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Curl v = 2w_k, help me

  1. Jun 19, 2005 #1
    A fluid rotates with an angular velocity w about the z-axis. The direction of rotation is related to the z-axis by the rigrt hand screw rule.
    a) Find the velocity v of a point in the fluid, and show that
    curl v = 2w_k

    b) If now w is a function of the radius r, show that curl v = zero if w= constant/r square
     
  2. jcsd
  3. Jun 19, 2005 #2

    HallsofIvy

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    First show us what you have done! Certainly if your teacher has asked you to do this, you must to how to find the curl of a vector function- what is the formula?
     
  4. Jun 20, 2005 #3
    what u are asking to find out is vorticity(curl of v). isnt it?

    vorticity=U(say)

    U= del X v...(i)

    find out the cross product using matrix and seperate i,j and k terms...

    take divergence of U but since divergence of curl =0 u will have

    del.U =0

    use stokes theorem to write the integral form of eqn (i)

    vel in theta direction i,e v(theta)=wo r...where w0 can be the angular vel of any point on the fluid and vel along vector r direction i,e vr=0

    find out delXv in polar coordinates and put v(theta)=wo r and vr=0...[note here theta,o and r (for vr)are suffixes]

    u will get the ans 2wk[reqd z component,i guess k is unit vector in z direction]

    I think it will be like this
     
  5. Jun 20, 2005 #4

    OlderDan

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    You still need to find the expression for v, and if you do that you can simply find the curl of v directly using del X v in cartesian coordinates as well as polar.
     
  6. Jun 20, 2005 #5
    question is to find the curl v and not v....what you have in mind?
     
  7. Jun 20, 2005 #6

    OlderDan

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    v is determined by w and r in both cases. You wrote two components of v in terms of w in cylindrical coordinates. You can write v in cartesian coordinates if you choose. However you write it, you can perform the operation del X v without using Stokes theorem or doing any integrals.
     
  8. Jun 20, 2005 #7

    I don't understand. please explain to me again. Thank you.
     
  9. Jun 20, 2005 #8

    OlderDan

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    The problem states that the fluid rotates about the z axis. In the first part, the angular velocity is constant. In the second part the angular velocity is inversely proportional to the square of the distance r from the z axis. In either case, a particle of water is in circular motion about the axis with velocity that depends on where it is in relation to the axis. For rotation about an axis, you can find the velocity if you know the angular velocity and the radius of rotation. In terms of vectors, the velocity under these conditions is

    [tex] \vec v = \vec \omega \times \vec r [/tex]

    In the first case

    [tex] \vec \omega = \omega_k\ \widehat k [/tex]

    In the second case

    [tex] \vec \omega = \frac {\omega_0}{r^2}\ \widehat k [/tex]

    In both cases

    [tex] \vec r = x\ \widehat i +y\ \widehat j +z\ \widehat k [/tex]

    The cross product is going to yield velocities with components parallel to the x-y plane, independent of the z coordinate, circulating about the z-axis. In the first case, velocity will increase with distance from the z-axis. In the second case, velocity will decrease with distance from the z-axis. Remembering that the distance r from the z-axis can be written as a function of x and y, you can find the curl of v

    [tex] \nabla \times \vec v [/tex]

    in Cartesian coordinates. If you prefer, you can write the velocity in cylindrical coordinates and do the problem that way.
     
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