(1)
p2+ q2+ r2= ω2.
(2)
If we draw a vector OJ to represent the angular velocity, then J traces out a certain curve in the body, called thepolhode, and a certain curve in space, called theherpolhode. The cones generated by the instantaneous axis in the body and in space are called the polhode and herpolhode cones, respectively; in the actual motion the former cone rolls on the latter (§ 7).
The special case where both cones are right circular and ω is constant is important in astronomy and also in mechanism (theory of bevel wheels). The “precession of the equinoxes” is due to the fact that the earth performs a motion of this kind about its centre, and the whole class of such motions has therefore been termedprecessional. In fig. 78, which shows the various cases, OZ is the axis of the fixed and OC that of the rolling cone, and J is the point of contact of the polhode and herpolhode, which are of course both circles. If αbe the semi-angle of the rolling cone, β the constant inclination of OC to OZ, and ψ̇ the angular velocity with which the plane ZOC revolves about OZ, then, considering the velocity of a point in OC at unit distance from O, we haveω sin α = ±ψ̇ sin β,(3)where the lower sign belongs to the third case. The earth’s precessional motion is of this latter type, the angles being α = .0087″, β = 23° 28′.
The special case where both cones are right circular and ω is constant is important in astronomy and also in mechanism (theory of bevel wheels). The “precession of the equinoxes” is due to the fact that the earth performs a motion of this kind about its centre, and the whole class of such motions has therefore been termedprecessional. In fig. 78, which shows the various cases, OZ is the axis of the fixed and OC that of the rolling cone, and J is the point of contact of the polhode and herpolhode, which are of course both circles. If αbe the semi-angle of the rolling cone, β the constant inclination of OC to OZ, and ψ̇ the angular velocity with which the plane ZOC revolves about OZ, then, considering the velocity of a point in OC at unit distance from O, we have
ω sin α = ±ψ̇ sin β,
(3)
where the lower sign belongs to the third case. The earth’s precessional motion is of this latter type, the angles being α = .0087″, β = 23° 28′.
If m be the mass of a particle at P, and PN the perpendicular to the instantaneous axis, the kinetic energy T is given by
2T = Σ {m (ω·PN)2} = ω2·Σ (m·PN2) = Iω2,
(4)
where I is the moment of inertia about the instantaneous axis. With the same notation for moments and products of inertia as in § 11 (38), we have
I = Al2+ Bm2+ Cn2− 2Fmn − 2Gnl − 2Hlm,
and therefore by (1),
2T = Ap2+ Bq2+ Cr2− 2Fqr − 2Grp − 2Hpq.
(5)
Again, if x, y, z be the co-ordinates of P, the component velocities of m are
qz − ry, rx − pz, py − qx,
(6)
by § 7 (5); hence, if λ, μ, ν be now used to denote the component angular momenta about the co-ordinate axes, we have λ = Σ {m (py − qx)y − m(rx − pz) z }, with two similar formulae, or
(7)
If the co-ordinate axes be taken to coincide with the principal axes of inertia at O, at the instant under consideration, we have the simpler formulae
2T = Ap2+ Bq2+ Cr2,
(8)
λ = Ap, μ = Bq, ν = Cr.
(9)
It is to be carefully noticed that the axis of resultant angular momentum about O does not in general coincide with the instantaneous axis of rotation. The relation between these axes may be expressed by means of the momental ellipsoid at O. The equation of the latter, referred to its principal axes, being as in § 11 (41), the co-ordinates of the point J where it is met by the instantaneous axis are proportional to p, q, r, and the direction-cosines of the normal at J are therefore proportional to Ap, Bq, Cr, or λ, μ, ν. The axis of resultant angular momentum is therefore normal to the tangent plane at J, and does not coincide with OJ unless the latter be a principal axis. Again, if Γ be the resultant angular momentum, so that
λ2+ μ2+ ν2= Γ2,
(10)
the length of the perpendicular OH on the tangent plane at J is
(11)
where ρ = OJ. This relation will be of use to us presently (§ 19).
The motion of a rigid body in the most general case may be specified by means of the component velocities u, v, w of any point O of it which is taken as base, and the component angular velocities p, q, r. The component velocities of any point whose co-ordinates relative to O are x, y, z are then
u + qz − ry, v + rx − pz, w + py − qx
(12)
by § 7 (6). It is usually convenient to take as our base-point the mass-centre of the body. In this case the kinetic energy is given by
2T = M0(u2+ v2+ w2) + Ap2+ Bq2+ Cr2− 2Fqr − 2Grp − 2Hpg,
(13)
where M0is the mass, and A, B, C, F, G, H are the moments and products of inertia with respect to the mass-centre; cf. § 15 (9).
The components ξ, η, ζ of linear momentum are
(14)
whilst those of the relative angular momentum are given by (7). The preceding formulae are sufficient for the treatment of instantaneous impulses. Thus if an impulse (ξ, η, ζ, λ, μ, ν) change the motion from (u, v, w, p, q, r) to (u′, v′, w′, p′, q′, r′) we have
(15)
where, for simplicity, the co-ordinate axes are supposed to coincide with the principal axes at the mass-centre. Hence the change of kinetic energy is
T′ − T = ξ ·1⁄2(u + u′) + η ·1⁄2(v + v′) + ζ ·1⁄2(w + w′),+ λ ·1⁄2(p + p′) + μ ·1⁄2(q + q′) + ν ·1⁄2(r + r′).
(16)
The factors of ξ, η, ζ, λ, μ, ν on the right-hand side are proportional to the constituents of a possible infinitesimal displacement of the solid, and the whole expression is proportional (on the same scale) to the work done by the given system of impulsive forces in such a displacement. As in § 9 this must be equal to the total work done in such a displacement by the several forces, whatever they are, which make up the impulse. We are thus led to the following statement: the change of kinetic energy due to any system of impulsive forces is equal to the sum of the products of the several forces into the semi-sum of the initial and final velocities of their respective points of application, resolved in the directions of the forces. Thus in the problem of fig. 77 the kinetic energy generated is1⁄2M (κ2+ Cq2)ω′2, if C be the instantaneous centre; this is seen to be equal to1⁄2F·ω′·CP, where ω′·CP represents the initial velocity of P.
The equations of continuous motion of a solid are obtained by substituting the values of ξ, η, ζ, λ, μ, ν from (14) and (7) in the general equations
(17)
where (X, Y, Z, L, M, N) denotes the system of extraneous forces referred (like the momenta) to the mass-centre as base, the co-ordinate axes being of course fixed in direction. The resulting equations are not as a rule easy of application, owing to the fact that the moments and products of inertia A, B, C, F, G, H are not constants but vary in consequence of the changing orientation of the body with respect to the co-ordinate axes.
An exception occurs, however, in the case of a solid which is kinetically symmetrical (§ 11) about the mass-centre,e.g.a uniform sphere. The equations then take the formsM0u̇ = X,M0v̇ = Y,M0ẇ = Z,Cṗ = L,Cq̇ = M,Cṙ = N,(18)where C is the constant moment of inertia about any axis throughthe mass-centre. Take, for example, the case of a sphere rolling on a plane; and let the axes Ox, Oy be drawn through the centre parallel to the plane, so that the equation of the latter is z = −a. We will suppose that the extraneous forces consist of a known force (X, Y, Z) at the centre, and of the reactions (F1, F2, R) at the point of contact. HenceM0u̇ = X + F1, M0v̇ = Y + F2, 0 = Z + R,Cṗ = F2a, Cq̇ = −F1a, Cṙ = 0.(19)The last equation shows that the angular velocity about the normal to the plane is constant. Again, since the point of the sphere which is in contact with the plane is instantaneously at rest, we have the geometrical relationsu + qa = 0, v + pa = 0, w = 0,(20)by (12). Eliminating p, q, we get(M0+ Ca−2) u̇ = X, (M0+ Ca−2) v̇ = Y.(21)The acceleration of the centre is therefore the same as if the plane were smooth and the mass of the sphere were increased by C/α2. Thus the centre of a sphere rolling under gravity on a plane of inclination a describes a parabola with an accelerationg sin α/(1 + C/Ma2)parallel to the lines of greatest slope.Take next the case of a sphere rolling on a fixed spherical surface. Let a be the radius of the rolling sphere, c that of the spherical surface which is the locus of its centre, and let x, y, z be the co-ordinates of this centre relative to axes through O, the centre of the fixed sphere. If the only extraneous forces are the reactions (P, Q, R) at the point of contact, we haveM0ẍ = P, M0ÿ = Q, M0z̈ = R,Cṗ = −a(yR − zQ), Cq̇ = −a(zP − xR), Cṙ = −a(xQ − yP),ccc(22)the standard case being that where the rolling sphere is outside the fixed surface. The opposite case is obtained by reversing the sign of a. We have also the geometrical relationsẋ = (a/c) (qz − ry), ẏ = (a/c) (rx − pz), ż = (a/c) (py − gx),(23)If we eliminate P, Q, R from (22), the resulting equations are integrable with respect to t; thusp = −M0a(yż − zẏ) + α, q = −M0a(zẋ − xż) + β, r = −M0a(xẏ − yẋ) + γ,CcCcCc(24)where α, β, γ are arbitrary constants. Substituting in (23) we find(1 +M0a2)ẋ =a(βz − γy),(1 +M0a2)ẏ =a(γx − αz),(1 +M0a2)ż =a(αy − βx).CcCcCc(25)Hence αẋ + βẏ + γż = 0, orαx + βy + γz = const.;(26)which shows that the centre of the rolling sphere describes a circle. If the axis of z be taken normal to the plane of this circle we have α = 0, β = 0, and(1 +M0a2)ẋ = −γay,(1 +M0a2)ẏ = γax.CcCc(27)The solution of these equations is of the typex = b cos (στ + ε), y = b sin (σt + ε),(28)where b, ε are arbitrary, andσ =γa/c.1 + M0a2/C(29)The circle is described with the constant angular velocity σ.When the gravity of the rolling sphere is to be taken into account the preceding method is not in general convenient, unless the whole motion of G is small. As an example of this latter type, suppose that a sphere is placed on the highest point of a fixed sphere and set spinning about the vertical diameter with the angular velocity n; it will appear that under a certain condition the motion of G consequent on a slight disturbance will be oscillatory. If Oz be drawn vertically upwards, then in the beginning of the disturbed motion the quantities x, y, p, q, P, Q will all be small. Hence, omitting terms of the second order, we findM0ẍ = P, M0ẏ = Q, R = M0g,Cṗ = −(M0ga/c) y + aQ, Cq̇ = (M0ga/c) x − aP, Cṙ = 0.(30)The last equation shows that the component r of the angular velocity retains (to the first order) the constant value n. The geometrical relations reduce toẋ = aq − (na/c) y, ẏ = −ap + (na/c) x.(31)Eliminating p, g, P, Q, we obtain the equations(C + M0a2) ẍ + (Cna/c) y − (M0ga2/c) x = 0,(C + M0a2) ÿ − (Cna/c) x − (M0ga2/c) y = 0,(32)which are both contained in{(C + M0a2)d2− iCnad−M0ga2}(x + iy) = 0.dt2cdtc(33)This has two solutions of the type x + iy = αei(σt + ε), where α, ε are arbitrary, and σ is a root of the quadratic(C + M0a2) σ2− (Cna/c) σ + M0ga2/c = 0.(34)Ifn2> (4Mgc/C) (1 + M0a2/C),(35)both roots are real, and have the same sign as n. The motion of G then consists of two superposed circular vibrations of the typex = α cos (σt + ε), y = α sin (σt + ε),(36)in each of which the direction of revolution is the same as that of the initial spin of the sphere. It follows therefore that the original position is stable provided the spin n exceed the limit defined by (35). The case of a sphere spinning about a vertical axis at the lowest point of a spherical bowl is obtained by reversing the signs of α and c. It appears that this position is always stable.It is to be remarked, however, that in the first form of the problem the stability above investigated is practically of a limited or temporary kind. The slightest frictional forces—such as the resistance of the air—even if they act in lines through the centre of the rolling sphere, and so do not directly affect its angular momentum, will cause the centre gradually to descend in an ever-widening spiral path.
An exception occurs, however, in the case of a solid which is kinetically symmetrical (§ 11) about the mass-centre,e.g.a uniform sphere. The equations then take the forms
(18)
where C is the constant moment of inertia about any axis throughthe mass-centre. Take, for example, the case of a sphere rolling on a plane; and let the axes Ox, Oy be drawn through the centre parallel to the plane, so that the equation of the latter is z = −a. We will suppose that the extraneous forces consist of a known force (X, Y, Z) at the centre, and of the reactions (F1, F2, R) at the point of contact. Hence
M0u̇ = X + F1, M0v̇ = Y + F2, 0 = Z + R,Cṗ = F2a, Cq̇ = −F1a, Cṙ = 0.
(19)
The last equation shows that the angular velocity about the normal to the plane is constant. Again, since the point of the sphere which is in contact with the plane is instantaneously at rest, we have the geometrical relations
u + qa = 0, v + pa = 0, w = 0,
(20)
by (12). Eliminating p, q, we get
(M0+ Ca−2) u̇ = X, (M0+ Ca−2) v̇ = Y.
(21)
The acceleration of the centre is therefore the same as if the plane were smooth and the mass of the sphere were increased by C/α2. Thus the centre of a sphere rolling under gravity on a plane of inclination a describes a parabola with an acceleration
g sin α/(1 + C/Ma2)
parallel to the lines of greatest slope.
Take next the case of a sphere rolling on a fixed spherical surface. Let a be the radius of the rolling sphere, c that of the spherical surface which is the locus of its centre, and let x, y, z be the co-ordinates of this centre relative to axes through O, the centre of the fixed sphere. If the only extraneous forces are the reactions (P, Q, R) at the point of contact, we have
M0ẍ = P, M0ÿ = Q, M0z̈ = R,
(22)
the standard case being that where the rolling sphere is outside the fixed surface. The opposite case is obtained by reversing the sign of a. We have also the geometrical relations
ẋ = (a/c) (qz − ry), ẏ = (a/c) (rx − pz), ż = (a/c) (py − gx),
(23)
If we eliminate P, Q, R from (22), the resulting equations are integrable with respect to t; thus
(24)
where α, β, γ are arbitrary constants. Substituting in (23) we find
(25)
Hence αẋ + βẏ + γż = 0, or
αx + βy + γz = const.;
(26)
which shows that the centre of the rolling sphere describes a circle. If the axis of z be taken normal to the plane of this circle we have α = 0, β = 0, and
(27)
The solution of these equations is of the type
x = b cos (στ + ε), y = b sin (σt + ε),
(28)
where b, ε are arbitrary, and
(29)
The circle is described with the constant angular velocity σ.
When the gravity of the rolling sphere is to be taken into account the preceding method is not in general convenient, unless the whole motion of G is small. As an example of this latter type, suppose that a sphere is placed on the highest point of a fixed sphere and set spinning about the vertical diameter with the angular velocity n; it will appear that under a certain condition the motion of G consequent on a slight disturbance will be oscillatory. If Oz be drawn vertically upwards, then in the beginning of the disturbed motion the quantities x, y, p, q, P, Q will all be small. Hence, omitting terms of the second order, we find
M0ẍ = P, M0ẏ = Q, R = M0g,Cṗ = −(M0ga/c) y + aQ, Cq̇ = (M0ga/c) x − aP, Cṙ = 0.
(30)
The last equation shows that the component r of the angular velocity retains (to the first order) the constant value n. The geometrical relations reduce to
ẋ = aq − (na/c) y, ẏ = −ap + (na/c) x.
(31)
Eliminating p, g, P, Q, we obtain the equations
(C + M0a2) ẍ + (Cna/c) y − (M0ga2/c) x = 0,(C + M0a2) ÿ − (Cna/c) x − (M0ga2/c) y = 0,
(32)
which are both contained in
(33)
This has two solutions of the type x + iy = αei(σt + ε), where α, ε are arbitrary, and σ is a root of the quadratic
(C + M0a2) σ2− (Cna/c) σ + M0ga2/c = 0.
(34)
If
n2> (4Mgc/C) (1 + M0a2/C),
(35)
both roots are real, and have the same sign as n. The motion of G then consists of two superposed circular vibrations of the type
x = α cos (σt + ε), y = α sin (σt + ε),
(36)
in each of which the direction of revolution is the same as that of the initial spin of the sphere. It follows therefore that the original position is stable provided the spin n exceed the limit defined by (35). The case of a sphere spinning about a vertical axis at the lowest point of a spherical bowl is obtained by reversing the signs of α and c. It appears that this position is always stable.
It is to be remarked, however, that in the first form of the problem the stability above investigated is practically of a limited or temporary kind. The slightest frictional forces—such as the resistance of the air—even if they act in lines through the centre of the rolling sphere, and so do not directly affect its angular momentum, will cause the centre gradually to descend in an ever-widening spiral path.
§ 19.Free Motion of a Solid.—Before proceeding to further problems of motion under extraneous forces it is convenient to investigate the free motion of a solid relative to its mass-centre O, in the most general case. This is the same as the motion about a fixed point under the action of extraneous forces which have zero moment about that point. The question was first discussed by Euler (1750); the geometrical representation to be given is due to Poinsot (1851).
The kinetic energy T of the motion relative to O will be constant. Now T =1⁄2Iω2, where ω is the angular velocity and I is the moment of inertia about the instantaneous axis. If ρ be the radius-vector OJ of the momental ellipsoid
Ax2+ By2+ Cz2= Mε4
(1)
drawn in the direction of the instantaneous axis, we have I = Mε4/ρ2(§ 11); hence ω varies as ρ. The locus of J may therefore be taken as the “polhode” (§ 18). Again, the vector which represents the angular momentum with respect to O will be constant in every respect. We have seen (§ 18) that this vector coincides in direction with the perpendicular OH to the tangent plane of the momental ellipsoid at J; also that
(2)
where Γ is the resultant angular momentum about O. Since ω varies as ρ, it follows that OH is constant, and the tangent plane at J is therefore fixed in space. The motion of the body relative to O is therefore completely represented if we imagine the momental ellipsoid at O to roll without sliding on a plane fixed in space, with an angular velocity proportional at each instant to the radius-vector of the point of contact. The fixed plane is parallel to the invariable plane at O, and the line OH is called theinvariable line. The trace of the point of contact J on the fixed plane is the “herpolhode.”
If p, q, r be the component angular velocities about the principal axes at O, we have
(A2p2+ B2q2+ C2r2) / Γ2= (Ap2+ Bq2+ Cr2) / 2T,
(3)
each side being in fact equal to unity. At a point on the polhode cone x : y : z = p : q : r, and the equation of this cone is therefore
(4)
Since 2AT − Γ2= B (A − B)q2+ C(A − C)r2, it appears that if A > B > C the coefficient of x2in (4) is positive, that of z2is negative, whilst that of y2is positive or negative according as 2BT ≷ Γ2. Hence the polhode cone surrounds the axis of greatest or least moment according as 2BT ≷ Γ2. In the critical case of 2BT = Γ2it breaks up into two planes through the axis of mean moment (Oy). The herpolhode curve in the fixed plane is obviously confined between two concentric circles which it alternately touches; it is not in general a re-entrant curve. It has been shown by De Sparre that, owing to the limitation imposed on the possible forms of the momental ellipsoid by the relation B + C > A, the curve has no points of inflexion. The invariable line OH describes another cone in thebody, called theinvariable cone. At any point of this we have x : y : z = Ap · Bq : Cr, and the equation is therefore
(5)
The signs of the coefficients follow the same rule as in the case of (4). The possible forms of the invariable cone are indicated in fig. 80 by means of the intersections with a concentric spherical surface. In the critical case of 2BT = Γ2the cone degenerates into two planes. It appears that if the body be sightly disturbed from a state of rotation about the principal axis of greatest or least moment, the invariable cone will closely surround this axis, which will therefore never deviate far from the invariable line. If, on the other hand, the body be slightly disturbed from a state of rotation about the mean axis a wide deviation will take place. Hence a rotation about the axis of greatest or least moment is reckoned as stable, a rotation about the mean axis as unstable. The question is greatly simplified when two of the principal moments are equal, say A = B. The polhode and herpolhode cones are then right circular, and the motion is “precessional” according to the definition of § 18. If α be the inclination of the instantaneous axis to the axis of symmetry, β the inclination of the latter axis to the invariable line, we have
Γ cos β = C ω cos α, Γ sin β = A ω sin α,
(6)
whence
(7)
Hence β ≷ α, and the circumstances are therefore those of the first or second case in fig. 78, according as A ≷ C. If ψ be the rate at which the plane HOJ revolves about OH, we have
(8)
by § 18 (3). Also if χ̇ be the rate at which J describes the polhode, we have ψ̇ sin (β − α) = χ̇ sin β, whence
(9)
If the instantaneous axis only deviate slightly from the axis of symmetry the angles α, β are small, and χ̇ = (A − C) A·ω; the instantaneous axis therefore completes its revolution in the body in the period
(10)
In the case of the earth it is inferred from the independent phenomenon of luni-solar precession that (C − A)/A = .00313. Hence if the earth’s axis of rotation deviates slightly from the axis of figure, it should describe a cone about the latter in 320 sidereal days. This would cause a periodic variation in the latitude of any place on the earth’s surface, as determined by astronomical methods. There appears to be evidence of a slight periodic variation of latitude, but the period would seem to be about fourteen months. The discrepancy is attributed to a defect of rigidity in the earth. The phenomenon is known as theEulerian nutation, since it is supposed to come under the free rotations first discussed by Euler.
In the case of the earth it is inferred from the independent phenomenon of luni-solar precession that (C − A)/A = .00313. Hence if the earth’s axis of rotation deviates slightly from the axis of figure, it should describe a cone about the latter in 320 sidereal days. This would cause a periodic variation in the latitude of any place on the earth’s surface, as determined by astronomical methods. There appears to be evidence of a slight periodic variation of latitude, but the period would seem to be about fourteen months. The discrepancy is attributed to a defect of rigidity in the earth. The phenomenon is known as theEulerian nutation, since it is supposed to come under the free rotations first discussed by Euler.
§ 20.Motion of a Solid of Revolution.—In the case of a solid of revolution, or (more generally) whenever there is kinetic symmetry about an axis through the mass-centre, or through a fixed point O, a number of interesting problems can be treated almost directly from first principles. It frequently happens that the extraneous forces have zero moment about the axis of symmetry, ase.g.in the case of the flywheel of a gyroscope if we neglect the friction at the bearings. The angular velocity (r) about this axis is then constant. For we have seen that r is constant when there are no extraneous forces; and r is evidently not affected by an instantaneous impulse which leaves the angular momentum Cr, about the axis of symmetry, unaltered. And a continuous force may be regarded as the limit of a succession of infinitesimal instantaneous impulses.
Suppose, for example, that a flywheel is rotating with angular velocity n about its axis, which is (say) horizontal, and that this axis is made to rotate with the angular velocity ψ̇ in the horizontal plane. The components of angular momentum about the axis of the flywheel and about the vertical will be Cn and A ψ̇ respectively, where A is the moment of inertia about any axis through the mass-centre (or through the fixed point O) perpendicular to that of symmetry. IfOK>be the vector representing the former component at time t, the vector which represents it at time t + δt will beOK′>, equal toOK>in magnitude and making with it an angle δψ. HenceKK′>(= Cn δψ) will represent the change in this component due to the extraneous forces. Hence, so far as this component is concerned, the extraneous forces must supply a couple of moment Cnψ̇ in a vertical plane through the axis of the flywheel. If this couple be absent, the axis will be tilted out of the horizontal plane in such a sense that the direction of the spin n approximates to that of the azimuthal rotation ψ̇. The remaining constituent of the extraneous forces is a couple Aψ̈ about the vertical; this vanishes if ψ̇ is constant. If the axis of the flywheel make an angle θ with the vertical, it is seen in like manner that the required couple in the vertical plane through the axis is Cn sin θ ψ̇. This matter can be strikingly illustrated with an ordinary gyroscope,e.g.by making the larger movable ring in fig. 37 rotate about its vertical diameter.
Suppose, for example, that a flywheel is rotating with angular velocity n about its axis, which is (say) horizontal, and that this axis is made to rotate with the angular velocity ψ̇ in the horizontal plane. The components of angular momentum about the axis of the flywheel and about the vertical will be Cn and A ψ̇ respectively, where A is the moment of inertia about any axis through the mass-centre (or through the fixed point O) perpendicular to that of symmetry. IfOK>be the vector representing the former component at time t, the vector which represents it at time t + δt will beOK′>, equal toOK>in magnitude and making with it an angle δψ. HenceKK′>(= Cn δψ) will represent the change in this component due to the extraneous forces. Hence, so far as this component is concerned, the extraneous forces must supply a couple of moment Cnψ̇ in a vertical plane through the axis of the flywheel. If this couple be absent, the axis will be tilted out of the horizontal plane in such a sense that the direction of the spin n approximates to that of the azimuthal rotation ψ̇. The remaining constituent of the extraneous forces is a couple Aψ̈ about the vertical; this vanishes if ψ̇ is constant. If the axis of the flywheel make an angle θ with the vertical, it is seen in like manner that the required couple in the vertical plane through the axis is Cn sin θ ψ̇. This matter can be strikingly illustrated with an ordinary gyroscope,e.g.by making the larger movable ring in fig. 37 rotate about its vertical diameter.
If the direction of the axis of kinetic symmetry be specified by means of the angular co-ordinates θ, ψ of § 7, then considering the component velocities of the point C in fig. 83, which are θ̇ and sin θψ̇ along and perpendicular to the meridian ZC, we see that the component angular velocities about the lines OA′, OB′ are −sin θ ψ̇ and θ̇ respectively. Hence if the principal moments of inertia at O be A, A, C, and if n be the constant angular velocity about the axis OC, the kinetic energy is given by
2T = A (θ̇2+ sin2θψ̇2) + Cn2.
(1)
Again, the components of angular momentum about OC, OA′ are Cn, −A sin θ ψ̇, and therefore the angular momentum (μ, say) about OZ is
μ = A sin2θψ̇ + Cn cos θ.
(2)
We can hence deduce the condition of steady precessional motion in a top. A solid of revolution is supposed to be free to turn about a fixed point O on its axis of symmetry, its mass-centre G being in this axis at a distance h from O. In fig. 83 OZ is supposed to be vertical, and OC is the axis of the solid drawn in the direction OG. If θ is constant the points C, A′ will in time δt come to positions C″, A″ such that CC″ = sin θ δψ, A′A″ = cos θ δψ, and the angular momentum about OB′ will become Cn sin θ δψ − A sin θ ψ̇ · cos θ δψ. Equating this to Mgh sin θ δt, and dividing out by sin θ, we obtain
A cos θ ψ̇2− Cnψ̇ + Mgh = 0,
(3)
as the condition in question. For given values of n and θ we have two possible values of ψ̇ provided n exceed a certain limit. With a very rapid spin, or (more precisely) with Cn large in comparison with √(4AMgh cos θ), one value of ψ̇ is small and the other large, viz. the two values are Mgh/Cn and Cn/A cos θ approximately. The absence of g from the latter expression indicates that the circumstances of the rapid precession are verynearly those of a free Eulerian rotation (§ 19), gravity playing only a subordinate part.
Again, take the case of a circular disk rolling in steady motion on a horizontal plane. The centre O of the disk is supposed to describe a horizontal circle of radius c with the constant angular velocity ψ̇, whilst its plane preserves a constant inclination θ to the horizontal. The components of the reaction of the horizontal lane will be Mcψ̇2at right angles to the tangent line at the point of contact and Mg vertically upwards, and the moment of these about the horizontal diameter of the disk, which corresponds to OB′ in fig. 83, is Mcψ̇2. α sin θ − Mgα cos θ, where α is the radius of the disk. Equating this to the rate of increase of the angular momentum about OB′, investigated as above, we find(C + Ma2+ Aacos θ)ψ̇2= Mga2cot θ,cc(4)where use has been made of the obvious relation nα = cψ̇. If c and θ be given this formula determines the value of ψ̇ for which the motion will be steady.
Again, take the case of a circular disk rolling in steady motion on a horizontal plane. The centre O of the disk is supposed to describe a horizontal circle of radius c with the constant angular velocity ψ̇, whilst its plane preserves a constant inclination θ to the horizontal. The components of the reaction of the horizontal lane will be Mcψ̇2at right angles to the tangent line at the point of contact and Mg vertically upwards, and the moment of these about the horizontal diameter of the disk, which corresponds to OB′ in fig. 83, is Mcψ̇2. α sin θ − Mgα cos θ, where α is the radius of the disk. Equating this to the rate of increase of the angular momentum about OB′, investigated as above, we find
(4)
where use has been made of the obvious relation nα = cψ̇. If c and θ be given this formula determines the value of ψ̇ for which the motion will be steady.
In the case of the top, the equation of energy and the condition of constant angular momentum (μ) about the vertical OZ are sufficient to determine the motion of the axis. Thus, we have
1⁄2A (θ̇2+ sin2θψ̇2) +1⁄2Cn2+ Mgh cos θ = const.,
(5)
A sin2θψ̇ + ν cos θ = μ,
(6)
where ν is written for Cn. From these ψ̇ may be eliminated, and on differentiating the resulting equation with respect to t we obtain
(7)
If we put θ̈ = 0 we get the condition of steady precessional motion in a form equivalent to (3). To find the small oscillation about a state of steady precession in which the axis makes a constant angle α with the vertical, we write θ = α + χ, and neglect terms of the second order in χ. The result is of the form
χ̈ + σ2χ = 0,
(8)
where
σ2= { (μ − ν cos α)2+ 2 (μ − ν cos α) (μ cos α − ν) cos α +(μ cos α − ν)2} / A2sin4α.
(9)
When ν is large we have, for the “slow” precession σ = ν/A, and for the “rapid” precession σ = A/ν cos α = ψ̇, approximately. Further, on examining the small variation in ψ̇, it appears that in a slightly disturbed slow precession the motion of any point of the axis consists of a rapid circular vibration superposed on the steady precession, so that the resultant path has a trochoidal character. This is a type of motion commonly observed in a top spun in the ordinary way, although the successive undulations of the trochoid may be too small to be easily observed. In a slightly disturbed rapid precession the superposed vibration is elliptic-harmonic, with a period equal to that of the precession itself. The ratio of the axes of the ellipse is sec α, the longer axis being in the plane of θ. The result is that the axis of the top describes a circular cone about a fixed line making a small angle with the vertical. This is, in fact, the “invariable line” of the free Eulerian rotation with which (as already remarked) we are here virtually concerned. For the more general discussion of the motion of a top seeGyroscope.
§ 21.Moving Axes of Reference.—For the more general treatment of the kinetics of a rigid body it is usually convenient to adopt a system of moving axes. In order that the moments and products of inertia with respect to these axes may be constant, it is in general necessary to suppose them fixed in the solid.
We will assume for the present that the origin O is fixed. The moving axes Ox, Oy, Oz form a rigid frame of reference whose motion at time t may be specified by the three component angular velocities p, q, r. The components of angular momentum about Ox, Oy, Oz will be denoted as usual by λ, μ, ν. Now consider a system of fixed axes Ox′, Oy′, Oz′ chosen so as to coincide at the instant t with the moving system Ox, Oy, Oz. At the instant t + δt, Ox, Oy, Oz will no longer coincide with Ox′, Oy′, Oz′; in particular they will make with Ox′ angles whose cosines are, to the first order, 1, −rδt, qδt, respectively. Hence the altered angular momentum about Ox′ will be λ + δλ + (μ + δμ) (−rδt) + (ν + δν) qδt. If L, M, N be the moments of the extraneous forces about Ox, Oy, Oz this must be equal to λ + Lδt. Hence, and by symmetry, we obtain
(1)
These equations are applicable to any dynamical system whatever. If we now apply them to the case of a rigid body moving about a fixed point O, and make Ox, Oy, Oz coincide with the principal axes of inertia at O, we have λ, μ, ν = Ap, Bq, Cr, whence
(2)
If we multiply these by p, q, r and add, we get
(3)
which is (virtually) the equation of energy.
As a first application of the equations (2) take the case of a solid constrained to rotate with constant angular velocity ω about a fixed axis (l, m, n). Since p, q, r are then constant, the requisite constraining couple is
L = (C − B) mnω2, M = (A − C) nlω2, N = (B − A) lmω2.