= Force-free pressure drop of an annular lubrication plug
{title2=$\Delta p=\frac{2\mu}{a}\int(4U/h-6Q/h^2)\,dx$}
The pressure force on a plug-and-film control volume is $\pi a^2\Delta p$. Its opposing wall traction is $2\pi a\mu\int w_y(0)\,dx$, where $w_y(0)=4U/h-6Q/h^2$. Equating them for a freely moving body proves the displayed formula. Writing $s=2Q/U$, $h=sH$ and $x=\sqrt{s/\kappa}X$ converts it into $H(-\widetilde L)-H(\widetilde L)=C\lambda\int[4/(3H)-1/H^2]dX$, with $C=s/a$ and $\lambda=6\mu U/(\alpha\sqrt\kappa s^{5/2})$.
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