Finding acceleration and streamlines for helical fluid flow

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sara_87
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Homework Statement



The velocity in a steady helical flow of a fluid is given by:
v1= -Ux2, v2=Ux1, v3=V
where U and V are constant.
show that divv is 0 and find the acceleration of the particle at x.
Also determine the streamlines.

Homework Equations





The Attempt at a Solution



to find the divergence, if we differentiate v1 wrt x1, we get 0, and differentiating v2 wrt x2, we get 0 and differentiating v3 wrt x3, we get 0 so the devergence is 0.
but i am having problem finding the acceleration. i know the acceleration is differentiating the velocity with respect to time but what do they mean 'find the acceleration of the particle at x.' ??
Thank you.
 
Ux1 and -Ux2 are the components of the velocity vector. U is just a constant.

By the way, the answer to the accelartion part is:
f1= -U^2(x1), f2= -U^2(x2), f3=0
where f1, f2, f3 are the components of the acceleration. But i don't understand how they came to this answer.
 
You have a position vector [tex]\vec{x}=(x_1,x_2,x_3)[/tex] and a velocity vector [tex]\vec{v}=(-Ux_1,Ux_2,V)[/tex], where U and V are constant.

What is the derivative of [itex]\vec{v}[/itex] with respect to time? That is, for the first component, what is the time derivative of [itex]-Ux_1[/itex] where U is a constant?

Then, how is [tex]\dot{\vec{x}}[/tex] related to [tex]\vec{v}[/tex]?
 
when we differentiate -Ux1 with respect to time, we get zero. but i think that is wrong because x1 might be depending on the value of t.

Is the derivative of vector x the integral of vector v?
 
sara_87 said:
but i think that is wrong because x1 might be depending on the value of t.

Exactly, so the derivative of something like [tex]-Ux_1[/tex] is [tex]-U\dot{x_1}[/tex], since x_1 is a function of time.

Is the derivative of vector x the integral of vector v?

No: the derivative of [itex]\vec{x}[/itex], the displacement vector is the velocity. that is, [tex]\dot{\vec{x}} \equiv \vec{v}[/tex].
 
Oh right i see. so the way i came to the conclusion that div(v)=0 was wrong.
so for the acceleration, i differentiate v with respect to time.
but then where does the U^2 come from (in the answer)?
 
sara_87 said:
Oh right i see. so the way i came to the conclusion that div(v)=0 was wrong.

Yea, that's wrong: I didn't read that part.

The divergence is defined as [tex]\vec{\nabla}\cdot \vec{A} = (\frac{\partial A_1}{\partial x_1}, \frac{\partial A_2}{\partial x_2},\frac{\partial A_3}{\partial x_3})[/tex]. Plugging [itex]\vec{v}[/itex] into this will give the result.

so for the acceleration, i differentiate v with respect to time.
but then where does the U^2 come from (in the answer)?

You use the fact that [tex]\dot{\vec{x}}\equiv\vec{v}[/tex] to substitute for the x dots in your differentiated expression.
 
oh right i see, ofcourse i have substitute, that makes a lot of sense.Thank you.

How do i find the streamlines?
The answer says: helices given parametrically by x1= Acos(Ut) + Bsin(Ut),
x2=Asin(Ut) - Bcos(Ut), x3=Vt+C
where A,B,C are constants.
I understand that x3 is the integral of v3 but how then did they derive x1 and x2?