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We've been learning about the derivation for moment of inertia as:

$$\int r^2 dm$$

However, for me, this looks like it's a bit backwards. As a first year calc student, I see the differential in the integration as almost being the "independent" variable and the function inside as being the "dependent" variable (when we usually do integration, we have something like $\int f(x) dx$, so I have developed this intuition).

However, going with this idea, it would then seem like radius is somehow a function of mass. I would think it would be the other way around, as in, I input radius and then the function gives me the differential mass at a given radius. Can you please give me intuition on why we are splitting up the object into differential masses and not differential distances from the axis of rotation?

rb612
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  • It's actually $\displaystyle \int r^2~\rho(r)~\mathrm d^3r,.$ –  Nov 11 '16 at 05:07
  • Ah can you please explain? – rb612 Nov 11 '16 at 05:07
  • $\rho(r)$ is the mass per unit volume, if that is what bothering you. –  Nov 11 '16 at 05:19
  • @MAFIA36790 -- Actually, that is incorrect. – David Hammen Nov 11 '16 at 08:37
  • Sure; I am not always right; so do let me know why that is incorrect @David. –  Nov 11 '16 at 09:15
  • @MAFIA36790 It is correct but I'm guessing he wants you to distinguish between the different $r$. The $r^2$ is the square of the perpendicular distance of any point in the volume to the axis of rotation, whereas the $r$ in $\rho(r)$ is just the radial coordinate, or represents the coordinates used. – JamalS Nov 11 '16 at 11:33
  • @JamalS, yes, I forgot the \mathbf sign as I wrote in the chat. –  Nov 11 '16 at 11:45
  • @MAFIA36790 That doesn't help; writing $\rho (\mathbf{r})$ does not change the fact that $r^2$ could still be interpreted as the $| \cdot |^2$ of it; every text usually doesn't make this distinction clear - one should be relabelled imo. – JamalS Nov 11 '16 at 12:22
  • yes, @JamalS, I very much agree; I should have cleared it; but JohnR already did it both in his post as well as in the chat. –  Nov 11 '16 at 12:24
  • @MAFIA36790 - A lesser problem is that $\mathrm d^3r$ is a bit sloppy. Still an abuse of notation, $\mathrm d^3 \mathbf r$ is a bit better. My main objection, however, was the use of $r^2$. This should be the square of the distance to the axis of rotation. If the axis of rotation passes through the origin, a better expression is $\int~\left(r^2-(\mathbf r \cdot \hat {\mathbf u})^2\right),\rho(\mathbf r),\mathrm d^3\mathbf r$ , where $\hat {\mathbf u}$ is a unit vector along the rotation axis. – David Hammen Nov 11 '16 at 14:52
  • I really do agree with you; I have seen many books using $\mathrm d^3r$ but I prefer $\mathrm d^3\mathbf r$ which I have mentioned in the chat while conversing about it with John and yes, $I_n = \mathbf n\cdot \mathsf I\cdot \mathbf n$ is the correct way to approach the very problem. Wrote it in haste; but it's good for future readers, I deem, after this followed-up discussion @davidH. –  Nov 11 '16 at 15:10
  • Related question here. – knzhou May 08 '20 at 04:05

2 Answers2

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Suppose you have a single point mass $m$ at a distance $r$, then the moment of inertia is of course just:

$$ I = mr^2 $$

But suppose now our mass is an extended object not a point mass, so the above equation doesn't apply. We could get an approximation for $I$ by splitting the mass up into $n$ smaller masses $m_i$ with centres of mass at $r_i$:

$$\begin{align} I &\approx m_0r_0^2 + m_1r_1^2 + \text{...} + m_nr_n^2 \\ &\approx \sum_{i~=~0}^n m_i r_i^2 \end{align}$$

To make this exactly we increase the splitting up into an infinite number of infinitesimally small masses $dm$, and the sum turns into an integral:

$$ I = \int~\mathrm dm\,r^2 $$

And that's what the equation you give means. But in practice we wouldn't express $r$ as a function of $m$ and integrate with respect to $\mathrm dm$. Instead we note that $\mathrm dm = \rho~\mathrm dV$, where $\mathrm dV$ is the infinitesimal volume of our infinitesimal mass. Suppose we are using Cartesian coordinates with the axis of rotation along the $z$ axis, then our volume element is $\mathrm dV = \mathrm dx\,\mathrm dy\,\mathrm dz$ and our integral becomes:

$$ I = \int \int \int r(x,y)^2 \rho(x,y,z)~\mathrm dx\,\mathrm dy\,\mathrm dz $$

where $r$ is the distance from the axis:

$$ r^2 = x^2 + y^2 $$

But writing $\displaystyle \int r^2~\mathrm dm$ is a lot shorter and means the same thing.

John Rennie
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The integral in the question is an acceptable definition; when we write,

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

we mean here that $r$ is the distance a point of the object is from the axis of rotation, as a function of mass. An equivalent definition of $I$ would be,

$$I = \int_V dV \, \rho \, r^2$$

where again $r$ is the distance from the axis, but this time as a function of the position of the point.

Illustrative Example

Consider a rod of infinitesimal thickness and length $\ell$ in the $xy$-plane, say, resting on the $x$ axis from the origin to $x= \ell$, and we are rotating it about the $z$-axis. Then, the mass of a point a distance $x$ from the origin will be, $m(x) = \lambda x$, where $\lambda$ is the linear density. Inverting this means,

$$r^2_{\mathrm{axis}} = \frac{m^2}{\lambda^2}$$

and using the first definition of $I$, by integrating over all masses, we have,

$$I = \int_0^{\lambda \ell} dm \, \frac{m^2}{\lambda^2} = \frac{1}{3}\lambda \ell^3 = \frac{1}{3}M\ell^2$$

where $M = \lambda \ell$ is the total mass. Now, proceeding with the second definition, note the density is $\rho(x,y) = \lambda \delta(y)\mathbb{1}_{[0,\ell]}( x)$ and that the distance from the axis is given by $r^2_{\mathrm{axis}} = x^2 + y^2$. Integrating, $$\int_V dy dx \, \rho (x^2 + y^2) = \left[ \frac{1}{3}\lambda x^3\right]^{\ell}_0 = \frac{1}{3}\lambda \ell^3 = \frac{1}{3}M\ell^2.$$

From this example, you can see why the definitions are equivalent, but we often use the second, particularly because expressing the distance from the axis as a function of mass is not always as simple, and it is rather odd to think of them relating in that way.

JamalS
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