_i[r_i^2\vec{\omega}-(\textbf{r}_i\cdot\vec{\omega})\textbf{r}_i].$$ Since $\vec{\omega}=\omega \hat{k},$ $$\textbf{L}=\sum\limits_{i}m_i\omega[-(z_ix_i)\hat{i}+(-z_iy_i)\hat{j}+(x_i^2+y_i^2)\hat{k}].$$ The Z-component of the angular momentum is $L_z=\sum\limits_{i}m_id_i^2\omega=I\omega$ is usually treated with a special importance, (also to $\tau_z=\frac{dL_z}{dt}=I\dot{\omega}$).

Why are the dynamics of the other components, $L_x$ and $L_y$, not considered (in school-level textbooks such as Halliday, Resnick and Walker) even though they are nonzero, and may change if a force $\textbf{F}$ in an arbitrary direction is applied (because the torque $\vec{\tau}=\textbf{r}\times\textbf{F}$ will, in general, have all components nonzero)?

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  • $\begingroup$ Probably because they won't apply any torques that aren't in the $z$ direction, so it's irrelevant at a pedagogical level $\endgroup$ – Aaron Oct 10 '17 at 16:46
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    $\begingroup$ Should you possibly stress you are talking about a randomly lumpy object, as we have all grown up with a bias towards symmetrical situations when learning physics? Also, in HRW and the like, they are not going into more details that your post illustrates if they think they have made their point and that exam questions will be simplified. $\endgroup$ – user167453 Oct 10 '17 at 16:55
  • $\begingroup$ All I need to stress that I'm talking about a 3-dimensional object. $\endgroup$ – mithusengupta123 Oct 10 '17 at 16:59
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    $\begingroup$ Ah sorry, I was confused and assumed that you chose the $z$ axis as an axis of symmetry or along an eigenvector of the inertia tensor. Can you be more precise on what exactly you mean by "the dynamics of the other components are not considered"? $\endgroup$ – Aaron Oct 10 '17 at 17:00
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    $\begingroup$ Without your specific textbook, I cannot comment whether or not there is an error in their book, but i would suspect in all their examples they only work with the inertia tensor after diagonalizing it, in which case the other components would decouple and be irrelevant. Hence, it would be sufficient to isolate a special axis for their purposes, namely the $z$-axis. In the general case, you are correct that the other pieces should matter, of course. $\endgroup$ – Aaron Oct 10 '17 at 17:08
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Rigid body dynamics in general is pretty complicated, so lower-level textbooks like HRK tend to simplify things. In particular, it's always possible to pick a set of axes (called "principal axes") such that $\vec L=I_xω_x\hat{i}+I_yω_y\hat{j}+I_zω_z\hat{k}$, which in your case gives $\vec L$ in the $\hat{k}$ direction. Your book is probably implicitly picking these principal axes as the coordinate system.

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