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
