Treat each short torus segment as a straight cylinder. For translation with velocity $U\mathbf e_x$, its components normal and tangent to the circular centreline are $U\cos\theta$ and $-U\sin\theta$. Integrating the local resistance per unit length around $ds=R\,d\theta$ gives
$$
\begin{aligned}
A_{xx}
&=\mu\varepsilon^{-1/2}R
\int_0^{2\pi}
(4\pi\cos^2\theta+2\pi\sin^2\theta)\,d\theta\\
&=\boxed{6\pi^2\mu R\varepsilon^{-1/2}}.
\end{aligned}
$$

Rotation at angular velocity $\Omega\mathbf e_z$ gives the everywhere tangential speed $\Omega R$. Only the axial-cylinder resistance $2\pi\mu\varepsilon^{-1/2}$ contributes. Multiplying the force by its moment arm $R$ and integrating around the centreline gives
$$
\boxed{D_{zz}=4\pi^2\mu R^3\varepsilon^{-1/2}}.
$$


Solution

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

 Solution

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