# The Box Drops

So the last piece I need for figuring out whether it’s easier to tip a box over by pushing on the middle of an edge or along one corner is to know the amount of torque applied by pushing with, presumably, the same force in both locations. Well, that’s almost the last bit. I also need to know how the torque and the moment of inertia connect together to say how fast an angular acceleration I can give the box.

I claimed the moment of inertia was the rotational equivalent of the inertia, the measure of how much effort it takes to make something start turning as opposed to what it takes to make it move. The angular acceleration is the rotational equivalent of the acceleration, a measure of how fast the speed of rotation is changing. And the torque serves a role akin to that of the force: the larger the torque, for the same moment of inertia, the larger the angular acceleration. In imitation of Newton’s third law, that the force exerted on an object equals the mass times the acceleration, in rotations, the torque applied to a rotating object equals the moment of inertia times the angular acceleration.

The torque is, if you’ll let me avoid a lot of vector stuff, the product of how far the point of application of the force is from the axis of rotation — that’s the radial distance, which I’ll bill as $d$ because that’s how it was in the textbook I learned it from — times the amount of force being applied perpendicular to this radial distance. The amount of force I’ll call $F$ since at this point someone not using $F$ for “amount of force” is just being a little perverse. The torque, $\tau$, is then the product $F d$. (Again, this is ignoring vector stuff that I don’t want to deal with.)

The moment of inertia is usually written as $I$, and the angular acceleration, how much the speed of rotation is changing, gets the Greek $\alpha$ since that evokes acceleration’s a without being too easily confused with it unless you write things out by hand. And with all this in mind, the rule for how much angular acceleration we have is:

$\tau = I \alpha \\ F d = I \alpha \\ \frac{F d}{I} = \alpha$

If I want to compare which is easier to tip over, the box pushed by the edge or the box pushed by the corner, I’ll suppose that the same force is applied both times. For the box pushed from the corner, the moment of inertia (previously worked out, for both kinds of rotation) was $I_e = \frac{4}{6}m a^2$, where $m$ is the mass of the box and $a$ is how long the box is on any side. And if we’re rotating around the edge, as implied here, then the distance between the axis of rotation and the point where we’re pushing is the diagonal $\sqrt{2}a$. So for a given force the angular acceleration this produces is:

$\alpha = \frac{F d}{I} = \frac{F \sqrt{2}a}{\frac{4}{6} m a^2} = \frac{3}{\sqrt{2}} \frac{F}{ma} \approx 2.12 \frac{F}{ma}$

When pushing the box by its corner the moment of inertia is $I_c = \frac{5}{6}m a^2$, while the distance between the axis of rotation and the corner where the pushing is going on is $\sqrt{3}a$. And this gives an angular acceleration of:

$\alpha = \frac{F d}{I} = \frac{F \sqrt{3}a}{\frac{5}{6} m a^2} = \frac{\sqrt{3} \cdot 6}{5} \frac{F}{ma} \approx 2.08 \frac{F}{ma}$

So the fact that we’d be pushing from farther away from the axis of rotation by pushing at the corner rather than the edge evens up the difference made by the moment of inertia.

So what to conclude from all this? It looks like the same force will produce pretty near the same rotational acceleration whether you push the box along an edge or along a corner. The $\frac{F}{ma}$ part in both expressions for angular acceleration are the same, by assumption, and I’m willing to bet you can’t feel the difference between 2.12 and 2.08. (I’d also bet the corner is harder to get the leverage just right for, but that’s a practical matter that falls too far outside my field of competence.) But you certainly have to put more energy into the push to get the center of mass above the pivot, and so to tip it over.

So this makes me think that there’s not a lot of difference in getting a box to start rolling, however you want to push it from the top, although getting it to tip over is a bit harder if you push from the corner rather than from the edge. And as mentioned, my niece doesn’t seem to have any problem making the thing fall over either way.