Page:EB1911 - Volume 12.djvu/803

these three pairs of equations may be replaced by the three equations

Aῶ.._k − Kῶ.._ki + 2Taῶ_k − Ta (σk+1 + σ_k) = 0,

(10)

Mῶ.._k − T (σk+1 − σ_k) = 0,

(11)

ωk+1 − ω_k − a(ῶk+1 + ῶ_k − 2lσk+1) = 0.

(12)

For a vibration of circular polarization assume a solution

ω_k, ῶ_k, σ_k = (L, P, Q) exp (nt + kc) i,

(13)

so that c/n is the time-lag between the vibration of one fly-wheel and the next; and the wave velocity is

U = 2 (a + l) n/c.

(14)

Then

P (−An² + Kn + 2Ta) − QTa (eci + 1) = 0,

(15)

−LMn² − QT (eci − 1) = 0,

(16)

L (eci − 1) − Pa (eci + 1) − 2Qleci = 0,

(17)

leading, on elimination of L, P, Q, to

cos c =	(2 Ta + Kn − An2) (1 − Mn2l/T) − Mna2	,
	2Ta + Kn − An2 + Mna2

(18)

2 sin2 1/2c =	Mn2 2Ta (a + l) + KNl − An2l	.
	T 2Ta + Kn − An2 + Mn2a2

(19)

With K = 0, A = 0, this reduces to Lagrange’s condition in the vibration of a string of beads.

Putting

ρ = M/2 (a + l),   the mass per unit length of the chain,⁠

(20)

κ = K/2 (a + l),   the gyrostatic angular momentum per unit length,

(21)

α = A/2 (a + l),   the transverse moment of inertia per unit length,

(22)

1/2c = (a + l) n/U,

(23)

equation (19) can be written

{sin (a + l) n/U}²

= (a + l)2n2	ρ	·	Ta + κnl − αn2l	,
	T		Ta + κn (a + l) − αn2 (a + l) + ρn2a2 (a + l)

(24)

{	(a + l) n	}	2
	sin (a + l) n/U

=	T	·	T + (κn − αn2) (1 + l/a) + ρn2a (a + l)	.
	ρ		T + (κn − an2) l/a

(25)

In a continuous chain of such gyrostatic links, with a and l infinitesimal,

U2 =	T	{ 1 +	κn − αn2	}
	ρ		T + (κn − αn2 l/a)

(26)

for the vibration of helical nature like circular polarization.

Changing the sign of n for circular polarization in the opposite direction

U′2 =	T	{ 1 −	κn + αn2	}
	ρ		T − (κn + αn2 l/a)

(27)

In this way a mechanical model is obtained of the action of a magnetized medium on polarized light, κ representing the equivalent of the magnetic field, while α may be ignored as insensible (J. Larmor, Proc. Lond. Math. Soc., 1890; Aether and Matter, Appendix E).

We notice that U² in (26) can be positive, and the gyrostatic chain stable, even when T is negative, and the chain is supporting a thrust, provided κn is large enough, and the thrust does not exceed

(κn − an²) (1 + l/a);

(28)

while U′² in (27) will not be positive and the straight chain will be unstable unless the tension exceeds

(κn + αn²) (1 + l/a).

(29)

15. Gyrostat suspended by a Thread.—In the discussion of the small vibration of a single gyrostat fly-wheel about the vertical position when suspended by a single thread of length 2l = b, the suffix k can be omitted in the preceding equations of § 14, and we can write

Aῶ.. − Kῶ.i + Taῶ − Taσ = 0,

(1)

Mw.. + Tσ = 0, with T = gM,

(2)

w − aῶ − bσ = 0.

(3)

Assuming a periodic solution of these equations

w, ῶ, σ, = (L, P, Q) exp nti,

(4)

and eliminating L, P, Q, we obtain

(−An² + Kn + gMa) (g − n²b) − gMn²a² = 0,

(5)

and the frequency of a vibration in double beats per second is n/2π, where n is a root of this quartic equation.

For upright spinning on a smooth horizontal plane, take b = ∞ and change the sign of a, then

An² − Kn + gMa = 0,

(6)

so that the stability requires

K² > 4gAMa.

(7)

Here A denotes the moment of inertia about a diametral axis through the centre of gravity; when the point of the fly-wheel is held in a small smooth cup, b = 0, and the condition becomes

(A + Ma²) n² − Kn + gMa = 0,

(8)

requiring for stability, as before in § 3,

K² > 4g (A + M²) Ma.

(9)

For upright spinning inside a spherical surface of radius b, the sign of a must be changed to obtain the condition at the lowest point, as in the gyroscopic horizon of Fleuriais.

For a gyrostat spinning upright on the summit of a sphere of radius b, the signs of a and b must be changed in (5), or else the sign of g, which amounts to the same thing.

Denoting the components of horizontal displacement of the point of the fly-wheel by ξ, η, then

br = ξ, bs = η, bσ = ξ + ηI = λ (suppose),

(10)

ω = αῶ + λ.

(11)

If the point is forced to take the motion (ξ, η, ζ) by components of force X, Y, Z, the equations of motion become

−Aq.. + Kp. =    Ya − Zaq,

(12)

Ap.. + Kq. =    −Xa + Zap,

(13)

Mw.. = X + Yi, M (ζ.. − g) = Z;

(14)

so that

Aῶ.. − Kῶ. i + gMaῶ + Maw.. = Maῶζ..,

(15)

or

(A + Ma²)ῶ.. − Kῶ. i + gMaῶ + Maλ = Maῶζ..

(16)

Thus if the point of the gyrostat is made to take the periodic motion given by λ = R exp nti, ζ = 0, the forced vibration of the axis is given by ῶ = P exp nti, where

P { −(A + Ma²) n² + Kn + gMa} − RMn²a = 0;

(17)

and so the effect may be investigated on the Fleuriais gyroscopic horizon of the motion of the ship.

Suppose the motion λ is due to the suspension of the gyrostat from a point on the axis of a second gyrostat suspended from a fixed point.

Distinguishing the second gyrostat by a suffix, then λ = bῶ₁, if b denotes the distance between the points of suspension of the two gyrostats; and the motion of the second gyrostat influenced by the reaction of the first, is given by

(A₁ + M₁h₁²)ῶ..1 − K₁ῶ.₁i

= −g (M₁h₁ + Mb) ῶ₁ − b (X + Yi) = −g (M₁h₁ + Mb) ῶ₁ − Mb(aῶ.. + λ..);

(18)

so that, in the small vibration,

R/b$\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$−(A₁ + M₁h₁²) n² + K₁n + g (M₁h₁ + Mb)$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$= Mn²b (aP + R),

(19)

R { −(A₁ + M₁h₁² + Mb²) n² + K₁n + g (M₁h₁ + Mb)} − PMn²ab² = 0.

(20)

Eliminating the ratio of P to R, we obtain

 { −(A + Ma²) n² + Kn + gMa} × { −(A₁ + M₁h₁² + Mb²) n² + K₁n + g (M₁h₁ + Mb)} − M²n⁴a²b² = 0,

(21)

a quartic for n, giving the frequency n/2π of a fundamental vibration.

Change the sign of g for the case of the gyrostats spinning upright, one on the top of the other, and so realize the gyrostat on the top of a gyrostat described by Maxwell.

In the gyrostatic chain of § 14, the tension T may change to a limited pressure, and U² may still be positive, and the motion stable; and so a motion is realized of a number of spinning tops, superposed in a column.

16. The Flexure Joint.—In Lord Kelvin’s experiment the gyrostats are joined up by equal light rods and short lengths of elastic wire with rigid attachment to the rod and case of a gyrostat, so as to keep the system still, and free from entanglement and twisting due to pivot friction of the fly-wheels.

When this gyrostatic chain is made to revolve with angular velocity n in relative equilibrium as a plane polygon passing through Oz the axis of rotation, each gyrostatic case moves as if its axis produced was attached to Oz by a flexure joint. The instantaneous axis of resultant angular velocity bisects the angle π − θ, if the axis of the case makes an angle θ with Oz, and, the components of angular velocity being n about Oz, and −n about the axis, the resultant angular velocity is 2n cos1/2 (π − θ) =2n sin1/2θ; and the components of this angular velocity are

(1) −2n sin 1/2θ sin 1/2θ = −n (1 − cos θ), along the axis, and

(2) −2n sin 1/2θ cos 1/2θ = −n sin θ, perpendicular to the axis of the case. The flexure joint behaves like a pair of equal bevel wheels engaging.

The component angular momentum in the direction Ox is therefore

L = −An sin θ cos θ − Cn (1 − cos θ) sin θ + K sin θ,

(3)

and Ln is therefore the couple acting on the gyrostat.

If α denotes the angle which a connecting link makes with Oz, and T denotes the constant component of the tension of a link parallel to Oz, the couple acting is

Ta cos θ_k (tan αk+1 + tan α_k) − 2Tα sin θ_k,

(4)

which is to be equated to Ln, so that

−An² sin θ_k cos θ_k − Cn (1 − cos θ_k) sin θ_k + Kn sin θ_k −Ta cos θ_k (tan αk+1 + tan α_k) + 2Tα sin θ_k = 0.

(5)

In addition

Mn²x_k + T (tan αk+1 − tan α_k) = 0,

(6)

with the geometrical relation

xk+1 − x_k − a (sin θk+1 + sin θ_k) − 2l sin αk+1 = 0.

(7)

When the polygon is nearly coincident with Oz, these equations can be replaced by