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Vorticity where angular velocity is function of r

  1. Oct 30, 2004 #1
    I am given, in each of three cases, an angular velocity [tex]\Omega(r)[/tex] and am told to assume no axial (z) velocity i.e., [tex]u_z = 0[/tex]. I am asked to

    (1) find the velocity field in cartesian coordinates
    (2) find the vorticity distribution in threee cases.

    (1) As setup, the problem asks me to "Show the velocity components are given by [tex](u_x,u_y,u_z) = (-\Omega(r), \Omega(r), 0)[/tex]".

    This doesn't seem possible to me....how can both x and y components of the velocity field be the same? I keep coming up with the like of, for a given r:

    [tex]u_x = - r * \Omega(r) * \sin(\theta)[/tex]​
    [tex]u_x = r * \Omega(r) * \cos(\theta)[/tex]​
    where [tex]\theta = \Omega(r) * t;
    Nevertheless, the supplied answer to (1) is [tex](u_x, u_y, u_z) = (-\Omega(r), \Omega(r), 0)[/tex] and it seems to contradict my later assertion that (in cyl coords) [tex]u_r = 0[/tex].

    For (2), I am asked to find the vorticity in three cases:

    a) [tex]\Omega = q/r[/tex] (typ. flow around strong concentrated vortex)

    b) [tex]\Omega r^2 = constant = k [/tex](fluid parcels slowly spiraling towards origin while conserving angular momentum)

    c) [tex]\Omega^2 r = G*M/r^2[/tex] (velocity distribution inside accretion disk in black hole or neutron star)

    So, I chose to ignore the bogus part (1) and solve in cyl coord where [tex]v = (u_r, u_{\theta}, u_z)[/tex]. For all three cases I assumed that [tex]u_z = 0[/tex] (given) and [tex]u_r = 0[/tex] ("flow is the form a circular 'swirl' about the origin in the x-y plane"). (If [tex]u_r[/tex] is not zero, I haven't a clue as to how to come up with a u_r.....)

    Then, blithely proceeding:

    * In cyl. coor, [tex]u_{\theta} = r * \Omega(r)[/tex], where [tex]\Omega(r) =[/tex] angular velocity.
    * Then, vorticity [tex]w = \nabla \times v[/tex] is [tex]w_z[/tex] only and reduces to
    [tex]w_z = \frac {1} {r} \frac {d} {dr} (r * u_{\theta}) = \frac {1} {r} \frac {d} {dr}(r * r * \Omega(r))[/tex]

    Using this approach I get,

    a) [tex]\Omega(r) = q/r[/tex] gives [tex]w = \frac {1} {r} \frac {d} {dr}(r * r * q/r) = \frac {1} {r} \frac {d} {dr}(r*q) = q/r[/tex]
    b) [tex]\Omega r^2 = k \Rightarrow \Omega(r) = k/ r^2 \Rightarrow
    w = \frac {1} {r} \frac {d} {dr}(r * r * k / r^2) = \frac {1} {r} \frac {d} {dr}(k) = 0[/tex]
    c) [tex]\Omega^2 r = G*M/r^2 \Rightarrow \Omega(r) = \frac {k_2} {r^{3/2}} \Rightarrow w = \frac {1} {r} \frac {d} {dr}(r * r * \frac {k_2} {r^{3/2}}) = \frac {k_2} {r} \frac {d} {dr}(r^{1/2}) = \frac {k_2} {2} r^{-3/2}[/tex]

    Am I even close here?

    Confused in Seattle,

    Last edited: Oct 30, 2004
  2. jcsd
  3. Oct 31, 2004 #2


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    Science Advisor
    Gold Member

    I'm not checked your solution, but the statement of your book (see the quoting) is wrong. That velocities haven't got dimensions of speed (they have dimensions of angular velocity). So I wouldn't pay much attention at that solution the book provides you.

    Desconfusing in Madrid. :biggrin:
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