Part 12
+ [Omega]z sin[theta](x - [rho] sin[theta]) - rz[rho] cos[theta] = 0,
/A \ dq h3 (C*) ( -- + x² + z²) -- + q²p(x cos[theta] - z sin[theta]) - [Omega] -- sin [theta] \M / dt M
/A \ + [Omega]² ( -- + z²) sin [theta] cos [theta] + [Omega]²xz sin²[theta] \M /
- [Omega]rx(x sin [theta] + z cos [theta]) - g(x cos [theta] - z sin [theta]) = 0.
The steady motion and nutation superposed may be expressed by
(1) [theta] = [alpha] + L, sin [theta] = sin[alpha] + L cos[alpha], cos[theta] = cos[alpha] - L sin [alpha], [Omega] = µ + N, r = R + Q,
where L, N, Q are small terms, involving a factor e^(nti), to express the periodic nature of the nutation; and then if a, c denote the mean value of x, z, at the point of contact
(2) x = a + L[rho] cos[alpha], z = c - L[rho] sin[alpha],
(3) x sin [theta] + z cos [theta] = a sin [alpha] + c cos [alpha] + L(a cos [alpha] - c sin [alpha]),
(4) x cos [theta] - z sin [theta] = a cos [alpha] - c sin [alpha] - L(a sin [alpha] + c cos [alpha] - [rho]).
Substituting these values in (C*) with dq/dt = -d²[theta]/dt² = n²L, and ignoring products of the small terms, such as L², LN, ...
/A \ /CR + K CQ\ (C**) ( -- + a² + c²) Ln² - (µ + N)( ------ + -- )(sin[alpha] + L cos[alpha]) \M / \ M M /
/A \ + (µ² + 2µN)( -- + c² - 2L[rho]c sin[alpha]) (sin [alpha] cos[alpha] + L cos[alpha]) \M /
+ (µ² + 2µN) [ac - L[rho](a sin [alpha] - c sin[alpha])] (sin² [alpha] + L sin 2[alpha]) - (µ + N)(R + Q)(a + L[rho]cos[alpha])[a sin [alpha] + c cos[alpha] + L(a cos[alpha] - c sin [alpha])] - g(a cos[alpha] - c sin[alpha]) + gL(a sin[alpha] + c cos[alpha] - [rho]) = 0,
which is equivalent to
CR + K /A \ (5) -µ ------ sin[alpha] + µ² ( -- + c²) sin [alpha] cos [alpha] M \M /
+ µ² ac sin²[alpha] - µRa(a sin[alpha] + c cos[alpha]) - g(a cos[alpha] - c sin[alpha]) = 0,
the condition of steady motion; and
(6) DL + EQ + FN = 0,
where
/A \ CK + K (7) D = ( -- + a² + c²) n² - µ ------ cos[alpha] \M / M
- 2µ²[rho]c sin²[alpha] cos[alpha]
/A \ + µ² ( -- + c²) cos [alpha] - µ²[rho](a sin [alpha] \M /
- c cos[alpha]) sin²[alpha] + µ²ac sin 2[alpha] - µR[rho] cos [alpha](a sin [alpha] + c cos[alpha]) - µRa(a cos [alpha] - c sin[alpha]) + g(a sin[alpha] + c cos[alpha] - [rho]),
C (8) E = -µ -- sin [alpha] - µa(a sin [alpha] + c cos [alpha]), M
CR + K /A \ (9) F = - ------ sin [alpha] + 2µ ( -- + c²) sin [alpha] cos [alpha] M \M /
+ 2µac sin² [alpha] - Ra(a sin [alpha] + c cos [alpha]).
With the same approximation (A*) and (B*) are equivalent to
/C \ Q N (A**) ( -- + a²) -- - ac sin [alpha]-- \M / L L
- µa(a sin [alpha] + 2c cos [alpha] - [rho] sin² [alpha]) + Ra[rho] cos [alpha] = 0,
Q /A \ N CR + K /A \ (B**) -ac -- + ( -- + c²) sin [alpha]-- - ------ + 2µ ( -- + c²) cos [alpha] L \M / L M \M /
+ µc sin [alpha](a - [rho] sin [alpha]) - Rc[rho] cos [alpha] = 0.
The elimination of L, Q, N will lead to an equation for the determination of n², and n² must be positive for the motion to be stable.
If b is the radius of the horizontal circle described by G in steady motion round the centre B,
(10) b = v/µ = (cP - aR)/µ = c sin [alpha] - aR/µ,
and drawing GL vertically upward of length [lambda] = g/µ², the height of the equivalent conical pendulum, the steady motion condition may be written
(11) (CR + K)µ sin[alpha] - µ² sin[alpha] cos[alpha] = -gM(a cos[alpha] - c sin[alpha]) + M(µ²c sin[alpha] - µRa) (a sin[alpha] + c cos[alpha]) = gM[b[lambda]^(-1) (a sin [alpha] + c cos [alpha]) - a cos [alpha] + c sin [alpha]] = gM·PT,
LG produced cuts the plane in T.
Interpreted dynamically, the left-hand side of this equation represents the velocity of the vector of angular momentum about G, so that the right-hand side represents the moment of the applied force about G, in this case the reaction of the plane, which is parallel to GA, and equal to gM·GA/GL; and so the angle AGL must be less than the angle of friction, or slipping will take place.
Spinning upright, with [alpha] = 0, a = 0, we find F = 0, Q = 0, and
CR + K /A \ (12) - ------ + 2µ( -- + c²) - Rcp = 0, M \M /
/A \ CR + K /A \ (13) ( -- + c²) n² = µ ------ - µ²( -- + c²) + µR[rho]c - g(c - [rho]), \M / M \M /
/A \² /CK + R \² /A \ (14) ( -- + c²) n² = ¼( ------ + Rc[rho] ) - g( -- + c²) (c - [rho]), \M / \ M / \M /
Thus for a top spinning upright on a rounded point, with K = 0, the stability requires that
(15) R > 2k´ [root]{g(c - [rho])}/(k² + c[rho]),
where k, k´ are the radii of gyration about the axis Gz, and a perpendicular axis at a distance c from G; this reduces to the preceding case of § 3 (7) when [rho] = 0.
Generally, with [alpha] = 0, but a ± 0, the condition (A) and (B) becomes
/C \ Q (16) ( - + a²) -- = 2µac - Ra[rho], \M / L
Q CR + K /A \ -ac -- = ------ + Rc[rho] - 2µ( -- + a²), L M \M /
so that, eliminating Q/L, _ _ | /A \ /C \ | /C \ /CR + K\ C (17) 2 | ( -- + c²)( -- + a²) - a²c²| µ = ( -- + a²)( ------ ) + -- Rc[rho], |_ \M / \M / _| \M / \ M / M
the condition when a coin or platter is rolling nearly flat on the table.
Rolling along in a straight path, with [alpha] = ½[pi], c = 0, µ = 0, E = 0; and
(18) N/L = (CR + K)/A,
/A \ (19) D = ( -- + a²) n² + g(a - [rho]), \M /
CR + K F = - ------ - Ra², M
/A \ ( -- + a²) n² + g(a - [rho]) N D \M / (20) -- = - -- = ----------------------------, L F /C \ K ( -- + a²) R + - \M / M _ _ /A \ (CR + K) | /C \ K | (21) ( -- + a²) n² = -------- | ( -- + a²) R + -- | - g(a - [rho]). \M / A |_ \M / M _|
Thus with K = 0, and rolling with velocity V = Ra, stability requires
V² a - [rho] A a - [rho] (22) -- > ---------------- > ½ -- ---------, 2g C / C \ C C 2 -- ( --- + 1 ) --- + 1 A \Ma² / Ma²
or the body must have acquired velocity greater than attained by rolling down a plane through a vertical height 1/2(a - [rho])A/C.
On a sharp edge, with [rho] = 0, a thin uniform disk or a thin ring requires
(23) V²/2g > a/6 or a/8.
The gyrostat can hold itself upright on the plane without advance when R = 0, provided
(24) K²/AM - g(a - [rho]) is positive.
For the stability of the monorail carriage of § 5 (6), ignoring the rotary inertia of the wheels by putting C = 0, and replacing K by G´ the theory above would require
G´ / G´\ (25) -- ( aV + -- ) > gh. A \ A /
For further theory and experiments consult Routh, _Advanced Rigid Dynamics_, chap. v., and Thomson and Tait, _Natural Philosophy_, § 345; also Bourlet, _Traité des bicycles_ (analysed in Appell, _Mécanique rationnelle_, ii. 297, and Carvallo, _Journal de l'école polytechnique_, 1900); Whipple, _Quarterly Journal of Mathematics_, vol. xxx., for mathematical theories of the bicycle, and other bodies.
Gyrostatic chain.
14. Lord Kelvin has studied theoretically and experimentally the vibration of a chain of stretched gyrostats (_Proc. London Math. Soc._, 1875; J. Perry, _Spinning Tops_, for a diagram). Suppose each gyrostat to be equivalent dynamically to a fly-wheel of axial length 2a, and that each connecting link is a light cord or steel wire of length 2l, stretched to a tension T.
Denote by x, y the components of the slight displacement from the central straight line of the centre of a fly-wheel; and let p, q, 1 denote the direction cosines of the axis of a fly-wheel, and r, s, 1 the direction cosines of a link, distinguishing the different bodies by a suffix.
Then with the previous notation and to the order of approximation required,
(1) [theta]1 = -dq/dt, [theta]2 = dp/dt,
(2) h1 = A[theta]1, h2 = A[theta]2, h3 = K,
to be employed in the dynamical equations
(3) dh1/dt - [theta]3 h2 + [theta]2 h3 = L, ...
in which [theta]3 h1 and [theta]3 h2 can be omitted.
For the kth fly-wheel
(4) -A :q_k + K .p_k = Ta(q_k - s_k) + Ta[q_k - s_(k+1)],
(5) A :p_k + K .q_k = -Ta(p_k - r_k) - Ta[p_k - r_(k+1)];
and for the motion of translation
(6) M :x_k = T[r_(k+1) - r_k], M :y_k = T[s(k+1) - s_k];
while the geometrical relations are
(7) x_(k+1) - x_k = a [p_(k+1) + p_k] + 2lr_(k+1),
(8) y_(k+1) - y_k = a [q_(k+1) + q_k] + 2ls_(k+1).
Putting
(9) x + yi = w, p + qi = [~omega], r + si = [sigma],
these three pairs of equations may be replaced by the three equations
(10) A :[~omega]_k - K .[~omega]_ki + 2Ta[~omega]_k - Ta([sigma]_(k+1) + [sigma]_k) = 0,
(11) M :[~omega]_k - T([sigma]_(k+1) - [sigma]_k) = 0,
(12) w_(k+1) - w_k - a([~omega]_(k+1) + [~omega]_k - 2l[sigma]_(k+1) = 0.
For a vibration of circular polarization assume a solution
(13) w_k, [~omega]_k, [sigma]_k = (L, P, Q) exp (nt + kc)i,
so that c/n is the time-lag between the vibration of one fly-wheel and the next; and the wave velocity is
(14) U = 2(a + l)n/c.
Then
(15) P(-An² + Kn + 2Ta) - QTa[e^(ci) + 1] = 0,
(16) -LMn² - QT[e^(ci) - 1] = 0,
(17) L[e^(ci) - 1] -Pa[e^(ci) + 1] - 2Qle^(ci) = 0,
leading, on elimination of L, P, Q, to
(2 Ta + Kn - An²) (1 - Mn²l/T) - Mn² (18) cos c = ------------------------------------, 2Ta + Kn - An² + Mna²
M n² 2Ta(a + l) + KNl - An²l (19) 2 sin² ½c = -- --------------------------. T 2Ta + Kn - An² + Mn²a²
With K = 0, A = 0, this reduces to Lagrange's condition in the vibration of a string of beads.
Putting
(20) [rho] = M/2(a + l), the mass per unit length of the chain,
(21) [kappa] = K/2(a + l), the gyrostatic angular momentum per unit length,
(22) [alpha] = A/2(a + l), the transverse moment of inertia per unit length,
(23) 1/2c = (a + l)n/U,
equation (19) can be written
(24) {sin (a + l)n/U}²
[rho] Ta + [kappa]nl - [alpha]n²l = (a + l)²n² ----- ----------------------------------------------------------, T Ta + [kappa]n(a + l) - [alpha]n²(a + l) + [rho]n²a²(a + l)
/ (a + l)n \² (25) ( --------------- ) \ sin (a + l)n/U /
T T + ([kappa]n - [alpha]n²) (1 + l/a) + [rho]n²a(a + l) = ----- . ------------------------------------------------------. [rho] T + ([kappa]n - an²)l/a
In a continuous chain of such gyrostatic links, with a and l infinitesimal,
T / [kappa]n - [alpha]n² \ (26) U² = ----- ( 1 + ------------------------------ ) [rho] \ T + ([kappa]n - [alpha]n² l/a) /
for the vibration of helical nature like circular polarization.
Changing the sign of n for circular polarization in the opposite direction
T / [kappa]n + [alpha]n² \ (27) U´² = ----- ( 1 - ------------------------------ ). [rho] \ T - ([kappa]n + [alpha]n² l/a) /
In this way a mechanical model is obtained of the action of a magnetized medium on polarized light, [kappa] representing the equivalent of the magnetic field, while [alpha] 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 [kappa]n is large enough, and the thrust does not exceed
(28) ([kappa]n - an²)(1 + l/a);
while U'² in (27) will not be positive and the straight chain will be unstable unless the tension exceeds
(29) ([kappa]n + [alpha]n²)(1 + l/a).
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
(1) A :[~omega] - K .[~omega]i + Ta[~omega] - Ta[sigma] = 0,
(2) M :w + T[sigma] = 0, with T = gM,
(3) w - a[~omega] - b[sigma] = 0.
Assuming a periodic solution of these equations
(4) w, [~omega], [sigma], = (L, P, Q) exp nti,
and eliminating L, P, Q, we obtain
(5) (-An² + Kn + gMa)(g - n²b) - gMn²a² = 0,
and the frequency of a vibration in double beats per second is n/2[pi], where n is a root of this quartic equation.
For upright spinning on a smooth horizontal plane, take b = [oo] and change the sign of a, then
(6) An² - Kn + gMa = 0,
so that the stability requires
(7) K² > 4gAMa.
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
(8) (A + Ma²)n² - Kn + gMa = 0,
requiring for stability, as before in § 3,
(9) K² > 4g(A + M²)Ma.
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 [xi], [eta], then
(10) br = [xi], bs = [eta], b[sigma] = [xi] + [eta]i = [lambda] (suppose),
(11) [omega] = [alpha][~omega] + [lambda].
If the point is forced to take the motion ([xi], [eta], [zeta]) by components of force X, Y, Z, the equations of motion become
(12) -A :q + K .p = Ya - Zaq,
(13) A :p + K .q = -Xa + Zap,
(14) M :w = X + Yi, M( :[zeta] - g) = Z;
so that
(15) A :[~omega] - K .[~omega]i + gMa[~omega] + Ma :w = Ma[~omega] :[zeta],
or
(16) (A + Ma²) :[~omega] - K .[~omega]i + gMa[~omega] + Ma[lambda] = Ma[~omega] :[zeta].
Thus if the point of the gyrostat is made to take the periodic motion given by [lambda] = R exp nti, [zeta] = 0, the forced vibration of the axis is given by [~omega] = P exp nti, where
(17) P{-(A + Ma²)n² + Kn + gMa} - RMn²a = 0;
and so the effect may be investigated on the Fleuriais gyroscopic horizon of the motion of the ship.
Suppose the motion [lambda] 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 [lambda] = b[~omega]1, 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
(18) (A1 + M1h1²) :[~omega]1 - K1 .[~omega]1 i = -g(M1h1 + Mb)[~omega]1 - b(X + Yi) = -g(M1h1 + Mb)[~omega]1 - Mb(a :[~omega] + :[lambda]);
so that, in the small vibration,
R / \ (19) -- ( -(A1 + M1h1²)n² + K1n + g(M1h1 + Mb) ) = Mn²b(aP + R), b \ /
(20) R { -(A1 + M1h1² + Mb²)n² + K1n + g(M1h1 + Mb)} - PMn²ab² = 0.
Eliminating the ratio of P to R, we obtain
(21) { -(A + Ma²)n² + Kn + gMa} × {-(A1 + M1h1² + Mb²)n² + K1n + g (M1h1 + Mb)} - M²n^4a²b² = 0,
a quartic for n, giving the frequency n/2[pi] 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 [pi] - [theta], if the axis of the case makes an angle [theta] with Oz, and, the components of angular velocity being n about Oz, and -n about the axis, the resultant angular velocity is 2n cos½([pi] - [theta]) = 2n sin½[theta]; and the components of this angular velocity are
(1) -2n sin ½[theta] sin ½[theta] = -n(1-cos[theta]), along the axis, and
(2) -2n sin ½[theta] cos ½[theta] = -n sin [theta], 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
(3) L = -An sin [theta] cos [theta] - Cn (1 - cos [theta]) sin [theta] + K sin [theta],
and Ln is therefore the couple acting on the gyrostat.
If [alpha] 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
(4) Ta cos [theta]_k(tan [alpha]_(k+1) + tan [alpha]_k) - 2T[alpha]sin[theta]_k,
which is to be equated to Ln, so that
(5) -An² sin [theta]_k cos [theta]_k - Cn(1 - cos [theta]_k) sin [theta]_k + Kn sin [theta]_k - T[alpha] cos [theta]_k(tan [alpha]_(k+1) + tan [alpha]_k) + 2T[alpha] sin [theta]_k = 0.
In addition
(6) Mn²[chi]_k + T(tan [alpha]_(k+1) - tan [alpha]_k) = 0,
with the geometrical relation
(7) [chi]_(k+1) - [chi]_k - [alpha](sin [theta]_(k+1) + sin [theta]_k) - 2l sin{k + 1} = 0.
When the polygon is nearly coincident with Oz, these equations can be replaced by
(8) (-An² + Kn + 2Ta)[theta]_k - Ta([alpha]_(k+1) + [alpha]_k) = 0,
(9) Mn²x_k + T([alpha]_(k+1) - [alpha]_k) = 0,
(10) x_(k+1) - x_k - a([theta]_(k+1) + [theta]_k) - 2la_k = 0,
and the rest of the solution proceeds as before in § 14, putting
(11) x_k, [theta]_k, [alpha]_k = (L, P, Q) exp cki.
A half wave length of the curve of gyrostats is covered when ck = [pi], so that [pi]/c is the number of gyrostats in a half wave, which is therefore of wave length 2[pi](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, [~omega], [sigma] of § 14 replace this x, [theta], [alpha].
_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, [theta], [phi], [psi], the kinetic energy is
(12) T = ½A( .[theta]² + sin² [theta] .[psi]²) + ½C´(1 - cos [theta])² .[psi]² + ½C( .[phi] + .[psi] cos [theta])²,
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 of .[psi] between the equation of conservation of angular momentum about the vertical, viz.
(13) A sin² [theta] .[psi] - C´(1 - cos [theta]) cos [theta] .[psi] + C( .[phi] + .[psi] cos [theta]) cos [theta] = G, a constant, and the equation of energy, viz.
(14) T - gMh cos [theta] = H, a constant, with [theta] measured from the downward vertical, and
(15) .[phi] + .[psi] cos [theta] = R, a constant, will lead to an equation for d[theta]/dt, or dz/dt, in terms of cos [theta] or z, the integral of which is of hyperelliptic character, except when A = C´.
In the suspension of fig. 8, the motion given by .[phi] is suppressed in the stalk, and for the fly-wheel .[phi] gives the rubbing angular velocity of the wheel on the stalk; the equations are now
(16) T = ½A( .[theta]² + sin² [theta] .[psi]²) + ½C´ cos² [theta] .[psi]² + ½CR² = H + gMh cos [theta],
(17) A sin² [theta][.[psi]] + C´ cos² [theta] .[psi] + CR cos [theta] = G,
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
(18) T = ½(A + Mh² sin² [theta]) .[theta]² + ½A sin² [theta] .[psi]² + ½CR² = H - gMh cos [theta],
with [theta] measured from the upward vertical, and
(19) A sin² [theta] .[psi] + CR cos [theta] = G,
where A now refers to a transverse axis through the centre of gravity. The elimination of [.[psi]] leads to an equation for z, = cos [theta], of the form
/dz\² g Z g (z1 - z)(z2 - z)(z3 - z) (20) ( -- ) = 2 -- -------------- = 2 -- ------------------------, \dt/ h 1 - z² + A/Mh² h (z4 - z)(z - z5)
with the arrangement
(21) z1, z4 > / > z2 > z > z3 > - / > z5;
so that the motion is hyperelliptic.