Question

Are velocity components Vr=5rcos(theta), Vtheta = -5rsin(theta) irrotational?

Answer 1

Umm, irrotational? What do you mean?

Answer 2

Irrotational flow

Answer 3

LIke the curl is 0?

Answer 4

Ya....is that all I have to do?

Answer 5

gradient X V....?

Answer 6

Hold on, I'm not sure what irrotational is. Can you give me a definition?

Answer 7

Well my textbook it as flows for which no particle rotation occurs

Answer 8

Okay so curl is 0... meaning we need to find if it has a potential function.

Answer 9

This is impossible in reality because all fluids have viscosity, but flows can be assumed irrotational in certain cases

Answer 10

\[
\frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)
\]

Answer 11

\[
\frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)
\]This let's us solve for \(g(\theta)\)

Answer 12

there might be another method, I'm not completely sure right now.

Answer 13

Okay. Well, I think it has to do with the gradient (del) cross velocity function (V)

Answer 14

The only thing I am confused about is whether there is a different between the irrotational flow and the incompressible flow. It seems like if I find that there is an incompressible flow, it's automatically going to give me an irrotational flow

Answer 15

My textbook defines the curl as (1/2)(del) X V.....but in rectangular coordinates it's simply defined as (1/2) (dv/dx - du/dx)

Answer 16

What's confusing me is if we take into consideration that it is in polar coordinates, and just treat it like Cartesian coordinates.

Answer 17

Ya. I mean every example I see in my book seems to look different ha....for example, for Vr = 0 and Vtheta = f(r) (1/r)* d/dr(r*Vtheta) = 0 for irrotational flow

Answer 18

I feel like I kind of understand what is going on, but just not quite able to piece it together.

Answer 19

\[

\frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)\\
g'(\theta) =(5r^2/2+r) (-\sin\theta)\implies g = (5r^2/2+r) (\cos\theta)+C
\]

Answer 20

\[
\frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)\\
f = 5r^2\cos(\theta)+r\cos(\theta)
\]

Answer 21

It has a potential function, that means it's curl is 0, it's conservative, irrotational, etc.

Answer 22

Another example my prof. gave in class was this:

Given: velocity field V= \[\frac{ -q }{2 \Pi r } e _{r} + \frac{ K }{ 2 \Pi r } e _{}\] is it irrotational

Answer 23

and what he did was gradient x V

Answer 24

and got 0.....so I guess that is the same thing as you are saying essentially, right?

Answer 25

Whoops, did a bit of algebra wrong there.

Answer 26

\[
\frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)\\
g'(\theta) =(5r^2/2-5r) (\sin\theta)\implies g(\theta) = -(5r^2/2+r) (\cos\theta)+C
\]

Answer 27

I keep messing up trying to find the damn potential function. Maybe it doesn't have one.

Answer 28

I think it is \[del = e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} \]

Answer 29

and then that cross the the velocity vector of vr and vtheta

Answer 30

\[e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} X [ e _{_{r}} 5\cos(\theta) + e _{_{\theta}} -5\sin(\theta)]\]

Answer 31

= 0

Answer 32

yeah, I don't know those formula unfortunately. The fact you bring them up makes me think we're dealing with polar coords though.

Answer 33

We are definitely dealing with polar coordinates, but I think it's still the curl

Answer 34

\[
\frac{dx}{dt} = \frac{dx}{dr}\frac{dr}{dt}
\]

Answer 35

anyway I'm going to bed.

Answer 36

Okay, thanks for your help