While writing an answer to this question, I encountered a puzzling conceptual problem when applying Newton's second law for rotation about an instantaneous axis of rotation when the moment of inertia is changing. I will restate the premise of a simplified version of the original problem.
A massless cylindrical tube of radius $R$ with infinitesimally thin walls has an infinitesimally narrow rod of mass $m$ attached to its inner surface, parallel to the tube. The tube is then laid on a floor so that the rod is in the uppermost position, and then is released to roll away from that unstable equilibrium position. The floor isn't slippery, so there's no sliding. The cross section of the tube looks like the following (from ja72's answer):
The square of angular velocity $\omega=d\theta/dt$ can be found from conservation of energy to be $$ \omega^2 = \frac{g}{R}\frac{1 - \cos\theta}{1 + \cos\theta}.\tag{1}$$ Differentiating with respect to time yields the angular acceleration $$\alpha=\frac{g}{R}\sin\theta\frac{1}{(1+\cos\theta)^2}.\tag{2}$$
Now my actual question is on how to apply Newton's second law to the rotation about the instantaneous line of contact of the tube with the floor. This axis is instantaneously at rest since the tube is not sliding on the floor. The total moment of inertia about this axis is $$ I = 2mR^2(1 + \cos\theta)\tag{3}$$ since its squared distance to the axis of rotation is $R^2\left[(1 + \cos\theta)^2 + \sin^2\theta\right] = 2R^2(1 + \cos\theta)$.
I can think of two different ways to apply Newton's second law for rotation:
- During the instantaneous pure rotation about the bottom of the tube, the distance of the rod to the axis of rotation doesn't change (since it is merely rotating about this axis), so the moment of inertia doesn't change.
- $\theta$ changes as the tube rolls, therefore the moment of inertia changes according to $(3)$.
So we can write Newton's second law for rotation as follows: $$mgR\sin\theta = \tau = \frac{d}{dt}(I\omega) = I\alpha + \epsilon\omega \frac{dI}{dt} = I\alpha + \epsilon \omega\frac{dI}{d\theta}\frac{d\theta}{dt} = I\alpha + \epsilon \omega^2\frac{dI}{d\theta}$$ where $\epsilon = 0$ for case 1 (we are considering $I$ fixed) and $\epsilon = 1$ for case 2 ($I$ changes according to $(3)$). Substituting $(1)$, solving for $\alpha$ and simplifying yields
$$\alpha = \frac{g}{2R}\sin\theta\frac{1 + 2\epsilon+(1 - 2\epsilon)\cos\theta}{(1 + \cos\theta)^2}. $$ This yields the correct angular acceleration $(2)$ only for $\epsilon = 1/2$! What is happening here? Is there just no straightforward way to apply Newton's second law for rotation about such an axis?