(23)(1 − cos θ) exp (φ − ψ − q′t) i= √ (sec β − cos θ) √ (cos β − cos θ) + i (√sec β − √ cos β) √ cos θ,(24)q, q′ = n√ (2 sec β) ± n√ (2 cos β);(25)and thence α, β, γ, δ can be inferred.The physical constants of a given symmetrical top have been denoted in § 1 by M, h, A, C, and l, n, T; to specify a given state of general motion we have G, G′ or CR, D, E, or F, which may be called the dynamical constants; or κ, v, w, v1, v2, or f, f′, f1, f2, the analytical constants; or the geometrical constants, such as α, β, δ, δ′, k of a given articulated hyperboloid.There is thus a triply infinite series of a state of motion; the choice of a typical state can be made geometrically on the hyperboloid, flattened in the plane of the local ellipse, of which κ is the ratio of the semiaxes α and β, and am(1 − f) K′ is the eccentric angle from the minor axis of the point of contact P of the generator HQ, so that two analytical constants are settled thereby; and the point H may be taken arbitrarily on the tangent line PQ, and HQ′ is then the other tangent of the focal ellipse; in which case θ3and θ2are the angles between the tangents HQ, HQ′, and between the focal distances HS, HS′, and k2will be HS·HS′, while HQ, HQ′ are δ, δ′.As H is moved along the tangent line HQ, a series of states of motion can be determined, and drawn with accuracy.Fig. 13.11. Equation (5) § 3 with slight modification will serve with the same notation for the steady rolling motion at a constant inclination α to the vertical of a body of revolution, such as a disk, hoop, wheel, cask, wine-glass, plate, dish, bowl, spinning top, gyrostat, or bicycle, on a horizontal plane, or a surface of revolution, as a coin in a conical lamp-shade.The point O is now the intersection of the axis GC′ with the vertical through the centre B of the horizontal circle described by the centre of gravity, and through the centre M of the horizontal circle described by P, the point of contact (fig. 13). Collected into a particle at G, the body swings round the vertical OB as a conical pendulum, of height AB or GL equal to g/μ2= λ, and GA would be the direction of the thread, of tension gM(GA/GL) dynes. The reaction with the plane at P will be an equal parallel force; and its moment round G will provide the couple which causes the velocity of the vector of angular momentum appropriate to the steady motion; and this moment will be gM·Gm dyne-cm. or ergs, if the reaction at P cuts GB in m.Draw GR perpendicular to GK to meet the horizontal AL in R, and draw RQC′K perpendicular to the axis Gz, and KC perpendicular to LG.The velocity of the vector GK of angular momentum is μ times the horizontal component, andhorizontal component /Aμ sin α = KC/KC′,(1)so thatgM·Gm = Aμ2sin α(KC/KC′),(2)A=KC′gGm = GQ·Gm.MKCμ2sin α(3)The instantaneous axis of rotation of the case of a gyrostat would be OP; drawing GI parallel to OP, and KK′ parallel to OG, making tan K′GC′ = (A/C) tan IGC’1; then if GK represents the resultant angular momentum, K′K will represent the part of it due to the rotation of the fly-wheel. Thus in the figure for the body rolling as a solid, with the fly-wheel clamped, the points m and Q move to the other side of G. The gyrostat may be supposed swung round the vertical at the end of a thread PA′ fastened at A′ where Pm produced cuts the vertical AB, and again at the point where it crosses the axis GO. The discussion of the small oscillation superposed on the state of steady motion requisite for stability is given in the next paragraph.12. In the theoretical discussion of the general motionGeneral motion of a gyrostat rolling on a plane.of a gyrostat rolling on a horizontal plane the safe and shortest plan apparently is to write down the most general equations of motion, and afterwards to introduce any special condition.Drawing through G the centre of gravity any three rectangular axes Gx, Gy, Gz, the notation employed isu, v, w,the components of linear velocity of G;p, q, r,the components of angular velocity about the axes;h1, h2, h3,the components of angular momentum;θ1, θ2, θ3,the components of angular velocity of the coordinate axes;x, y, z,the co-ordinates of the point of contact with the horizontal plane;X, Y, Z,the components of the reaction of the plane;α, β, γ,the direction cosines of the downward vertical.The geometrical equations, expressing that the point of contact is at rest on the plane, areu − ry + qz = 0,(1)v − pz + rx = 0,(2)w − qx + py = 0.(3)The dynamical equations aredu/dt − θ3v + θ2w = gα + X/M,(4)dv/dt − θ1w + θ2u = gβ + Y/M,(5)dw/dt − θ2u + θ1v = gγ + Z/M,(6)anddh1/dt − θ3h2+ θ2h3= yZ − zY,(7)dh2/dt − θ1h3+ θ3h1= zX − xZ,(8)dh3/dt − θ2h1+ θ1h2= xY − yX.(9)In the special case of the gyrostat where the surface is of revolution round Gz, and the body is kinetically symmetrical about Gz, we take Gy horizontal and Gzx through the point of contact so that y = 0; and denoting the angle between Gz and the downward vertical by θ (fig. 13)α = sin θ, β = 0, γ = cos θ.(10)The components of angular momentum areh1= Ap, h2= Aq, h3= Cr + K,(11)where A, C denote the moment of inertia about Gx, Gz, and K is the angular momentum of a fly-wheel fixed in the interior with its axis parallel to Gz; K is taken as constant during the motion.The axis Gz being fixed in the body,θ1= p, θ2= q = −dθ/dt, θ3= p cot θ.(12)With y = 0, (1), (2), (3) reduce tou = −qz, v = pz − rx, w = qx;(13)and, denoting the radius of curvature of the meridian curve of the rolling surface by ρ,dx= ρ cos θdθ= −q ρ cos θ,dz= −ρ sin θdθ= q ρ sin θ;dtdtdtdt(14)so thatdu= −dqz − q2ρ sin θ,dtdt(15)dv=dpz −drx + pqρ sin θ + qrρ sin θ,dtdtdt(16)dw=dqx − q2ρ cos θ.dtdt(17)The dynamical equations (4)...(9) can now be reduced toX= −dqz − p2z cotθ + q2(x − ρ sin θ) + prx cot θ − g sin θ,Mdt(18)Y=dpz −drx − pq (x + z cot θ − ρ sin θ) + qrp cos θ,Mdtdt(19)Z=dqx + q2(z − ρ cos θ) + p2z − prx − g cos θ,Mdt(20)−zY = Adp− Apq cot θ + qh3,dt(21)−zX − xZ = Adq+ Ap2cot θ − ph3,dtxY =dh3= Cdr= −Cqd.dtdtdθ(23)Eliminating Y between (19) and (23),(C+ x2)dr− xzdp+ pqx (x + z cot θ − ρ sin θ) − qrxρ cos θ = 0,Mdtdt(24)(C+ x2)dr− xzdp− px (x + z cot θ − ρ sin θ) + rxρ cos θ = 0.Mdθdθ(A)Eliminating Y between (19) and (21)(A+ z2)dp− xzdr−Apq cot θ + qh3MdtdtMM− pqz (x + z cot θ − ρ sin θ) + qrzρ cos θ = 0,(25)−xzdr+(A+ z2)dp+Ap cot θ −h3dθMdθMM+ pz (x + z cot θ − ρ sin θ) + rzρ cos θ = 0.(B)In the special case of a gyrostat rolling on the sharp edge of a circle passing through G, z = 0, ρ = 0, (A) and (B) reduce top =(C+ 1)dr=(1+1)dh3,Mx2dθMx2Cdθ(26)dp+ p cot θ =h3,d·p sin θ=h3sin θ;dθAdθA(27)d2h3+dh3cot θ =CMx2h3,dθ2dθA (Mx2+ C)(28)a differential equation of a hypergeometric series, of the form of Legendre’s zonal harmonic of fractional order n, given byn (n + 1) = CMx2/ A (Mx2+ C).(29)For a sharp point, x = 0, ρ = 0, and the previous equations are obtained of a spinning top.The elimination of X and Z between (18) (20) (22), expressed symbolically as(22) − z(18) + x(20) = 0,(30)gives(A+ x2+ z2)dq− ph3+(A+ z2)p2cot θ + p2xzMdtMM+ q2ρ (x cos θ − z sin θ) − prx (x + z cot θ) − g (x cos θ − z sin θ) = 0,(C)and this combined with (A) and (B) will lead to an equation the integral of which is the equation of energy.13. The equations (A) (B) (C) are intractable in this general form; but the restricted case may be considered when the axis moves in steady motion at a constant inclination α to the vertical; and the stability is secured if a small nutation of the axis can be superposed.It is convenient to put p = Ω sin θ, so that Ω is the angular velocity of the plane Gzx about the vertical; (A) (B) (C) become(C+ x2)dr− xz sin θdΩMdθdθ− Ωx (x sin θ − 2z cos θ − ρ sin2θ) + rxρ cos θ = 0,(A*)−xzdr+(A+ z2)sin θdΩ−h3+ 2Ω(A+ z2)cos θdθMdθMM+ Ωz sin θ (x − ρ sin θ) − rzρ cos θ = 0,(B*)(A+ x2+ z2)dq+ q2p (x cos θ − z sin θ) − Ωh3sin θMdtM+ Ω2(A+ z2)sin θ cos θ + Ω2xz sin2θM− Ωrx (x sin θ + z cos θ) − g (x cos θ − z sin θ) = 0.(C*)The steady motion and nutation superposed may be expressed byθ = α + L, sin θ = sin α + L cos α, cos θ = cos α − L sin α, Ω = μ + N, r = R + Q,(1)where L, N, Q are small terms, involving a factor enti, to express the periodic nature of the nutation; and then if a, c denote the mean value of x, z, at the point of contactx = a + Lρ cos α, z = c − Lρ sin α,(2)x sin θ + z cos θ = a sin α + c cos α + L (a cos α − c sin α),(3)x cos θ − z sin θ = a cos α − c sin α − L (a sin α + c cos α − ρ).(4)Substituting these values in (C*) with dq/dt = −d2θ/dt2= n2L, and ignoring products of the small terms, such as L2, LN, ...(A+ a2+ c2)Ln2− (μ + N)(CR + K+CQ)(sin α + L cos α)MMM+ (μ2+ 2μN) (A/M + c2− 2Lρc sin α) (sin α cos α + L cos α)+ (μ2+ 2μN) [ac − Lρ (a sin α − c sin α)] (sin2α + L sin 2α)− (μ + N) (R + Q) (a + Lρcos α) [a sin α + c cos α + L (a cos α − c sin α)]− g (a cos α − c sin α) + gL (a sin α + c cos α − ρ) = 0,(C**)which is equivalent to−μCR + Ksin α + μ2(A+ c2)sin αcos αMM+ μ2ac sin2α − μRa (a sin α + c cos α) − g (a cos α − c sin α) = 0,(5)the condition of steady motion; andDL + EQ + FN = 0,(6)whereD =(A+ a2+ c2)n2− μCK + Kcos α − 2μ2ρc sin2α cos αMM+ μ2(A/M + c2) cos α − μ2ρ (a sin α − c cos α) sin2α+ μ2ac sin 2α − μRρ cos α (a sin α + c cos α)− μRa (a cos α − c sin α) + g (a sin α + c cos α − ρ),(7)E = −μCsin α − μa (a sin α + c cos α),M(8)F = −CR + Ksin α + 2μ(A+ c2)sin α cos αMM+ 2μac sin2α − Ra (a sin α + c cos α).(9)With the same approximation (A*) and (B*) are equivalent to(C+ a2)Q− ac sin αN− μa (a sin α + 2c cos α − ρ sin2α) + Raρ cos α = 0,MLL(A**)−acQ+(A+ c2)sin αN−CR + K+ 2μ(A+ c2)cos αLMLMM+ μc sin α (a − ρ sin α) − Rcρ cos α = 0.(B**)The elimination of L, Q, N will lead to an equation for the determination of n2, and n2must 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,b = v/μ = (cP − aR) / μ = c sin α − aR / μ,(10)and drawing GL vertically upward of length λ = g/μ2, the height of the equivalent conical pendulum, the steady motion condition may be written(CR + K) μ sin α − μ2sin α cos α = −gM (a cos α − c sin α)+ M (μ2c sin α − μRa) (a sin α + c cos α)= gM [bλ−1(a sin α + c cos α) − a cos α + c sin α]= gM·PT,(11)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 α = 0, a = 0, we find F = 0, Q = 0, and−CR + K+ 2μ(A+ c2)− Rcp = 0,MM(12)(A+ c2)n2= μCR + K− μ2(A+ c2)+ μRρc − g (c − ρ),MMM(13)(A+ c2)2n2= ¼(CK + R+ Rcρ)2− g(A+ c2)(c − ρ).MMM(14)Thus for a top spinning upright on a rounded point, with K = 0, the stability requires thatR > 2k′√ {g (c − ρ)} / (k2+ cρ),(15)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 ρ = 0.Generally, with α = 0, but a ± 0, the condition (A) and (B) becomes(C+ a2)Q= 2μac − Raρ,ML−acQ=CR + K+ Rcρ − 2μ(A+ a2),LMM(16)so that, eliminating Q/L,2[(A+ c2)(C+ a2)− a2c2]μ =(C+ a2)(CR + K)+CRcρ,MMMMM(17)the condition when a coin or platter is rolling nearly flat on the table.Rolling along in a straight path, with α = ½π, c = 0, μ = 0, E = 0; andN/L = (CR + K)/A,(18)D =(A+ a2)n2+ g (a − ρ),MF = −CR + K− Ra2,M(19)N= −D=(A+ a2)n2+ g (a − ρ)M,LF(C+ a2)R +KMM(20)(A+ a2)n2=(CR + K)[(C+ a2)R +K]− g (a − ρ).MAMM(21)Thus with K = 0, and rolling with velocity V = Ra, stability requiresV2>a − ρ> ½Aa − ρ,2g2C/A (C/Ma2+ 1)CC/Ma2+ 1(22)or the body must have acquired velocity greater than attained by rolling down a plane through a vertical height ½ (a − ρ) A/C.On a sharp edge, with ρ = 0, a thin uniform disk or a thin ring requiresV2/2g > a/6 or a/8.(23)The gyrostat can hold itself upright on the plane without advance when R = 0, providedK2/AM − g (a − ρ) is positive.(24)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 requireG′(aV +G′)> gh.AAFor 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.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,Gyrostatic chain.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,θ1= −dq/dt, θ2= dp/dt,(1)h1= Aθ1, h2= Aθ2, h3= K,(2)to be employed in the dynamical equationsdh1− θ3h2+ θ2h3= L, ...dt(3)in which θ3h1and θ3h2can be omitted.For the kth fly-wheel−Aqk+ Kpk= Ta (qk− sk) + Ta (qk− sk+1),(4)Apk+ Kqk= −Ta (pk− rk) − Ta (pk− rk+1);(5)and for the motion of translationMxk= T (rk+1− rk), Myk= T (sk+1− sk);(6)while the geometrical relations arexk+1− xk= a (pk+1+ pk) + 2lrk+1,(7)yk+1− yk= a (qk+1+ qk) + 2lsk+1.(8)Puttingx + yi = w, p + qi = ω, r + si = σ,(9)these three pairs of equations may be replaced by the three equationsAῶk− Kῶki + 2Taῶk− Ta (σk+1+ σk) = 0,(10)Mῶk− T (σk+1− σk) = 0,(11)ωk+1− ωk− a(ῶk+1+ ῶk− 2lσk+1) = 0.(12)For a vibration of circular polarization assume a solutionωk, ῶk, σk= (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 isU = 2 (a + l) n/c.(14)ThenP (−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, tocos c =(2 Ta + Kn − An2) (1 − Mn2l/T) − Mna2,2Ta + Kn − An2+ Mna2(18)2 sin2½c =Mn22Ta (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.Puttingρ = M/2 (a + l), the mass per unit length of the chain,
(23)
(1 − cos θ) exp (φ − ψ − q′t) i= √ (sec β − cos θ) √ (cos β − cos θ) + i (√sec β − √ cos β) √ cos θ,
(24)
q, q′ = n√ (2 sec β) ± n√ (2 cos β);
(25)
and thence α, β, γ, δ can be inferred.
The physical constants of a given symmetrical top have been denoted in § 1 by M, h, A, C, and l, n, T; to specify a given state of general motion we have G, G′ or CR, D, E, or F, which may be called the dynamical constants; or κ, v, w, v1, v2, or f, f′, f1, f2, the analytical constants; or the geometrical constants, such as α, β, δ, δ′, k of a given articulated hyperboloid.
There is thus a triply infinite series of a state of motion; the choice of a typical state can be made geometrically on the hyperboloid, flattened in the plane of the local ellipse, of which κ is the ratio of the semiaxes α and β, and am(1 − f) K′ is the eccentric angle from the minor axis of the point of contact P of the generator HQ, so that two analytical constants are settled thereby; and the point H may be taken arbitrarily on the tangent line PQ, and HQ′ is then the other tangent of the focal ellipse; in which case θ3and θ2are the angles between the tangents HQ, HQ′, and between the focal distances HS, HS′, and k2will be HS·HS′, while HQ, HQ′ are δ, δ′.As H is moved along the tangent line HQ, a series of states of motion can be determined, and drawn with accuracy.
11. Equation (5) § 3 with slight modification will serve with the same notation for the steady rolling motion at a constant inclination α to the vertical of a body of revolution, such as a disk, hoop, wheel, cask, wine-glass, plate, dish, bowl, spinning top, gyrostat, or bicycle, on a horizontal plane, or a surface of revolution, as a coin in a conical lamp-shade.
The point O is now the intersection of the axis GC′ with the vertical through the centre B of the horizontal circle described by the centre of gravity, and through the centre M of the horizontal circle described by P, the point of contact (fig. 13). Collected into a particle at G, the body swings round the vertical OB as a conical pendulum, of height AB or GL equal to g/μ2= λ, and GA would be the direction of the thread, of tension gM(GA/GL) dynes. The reaction with the plane at P will be an equal parallel force; and its moment round G will provide the couple which causes the velocity of the vector of angular momentum appropriate to the steady motion; and this moment will be gM·Gm dyne-cm. or ergs, if the reaction at P cuts GB in m.
Draw GR perpendicular to GK to meet the horizontal AL in R, and draw RQC′K perpendicular to the axis Gz, and KC perpendicular to LG.
The velocity of the vector GK of angular momentum is μ times the horizontal component, and
horizontal component /Aμ sin α = KC/KC′,
(1)
so that
gM·Gm = Aμ2sin α(KC/KC′),
(2)
(3)
The instantaneous axis of rotation of the case of a gyrostat would be OP; drawing GI parallel to OP, and KK′ parallel to OG, making tan K′GC′ = (A/C) tan IGC’1; then if GK represents the resultant angular momentum, K′K will represent the part of it due to the rotation of the fly-wheel. Thus in the figure for the body rolling as a solid, with the fly-wheel clamped, the points m and Q move to the other side of G. The gyrostat may be supposed swung round the vertical at the end of a thread PA′ fastened at A′ where Pm produced cuts the vertical AB, and again at the point where it crosses the axis GO. The discussion of the small oscillation superposed on the state of steady motion requisite for stability is given in the next paragraph.
12. In the theoretical discussion of the general motionGeneral motion of a gyrostat rolling on a plane.of a gyrostat rolling on a horizontal plane the safe and shortest plan apparently is to write down the most general equations of motion, and afterwards to introduce any special condition.
Drawing through G the centre of gravity any three rectangular axes Gx, Gy, Gz, the notation employed is
The geometrical equations, expressing that the point of contact is at rest on the plane, are
u − ry + qz = 0,
(1)
v − pz + rx = 0,
(2)
w − qx + py = 0.
(3)
The dynamical equations are
du/dt − θ3v + θ2w = gα + X/M,
(4)
dv/dt − θ1w + θ2u = gβ + Y/M,
(5)
dw/dt − θ2u + θ1v = gγ + Z/M,
(6)
and
dh1/dt − θ3h2+ θ2h3= yZ − zY,
(7)
dh2/dt − θ1h3+ θ3h1= zX − xZ,
(8)
dh3/dt − θ2h1+ θ1h2= xY − yX.
(9)
In the special case of the gyrostat where the surface is of revolution round Gz, and the body is kinetically symmetrical about Gz, we take Gy horizontal and Gzx through the point of contact so that y = 0; and denoting the angle between Gz and the downward vertical by θ (fig. 13)
α = sin θ, β = 0, γ = cos θ.
(10)
The components of angular momentum are
h1= Ap, h2= Aq, h3= Cr + K,
(11)
where A, C denote the moment of inertia about Gx, Gz, and K is the angular momentum of a fly-wheel fixed in the interior with its axis parallel to Gz; K is taken as constant during the motion.
The axis Gz being fixed in the body,
θ1= p, θ2= q = −dθ/dt, θ3= p cot θ.
(12)
With y = 0, (1), (2), (3) reduce to
u = −qz, v = pz − rx, w = qx;
(13)
and, denoting the radius of curvature of the meridian curve of the rolling surface by ρ,
(14)
so that
(15)
(16)
(17)
The dynamical equations (4)...(9) can now be reduced to
(18)
(19)
(20)
(21)
(23)
Eliminating Y between (19) and (23),
(24)
(A)
Eliminating Y between (19) and (21)
− pqz (x + z cot θ − ρ sin θ) + qrzρ cos θ = 0,
(25)
+ pz (x + z cot θ − ρ sin θ) + rzρ cos θ = 0.
(B)
In the special case of a gyrostat rolling on the sharp edge of a circle passing through G, z = 0, ρ = 0, (A) and (B) reduce to
(26)
(27)
(28)
a differential equation of a hypergeometric series, of the form of Legendre’s zonal harmonic of fractional order n, given by
n (n + 1) = CMx2/ A (Mx2+ C).
(29)
For a sharp point, x = 0, ρ = 0, and the previous equations are obtained of a spinning top.
The elimination of X and Z between (18) (20) (22), expressed symbolically as
(22) − z(18) + x(20) = 0,
(30)
gives
+ q2ρ (x cos θ − z sin θ) − prx (x + z cot θ) − g (x cos θ − z sin θ) = 0,
(C)
and this combined with (A) and (B) will lead to an equation the integral of which is the equation of energy.
13. The equations (A) (B) (C) are intractable in this general form; but the restricted case may be considered when the axis moves in steady motion at a constant inclination α to the vertical; and the stability is secured if a small nutation of the axis can be superposed.
It is convenient to put p = Ω sin θ, so that Ω is the angular velocity of the plane Gzx about the vertical; (A) (B) (C) become
− Ωx (x sin θ − 2z cos θ − ρ sin2θ) + rxρ cos θ = 0,
(A*)
+ Ωz sin θ (x − ρ sin θ) − rzρ cos θ = 0,
(B*)
− Ωrx (x sin θ + z cos θ) − g (x cos θ − z sin θ) = 0.
(C*)
The steady motion and nutation superposed may be expressed by
θ = α + L, sin θ = sin α + L cos α, cos θ = cos α − L sin α, Ω = μ + N, r = R + Q,
(1)
where L, N, Q are small terms, involving a factor enti, 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
x = a + Lρ cos α, z = c − Lρ sin α,
(2)
x sin θ + z cos θ = a sin α + c cos α + L (a cos α − c sin α),
(3)
x cos θ − z sin θ = a cos α − c sin α − L (a sin α + c cos α − ρ).
(4)
Substituting these values in (C*) with dq/dt = −d2θ/dt2= n2L, and ignoring products of the small terms, such as L2, LN, ...
+ (μ2+ 2μN) (A/M + c2− 2Lρc sin α) (sin α cos α + L cos α)+ (μ2+ 2μN) [ac − Lρ (a sin α − c sin α)] (sin2α + L sin 2α)− (μ + N) (R + Q) (a + Lρcos α) [a sin α + c cos α + L (a cos α − c sin α)]− g (a cos α − c sin α) + gL (a sin α + c cos α − ρ) = 0,
+ (μ2+ 2μN) (A/M + c2− 2Lρc sin α) (sin α cos α + L cos α)
+ (μ2+ 2μN) [ac − Lρ (a sin α − c sin α)] (sin2α + L sin 2α)
− (μ + N) (R + Q) (a + Lρcos α) [a sin α + c cos α + L (a cos α − c sin α)]
− g (a cos α − c sin α) + gL (a sin α + c cos α − ρ) = 0,
(C**)
which is equivalent to
+ μ2ac sin2α − μRa (a sin α + c cos α) − g (a cos α − c sin α) = 0,
(5)
the condition of steady motion; and
DL + EQ + FN = 0,
(6)
where
+ μ2(A/M + c2) cos α − μ2ρ (a sin α − c cos α) sin2α+ μ2ac sin 2α − μRρ cos α (a sin α + c cos α)− μRa (a cos α − c sin α) + g (a sin α + c cos α − ρ),
+ μ2(A/M + c2) cos α − μ2ρ (a sin α − c cos α) sin2α
+ μ2ac sin 2α − μRρ cos α (a sin α + c cos α)
− μRa (a cos α − c sin α) + g (a sin α + c cos α − ρ),
(7)
(8)
+ 2μac sin2α − Ra (a sin α + c cos α).
(9)
With the same approximation (A*) and (B*) are equivalent to
(A**)
+ μc sin α (a − ρ sin α) − Rcρ cos α = 0.
(B**)
The elimination of L, Q, N will lead to an equation for the determination of n2, and n2must 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,
b = v/μ = (cP − aR) / μ = c sin α − aR / μ,
(10)
and drawing GL vertically upward of length λ = g/μ2, the height of the equivalent conical pendulum, the steady motion condition may be written
(CR + K) μ sin α − μ2sin α cos α = −gM (a cos α − c sin α)
+ M (μ2c sin α − μRa) (a sin α + c cos α)
= gM [bλ−1(a sin α + c cos α) − a cos α + c sin α]= gM·PT,
= gM [bλ−1(a sin α + c cos α) − a cos α + c sin α]
= gM·PT,
(11)
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 α = 0, a = 0, we find F = 0, Q = 0, and
(12)
(13)
(14)
Thus for a top spinning upright on a rounded point, with K = 0, the stability requires that
R > 2k′√ {g (c − ρ)} / (k2+ cρ),
(15)
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 ρ = 0.
Generally, with α = 0, but a ± 0, the condition (A) and (B) becomes
(16)
so that, eliminating Q/L,
(17)
the condition when a coin or platter is rolling nearly flat on the table.
Rolling along in a straight path, with α = ½π, c = 0, μ = 0, E = 0; and
N/L = (CR + K)/A,
(18)
(19)
(20)
(21)
Thus with K = 0, and rolling with velocity V = Ra, stability requires
(22)
or the body must have acquired velocity greater than attained by rolling down a plane through a vertical height ½ (a − ρ) A/C.
On a sharp edge, with ρ = 0, a thin uniform disk or a thin ring requires
V2/2g > a/6 or a/8.
(23)
The gyrostat can hold itself upright on the plane without advance when R = 0, provided
K2/AM − g (a − ρ) is positive.
(24)
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
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.
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,Gyrostatic chain.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,
θ1= −dq/dt, θ2= dp/dt,
(1)
h1= Aθ1, h2= Aθ2, h3= K,
(2)
to be employed in the dynamical equations
(3)
in which θ3h1and θ3h2can be omitted.
For the kth fly-wheel
−Aqk+ Kpk= Ta (qk− sk) + Ta (qk− sk+1),
(4)
Apk+ Kqk= −Ta (pk− rk) − Ta (pk− rk+1);
(5)
and for the motion of translation
Mxk= T (rk+1− rk), Myk= T (sk+1− sk);
(6)
while the geometrical relations are
xk+1− xk= a (pk+1+ pk) + 2lrk+1,
(7)
yk+1− yk= a (qk+1+ qk) + 2lsk+1.
(8)
Putting
x + yi = w, p + qi = ω, r + si = σ,
(9)
these three pairs of equations may be replaced by the three equations
Aῶk− Kῶki + 2Taῶk− Ta (σk+1+ σk) = 0,
(10)
Mῶk− T (σk+1− σk) = 0,
(11)
ωk+1− ωk− a(ῶk+1+ ῶk− 2lσk+1) = 0.
(12)
For a vibration of circular polarization assume a solution
ωk, ῶk, σk= (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
(18)
(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,