 Solution

For a uniform circular cross-section, averaging gives $⟨x^2⟩=⟨y^2⟩=R^2/4$; for a uniform axial coordinate in $[-h,h]$, $⟨z^2⟩=h^2/3$. Symmetry makes the coordinate axes principal axes, since all products of distinct coordinates integrate to zero. Thus the principal moments of inertia about the centre areThe inertia tensor is a scalar multiple of identity exactly when $h^2/3=R^2/4$. Thereforeis the height-to-radius ratio for sphere-like rotational inertia; the height-to-diameter ratio is $\sqrt{3}/2$. The geometric shape remains cylindrical, but its angular momentum is parallel to angular velocity for every axis, just as for a sphere.