When computing the moment of inertia, I observed that people usually use the following logic:

$$d I=r^2 dm,\\ \therefore I=\int r^2 dm$$

My question here is, why not use $dI=m ~d(r^2)$?

I acknowledge that integrating $dI$ means that I am summing up all the small I's ($\Sigma I_i$). What I find strange is that viewing $dI$ as $md(r^2)$ seems to make no sense when viewed from such perspective.

Is there a reason why no textbook seems to view $dI$ that way?

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    $\begingroup$ If it as you say "seems to make no sense" then why are you suggesting it? (Are you under the impression that $r^2 dm = md(r^2)$? Otherwise I can't work out why this is being suggested.) $\endgroup$
    – jacob1729
    Jun 13, 2018 at 12:37
  • $\begingroup$ Because I'm not sure if it actually makes no sense. Also, I want to know why it should be dm instead of something else by examining why $d r^2$ does not make sense $\endgroup$
    – Danny Han
    Jun 13, 2018 at 12:39
  • $\begingroup$ Think of the moment of inertia as being the second moment of the mass distribution. In this way, it is entirely intuitive why it cannot be as you suggest. $\endgroup$
    – JamalS
    Jun 13, 2018 at 12:40
  • $\begingroup$ This might be helpful reading. As will the Wikipedia page on moment of inertia $\endgroup$
    – jacob1729
    Jun 13, 2018 at 12:49
  • $\begingroup$ Think of differentials as small increments. That might not be very rigorous, but it's the way physicits are. A small mass, $dm$, as a "small" momento f Inertia given by the product mass times its squared radius. $dI=r_m^2 \cdot dm$. $\endgroup$
    – FGSUZ
    Jun 13, 2018 at 12:57

5 Answers 5


If the body were a discrete set of point mass objects rotating around an axis, you would write $I=\sum_j m_j r_j^2$

For a continuous body the sum goes over into an integral. You consider it as smaller and smaller mass objects. You think of, say, a brick as being made up of grains of rock, then atoms... The objects get smaller $m \to \delta m$ but their radii do not. If the brick has $r=1 m$ then the grains that comprise the brick have $r$ in the range $0.9 - 1.1 m$. So you consider lots of small masses at non-small radii, not the other way round.


Conceptually, each little piece of an object contributes to the total moment of inertia. Each little piece of an object consists of a volume of matter so that each piece has some small mass, $$\delta m = \rho\left(\vec{r}\right) \delta V.$$ $\rho\left(\vec{r}\right)$ is a density function whose value depends on the position of each specific $\delta m$.

Without getting into tensor calculations, i.e., keeping things simplistic, each $\delta m$ has a contribution to the moment of inertia about some axis, $$\delta I = \delta m R_{\perp}^2$$ where $R_{\perp}$ is the perpendicular distance from the axis of interest. (It's actually a little more complicated than that, but the difference is not important to this question.) $R_{\perp}$ is a certain value depending on each $\delta m$. The sum you want to do considers each $\delta m$, not each $R_{\perp}$. (In certain specific symmetries, the sum can be simplified, but you asked about the conceptual reason for summing over the masses.)

As you shrink the volume $\delta V$ to an infinitesimal d$V$, the mass $\delta m$ becomes d$m$, but the specific distance from the axis $R_{\perp}$ doesn't change at all! And so, $$\delta I \to \text{d}I = \text{d}m R_{\perp}^2 = \text{d}V \rho\left(\vec{r}\right) R_{\perp}^2.$$

Integrating this over the complete volume adds the individual contributions of each little volume of matter.


Answer: For a system of continuously distributed total mass $m$, $r$ is constant for a differential (point) mass $dm$ so we use $dI=r^2dm$ but mass $m$ is not constant over a differential distance $dr$ so we don't use $dI=md(r^2)$


When a system of continuously distributed (total) mass $m$ rotates about a point then each point of the system rotates at a certain normal distance $r$ from the axis of rotation.

We should consider a differential mass $dm$ at a normal distance $r$ from the axis of rotation to compute the moment of inertia of differential mass $dI$ which is $$dI=(\text{Normal distance from axis of rotation})^2\times (\text{Point mass})=r^2dm$$
The moment of inertial $I$ of the system of continuously distributed mass is obtained by integrating $dI$ with proper limits $$I=\int r^2dm$$

For a system consisting of $n$ number of discrete point masses $m_1, m_2, m_3, \ldots m_n$ at normal distances $r_1, r_2, r_3, \ldots r_n$ respectively the moment of inertia $I$ is obtained taking the discrete summation of individual moments of inertia as follows $$I=m_1r_1^2+m_2r_2^2+m_3r_3^2+\ldots +m_nr_n^2=\Sigma_{i=1}^{n}m_ir_i^2 $$

  • 1
    $\begingroup$ I don't forget respectful friends like you. Sometimes look my profile :-) and my recent questions :-) $\endgroup$
    – Sebastiano
    Sep 13, 2021 at 22:12

This was also a bit confusing for me at the beginning - and what I think the real problem here is is how that integrals are notated and explained and, in particular, the stereotypical "calculus textbook" idea doesn't really make these things work very well. The problem is on the maths side, not the physics.

The integral expression is correct because you are adding up each tiny little bit of mass. Think about grinding down the object into a bunch of tiny grains, and then you add up the contributions from each little grain. That's why it is $dm$.

However, the trouble arises when this "strange" integral runs headlong into a dogma that is taught in beginning calculus which is "the 'd' thing just tells you what variable you're integrating with respect to". Hence, when you run into

$$I = \int_\text{obj} r^2\ dm$$

you're left with an odd puzzle: there is no $m$ in $r^2$, so $r$ is just a constant! Hence, why is this not equal to

$$I = r^2 \left(\int_\text{obj}\ dm\right)$$

? Or, if $r$ is to be taken as a function of $m$, that clearly becomes very bad because many different points may have the same mass $m$ and hence this function would be extremely multivalued, very ill-defined and so it would be likewise very hard to make sense of the integral.

But the real trick is this: calculus books, and the integration notation itself, lie to you. You see, in fact the $'d'$ bit in an integral is better understood - however you want to actually formalize this, as being an infinitesimal increment, as you might suggest. $d(\text{variable})$ means the infinitesimal increment when the named variable changes. It does not mean that that is directly the changing variable used when integrating. Instead, to make the distinction clear, one would do better to write integrals much more like a summation because, in effect, they are: from the infinitesimal perspective and integral is a summation of uncountably infinitely many contributions, while a summation is only countably many - with, of course, attendant increased sophistication in how to make that rigorous. Hence, this:

$$\int_{a}^{b}\ f(x)\ dx$$

should better be understood as this:

$$\int_{x=a}^{b}\ f(x)\ dx$$

where the variable being integrated over is now named explicitly, and there is no reason that $dx$ need have anything at all to do with this. Instead, this is to be understood by direct analogy to the sum expression

$$\sum_{n=a}^{b} a_n$$

where $a_n$ is a complete expression. The point then is that the analogy of $a_n$ for the integral is not $f(x)$, but rather all of $f(x)\ dx$ - an expression in its own right which involves this "infinitesimal value" (make that a Robinsonian nonstandard number, make it a differential form, make it the naive "limited sum" approach in elementary calc, or whatever other formalism you like best) $dx$ in it. The $dx$ there is not really to "tell us what variable to integrate with respect to", but is to instead, in an informal sense, keep the uncountable sum from blowing up. To drive that further, note that in this framework, it makes absolute sense to just write

$$\int_{x=a}^{b}\ f(x)$$

and then, unless $f$ is zero at all but a countable number of points, this integral will be infinite or divergent. Hence you need to include a compensating factor, and it's got to be very strong: infinitely strong, in fact, so it should be infinitesimal. And the most common infinitesimal to use simply happens to be the direct change in the variable $x$ as it smoothly is moved from $a$ to $b$ just as the index variable $n$ in a discrete summation

$$\sum_{n=a}^{b} a_n$$

is ticked discretely from $a$ to $b$ in steps of 1. And this infinitesimal is $dx$.

So what is happening in your integral, then? Well, $dm$ is an infinitesimal, $m$ is a variable, but such $m$ is NOT directly the variable of integration, i.e. the variable being incremented. Instead, the incrementation variable is a position, or point within the object, and so we should perhaps better write the integral as

$$I = \int_{P\ \in\ \text{obj}} r(P)^2\ dm$$

where now $r(P)$ is the radial coordinate of $P$ away from the axis, as in cylindrical coordinates, and we are then sweeping that point to and fro hither and thither all thence and around the interior of the object, at each point getting contributions $[r(P)]^2\ dm$ to the moment of inertia. The $dm$ is now just an infinitesimal chunk of mass centered at $P$. $P$ is changing, and as a result of that, so are both $r$ and $dm$, together.

Finally, to make it more explicit, we have

$$dm = \rho(P)\ dV$$

and the volume element $dV$ depends on the coordinate system we are using. If we are using the most natural choice, cylindrical coordinates, then $P = (r, \theta, z)$, $dV = r\ dr\ d\theta\ dz$, and "sweeping $P$ all over the inside of the object" now corresponds to the more rigorous "scanning" of a multilayered integral:

$$I = \int_{P\ \in\ \text{obj}} r^2\ dm = \int_{z=-\infty}^{\infty} \int_{\theta=0}^{2\pi} \int_{r=0}^{\infty} r^2\ [\rho(r, \theta, z)\ r\ dr\ d\theta\ dz]$$

where again, note the notation!

Conclusion: It is $dm$ because we are breaking up into little chunks of mass, at small points in space. The confusion results from bad maths teaching and historical problems with notation that don't match actual practice best.

  • $\begingroup$ I read this just nopw after all these years, and wow your answers are quite illuminating! Thank you:) $\endgroup$
    – Danny Han
    Apr 5 at 8:34

The division is done over the body (and its mass) in order to sum up all the effects needed.

It is not just a mathematical trick, but a physical measurement taken over the entirety of a body.


The general rule is

$$ \text{(result)} = \iiint \text{(measurement)} \,{\rm d}V $$

as in mass, center of mass and mass moment of inertia.

$$ \begin{aligned} m & = \iiint \rho(\boldsymbol{r}) \,{\rm d}V \\ m \,\boldsymbol{r}_{\rm CM} & = \iiint \boldsymbol{r}\, \rho(\boldsymbol{r})\,{\rm d}V \\ \mathbf{I} & = \iiint \pmatrix{y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & - y z & x^2+y^2} \rho(\boldsymbol{r})\,{\rm d}V \end{aligned} $$


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