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Question: We need to find, the angular momentum of the assembly, about the Center of mass.

As per Kleppner and Kolenkow, the general Expression for $\vec{L}$ about any point is: $$\vec{L_{p}}=I_1\vec\omega_1+I_2\vec\omega_2+I_3\vec\omega_3$$ where $I_1,I_2,I_3$ are moments of inertia about the principal axes. As far as I understand, these "principal axes" pass through though the point P.

However, an (although excellent) blog post:https://crazycosmos.wordpress.com/2017/12/08/rigid-body-motion-the-iit-jee-saga-i/, under the heading truth of part A, selects the principal axes in such a way that two of them dont pass through the center of mass!

Am I incorrect in my understanding that all 3 principal axes must pass through the point? Was there any reason to chose the principal axes (the 2 except the axis symmetry), that dont pass through the point?

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Sorry for the diagrammatic confusion. The 1, 2 and 3 axes in part A are assumed to be parallel to the 1,2,3 axes in the diagram, but chosen to be passing through CM instead. I should've mentioned this explicitly. This is reflected in the moment of inertia tensor (diagonal) values - viz. 1,2 - being different from the ones used in part B, in which 1,2,3 are exactly the ones shown in the diagram.

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The three principal axes all pass through the center of mass of the structure and are at right angles to each other.

The point P doesn't seem to be on your picture so I have no idea where it is.

For the two disks shown, one principal axis will be along the rod, and the other two are arbitrary directions perpendicular to it (and to each other).

The blog post seems to be mostly a rant.

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  • $\begingroup$ point P corresponds to the Center of mass of the system $\endgroup$
    – satan 29
    Commented Aug 17, 2020 at 19:58

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