= Solution
Away from the rim's $O(h)$ edge region, the geometry and uniform injection have no radial scale other than the factor required by continuity. It is therefore consistent to take $w=w(z)$. Regularity at $r=0$ then lets the continuity equation integrate to
$$
u(r,z)=-\frac r2w'(z).
$$
The <no-slip boundary condition> and prescribed normal velocities are
$$
w(0)=V,\quad w'(0)=0,\qquad
w(h)=0,\quad w'(h)=0.
$$
Since $p$ is independent of $z$, the radial equation implies that $w'''$ is constant. The four boundary conditions give
$$
w(z)=V\left[1-3\left(\frac zh\right)^2+2\left(\frac zh\right)^3\right],
$$
and hence
$$
u(r,z)=\frac{3Vr}{h}\frac zh\left(1-\frac zh\right).
$$
Using $w'''=12V/h^3$ in $p_r=\mu u_{zz}$ gives
$$
p_r=-\frac{6\mu Vr}{h^3}.
$$
Taking the pressure at the rim to be atmospheric, $p(R)=p_0$, produces the <porous-plate lubrication cushion> pressure
$$
\boxed{p(r)-p_0=\frac{3\mu V}{h^3}(R^2-r^2).}
$$
The upward pressure force balances the disc's weight:
$$
W=\int_0^R[p(r)-p_0],2\pi r\,dr
=\frac{3\pi\mu VR^4}{2h^3}.
$$
Since $\varepsilon=h/R$,
$$
\boxed{W=\frac{3\pi\mu VR}{2\varepsilon^3}.}
$$
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