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anonymous
 3 years ago
Are velocity components Vr=5rcos(theta), Vtheta = 5rsin(theta) irrotational?
anonymous
 3 years ago
Are velocity components Vr=5rcos(theta), Vtheta = 5rsin(theta) irrotational?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Umm, irrotational? What do you mean?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ya....is that all I have to do?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hold on, I'm not sure what irrotational is. Can you give me a definition?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well my textbook it as flows for which no particle rotation occurs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay so curl is 0... meaning we need to find if it has a potential function.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta) \]This let's us solve for \(g(\theta)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there might be another method, I'm not completely sure right now.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0My 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
 3 years ago
Best ResponseYou've already chosen the best response.0What's confusing me is if we take into consideration that it is in polar coordinates, and just treat it like Cartesian coordinates.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ya. 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
 3 years ago
Best ResponseYou've already chosen the best response.0I feel like I kind of understand what is going on, but just not quite able to piece it together.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta)\\ g'(\theta) =(5r^2/2+r) (\sin\theta)\implies g = (5r^2/2+r) (\cos\theta)+C \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)\\ f = 5r^2\cos(\theta)+r\cos(\theta) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It has a potential function, that means it's curl is 0, it's conservative, irrotational, etc.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Another 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
 3 years ago
Best ResponseYou've already chosen the best response.0and what he did was gradient x V

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Whoops, did a bit of algebra wrong there.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta)\\ g'(\theta) =(5r^2/25r) (\sin\theta)\implies g(\theta) = (5r^2/2+r) (\cos\theta)+C \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I keep messing up trying to find the damn potential function. Maybe it doesn't have one.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then that cross the the velocity vector of vr and vtheta

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, I don't know those formula unfortunately. The fact you bring them up makes me think we're dealing with polar coords though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We are definitely dealing with polar coordinates, but I think it's still the curl

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{dx}{dt} = \frac{dx}{dr}\frac{dr}{dt} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway I'm going to bed.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, thanks for your help
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