Angular momentum of a rigid body 105 the notation being as in Problem 7. The instantaneous axis of rotation is the generator OA which passes through the point where the cone touches the plane. The centre of mass is at a distance a sin a from this axis, and so l = V/a sin a = O/sin a. The components of the vector Ca along the principal axes of inertia are, if the x2-axis is taken perpendicular to the axis of the cone and to the line OA, O sin a = 6, 0, N cos a = è cot a. The kinetic energy is therefore T cot2a = 314h282(sec2x+5)/40. Z 0 Y A FIG. 43 PROBLEM 9. Find the kinetic energy of a homogeneous ellipsoid which rotates about one of its axes (AB in Fig. 44) while that axis itself rotates about a line CD perpendicular to it and passing through the centre of the ellipsoid. SOLUTION. Let the angle of rotation about CD be 0, and that about AB (i.e. the angle between CD and the x1-axis of inertia, which is perpendicular to AB) be . Then the com- ponents of S along the axes of inertia are é cos , 0 sin , b, if the x3-axis is AB. Since the centre of mass, at the centre of the ellipsoid, is at rest, the kinetic energy is = D B D A Do A 1a C of 8 FIG. 44 FIG. 45 PROBLEM 10. The same as Problem 9, but for the case where the axis AB is not perpendicu- lar to CD and is an axis of symmetry of the ellipsoid (Fig. 45). SOLUTION. The components of Ca along the axis AB and the other two principal axes of inertia, which are perpendicular to AB but otherwise arbitrary, are è cos a cos , è cos a X sin , o to sin a. The kinetic energy is T = 11102 a)2. $33. Angular momentum of a rigid body The value of the angular momentum of a system depends, as we know, on the point with respect to which it is defined. In the mechanics of a rigid body, 106 Motion of a Rigid Body §33 the most appropriate point to choose for this purpose is the origin of the moving system of co-ordinates, i.e. the centre of mass of the body, and in what follows we shall denote by M the angular momentum SO defined. According to formula (9.6), when the origin is taken at the centre of mass of the body, the angular momentum M is equal to the “intrinsic” angular momentum resulting from the motion relative to the centre of mass. In the definition M = Emrxv we therefore replace V by Sxr: M = = or, in tensor notation, Mi = OK Finally, using the definition (32.2) of the inertia tensor, we have (33.1) If the axes X1, X2, X3 are the same as the principal axes of inertia, formula (33.1) gives M2 = I2DQ, M3 = I303. = (33.2) In particular, for a spherical top, where all three principal moments of inertia are equal, we have simply M = IS, (33.3) i.e. the angular momentum vector is proportional to, and in the same direc- tion as, the angular velocity vector. For an arbitrary body, however, the vector M is not in general in the same direction as S; this happens only when the body is rotating about one of its principal axes of inertia. Let us consider a rigid body moving freely, i.e. not subject to any external forces. We suppose that any uniform translational motion, which is of no interest, is removed, leaving a free rotation of the body. As in any closed system, the angular momentum of the freely rotating body is constant. For a spherical top the condition M = constant gives C = con- stant; that is, the most general free rotation of a spherical top is a uniform rotation about an axis fixed in space. The case of a rotator is equally simple. Here also M = IS, and the vector S is perpendicular to the axis of the rotator. Hence a free rotation of a rotator is a uniform rotation in one plane about an axis perpendicular to that plane. The law of conservation of angular momentum also suffices to determine the more complex free rotation of a symmetrical top. Using the fact that the principal axes of inertia X1, X2 (perpendicular to the axis of symmetry (x3) of the top) may be chosen arbitrarily, we take the x2-axis perpendicular to the plane containing the constant vector M and the instantaneous position of the x3-axis. Then M2 = 0, and formulae (33.2) show that Q2 = 0. This means that the directions of M, St and the axis of the top are at every instant in one plane (Fig. 46). Hence, in turn, it follows that the velocity V = Sxr of every point on the axis of the top is at every instant perpendicular to that