# Radial component of velocity in a Rankine-vortex

A Rankine vortex is a model of a vortex that assumes an inner region ($$r) with constant vorticity, $$\omega_0$$, which rotates as a rigid solid, and an outer region ($$r>R$$) with zero vorticity, $$\omega = 0$$, which behaves as an irrotational flow.

I'm trying to derive the velocity field of this model in two dimensions through the expression of the vorticity in plane polar coordinates,

$$\omega=\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{\theta}\right)-\frac{1}{r} \frac{\partial v_{r}}{\partial \theta}$$

I have seen that the radial component of the velocity field in this model is zero: $$v_r=0$$. Is this part of the definition of the model or a consequence of its postulates about the vorticity?

Generally I'd say it is just defined by the velocity field $$\boldsymbol{v}$$, but even so you can derive it if you are willing to accept that it is cylindrically symmetric.
Cylindrical symmetry means that $$\boldsymbol{v}$$ must be independent of $$\theta$$ which gives $$\omega=\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{\theta}\right).$$ In the inner region, $$\omega=\omega_0$$ and so we need $$v_{\theta}=\frac{1}{2}\omega_0r+\frac{C}{r}$$ where $$C$$ is a constant. However, the velocity must be finite as $$r\rightarrow 0$$ and so $$C=0$$. In the outer region, $$\omega=0$$ which gives $$v_{\theta}\propto{1/r}$$.
Now we just need to argue that $$v_r$$ is zero. We can do this by mass conservation. The mass flux through a cylinder of unit length and radius $$r$$ is $$F=2\pi r v_r$$ where we have used that $$v_r$$ is independent of $$\theta$$. $$F$$ must be the same for all $$r$$, so by considering $$r\rightarrow0$$ with $$v_r$$ finite, we see that $$F=0$$ and so $$v_r=0$$ for all $$r$$.