= Solution
Use the quasisteady <lubrication approximation> for the <premelting> film between the disk and the underlying <ice>. The <ice> is anchored to the lower plane; new <ice> formed from <water> supplied through this upper film raises the disk. Let $U$ be its upward separation rate, $d(r)$ the premelted thickness, $\mu$ the liquid <dynamic viscosity>, $\rho$ a common <ice>/<water> <mass density>, $L$ the specific <latent heat> of fusion, and $T_m$ the absolute bulk <melting> <temperature>. As usual in this leading model, neglect the <ice>–<water> density difference, pressure-dependent changes in $T_m$ due to a common phase <pressure>, inertia and edge corrections. The imposed thermal field is taken as maintained by an external heat supply/removal system, so no additional heat-conduction bottleneck is introduced.
Write $\Delta=T_m-T_0>0$, $K_T=\rho L/T_m$ and $\Delta(r)=\Delta(1-r^2/a^2)$. The <Clapeyron pressure relation for equal-density phases> makes the <pressure> difference across the solid–liquid interface
$$
p_i-p_l=K_T\Delta(r)\equiv P_T(r).
$$
This <thermomolecular pressure> is the repulsive stress transmitted through the film. To calculate a definite rate one also needs a film-thickness constitutive law. In the standard nonretarded dispersion-force model, define $A>0$ as the magnitude of the repulsive effective <Hamaker constant>, with its sign convention specified by
$$
\Pi_d(d)=\frac{A}{6\pi d^3}=P_T(r),\qquad
\boxed{d^3(r)=\frac{A}{6\pi K_T\Delta(1-r^2/a^2)}.}
$$
Here $\Pi_d$ is <disjoining pressure>, not the permeability used in the preceding question. The sign of an effective Hamaker coefficient depends on the three materials; the positive parameter here denotes an interaction that stabilizes a premelted film. The problem does not specify $A$ numerically, so it must remain a defined material parameter.
Take the reservoir <pressure> at the rim to be $p_\infty$. With <no-slip boundary conditions> on both sides of the thin film, the radial <volume flux per unit width> is
$$
q_r=-\frac{d^3}{12\mu}\frac{dp_l}{dr}.
$$
A disk moving upward at rate $U$ needs <water> volume $\pi r^2U$ per unit time to freeze below radius $r$. Regularity on the axis and <mass conservation> therefore give $2\pi r q_r=-\pi r^2U$, or $q_r=-Ur/2$. Thus
$$
\frac{dp_l}{dr}=\frac{6\mu Ur}{d^3(r)}.
$$
Define $B=A/(6\pi K_T)$, so $d^3=B/[\Delta(1-r^2/a^2)]$. Integrating inward from $p_l(a)=p_\infty$ gives
$$
\boxed{p_l(r)-p_\infty=-\frac{3\mu U\Delta a^2}{2B}\left(1-\frac{r^2}{a^2}\right)^2.}
$$
<Water> is drawn inward by a <pressure> deficit; its viscous resistance reduces the load that the thermal force can support. The net normal stress on the disk is $p_l-p_\infty+P_T$. Let $W$ be the downward apparent load after subtracting ordinary <buoyancy>. For an ideal thin disk with negligible buoyant volume, $W=Mg$, where $g$ is <gravitational acceleration>. If a finite displaced volume is known, use its submerged weight instead; the stated mass and radius alone do not determine that volume.
The disk <force balance> is
$$
W=2\pi\int_0^a r\,[p_l(r)-p_\infty+P_T(r)]\,dr
=\frac{\pi a^2K_T\Delta}{2}-\frac{\pi\mu U\Delta a^4}{2B}.
$$
Consequently the <premelted-film lifting of a disk> rate is
$$
\boxed{U=\frac{A}{6\pi\mu a^2}\left(1-\frac{2WT_m}{\pi a^2\rho L\Delta}\right).}
$$
For $W=Mg$ this gives the requested expression in terms of the mass. Every additional symbol in this formula has been defined above. The growing-ice lifting branch exists only when the bracket is positive.
At the threshold the required <water> flux and viscous <pressure> deficit vanish. The maximum load supportable at the specified radial thermal profile is therefore
$$
\boxed{W_{\max}=\frac{\pi a^2\rho L\Delta}{2T_m},\qquad M_{\max}=\frac{\pi a^2\rho L\Delta}{2gT_m}\quad\text{if buoyancy is negligible}.}
$$
The factor $1/2$ is the area average of the parabolic undercooling, not an arbitrary geometrical correction. At equality the lift rate is zero; strictly positive lift requires $W<W_{\max}$. There is no radius-only, temperature-independent maximum load in this ideal model: one must specify the imposed undercooling or a physically admissible range of it.
For a fixed positive load define $\Delta_c=2WT_m/(\pi a^2\rho L)$ and $U_\infty=A/(6\pi\mu a^2)$. The lifting branch is
$$
\frac{U}{U_\infty}=1-\frac{\Delta_c}{\Delta},\qquad\Delta>\Delta_c.
$$
It begins at zero, increases with $dU/d\Delta=U_\infty\Delta_c/\Delta^2>0$, and is concave downward because $d^2U/d\Delta^2=-2U_\infty\Delta_c/\Delta^3<0$. Below threshold no positive steady heave is possible: the thermomolecular force is too weak to support the load. The mathematical continuation to negative $U$ is not a prediction of upward ice-driven separation there.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-75-premelted-lift.png]
{title=Premelted-film disk lifting rate: load threshold and limiting speed as undercooling increases}
Increasing undercooling increases the available thermal <pressure>, but it also thins the <premelting> film. In the inverse-cube <disjoining pressure> model, $d^3\propto\Delta^{-1}$, so the hydraulic conductance decreases in exact inverse proportion to the driving scale. The remaining load correction tends to zero and the rate approaches a plateau:
$$
\boxed{\sup_{\Delta>\Delta_c}U=U_\infty=\frac{A}{6\pi\mu a^2}.}
$$
For positive load this is approached from below, not attained at finite undercooling. With zero load the reduced model gives the same constant rate wherever its thin-film assumptions hold.
Two limits of the calculation are worth making explicit. First, the formal $d(r)$ diverges at the warm rim, so a narrow outer connection to bulk <water> replaces the thin-film solution there; this leading model assumes that its contribution to the <integrals> is negligible. Second, very large undercooling produces a molecularly thin film and eventually invalidates continuum <lubrication theory> and the simple dispersion-force law. Thus the plateau is the maximum within the stated ideal constitutive model, not an unrestricted extrapolation for real <water>.
If a different molecular interaction is intended, the result is not universal. For any supplied equilibrium $d(r)$ in the equal-density model the direct force calculation instead gives
$$
U=\frac{\pi a^2K_T\Delta/2-W}{6\pi\mu\displaystyle\int_0^a r^3/d^3(r)\,dr}.
$$
This displays exactly where the film law enters. The closed formula and plateau above use the conventional repulsive inverse-cube law; <premelting> alone, without that material law, would not determine a unique rate curve.
Back to article page
