Axisymmetric <mass conservation> in the spherical gap gives
$$
\frac{\partial h}{\partial t}
+\frac1{a\sin\theta}
\frac{\partial}{\partial\theta}(q\sin\theta)=0.
$$
Since $h_t=-V\cos\theta$ and regularity requires $q\sin\theta=0$ at $\theta=0$, integration gives
$$
q\sin\theta
=aV\int_0^\theta\sin\vartheta\cos\vartheta\,d\vartheta
=\frac12aV\sin^2\theta,
\qquad
\boxed{q=\frac12Va\sin\theta}.
$$

The leading <Poiseuille flow>[pressure-driven lubrication flux] is
$$
q=-\frac{h^3}{12\mu a}\frac{dp}{d\theta}.
$$
Substitution of $h$ and $q$ gives
$$
\frac{dp}{d\theta}
=-\frac{6\mu a^2V\sin\theta}
{\Delta^3(1-\lambda\cos\theta)^3},
$$
and therefore
$$
\boxed{
p(\theta)=
\frac{3\mu a^2V}
{\lambda\Delta^3(1-\lambda\cos\theta)^2}+p_0}.
$$

Take downward as the positive vertical direction. The constant pressure contributes no resultant, while the pressure force on the inner sphere is opposite its outward normal. Thus
$$
F_z=-2\pi a^2\int_0^\pi
(p-p_0)\cos\theta\sin\theta\,d\theta.
$$
With $t=\lambda\cos\theta$ and the supplied integral,
$$
\boxed{
F_z=-\frac{6\pi\mu a^4V}{\lambda^3\Delta^3}
\left[
\frac{2\lambda}{1-\lambda^2}
+\log\left(\frac{1-\lambda}{1+\lambda}\right)
\right]}.
$$
The sign is upward for $V>0$, so this is a <drag force>. As $\lambda\to0$, the bracket is $4\lambda^3/3+O(\lambda^5)$ and $F_z\to-8\pi\mu a^4V/\Delta^3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 1 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-329/1/i/solution