Chapter 10

(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)2n2ρ·Ta + κnl − αn2l,TTa + κn (a + l) − αn2(a + l) + ρn2a2(a + l)(24){(a + l) n}2sin (a + l) n/U=T·T + (κn − αn2) (1 + l/a) + ρn2a (a + l).ρT + (κn − an2) l/a(25)In a continuous chain of such gyrostatic links, with a and l infinitesimal,U2=T{1 +κn − αn2}ρT + (κn − αn2l/a)(26)for the vibration of helical nature like circular polarization.Changing the sign of n for circular polarization in the opposite directionU′2=T{1 −κn + αn2}ρT − (κn + αn2l/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 U2in (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′2in (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 writeAῶ− Kῶi + Taῶ − Taσ = 0,(1)Mw+ Tσ = 0, with T = gM,(2)w − aῶ − bσ = 0.(3)Assuming a periodic solution of these equationsw, 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, thenAn2− Kn + gMa = 0,(6)so that the stability requiresK2> 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, thenbr = 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)Mῶ= X + Yi, M (ζ− g) = Z;(14)so thatAῶ− Kῶi + gMaῶ + Maw= Maῶζ,(15)or(A + Ma2)ῶ− Kῶi + 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, whereP { −(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)ῶ1 − K1ῶ1i= −g (M1h1+ Mb) ῶ1− b (X + Yi)= −g (M1h1+ Mb) ῶ1− Mb(aῶ+λ);(18)so that, in the small vibration,R{−(A1+ M1h12) n2+ K1n + g (M1h1+ Mb)}= Mn2b (aP + R),b(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 U2may 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 cos½ (π − θ) =2n sin½θ; and the components of this angular velocity are(1) −2n sin ½θ sin ½θ = −n (1 − cos θ), along the axis, and(2) −2n sin ½θ cos ½θ = −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 thereforeL = −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 isTa cos θk(tan αk+1+ tan αk) − 2Tα sin θk,(4)which is to be equated to Ln, so that−An2sin θkcos θk− Cn (1 − cos θk) sin θk+ Kn sin θk−Ta cos θk(tan αk+1+ tan αk) + 2Tα sin θk= 0.(5)In additionMn2xk+ T (tan αk+1− tan αk) = 0,(6)with the geometrical relationxk+1− xk− a (sin θk+1+ sin θk) − 2l sin αk+1= 0.(7)When the polygon is nearly coincident with Oz, these equations can be replaced by(−An2+ Kn + 2Ta) θk− Ta (αk+1+ αk) = 0,(8)Mn2xk+ T (αk+1− αk) = 0,(9)xk+1− xk− a (θk+1+ θk) − 2lak= 0,(10)and the rest of the solution proceeds as before in § 14, puttingxk, θk, αk= (L, P, Q) exp cki.(11)A half wave length of the curve of gyrostats is covered when ck = π, so that π/c is the number of gyrostats in a half wave, which is therefore of wave length 2π (a + l)/c.A plane polarized wave is given when exp cki is replaced by exp (nt + ck) i, and a wave circularly polarized when w, ῶ, σ of § 14 replace this x, θ, α.Gyroscopic Pendulum.—The elastic flexure joint is useful for supporting a rod, carrying a fly-wheel, like a gyroscopic pendulum.Expressed by Euler’s angles, θ, φ, ψ, the kinetic energy isT = ½A (θ2+ sin2θψ2) + ½C′ (1 − cos θ)2ψ2+ ½C (φ+ψcos θ)2,(12)where A refers to rod and gyroscope about the transverse axis at the point of support, C′ refers to rod about its axis of length, and C refers to the revolving fly-wheel.The elimination ofψbetween the equation of conservation of angular momentum about the vertical, viz.(13) A sin2θψ− C′ (1 − cos θ) cos θψ+ C(φ+ψcos θ) cos θ = G, a constant, and the equation of energy, viz.(14) T − gMh cos θ = H, a constant, with θ measured from the downward vertical, and(15)φ+ψcos θ = R, a constant, will lead to an equation for dθ/dt, or dz/dt, in terms of cos θ or z, the integral of which is of hyperelliptic character, except when A = C′.In the suspension of fig. 8, the motion given byφis suppressed in the stalk, and for the fly-wheelφgives the rubbing angular velocity of the wheel on the stalk; the equations are nowT = ½A (θ2+ sin2θψ2) + ½C′ cos2θψ2+ ½CR2= H + gMh cos θ,(16)A sin2θψ+ C′ cos2θψ+ CR cos θ = G,(17)and the motion is again of hyperelliptic character, except when A = C′, or C′ = 0. To realize a motion given completely by the elliptic function, the suspension of the stalk must be made by a smooth ball and socket, or else a Hooke universal joint.Finally, there is the case of the general motion of a top with a spherical rounded point on a smooth plane, in which the centre of gravity may be supposed to rise and fall in a vertical line. HereT = ½ (A + Mh2sin2θ)θ2+ ½A sin2θψ2+ ½CR2= H − gMh cos θ,(18)with θ measured from the upward vertical, andA sin2θψ+ CR cos θ = G,(19)where A now refers to a transverse axis through the centre of gravity. The elimination ofψleads to an equation for z, = cos θ, of the form(dz)2= 2gZ= 2g(z1− z) (z2− z) (z3− z),dth1 − z2+ A/Mh2h(z4− z) (z − z5)(20)with the arrangementz1, z4> / > z2> z > z3> − / > z5;(21)so that the motion is hyperelliptic.Authorities.—In addition to the references in the text the following will be found useful:—Ast. Notices, vol. i.;Comptes rendus, Sept. 1852; Paper by Professor Magnus translated in Taylor’sForeign Scientific Memoirs, n.s., pt. 3, p. 210;Ast. Notices, xiii. 221-248;Theory of Foucault’s Gyroscope Experiments, by the Rev. Baden Powell, F.R.S.;Ast. Notices, vol. xv.; articles by Major J. G. Barnard inSilliman’s Journal, 2nd ser., vols. xxiv. and xxv.; E. Hunt on “Rotatory Motion,”Proc. Phil. Soc. Glasgow, vol. iv.; J. Clerk Maxwell, “On a Dynamical Top,”Trans. R.S.E.vol. xxi.;Phil. Mag.4th ser. vols. 7, 13, 14;Proc. Royal Irish Academy, vol. viii.; Sir William Thomson on “Gyrostat,”Nature, xv. 297; G. T. Walker, “The Motion of a Celt,”Quar. Jour. Math., 1896; G. T. Walker,Math. Ency.iv. 1, xi. 1; Gallop,Proc. Camb. Phil. Soc.xii. 82, pt. 2, 1903, “Rise of a Top”; Price’sInfinitesimal Calculus, vol. iv.; Worms,The Earth and its Mechanism; Routh,Rigid Dynamics; A. G. Webster,Dynamics(1904); H. Crabtree,Spinning Tops and Gyroscopic Motion(1909). For a complete list of the mathematical works on the subject of the Gyroscope and Gyrostat from the outset, Professor Cayley’s Report to the British Association (1862) on theProgress of Dynamicsshould be consulted. Modern authors will be found cited in Klein and Sommerfeld,Theorie des Kreisels(1897), and in theEncyclopädie der mathematischen Wissenschaften.

(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

(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, 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 U2in (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′2in (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− Kῶi + 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/2Ï€, 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)

Mῶ= X + Yi, M (ζ− g) = Z;

(14)

so that

Aῶ− Kῶi + gMaῶ + Maw= Maῶζ,

(15)

or

(A + Ma2)ῶ− Kῶi + 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)ῶ1 − K1ῶ1i

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

= −g (M1h1+ Mb) ῶ1− b (X + Yi)

= −g (M1h1+ Mb) ῶ1− Mb(aῶ+λ);

(18)

so that, in the small vibration,

(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,

{ −(A + Ma2) n2+ Kn + gMa}

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

(21)

a quartic for n, giving the frequency n/2Ï€ 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 U2may 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 cos½ (π − θ) =2n sin½θ; and the components of this angular velocity are

(1) −2n sin ½θ sin ½θ = −n (1 − cos θ), along the axis, and

(2) −2n sin ½θ cos ½θ = −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 θk(tan αk+1+ tan αk) − 2Tα sin θk,

(4)

which is to be equated to Ln, so that

−An2sin θkcos θk− Cn (1 − cos θk) sin θk+ Kn sin θk−Ta cos θk(tan αk+1+ tan αk) + 2Tα sin θk= 0.

(5)

In addition

Mn2xk+ T (tan αk+1− tan αk) = 0,

(6)

with the geometrical relation

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

(7)

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

(−An2+ Kn + 2Ta) θk− Ta (αk+1+ αk) = 0,

(8)

Mn2xk+ T (αk+1− αk) = 0,

(9)

xk+1− xk− a (θk+1+ θk) − 2lak= 0,

(10)

and the rest of the solution proceeds as before in § 14, putting

xk, θk, αk= (L, P, Q) exp cki.

(11)

A half wave length of the curve of gyrostats is covered when ck = π, so that π/c is the number of gyrostats in a half wave, which is therefore of wave length 2π (a + l)/c.

A plane polarized wave is given when exp cki is replaced by exp (nt + ck) i, and a wave circularly polarized when w, ῶ, σ of § 14 replace this x, θ, α.

Gyroscopic Pendulum.—The elastic flexure joint is useful for supporting a rod, carrying a fly-wheel, like a gyroscopic pendulum.

Expressed by Euler’s angles, θ, φ, ψ, the kinetic energy is

T = ½A (θ2+ sin2θψ2) + ½C′ (1 − cos θ)2ψ2+ ½C (φ+ψcos θ)2,

(12)

where A refers to rod and gyroscope about the transverse axis at the point of support, C′ refers to rod about its axis of length, and C refers to the revolving fly-wheel.

The elimination ofψbetween the equation of conservation of angular momentum about the vertical, viz.

(13) A sin2θψ− C′ (1 − cos θ) cos θψ+ C(φ+ψcos θ) cos θ = G, a constant, and the equation of energy, viz.

(14) T − gMh cos θ = H, a constant, with θ measured from the downward vertical, and

(15)φ+ψcos θ = R, a constant, will lead to an equation for dθ/dt, or dz/dt, in terms of cos θ or z, the integral of which is of hyperelliptic character, except when A = C′.

In the suspension of fig. 8, the motion given byφis suppressed in the stalk, and for the fly-wheelφgives the rubbing angular velocity of the wheel on the stalk; the equations are now

T = ½A (θ2+ sin2θψ2) + ½C′ cos2θψ2+ ½CR2= H + gMh cos θ,

(16)

A sin2θψ+ C′ cos2θψ+ CR cos θ = G,

(17)

and the motion is again of hyperelliptic character, except when A = C′, or C′ = 0. To realize a motion given completely by the elliptic function, the suspension of the stalk must be made by a smooth ball and socket, or else a Hooke universal joint.

Finally, there is the case of the general motion of a top with a spherical rounded point on a smooth plane, in which the centre of gravity may be supposed to rise and fall in a vertical line. Here

T = ½ (A + Mh2sin2θ)θ2+ ½A sin2θψ2+ ½CR2= H − gMh cos θ,

(18)

with θ measured from the upward vertical, and

A sin2θψ+ CR cos θ = G,

(19)

where A now refers to a transverse axis through the centre of gravity. The elimination ofψleads to an equation for z, = cos θ, of the form

(20)

with the arrangement

z1, z4> / > z2> z > z3> − / > z5;

(21)

so that the motion is hyperelliptic.

Authorities.—In addition to the references in the text the following will be found useful:—Ast. Notices, vol. i.;Comptes rendus, Sept. 1852; Paper by Professor Magnus translated in Taylor’sForeign Scientific Memoirs, n.s., pt. 3, p. 210;Ast. Notices, xiii. 221-248;Theory of Foucault’s Gyroscope Experiments, by the Rev. Baden Powell, F.R.S.;Ast. Notices, vol. xv.; articles by Major J. G. Barnard inSilliman’s Journal, 2nd ser., vols. xxiv. and xxv.; E. Hunt on “Rotatory Motion,”Proc. Phil. Soc. Glasgow, vol. iv.; J. Clerk Maxwell, “On a Dynamical Top,”Trans. R.S.E.vol. xxi.;Phil. Mag.4th ser. vols. 7, 13, 14;Proc. Royal Irish Academy, vol. viii.; Sir William Thomson on “Gyrostat,”Nature, xv. 297; G. T. Walker, “The Motion of a Celt,”Quar. Jour. Math., 1896; G. T. Walker,Math. Ency.iv. 1, xi. 1; Gallop,Proc. Camb. Phil. Soc.xii. 82, pt. 2, 1903, “Rise of a Top”; Price’sInfinitesimal Calculus, vol. iv.; Worms,The Earth and its Mechanism; Routh,Rigid Dynamics; A. G. Webster,Dynamics(1904); H. Crabtree,Spinning Tops and Gyroscopic Motion(1909). For a complete list of the mathematical works on the subject of the Gyroscope and Gyrostat from the outset, Professor Cayley’s Report to the British Association (1862) on theProgress of Dynamicsshould be consulted. Modern authors will be found cited in Klein and Sommerfeld,Theorie des Kreisels(1897), and in theEncyclopädie der mathematischen Wissenschaften.

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