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Since the moment of inertia is defined as $\int r^2 dm$ where $r$ is the distance of the mass element from the axis, one can move a mass element parallel to the axis, without changing the moment of inertia.

For example, consider a thin rod of length $L$, inclined at an angle $\theta$ to the axis passing through its center of mass. If you move all its particles parallel to its axis to a plane perpendicular to the axis, it will make another rod of length $L \sin \theta$, and you would expect its moment of inertia to be $\frac {mL^2 \sin ^2 \theta}{12}$, in accordance to the formula of the moment of inertia of a rod perpendicular to the axis passing through its COM. After calculation, this does turn out to be the case.

Another example, is a hollow cone of radius $R$, open at its base, with an axis passing through its apex, perpendicular to its base. If you parallelly shift all its particles (rings) to one plane perpendicular to the axis, you get a disc, and the cone's moment of inertia comes out to be $\frac {mR^2}{2}$ by the formula of the moment of inertia for a disc.

However, this only seems to work for linearly varying objects. If you take a circular arc and shift all its particles to a plane, you get a rod again. However, the moment of inertia of the rod is not the same as the moment of inertia of the arc. Similarly, if you take a hemispherical shell and shift its particles to a plane, you get a disc of the same radius, however, the moment of inertia of the hemisphere is not $\frac {mR^2}{2}$

Why is this the case? What is special about linearly varying objects?

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  • $\begingroup$ "particles parallel to its axis to a plane perpendicular to the axis"... I don't understand this, can you phrase it differently? $\endgroup$ – Buraian Oct 13 '20 at 14:43
  • $\begingroup$ "make another rod of length Lsinθ,"... what?? how did the rod's length change when you moved it..? $\endgroup$ – Buraian Oct 13 '20 at 14:45
  • $\begingroup$ @Buraian I didnt move the rod, I moved all the particles of the rod to one specific height. Imagine a plane perpendicular to the axis passing through the COM. When all the particles of the rod are shifted onto that plane, while keeping their distance from the axis constant, the length reduces, but the mass remains constant. It is like taking a component. of the rod. $\endgroup$ – dnaik Oct 13 '20 at 16:29
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This happens because in linearly varying objects the distribution of mass is uniform before you shift the particles and after you shift the particles. However, for a nonlinear object, mass distribution is not uniform. For a hollow hemisphere, after shifting the particles the surface is a flat disk however its mass density is not uniform. The mass will be more concentrated on the outer region of the disk than the center of the disk. While the moment of inertia of this non-uniform disk will still be the same as the hemisphere, it will be different than the standard moment of inertia of a uniformly distributed disk.

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