anonymous
  • anonymous
Are velocity components Vr=5rcos(theta), Vtheta = -5rsin(theta) irrotational?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Umm, irrotational? What do you mean?
anonymous
  • anonymous
Irrotational flow
anonymous
  • anonymous
LIke the curl is 0?

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anonymous
  • anonymous
Ya....is that all I have to do?
anonymous
  • anonymous
gradient X V....?
anonymous
  • anonymous
Hold on, I'm not sure what irrotational is. Can you give me a definition?
anonymous
  • anonymous
Well my textbook it as flows for which no particle rotation occurs
anonymous
  • anonymous
Okay so curl is 0... meaning we need to find if it has a potential function.
anonymous
  • anonymous
This is impossible in reality because all fluids have viscosity, but flows can be assumed irrotational in certain cases
anonymous
  • anonymous
\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta) \]
anonymous
  • anonymous
\[ \frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta) \]This let's us solve for \(g(\theta)\)
anonymous
  • anonymous
there might be another method, I'm not completely sure right now.
anonymous
  • anonymous
Okay. Well, I think it has to do with the gradient (del) cross velocity function (V)
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
What's confusing me is if we take into consideration that it is in polar coordinates, and just treat it like Cartesian coordinates.
anonymous
  • anonymous
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
anonymous
  • anonymous
I feel like I kind of understand what is going on, but just not quite able to piece it together.
anonymous
  • anonymous
\[ \frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)\\ g'(\theta) =(5r^2/2+r) (-\sin\theta)\implies g = (5r^2/2+r) (\cos\theta)+C \]
anonymous
  • anonymous
\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)\\ f = 5r^2\cos(\theta)+r\cos(\theta) \]
anonymous
  • anonymous
It has a potential function, that means it's curl is 0, it's conservative, irrotational, etc.
anonymous
  • anonymous
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
anonymous
  • anonymous
and what he did was gradient x V
anonymous
  • anonymous
and got 0.....so I guess that is the same thing as you are saying essentially, right?
anonymous
  • anonymous
Whoops, did a bit of algebra wrong there.
anonymous
  • anonymous
\[ \frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)\\ g'(\theta) =(5r^2/2-5r) (\sin\theta)\implies g(\theta) = -(5r^2/2+r) (\cos\theta)+C \]
anonymous
  • anonymous
I keep messing up trying to find the damn potential function. Maybe it doesn't have one.
anonymous
  • anonymous
I think it is \[del = e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} \]
anonymous
  • anonymous
and then that cross the the velocity vector of vr and vtheta
anonymous
  • anonymous
\[e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} X [ e _{_{r}} 5\cos(\theta) + e _{_{\theta}} -5\sin(\theta)]\]
anonymous
  • anonymous
= 0
anonymous
  • anonymous
yeah, I don't know those formula unfortunately. The fact you bring them up makes me think we're dealing with polar coords though.
anonymous
  • anonymous
We are definitely dealing with polar coordinates, but I think it's still the curl
anonymous
  • anonymous
\[ \frac{dx}{dt} = \frac{dx}{dr}\frac{dr}{dt} \]
anonymous
  • anonymous
anyway I'm going to bed.
anonymous
  • anonymous
Okay, thanks for your help

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