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Are velocity components Vr=5rcos(theta), Vtheta = 5rsin(theta) irrotational?
 one year ago
 one year ago
Are velocity components Vr=5rcos(theta), Vtheta = 5rsin(theta) irrotational?
 one year ago
 one year ago

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wioBest ResponseYou've already chosen the best response.1
Umm, irrotational? What do you mean?
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
Ya....is that all I have to do?
 one year ago

wioBest ResponseYou've already chosen the best response.1
Hold on, I'm not sure what irrotational is. Can you give me a definition?
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
Well my textbook it as flows for which no particle rotation occurs
 one year ago

wioBest ResponseYou've already chosen the best response.1
Okay so curl is 0... meaning we need to find if it has a potential function.
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
This is impossible in reality because all fluids have viscosity, but flows can be assumed irrotational in certain cases
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta) \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta) \]This let's us solve for \(g(\theta)\)
 one year ago

wioBest ResponseYou've already chosen the best response.1
there might be another method, I'm not completely sure right now.
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
Okay. Well, I think it has to do with the gradient (del) cross velocity function (V)
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
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
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
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)
 one year ago

wioBest ResponseYou've already chosen the best response.1
What's confusing me is if we take into consideration that it is in polar coordinates, and just treat it like Cartesian coordinates.
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
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
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
I feel like I kind of understand what is going on, but just not quite able to piece it together.
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta)\\ g'(\theta) =(5r^2/2+r) (\sin\theta)\implies g = (5r^2/2+r) (\cos\theta)+C \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)\\ f = 5r^2\cos(\theta)+r\cos(\theta) \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
It has a potential function, that means it's curl is 0, it's conservative, irrotational, etc.
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
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
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
and what he did was gradient x V
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
and got 0.....so I guess that is the same thing as you are saying essentially, right?
 one year ago

wioBest ResponseYou've already chosen the best response.1
Whoops, did a bit of algebra wrong there.
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta)\\ g'(\theta) =(5r^2/25r) (\sin\theta)\implies g(\theta) = (5r^2/2+r) (\cos\theta)+C \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
I keep messing up trying to find the damn potential function. Maybe it doesn't have one.
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
I think it is \[del = e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} \]
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
and then that cross the the velocity vector of vr and vtheta
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
\[e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} X [ e _{_{r}} 5\cos(\theta) + e _{_{\theta}} 5\sin(\theta)]\]
 one year ago

wioBest ResponseYou've already chosen the best response.1
yeah, I don't know those formula unfortunately. The fact you bring them up makes me think we're dealing with polar coords though.
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
We are definitely dealing with polar coordinates, but I think it's still the curl
 one year ago

wioBest ResponseYou've already chosen the best response.1
\[ \frac{dx}{dt} = \frac{dx}{dr}\frac{dr}{dt} \]
 one year ago

flumech2Best ResponseYou've already chosen the best response.0
Okay, thanks for your help
 one year ago
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