Solution
= Solution
Lubrication gives $u_r=(p_r/2\mu)z(z-h)$ and continuity gives $\int_0^h u_rdz=-Vr/2$. Hence $p_r=6\mu Vr/h^3$ and $p-p_0=3\mu V(r^2-a^2)/h^3$. Balancing the upward <pressure> force with $W$ yields
$$
\dot h=-\frac{2Wh^3}{3\pi\mu a^4}.
$$
\b[Therefore $h^{-2}=h_0^{-2}+4Wt/(3\pi\mu a^4)$ and $h\sim(3\pi\mu a^4/4Wt)^{1/2}$.]