
§32
The inertia tensor
101
two of the principal moments of inertia are equal and the third is zero:
I3 = 0.
(32.11)
Such a system is called a rotator. The characteristic property which distin-
guishes a rotator from other bodies is that it has only two, not three, rotational
degrees of freedom, corresponding to rotations about the X1 and X2 axes: it
is clearly meaningless to speak of the rotation of a straight line about itself.
Finally, we may note one further result concerning the calculation of the
inertia tensor. Although this tensor has been defined with respect to a system
of co-ordinates whose origin is at the centre of mass (as is necessary if the
fundamental formula (32.3) is to be valid), it may sometimes be more con-
veniently found by first calculating a similar tensor I' =
defined with respect to some other origin O'. If the distance OO' is repre-
sented by a vector a, then r = r'+a, Xi = x'i+ai; since, by the definition
of O, Emr = 0, we have
I'ikIk(a2ik-aiak).
(32.12)
Using this formula, we can easily calculate Iik if I'ik is known.
PROBLEMS
PROBLEM 1. Determine the principal moments of inertia for the following types of mole-
cule, regarded as systems of particles at fixed distances apart: (a) a molecule of collinear
atoms, (b) a triatomic molecule which is an isosceles triangle (Fig. 36), (c) a tetratomic
molecule which is an equilateral-based tetrahedron (Fig. 37).
m2
X2
m2
h
x
m
a
a
m
a
m
m
a
FIG. 36
FIG. 37
SOLUTION. (a)
Is = 0,
where Ma is the mass of the ath atom, lao the distance between the ath and bth atoms, and
the summation includes one term for every pair of atoms in the molecule.
For a diatomic molecule there is only one term in the sum, and the result is obvious it is
the product of the reduced mass of the two atoms and the square of the distance between
them: I1 = I2 = m1m2l2((m1+m2).
(b) The centre of mass is on the axis of symmetry of the triangle, at a distance X2 = mzh/u
from its base (h being the height of the triangle). The moments of inertia are I1 = 2m1m2h2/u,
I2 = 1m1a2, I3 = I+I2.
(c) The centre of mass is on the axis of symmetry of the tetrahedron, at a distance
X3 = mgh/u from its base (h being the height of the tetrahedron). The moments of inertia
102
Motion of a Rigid Body
§32
are I1 = = I3 = mia². If M1 = M2, h = (2/3)a, the molecule is a
regular tetrahedron and I1=I2 = I3 = mia2.
PROBLEM 2. Determine the principal moments of inertia for the following homogeneous
bodies: (a) a thin rod of length l, (b) a sphere of radius R, (c) a circular cylinder of radius R
and height h, (d) a rectangular parallelepiped of sides a, b, and c, (e) a circular cone of height
h and base radius R, (f) an ellipsoid of semiaxes a, b, c.
SOLUTION. (a) I1 = I2 = Trul2, I3 = 0 (we neglect the thickness of the rod).
(b) I1 = I2 = I3 = zuR2 (found by calculating the sum I+I+I3 = 2p dV).
(c) I1 = I2 = tu(R2+th2), I3 = tuR2 (where the x3-axis is along the axis of the cylinder).
(d) I1 = (2+c2), I2 = (a2+cc), I3 = 121(a2++b) (where the axes X1, x2, X3 are
along the sides a, b, c respectively).
(e) We first calculate the tensor I'ik with respect to axes whose origin is at the vertex of
the cone (Fig. 38). The calculation is simple if cylindrical co-ordinates are used, and the result
is I'1 = I'2 = I'3 = 2 The centre of mass is easily shown to be on the
axis of the cone and at a distance a = 3h from the vertex. Formula (32.12) therefore gives
I1 = I2 = I'1-2 = I3 = I'3 = TouR2.
X3,X3'
X2
xi
x2
FIG. 38
(f) The centre of mass is at the centre of the ellipsoid, and the principal axes of inertia are
along the axes of the ellipsoid. The integration over the volume of the ellipsoid can be reduced
to one over a sphere by the transformation x = as,y = bn, 2 = c5, which converts the equa-
tion of the surface of the ellipsoid 1 into that of the unit sphere
st+24's = 1.
For example, the moment of inertia about the x-axis is
dz
= tabcI'(b2 tc2),
where I' is the moment of inertia of a sphere of unit radius. Since the volume of the ellipsoid
is 4nabc/3, we find the moments of inertia I = tu(b2+c2), I2 = tu(a2+c2), I3 = tu(a2+b2).
PROBLEM 3. Determine the frequency of small oscillations of a compound pendulum (a
rigid body swinging about a fixed horizontal axis in a gravitational field).
SOLUTION. Let l be the distance between the centre of mass of the pendulum and the axis
about which it rotates, and a, B, y the angles between the principal axes of inertia and the
axis of rotation. We take as the variable co-ordinate the angle between the vertical
and a line through the centre of mass perpendicular to the axis of rotation. The velocity of
the centre of mass is V = 10, and the components of the angular velocity along the principal
§32
The inertia tensor
103
axes of inertia are o cos a, b cos B, b cos y. Assuming the angle to be small, we find the
potential energy U = ugl(1-cos 9 22 12. The Lagrangian is therefore
=
The frequency of the oscillations is consequently
w2 = cos2y).
PROBLEM 4. Find the kinetic energy of the system shown in Fig. 39: OA and AB are thin
uniform rods of length l hinged together at A. The rod OA rotates (in the plane of the diagram)
about O, while the end B of the rod AB slides along Ox.
A
l
l
x
B
FIG. 39
SOLUTION. The velocity of the centre of mass of the rod OA (which is at the middle of
the rod) is 110, where is the angle AOB. The kinetic energy of the rod OA is therefore
T1 = where u is the mass of each rod.
The Cartesian co-ordinates of the centre of mass of the rod AB are X = sl cos o, Y
= 1/ sin b. Since the angular velocity of rotation of this rod is also b, its kinetic energy is
T2 = = tul2(1- +8 sin2o)62 +1162. The total kinetic energy of this
system is therefore = I = Tzul2 (see Problem 2(a)).
PROBLEM 5. Find the kinetic energy of a cylinder of radius R rolling on a plane, if the mass
of the cylinder is so distributed that one of the principal axes of inertia is parallel to the axis
of the cylinder and at a distance a from it, and the moment of inertia about that principal
axis is I.
SOLUTION. Let be the angle between the vertical and a line from the centre of mass
perpendicular to the axis of the cylinder (Fig. 40). The motion of the cylinder at any instant
R
FIG. 40
may be regarded as a pure rotation about an instantaneous axis which coincides with the
line where the cylinder touches the plane. The angular velocity of this rotation is o, since
the angular velocity of rotation about all parallel axes is the same. The centre of mass is at a
distance V(a2+R2-2aR cos ) from the instantaneous axis, and its velocity is therefore
V = bv /(a2+R2-2aR cos ). The total kinetic energy is
T = cos
104
Motion of a Rigid Body
§32
PROBLEM 6. Find the kinetic energy of a homogeneous cylinder of radius a rolling inside
a cylindrical surface of radius R (Fig. 41).
R
FIG. 41
SOLUTION. We use the angle between the vertical and the line joining the centres of the
cylinders. The centre of mass of the rolling cylinder is on the axis, and its velocity is V =
o(R-a). We can calculate the angular velocity as that of a pure rotation about an instantaneous
axis which coincides with the line of contact of the cylinders it is Q = V/a = p(R-a)/a.
If I3 is the moment of inertia about the axis of the cylinder, then
T =
I3 being given by Problem 2(c).
PROBLEM 7. Find the kinetic energy of a homogeneous cone rolling on a plane.
SOLUTION. We denote by 0 the angle between the line OA in which the cone touches the
plane and some fixed direction in the plane (Fig. 42). The centre of mass is on the axis of the
cone, and its velocity V = a0 cos a, where 2a is the vertical angle of the cone and a the
Z
Y
A
FIG. 42
distance of the centre of mass from the vertex. The angular velocity can be calculated as
that of a pure rotation about the instantaneous axis OA: S2 = V/a sin a = é cot a. One of
the principal axes of inertia (x3) is along the axis of the cone, and we take another (x2) perpen-
dicular to the axis of the cone and to the line OA. Then the components of the vector S
(which is parallel to OA) along the principal axes of inertia are O sin a, 0, O cos a. The kinetic
energy is thus
=
= 3uh202(1+5 cos2x)/40,
where h is the height of the cone, and I1, I3 and a have been given in Problem 2(e).
PROBLEM 8. Find the kinetic energy of a homogeneous cone whose base rolls on a plane
and whose vertex is fixed at a height above the plane equal to the radius of the base, so that
the axis of the cone is parallel to the plane.
SOLUTION. We use the angle 0 between a fixed direction in the plane and the projection
of the axis of the cone on the plane (Fig. 43). Then the velocity of the centre of mass is V = aß,
§33
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
§34
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
§35
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.