The equations of motion of a rigid body 107 plane. That is, the axis of the top rotates uniformly (see below) about the direction of M, describing a circular cone. This is called regular precession of the top. At the same time the top rotates uniformly about its own axis. M n x3 22pr x1 FIG. 46 The angular velocities of these two rotations can easily be expressed in terms of the given angular momentum M and the angle 0 between the axis of the top and the direction of M. The angular velocity of the top about its own axis is just the component S3 of the vector S along the axis: Q3 = M3/I3 = (M/I3) cos 0. (33.4) To determine the rate of precession Spr, the vector S must be resolved into components along X3 and along M. The first of these gives no displacement of the axis of the top, and the second component is therefore the required angular velocity of precession. Fig. 46 shows that Spr sin 0 = Q1, and, since S21 = M1/I1 = (M/I1) sin 0, we have Spr r=M/I1. (33.5) $34. The equations of motion of a rigid body Since a rigid body has, in general, six degrees of freedom, the general equations of motion must be six in number. They can be put in a form which gives the time derivatives of two vectors, the momentum and the angular momentum of the body. The first equation is obtained by simply summing the equations p = f for each particle in the body, p being the momentum of the particle and f the 108 Motion of a Rigid Body §34 force acting on it. In terms of the total momentum of the body P = and total force acting on it F = Ef, we have dP/dt = F. (34.1) Although F has been defined as the sum of all the forces f acting on the various particles, including the forces due to other particles, F actually includes only external forces: the forces of interaction between the particles composing the body must cancel out, since if there are no external forces the momentum of the body, like that of any closed system, must be conserved, i.e. we must have F = 0. If U is the potential energy of a rigid body in an external field, the force F is obtained by differentiating U with respect to the co-ordinates of the centre of mass of the body: F = JUIR. (34.2) For, when the body undergoes a translation through a distance SR, the radius vector r of every point in the body changes by SR, and so the change in the potential energy is SU = (U/dr) Sr = RR Couldr = SR SR. It may be noted that equation (34.1) can also be obtained as Lagrange’s equation for the co-ordinates of the centre of mass, (d/dt)àL/JV = aL/JR, with the Lagrangian (32.4), for which OL/OV=,MV=P, 0L/JR = JU/OR = F. Let us now derive the second equation of motion, which gives the time derivative of the angular momentum M. To simplify the derivation, it is convenient to choose the “fixed” (inertial) frame of reference in such a way that the centre of mass is at rest in that frame at the instant considered. We have M = (d/dt) Erxp = Eixp+. Our choice of the frame of reference (with V = 0) means that the value of i at the instant considered is the same as V = i. Since the vectors V and p = mv are parallel, ixp = 0. Replacing p by the force f, we have finally dM/dt = K, (34.3) where K = . (34.4) Since M has been defined as the angular momentum about the centre of mass (see the beginning of $33), it is unchanged when we go from one inertial frame to another. This is seen from formula (9.5) with R = 0. We can there- fore deduce that the equation of motion (34.3), though derived for a particular frame of reference, is valid in any other inertial frame, by Galileo’s relativity principle. The vector rxf is called the moment of the force f, and so K is the total torque, i.e. the sum of the moments of all the forces acting on the body. Like §34 The equations of motion of a rigid body 109 the total force F, the sum (34.4) need include only the external forces: by the law of conservation of angular momentum, the sum of the moments of the internal forces in a closed system must be zero. The moment of a force, like the angular momentum, in general depends on the choice of the origin about which it is defined. In (34.3) and (34.4) the moments are defined with respect to the centre of mass of the body. When the origin is moved a distance a, the new radius vector r’ of each point in the body is equal to r-a. Hence K = Erxf = Er’xf+ Eaxf or K = K’+axF. (34.5) Hence we see, in particular, that the value of the torque is independent of the choice of origin if the total force F = 0. In this case the body is said to be acted on by a couple. Equation (34.3) may be regarded as Lagrange’s equation (d/dt) OL/OS = 0L/dd for the “rotational co-ordinates”. Differentiating the Lagrangian (32.4) with respect to the components of the vector S2, we obtain = IikOk = Mi. The change in the potential energy resulting from an infinitesimal rotation SO of the body is SU = - Ef.Sr = - = So. Erxf = -K.SO, whence K =-20/00, = (34.6) so that aL/dd = 00/08 = K. Let us assume that the vectors F and K are perpendicular. Then a vector a can always be found such that K’ given by formula (34.5) is zero and K a x F. (34.7) The choice of a is not unique, since the addition to a of any vector parallel to F does not affect equation (34.7). The condition K’ = 0 thus gives a straight line, not a point, in the moving system of co-ordinates. When K is perpendi- cular to F, the effect of all the applied forces can therefore be reduced to that of a single force F acting along this line. Such a case is that of a uniform field of force, in which the force on a particle is f = eE, with E a constant vector characterising the field and e characterising the properties of a particle with respect to the field. Then F = Ee, K = erxE. Assuming that # 0, we define a radius vector ro such that (34.8) Then the total torque is simply =roxF (34.9) Thus, when a rigid body moves in a uniform field, the effect of the field reduces to the action of a single force F applied at the point whose radius vector is (34.8). The position of this point is entirely determined by the t For example, in a uniform electric field E is the field strength and e the charge; in a uniform gravitational field E is the acceleration g due to gravity and e is the mass m. 110 Motion of a Rigid Body