/d[phi]\^2
(17) 1/2A ( ------ ) sin [gamma]
\ dt /
+ 1/2A[mu]^2 sin [gamma] sin^2 [theta] sin^2 [phi]
- A[mu]^2 cos [gamma] sin [theta] cos [theta] sin [phi]
+ (K[mu] + gMh) sin [theta] sin [phi] - H = 0,
and putting tan (1/4[pi] + 1/2[phi]) = z, this reduces to
dz
(18) -- = n [root]Z
dt
where Z is a quadratic in z^2, so that z is a Jacobian elliptic
function of t, and we have
(19) tan (1/4[pi] + 1/2[phi]) = C(tn, dn, nc, or cn)nt,
according as the ring ZC performs complete revolutions, or oscillates
about a sidelong position of equilibrium, or oscillates about the
stable position of equilibrium [phi] = [+-]1/2[pi].
Suppose Oz is parallel to the earth's axis, and [mu] is the diurnal
rotation, the square of which may be neglected, then if Gilbert's
barogyroscope of S 6 has the knife-edges turned in azimuth to make an
angle [beta] with E. and W., so that OZ lies in the horizon at an
angle E.[beta].N., we must put [gamma] = 1/2[pi], cos [theta] = sin
[alpha] sin [beta]; and putting [phi] = 1/2[pi] - [delta] + E, where
[delta] denotes the angle between Zz and the vertical plane Z[zeta]
through the zenith [zeta],
(20) sin [theta] cos [delta] = cos [alpha], sin [theta] sin [delta]
= sin [alpha] cos [beta];
so that equations (9) and (10) for relative equilibrium reduce to
(21) gMh sin E = KQ = K[mu] sin [theta] cos [phi]
= K[mu] sin [theta] sin ([delta] - E),
and will change (3) S 6 into
K[mu] sin [alpha] cos [beta]
(22) tan E = ----------------------------,
gMh + K[mu] cos [alpha]
a multiplication of (3) S 6 by cos [beta] (Gilbert, _Comptes rendus_,
1882).
Changing the sign of K or h and E and denoting the revolutions/second
of the gyroscope wheel by F, then in the preceding notation, T
denoting the period of vibration as a simple pendulum,
K[mu] sin [alpha] cos [beta] F sin [alpha] cos [beta]
(23) tan E = ----------------------------- = ----------------------------,
gMh - K[mu] cos [alpha] 86400 A/T^2C - F cos [alpha]
so that the gyroscope would reverse if it were possible to make F cos
[alpha] > 86400 A/T^2C (Foppl, _Munch. Ber_, 1904).
A gyroscopic pendulum is made by the addition to it of a fly-wheel,
balanced and mounted, as in Gilbert's barogyroscope, in a ring movable
about an axis fixed in the pendulum, in the vertical plane of motion.
As the pendulum falls away to an angle [theta] with the upward
vertical, and the axis of the fly-wheel makes an angle [phi] with the
vertical plane of motion, the three components of angular momentum are
(24) h1 = K cos [phi], h2 = A .[theta] + K sin [phi], h3 = A .[phi],
where h3 is the component about the axis of the ring and K of the
fly-wheel about its axis; and if L, M', N denote the components of the
couple of reaction of the ring, L may be ignored, while N is zero,
with P = 0, Q = [.[theta]], R = 0, so that
(25) M' = h2 = A :[theta] + K .[phi] cos [phi],
(26) 0 = h3 - h1 .[theta] = A :[phi] - K .[theta] cos [phi].
For the motion of the pendulum, including the fly-wheel,
(27) MK^2 :[theta] = gMH sin [theta] - M'
= gMH sin [theta] - A :[theta] - K .[phi] cos [phi].
If [theta] and [phi] remain small,
(28) A :[phi] = K .[theta], A .[phi] = K([theta] - [alpha]),
(29) (MK^2 + A) :[theta] + (K^2/A) ([theta] - [alpha]) - gMH[theta] = 0;
so that the upright position will be stable if K^2 > gMHA, or the
rotation energy of the wheel greater than 1/2A/C times the energy
acquired by the pendulum in falling between the vertical and
horizontal position; and the vibration will synchronize with a simple
pendulum of length
(30) (MK^2 + A)/[(K^2/gA) - MH].
This gyroscopic pendulum may be supposed to represent a ship among
waves, or a carriage on a monorail, and so affords an explanation of
the gyroscopic action essential in the apparatus of Schlick and
Brennan.
General motion of the top.
8. Careful scrutiny shows that the steady motion of a top is not
steady absolutely; it reveals a small nutation superposed, so that a
complete investigation requires a return to the equations of unsteady
motion, and for the small oscillation to consider them in a
penultimate form.
In the general motion of the top the vector OH of resultant angular
momentum is no longer compelled to lie in the vertical plane COC'
(fig. 4), but since the axis Oh of the gravity couple is always
horizontal, H will describe a curve in a fixed horizontal plane
through C. The vector OC' of angular momentum about the axis will be
constant in length, but vary in direction; and OK will be the
component angular momentum in the vertical plane COC', if the planes
through C and C' perpendicular to the lines OC and OC' intersect in
the line KH; and if KH is the component angular momentum perpendicular
to the plane COC', the resultant angular momentum OH has the three
components OC', C'K, KH, represented in Euler's angles by
(1) KH = A d[theta]/dt, C'K = A sin [theta] d[psi]/dt, OC' = G'.