properties of the body itself. In a gravitational field, for example, it is the centre of mass. $35. Eulerian angles As has already been mentioned, the motion of a rigid body can be described by means of the three co-ordinates of its centre of mass and any three angles which determine the orientation of the axes X1, X2, X3 in the moving system of co-ordinates relative to the fixed system X, Y, Z. These angles may often be conveniently taken as what are called Eulerian angles. Z X2 Y FIG. 47 Since we are here interested only in the angles between the co-ordinate axes, we may take the origins of the two systems to coincide (Fig. 47). The moving x1x2-plane intersects the fixed XY-plane in some line ON, called the line of nodes. This line is evidently perpendicular to both the Z-axis and the x3-axis; we take its positive direction as that of the vector product ZXX3 (where Z and X3 are unit vectors along the Z and X3 axes). We take, as the quantities defining the position of the axes x1, X2, X3 relative to the axes X, Y, Z the angle 0 between the Z and X3 axes, the angle between the X-axis and ON, and the angle as between the x1-axis and ON. The angles and 4 are measured round the Z and X3 axes respectively in the direction given by the corkscrew rule. The angle 0 takes values from 0 to TT, and and 4 from 0 to 2n.t t The angles 0 and - are respectively the polar angle and azimuth of the direction X3 with respect to the axes X, Y, Z. The angles 0 and 12– are respectively the polar angle and azimuth of the direction Z with respect to the axes X1, X2, X3. §35 Eulerian angles 111 Let us now express the components of the angular velocity vector S along the moving axes X1, X2, X3 in terms of the Eulerian angles and their derivatives. To do this, we must find the components along those axes of the angular velocities 6, b, 4. The angular velocity è is along the line of nodes ON, and its components are 1 = O cos 4/5, = - O sin 4/5, = 0. The angular velo- city is along the Z-axis; its component along the x3-axis is 03 = cos 0, and in the x1x2-plane sin A. Resolving the latter along the X1 and X2 axes, we have 01 = sin 0 sin 4/s, O2 = sin 0 cos 4. Finally, the angular velocity is is along the x3-axis. Collecting the components along each axis, we have S21 = 0 COS 4, Q2 = sin 0 cosy-osiny, (35.1) S23 = o cos0+4. = If the axes X1, X2, X3 are taken to be the principal axes of inertia of the body, the rotational kinetic energy in terms of the Eulerian angles is obtained by substituting (35.1) in (32.8). For a symmetrical top (I1 = I2 # I3), a simple reduction gives Trot = (35.2) This expression can also be more simply obtained by using the fact that the choice of directions of the principal axes X1, X2 is arbitrary for a symmetrical top. If the X1 axis is taken along the line of nodes ON, i.e. 4 = 0, the compo- nents of the angular velocity are simply O2 = o sin A, (35.3) As a simple example of the use of the Eulerian angles, we shall use them to determine the free motion of a symmetrical top, already found in $33. We take the Z-axis of the fixed system of co-ordinates in the direction of the constant angular momentum M of the top. The x3-axis of the moving system is along the axis of the top; let the x1-axis coincide with the line of nodes at the instant considered. Then the components of the vector M are, by formulae (35.3), M1 = I1 = I, M2 = IS2 = sin 0, M3 = I3Q3 = I3( cos 0+4). Since the x1-axis is perpendicular to the Z-axis, we have M1 = 0, M2 = M sin 0, M3 = M cos 0. Comparison gives 0=0, I = M, = (35.4) The first of these equations gives 0 = constant, i.e. the angle between the axis of the top and the direction of M is constant. The second equation gives the angular velocity of precession = M/I1, in agreement with (33.5). Finally, the third equation gives the angular velocity with which the top rotates about its own axis: S3 = (M/I3) cos 0. 112 Motion of a Rigid Body §35 PROBLEMS PROBLEM 1. Reduce to quadratures the problem of the motion of a heavy symmetrical top whose lowest point is fixed (Fig. 48). SOLUTION. We take the common origin of the moving and fixed systems of co-ordinates at the fixed point O of the top, and the Z-axis vertical. The Lagrangian of the top in a gravita- tional field is L = (02 +02 sin ²0 + 1/3(1- cos 0)2-ugl - cos 0, where u is the mass of the top and l the distance from its fixed point to the centre of mass. Z X3 x2 a ug Y x1 N FIG. 48 The co-ordinates 4 and are cyclic. Hence we have two integrals of the motion: P4 = = cos 0) = constant = M3 (1) = = cos 0 = constant III M2, (2) where I’1 = I1+ul2; the quantities P4 and Po are the components of the rotational angular momentum about O along the X3 and Z axes respectively. The energy E = cos 0 (3) is also conserved. From equations (1) and (2) we find = 0)/I’1 sin 20, (4) (5) Eliminating b and of from the energy (3) by means of equations (4) and (5), we obtain E’ = where E’ = (6) §35 Eulerian angles 113 Thus we have t= (7) this is an elliptic integral. The angles 4 and are then expressed in terms of 0 by means of integrals obtained from equations (4) and (5). The range of variation of 0 during the motion is determined by the condition E’ Ueff(0). The function Uett(8) tends to infinity (if M3 # M2) when 0 tends to 0 or II, and has a minimum between these values. Hence the equation E’ = Ueff(0) has two roots, which determine the limiting values 01 and O2 of the inclination of the axis of the top to the vertical. When 0 varies from 01 to O2, the derivative o changes sign if and only if the difference M-M3 cos 0 changes sign in that range of 0. If it does not change sign, the axis of the top precesses monotonically about the vertical, at the same time oscillating up and down. The latter oscillation is called nutation; see Fig. 49a, where the curve shows the track of the axis on the surface of a sphere whose centre is at the fixed point of the top. If does change sign, the direction of precession is opposite on the two limiting circles, and so the axis of the top describes loops as it moves round the vertical (Fig. 49b). Finally, if one of 01, O2 is a zero of M2-M3 cos 0, of and è vanish together on the corresponding limiting circle, and the path of the axis is of the kind shown in Fig. 49c. O2 O2 O2 (a) (b) (c) FIG. 49 PROBLEM 2. Find the condition for the rotation of a top about a vertical axis to be stable. SOLUTION. For 0 = 0, the X3 and Z axes coincide, so that M3 = Mz, E’ = 0. Rotation about this axis is stable if 0 = 0 is a minimum of the function Ueff(9). For small 0 we have Ueff 22 whence the condition for stability is M32 > 41’1ugl or S232 > 41’1ugl/I32. PROBLEM 3. Determine the motion of a top when the kinetic energy of its rotation about its axis is large compared with its energy in the gravitational field (called a “fast” top). SOLUTION. In a first approximation, neglecting gravity, there is a free precession of the axis of the top about the direction of the angular momentum M, corresponding in this case to the nutation of the top; according to (33.5), the angular velocity of this precession is Sunu = M/I’ 1. (1) In the next approximation, there is a slow precession of the angular momentum M about the vertical (Fig. 50). To determine the rate of this precession, we average the exact equation of motion (34.3) dM/dt = K over the nutation period. The moment of the force of gravity on the top is K=uln3xg, where n3 is a unit vector along the axis of the top. It is evident from symmetry that the result of averaging K over the “nutation cone” is to replace n3 by its component (M/M) cos a in the direction of M, where a is the angle between M and the axis of the top. Thus we have dM/dt = -(ul/M)gxM cos a. This shows that the vector M 114 Motion of a Rigid Body