# 24. Moments of Inertia and Rolling Motion

*Michael Fowler*

## Examples of Moments of Inertia

### Molecules

The moment of inertia of the hydrogen molecule was historically important. It’s trivial to find: the nuclei (protons) have 99.95% of the mass, so a classical picture of two point masses $m$ a fixed distance $a$ apart gives $I={\scriptscriptstyle \frac{1}{2}}m{a}^{2}.$ In the nineteenth century, the mystery was that equipartition of energy, which gave an excellent account of the specific heats of almost all gases, didn’t work for hydrogen$\u2014$at low temperatures, apparently these diatomic molecules didn’t spin around, even though they constantly collided with each other. The resolution was that the moment of inertia was so low that a lot of energy was needed to excite the first quantized angular momentum state, $L=\hslash $. This was not the case for heavier diatomic gases, since the energy of the lowest angular momentum state $E={L}^{2}/2I={\hslash}^{2}/2I,$ is lower for molecules with bigger moments of inertia .

Here’s a simple *planar* molecule:

Obviously, one principal axis is through the centroid, perpendicular to the plane. We’ve also established that any axis of symmetry is a principal axis, so there are evidently three principal axes in the plane, one along each bond! The only interpretation is that there is a degeneracy: there are two equal-value principal axes in the plane, and any two perpendicular axes will be fine. The moment of inertial about either of these axes will be one-half that about the perpendicular-to-the-plane axis.

What about a symmetrical three dimensional molecule?

Here we have four obvious principal axes: only possible if we have spherical degeneracy, meaning all three principal axes have the same moment of inertia.

### Various Shapes

A thin rod, linear mass density $\lambda ,$ length $\ell $: $$I=2{\displaystyle \underset{0}{\overset{\ell /2}{\int}}\lambda {x}^{2}dx}=2\lambda {\ell}^{3}/24={\scriptscriptstyle \frac{1}{12}}m{\ell}^{2}.$$

A square of mass $m$, side $\ell $, about an axis in its plane, through the center, perpendicular to a side: $$I=\frac{1}{12}m{\ell}^{2}.$$ (It’s just a row of rods.) in fact, the moment is the same about any line in the plane through the center, from the symmetry, and the moment about a line perpendicular to the plane through the center is twice this$\u2014$that formula will then give the moment of inertia of a cube, about any axis through its center.

A disc of mass $M,$ radius $a$ and surface density $\sigma $ has $$I={\displaystyle \underset{0}{\overset{a}{\int}}{r}^{2}\cdot \sigma \cdot 2\pi rdr}={\scriptscriptstyle \frac{1}{2}}\pi {a}^{4}\sigma ={\scriptscriptstyle \frac{1}{2}}M{a}^{2}.$$ This is also correct for a cylinder (think of it as a stack of discs) about its axis.

A disc about a line through its center* in its plane *must be ${\scriptscriptstyle \frac{1}{4}}M{a}^{2}$ from the perpendicular axis theorem. A solid
cylinder about a line through its center *perpendicular* to its main axis can be
regarded as a stack of discs, of radius $a$,
height $h$,
taking the mass of a disc as $\rho dz$,
and using the parallel axes theorem,

$$I=2{\displaystyle \underset{0}{\overset{h/2}{\int}}\rho dz\left({\scriptscriptstyle \frac{1}{4}}{a}^{2}+{z}^{2}\right)=}{\scriptscriptstyle \frac{1}{4}}M{a}^{2}+{\scriptscriptstyle \frac{1}{12}}M{h}^{2}.$$

For a sphere, a stack of discs of varying radii, $$I={\displaystyle \underset{-a}{\overset{a}{\int}}dz{\scriptscriptstyle \frac{1}{2}}\rho \pi {\left({a}^{2}-{z}^{2}\right)}^{2}}={\scriptscriptstyle \frac{8}{15}}\rho \pi {a}^{5}={\scriptscriptstyle \frac{2}{5}}M{a}^{2}.$$

An ellipsoid of
revolution and a sphere of the *same mass and radius* clearly have the same
motion of inertial about their common axis (shown).

### Moments of Inertia of a Cone

Following Landau, we take height $h$ and base radius $R$ and semivertical angle $\alpha $ so that $R=h\mathrm{tan}\alpha $.

As a preliminary, the volume of the cone is

$$V={\displaystyle \underset{0}{\overset{h}{\int}}\pi {r}^{2}dz}={\displaystyle \underset{0}{\overset{h}{\int}}\pi}{\left(\frac{Rz}{h}\right)}^{2}dz={\scriptscriptstyle \frac{1}{3}}\pi {R}^{2}h.$$ The center of mass is distance $a$ from the vertex, where

$$aV=a\cdot {\scriptscriptstyle \frac{1}{3}}\pi {R}^{2}h={\displaystyle \underset{0}{\overset{h}{\int}}zdV=}{\displaystyle \underset{0}{\overset{h}{\int}}\pi}z{\left(\frac{Rz}{h}\right)}^{2}dz={\scriptscriptstyle \frac{1}{4}}\pi {R}^{2}{h}^{2},\text{\hspace{1em}}a={\scriptscriptstyle \frac{3}{4}}h.$$The moment of inertia about the central axis of the cone is (taking density $\rho $ ) that of a stack of discs each having mass $$m\left(dz\right)=\pi {r}^{2}\rho dz=\pi {\left(\frac{Rz}{h}\right)}^{2}\rho dz$$ and moment of inertia $I\left(dz\right)={\scriptscriptstyle \frac{1}{2}}m\left(dz\right){r}^{2}$:

$$\underset{0}{\overset{h}{\int}}{\scriptscriptstyle \frac{1}{2}}\pi}\rho {\left(\frac{Rz}{h}\right)}^{4}dz={\scriptscriptstyle \frac{1}{10}}\pi \rho {R}^{4}h={\scriptscriptstyle \frac{3}{10}}M{R}^{2}.$$

The moment of inertia about the axis ${{x}^{\prime}}_{1}$ through the vertex, perpendicular to the central axis, can be calculated using the stack-of-discs parallel axis approach, the discs having mass $\pi \rho {\left(\frac{Rz}{h}\right)}^{2}dz$, it is

$$\underset{0}{\overset{h}{\int}}\pi}\rho {\left(\frac{Rz}{h}\right)}^{2}\left[{\scriptscriptstyle \frac{1}{4}}{\left(\frac{Rz}{h}\right)}^{2}+{z}^{2}\right]dz={\scriptscriptstyle \frac{1}{20}}\pi \rho {R}^{4}h+{\scriptscriptstyle \frac{1}{5}}\pi \rho {R}^{2}{h}^{3}={\scriptscriptstyle \frac{3}{20}}M{R}^{2}+{\scriptscriptstyle \frac{3}{5}}M{h}^{2}.$$

## Analyzing Rolling Motion

### Kinetic Energy of a Cone Rolling on a Plane

(*This is from Landau*.)

The cone rolls without slipping on the horizontal $XY$ plane. The momentary line of contact with the plane is $OA$, at an angle $\theta $ in the horizontal plane from the $X$ axis.

The important point is that this line of contact, *regarded as part of the rolling cone*, is
momentarily at rest when it’s in contact with the plane. This means that, *at that moment*, the cone is rotating about the stationary line $OA.$ Therefore, the angular velocity vector $\overrightarrow{\Omega}$ points along $OA.$

Taking the cone to
have semi-vertical angle $\alpha $ (meaning this is the angle between $OA$ and the central axis of the cone) the center
of mass, which is a distance $a$ from the vertex, and on the central line,
moves along a circle at height $a\mathrm{sin}\alpha $ above the plane, this circle being centered on
the $Z$ axis, and having radius $a\mathrm{cos}\alpha $. The center of mass moves at velocity $V=\dot{\theta}a\mathrm{cos}\alpha $,
so contributes *translational* kinetic energy $${\scriptscriptstyle \frac{1}{2}}M{V}^{2}={\scriptscriptstyle \frac{1}{2}}M{\dot{\theta}}^{2}{a}^{2}{\mathrm{cos}}^{2}\alpha .$$Now visualize the rolling cone turning around the
momentarily fixed line $OA$:
the center of mass, at height $a\mathrm{sin}\alpha $,
moves at $V$,
so the angular velocity $\Omega =\frac{V}{a\mathrm{sin}\alpha}=\dot{\theta}\mathrm{cot}\alpha .$

Next, we first define a new set of axes with origin $O$:
one, ${x}_{3}$,
is the cone’s own center line, another, ${x}_{2}$,
is perpendicular to that *and* to $OA$,
this determines ${x}_{1}$.
(For
these last two, since they’re through the vertex, the moment of inertia is the one worked out at the end of the previous section, see above.)

Since $\overrightarrow{\Omega}$ is along $OA$, its components with respect to these axes $\left({x}_{1},{x}_{2},{x}_{3}\right)$ are $\left(\Omega \mathrm{sin}\alpha ,\text{\hspace{0.33em}}\text{\hspace{0.33em}}0,\text{\hspace{0.33em}}\text{\hspace{0.33em}}\Omega \mathrm{cos}\alpha \right)$.

However, to compute the total kinetic energy, for the rotational contribution we need to use a parallel set of axes *through the center of mass*. This just means subtracting from the vertex perpendicular moments of inertia found above a factor $M{a}^{2}.$

The total kinetic energy is

$$\begin{array}{c}T={\scriptscriptstyle \frac{1}{2}}M{\dot{\theta}}^{2}{a}^{2}{\mathrm{cos}}^{2}\alpha +{\scriptscriptstyle \frac{1}{2}}{I}_{1}{\dot{\theta}}^{2}{\mathrm{cos}}^{2}\alpha +{\scriptscriptstyle \frac{1}{2}}{I}_{3}{\dot{\theta}}^{2}\frac{{\mathrm{cos}}^{4}\alpha}{{\mathrm{sin}}^{2}\alpha}\\ =3M{h}^{2}{\dot{\theta}}^{2}\left(1+5{\mathrm{cos}}^{2}\alpha \right)/40,\end{array}$$

using $${I}_{1}={\scriptscriptstyle \frac{3}{20}}M{R}^{2}+{\scriptscriptstyle \frac{3}{80}}M{h}^{2},\text{\hspace{0.33em}}\text{\hspace{0.33em}}{I}_{3}={\scriptscriptstyle \frac{3}{10}}M{R}^{2},\text{\hspace{0.33em}}\text{\hspace{0.33em}}a={\scriptscriptstyle \frac{3}{4}}h,\text{\hspace{0.33em}}\text{\hspace{0.33em}}R=h\mathrm{tan}\alpha .$$

## Rolling Without Slipping: Two Views

Think of a hoop, mass $M$ radius $R$, rolling along a flat plane at speed $V$. It has translational kinetic energy ${\scriptscriptstyle \frac{1}{2}}M{V}^{2},$ angular velocity $\Omega =V/R,$ and moment of inertia $I=M{R}^{2}$ so its angular kinetic energy ${\scriptscriptstyle \frac{1}{2}}I{\Omega}^{2}={\scriptscriptstyle \frac{1}{2}}M{V}^{2}$, and its total kinetic energy is $M{V}^{2}$.

But we could also have thought of it as *rotating about the point of contact*$\u2014$remember,
that point of the hoop is momentarily at rest.
The angular velocity would again be $\Omega $,
but now with moment of inertia, from the parallel axes theorem, $I=M{R}^{2}+M{R}^{2}=2M{R}^{2}$,
giving same total kinetic energy, but now all rotational .

## Cylinder Rolling Inside another Cylinder

Now consider a solid cylinder radius $a$ rolling inside a hollow cylinder radius $R$, angular distance from the lowest point $\theta $, the solid cylinder axis moving at $V=\left(R-a\right)\dot{\theta}$, and therefore having angular velocity (compute about the point of contact) $\Omega =V/a.$

The kinetic energy is ${\scriptscriptstyle \frac{1}{2}}M{V}^{2}+{\scriptscriptstyle \frac{1}{2}}I{\left(V/a\right)}^{2}={\scriptscriptstyle \frac{1}{2}}\left(M+\frac{I}{{a}^{2}}\right){\left(R-a\right)}^{2}{\dot{\theta}}^{2}$

The potential energy is $-Mg\left(R-a\right)\mathrm{cos}\theta $

The Lagrangian $L=T-V$, the equation of motion is $\left(M+\frac{I}{{a}^{2}}\right){\left(R-a\right)}^{2}\ddot{\theta}=-Mg\left(R-a\right)\mathrm{sin}\theta \cong -Mg\left(R-a\right)\theta ,$

so small oscillations are at frequency $\omega =\sqrt{\frac{g}{\left(1+\frac{I}{M{a}^{2}}\right)\left(R-a\right)}}$.