Theory of the Symmetrical Top.1. The physical constants of a given symmetrical top, expressed in C.G.S. units, which are employed in the subsequent formulae, are denoted by M, h, C and A. M is the weight in grammes (g) as given by the number of gramme weights which equilibrate the top when weighed in a balance; h is the distance OG in centimetres (cm.) between G the centre of gravity and O the point of support, and Mh may be called the preponderance in g.-cm.; Mh and M can be measured by a spring balance holding up in a horizontal position the axis OC in fig. 8 suspended at O. Then gMh (dyne-cm. or ergs) is the moment of gravity about O when the axis OG is horizontal, gMh sin θ being the moment when the axis OG makes an angle θ with the vertical, and g = 981 (cm./s2) on the average; C is the moment of inertia of the top about OG, and A about any axis through O at right angles to OG, both measured in g-cm.2.To measure A experimentally, swing the top freely about O in small plane oscillation, and determine the length, l cm., of the equivalent simple pendulum; thenl = A/Mh, A = Mhl.(1)Next make the top, or this simple pendulum, perform small conical revolutions, nearly coincident with the downward vertical position of equilibrium, and measure n, the mean angular velocity of the conical pendulum in radians / second; and T its period in seconds; then4π2/T2= n2= g/l = gMh/A;(2)and f = n/2π is the number of revolutions per second, called thefrequency, T = 2π/n is the period of a revolution, in seconds.2. In the popular explanation of the steady movement of the top at a constant inclination to the vertical, depending on the composition of angular velocity, such as given in Perry’sSpinning Tops, or Worthington’sDynamics of Rotation,Steady motion of the top.it is asserted that the moment of gravity is always generating an angular velocity about an axis OB perpendicular to the vertical plane COC′ through the axis of the top OC′; and this angular velocity, compounded with the resultant angular velocity about an axis OI, nearly coincident with OC′, causes the axes OI and OC′ to keep taking up a new position by moving at right angles to the plane COC′, at a constant precessional angular velocity, say μ rad./sec., round the vertical OC (fig. 4).If, however, the axis OC′ is prevented from taking up this precessional velocity, the top at once falls down; thence all the ingenious attempts—for instance, in the swinging cabin of the Bessemer ship—to utilise the gyroscope as a mechanical directive agency have always resulted in failure (Engineer, October 1874), unless restricted to actuate a light relay, which guides the mechanism, as in steering a torpedo.An experimental verification can be carried out with the gyroscope in fig. 1; so long as the vertical spindle is free to rotate in its socket, the rapidly rotating wheel will resist the impulse of tapping on the gimbal by moving to one side; but when the pinch screw prevents the rotation of the vertical spindle in the massive pedestal, this resistance to the tapping at once disappears, provided the friction of the table prevents the movement of the pedestal; and if the wheel has any preponderance, it falls down.Familiar instances of the same principles are observable in the movement of a hoop, or in the steering of a bicycle; it is essential that the handle of the bicycle should be free to rotate to secure the stability of the movement.The bicycle wheel, employed as a spinning top, in fig. 4, can also be held by the stalk, and will thus, when rotated rapidly, convey a distinct muscular impression of resistance to change of direction, if brandished.3. A demonstration, depending on the elementary principles of dynamics, of the exact conditions required for theElementary demonstration of the condition of steady motion.axis OC′ of a spinning top to spin steadily at a constant inclination θ to the vertical OC, is given here before proceeding to the more complicated question of the general motion, when θ, the inclination of the axis, is varying by nutation.It is a fundamental principle in dynamics that if OH is a vector representing to scale the angular momentum of a system, and if Oh is the vector representing the axis of the impressed couple or torque, then OH will vary so that the velocity of H is represented to scale by the impressed couple Oh, and if the top is moving freely about O, Oh is at right angles to the vertical plane COC′, andOh = gMh sin θ.(1)In the case of the steady motion of the top, the vector OH lies in the vertical plane COC′, in OK suppose (fig. 4), and has a component OC = G about the vertical and a component OC′ = G′, suppose, about the axis OC; and G′ = CR, if R denotes the angular velocity of the top with which it is spun about OC′.If μ denotes the constant precessional angular velocity of the vertical plane COC′ the components of angular velocity and momentum about OA are μ sin θ and Aμ sin θ, OA being perpendicular to OC′ in the plane COC′; so that the vector OK has the componentsOC′ = G′, and C′K = Aμ sin θ,(2)and the horizontal componentCK = OC′ sin θ − C′K cos θ= G′ sin θ − Aμ sin θ cos θ.(3)The velocity of K being equal to the impressed couple Oh,gMh sin θ = μ·CK = sin θ (G′μ − Aμ2cos θ),(4)and dropping the factor sin θ,Aμ2cos θ − G′μ + gMh = 0, or Aμ2cos θ − CRμ + An2= 0,(5)the condition for steady motion.Solving this as a quadratic in μ, the roots μ1, μ2are given byμ1, μ2=G′sec θ[1 ± √ (1 −4A2n2cos θ)];2AG′2(6)and the minimum value of G′ = CR for real values of μ is given byG′2= cos θ,CR2√(cos θ);4A2n2An(7)for a smaller value of R the top cannot spin steadily at the inclination θ to the upward vertical.Interpreted geometrically in fig. 4μ = gMh sin θ/CK = An2/KN, and μ = C′K/A sin θ = KM/A,(8)KM·KN = A2n2,(9)so that K lies on a hyperbola with OC, OC′ as asymptotes.4. Suppose the top or gyroscope, instead of moving freely about the point O, is held in a ring or frame which is compelledConstrained motion of the gyroscope.to rotate about the vertical axis OC with constant angular velocity μ; then if N denotes the couple of reaction of the frame keeping the top from falling, acting in the plane COC’, equation (4) § 3 becomes modified intogMh sin θ − N = μ·CK = sin θ G′μ − Aμ2cos θ,(1)N = sin θ (Aμ2cos θ − G′μ + gMh)= A sin θ cos θ (μ − μ1) (μ − μ2);(2)and hence, as μ increases through μ2and μ1, the sign of N can be determined, positive or negative, according as the tendency of the axis is to fall or rise.When G′ = CR is large, μ2is large, andμ1≈ gMh/G′ = An2/CR,(3)Fig. 10.the same for all inclinations, and this is the precession observed in the spinning top and centrifugal machine of fig. 10 This is true accurately when the axis OC′ is horizontal, and then it agrees with the result of the popular explanation of § 2.If the axis of the top OC′ is pointing upward, the precession is in the same direction as the rotation, and an increase of μ from μ1makes N negative, and the top rises; conversely a decrease of the procession μ causes the axis to fall (Perry,Spinning Tops, p. 48).If the axis points downward, as in the centrifugal machine with upper support, the precession is in the opposite direction to the rotation, and to make the axis approach the vertical position the precession must be reduced.This is effected automatically in the Weston centrifugal machine (fig. 10) used for the separation of water andCentrifugal machine.molasses, by the friction of the indiarubber cushions above the support; or else the spindle is produced downwards below the drum a short distance, and turns in a hole in a weight resting on the bottom of the case, which weight is dragged round until the spindle is upright; this second arrangement is more effective when a liquid is treated in the drum, and wave action is set up (The Centrifugal Machine, C. A. Matthey).Similar considerations apply to the stability of the whirling bowl in a cream-separating machine.We can write equation (1)N = An2sin θ − μ·CK = (A2n2− KM·KN) sin θ/A,(4)so that N is negative or positive, and the axis tends to rise or fall according as K moves to the inside or outside of the hyperbola of free motion. Thus a tap on the axis tending to hurry the precession is equivalent to an impulse couple giving an increase to C′K, and will make K move to the interior of the hyperbola and cause the axis to rise; the steering of a bicycle may be explained in this way; but K1will move to the exterior of the hyperbola, and so the axis will fall in this second more violent motion.Friction on the point of the top may be supposed to act like a tap in the direction opposite to the precession; and so the axis of a top spun violently rises at first and up to the vertical position, but falls away again as the motion dies out. Friction considered as acting in retarding the rotation may be compared to an impulse couple tending to reduce OC′, and so make K and K1both move to the exterior of the hyperbola, and the axis falls in both cases. The axis may rise or fall according to the direction of the frictional couple, depending on the shape of the point; an analytical treatment of the varying motion is very intractable; a memoir by E. G. Gallop may be consulted in theTrans. Camb. Phil. Soc., 1903.The earth behaves in precession like a large spinning top, of which the axis describes a circle round the pole of the ecliptic of mean angular radius θ, about 23½°, in a period of 26,000 years, so that R/μ = 26000 × 365; and the mean couple producing precession isCRμ sin θ = CR2sin 23½° /26000 × 365,(5)one 12 millionth part of ½CR2, the rotation energy of the earth.5. If the preponderance is absent, by making the C·G coincide with O, and if Aμ is insensible compared with G′,N = −G′μ sin θ,(1)the formula which suffices to explain most gyroscopic action.Thus a carriage running round a curve experiences, in consequenceGyroscopic action of railway wheels.of the rotation of the wheels, an increase of pressure Z on the outer track, and a diminution Z on the inner, giving a couple, if a is the gauge,Za = G′μ,(2)tending to help the centrifugal force to upset the train; and if c is the radius of the curve, b of the wheels, C their moment of inertia, and v the velocity of the train,μ = v/c, G′ = Cv/b,(3)Z = Cv2/abc (dynes),(4)so that Z is the fraction C/Mabof the centrifugal force Mv2/c, or the fraction C/Mh of its transference of weight, with h the height of the centre of gravity of the carriage above the road. A Brennan carriage on a monorail would lean over to the inside of the curve at an angle α, given bytan α = G′μ/gMh = G′v/gMhc.(6)The gyroscopic action of a dynamo, turbine, and other rotating machinery on a steamer, paddle or screw, due to its rolling and pitching, can be evaluated in a similar elementary manner (Worthington,Dynamics of Rotation), and Schlick’s gyroscopic apparatus is intended to mitigate the oscillation.6. If the axis OC in fig. 4 is inclined at an angle α to the vertical, the equation (2) § 4 becomesN = sin θ (Aμ2cos θ − G′μ) + gMh sin (α − θ).(1)Suppose, for instance, that OC is parallel to the earth’s axis, and that the frame is fixed in the meridian; then α is the co-latitude, and μ is the angular velocity of the earth, the square of which may be neglected; so that, putting N = 0, α − θ = E,gMh sin E − G′μ sin (α − E) = 0,(2)tan E =G′μ sin α≈G′μsin α.gMh + G′μ cos αgMh(3)This is the theory of Gilbert’s barogyroscope, described in Appell’sMécanique rationnelle, ii. 387: it consists essentially of a rapidlyThe barogyroscope.rotated fly-wheel, mounted on knife-edges by an axis perpendicular to its axis of rotation and pointing east and west; spun with considerable angular momentum G′, and provided with a slight preponderance Mh, it should tilt to an angle E with the vertical, and thus demonstrate experimentally the rotation of the earth.In Foucault’s gyroscope (Comptes rendus, 1852; Perry, p. 105)Foucault’s gyroscope.the preponderance is made zero, and the axis points to the pole, when free to move in the meridian.Generally, if constrained to move in any other plane, the axis seeks the position nearest to the polar axis, like a dipping needle with respect to the magnetic pole. (A gyrostatic working model of the magnetic compass, by Sir W. Thomson. British Association Report, Montreal, 1884. A. S. Chessin, St Louis Academy of Science, January 1902.)A spinning top with a polished upper plane surface will provide an artificial horizon at sea, when the real horizon is obscured. The first instrument of this kind was constructed by Serson, and is described in theGentleman’s Magazine,Gyroscopic horizon.vol. xxiv., 1754; also by Segner in hisSpecimen theoriae turbinum(Halae, 1755). The inventor was sent to sea by the Admiralty to test his instrument, but he was lost in the wreck of the “Victory,” 1744. A copy of the Serson top, from the royal collection, is now in the Museum of King’s College, London. Troughton’s Nautical Top (1819) is intended for the same purpose.The instrument is in favour with French navigators, perfected byAdmiral Fleuriais (fig. 9); but it must be noticed that the horizon given by the top is inclined to the true horizon at the angle E given by equation (3) above; and if μ1is the precessional angular velocity as given by (3) § 4, and T = 2π/μ, its period in seconds,tan E =μcos lat =T cos lat, or E =T cos lat,μ1864008π(4)if E is expressed in minutes, taking μ = 2π/86400; thus making the true latitude E nautical miles to the south of that given by the top (Revue maritime, 1890;Comptes rendus, 1896).This can be seen by elementary consideration of the theory above, for the velocity of the vector OC′ of the top due to the rotation of the earth isμ·OC′ cos lat = gMh sin E = μ1·OC′ sin E,sin E =μcos lat, E =T cos lat,μ18π(5)Fig. 11.in which 8π can be replaced by 25, in practice; so that the Fleuriais gyroscopic horizon is an illustration of the influence of the rotation of the earth and of the need for its allowance.7. In the ordinary treatment of the general theory of the gyroscope, the motion is referred to two sets of rectangular axes; theEuler’s coordinate angles.one Ox, Oy, Oz fixed in space, with Oz vertically upward and the other OX, OY, OZ fixed in the rotating wheel with OZ in the axis of figure OC.The relative position of the two sets of axes is given by means of Euler’s unsymmetrical angles θ, φ, ψ, such that the successive turning of the axes Ox, Oy, Oz through the angles (i.) ψ about Oz, (ii.) θ about OE, (iii.) φ about OZ, brings them into coincidence with OX, OY, OZ, as shown in fig. 11, representing theconcaveside of a spherical surface.The component angular velocities about OD, OE, OZ areψsin θ,θ,φ+ψcos θ;(1)so that, denoting the components about OX, OY, OZ by P, Q, R,P =θcos φ+ψsin θ sin φ,Q =−θsin φ+ψsin θ cos φ,R =φ+ψcos θ.(2)Consider, for instance, the motion of a fly-wheel of preponderance Mh, and equatoreal moment of inertia A, of which the axis OC is held in a light ring ZCX at a constant angle γ with OZ, while OZ is held by another ring zZ, which constrains it to move round the vertical Oz at a constant inclination θ with constant angular velocity μ, so thatθ= 0,ψ= μ;(3)P = μ sin θ sin φ, Q = μ sin θ cos φ, R =φ+ μ cos θ.(4)With CXF a quadrant, the components of angular velocity and momentum about OF, OY, areP cos γ − R sin γ, Q, and A (P cos γ − R sin γ), AQ,(5)so that, denoting the components of angular momentum of the fly-wheel about OC, OX, OY, OZ by K or G′, h1, h2, h3,h1= A (P cos γ − R sin γ) cos γ + K sin γ,(6)h2= AQ,(7)h3= −A (P cos γ − R sin γ) sin γ + K cos γ;(8)and the dynamical equationdh3− h1Q + h2P = N,dt(9)with K constant, and with preponderance downwardN = gMh cos zY sin γ = gMh sin γ sin θ cos φ,(10)reduces toAd2φsin γ + Aμ2sin γ sin2θ sin φ cos φdt2+ Aμ2cos γ sin θ cos θ cos φ − (Kμ + gMh) sin θ cos φ = 0.(11)The position of relative equilibrium is given bycos φ = 0, and sin φ =Kμ + gMh − Aμ2cos γ cos θ.Aμ2sin γ sin θ(12)For small values of μ the equation becomesAd2φsin γ − (Kμ + gMh) sin θ cos φ = 0,dt2(13)so that φ = ½π gives the position of stable equilibrium, and the period of a small oscillation is 2π √{A sin γ/(Kμ + gMh) sin θ}.In the general case, denoting the periods of vibration about φ = ½π, −½π, and the sidelong position of equilibrium by 2π/(n1, n2, or n3), we shall findn12=sin θ{ gMh + Kμ − Aμ2cos (γ − θ) },A sin γ(14)n22=sin θ{ −gMh − Kμ + Aμ2cos (γ + θ) },A sin γ(15)n3= n1n2/μ sin θ.(16)The first integral of (11) gives½A(dφ)2sin γ + ½Aμ2sin γ sin2θ sin2φdt− Aμ2cos γ sin θ cos θ sin φ + (Kμ + gMh) sin θ sin φ − H = 0,(17)and putting tan (¼π + ½φ) = z, this reduces todzn √Zdt(18)where Z is a quadratic in z2, so that z is a Jacobian elliptic function of t, and we havetan (¼π + ½φ) = C (tn, dn, nc, or cn) nt,(19)according as the ring ZC performs complete revolutions, or oscillates about a sidelong position of equilibrium, or oscillates about the stable position of equilibrium φ = ±½π.Suppose Oz is parallel to the earth’s axis, and μ is the diurnal rotation, the square of which may be neglected, then if Gilbert’s barogyroscope of § 6 has the knife-edges turned in azimuth to make an angle β with E. and W., so that OZ lies in the horizon at an angle E·β·N., we must put γ = ½π, cos θ = sin α sin β; and putting φ = ½π − δ + E, where δ denotes the angle between Zz and the vertical plane Zζ through the zenith ζ,sin θ cos δ = cos α, sin θ sin δ = sin α cos β;(20)so that equations (9) and (10) for relative equilibrium reduce togMh sin E = KQ = Kμ sin θ cos φ = Kμ sin θ sin (δ − E),(21)and will change (3) § 6 intotan E =Kμ sin α cos β,gMh + Kμ cos α(22)a multiplication of (3) § 6 by cos β (Gilbert,Comptes rendus, 1882).Changing the sign of K or h and E and denoting the revolutions/second of the gyroscope wheel by F, then in the preceding notation, T denoting the period of vibration as a simple pendulum,tan E =Kμ sin α cos β=F sin α cos β,gMh − Kμ cos α86400 A/T2C − F cos α(23)so that the gyroscope would reverse if it were possible to make F cos α > 86400 A/T2C (Föppl,Münch. Ber, 1904).A gyroscopic pendulum is made by the addition to it of a fly-wheel, balanced and mounted, as in Gilbert’s barogyroscope, in a ring movable about an axis fixed in the pendulum, in the vertical plane of motion.As the pendulum falls away to an angle θ with the upward vertical, and the axis of the fly-wheel makes an angle φ with the vertical plane of motion, the three components of angular momentum areh1= K cos φ, h2= Aθ+ K sin φ, h3= Aφ,(24)where h3is the component about the axis of the ring and K of the fly-wheel about its axis; and if L, M′, N denote the components of the couple of reaction of the ring, L may be ignored, while N is zero, with P = 0, Q =θ, R = 0, so thatM′ = h2= Aθ+ Kφcos φ,(25)0 = h3− h1θ= Aφ− Kθcos φ.(26)For the motion of the pendulum, including the fly-wheel,MK2θ= gMH sin θ − M′ = gMH sin θ − Aθ− Kφcos φ.(27)If θ and φ remain small,Aφ= Kθ, Aφ= K(θ − α),(28)(MK2+ A)θ+ (K2/A) (θ − α) − gMHθ = 0;(29)so that the upright position will be stable if K2> gMHA, or the rotation energy of the wheel greater than ½A/C times the energy acquired by the pendulum in falling between the vertical and horizontal position; and the vibration will synchronize with a simple pendulum of length(MK2+ A) / [(K2/gA) − MH].(30)This gyroscopic pendulum may be supposed to represent a ship among waves, or a carriage on a monorail, and so affords an explanation of the gyroscopic action essential in the apparatus of Schlick and Brennan.8. Careful scrutiny shows that the steady motion of a top is not steady absolutely; it reveals a small nutationGeneral motion of the top.superposed, so that a complete investigation requires a return to the equations of unsteady motion, and for the small oscillation to consider them in a penultimate form.In the general motion of the top the vector OH of resultant angular momentum is no longer compelled to lie in the vertical plane COC′ (fig. 4), but since the axis Oh of the gravity couple is always horizontal, H will describe a curve in a fixed horizontal plane through C. The vector OC′ of angular momentum about the axis will be constant in length, but vary in direction; and OK will be the component angular momentum in the vertical plane COC′, if the planes through C and C′ perpendicular to the lines OC and OC′ intersect in the line KH; and if KH is the component angular momentum perpendicular to the plane COC′, the resultant angular momentum OH has the three components OC′, C′K, KH, represented in Euler’s angles byKH = A dθ/dt, C′K = A sin θd ψ/dt, OC′ = G′.(1)Drawing KM vertical and KN parallel to OC′, thenKM = A dψ/dt, KN = CR − A cos θ dψ/dt = (C − A) R + A dφ/dt(2)so that in the spherical top, with C = A, KN = A dφ/dt.The velocity of H is in the direction KH perpendicular to the plane COC′, and equal to gMh sin θ or An2sin θ, so that if a point in the axis OC′ at a distance An2from O is projected on the horizontal plane through C in the point P on CK, the curve described by P, turned forwards through a right angle, will be the hodograph of H; this is expressed byAn2sin θ e(ψ + 1/2π)i= iAn2sin θ eψi=d(ρeῶi)dt(3)where ρeῶiis the vector CH; and so the curve described by P and the motion of the axis of the top is derived from the curve described by H by a differentiation.Resolving the velocity of H in the direction CH,d·CH/dt = An2sin θ sin KCH = An2sin θ KH/CH,(4)d·½CH2/dt = A2n2sin θ dθ/dt.(5)and integrating½CH2= A2n2(E − cos θ),(6)½OH2= A2n2(F − cos θ),(7)½C′H2= A2n2(D − cos θ),(8)where D, E, F are constants, connected byF = E + G2/2A2n2= D + G′2/2A2n2.(9)ThenKH2= OH2− OK2,(10)OK2sin2θ = CC′2= G2− 2GG′ cos θ + G′2,(11)A2sin2θ (dθ/dt)2= 2A2n2(F − cos θ) sin2θ − G2+ 2GG′ cos θ − G′2;(12)and putting cos θ = z,(dz)= 2n2(F − z) (1 − z2) − (G2− 2GG′z + G′2) /A2dt(13)= 2n2(E − z) (1 − z2) − (G′ − Gz)2/A2= 2n2(D − z) (1 − z2) − (G − G′z)2/A2= 2n2Z suppose.Denoting the roots of Z = 0 by z1, z2, z3, we shall have them arranged in the orderz1> 1 > z2> z > z3> −1.(14)(dz/dt)2= 2n2(z1− z) (z2− z) (z − z3).(15)nt =∫zz3dz/ √(2Z),(16)an elliptic integral of the first kind, which withm = n√z1− z3, κ2=z2− z3,2z1− z2(17)can be expressed, when normalized by the factor √(z1− z3)/2, by the inverse elliptic function in the formmt =∫zz3√ (z1− z3) dz√ [4 (z1− z) (z2− z) (z − z3)]= sn−1√z − z3= cn−1√z2− z= dn−1√z1− z.z2− z3z2− z3z1− z3(18)z − z3= (z2− z3) sn2mt, z2− z = (z2− z3) cn2mt, z1− z = (z1− z3) dn2mt.(19)z = z2sn2mt + z3cn2mt.(20)Interpreted dynamically, the axis of the top keeps time with the beats of a simple pendulum of lengthL = l/½ (z1− z3),(21)suspended from a point at a height ½ (z1+ z3)l above O, in such a manner that a point on thependulumat a distance½ (z1− z3) l = l2/L(22)from the point of suspension moves so as to be always at the same level as the centre of oscillation of the top.The polar co-ordinates of H are denoted by ρ, ῶ in the horizontal plane through C; and, resolving the velocity of H perpendicular to CH,ρdῶ/dt = An2sin θ cos KCH.(23)ρ2dῶ/dt = An2sin θ·CK = An2(G′ − G cos θ)(24)ῶ = ½∫G′ − Gzdt=∫z3(G′ − Gz) / 2Andz,E − zAE − z√ (2Z)
Theory of the Symmetrical Top.
1. The physical constants of a given symmetrical top, expressed in C.G.S. units, which are employed in the subsequent formulae, are denoted by M, h, C and A. M is the weight in grammes (g) as given by the number of gramme weights which equilibrate the top when weighed in a balance; h is the distance OG in centimetres (cm.) between G the centre of gravity and O the point of support, and Mh may be called the preponderance in g.-cm.; Mh and M can be measured by a spring balance holding up in a horizontal position the axis OC in fig. 8 suspended at O. Then gMh (dyne-cm. or ergs) is the moment of gravity about O when the axis OG is horizontal, gMh sin θ being the moment when the axis OG makes an angle θ with the vertical, and g = 981 (cm./s2) on the average; C is the moment of inertia of the top about OG, and A about any axis through O at right angles to OG, both measured in g-cm.2.
To measure A experimentally, swing the top freely about O in small plane oscillation, and determine the length, l cm., of the equivalent simple pendulum; then
l = A/Mh, A = Mhl.
(1)
Next make the top, or this simple pendulum, perform small conical revolutions, nearly coincident with the downward vertical position of equilibrium, and measure n, the mean angular velocity of the conical pendulum in radians / second; and T its period in seconds; then
4π2/T2= n2= g/l = gMh/A;
(2)
and f = n/2π is the number of revolutions per second, called thefrequency, T = 2π/n is the period of a revolution, in seconds.
2. In the popular explanation of the steady movement of the top at a constant inclination to the vertical, depending on the composition of angular velocity, such as given in Perry’sSpinning Tops, or Worthington’sDynamics of Rotation,Steady motion of the top.it is asserted that the moment of gravity is always generating an angular velocity about an axis OB perpendicular to the vertical plane COC′ through the axis of the top OC′; and this angular velocity, compounded with the resultant angular velocity about an axis OI, nearly coincident with OC′, causes the axes OI and OC′ to keep taking up a new position by moving at right angles to the plane COC′, at a constant precessional angular velocity, say μ rad./sec., round the vertical OC (fig. 4).
If, however, the axis OC′ is prevented from taking up this precessional velocity, the top at once falls down; thence all the ingenious attempts—for instance, in the swinging cabin of the Bessemer ship—to utilise the gyroscope as a mechanical directive agency have always resulted in failure (Engineer, October 1874), unless restricted to actuate a light relay, which guides the mechanism, as in steering a torpedo.
An experimental verification can be carried out with the gyroscope in fig. 1; so long as the vertical spindle is free to rotate in its socket, the rapidly rotating wheel will resist the impulse of tapping on the gimbal by moving to one side; but when the pinch screw prevents the rotation of the vertical spindle in the massive pedestal, this resistance to the tapping at once disappears, provided the friction of the table prevents the movement of the pedestal; and if the wheel has any preponderance, it falls down.
Familiar instances of the same principles are observable in the movement of a hoop, or in the steering of a bicycle; it is essential that the handle of the bicycle should be free to rotate to secure the stability of the movement.
The bicycle wheel, employed as a spinning top, in fig. 4, can also be held by the stalk, and will thus, when rotated rapidly, convey a distinct muscular impression of resistance to change of direction, if brandished.
3. A demonstration, depending on the elementary principles of dynamics, of the exact conditions required for theElementary demonstration of the condition of steady motion.axis OC′ of a spinning top to spin steadily at a constant inclination θ to the vertical OC, is given here before proceeding to the more complicated question of the general motion, when θ, the inclination of the axis, is varying by nutation.
It is a fundamental principle in dynamics that if OH is a vector representing to scale the angular momentum of a system, and if Oh is the vector representing the axis of the impressed couple or torque, then OH will vary so that the velocity of H is represented to scale by the impressed couple Oh, and if the top is moving freely about O, Oh is at right angles to the vertical plane COC′, and
Oh = gMh sin θ.
(1)
In the case of the steady motion of the top, the vector OH lies in the vertical plane COC′, in OK suppose (fig. 4), and has a component OC = G about the vertical and a component OC′ = G′, suppose, about the axis OC; and G′ = CR, if R denotes the angular velocity of the top with which it is spun about OC′.
If μ denotes the constant precessional angular velocity of the vertical plane COC′ the components of angular velocity and momentum about OA are μ sin θ and Aμ sin θ, OA being perpendicular to OC′ in the plane COC′; so that the vector OK has the components
OC′ = G′, and C′K = Aμ sin θ,
(2)
and the horizontal component
CK = OC′ sin θ − C′K cos θ= G′ sin θ − Aμ sin θ cos θ.
(3)
The velocity of K being equal to the impressed couple Oh,
gMh sin θ = μ·CK = sin θ (G′μ − Aμ2cos θ),
(4)
and dropping the factor sin θ,
Aμ2cos θ − G′μ + gMh = 0, or Aμ2cos θ − CRμ + An2= 0,
(5)
the condition for steady motion.
Solving this as a quadratic in μ, the roots μ1, μ2are given by
(6)
and the minimum value of G′ = CR for real values of μ is given by
(7)
for a smaller value of R the top cannot spin steadily at the inclination θ to the upward vertical.
Interpreted geometrically in fig. 4
μ = gMh sin θ/CK = An2/KN, and μ = C′K/A sin θ = KM/A,
(8)
KM·KN = A2n2,
(9)
so that K lies on a hyperbola with OC, OC′ as asymptotes.
4. Suppose the top or gyroscope, instead of moving freely about the point O, is held in a ring or frame which is compelledConstrained motion of the gyroscope.to rotate about the vertical axis OC with constant angular velocity μ; then if N denotes the couple of reaction of the frame keeping the top from falling, acting in the plane COC’, equation (4) § 3 becomes modified into
gMh sin θ − N = μ·CK = sin θ G′μ − Aμ2cos θ,
(1)
N = sin θ (Aμ2cos θ − G′μ + gMh)= A sin θ cos θ (μ − μ1) (μ − μ2);
(2)
and hence, as μ increases through μ2and μ1, the sign of N can be determined, positive or negative, according as the tendency of the axis is to fall or rise.
When G′ = CR is large, μ2is large, and
μ1≈ gMh/G′ = An2/CR,
(3)
the same for all inclinations, and this is the precession observed in the spinning top and centrifugal machine of fig. 10 This is true accurately when the axis OC′ is horizontal, and then it agrees with the result of the popular explanation of § 2.
If the axis of the top OC′ is pointing upward, the precession is in the same direction as the rotation, and an increase of μ from μ1makes N negative, and the top rises; conversely a decrease of the procession μ causes the axis to fall (Perry,Spinning Tops, p. 48).
If the axis points downward, as in the centrifugal machine with upper support, the precession is in the opposite direction to the rotation, and to make the axis approach the vertical position the precession must be reduced.
This is effected automatically in the Weston centrifugal machine (fig. 10) used for the separation of water andCentrifugal machine.molasses, by the friction of the indiarubber cushions above the support; or else the spindle is produced downwards below the drum a short distance, and turns in a hole in a weight resting on the bottom of the case, which weight is dragged round until the spindle is upright; this second arrangement is more effective when a liquid is treated in the drum, and wave action is set up (The Centrifugal Machine, C. A. Matthey).
Similar considerations apply to the stability of the whirling bowl in a cream-separating machine.
We can write equation (1)
N = An2sin θ − μ·CK = (A2n2− KM·KN) sin θ/A,
(4)
so that N is negative or positive, and the axis tends to rise or fall according as K moves to the inside or outside of the hyperbola of free motion. Thus a tap on the axis tending to hurry the precession is equivalent to an impulse couple giving an increase to C′K, and will make K move to the interior of the hyperbola and cause the axis to rise; the steering of a bicycle may be explained in this way; but K1will move to the exterior of the hyperbola, and so the axis will fall in this second more violent motion.
Friction on the point of the top may be supposed to act like a tap in the direction opposite to the precession; and so the axis of a top spun violently rises at first and up to the vertical position, but falls away again as the motion dies out. Friction considered as acting in retarding the rotation may be compared to an impulse couple tending to reduce OC′, and so make K and K1both move to the exterior of the hyperbola, and the axis falls in both cases. The axis may rise or fall according to the direction of the frictional couple, depending on the shape of the point; an analytical treatment of the varying motion is very intractable; a memoir by E. G. Gallop may be consulted in theTrans. Camb. Phil. Soc., 1903.
The earth behaves in precession like a large spinning top, of which the axis describes a circle round the pole of the ecliptic of mean angular radius θ, about 23½°, in a period of 26,000 years, so that R/μ = 26000 × 365; and the mean couple producing precession is
CRμ sin θ = CR2sin 23½° /26000 × 365,
(5)
one 12 millionth part of ½CR2, the rotation energy of the earth.
5. If the preponderance is absent, by making the C·G coincide with O, and if Aμ is insensible compared with G′,
N = −G′μ sin θ,
(1)
the formula which suffices to explain most gyroscopic action.
Thus a carriage running round a curve experiences, in consequenceGyroscopic action of railway wheels.of the rotation of the wheels, an increase of pressure Z on the outer track, and a diminution Z on the inner, giving a couple, if a is the gauge,
Za = G′μ,
(2)
tending to help the centrifugal force to upset the train; and if c is the radius of the curve, b of the wheels, C their moment of inertia, and v the velocity of the train,
μ = v/c, G′ = Cv/b,
(3)
Z = Cv2/abc (dynes),
(4)
so that Z is the fraction C/Mabof the centrifugal force Mv2/c, or the fraction C/Mh of its transference of weight, with h the height of the centre of gravity of the carriage above the road. A Brennan carriage on a monorail would lean over to the inside of the curve at an angle α, given by
tan α = G′μ/gMh = G′v/gMhc.
(6)
The gyroscopic action of a dynamo, turbine, and other rotating machinery on a steamer, paddle or screw, due to its rolling and pitching, can be evaluated in a similar elementary manner (Worthington,Dynamics of Rotation), and Schlick’s gyroscopic apparatus is intended to mitigate the oscillation.
6. If the axis OC in fig. 4 is inclined at an angle α to the vertical, the equation (2) § 4 becomes
N = sin θ (Aμ2cos θ − G′μ) + gMh sin (α − θ).
(1)
Suppose, for instance, that OC is parallel to the earth’s axis, and that the frame is fixed in the meridian; then α is the co-latitude, and μ is the angular velocity of the earth, the square of which may be neglected; so that, putting N = 0, α − θ = E,
gMh sin E − G′μ sin (α − E) = 0,
(2)
(3)
This is the theory of Gilbert’s barogyroscope, described in Appell’sMécanique rationnelle, ii. 387: it consists essentially of a rapidlyThe barogyroscope.rotated fly-wheel, mounted on knife-edges by an axis perpendicular to its axis of rotation and pointing east and west; spun with considerable angular momentum G′, and provided with a slight preponderance Mh, it should tilt to an angle E with the vertical, and thus demonstrate experimentally the rotation of the earth.
In Foucault’s gyroscope (Comptes rendus, 1852; Perry, p. 105)Foucault’s gyroscope.the preponderance is made zero, and the axis points to the pole, when free to move in the meridian.
Generally, if constrained to move in any other plane, the axis seeks the position nearest to the polar axis, like a dipping needle with respect to the magnetic pole. (A gyrostatic working model of the magnetic compass, by Sir W. Thomson. British Association Report, Montreal, 1884. A. S. Chessin, St Louis Academy of Science, January 1902.)
A spinning top with a polished upper plane surface will provide an artificial horizon at sea, when the real horizon is obscured. The first instrument of this kind was constructed by Serson, and is described in theGentleman’s Magazine,Gyroscopic horizon.vol. xxiv., 1754; also by Segner in hisSpecimen theoriae turbinum(Halae, 1755). The inventor was sent to sea by the Admiralty to test his instrument, but he was lost in the wreck of the “Victory,” 1744. A copy of the Serson top, from the royal collection, is now in the Museum of King’s College, London. Troughton’s Nautical Top (1819) is intended for the same purpose.
The instrument is in favour with French navigators, perfected byAdmiral Fleuriais (fig. 9); but it must be noticed that the horizon given by the top is inclined to the true horizon at the angle E given by equation (3) above; and if μ1is the precessional angular velocity as given by (3) § 4, and T = 2π/μ, its period in seconds,
(4)
if E is expressed in minutes, taking μ = 2π/86400; thus making the true latitude E nautical miles to the south of that given by the top (Revue maritime, 1890;Comptes rendus, 1896).
This can be seen by elementary consideration of the theory above, for the velocity of the vector OC′ of the top due to the rotation of the earth is
μ·OC′ cos lat = gMh sin E = μ1·OC′ sin E,
(5)
in which 8π can be replaced by 25, in practice; so that the Fleuriais gyroscopic horizon is an illustration of the influence of the rotation of the earth and of the need for its allowance.
7. In the ordinary treatment of the general theory of the gyroscope, the motion is referred to two sets of rectangular axes; theEuler’s coordinate angles.one Ox, Oy, Oz fixed in space, with Oz vertically upward and the other OX, OY, OZ fixed in the rotating wheel with OZ in the axis of figure OC.
The relative position of the two sets of axes is given by means of Euler’s unsymmetrical angles θ, φ, ψ, such that the successive turning of the axes Ox, Oy, Oz through the angles (i.) ψ about Oz, (ii.) θ about OE, (iii.) φ about OZ, brings them into coincidence with OX, OY, OZ, as shown in fig. 11, representing theconcaveside of a spherical surface.
The component angular velocities about OD, OE, OZ are
ψsin θ,θ,φ+ψcos θ;
(1)
so that, denoting the components about OX, OY, OZ by P, Q, R,
(2)
Consider, for instance, the motion of a fly-wheel of preponderance Mh, and equatoreal moment of inertia A, of which the axis OC is held in a light ring ZCX at a constant angle γ with OZ, while OZ is held by another ring zZ, which constrains it to move round the vertical Oz at a constant inclination θ with constant angular velocity μ, so that
θ= 0,ψ= μ;
(3)
P = μ sin θ sin φ, Q = μ sin θ cos φ, R =φ+ μ cos θ.
(4)
With CXF a quadrant, the components of angular velocity and momentum about OF, OY, are
P cos γ − R sin γ, Q, and A (P cos γ − R sin γ), AQ,
(5)
so that, denoting the components of angular momentum of the fly-wheel about OC, OX, OY, OZ by K or G′, h1, h2, h3,
h1= A (P cos γ − R sin γ) cos γ + K sin γ,
(6)
h2= AQ,
(7)
h3= −A (P cos γ − R sin γ) sin γ + K cos γ;
(8)
and the dynamical equation
(9)
with K constant, and with preponderance downward
N = gMh cos zY sin γ = gMh sin γ sin θ cos φ,
(10)
reduces to
+ Aμ2cos γ sin θ cos θ cos φ − (Kμ + gMh) sin θ cos φ = 0.
(11)
The position of relative equilibrium is given by
(12)
For small values of μ the equation becomes
(13)
so that φ = ½π gives the position of stable equilibrium, and the period of a small oscillation is 2π √{A sin γ/(Kμ + gMh) sin θ}.
In the general case, denoting the periods of vibration about φ = ½π, −½π, and the sidelong position of equilibrium by 2π/(n1, n2, or n3), we shall find
(14)
(15)
n3= n1n2/μ sin θ.
(16)
The first integral of (11) gives
− Aμ2cos γ sin θ cos θ sin φ + (Kμ + gMh) sin θ sin φ − H = 0,
(17)
and putting tan (¼π + ½φ) = z, this reduces to
(18)
where Z is a quadratic in z2, so that z is a Jacobian elliptic function of t, and we have
tan (¼π + ½φ) = C (tn, dn, nc, or cn) nt,
(19)
according as the ring ZC performs complete revolutions, or oscillates about a sidelong position of equilibrium, or oscillates about the stable position of equilibrium φ = ±½π.
Suppose Oz is parallel to the earth’s axis, and μ is the diurnal rotation, the square of which may be neglected, then if Gilbert’s barogyroscope of § 6 has the knife-edges turned in azimuth to make an angle β with E. and W., so that OZ lies in the horizon at an angle E·β·N., we must put γ = ½π, cos θ = sin α sin β; and putting φ = ½π − δ + E, where δ denotes the angle between Zz and the vertical plane Zζ through the zenith ζ,
sin θ cos δ = cos α, sin θ sin δ = sin α cos β;
(20)
so that equations (9) and (10) for relative equilibrium reduce to
gMh sin E = KQ = Kμ sin θ cos φ = Kμ sin θ sin (δ − E),
(21)
and will change (3) § 6 into
(22)
a multiplication of (3) § 6 by cos β (Gilbert,Comptes rendus, 1882).
Changing the sign of K or h and E and denoting the revolutions/second of the gyroscope wheel by F, then in the preceding notation, T denoting the period of vibration as a simple pendulum,
(23)
so that the gyroscope would reverse if it were possible to make F cos α > 86400 A/T2C (Föppl,Münch. Ber, 1904).
A gyroscopic pendulum is made by the addition to it of a fly-wheel, balanced and mounted, as in Gilbert’s barogyroscope, in a ring movable about an axis fixed in the pendulum, in the vertical plane of motion.
As the pendulum falls away to an angle θ with the upward vertical, and the axis of the fly-wheel makes an angle φ with the vertical plane of motion, the three components of angular momentum are
h1= K cos φ, h2= Aθ+ K sin φ, h3= Aφ,
(24)
where h3is the component about the axis of the ring and K of the fly-wheel about its axis; and if L, M′, N denote the components of the couple of reaction of the ring, L may be ignored, while N is zero, with P = 0, Q =θ, R = 0, so that
M′ = h2= Aθ+ Kφcos φ,
(25)
0 = h3− h1θ= Aφ− Kθcos φ.
(26)
For the motion of the pendulum, including the fly-wheel,
MK2θ= gMH sin θ − M′ = gMH sin θ − Aθ− Kφcos φ.
(27)
If θ and φ remain small,
Aφ= Kθ, Aφ= K(θ − α),
(28)
(MK2+ A)θ+ (K2/A) (θ − α) − gMHθ = 0;
(29)
so that the upright position will be stable if K2> gMHA, or the rotation energy of the wheel greater than ½A/C times the energy acquired by the pendulum in falling between the vertical and horizontal position; and the vibration will synchronize with a simple pendulum of length
(MK2+ A) / [(K2/gA) − MH].
(30)
This gyroscopic pendulum may be supposed to represent a ship among waves, or a carriage on a monorail, and so affords an explanation of the gyroscopic action essential in the apparatus of Schlick and Brennan.
8. Careful scrutiny shows that the steady motion of a top is not steady absolutely; it reveals a small nutationGeneral motion of the top.superposed, so that a complete investigation requires a return to the equations of unsteady motion, and for the small oscillation to consider them in a penultimate form.
In the general motion of the top the vector OH of resultant angular momentum is no longer compelled to lie in the vertical plane COC′ (fig. 4), but since the axis Oh of the gravity couple is always horizontal, H will describe a curve in a fixed horizontal plane through C. The vector OC′ of angular momentum about the axis will be constant in length, but vary in direction; and OK will be the component angular momentum in the vertical plane COC′, if the planes through C and C′ perpendicular to the lines OC and OC′ intersect in the line KH; and if KH is the component angular momentum perpendicular to the plane COC′, the resultant angular momentum OH has the three components OC′, C′K, KH, represented in Euler’s angles by
KH = A dθ/dt, C′K = A sin θd ψ/dt, OC′ = G′.
(1)
Drawing KM vertical and KN parallel to OC′, then
KM = A dψ/dt, KN = CR − A cos θ dψ/dt = (C − A) R + A dφ/dt
(2)
so that in the spherical top, with C = A, KN = A dφ/dt.
The velocity of H is in the direction KH perpendicular to the plane COC′, and equal to gMh sin θ or An2sin θ, so that if a point in the axis OC′ at a distance An2from O is projected on the horizontal plane through C in the point P on CK, the curve described by P, turned forwards through a right angle, will be the hodograph of H; this is expressed by
(3)
where ρeῶiis the vector CH; and so the curve described by P and the motion of the axis of the top is derived from the curve described by H by a differentiation.
Resolving the velocity of H in the direction CH,
d·CH/dt = An2sin θ sin KCH = An2sin θ KH/CH,
(4)
d·½CH2/dt = A2n2sin θ dθ/dt.
(5)
and integrating
½CH2= A2n2(E − cos θ),
(6)
½OH2= A2n2(F − cos θ),
(7)
½C′H2= A2n2(D − cos θ),
(8)
where D, E, F are constants, connected by
F = E + G2/2A2n2= D + G′2/2A2n2.
(9)
Then
KH2= OH2− OK2,
(10)
OK2sin2θ = CC′2= G2− 2GG′ cos θ + G′2,
(11)
A2sin2θ (dθ/dt)2= 2A2n2(F − cos θ) sin2θ − G2+ 2GG′ cos θ − G′2;
(12)
and putting cos θ = z,
(13)
= 2n2(E − z) (1 − z2) − (G′ − Gz)2/A2= 2n2(D − z) (1 − z2) − (G − G′z)2/A2= 2n2Z suppose.
= 2n2(E − z) (1 − z2) − (G′ − Gz)2/A2
= 2n2(D − z) (1 − z2) − (G − G′z)2/A2
= 2n2Z suppose.
Denoting the roots of Z = 0 by z1, z2, z3, we shall have them arranged in the order
z1> 1 > z2> z > z3> −1.
(14)
(dz/dt)2= 2n2(z1− z) (z2− z) (z − z3).
(15)
nt =∫zz3dz/ √(2Z),
(16)
an elliptic integral of the first kind, which with
(17)
can be expressed, when normalized by the factor √(z1− z3)/2, by the inverse elliptic function in the form
(18)
z − z3= (z2− z3) sn2mt, z2− z = (z2− z3) cn2mt, z1− z = (z1− z3) dn2mt.
(19)
z = z2sn2mt + z3cn2mt.
(20)
Interpreted dynamically, the axis of the top keeps time with the beats of a simple pendulum of length
L = l/½ (z1− z3),
(21)
suspended from a point at a height ½ (z1+ z3)l above O, in such a manner that a point on thependulumat a distance
½ (z1− z3) l = l2/L
(22)
from the point of suspension moves so as to be always at the same level as the centre of oscillation of the top.
The polar co-ordinates of H are denoted by ρ, ῶ in the horizontal plane through C; and, resolving the velocity of H perpendicular to CH,
ρdῶ/dt = An2sin θ cos KCH.
(23)
ρ2dῶ/dt = An2sin θ·CK = An2(G′ − G cos θ)
(24)