Near the closest point, the cylinder–plane gap is
$$
h(x)=\frac a2\varepsilon+\frac{x^2}{2a}
=\frac a2(\varepsilon+\xi^2),
\qquad
\xi=\frac xa.
$$
For translation parallel to the cylinder axis, the leading flow is <Couette flow>. Its shear traction is $\mu U/h$, so
$$
\frac{F_y}{U}
=\mu L\int_{-\infty}^{\infty}\frac{dx}{h(x)}
=2\mu L\int_{-\infty}^{\infty}
\frac{d\xi}{\varepsilon+\xi^2}
=\boxed{2\pi\varepsilon^{-1/2}\mu L}.
$$
Hence $A_{yy}=2\pi\varepsilon^{-1/2}\mu L$.

For transverse translation, the local <Couette-Poiseuille flow in a thin gap> and mass conservation give the <Reynolds lubrication equation>. Writing $x=a\sqrt\varepsilon\,X$ and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce to
$$
\frac{F_x}{U}
=\frac{\mu L}{\sqrt\varepsilon}
\left(16I_2-16I_3\right)
=\frac{\mu L}{\sqrt\varepsilon}
\left(8\pi-6\pi\right)
+\frac{2\pi\mu L}{\sqrt\varepsilon},
$$
where the final term is the direct Couette shear contribution. Therefore
$$
\boxed{A_{xx}=4\pi\varepsilon^{-1/2}\mu L}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 329 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-329/3/a/solution