Previously, we solved Laplace's equation using separation of variables for the velocity potential about a cylinder with circulation. The solution was of the form:

\[\phi=c_{1} \theta+v_{o}\left(r+\frac{r_{o}^{2}}{r}\right) \cos \theta\]

and \(c_{1}\) was effectively the strength of the circulation.

a. Show that the transformation below yields the same result for the velocity potential.

What does the equation for the streamlines look like?

\[w=v_{o}\left(z+\frac{r_{o}^{2}}{z}\right)+i c_{1} \log \left(\frac{z}{r_{o}}\right) \quad z=x+i y\]

b. The lift force on the cylinder is given by:

\[F_{x}=-\int_{0}^{2 \pi}\left(P_{s} \cos \theta\right) r d \theta \quad F_{y}=-\int_{0}^{2 \pi}\left(P_{s} \sin \theta\right) r d \theta\]

where \(P_{s}\) is the surface pressure. The surface pressure can be obtained from Bernoulli's equation (see problem 18):

\[P_{s}+\frac{1}{2} ho v_{s}^{2}=P_{o}+\frac{1}{2} ho v_{o}^{2}\]

where \(P_{o}\) is a reference pressure and \(v_{s}\) is the fluid velocity on the surface of the cylinder. Show that the lift force is given by:

\[F_{l i f t}=-2 \pi ho v_{o} c_{1}\]