# Angular momentum in Uniform circular flow

1. Jan 10, 2015

### Jzhang27143

My book says that for uniform rotational flow, the velocity at any point is proportional to r (v = wr.) In vortex flow, the velocity at any point is proportional to 1/r (angular momentum is conserved.) However, in uniform rotational flow, isn't angular momentum also conserved so the same logic applies here? If not, what is changing the angular momentum? I know that physically, uniform rotational flow follows v = wr, but I am confused about angular momentum here.

2. Jan 10, 2015

### Stephen Tashi

It's difficult to parse the phrase "uniform circular flow" because "uniform flow" (I think) means that the velocity is the same at every point in the flow. The velocity is a vector and "circular" suggests a change in direction of velocity as we move to different points in the flow. Is your book talking about total angle momentum being conserved or is it talking about momentum being constant at all points in the flow?

3. Jan 12, 2015

### vanhees71

I guess, it's a good idea to point out the whole problem described in the book. So I don't know, what's really meant.

Obviously there are two different kinds of flow discussed, namely

(a) rigid rotation

The velocity field then is
$\vec{v}(\vec{r})=\vec{\omega} \times \vec{r}$
with $\vec{\omega}=\text{const}$.

(b) "potential vortex"
$\vec{v}(\vec{r})=\frac{C}{2 \pi \rho^2} \begin{pmatrix}-y \\ x \\ 0 \end{pmatrix},$
where $\rho=\sqrt{x^2+y^2}$.

4. Jan 12, 2015

### Stephen Tashi

One can examine whether momentum is constant over all points in these fields (for a uniform distribution of mass), but I don't see any information about conservation or non-conservation of total momentum.