Force-free pressure drop of an annular lubrication plug ($\Delta p=\frac{2\mu }{a}\int (4U/h-6Q/h^2)dx$)

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 $H(-\tilde{L})-H(\tilde{L})=C\lambda \int [4/(3H)-1/H^2]dX$, with $C=s/a$ and $\lambda =6\mu U/(\alpha \sqrt{\kappa }{s}^{5/2})$.