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
| 2π | = | A − C | ω. |
| χ̇ | A |
(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 the Eulerian 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, as e.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.
| Fig. 82. |
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. If OK> be the vector representing the former component at time t, the vector which represents it at time t + δt will be OK′>, equal to OK> in magnitude and making with it an angle δψ. Hence KK′> (= 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.
| Fig. 83. |
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 + sin2 θψ̇2) + Cn2.