= Solution
\b[Flow visualization and the idealization.] A <streamline> is everywhere tangent to the instantaneous <velocity field> at one fixed time. A <streakline> is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a <pathline>. These coincide geometrically in <steady flow>, but need not do so in an unsteady draining flow.
The intended ideal liquid is approximately incompressible and inviscid: <viscosity> and associated shear stresses are negligible on the scales and times being considered, the <Mach number> is small, and the density is effectively constant. This does not require zero <vorticity>; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.
\b[Spiralling flow and angular momentum.] In an axisymmetric <inviscid flow>, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial <specific angular momentum>,
$$
\frac{D}{Dt}(ru_\phi)=0,\qquad
r^2\omega=r_0^2\omega_0.
$$
The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The <slow-drain angular-momentum amplification> then gives
$$
\boxed{\omega(r)\simeq\omega_0\frac{R^2}{r^2},\qquad
u_\phi(r)\simeq\frac{\omega_0R^2}{r}.}
$$
The assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-1-c12-draining-streaklines.png]
{title=Original quasisteady illustration of inward-spiralling dye streaks outside a finite drain core; the azimuthal speed increases as the radius decreases}
A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_\phi=C/r$ with positive constants $D,C$, giving $\phi=\phi_0-(C/D)\ln(r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.
\b[The free-surface depression.] Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal <centripetal acceleration>. The radial <Euler equation> and vertical <hydrostatic equilibrium> give
$$
\frac1\rho\frac{\partial p}{\partial r}=\frac{u_\phi^2}{r},
\qquad p(r,z)=p_{\rm atm}+\rho g[H(r)-z].
$$
Thus $gH'(r)=u_\phi^2/r=C^2/r^3$, with $C=\omega_0R^2$. Integrating from $r$ to the rim gives
$$
h=H(R)-H(r)=\frac{C^2}{2g}\left(\frac1{r^2}-\frac1{R^2}\right).
$$
For $r\ll R$ this becomes
$$
\boxed{h\simeq\frac{\omega_0^2R^4}{2gr^2}.}
$$
The surface rises outward because the inward <pressure gradient> force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.
\b[The terrestrial estimate.] Use the vertical component of <Earth's rotation> as the ideal initial angular speed, $\omega_0\simeq\Omega_E\sin\lambda$. At Cambridge, taking $\lambda\simeq52.2^\circ$ and $\Omega_E=7.292\times10^{-5}\ \mathrm{s^{-1}}$, this gives $\omega_0\simeq5.76\times10^{-5}\ \mathrm{s^{-1}}$. Hence, for $R=1\ \mathrm m$, $r=0.01\ \mathrm m$ and $g=9.81\ \mathrm{m\,s^{-2}}$,
$$
\boxed{h\simeq1.7\times10^{-6}\ \mathrm m=1.7\ \mathrm{\mu m}.}
$$
Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\ \mathrm{\mu m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.
\b[Curl of the azimuthal velocity.] For axisymmetric $u_\phi=r\omega(r)$ with no vertical dependence, the rotational part of the <vorticity> is vertical:
$$
\boldsymbol\zeta=\nabla\times\mathbf u
=\frac1r\frac{d(ru_\phi)}{dr}\widehat{\mathbf z}
=\boxed{\left(2\omega+r\frac{d\omega}{dr}\right)\widehat{\mathbf z}.}
$$
The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.
\b[The two limiting states.] For initial uniform rotation, $\omega=\omega_0$, whereas the late approximation has $\omega=C/r^2$. Therefore
$$
\boxed{\Omega_{\rm initial}=2\omega_0,\qquad
\Omega_{\rm late}\simeq0\quad(r\ \text{outside the drain core}).}
$$
The late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.
\b[What is actually conserved.] Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the <vorticity equation> is
$$
\frac{D\boldsymbol\zeta}{Dt}=(\boldsymbol\zeta\cdot\nabla)\mathbf u.
$$
Vertical compression of a material column changes its vorticity through <vortex stretching>, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, <vorticity dilution in a shallow draining layer> gives
$$
\boxed{\frac{D}{Dt}\left(\frac{\zeta}{H}\right)=0.}
$$
Thus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.
<Kelvin's circulation theorem> conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi\omega_0R^2$, consistent with the late $u_\phi=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding <shallow-water potential vorticity> is $(\zeta+f)/H$, with $f=2\Omega_E\sin\lambda$; the conserved quantity involves <absolute vorticity>.
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