I was recently working on a physics problem from edx.org. I think I can show below that the center of mass of an infinitesimal pie slice of a disk is $\frac{R}{\sqrt2}$ from the center. ($R$ is the radius). I'm a little surprised by that, and I wonder if I made a mistake.

The task in the physics problem is to find the moment of inertia, about the center of mass, of a spinning disk of uniform density. The disc's radius is $R$, and its mass is $m$. This moment of inertia $I_{cm}$ is:

$$ I_{cm} = \int_{disc} r_{cm}^2 \ dm $$

That's the integral over all the infinitesimal mass elements of the disc. $dm$ is the mass of each element, and $r_{cm}$ is the element's distance from the disc's center of mass.

The professor shows an answer divides the disc into rings, integrating $ \int_{r=0}^{r=R}r^2dm $, and $dm$ there is the density times the ring area: $dm = \frac{m}{\pi*R^2}2\pi r \ dr$. Integrating that gives:

$$ \int_{r=0}^{r=R} \frac{m}{\pi R^2} 2 \pi r \ dr \ r^2 = \frac{2m}{R^2}\int_{r=0}^{r=R} r^3 dr= \frac{m R^2}{2} $$

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I wondered if I could do with pie slices:

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$$ dm = \frac{d\theta}{2\pi}m \\ \int_{\theta = 0}^{\theta = 2\pi} \frac{d\theta}{2\pi}m \ r_{cm}^2 $$

I'm not sure what $r_{cm}$ is for the infinitesimal pie slice. But I can leave it unknown, integrate, and set it equal to the other solution...

$$ \frac{mr_{cm}^2}{2\pi} \int_0^{2\pi} d\theta = m r_{cm}^2 \\ m r_{cm}^2 = \frac{m R^2}{2} \\ r_{cm} = \frac{R}{\sqrt2} $$

That's interesting to me because I might have guessed the center of mass of the infinitely thin slice would be at $R/2$, because as the slice gets thinner, its sides become closer to parallel, so it approaches the shape of a rectangle, which would have its center of mass at $R/2$.

  • 1
    $\begingroup$ How about this: a narrow sector is approximately a triangle. The centre of mass of a triangle is its centroid, which is two-thirds of the way from a vertex to the opposite side, so the centre of mass is of a narrow sector should be at distance $2R/3$ from the centre of the circle. $\endgroup$ Commented Apr 27, 2018 at 5:05

2 Answers 2


The conceptual error is this: The moment of inertia of the pie slice is not the same as the moment of inertia of an equivalent point mass at its center of mass. Refer to the parallel axis theorem; you're missing one of the two terms.

The mathematical error is that $r_{\mathrm{cm}}$ is neither a constant nor a function of $\theta$, so the expression $\int_{\theta = 0}^{\theta = 2\pi} \frac{d\theta}{2\pi}m \ r_{cm}^2$ doesn't really make sense.

As for the correct approximation, Lord Shark the Unknown is right in the comments. To leading order in $\theta$, a pie slice with angle $\theta$ is a triangle, which is why the slice's center of mass is at $2R/3$.

  • $\begingroup$ Thanks. The professor in class wrote $\int dm \ r_{cm}^2$. The "cm" should not be there, correct? $\endgroup$
    – Rob N
    Commented Apr 27, 2018 at 14:49
  • $\begingroup$ The definition "$dm$ is the mass of each element, and $r_{cm}$ is the element's distance from the disc's center of mass" makes sense. Since the disc's center of mass is placed at the origin of the coordinate system, it so happens that $r_{cm}=r$ identically. So if you prefer, you can write it as just $r$. $\endgroup$ Commented Apr 27, 2018 at 16:44

yes you could also find moment of inertia of circular disc about it's centre of mass using pie slices :

but you've done one thing wrong ..mass is distributed in whole area of disc rather than only on periphery so you need to perform double integration to come up with answer one for variable $\theta$ and other for radial distance

let $\sigma$ be the surface mass density of disc of radius $R$ and mass $M$ i.e,

$\sigma=\dfrac{M}{\pi R^2}$

go to an angle $\theta$(in ACW sense from horizontal) and radial distance $r(<R)$ and then, slowly increase them by small angle $d\theta$ and infinitisimal distance $dr$ in order to swept differential area


$I_{y}=\sigma\displaystyle\int_{r=0}^{r=R}\displaystyle\int_{\theta=0}^{\theta=2\pi} (rcos\theta)^2.rdrd\theta=\dfrac{\sigma}{2}\displaystyle\int_{r=0}^{r=R}r^3dr\displaystyle\int_{\theta=0}^{\theta=2\pi} (1+cos2\theta)d\theta=\dfrac{\pi\sigma R^4}{4}=\dfrac{MR^2}{4}$

similarly ,


since, centre of disc is at origin and it is symmetrical about vertical and horizontal axis so, it's centre of mass will lie at co-ordinates $(0,0)$

by parallel axis theoram we know,


but , $I_{z}=I_{cm}=2\times \dfrac{MR^2}{4}=\dfrac{MR^2}{2}$



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