Page:EB1911 - Volume 17.djvu/1007

988
[KINETICS
MECHANICS

body, called the invariable cone. At any point of this we have x: y:z = Ap. Bq: Cr, and the equation is therefore 2 2

<5>

The signs of the coellicients follow the same rule as in the case of (4). The possible forms of the invariable cone are indicated in fig. 80 by means of the intersections with a concentric spherical surface. In the critical case of

2 BT = 1' 2 the cone degenerates

into two planes. It appears

that if the body be sightly disturbed from a state of rotation

greatest or least moment, the

invariable cone will closely surround this axis, which will

therefore never deviate far

from the invariable line. If,

on the other hand, the body be

slightly disturbed from a state

of rotation about the mean axis

awide deviation willtake place.

of greatest or least moment is

reckoned as stable, a rotation about the mean axis as unstable. The question is greatly simplified when two of the principal moments are equal, say A=B. The polhode and herpolhode cones are then right circular, and the motion is “ processional ” according to the definition of § 18. If a be the inclination of the instantaneous axis to the axis of symmetry, B the inclination of the latter axis to the invariable line, we have I' cos;3=Cw cos a, I' sin 13 = Aw sin a, (6) FIG. So.

Hence a rotation about the axis

whence

tan /3 =-2 tan a. (7)

Hence B 2 a, and the circumstances are therefore those of the first or second case in fig. 78, according as A 2 C. If xl/ be the H - H C

3 c.

Fig. 81.

rate at which the plane HO] revolves about OH, we have ¢ sin aw

sinB

by § 18 (3). Also if 72 be the

C cos a

same ~ <8>

rate at which ] describes the

polhode, we have ip sin (5 -a) =X sin B, whence, sin (a-13)

X sin a. w' (9)

If the instantaneous axis only deviate slightly from the axis of symmetry the angles a, B are small, and X = (A-C) A, .o.>; the instantaneous axis therefore completes its revolution in the body in the period

21- = A - C w, (0

X — A . 1)

at time t, the vector which represents it at time t-l-Bt will be OK', 6

equal to OK in magnitude and making with it an angle 650. Hence KH' (=Cn5p) will represent the change in this component due to the extraneous forces. Hence, so far as this component is concerned, the extraneous forces must supply a couple of moment Cntb in a vertical plane through the axis of the flywheel. If this couple be absent, the axis will be tilted out of the horizontal plane in such a sense that the direction of the spin n approximates to that of the azimuthal rota- K

tion ul/ The remaining constituent of, the extraneous forces is a couple AQ 0 about the vertical; this vanishes if gb is constant. If the axis of the flywheel make an angle 9 with the vertical, it is seen in like manner that the required couple in the vertical plane through the axis is Cn sin H gb. This matter can be strikingly illustrated with an ordinary gyroscope, e.g. by making the larger movable ring in fig. 37 rotate about its vertical diameter.-FIG.

82.

If the direction of the axis of kinetic symmetry be specified by means of the angular co-ordinates 0,111 of § 7, then considering the component velocities of the point C in iig. 83, which are 0 and sin 91/1 along and perpendicular to the meridian ZC, we see that the component angular velocities about the lines

OA', OB' are-sin 0 1,0 and 0 respectively. Hence if the principal moments of inertia at O be A, A, C, and if n be the constant angular velocity about the axis OC, the kinetic energy is given by

2T=A(@2-|-sin”0 552)-I-Cnz. i (I)

Again, the components of angular momentum about OC, OA' are Cn, ~A sin 0¢, and therefore the angular momentum (, u., say) about OZ is

p=A sin' 0 \$4-Cn cos H. (2)

We can hence deduce the condition of steady processional motion in a top. A solid of revolution is supposed to be free to turn about a fixed point O on its axis of symmetry, its mass centre G being in this axis at a distance h from O. In fig. 83 OZ is supposed to be vertical, and OC is the axis of the solid drawn in the direction OG. If 0 is constant the points C, A' will in timeét come to positions C", A” such that CC”=sin 0 61, b, A'A"= cos 6 51//, and the angular momentum about OB' will become Cn sin 6 6¢-A sin 6' //. cos 0 6¢. Equating this to M gh sin 0 iii, and dividing out by sin 0, we obtain A cos 0 1,52-Cn:/;+Mgh=o, (3)

as the condition in question. For given values of n and 0 we have two possible values of 1,0 provided n exceed a certain limit. With a very rapid spin, or (more precisely) with Cn large in comparison with 1/ (4AMgh cos 0), one value of 1,0 is small and the other large, viz. the two values are Mgh/Cn and Cn/A cos 0 approximately. The absence of g from the latter expression indicates that the circumstances of the rapid precession are very