# Is it always true that a solid object(geometrical figure) has less moment of inertia than its corresponding hollow one? [duplicate]

what I mean is that,if you have two objects(one hollow and other solid):

Lets say, a solid sphere and a hollow sphere and if you calculate moment of inertia of the two of them, you would find that moment of inertia is more in the hollow one than in the solid one.

So, is this concept i.e moment of inertia of a hollow object is more than its corresponding solid one always true?

Given:mass of both the bodies are equal

NOTE:THIS ISN'T A DUPLICATE QUESTION,ALL THE PREVIOUS QUESTION EXPLAIN IT FOR A PARTICULAR BODY ,BUT I'M QUESTIONING ITS VALIDITY FOR EVERYTHING(UNIVERSAL OR NOT)

• 2 objects of same mass and density? – Qmechanic Jul 7 '18 at 13:49
• Yeah,sorry I forgot to mention that! – Banchin Jul 7 '18 at 13:55
• It makes sense, except a hollow sphere is a solid shell, and a hollow shell is 2 concentric solid shells...and so on---so for every shape, idk. – JEB Jul 7 '18 at 13:57
• – AccidentalFourierTransform Jul 7 '18 at 15:30
• To reopen this post, for starters consider to insert ALL assumptions in main text. Comments don't count. – Qmechanic Jul 8 '18 at 9:01

It is always true and you can understand why from the physical interpretation of moment of inertia. It is a measure of how easily you can rotate an object in the same sense that the mass is a measure of how easily you can move an object. After all, the moment of inertia is the precise analogue of the mass for rotational motion.

And it is not hard to convince yourself that the further from the center of mass a mass is distributed, the harder it is to rotate it just like the heavier an object is, the harder it is to move it. Therefore, since any hollow geometrical object is always harder to rotate than its equal mass compact analogue, its moment of inertia is always bigger.

Hope this makes things more intuitive for you.

• I liked the answer ,moreover, could you explain me why moment of inertia is defined as 'mr^2'(any mathematical proof is appreciable). – Banchin Jul 7 '18 at 14:27
• @Banchin that is the unit of torque divided by the unit of angular acceleration – Henricus V. Jul 7 '18 at 15:47
• That's right. Just like the mass is defined through Newton's law of motion F=ma with F the force and a the linear acceleration, the moment of inertia is defined through Newton's law of rotational motion N=Ig with N the torque and and g the angular acceleration (sorry for the bad symbol 'g'). You can see that the r^2 term comes from the definition of torque N=Fxr, and the fact that the angular acceleration can be written in terms of centrifugal acceleration a as g=a/r so the moment of inertia must be something like I=mr^2 in order to restore units, N=Ig => Fr=(mr^2)(a/r) = mar. – Panos C. Jul 7 '18 at 18:11
• This answer appears to be incorrect, as illustrated by the counterexample given in A.V.S.'s answer. Basically, for non-convex objects, it's not guaranteed that (most of) the surface is further from the center than the interior. – Ilmari Karonen Jul 7 '18 at 19:40

That is not always true.

Consider the following figures:

Let us assume that this is a section of a cylindrical shape that stretches along the third dimension long enough that the effects of bottom and top are negligible for the overall moment.

We see that most of the mass of hollow shape is concentrated in the star-shaped curve which has a relatively small average distance from the center. While for the solid shape the average distance would be larger.

By simultaneously increasing the number of rays in the star and radius of outer circle (and keeping the overall mass the same) we can make the ratio $I_\text{solid}/I_\text{hollow}$ arbitrary high.

• I am not convinced that $I_\mathrm{solid}>I_\mathrm{hollow}$. Do you think you could back this up with some math or a reference? – AccidentalFourierTransform Jul 7 '18 at 23:37
• But,here you considered that the cylinders are long enough and you said that the effects of top and bottom are negligible .I don't think that it's a good approximation because if you increase the cylinder even to infinity the distance of the point(or the part of the cylinder), which you increased to infinity, would be same i.e the perpendicular distance of the point from axis of rotation would still remain same and we cannot consider its effect as negligible( just because stretch the point in a direction parallel to axis of rotation). Hope you get it @A.V.S – Banchin Jul 7 '18 at 23:40
• @AccidentalFourierTransform: Let us assume $r_1$ ($r_2$) is an inner (outer) radius of star cuve, $r_3$ is a radius of outer circle. If we increase the number of rays in a star (keeping $r_{1,2}$ intact) $I_\text{hollow}$ would be dominated by a star curve: $I_\text{hollow}\approx \frac m2 (r_1^2+r_2^2)<m r_2^2$. But for solid shape it could (if $r_1$ and $r_2$ are close enough) $I_\text{solid}$ is approximately the inertia moment of a solid cylinder $I_\text{solid}\ge \frac m2 (r_1^2+r_3^2) > I_\text{hollow}$ (for a large enough $r_3$). – A.V.S. Jul 8 '18 at 7:31
• @Banchin: If we stretch the cylindrical shape long enough $L\gg r$ the mass of the bottom and top for the hollow shape would be only a tiny fraction of the mass of the sides and so its contribution to the moment of inertia would be small. For the solid shape, of course, the length of the cylinder does not enter into the moment of inertia around the cylinder axis. – A.V.S. Jul 8 '18 at 7:55