The inviscid irrotational flow around a circular cylinder of radius $a$ is described by the complex potential $$w =Uz+ \frac{Ua^2}{z}$$ where $U$ is a positive constant.
I found $\psi=U \cos \theta (r+a^2/r)$ and $\phi= U\sin \theta (r-a^2/r)$
And $u_r=U\cos \theta (a^2/r^2 +1)$ and $u_{\theta}=U\sin \theta (a^2/r^2 -1)$
Use Bernoulli’s theorem to determine the pressure on $r =a$. Explain why this result in unphysical.
Is this what we are supposed to do: By Bernoulli theorem, we have $$\frac p{\rho} +\textbf u ^2/2= \text{constant}$$ Are we then meant to sub $\textbf u $ as $u_{r}$ for some reason to get: $$p=\frac{\rho}2 u_r +C$$ $$p(r=a, \theta)=2U^2 \rho \cos^2 \theta+C$$ Is that what we are meant to do to find the pressure on $a$?
I am unsure about the second part of the question.
You have the potential and stream function switched. The correct velocity components are
$$u_r = U \cos \theta(1 - a^2/r^2), \\ u_\theta = -U \sin \theta(1 +a^2/r^2).$$
At the surface of the cylinder $r = a$ we have $u_r|_{r=a} = 0$ and $u_\theta|_{r=a} = -2U \sin \theta$.
Using Bernoulli's equation the pressure on the surface is
$$p = C + \frac{\rho}{2}|\mathbb{u}|^2|_{r=a} = C + \frac{\rho}{2}(u_r^2|_{r=a} +u_\theta^2|_{r = a}) = C- 2 \rho U^2 \sin^2 \theta.$$
The result is unphysical because the pressure distribution is symmetrical. The values of pressure at the front ($\theta = 0$) and rear ($\theta = \pi$) stagnation points are equal. In reality for viscous flow, the pressure is lower in the wake behind the cylinder resulting in a net resultant force (form drag).