Example: Cube tipping
A cube of mass \( m \) and side length \( a \) is balanced perfectly on one edge. At \( t=0 \), we nudge it so that it will tip in the \( +\hat{x} \) direction. Assuming the initial speed is roughly zero, what is the angular speed of the cube when it hits the ground?
This time, conservation of energy can be applied: the only force acting is gravity, which is conservative. The gravitational potential depends only on the height of the center of mass:
\[ \begin{aligned} U_g = mgh_{\rm CM}, \end{aligned} \]
which starts at a height of \( h_{\rm CM} = a/\sqrt{2} \), and ends at \( h_{\rm CM} = a/2 \). The kinetic energy is zero beforehand, and after is due to the rotation and due to the rotation of the cube, so we have
\[ \begin{aligned} \frac{mga}{\sqrt{2}} = \frac{mga}{2} + \frac{1}{2} m v_{\rm CM}^2 + \frac{1}{2} I \omega^2. \end{aligned} \]
The center of mass is always \( a/\sqrt{2} \) from the edge, so its linear speed is related to the angular speed by
\[ \begin{aligned} v_{\rm CM} = \frac{a}{\sqrt{2}} \omega. \end{aligned} \]
Finally, we need the moment of inertia for a cube rotating about its CM, which (as we found before) is \( I = 1/6 ma^2 \). Plugging back in and rearranging, then,
\[ \begin{aligned} mga (\sqrt{2} - 1) = \frac{ma^2 \omega^2}{2} + \frac{1}{6} ma^2 \omega^2 = \frac{2}{3} ma^2 \omega^2 \end{aligned} \]
or inverting,
\[ \begin{aligned} \omega^2 = \frac{3g}{2a} \left( \sqrt{2} - 1 \right). \end{aligned} \]
Now, that all went pretty quickly, so let's stop and think about the details. First of all, I'm cheating a little bit here by going back to the intro physics formula; we know that the full expression for the kinetic energy should be
\[ \begin{aligned} T = \frac{1}{2} \vec{\omega}^T \mathbf{I} \vec{\omega}. \end{aligned} \]
However, in this problem we "know" what the motion of the cube is already. If you imagine this as an experiment, once you've seen the cube tip over, you know what the motion looks like - it is only rotating about one axis as it falls over. As long as we choose our coordinates properly (which is especially easy for the cube, as a spherical top!), this will be a principal axis, so we just have \( Lz = I{zz} \omega_z \), or \( L = I \omega \) for short. Basically, if there were non-trivial off-diagonal contributions from \( \mathbf{I} \) here, we would know about it because the rotation would be more complicated!
Another burning question you might have: the cube is pivoting about its edge, so why aren't we using the elements of that inertia tensor? The inertia tensor for a cube rotating about its edge is, in fact, different, and in fact it isn't even diagonal. At the bottom of the fall when the cube is lying flat, it will be equal to
\[ \begin{aligned} \mathbf{I}_{\rm edge} = Ma^2 \left( \begin{array}{ccc} 5/12 & -1/4 & 0 \\ -1/4 & 5/12 & 0 \\ 0 & 0 & 2/3 \end{array} \right). \end{aligned} \]
The axis we're rotating about, \( \hat{z} \), is at least still principal in this way of looking at it. But the value of \( I \) here would be \( 8/3 Ma^2 \), not \( 1/6 Ma^2 \). Why shouldn't we use that number?
In the equations I wrote above, we've silently invoked one of the main results of this chapter, which is separation of kinetic energy between motion of the CM and motion relative to it:
\[ \begin{aligned} T = T_{CM} + T_{\rm rot}. \end{aligned} \]
To divide things up like this, the rotation is explicitly taken to be about the center of mass. Hence, we use \( I \) for rotation about the CM when we're accounting for the center of mass's translational motion separately. Of course, once we have the inertia tensor about the edge, nothing stops us from using that! Now the rotation includes all of the motion of the cube:
\[ \begin{aligned} mga (\sqrt{2} - 1) = \left( \frac{2}{3} ma^2 \right) \omega^2 \end{aligned} \]
and we find exactly the same answer for \( \omega \)!
You could object one more time at this point: that inertia tensor is only valid at the bottom of the fall. Why aren't we keeping track of how \( \mathbf{I} \) changes with time, in this approach? Really we should be, which tends to make it dangerous to work in the space frame, and was the motivation for developing the formalism to switch to the body frame in the first place. But we now know that we can describe the inertia matrix as it rotates, by using a rotation matrix. The rotation will be about the \( \hat{z} \) axis, so the matrix will take the form
\[ \begin{aligned} \mathbf{R} = \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right), \end{aligned} \]
and then at some angle \( \theta \) between 0 and \( \pi/4 \), the inertia tensor as the cube falls is
\[ \begin{aligned} \mathbf{I}'(\theta) = \mathbf{R} \mathbf{I} \mathbf{R}^T. \end{aligned} \]
The bad news is that the expression for \( \mathbf{I}' \) is very messy, too horrible to write on the board. The good news is that nothing changes in the \( \hat{z} \) direction! Since we're rotating about \( \hat{z} \), which is already principal, the rotation just scrambles the components in \( \hat{x} \) and \( \hat{y} \), but has no effect through the motion we're considering. This is a simple, but important observation: any rotation about a principal axis will leave it principal.