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132
[HYDROMECHANICS
HYDROMECHANICS

The velocity function of the liquid inside the ellipsoid λ = 0 due to the same angular velocity will be

φ1 = Rxy (a2b2) / (a2 + b2),
(7)

and on the surface outside

φ0 = xyχ0 = xy N   B0 − A0 ,
abc a2b2
(8)

so that the ratio of the exterior and interior value of φ at the surface is

φ0 = B0 − A0 ,
φ1 (a2b2) / (a2 + b2) − (B0 − A0)
(9)

and this is the ratio of the effective angular inertia of the liquid, outside and inside the ellipsoid λ = 0.

The extension to the case where the liquid is bounded externally by a fixed ellipsoid λ = λ1 is made in a similar manner, by putting

φ = xy (χ + M),
(10)

and the ratio of the effective angular inertia in (9) is changed to

(B0 − A0) − (B1 − A1) + a12b12   abc
a12 + b12 a1b1c1
.
a2b2 a12b12   abc − (B0 − A0) + (B1 − A1)
a2 + b2 a12 + b12 a1b1c1
(11)

Make c = ∞ for confocal elliptic cylinders; and then

Aλ = λ ab = ab ( 1 − b2 + λ ),
(a2 + λ) √ (4a2 + λb2 + λ) a2b2 a2 + λ
(12)
Bλ = ab ( √ a2 + λ − 1 ),   Cλ = 0;
a2b2 b2 + λ

and then as above in § 31, with

a = c ch α, b = c sh α, a1 = √ (a2 + λ) = c ch α1, b1 = c sh α1
(13)

the ratio in (11) agrees with § 31 (6).

As before in § 31, the rotation may be resolved into a shear-pair, in planes perpendicular to Ox and Oy.

A torsion of the ellipsoidal surface will give rise to a velocity function of the form φ = xyzΩ, where Ω can be expressed by the elliptic integrals Aλ, Bλ, Cλ, in a similar manner, since

Ω = L λ dλ / P3.

48. The determination of the φ’s and χ’s is a kinematical problem, solved as yet only for a few cases, such as those discussed above.

But supposing them determined for the motion of a body through a liquid, the kinetic energy T of the system, liquid and body, is expressible as a quadratic function of the components U, V, W, P, Q, R. The partial differential coefficient of T with respect to a component of velocity, linear or angular, will be the component of momentum, linear or angular, which corresponds.

Conversely, if the kinetic energy T is expressed as a quadratic function of x1, x2, x3, y1, y2, y3, the components of momentum, the partial differential coefficient with respect to a momentum component will give the component of velocity to correspond.

These theorems, which hold for the motion of a single rigid body, are true generally for a flexible system, such as considered here for a liquid, with one or more rigid bodies swimming in it; and they express the statement that the work done by an impulse is the product of the impulse and the arithmetic mean of the initial and final velocity; so that the kinetic energy is the work done by the impulse in starting the motion from rest.

Thus if T is expressed as a quadratic function of U, V, W, P, Q, R, the components of momentum corresponding are

x1 = dT , x2 = dT , x3 = dT ,
dU dV dW
(1)
y1 = dT , y2 = dT , y3 = dT ;
dP dQ dR

but when it is expressed as a quadratic function of x1, x2, x3, y1, y2, y3,

U = dT , V = dT , W = dT ,
dx1 dx2 dx3
(2)
P = dT , Q = dT , R = dT .
dy1 dy2 dy3

The second system of expression was chosen by Clebsch and adopted by Halphen in his Fonctions elliptiques; and thence the dynamical equations follow

X = dx1 x2 dT + x3 dT , Y = ..., Z = ...,
dt dy3 dy2
(3)
L = dy1 y2 dT + y3 dT x2 dT + x3 dT , M = ..., N = ...,
dt dy3 dy2 dx3 dx2
(4)

where X, Y, Z, L, M, N denote components of external applied force on the body.

These equations are proved by taking a line fixed in space, whose direction cosines are l, m, n, then

dl = mR − nQ,   dm = nP − lR,   dn = lQ − mP.
dt dt dt
(5)

If P denotes the resultant linear impulse or momentum in this direction

P = lx1 + mx2 + nx3,
(6)
dP = dl x1 + dm x2 + dn x3
dt dt dt dt
+ l dx1 + m dx2 + n dx3 ,
dt dt dt
= l ( dx1 x2R + x3Q )
dt
+ m ( dx2 x3P + x1R )
dt
+ n ( dx3 x1Q + x2P )
dt
= lX + mY + nZ,
(7)

for all values of l, m, n.

Next, taking a fixed origin Ω and axes parallel to Ox, Oy, Oz through O, and denoting by x, y, z the coordinates of O, and by G the component angular momentum about Ω in the direction (l, m, n)

G = l (y1x2z + x3y) + m (y2x3x + x1z) + n (y3x1y + x2x).

(8)

Differentiating with respect to t, and afterwards moving the fixed origin up to the moving origin O, so that

x = y = z = 0, but dx = U, dy = V, dz = W,
dt dt dt
dG = l ( dy1 y2R + y3Q − x2W + x3V )
dt dt
+ m ( dy2 y3P + y1R − x3U + x1W )
dt
+ n ( dy3 y1Q + y2P − x1V + x2U )
dt
= lL + mM + nN,
(9)

for all values of l, m, n.

When no external force acts, the case which we shall consider, there are three integrals of the equations of motion

(i.) T = constant, (ii.) x12 + x22+ x32 = F2, a constant, (iii.) x1y1 + x2y2 + x3y3 = n = GF, a constant;

and the dynamical equations in (3) express the fact that x1, x2, x3 are the components of a constant vector having a fixed direction; while (4) shows that the vector resultant of y1, y2, y3 moves as if subject to a couple of components

x2W − x3V, x3U − x1W, x1V − x2U,
(10)

and the resultant couple is therefore perpendicular to F, the resultant of x1, x2, x3, so that the component along OF is constant, as expressed by (iii).

If a fourth integral is obtainable, the solution is reducible to a quadrature, but this is not possible except in a limited series of cases, investigated by H. Weber, F. Kötter, R. Liouville, Caspary, Jukovsky, Liapounoff, Kolosoff and others, chiefly Russian mathematicians; and the general solution requires the double-theta hyperelliptic function.

49. In the motion which can be solved by the elliptic function, the most general expression of the kinetic energy was shown by A. Clebsch to take the form

T = 1/2p (x12 + x22) + 1/2px32+ q (x1y1 + x2y2) + qx3y3+ 1/2r (y12 + y22) + 1/2ry32

(1)

so that a fourth integral is given by

dy3 / dt = 0, y3 = constant;
(2)
dx3 = x1 (qx2 + ry2) − x2 (qx1 + ry1) = r (x1y2x2y1),
dt
(3)
1 ( dx3 ) 2 = (x12 + x22) (y12 + y22) − (x1y1 + x2y2)2
r2 dt  

= (x12 + x22) (y12 + y22) − (FG − x3y3)2 = (x12 + x22) (y12 + y22 + y32 − G2) − (Gx3 − Fy3)2,

(4)

in which

x12 + x22 = F2x32, x1y1 + x2y2 = FG − x3y3,
(5)

r (y12 + y22) = 2T − p(x12 + x22) − px32− 2q (x1y1 + x2y2) − 2qx3y3ry32= (pp′) x32 + 2 (qq′) x3y3 + m1,

(6)
m1 − 2T − pF2 − 2qFG − r1y32
(7)

so that

1 ( dx3 ) 2 = X3
r2 dt  
(8)

where X3 is a quartic function of x3, and thus t is given by an elliptic