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The wiki says that rotational inertia is defined for point-masses, and by extension continuous bodies. It says:

This simple formula generalizes to define moment of inertia for an arbitrarily shaped body as the sum of all the elemental point masses $dm$ each multiplied by the square of its perpendicular distance $r$ to an axis $k$. Of course, we can use that to calculate the rotational inertia for discreet-distribution bodies. We can also use that for continuous mass bodies if we treat the differentials as point masses.

While the standard area and volume differentials are small enough to be approximated as points. How will you approximate a cylindrical shell as a point (we use cylindrical shell differentials when we find the moment of inertia of a rod with the axis of rotation being the axis of symmetry of the rod)?

This is the rod I'm talking about:

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  • $\begingroup$ Does the rod have a finite radius? $\endgroup$ – Qmechanic Sep 10 '20 at 14:23
  • $\begingroup$ I think that it does. $\endgroup$ – KaceEnigma Sep 10 '20 at 14:23
  • $\begingroup$ Remember that cylindrical shell method is a shortcut to doing 3D calculus integrals that have rotational symmetry. You can write all these integrals as triple integrals if you'd like but you'll get the same result. $\endgroup$ – Triatticus Sep 10 '20 at 14:25
  • $\begingroup$ @Triatticus , could you please send a link/post an answer about this shortcut? I haven't heard of it before. $\endgroup$ – KaceEnigma Sep 10 '20 at 14:26
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For a very thin cylindrical shell, all of the very small segment of mass are at the same radius from the axis of symmetry. In terms of moment of inertia, they can all be considered as part of one (point) mass.

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