# Homework Help: Translating cylinder in viscous fluid

1. May 1, 2010

### MichielM

1. The problem statement, all variables and given/known data
I have a cylinder of radius R translating (in radial direction) with velocity U in a viscous fluid at Re<<1. Find the force exerted on the cylinder per unit length

2. Relevant equations
Re<<1 so the Navier-Stokes equation simplifies to the Stokes equation:
$$\nabla P=\mu \nabla^2 v$$
The cylinder makes use of polar coordinates useful so the stream function for this problem is defined as:
$$\frac{1}{r}\frac{\partial \Psi}{\partial \theta}=u_r$$
$$-\frac{\partial \Psi}{\partial r}=u_{\theta}$$

3. The attempt at a solution
Given the problem statement and the choice for polar coordinates, the boundary conditions become:
$$u_r(R)=U \cos\theta$$
$$u_{\theta}(R)=-U \sin\theta$$
$$u_r(\infty)=u_{\theta}(\infty)=0$$

With help of the continuity equation the Stokes equation can be turned into:
$$\nabla^4 \Psi=0$$
And taking $$\Psi=\sin \theta f(r)$$ the equation further 'simplifies' to:
$$\left[\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}-\frac{1}{r^2}\right]^2f(r)=0$$

A general solution to this equation is $$f(r)=A r^3+Br \ln r + C r+\frac{D}{r}$$

Using the definition of the stream function it follows that:
$$u_r=\frac{f(r)}{r}\cos \theta$$ and $$u_{\theta}=-\sin\theta \frac{\partial f(r)}{\partial r}$$.

The boundary conditions for f(r) then become:
$$f(R)=U R$$
$$\frac{\partial f(R)}{\partial r}=U$$
$$\frac{f(\infty)}{\infty}\cos \theta=0$$
$$-\sin \theta \frac{\partial f(\infty)}{\partial r}=0$$

The two infinity conditions yield $$A=B=C=0$$ and for the other two I find:

$$\frac{D}{R}=U R$$ and $$\frac{-D}{R^2}=U$$

And that is where I get stuck, because I will not be able to satisfy these conditions, so what did I do wrong?

Last edited: May 1, 2010
2. May 20, 2010

### MichielM

mmm, seems like I found the famous Stokes' paradox. For other people reading this and wondering how to solve it: take a look at Oseen's approximation method