Copying is never allowed, even when working together.
Differentiate the equation obtained in the previous question
twice; if your equation is correct, should satisfy the same
equation as for Stokes' first problem. So
too should be a combination of a multiple of the complementary error
function and a constant. Then two integrations should produce your
desired function , if you take account of the boundary
conditions. Your task is now to find a simple expression for in
terms of the (complementary) error function and elementary
functions. Note: you might want to google “repeated
integrals of the error function.” Note: Another way of doing
this is to change the order of integration in the double integrals
you encounter. For example, to find from , the integral
of erfc can be written:
the second equality from the definition of the complementary error
function. If you change the order of integration to integrate
first, you can do the first integral since the
integrand does not depend on . To figure out the new limits
of integration, simply draw the original region of integration in
the plane and look at it. This trick will give you
in terms of a single integral. Repeat the trick to find
from and identify in terms of erfc. Write the velocity field.
For an ideal point vortex at the origin, the velocity field is
given in cylindrical coordinates by
Show that the vorticity of this flow
is everywhere zero. Now sketch a contour (closed curve) that
loops once around the vortex at the origin, in the
counter-clockwise direction. In fluid mechanics, (for any
flow, not just this one), the circulation of
a contour is defined as
Here the integration starts from an arbitrary point on the contour
and loops back to that point in the counter-clocwise direction.
Evaluate the circulation of your contour around the vortex. Do
not take a circle as contour ; take a square or a triangle or
an arbitrary curve. Of course you know that in polar coordinates
an infinitesimal change in position is given by
(If not, you better also figure out what it is in spherical.) You
should find that has a nonzero value for your contour.
So far so good. But the Stokes theorem of Calculus III says
where is an area bounded by contour . You just showed that
the left hand side in this equation is not zero, but that the
right hand side is because is. Something is
horribly wrong???! To figure out what is going on, instead of using
an ideal vortex, use the Oseen vortex from your notes. To simplify
this, now take your contour C to be (the perimeter of) a circle
around the origin in the -plane, and take area to be the
inside of that circle in the -plane. Do both the contour
integral and the area integral. In this case, they should indeed be
equal. Now in the limit , the Oseen vortex becomes
an ideal vortex. So if you look at a very small time, you should be
able to figure out what goes wrong for the ideal vortex with the
Stokes theorem. You might want to plot the vorticity versus for
a few times that become smaller and smaller. Based on that, explain
what goes wrong for . Is the area integral of the
ideal vortex really zero?
Do bathtub vortices have opposite spin in the southern
hemisphere as they have in the northern one? Derive some ballpark
number for the exit speed and angular velocity of a bathtub vortex
at the north pole and one at the south pole, assuming that the bath
water is initially at rest compared to the rotating earth. Use
Kelvin’s theorem. Note that the theorem applies to an inertial
frame, not that of the rotating earth. So assume you look at the
entire thing from a passing star ship. (But define the direction of
rotation as the one someone on earth looking at the bathtub sees.)
What do you conclude about the starting question? In particular,
how do you explain the bathtub vortices that we observe?
Consider a two-dimensional cylindrical balloon
of radius surrounded by an incompressible fluid with an ideal
vortex flow field. If we lower pressure inside the balloon, its
radius decreases. Then there is also an ideal
sink flow field proportional to . The
complete ideal flow field is then:
Now if this is a Newtonian fuid, over time viscous boundary layers
would develop around the surface of the balloon that would propagate
outwards and the rotational motion would slow down. But suppose we
apply just enough of an axial moment on the balloon to keep it
rotating with the ideal flow fluid velocity? Then we have a viscous no-slip flow around a body that is also an ideal
one, i.e. an irrotational flow, i.e. one with zero vorticity.
Sounds interesting?
Is the above flow indeed irrotational?
Integrate the circulation along a circular fluid contour
around the cylinder. What is it?
Is the ring of fluid particles right at the expanding balloon
surface a material contour? Why?
Suppose decreases in time like , what happens
when time increases from to 10 ? In particular, what
does the Kelvin theorem say about what velocity component of the
fluid particles at the surface? And what about the other velocity
component? What happens to the angular velocity at
which the cylinder must rotate?
What is the moment per unit axial length that must be exerted
on the cylinder to keep it rotating at the right speed? (Ignore
the inertia of the balloon.) Does the moment become infinite when
tends to zero? Should it not take more and more effort to
keep the flow rotating when total dissipation increases.