Page:EB1911 - Volume 12.djvu/803

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

A .. k − K .. ki + 2Tak − Ta ( k+1 + k) = 0, (10)

M .. k − T ( k+1 − k) = 0, (11)

k+1 − k − a( k+1 + k − 2l k+1 ) = 0. (12)

For a vibration of circular polarization assume a solution

k, k, 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 (e ci + 1) = 0, (15)

−LMn2 − QT (e ci − 1) = 0, (16)

L (e ci − 1) − Pa (e ci + 1) − 2Qle ci = 0, (17)

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

(18)

(19)

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

Putting

= M/2 (a + l), &emsp; the mass per unit length of the chain, (20)

= K/2 (a + l), &emsp; the gyrostatic angular momentum per unit length, (21)

= A/2 (a + l), &emsp; 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

(24)

(25)

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

(26)

for the vibration of helical nature like circular polarization.

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

(27)

In this way a mechanical model is obtained of the action of a magnetized medium on polarized light, representing the equivalent of the magnetic field, while  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 n is large enough, and the thrust does not exceed

(n − 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

(n + n2) (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

A .. − K. i + Ta − Ta = 0, (1)

Mw .. + T = 0, with T = gM, (2)

w − a − b = 0. (3)

Assuming a periodic solution of these equations

w,, , = (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/2, where n is a root of this quartic equation.

For upright spinning on a smooth horizontal plane, take b = &infin; 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, , then

br =, bs = , b = + I =  (suppose), (10)

= + .  (11)

If the point is forced to take the motion by components of force X, Y, Z, the equations of motion become

−Aq .. + Kp. = &emsp;&emsp; Ya − Zaq, (12)

Ap .. + Kq. = &emsp;&ensp; −Xa + Zap, (13)

Mw .. = X + Yi, M ( .. − g) = Z; (14)

so that

A .. − K. &thinsp;i + gMa + Maw .. = Ma .. , (15)

or

(A + Ma2) .. − K. &thinsp;i + gMa + Ma = Ma. . (16)

Thus if the point of the gyrostat is made to take the periodic motion given by = R exp nti,  = 0, the forced vibration of the axis is given by = 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 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 = b1, 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) .. 1 − K1. &thinsp;1 i

(18)

so that, in the small vibration,

$R⁄b$−(A1 + M1h12) n2 + K1n + g (M1h1 + Mb)= 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

(21)

a quartic for n, giving the frequency n/2 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 −, if the axis of the case makes an angle with Oz, and, the components of angular velocity being n about Oz, and −n about the axis, the resultant angular velocity is 2n cos ( − ) =2n sin; and the components of this angular velocity are

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

(2) −2n sin cos  = −n sin, 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 cos  − Cn (1 − cos ) sin  + K sin , (3)

and Ln is therefore the couple acting on the gyrostat.

If 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 k (tan k+1 + tan k) − 2T sin k,  (4)

which is to be equated to Ln, so that

−An2 sin k cos k − Cn (1 − cos k) sin k + Kn sin k −Ta cos k (tan k+1 + tan k) + 2T sin k = 0. (5)

In addition

Mn2xk + T (tan k+1 − tan k) = 0, (6)

with the geometrical relation

x k+1 − xk − a (sin k+1 + sin k) − 2l sin  k+1 = 0. (7)

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