# A Second Way To Fall Over

I admit not being perfectly satisfied with my answer, about whether a box is easier to tip over by pushing on the middle of one of its top edges or by pushing on its corner, just by looking at it from the energy both approaches need to raise the box’s center of mass above the pivot. It’s straightforward enough, but I don’t do this sort of calculation often, so maybe I’m looking at the wrong things. Can I find another, independent, line of argument? If I can, does that get to the same answer? If it does, good. If it doesn’t, then I get to wonder which line of argument I believe in more. So here’s one.

If I’m to make a box tip over, I have to be rolling it. There’s a measure of how hard it is to make something roll; this is known as the moment of inertia. It’s the rotational equivalent of inertia, the measure of how hard it is to make something start moving (or to accelerate, or to decelerate it). The larger the moment of inertia is, the harder it is to make it roll. If it’s harder to roll something a particular way, it’s probably harder to tip it over that way.

Inertia, the nice familiar property, is just the mass of something. That’s very easy to work with. The moment of inertia is harder to work with. The moment of inertia of a little bit of mass — call it the amount $\delta m$, with the $\delta$ a common notation for “a tiny bit of” and “m” here evoking “mass” — being spun along an axis a distance $r$ away equal to $\delta m r^2$. This looks easy enough to calculate, but it assumes we have a little dot of mass. For a big solid object like a cube we need to imagine breaking the cube into incredibly many — no, even more than that — tiny bits of mass, each at different positions relative to the axis around which the cube is rotating, and add up all the man $\delta m r^2$ contributions from all these tiny pieces. This is pretty much what integral calculus was invented for, and I’m not crazy enough to explain that in 700 words.

I don’t need to, either. Any good reference page for physics stuff — including, of course, Wikipedia — will have the moments of inertia for many common shapes and many common axes of rotation already worked out. We maybe can’t use that in an exam, but we’re happily free of exams here.

Using Wikipedia’s table of moments of inertia of various things, then, it appears that if we want to roll a rectangular box, of length $l$, width $w$, and height $d$ along the center of its shorter faces (the ones of width $w$ and height $d$), then that moment of inertia is $I_e = \frac{1}{12}m\left(w^2 + d^2\right)$. (The moment of inertia is usually written $I$ and if that’s not for “inertia” then I can make no sense of it. I use the “e” subscript because this would represent the moment of inertia rolling by pushing along the center of the edge.) For our cube, both the width and height — and for that matter the length — are the same measure, which last time we settled on calling $a$. So that gives us a moment of inertia of $I_c = \frac{1}{12} m \left(a^2 + a^2\right) = \frac{1}{6}m a^2$.

Except … if we’re tipping the cube over, we aren’t rotating it around the center of the cube, we’re rotating it around the edge, along one of the edges of the bottom face. The moment of inertia is different if the axis of rotation is different.

Fortunately there’s a theorem — the parallel axis theorem — which says how we can convert the moment of inertia around one axis to the moment of inertia around another, parallel, axis. If we know the moment of inertia around an axis going through the center of mass, and want to know the moment of inertia rotating around an axis a distance $r$ from the first, then we can find the moment of inertia around that new axis by taking the sum of the first moment of inertia and adding to it $m r^2$. This is (probably) why the Wikipedia table only bothers giving the moment of inertia for an axis going through the center of this box; since that’s an axis through the center of mass, we can find what the rotations around other axes are just by adding that nice simple formula there.

The axis going through the center of these faces is a distance of $\frac{1}{\sqrt{2}} a$ away from any edge of the cube. So the moment of inertia for rotating around an edge of the cube — as we must be doing if we’re tipping the cube over, that is, making it roll — will be the sum of the center of mass’s moment of inertia, $\frac{1}{6} m a^2$, and this parallel axis value, $m r^2 = m \left(\frac{1}{\sqrt{2}} a\right)^2 = \frac{1}{2} ma^2$. Therefore the moment of inertia for the cube rotating around this edge is a total of $I_e = \frac{1}{6}ma^2 + \frac{1}{2}ma^2 = \frac{4}{6}ma^2$. (And I don’t change that to two-thirds because I peeked at the next part and I want to make the comparison easier.)

Now, how about the cube rotating along its longest diagonal, the case where the cube is pushed over from a corner? For a rotation along the longest diagonal, the moment of inertia is an impressive and slightly scary form that mercifully simplifies when we remember the length, width, and height are all the same value:

$I_c = \frac{1}{6}m\left( \frac{w^2\cdot d^2 + l^2\cdot d^2 + l^2\cdot w^2}{l^2 + w^2 + d^2}\right) \\ I_c = \frac{1}{6}m\left( \frac{a^2\cdot a^2 + a^2\cdot a^2 + a^2\cdot a^2}{a^2 + a^2 + a^2}\right) = \frac{1}{6}m\left(\frac{3 a^4}{3 a^2}\right) = \frac{1}{6} m a^2$

Again, that’s the rotation through the center of mass. But the axis going through the edge — which is our pivot — is a distance $r = \sqrt{\frac{2}{3}}a$ away from that center. So we have to add to the above figure $m\left(\sqrt{\frac{2}{3}} a\right)^2 = \frac{2}{3}m a^2$. And that gives us a total angular momentum, for rotating around the corner, of:

$I_c = \frac{1}{6} m a^2 + \frac{2}{3}m a^2 = \frac{5}{6}m a^2$

I don’t know the mass of the box, or the size, but I do know which is larger, $\frac{4}{6}$ or $\frac{5}{6}$. Rolling the box by the corner has a higher moment of inertia than rolling the box by the edge, so it’s harder to do. So this separate line of reasoning says the same, that it’s easier to tip the box over pushing along the center than by the corner.

Well, except …