= Solution
{c}
Let $F_x,F_y$ denote the <force per unit length> exerted by the fluid on the <cylinder>, and let $\rho$ be the constant <mass density>. For a counterclockwise contour $C$ around the unit <circle>, the <Blasius force theorem> gives
$$
F_x-iF_y=\frac{i\rho}{2}\oint_C q(z)^2\,dz,
\qquad
q(z)=U-\frac U{z^2}+\frac1{z-b}+\frac1{z-a}-\frac1z,
\quad a=\frac1b.
$$
An efficient use of the <residue theorem> finds the sum of the enclosed <residues> by subtracting the external <pole> at $b$ from the sum of all finite <residues>. At large $z$,
$$
q(z)=U+\frac1z+O(z^{-2}),\qquad
q(z)^2=U^2+\frac{2U}{z}+O(z^{-2}).
$$
The sum of all finite <residues> of $q^2$ is therefore $2U$. At $b$, write $q=(z-b)^{-1}+H(z)$, with $H$ a <holomorphic function> near $b$. Then
$$
\operatorname{Res}_{z=b}q^2=2H(b),\qquad
H(b)=U-\frac U{b^2}+\frac1{b-a}-\frac1b.
$$
Thus
$$
\oint_Cq^2\,dz=2\pi i\,[2U-2H(b)],
\qquad
F_x-iF_y=2\pi\rho[H(b)-U].
$$
Using $a=1/b$ and $1/(b-a)-1/b=1/[b(b^2-1)]$ gives
$$
\boxed{F_x=2\pi\rho\left(\frac1{b(b^2-1)}-\frac U{b^2}\right),
\qquad F_y=0}.
$$
The vanishing vertical <force> also follows from <reflection symmetry> about the real axis. This <force on a unit cylinder due to a uniform flow and an external source> uses the counterclockwise orientation in the <Blasius force theorem> to fix the sign of the horizontal <force>.
Back to article page
