I read in my textbook about the various results of moment of inertia for different geometrical shapes like solid and hollow cylinder, sphere, disc and ring etc. Something general I noted is that $M.I$ for a hollow body is always greater than a solid body, say for a sphere $$MI_{hollow}=\frac{2}{3}MR^2,$$ $$MI_{solid}=\frac{2}{5}MR^2.$$ So clearly $MI_{hollow}$ is greater. And this works even for 2D shapes like ring and disc for which $MI_{ring}$ is twice $MI_{disc}$. My question is is true for any irregular shape that the hollow will have more MI than the solid version of the body. And secondly, is there any proof to it?

  • $\begingroup$ Things will become far clearer if you actually derive the moments of inertia for, say, a half-dozen different geometries. It’s just a matter of working through the integrals, if you’ve taken calculus. It would be easy to simply tell you that the moment of inertia is higher for mass farther from the revolution axis, but you’ll really understand the concept by working through the calculations. $\endgroup$ May 28, 2023 at 15:01
  • $\begingroup$ and, it's just the 2nd moment of the density distribution $\endgroup$
    – JEB
    May 28, 2023 at 15:29
  • $\begingroup$ You are making comparison using the same mass for two different shapes. Of course if your hollow object has more mass concentrated in the periphery it will have a larger moment of inertia compared to a solid shaped object of similar radius and same mass because of the physicial meaning of moment of inertia. $\endgroup$
    – Rescy_
    May 28, 2023 at 17:22
  • $\begingroup$ MMOI measures how far away from the axis of rotation is mass concentrated. For a hollow object the mass is obviously as far away as possible since the middle is hollow. $\endgroup$ May 29, 2023 at 0:14

2 Answers 2


You can find the general approach to calculating the moment of inertia here: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#:~:text=For%20a%20point%20mass%2C%20the,a%20collection%20of%20point%20masses.

The moment of inertia is a measure of an objects resistance to angular acceleration. For an object of given mass the moment of inertia will be greatest when the mass is concentrated further away from the center of rotation. This is why the moment of inertia is greater for the hollow sphere than the solid sphere of the same mass.

Regarding any two irregular shapes, one hollow and one solid but having the same mass, the hollow shape will have the greater moment of inertia about the center of mass. This follows from the moment of inertia of a particle rotating about a point being $I=MR^2$. The moment of inertia varies linearly with mass but as the square of the distance away from the center of rotation.

Hope this helps.


It's important to carefully define the objects we are comparing, in particular how the mass is distributed in the hollow object. If the hollow object has a constant mass per unit area, the filled object has a constant mass per unit volume, and the two objects have the same total mass, then the hollow object doesn't necessarily have a higher moment of inertia. Here is a 2D counterexample.

enter image description here

The filled shape is a gear and a disk, both with radii $r$, connected by a very thin rectangle of length $L$. The gear has many narrow teeth on its periphery so that while the outer and inner radii that bound these teeth are both close to $r$, the perimeter of this shape is many times greater than that of the disk to the right, by a factor we will call $\alpha$. The rectangle is narrow enough that its mass is negligible. Since the gear teeth are short, both the gear and the disk have approximately the same mass $M/2$, where $M$ is the total mass of the shape. The axis of rotation is at the center of the gear. Using the parallel axis theorem, $$I_\text{filled} \approx \frac{1}{2}\frac M2r^2+\frac M2\left[\frac12r^2+(L+2r)^2\right]= \frac{5}{2}Mr^2+\frac12 M(L^2 + 2Lr). $$

In the hollow shape, we can no longer neglect the mass of the rectangle. The mass per unit length of the shape is $$\lambda = \frac{M}{\alpha2\pi r + 2\pi r + 2L}.$$ The hollow gear has all its mass distributed at a distance approximately $r$ away from the axis of rotation, so just like a circle it has $I_\text{gear} = m_\text{gear}r^2$. The total moment of inertia is $$I_\text{hollow} \approx \lambda\left\{\alpha2\pi r^3 + 2\pi r\left[r^2 + (L + 2r)^2\right] + 2L\left[\frac{1}{12}L^2 + (r+L/2)^2\right]\right\}$$ $$= M\frac{\alpha2\pi r^3 + 2\pi r\left[r^2 + (L + 2r)^2\right] + 2L\left[\frac{1}{12}L^2 + (r+L/2)^2\right]}{\alpha2\pi r + 2\pi r + 2L}$$

We can make $\alpha$ arbitrarily large by increasing the number of teeth on the gear (while making the teeth narrower). As $\alpha \to\infty$, $I_\text{hollow}\to Mr^2<I_\text{filled}$.

If we are comparing moments of inertia about the centers of mass, we can modify the objects by putting another disk to the left of the gear so that the object is symmetric. Then the centers of mass of both the filled and hollow objects are still the center of the gear. The calculation proceeds the same way and the conclusion is still the same.


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