I just came back from my Introduction to Rotational Kinematics class, and one of the important concepts they described was Rotational Inertia, or Moment of Inertia.

It's basically the equivalent of mass in Netwon's $F = m a$ in linear motion. The equivalent rotational equation is $\tau = I \alpha$, where $\tau$ is rotational force, $\alpha$ is rotational acceleration, and $I$ is rotational inertia.

For a point about an axis, $I$ is $m r^2$, where $r$ is the distance from the point to the axis of rotation.

For a continuous body, this is an integral -- $I = \int r^2 \,dm$.

This really doesn't make any sense to me...you have two independent variables? I am only used to having one independent variable and one constant. So I would solve this, using my experience with calculus (which encompasses a read through the Sparks Notes packet) as $ I = m r^2 $.

But obviously, this is wrong? $r$ is not a constant! How do I deal with it? Do I need to replace $r$ with an expression that varies with $m$? But how could $r$ possibly vary with $m$? Isn't it more likely the other way around? But how can $m$ vary with $r$? It's all rather confusing me.

Could someone help me figure out what to do with all these substitutions for, example, figuring out the Moment of Inertia of a hoop with no thickness and width $w$, with the axis of rotation running through its center orthogonal to its plane?


2 Answers 2


Seeing an expression like $I = \int r^2 dm$ is certainly confusing the first time you see it. When you see an expression like $I = \int x^2 dx$ we effectively iterating over a range on the x axis and adding up the area of an infinite amount of infinitesimally small strips. Note that the area of each strip is approximated as the function value (x^2) times the strip length (dx). When you see an expression like dm, we are iterating over masses instead. We still sum up the function value (r^2), but this time we multiply it by the strip mass (dm). To solve the problem, we usually put m in terms of another variable which we can iterate over more easily.

For example, consider the moment of inertia of a rod of length L around its center with total mass of L. Each bit of length (dx) has mass (dm) and r=|x|. Solving for $I = \int r^2 dm = \int |x|^2 dx = \int x^2 dx = (x^3)/3+c$. Now, we have definite values for x to sub in (-L/2 and L/2), so we write $$ I = \frac{(L/8)}{3}-\frac{(-L/8)}{3}=\frac{L}{12} $$

Now lets calculate the moment of inertia of the hoop case you described. We break the hoop up into infinitesimally small rings the same distance from the center. Let the hoop have inner thickness r and outer thickness R. The area is $R^2-r^2$. The area density, d, is therefore $\frac{m}{R^2-r^2}$. A ring at radius k with thickness dk has area $2\pi~k~\mathrm{d}k$, mass is $2\pi~k~\mathrm{d}k\frac{m}{R^2-r^2}$ and moment of inertia around the central axis $2\pi~k^3~\mathrm{d}k\frac{m}{R^2-r^2}$. The integral is $\frac{\pi}{2}k^4\frac{m}{R^2-r^2}+c$. Subbing in the exact values, we get $$ \frac{\pi}{2}(R^4-r^4)\frac{m}{R^2-r^2} = \frac{\pi}{2}(R^2+r^2)m $$

  • 1
    $\begingroup$ In the hoop case, the area should be $\pi (R^2-r^2)$. This divides the area density by $\pi$ and flows through to divide the final answer by $\pi\ $\endgroup$ Commented Nov 9, 2010 at 14:40
  • $\begingroup$ That $L$ should be $L^3$. And $dm\neq dx$, but $\frac{dm}{m}=\frac{dx}{L}$. $\endgroup$
    – mr_e_man
    Commented Jun 3, 2018 at 8:16

$dm$ takes density fluctuations into account, it's just $dm = \rho(\vec r) d^3 \mathbf r$.

E.g. for a homogeneous cylinder with the rotational axis align parallel to the z-axis it is $\rho = \frac{m}{V}\cdot\theta(R-r)\theta(a^2-z^2)$, where $\theta(x) = \begin{cases} 0 \;\text{ for } x<0 \text{ and}\\ 1 \;\text{ for } x>0 \end{cases}$ and $r = \sqrt{x^2+y^2}$ is the radial coordinate in cylindrical coordinates. The result is that your integration over $r$ is from $0$ to $R$ and for $z$ from $-a$ to $a$. $V=\pi R^2 \cdot 2a$ is the volume of the cylinder, $m$ its weight. In cylindrical coordinates, $d^3\mathbf r = r\, dr \,d\phi\, dz$, so you got

$$I = \int_{r=0}^{R} \int_{z=-a}^a \int_{\phi=0}^{2\pi} r^2 \frac{m}{V} \cdot r\, d\phi\, dz\, dr = 4a\pi \frac{m}{V} \int_{r=0}^R r^3 dr = \frac{am\pi}{V}R^4 = \frac{1}{2}mR^2.$$


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