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Particle kinematics

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Aug 4, 2009 #2

    Mark44

    Staff: Mentor

    Can you elaborate a bit on what Ux2 and Ux1 are, and how they relate to U?
     
  4. Aug 5, 2009 #3
    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 dont understand how they came to this answer.
     
  5. Aug 5, 2009 #4

    cristo

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    Staff Emeritus
    Science Advisor

    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]?
     
  6. Aug 5, 2009 #5
    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?
     
  7. Aug 5, 2009 #6

    cristo

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    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.

    No: the derivative of [itex]\vec{x}[/itex], the displacement vector is the velocity. that is, [tex]\dot{\vec{x}} \equiv \vec{v}[/tex].
     
  8. Aug 5, 2009 #7
    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)?
     
  9. Aug 5, 2009 #8

    cristo

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    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.

    You use the fact that [tex]\dot{\vec{x}}\equiv\vec{v}[/tex] to substitute for the x dots in your differentiated expression.
     
  10. Aug 5, 2009 #9
    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?
     
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