# Moment of inertia: why $\mathrm dI=r^2\mathrm dm$ instead of $\mathrm dI=m\mathrm dr^2$?

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?

• 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.) – jacob1729 Jun 13 '18 at 12:37
• 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 – Danny Han Jun 13 '18 at 12:39
• 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. – JamalS Jun 13 '18 at 12:40
• This might be helpful reading. As will the Wikipedia page on moment of inertia – jacob1729 Jun 13 '18 at 12:49
• 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. $dI=r_m^2 \cdot dm$. – FGSUZ Jun 13 '18 at 12:57

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, $$\delta I \to \text{d}I = \text{d}m R_{\perp}^2 = \text{d}V \rho\left(\vec{r}\right) R_{\perp}^2.$$