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Set $a=2\pi G{\rho }_0$. The axisymmetric Poisson equation for Newtonian gravity integrates to $(R{\Phi }_R{)}^{\prime }=4\pi G{\rho }_{0}R$. Regularity on the axis fixes the integration constant, so ${\Phi }_R=aR$. Radial force balance in the rotating fluid givesWith the stated central pressure, $p_c={\rho }_0{a}^2/(4k^2)$, this is a perfect square:The free surface lies at the first zero of pressure. For the source's positive rotation parameter, $R_0=\sqrt{2\pi G{\rho }_0}/k$; allowing either rotation direction requires $|k|$ in the denominator. The inward effective gravity is ${\Phi }_R-R{\Omega }^2=R(a-k^2{R}^2)$, so $\boxed{g_{\mathrm{eff}}(R_0)=0}$. In particular $p_R(R_0)=0$. This is the critically rotating self-gravitating cylinder.

For a normal mode proportional to $e^{i\omega t+im\varphi }$, the background material derivative becomes $i\sigma$, where $\sigma =\omega +m\Omega (R)$. Linearizing the centrifugal term gives $-2\Omega u_{\varphi }$ in the radial equation. In the azimuthal equation the background velocity gradient and cylindrical geometric term together give $(2\Omega +R{\Omega }^{\prime })u_R=3\Omega u_R$. Since the interior mass density is constant, write $W=p^{\prime }/{\rho }_0+{\Phi }^{\prime }$. The two momentum equations and incompressible flow condition are consequentlyFor $m\ne 0$, the last two equations express the other amplitudes in terms of $u=u_R$:Differentiate the second expression using $R{\Omega }^{\prime }=\Omega$ and $R{\sigma }^{\prime }=m\Omega$, then substitute into the radial momentum equation. Before any division by $\sigma$, the result isAway from corotation this gives the required ordinary differential equationAt corotation the original coupled equations, or the undivided equation, must be used. A regular axisymmetric radial velocity would vanish by the incompressible flow condition; the requested dipolar calculation has $m=1$.

For $m=1$, use the amplitude normalization $u=A(\omega +kR)$, which remains useful when $\omega$ tends to zero. Substitution into the undivided equation gives zero identically. For nonzero $\omega$ this is the source's $1+kR/\omega$ after choosing $A=1/\omega$. The other amplitudes areThese formulas also satisfy the three original equations directly. By the regularity assumption in the question, this is the interior velocity family to which the free surface conditions must be applied.

The gravitational potential of a displaced cylindrical interface supplies an essential part of those conditions. For surface fluid displacement $\eta e^{im\varphi }$, the bulk mass density perturbation vanishes but the displaced mass density step contributes ${\rho }_0\eta \delta (R-R_0)$. The interior and exterior gravitational potential amplitudes solve Laplace's equation, are proportional to $R^m$ and $R^{-m}$ for $m\ge 1$, and are continuous at the surface. Integrating the Poisson equation for Newtonian gravity across the mass density step givesSince the two radial derivatives at the surface are $-m{\Phi }_s^{\prime }/R_0$ and $m{\Phi }_s^{\prime }/R_0$, respectively,It would be incorrect to set the entire potential perturbation to zero merely because the bulk mass density is unchanged.

Let ${\Omega }_s=k{R}_0$, so $a={\Omega }_s^2$. Zero fluid Lagrangian perturbation of surface pressure gives $p_s^{\prime }+\eta p_R(R_0)=0$, hence $p_s^{\prime }=0$. The surface momentum potential and kinematic condition for $m=1$ are thereforeIf $\omega +{\Omega }_s\ne 0$, the kinematic condition fixes $\eta =-iA$. The same value follows if the radial fluid Lagrangian displacement $u_R/[i(\omega +kR)]=-iA$ is required to extend continuously to the free surface, including a possible surface-corotation limit. Comparing the gravitational condition with the interior value of $W_s$ then givesfor a nontrivial perturbation. These dipolar perturbations of a critically rotating cylinder describe a displacement of the whole cylinder's axis. Translating an isolated self-gravitating system gives another equilibrium, so there is no external restoring force and the translation is neutral. A uniform drift of the centre also follows from invariance under a Galilean transformation, explaining the double zero of the frequency equation. The singular $1/\omega$ in the originally chosen velocity normalization does not obstruct the finite-amplitude limiting solution.

There is an admissibility qualification to the frequency conclusion as printed. Regular interior fluid Eulerian perturbation of velocity alone does not justify cancelling the surface factor $\omega +{\Omega }_s$. Eliminating $\eta$ without that cancellation instead givesThe additional surface-corotation case $\omega =-{\Omega }_s$ formally satisfies every displayed velocity equation and the usual linearized surface conditions withHere $u_R(R_0)=p^{\prime }(R_0)=0$, and the velocity is finite on the axis. However, its interior radial fluid Lagrangian displacement tends to $-iA$, whereas the undisplaced surface has $\eta =0$. Thus a regular single-frequency fluid Lagrangian displacement continuous up to the surface excludes this degenerate branch. The intended ${\omega }^2=0$ conclusion holds with that regular-displacement condition, or with surface corotation excluded; the printed velocity regularity assumption by itself leaves this formal exception.