What is the maximum length of a cylindrical space station for stability versus small asymmetries in the distribution of mass?

Long cylinders, initially rotating around their physical axis of symmetry (a line running down the length through the middle) are rotationally unstable versus small deviations in the internal distribution of mass. They start to precess, and then they tumble end-over-end, just as the Explorer 1 satellite did.

Short, squat cylinders are, on the other hand, rotationally stable around their physical axis of symmetry.

Where's the break-even point in frustum length between cylinders that rotate stably around their physical axis of symmetry, and those that are unstable? Rotating objects tend to seek the rotational axis having the greatest moment of inertia, which is usually the shortest axis.

The moments of inertia for a hollow cylinder of inner radius Rᵢ , outer radius Rₒ , and length h are

Rotation around the physical axis of symmetry (desired)

I∥ = (M/2)(Rᵢ²+Rₒ²)

Rotation perpendicular to the physical axis of symmetry (to be avoided)

I⟂ = (M/12) [ 3(Rₒ²+Rᵢ²) + h² ]

Set them equal to each other.

(M/2)(Rᵢ²+Rₒ²) = (M/12) [ 3(Rₒ²+Rᵢ²) + h² ]

6Rᵢ² + 6Rₒ² = 3Rₒ² + 3Rᵢ² + h²

3Rᵢ² + 3Rₒ² = h²

Assuming that the thickness of the frustum wall (Rₒ−Rᵢ) is negligible compared with the diameter of the cylinder,

6R² ≈ h²

h = √6 R = 2.449489743 R

When the length, h, of the cylinder is less than √6 R, the cylinder is rotationally stable versus small asymmetries in mass distribution around the frustum. But when h is greater than √6 R, the cylinder is rotationally unstable.

Wasn't that simple?