Oscillations of a cylinder about its midpoint

Question

Consider a uniform cylinder rotating about an axis perpendicular to it through its midpoint.

Because the rotation is around the center of mass of the cylinder, the equation of motion neglecting friction is

$$\ddot{\theta}=0$$

where $\theta$ is the angle the cylinder makes with the vertical.

The solution to this is

$$\theta=At+B$$

This seems to indicate that if we lift the cylinder from its equilibrium position to some initial angle $\theta_{0}$ and let it go without any initial velocity, then $\theta=B$, i.e. $\theta$ would be constant which doesnt make sense.

I think there is an error in my analysis somewhere but not sure where.

Answer 1

Your analysis is correct. It is your intuition ("doesn't make sense") which is at fault.

If the cylinder is pivoted at its centre of mass, then it will not oscillate or accelerate. It is in a position of neutral equilibrium, like a cylinder on a flat surface. There is no equilibrium position to which the cylinder returns. If placed at rest in a certain angular position $(A=0, B=\theta_0)$ it will stay there. If given an initial angular velocity $(A=\omega_0, B=\theta_0)$ it will continue rotating with that velocity. (Newton's 1st Law.)