V1 2a² b² - a²
X1 = 0, -- = -------, X = [pi][rho]a²U -------. (22)
U b² + a² b² + a²
But if the outside cylinder is moved with velocity U1, and the inside
cylinder is solid or filled with liquid of density [sigma],
U1 2[rho]b²
X = -[pi][rho]a²U, -- = --------------------------------,
U [rho](b² + a²) + [sigma](b² - a²)
U - U1 ([rho] - [sigma])(b² - a²)
------ = ---------------------------------, (23)
U1 [rho](b² + a²) + [sigma](b² - a²)
and the inside cylinder starts forward or backward with respect to the
outside cylinder, according as [rho] > or < [sigma].
30. The expression for [omega] in (1) § 29 may be increased by the
addition of the term
im log z = -m[theta] + im log r, (1)
representing vortex motion circulating round the annulus of liquid.
Considered by itself, with the cylinders held fixed, the vortex sets
up a circumferential velocity m/r on a radius r, so that the angular
momentum of a circular filament of annular cross section dA is
[rho]mdA, and of the whole vortex is [rho]m[pi](b² - a²).
Any circular filament can be started from rest by the application of a
circumferential impulse [pi][rho]mdr at each end of a diameter; so
that a mechanism attached to the cylinders, which can set up a uniform
distributed impulse [pi][rho]m across the two parts of a diameter in
the liquid, will generate the vortex motion, and react on the cylinder
with an impulse couple -[rho]m[pi]a² and [rho]m[pi]b², having
resultant [rho]m[pi](b² - a²), and this couple is infinite when b =
[oo], as the angular momentum of the vortex is infinite. Round the
cylinder r = a held fixed in the U current the liquid streams past
with velocity
q´ = 2U sin [theta] + m/a; (2)
and the loss of head due to this increase of velocity from U to q´ is
q´² - U² (2U sin [theta] + m/a)² - U²
-------- = ----------------------------, (3)
2g 2g
so that cavitation will take place, unless the head at a great
distance exceeds this loss.
The resultant hydrostatic thrust across any diametral plane of the
cylinder will be modified, but the only term in the loss of head which
exerts a resultant thrust on the whole cylinder is 2mU sin[theta]/ga,
and its thrust is 2[pi][rho]mU absolute units in the direction Cy, to
be counteracted by a support at the centre C; the liquid is streaming
past r = a with velocity U reversed, and the cylinder is surrounded by
a vortex. Similarly, the streaming velocity V reversed will give rise
to a thrust 2[pi][rho]mV in the direction xC.
Now if the cylinder is released, and the components U and V are
reversed so as to become the velocity of the cylinder with respect to
space filled with liquid, and at rest at infinity, the cylinder will
experience components of force per unit length
(i.) - 2[pi][rho]mV, 2[pi][rho]mU, due to the vortex motion;
(ii.) - [pi][rho]a² dU/dt, -[pi][rho]a² dV/dt, due to the kinetic
reaction of the liquid;
(iii.) 0, -[pi]([sigma] - [rho])a²g, due to gravity,
taking Oy vertically upward, and denoting the density of the cylinder
by [sigma]; so that the equations of motion are
dU dU
[pi][rho]a²-- = - [pi][rho]a²-- - 2[pi][rho]mV, (4)
dt dt
dV dV
[pi][rho]a²-- = - [pi][rho]a²-- + 2[pi][rho]mV - [pi]([sigma] - [rho])a²g, (5)
dt dt
or, putting m = a²[omega], so that the vortex velocity is due to an
angular velocity [omega] at a radius a,
([sigma] + [rho])dU/dt + 2[rho][omega]V = 0, (6)
([sigma] + [rho])dV/dt - 2[rho][omega]U + ([sigma] - [rho])g = 0. (7)
Thus with g = 0, the cylinder will describe a circle with angular
velocity 2[rho][omega]/([sigma] + [rho]), so that the radius is
([sigma] + [rho])v/2[rho][omega], if the velocity is v. With [sigma] =
0, the angular velocity of the cylinder is 2[omega]; in this way the
velocity may be calculated of the propagation of ripples and waves on
the surface of a vertical whirlpool in a sink.
Restoring [sigma] will make the path of the cylinder a trochoid; and
so the swerve can be explained of the ball in tennis, cricket,
baseball, or golf.
Another explanation may be given of the sidelong force, arising from
the velocity of liquid past a cylinder, which is encircled by a
vortex. Taking two planes x = ± b, and considering the increase of
momentum in the liquid between them, due to the entry and exit of
liquid momentum, the increase across dy in the direction Oy, due to
elements at P and P´ at opposite ends of the diameter PP´, is
[rho]dy (U - Ua²r^(-2) cos 2[theta] + mr^(-1) sin [theta])(Ua²r^(-2) sin 2[theta] + mr^(-1) cos [theta])
+ [rho]dy (- U + Ua²r^(-2) cos 2[theta] + mr^(-1) sin [theta])(Ua²r^(-2) sin 2[theta] - mr^(-1) cos [theta])
= 2[rho]dymUr^(-1)(cos [theta] - a^2r^(-2)cos 3[theta]), (8)
and with y = b tan [theta], r = b sec [theta], this is