so that the increase of the components of momentum, X1, Y1, and N1,
linear and angular, are
X1 = ([alpha]M´U´ + [xi])[cos] Rt - [alpha]M´U - [xi] - [beta]M´V´ sin Rt
= ([alpha] - [beta])M´Q sin ([theta] - Rt) sin Rt - [xi] ver Rt (3)
Y1 = ([alpha]M´U´ + [xi]) sin Rt + [beta]M´V´ cos Rt - [beta]M´V
= ([alpha] - [beta])M´Q cos ([theta] - Rt) sin Rt + [xi] sin RT, (4)
N1 = [ -([alpha]M´U´ + [xi]) sin ([theta] - Rt) + [beta]M´V´ cos ([theta] - Rt)OO´
= [ -([alpha] - [beta])´Q cos ([theta] - Rt) sin ([theta] - Rt) - [xi] sin ([theta] - Rt)]Qt. (5)
The components of force, X, Y, and N, acting on the liquid at O, and
reacting on the body, are then
X = lt. X1/t = ([alpha] - [beta])M´QR sin [theta] = ([alpha] - [beta])M´VR, (6)
Y = lt. Y1/t = ([alpha] - [beta])M´QR cos [theta] + [xi]R = ([alpha] - [beta])M´UR + [xi]R, (7)
Z = lt. Z1/t = -([alpha] - [beta])M´Q² sin [theta]cos[theta] - [xi]Q sin [theta]
= [-([alpha] - [beta])M´U + [xi]]V. (8)
Now suppose the cylinder is free; the additional forces acting on the
body are the components of kinetic reaction of the liquid
/ dU \ / dV \ dR
-[alpha]M´ ( -- - VR ), -[beta]M´ ( -- + UR ), [epsilon]C´--, (9)
\ dt / \ dt / dt
so that its equations of motion are
/ dU \ / dU \
M ( -- -VR ) = -[alpha]M´ ( -- - VR ) - ([alpha] - [beta])M´VR, (10)
\ dt / \ dt /
/ dV \ / dV \
M ( -- + UR ) = -[beta]M´ ( -- + UR ) - ([alpha] - [beta])M´UR - [xi]R, (11)
\ dt / \ dt /
dR dR
C-- = -[epsilon]C´-- + ([alpha] - [beta])M´UV + [xi]V; (12)
dt dt
and putting as before