9 rigid body dynamics.pdf

Rigid Body Dynamics
9
CHAPTER OUTLINE
9.1 Introduction ..................................................................................................................................459
9.2 Kinematics....................................................................................................................................460
9.3 Equations of translational motion ...................................................................................................469
9.4 Equations of rotational motion........................................................................................................470
9.5 Moments of inertia ........................................................................................................................475
Parallel axis theorem ...........................................................................................................490
9.6 Euler’s equations...........................................................................................................................496
9.7 Kinetic energy...............................................................................................................................502
9.8 The spinning top ...........................................................................................................................504
9.9 Euler angles..................................................................................................................................510
9.10 Yaw, pitch, and roll angles ..........................................................................................................520
9.11 Quaternions ................................................................................................................................523
Problems.............................................................................................................................................532
Section 9.2 .........................................................................................................................................532
Section 9.5 .........................................................................................................................................535
Section 9.7 .........................................................................................................................................540
Section 9.8 .........................................................................................................................................540
Section 9.9 .........................................................................................................................................542
9.1 Introduction
Just as Chapter 1 provides a foundation for the dof the equations of orbital mechanics, this chapter
serves as a basis for developing the equations of satellite attitude dynamics. Chapter 1 deals with
particles, whereas here we are concerned with rigid bodies. Those familiar with rigid body dynamics
can move on to the next chapter, perhaps returning from time to time to review concepts.
The kinematics of rigid bodies is presented first. The subject depends on a theorem of the French
mathematician Michel Chasles (1793–1880). Chasles’ theorem states that the motion of a rigid body
can be described by the displacement of any point of the body (the base point) plus a rotation about a
unique axis through that point. The magnitude of the rotation does not depend on the base point. Thus,
at any instant, a rigid body in a general state of motion has an angular velocity vector whose direction
is that of the instantaneous axis of rotation. Describing the rotational component of the motion a rigid
body in three dimensions requires taking advantage of the vector nature of angular velocity and
knowing how to take the time derivative of moving vectors, which is explained in Chapter 1. Several
examples illustrate how this is done.
CHAPTER
Orbital Mechanics for Engineering Students. http://dx.doi.org/10.1016/B978-0-08-097747-8.00009-8
Copyright Ó 2014 Elsevier Ltd. All rights reserved.
459

We then move on to study the interaction between the motion of a rigid body and the forces acting on
it. Describing the translational component of the motion requires simply concentrating all of the mass at a
point, thecenter of mass, andapplying themethodsof particle mechanics to determineits motion.Indeed,
our study of the two-body problem up to this point has focused on the motion of their centers of mass
without regard to the rotational aspect. Analyzing the rotational dynamics requires computing the body’s
angular momentum, and that in turn requires accounting for how the mass is distributed throughout the
body. The mass distribution is described by the six components of the moment of inertia tensor.
Writing the equations of rotational motion relative to coordinate axes embedded in the rigid body
and aligned with the principal axes of inertia yields the nonlinear Euler equations of motion, which are
applied to a study of the dynamics of a spinning top (or one-axis gyro).
The expression for the kinetic energy of a rigid body is derived because it will be needed in the
following chapter.
The chapter next describes the two sets of three angles commonly employed to specify the orien-
tation of a body in three-dimensional space. One of these comprises the Euler angles, which are the same
as the right ascension of the node (U), argument of periapsis (u), and inclination (i) introduced in
Chapter 4 to orient orbits in space. The other set comprises the yaw, pitch, and roll angles, which are
suitable for describing the orientation of an airplane. Both the Euler angles and yaw, pitch, and roll
angles will be employed in Chapter 10.
The chapter concludes with a brief discussion of quaternions and an example of how they are used
to describe the evolution of the attitude of a rigid body.
9.2 Kinematics
Figure 9.1 shows a moving rigid body and its instantaneous axis of rotation, which defines the direction
of the absolute angular velocity vector u. The XYZ axes are a fixed, inertial frame of reference. The
position vectors RA and RB of two points on the rigid body are measured in the inertial frame. The
vector RB/A drawn from point A to point B is the position vector of B relative to A. Since the body is
rigid, RB/A has a constant magnitude even though its direction is continuously changing. Clearly,
RB ¼ RA þ RB=A
Differentiating this equation through with respect to time, we get
_
RB ¼ _
RA þ
dRB=A
dt
(9.1)
_
RA and _
RB are the absolute velocities vA and vB of points A and B. Because the magnitude of RB/A does
not change, its time derivative is given by Eqn (1.61), that is,
dRB=A
dt
¼ u RB=A
Thus, Eqn (9.1) becomes
vB ¼ vA þ u RB=A (9.2)
Taking the time derivative of Eqn (9.1) yields
€
RB ¼ €
RA þ
d2
RB=A
dt2
(9.3)
460 CHAPTER 9 Rigid Body Dynamics

€
RA and €
RB are the absolute accelerations aA and aB of the two points of the rigid body, while from Eqn
(1.62) we have
d2
RB=A
dt2
¼ a RB=A þ u

u RB=A

in which a is the angular acceleration, a ¼ du=dt. Therefore, Eqn (9.3) can be written
aB ¼ aA þ a RB=A þ u

u RB=A

(9.4)
Equations (9.2) and (9.4) are the relative velocity and acceleration formulas. Note that all quantities in
these expressions are measured in the same inertial frame of reference.
When the rigid body under consideration is connected to and moving relative to another rigid body,
computation of its inertial angular velocity u and angular acceleration a must be done with care. The
key is to remember that angular velocity is a vector. It may be found as the vector sum of a sequence of
angular velocities, each measured relative to another, starting with one measured relative to an ab-
solute frame, as illustrated in Figure 9.2. In that case, the absolute angular velocity u of body 4 is
u ¼ u1 þ u2=1 þ u3=2 þ u4=3 (9.5)
Each of these angular velocities is resolved into components along the axes of the moving frame of
reference xyz as shown in Figure 9.2, so that the vector sum may be written as
u ¼ ux
^
i þ uy
^
j þ uz
^
k (9.6)
The moving frame is chosen for convenience of the analysis, and its own inertial angular velocity is
denoted as U, as discussed in Section 1.6. According to Eqn (1.65), the absolute angular acceleration
a ¼ du=dt is obtained from Eqn (9.6) by means of the following calculation:
a ¼
du
dt

rel
þ U u (9.7)
X
Y
Z
A
RB
RA
B
RB/A
FIGURE 9.1
A rigid body and its instantaneous axis of rotation.
9.2 Kinematics 461

where
du
dt

rel
¼ _
ux
^
i þ _
uy
^
j þ _
uz
^
k (9.8)
and _
ux ¼ dux=dt, etc.
Being able to express the absolute angular velocity vector in an appropriately chosen moving
reference frame, as in Eqn (9.6), is crucial to the analysis of rigid body motion. Once we have the
components of u, we simply differentiate them with respect to time to arrive at Eqn (9.8). Observe that
the absolute angular acceleration a and du=dtÞrel, the angular acceleration relative to the moving
frame, are the same if and only if U ¼ u. That occurs if the moving reference is a body-fixed frame,
that is, a set of xyz axes imbedded in the rigid body itself.
EXAMPLE 9.1
The airplane in Figure 9.3 flies at a constant speed v while simultaneously undergoing a constant yaw rate uyaw
about a vertical axis and describing a circular loop in the vertical plane with a radius r. The constant propeller spin
rate is uspin relative to the airframe. Find the velocity and acceleration of the tip P of the propeller relative to the
hub H, when P is directly above H. The propeller radius is l.
Solution
The xyz axes are rigidly attached to the airplane. The x-axis is aligned with the propeller’s spin axis. The y axis is
vertical, and the z axis is in the spanwise direction, so that xyz forms a right-handed triad. Although the xyz frame is
not inertial, we can imagine it to instantaneously coincide with an inertial frame.
The absolute angular velocity of the airplane has two components, the yaw rate and the counterclockwise pitch
angular velocity v/r of its rotation in the circular loop,
uairplane ¼ uyaw
^
j þ upitch
^
k ¼ uyaw
^
j þ
v
r
^
k
FIGURE 9.2
Angular velocity is the vector sum of the relative angular velocities starting with u1, measured relative to the
inertial frame.

The angular velocity of the body-fixed moving frame is that of the airplane, U ¼ uairplane, so that
U ¼ uyaw
^
j þ
v
r
^
k (a)
The absolute angular velocity of the propeller is that of the airplane plus the angular velocity of the propeller
relative to the airplane
uprop ¼ uairplane þ uspin
^
i ¼ U þ uspin
^
i
which means
uprop ¼ uspin
^
i þ uyaw
^
j þ
v
r
^
k (b)
From Eqn (9.2), the velocity of point P on the propeller relative to H on the hub, vP/H, is given by
vP=H ¼ vP vH ¼ uprop rP=H
where rP/H is the position vector of P relative to H at this instant,
rP=H ¼ l^
j (c)
Thus, using Eqns (b) and (c),
vP=H ¼

uspin
^
i þ uyaw
^
j þ
v
r
^
k

l^
j

from which
vP=H ¼
v
r
l^
i þ uspinl^
k
The absolute angular acceleration of the propeller is found by substituting Eqns (a) and (b) into Eqn (9.7),
aprop ¼
duprop
dt

rel
þ U uprop
¼

duspin
dt
^
i þ
duyaw
dt
^
j þ
dðv=rÞ
dt
^
k

þ

uyaw
^
j þ
v
r
^
k

uspin
^
i þ uyaw
^
j þ
v
r
^
k

Since uspin, uyaw, v, and r are all constant, this reduces to
aprop ¼

uyaw
^
j þ
v
r
^
k

uspin
^
i þ uyaw
^
j þ
v
r
^
k

FIGURE 9.3
Airplane with an attached xyz body frame.
9.2 Kinematics 463

Carrying out the cross product yields
aprop ¼
v
r
uspin
^
j uyawuspin
^
k (d)
From Eqn (9.4), the acceleration of P relative to H, aP/H, is given by
aP=H ¼ aP aH ¼ aprop rP=H þ uprop

uprop rP=H

Substituting Eqns (b), (c) and (d) into this expression yields
aP=H ¼

v
r
uspin
^
j uyawuspin
^
k

l^
j

þ

uspin
^
i þ uyaw
^
j þ
v
r
^
k

uspin
^
i þ uyaw
^
j þ
v
r
^
k

l^
j

From this we find that
aP=H ¼

uyawuspinl^
i

þ

uspin
^
i þ uyaw
^
j þ
v
r
^
k

v
r
l^
i þ uspinl^
k
¼

uyawuspinl^
i

þ

uyawuspinl^
i

v2
r2
þ u2
spin

l^
j þ uyaw
v
r
l^
k
so that finally,
aP=H ¼ 2uyawuspinl^
i

v2
r2
þ u2
spin

l^
j þ uyaw
v
r
l^
k
EXAMPLE 9.2
The satellite in Figure 9.4 is rotating about the z axis at a constant rate N. The xyz axes are attached to the
spacecraft, and the z-axis has a fixed orientation in inertial space. The solar panels rotate at a constant rate _
q in the
direction shown. Relative to point O, which lies at the center of the spacecraft and on the centerline of the panels,
calculate for point A on the panel
(a) Its absolute velocity and
(b) Its absolute acceleration.
FIGURE 9.4
Rotating solar panel on a rotating satellite.

Solution
(a) Since the moving xyz frame is attached to the body of the spacecraft, its angular velocity is
U ¼ N^
k (a)
The absolute angular velocity of the panel is the absolute angular velocity of the spacecraft plus the angular
velocity of the panel relative to the spacecraft,
upanel ¼ _
q^
j þ N^
k (b)
The position vector of A relative to O is
rA=O ¼
w
2
sin q^
i þ d^
j þ
w
2
cos q^
k (c)
According to Eqn (9.2), the velocity of A relative to O is
vA=O ¼ vA vO ¼ upanel rA=O ¼
^
i ^
j ^
k
0 _
q N

w
2
sin q d
w
2
cos q
from which we get
vA=O ¼
w
2
_
q cos q þ Nd

^
i
w
2
N sin q^
j
w
2
_
q sin q^
k
(b) The absolute angular acceleration of the panel is found by substituting Eqns (a) and (b) into Eqn (9.7),
apanel ¼
dupanel
dt

rel
þ U upanel
¼

d

_
q

dt
^
j þ
dN
dt
^
k
#
þ

N^
k

_
q^
j þ N^
k

Since N and _
q are constants, this reduces to
apanel ¼ _
qN^
i (d)
To find the acceleration of A relative to O, we substitute Eqns (b)e(d) into Eqn (9.4),
aA=O ¼ aA aO ¼ apanel rA=O þ upanel

upanel rA=O

¼
^
i ^
j ^
k
_
qN 0 0

w
2
sin q d
w
2
cos q
þ

_
q^
j þ N^
k

^
i ^
j ^
k
0 _
q N

w
2
sin q d
w
2
cos q
¼

w
2
N _
q cos q^
j þ N _
qd^
k

þ
^
i ^
j ^
k
0 _
q N

w
2
_
q cos q Nd N
w
2
sin q
w
2
_
q sin q
9.2 Kinematics 465

which leads to
aA=O ¼
w
2

N2
þ _
q
2

sin q^
i N

Nd þ w _
q cos q
^
j
w
2
_
q
2
cos q^
k
EXAMPLE 9.3
The gyro rotor illustrated in Figure 9.5 has a constant spin rate uspin around axis bea in the direction shown. The
XYZ axes are fixed. The xyz axes are attached to the gimbal ring, whose angle q with the vertical is increasing at a
constant rate _
q in the direction shown. The assembly is forced to precess at a constant rate N around the vertical,
For the rotor in the position shown, calculate
(a) The absolute angular velocity and
(b) The absolute angular acceleration.
Express the results in both the fixed XYZ frame and the moving xyz frame.
Solution
(a) We will need the instantaneous relationship between the unit vectors of the inertial XYZ axes and the comoving
xyz frame, which on inspecting Figure 9.6 can be seen to be
^
I ¼ cos q^
j þ sin q^
k
^
J ¼ ^
i
^
K ¼ sin q^
j þ cos q^
k
(a)
so that the matrix of the transformation from xyz to XYZ is (Section 4.5)
½QxX ¼
2
6
6
4
0 cos q sin q
1 0 0
0 sin q cos q
3
7
7
5 (b)
The absolute angular velocity of the gimbal ring is that of the base plus the angular velocity of the gimbal
relative to the base given as
ugimbal ¼ N^
K þ _
q^
i ¼ N

sin q^
j þ cos q^
k

þ _
q^
i ¼ _
q^
i þ N sin q^
j þ N cos q^
k (c)
where we made use of Eqn (a) above. Since the moving xyz frame is attached to the gimbal, U ¼ ugimbal,
so that
U ¼ _
q^
i þ N sin q^
j þ N cos q^
k (d)
The absolute angular velocity of the rotor is its spin relative to the gimbal, plus the angular velocity of the
gimbal,
urotor ¼ ugimbal þ uspin
^
k (e)
From Eqn (c), it follows that
urotor ¼ _
q^
i þ N sin q^
j þ

N cos q þ uspin

^
k (f)

θ
θ
Z
FIGURE 9.6
Orientation of the fixed XZ axes relative to the rotating xz axes.
FIGURE 9.5
Rotating, precessing, nutating gyro.
9.2 Kinematics 467

Because^
i, ^
j and ^
k move with the gimbal, expression (f) is valid for any time, not just the instant shown in
Figure 9.5. Alternatively, applying the vector transformation
furotorgXYZ ¼ ½QxX furotorgxyz (g)
we obtain the angular velocity of the rotor in the inertial frame, but only at the instant shown in the figure, that
is, when the x-axis aligns with the Y-axis.
8

:
uX
uY
uZ
9

=

;
¼
2
6
6
6
4
0 cos q sin q
1 0 0
0 sin q cos q
3
7
7
7
5
8

:
_
q
N sin q
N cos q þ uspin
9

=

;
¼
8

:
N sin q cos q þ N sin q cos q þ uspin sin q
_
q
N sin2
q þ N cos2q þ uspin cos q
9

=

;
or
urotor ¼ uspin sin q^
I þ _
q^
J þ

N þ uspin cos q

^
K (h)
(b) The angular acceleration of the rotor can be found by substituting Eqns (d) and (f) into Eqn (9.7), recalling that
N, _
q, and uspin are independent of time:
arotor ¼
durotor
dt

rel
þ U urotor
¼

d

_
q

dt
^
i þ
dðN sin qÞ
dt
^
j þ
d

N cos q þ uspin

dt
^
k
#
þ
^
i ^
j ^
k
_
q N sin q N cos q
_
q N sin q N cos q þ uspin
¼

N _
q cos q^
j N _
q sin q^
k

þ
h
^
i

Nuspin sin q

^
j

uspin
_
q

þ ^
kð0Þ
i
Upon collecting terms, we get
arotor ¼ Nuspin sin q^
i þ _
q

N cos q uspin
^
j N _
q sin q^
k (i)
This expression, like Eqn (f), is valid at any time.
The components of arotor along the XYZ axes are found in the same way as for urotor,
farotorgXYZ ¼ ½QxX farotorgxyz
which means
8

:
aX
aY
aZ
9

=

;
¼
2
6
6
6
4
0 cos q sin q
1 0 0
0 sin q cos q
3
7
7
7
5
8

:
Nuspin sin q
_
q

N cos q uspin

N _
q sin q
9

=

;
¼
8

:
N _
qcos2q þ _
quspin cos q N _
q sin2
q
Nuspin sin q
N _
q sin q cos q _
quspin sin q N _
q sin q cos q
9

=

;

or
arotor ¼ _
q

uspin cos q N

^
I þ Nuspin sin q^
J _
quspin sin q^
K (j)
Note carefully that Eqn (j) is not simply the time derivative of Eqn (h). Equations (h) and (j) are valid only at the
instant that the xyz and XYZ axes have the alignments shown in Figure 9.5.
9.3 Equations of translational motion
Figure 9.7 again shows an arbitrary, continuous, three-dimensional body of mass m. “Continuous”
means that as we zoom in on a point it remains surrounded by a continuous distribution of matter
having the infinitesimal mass dm in the limit. The point never ends up in a void. In particular, we ignore
the actual atomic and molecular microstructure in favor of this continuum hypothesis, as it is called.
Molecular microstructure does bear upon the overall dynamics of a finite body. We will use G to denote
the center of mass. The position vectors of points relative to the origin of the inertial frame will be
designated by capital letters. Thus, the position of the center of mass is RG defined as
mRG ¼
Z
m
Rdm (9.9)
R is the position of a mass element dm within the continuum. Each element of mass is acted upon by a
net external force dFnet and a net internal force dfnet. The external force comes from direct contact with
other objects and from action at a distance, such as gravitational attraction. The internal forces are
those exerted from within the body by neighboring particles. These are the forces that hold the body
together. For each mass element, Newton’s second law, Eqn (1.47), is written as
dFnet þ dfnet ¼ dm€
R (9.10)
Writing this equation for the infinite number of mass elements of which the body is composed, and
then summing them all together leads, to the integral
Z
dFnet þ
Z
dfnet ¼
Z
m
€
Rdm
FIGURE 9.7
Forces on the mass element dm of a continuous medium.
9.3 Equations of translational motion 469

Because the internal forces occur in action–reaction pairs,
R
dfnet ¼ 0. (External forces on the body are
those without an internal reactant; the reactant lies outside the body and, hence, is outside our pur-
view.) Thus,
Fnet ¼
Z
m
€
Rdm (9.11)
where Fnet is the resultant external force on the body, Fnet ¼
R
dFnet. From Eqn (9.9),
Z
m
€
Rdm ¼ m€
RG
where €
RG ¼ aG, the absolute acceleration of the center of mass. Therefore, Eqn (9.11) can be
written as
Fnet ¼ m€
RG (9.12)
We are therefore reminded that the motion of the center of mass of a body is determined solely by the
resultant of the external forces acting on it. So far, our study of orbiting bodies has focused exclusively
on the motion of their centers of mass. In this chapter, we will turn our attention to rotational motion
around the center of mass. To simplify things, we will ultimately assume that the body is not only
continuous but that it is also rigid. This means all points of the body remain at a fixed distance from
each other and there is no flexing, bending, or twisting deformation.
9.4 Equations of rotational motion
Our development of the rotational dynamics equations does not require at the outset that the body
under consideration be rigid. It may be a solid, fluid, or gas.
Point P in Figure 9.8 is arbitrary; it need not be fixed in space nor attached to a point on the body.
Then, the moment about P of the forces on mass element dm (cf. Figure 9.7) is
dMP ¼ r dFnet þ r dfnet
where r is the position vector of the mass element dm relative to the point P. Writing the right-hand side
as r ðdFnet þ dfnetÞ, substituting Eqn (9.10), and integrating over all the mass elements of the body
yields
MPnet
¼
Z
m
r €
Rdm (9.13)
where €
R is the absolute acceleration of dm relative to the inertial frame and
MPnet
¼
Z
r dFnet þ
Z
r dfnet

But
R
r dfnet ¼ 0 because the internal forces occur in action–reaction pairs. Thus,
MPnet
¼
Z
r dFnet
which means the net moment includes only the moment of all of the external forces on the body.
From the product rule of calculus, we know that dðr _
RÞ=dt ¼ r €
R þ _
r _
R, so that the inte-
grand in Eqn (9.13) may be written as
r €
R ¼
d
dt

r _
R

_
r _
R (9.14)
Furthermore, Figure 9.8 shows that r ¼ R RP, where RP is the absolute position vector of P. It
follows that
_
r _
R ¼

_
R _
RP

_
R ¼ _
RP _
R (9.15)
Substituting Eqn (9.15) into Eqn (9.14), then moving that result into Eqn (9.13), yields
MPnet
¼
d
dt
Z
m
r _
Rdm þ _
RP
Z
m
_
Rdm (9.16)
Now, r _
Rdm is the moment of the absolute linear momentum of mass element dm about P.
The moment of momentum, or angular momentum, of the entire body is the integral of this cross-
product over all of its mass elements. That is, the absolute angular momentum of the body relative to
point P is
HP ¼
Z
m
r _
Rdm (9.17)
FIGURE 9.8
Position vectors of a mass element in a continuum from several key reference points.
9.4 Equations of rotational motion 471

Observing from Figure 9.8 that r ¼ rG/P þ r, we can write Eqn (9.17) as
HP ¼
Z
m

rG=P þ r

_
Rdm ¼ rG=P
Z
m
_
Rdm þ
Z
m
r _
Rdm (9.18)
The last term is the absolute angular momentum relative to the center of mass G,
HG ¼
Z
r _
Rdm (9.19)
Furthermore, by the definition of center of mass, Eqn (9.9),
Z
m
_
Rdm ¼ m _
RG (9.20)
Equations (9.19) and (9.20) allow us to write Eqn (9.18) as
HP ¼ HG þ rG=P mvG (9.21)
This useful relationship shows how to obtain the absolute angular momentum about any point P once
HG is known.
For calculating the angular momentum about the center of mass, Eqn (9.19) can be cast in a much
more useful form by making the substitution (cf. Figure 9.8) R ¼ RG þ r, so that
HG ¼
Z
m
r

_
RG þ _
r

dm ¼
Z
m
r _
RGdm þ
Z
m
r _
rdm
In the two integrals on the right, the variable is r. _
RG is fixed and can therefore be factored out of the
first integral to obtain
HG ¼
0
@
Z
m
rdm
1
A _
RG þ
Z
m
r _
rdm
By definition of the center of mass,
R
m
rdm ¼ 0 (the position vector of the center of mass relative to
itself is zero), which means
HG ¼
Z
m
r _
rdm (9.22)
Since r and _
r are the position and velocity relative to the center of mass G,
R
m
r _
rdm is the total
moment about the center of mass of the linear momentum relative to the center of mass, HGrel
. In other
words,
HG ¼ HGrel
(9.23)
This is a rather surprising fact, hidden in Eqn (9.19), and is true in general for no other point of the body.

Another useful angular momentum formula, similar to Eqn (9.21), may be found by substituting
R ¼ RP þ r into Eqn (9.17),
HP ¼
Z
m
r

_
RP þ _
r

dm ¼
0
@
Z
m
rdm
1
A _
RP þ
Z
m
r _
rdm (9.24)
The term on the far right is the net moment of relative linear momentum about P,
HPrel
¼
Z
m
r _
rdm (9.25)
Also,
R
m
rdm ¼ mrG=P, where rG/P is the position of the center of mass relative to P. Thus, Eqn (9.24)
can be written as
HP ¼ HPrel
þ rG=P mvP (9.26)
Finally, substituting this into Eqn (9.21), solving for HPrel
, and noting that vG vP ¼ vG/P yields
HPrel
¼ HG þ rG=P mvG=P (9.27)
This expression is useful when the absolute velocity vG of the center of mass, which is required in Eqn
(9.21), is not available.
So far, we have written down some formulas for calculating the angular momentum about an
arbitrary point in space and about the center of mass of the body itself. Let us now return to the
problem of relating angular momentum to the applied torque. Substituting Eqns (9.17) and (9.20) into
Eqn (9.16), we obtain
MPnet
¼ _
HP þ _
RP m _
RG
Thus, for an arbitrary point P,
MPnet
¼ _
HP þ vP mvG (9.28)
where vP and vG are the absolute velocities of points P and G, respectively. This expression is
applicable to two important special cases.
If the point P is at rest in inertial space ðvP ¼ 0Þ, then Eqn (9.28) reduces to
MPnet
¼ _
HP (9.29)
This equation holds as well if vP and vG are parallel. If P is the point of contact of a wheel rolling while
slipping in the plane. Note that the validity of Eqn (9.29) depends neither on the body’s being rigid nor
on its being in pure rotation about P. If point P is chosen to be the center of mass, then, since
vG vG ¼ 0, Eqn (9.28) becomes
MGnet
¼ _
HG (9.30)
This equation is valid for any state of motion.
9.4 Equations of rotational motion 473

If Eqn (9.30) is integrated over a time interval, then we obtain the angular impulse–momentum
principle,
Zt2
t1
MGnet
dt ¼ HG2
HG1
(9.31)
A similar expression follows from Eqn (9.29).
R
Mdt is the angular impulse. If the net angular impulse
is zero, then DH ¼ 0, which is a statement of the conservation of angular momentum. Keep in mind
that the angular impulse–momentum principle is not valid for just any reference point.
Additional versions of Eqns (9.29) and (9.30) can be obtained which may prove useful in special
circumstances. For example, substituting the expression for HP (Eqn (9.21)) into Eqn (9.28) yields
MPnet
¼

_
HG þ
d
dt

rG=P mvG

þ vP mvG
¼ _
HG þ
d
dt
½ðrG rPÞ mvG þ vP mvG
¼ _
HG þ ðvG vPÞ mvG þ rG=P maG þ vP mvG
or, finally,
MPnet
¼ _
HG þ rG=P maG (9.32)
This expression is useful when it is convenient to compute the net moment about a point other than the
center of mass. Alternatively, by simply differentiating Eqn (9.27) we get
_
HPrel
¼ _
HG þ vG=P mvG=P
zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{
¼0
þ rG=P maG=P
Solving for _
HG, invoking Eqn (9.30), and using the fact that aP/G ¼ aG/P leads to
MGnet
¼ _
HPrel
þ rG=P maP=G (9.33)
Finally, if the body is rigid, the magnitude of the position vector r of any point relative to the center
of mass does not change with time. Therefore, Eqn (1.52) requires that _
r ¼ u r, leading us to
conclude from Eqn (9.22) that
HG ¼
Z
m
r ðu rÞdm

Rigid body

(9.34)
Again, the absolute angular momentum about the center of mass depends only on the absolute angular
velocity and not on the absolute translational velocity of any point of the body.

No such simplification of Eqn (9.17) exists for an arbitrary reference point P. However, if the
point P is fixed in inertial space and the rigid body is rotating about P, then the position vector r from
P to any point of the body is constant. It follows from Eqn (1.52) that _
r ¼ u r. According to
Figure 9.8,
R ¼ RP þ r
Differentiating with respect to time gives
_
R ¼ _
RP þ _
r ¼ 0 þ u r ¼ u r
Substituting this into Eqn (9.17) yields the formula for angular momentum in this special case as
HP ¼
Z
m
r ðu rÞdm ðRigid body rotating about fixed point PÞ (9.35)
Although Eqns (9.34) and (9.35) are mathematically identical, one must keep in mind the notation
of Figure 9.8. Eqn (9.35) applies only if the rigid body is in pure rotation about a stationary point in
inertial space, whereas Eqn (9.34) applies unconditionally to any situation.
9.5 Moments of inertia
To use Eqn (9.29) or Eqn (9.30) to solve problems, the vectors within them have to be resolved
into components. To find the components of angular momentum, we must appeal to its definition.
We will focus on the formula for angular momentum of a rigid body about its center of mass
(Eqn (9.34)) because the expression for fixed-point rotation (Eqn (9.35)) is mathematically the
same. The integrand of Eqn (9.34) can be rewritten using the bac–cab vector identity presented in
Eqn (2.33),
r ðu rÞ ¼ ur2
rðu$rÞ (9.36)
Let the origin of a comoving xyz coordinate system be attached to the center of mass G, as shown in
Figure 9.9. The unit vectors of this frame are ^
i, ^
j, and ^
k. The vectors r and u can be resolved into
components in the xyz directions to get r ¼ x^
i þ y^
j þ z^
k and u ¼ ux
^
i þ uy
^
j þ uz
^
k. Substituting these
vector expressions into the right side of Eqn (9.36) yields
r ðu rÞ ¼

ux
^
i þ uy
^
j þ uz
^
k

x2
þ y2
þ z2

x^
i þ y^
j þ z^
k

uxx þ uyy þ uzz

Expanding the right side and collecting the terms having the unit vectors ^
i, ^
j, and ^
k in common,
we get
r ðu rÞ ¼

y2
þ z2

ux xyuy xzuz
^
i þ yxux þ

x2
þ z2

uy yzuz
^
j
þ zxux zyuy þ

x2
þ y2

uz
^
k (9.37)
9.5 Moments of inertia 475

We put this result into the integrand of Eqn (9.34) to obtain
HG ¼ Hx
^
i þ Hy
^
j þ Hz
^
k (9.38)
where
8

:
Hx
Hy
Hz
9

=

;
¼
2
6
6
6
4
Ix Ixy Ixz
Iyx Iy Iyz
Izx Izy Iz
3
7
7
7
5
8

:
ux
uy
uz
9

=

;
(9.39a)
or, in matrix notation,
fHg ¼ ½Ifug (9.39b)
The nine components of the moment of inertia matrix [I] about the center of mass are
Ix ¼
R
y2 þ z2

dm Ixy ¼
R
xydm Ixz ¼
R
xzdm
Iyx ¼
R
yxdm Iy ¼
R
x2 þ z2

dm Iyz ¼
R
yzdm
Izx ¼
R
zxdm Izy ¼
R
zydm Iz ¼
R
x2 þ y2

dm
(9.40)
Since Iyx ¼ Ixy, Izx ¼ Ixz, and Izy ¼ Iyz, it follows that [I] is a symmetric matrix ð½IT
¼ ½IÞ. Therefore,
[I] has six independent components instead of nine. Observe that, whereas the products of inertia
Ixy, Ixz, and Iyz can be positive, negative, or zero, the moments of inertia Ix, Iy, and Iz are always
positive (never zero or negative) for bodies of finite dimensions. For this reason, [I] is a symmetric
positive-definite matrix. Keep in mind that Eqns (9.38) and (9.39) are valid as well for axes attached to
a fixed point P about which the body is rotating.
The moments of inertia reflect how the mass of a rigid body is distributed. They manifest a body’s
rotational inertia, that is, its resistance to being set into rotary motion or stopped once rotation is
FIGURE 9.9
A comoving xyz frame used to compute the moments of inertia.

underway. It is not an object’s mass alone but how that mass is distributed that determines how the
body will respond to applied torques.
If the xy plane is a plane of symmetry, then for any x and y within the body there are identical mass
elements located at þz and z. This means the products of inertia with z in the integrand vanish.
Similar statements are true if xz or yz are symmetry planes. In summary, we conclude
If the xy plane is a plane of symmetry of the body, then Ixz ¼ Iyz ¼ 0.
If the xz plane is a plane of symmetry of the body, then Ixy ¼ Iyz ¼ 0.
If the yz plane is a plane of symmetry of the body, then Ixy ¼ Ixz ¼ 0.
It follows that if the body has two planes of symmetry relative to the xyz frame of reference, then all
three products of inertia vanish, and [I] becomes a diagonal matrix such that,
½I ¼
2
6
6
4
A 0 0
0 B 0
0 0 C
3
7
7
5 (9.41)
A, B, and C are the principal moments of inertia (all positive), and the xyz axes are the body’s
principal axes of inertia. In this case, relative to either the center of mass or a fixed point of rotation,
we have
Hx ¼ Aux Hy ¼ Buy Hz ¼ Cuz (9.42)
In general, the angular velocity u and the angular momentum H are not parallel. However, if, for
example, u ¼ u^
i, then according to Eqn (9.42), H ¼ Au. In other words, if the angular velocity is
aligned with a principal direction, so is the angular momentum. In that case, the two vectors u and H
are indeed parallel.
Each of the three principal moments of inertia can be expressed as follows:
A ¼ mk2
x B ¼ mk2
y C ¼ mk2
z (9.43)
where m is the mass of the body and kx, ky, and kz are the three radii of gyration. One may imagine
the mass of a body to be concentrated around a principal axis at a distance equal to the radius of
gyration.
The moments of inertia for several common shapes are listed in Figure 9.10. By symmetry, their
products of inertia vanish for the coordinate axes used. Formulas for other solid geometries can be
found in engineering handbooks and in dynamics textbooks.
For a mass concentrated at a point, the moments of inertia in Eqn (9.40) are just the mass times
the integrand evaluated at the point. That is, the moment of inertia matrix [I(m)
] of a point mass m is
given by
h
IðmÞ
i
¼
2
6
6
4
m

y2 þ z2

mxy mxz
mxy m

x2 þ z2

myz
mxz myz m

x2 þ y2

3
7
7
5 (9.44)

EXAMPLE 9.4
The following table lists out the mass and coordinates of seven point masses. Find the center of mass of the system
and the moments of inertia about the origin.
Solution
The total mass of this system is
m ¼
X
7
i¼1
mi ¼ 28 kg
For concentrated masses, the integral in Eqn (9.9) is replaced by the mass times its position vector. Therefore, in
this case, the three components of the position vector of the center of mass are xG ¼ ð1=mÞ
P7
i¼1 mixi,
yG ¼ ð1=mÞ
P7
i¼1 miyi, and zG ¼ ð1=mÞ
P7
i¼1 mizi, so that
xG ¼ 0:35 m yG ¼ 0:01964 m zG ¼ 0:4411 m
Point, i Mass mi (kg) xi (m) yi (m) zi (m)
1 3 0.5 0.2 0.3
2 7 0.2 0.75 0.4
3 5 1 0.8 0.9
4 6 1.2 1.3 1.25
5 2 1.3 1.4 0.8
6 4 0.3 1.35 0.75
7 1 1.5 1.7 0.85
(b)
(a) (c)
FIGURE 9.10
Moments of inertia for three common homogeneous solids of mass m. (a) Solid circular cylinder. (b) Circular
cylindrical shell. (c) Rectangular parallelepiped.

The total moment of inertia is the sum over all the particles of Eqn (9.44) evaluated at each point. Thus,
½I ¼
2
4
0:39 0:3 0:45
0:3 1:02 0:18
0:45 0:18 0:87
3
5
zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{
ð1Þ
þ
2
4
5:0575 1:05 0:56
1:05 1:4 2:1
0:56 2:1 4:2175
3
5
zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{
ð2Þ
þ
2
4
7:25 4 4:5
4 9:05 3:6
4:5 3:6 8:2
3
5
zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{
ð3Þ
þ
2
4
19:515 9:36 9
9:36 18:015 9:75
9 9:75 18:78
3
5
zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{
ð4Þ
þ
2
4
5:2 3:64 2:08
3:64 4:66 2:24
2:08 2:24 7:3
3
5
ð5Þ
þ
2
4
9:54 1:62 0:9
1:62 2:61 4:05
0:9 4:05 7:65
3
5
ð6Þ
þ
2
4
3:6125 2:55 1:275
2:55 2:9725 1:445
1:275 1:445 5:14
3
5
zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{
ð7Þ
or
½I ¼
2
6
6
4
50:56 20:42 14:94
20:42 39:73 14:90
14:94 14:90 52:16
3
7
7
5

kg$m2

EXAMPLE 9.5
Calculate the moments of inertia of a slender, homogeneous straight rod of length [ and mass m shown in
Figure 9.11. One end of the rod is at the origin and the other has coordinates (a, b, c).
Solution
A slender rod is one whose cross-sectional dimensions are negligible compared with its length. The mass is
concentrated along its centerline. Since the rod is homogeneous, the mass per unit length r is uniform and given by
r ¼
m
‘
(a)
The length of the rod is
‘ ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2 þ c2
p
(a, b, c)
(0, 0, 0)
A
B
FIGURE 9.11
A uniform slender bar of mass m and length [.

Starting with Ix, we have from Eqn (9.40),
Ix ¼
Z‘
0

y2
þ z2

rds
in which we replaced the element of mass dm by rds, where ds is the element of length along the rod. The distance
s is measured from end A of the rod, so that the x, y, and z coordinates of any point along it are found in terms of s by
the following relations:
x ¼
s
‘
a y ¼
s
‘
b z ¼
s
‘
c
Thus,
Ix ¼
Z‘
0

s2
‘2
b2
þ
s2
‘2
c2

rds ¼ r
b2 þ c2
‘2
Z‘
0
s2
ds ¼
1
3
r

b2
þ c2

‘
Substituting Eqn (a) yields
Ix ¼
1
3
m

b2
þ c2

In precisely the same way, we find
Iy ¼
1
3
m

a2
þ c2

Iz ¼
1
3
m

a2
þ b2

For Ixy we have
Ixy ¼
Z‘
0
xyrds ¼
Z‘
0
s
‘
a $
s
‘
brds ¼ r
ab
‘2
Z‘
0
s2
ds ¼
1
3
rab‘
Once again using Eqn (a),
Ixy ¼
1
3
mab
Likewise,
Ixz ¼
1
3
mac Iyz ¼
1
3
mbc
EXAMPLE 9.6
The gyro rotor (Figure 9.12) in Example 9.3 has a mass m of 5 kg, radius r of 0.08 m, and thickness t of 0.025 m.
If N ¼ 2.1 rad/s, _
q ¼ 4 rad=s, u ¼ 10.5 rad/s, and q ¼ 60, calculate
(a) The angular momentum of the rotor about its center of mass G in the body-fixed xyz frame.
(b) The angle between the rotor’s angular velocity vector and its angular momentum vector.

Solution
Equation (f) from Example 9.3 gives the components of the absolute angular velocity of the rotor in the moving xyz
frame.
ux ¼ _
q ¼ 4 rad=s
uy ¼ N sin q ¼ 2:1$sin 60 ¼ 1:819 rad=s
uz ¼ uspin þ N cos q ¼ 10:5 þ 2:1$cos 60 ¼ 11:55 rad=s
(a)
Therefore,
u ¼ 4^
i þ 1:819^
j þ 11:55^
k ðrad=sÞ (b)
All three coordinate planes of the body-fixed xyz frame contain the center of mass G and all are planes of symmetry
of the circular cylindrical rotor. Therefore, [xy ¼ [zx ¼ [yz ¼ 0.
From Figure 9.10(a), we see that the nonzero diagonal entries in the moment of inertia tensor are
A ¼ B ¼
1
12
mt2
þ
1
4
mr2
¼
1
12
$5$0:0252
þ
1
4
$5$0:082
¼ 0:008260 kg$m2
C ¼
1
2
mr2
¼
1
2
$5$0:082
¼ 0:0160 kg$m2
(c)
We can use Eqn (9.42) to calculate the angular momentum, because the origin of the xyz frame is the rotor’s
center of mass (which in this case also happens to be a fixed point of rotation, which is another reason why we can
use Eqn (9.42)). Substituting Eqns (a) and (c) in Eqn (9.42) yields
Hx ¼ Aux ¼ 0:008260$4 ¼ 0:03304 kg$m2=s
Hy ¼ Buy ¼ 0:008260$1:819 ¼ 0:0150 kg$m2=s
Hz ¼ Cuz ¼ 0:0160$11:55 ¼ 0:1848 kg$m2=s
(d)
so that
H ¼ 0:03304^
i þ 0:0150^
j þ 0:1848^
k

kg$m2=s

(e)
θ
ω
FIGURE 9.12
Rotor of the gyroscope in Figure 9.4.

The angle f between H and u is found by taking the dot product of the two vectors,
f ¼ cos1

H$u
Hu

¼ cos1

2:294
0:1883$12:36

¼ 9:717 (f)
As this problem illustrates, the angular momentum and the angular velocity are in general not collinear.
Consider a coordinate system x0y0z0 with the same origin as xyz but a different orientation. Let [Q]
be the orthogonal matrix ð½Q1
¼ ½QT
Þ that transforms the components of a vector from the xyz
system to the x0y0z0 frame. Recall from Section 4.5 that the rows of [Q] are the direction cosines of
the x0y0z0 axes relative to xyz. If fH0
g comprises the components of the angular momentum vector
along the x0y0z0 axes, then fH0
g is obtained from its components {H} in the xyz frame by the
relation
fH0
g ¼ ½QfHg
From Eqn (9.39), we can write this as
fH0
g ¼ ½Q½Ifug (9.45)
where [I] is the moment of inertia matrix (Eqn (9.39)) in the xyz coordinates. Like the angular mo-
mentum vector, the components {u} of the angular velocity vector in the xyz system are related to
those in the primed system ðfu0gÞ by the expression
fu0
g ¼ ½Qfug
The inverse relation is simply
fug ¼ ½Q1
fu0
g ¼ ½QT
fu0
g (9.46)
Substituting this into Eqn (9.45), we get
fH0
g ¼ ½Q½I½QT
fu0
g (9.47)
But the components of angular momentum and angular velocity in the x0y0z0 frame are related by an
equation of the same form as Eqn (9.39), so that
fH0
g ¼ ½I0
fu0
g (9.48)
where ½I0
comprises the components of the inertia matrix in the primed system. Comparing the right-
hand sides of Eqns (9.47) and (9.48), we conclude that
½I0
¼ ½Q½I½QT
(9.49a)
that is,
2
6
6
4
Ix0 Ix0y0 Ix0z0
Iy0x0 Iy0 Iy0z0
Iz0x0 Iz0y0 Iz0
3
7
7
5 ¼
2
6
6
4
Q11 Q12 Q13
Q21 Q22 Q23
Q31 Q32 Q33
3
7
7
5
2
6
6
4
Ix Ixy Ixz
Iyx Iy Iyz
Izx Izy Iz
3
7
7
5
2
6
6
4
Q11 Q21 Q31
Q12 Q22 Q32
Q13 Q32 Q33
3
7
7
5 (9.49b)

This shows how to transform the components of the inertia matrix from the xyz coordinate system to
any other orthogonal system with a common origin. Thus, for example,
Ix0 ¼ P Q11 Q12 Q13 R
zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{
PRow 1R
2
6
6
4
Ix Ixy Ixz
Iyx Iy Iyz
Izx Izy Iz
3
7
7
5
8

:
Q11
Q12
Q13
9

=

;
zfflfflfflfflffl}|fflfflfflfflffl{
PRow 1R
T
Iy0z0 ¼ P Q21 Q22 Q23 R
zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{
PRow 2R
2
6
6
4
Ix Ixy Ixz
Iyx Iy Iyz
Izx Izy Izz
3
7
5
8

:
Q31
Q32
Q33
9

=

;
zfflfflfflfflffl}|fflfflfflfflffl{
PRow 3R
T
(9.50)
etc.
Any object represented by a square matrix whose components transform according to Eqn (9.49) is
called a second-order tensor. We may therefore refer to [I] as the inertia tensor.
EXAMPLE 9.7
Find the mass moment of inertia of the system of point masses in Example 9.4 about an axis from the origin
through the point with coordinates (2 m, 3 m, 4 m).
Solution
From Example 9.4, the moment of inertia tensor for the system of point masses is
½I ¼
2
6
6
4
50:56 20:42 14:94
20:42 39:73 14:90
14:94 14:90 52:16
3
7
7
5

kg$m2

The vector V connecting the origin with (2 m, 3 m, 4 m) is
V ¼ 2^
i 3^
j þ 4^
k
The unit vector in the direction of V is
^
uV ¼
V
kVk
¼ 0:3714^
i 0:5571^
j þ 0:7428^
k
We may consider ^
uV as the unit vector along the x0-axis of a rotated Cartesian coordinate system. Then, from Eqn
(9.50) we get
‘V 0 ¼ P 0:3714 0:5571 0:7428 R
2
6
6
6
6
4
50:56 20:42 14:94
20:42 39:73 14:90
14:94 14:90 52:16
3
7
7
7
7
5
8

:
0:3714
0:5571
0:7428
9

=

;
¼ P 0:3714 0:5571 0:7428 R
8

:
3:695
3:482
24:90
9

=

;
¼ 19:06 kg$m2

EXAMPLE 9.8
For the satellite of Example 9.2, which is reproduced in Figure 9.13, the data are as follows: N ¼ 0.1 rad/s and
_
q ¼ 0:01 rad=s, in the directions shown: q ¼ 40 and d0 ¼ 1.5 m. The length, width, and thickness of the panel
are [ ¼ 6 m, w ¼ 2 m, and t ¼ 0.025 m. The uniformly distributed mass of the panel is 50 kg. Find the angular
momentum of the panel relative to the center of mass O of the satellite.
Solution
We can treat the panel as a thin parallelepiped. The panel’s xyz axes have their origin at the center of mass G of the
panel and are parallel to its three edge directions. According to Figure 9.10(c), the moments of inertia relative to
the xyz coordinate system are
‘Gx ¼
1
12
m

‘2
þ t2

¼
1
12
$50$

62
þ 0:0252

¼ 150:0 kg$m2
‘Gy ¼
1
12
m

w2
þ t2

¼
1
12
$50$

22
þ 0:0252

¼ 16:67 kg$m2
‘Gz ¼
1
12
m

w2
þ ‘2

¼
1
12
$50$

22
þ 62

¼ 166:7 kg$m2
‘Gxy ¼ ‘Gxz ¼ ‘Gyz ¼ 0
(a)
In matrix notation,
½lG ¼
2
6
6
4
150:0 0 0
0 16:67 0
0 0 166:7
3
7
7
5

kg$m2

(b)
The unit vectors of the satellite’s x0y0z0 system are related to those of the panel’s xyz frame. By inspection,
^
i
0
¼ sin q^
i þ cos q^
k ¼ 0:6428^
i þ 0:7660^
k
^
j
0
¼ ^
j
^
k
0
¼ cos q^
i þ sin q^
k ¼ 0:7660^
i þ 0:6428^
k
(c)
θ
FIGURE 9.13
A satellite and solar panel.

The matrix [Q] of the transformation from xyz to x0y0z0 comprises the direction cosines of ^
i
0
, ^
j
0
, and ^
k
0
:
½Q ¼
2
6
6
4
0:6428 0 0:7660
0 1 0
0:7660 0 0:6428
3
7
7
5 (d)
In Example 9.2, we found that the absolute angular velocity of the panel, in the satellite’s x0y0z0 frame of
reference, is
u ¼ _
q^
j
0
þ N^
k
0
¼ 0:01^
j
0
þ 0:1^
k
0
ðrad=sÞ
That is,
fu0
g ¼
8

:
0
0:01
0:1
9

=

;
ðrad=sÞ (e)
To find the absolute angular momentum fH0
Gg in the satellite system requires the use of Eqn (9.39),
fH0
Gg ¼ ½I0
Gfu0
g (f)
Before doing so, we must transform the components of the moments of the inertia tensor in Eqn (b) from the
unprimed system to the primed system, by means of Eqn (9.49),
½I0
G ¼ ½Q½IG½QT
¼
2
6
6
4
0:6428 0 0:7660
0 1 0
0:7660 0 0:6428
3
7
7
5
2
6
6
4
150:0 0 0
0 16:67 0
0 0 166:7
3
7
7
5
2
6
6
4
0:6428 0 0:7660
0 1 0
0:7660 0 0:6428
3
7
7
5
so that
½I0
G ¼
2
6
6
4
159:8 0 8:205
0 16:67 0
8:205 0 156:9
3
7
7
5

kg$m2

(g)
Then Eqn (f) yields
fH0
Gg ¼
2
6
6
4
159:8 0 8:205
0 16:67 0
8:205 0 156:9
3
7
7
5
8

:
0
0:01
0:1
9

=

;
¼
8

:
0:8205
0:1667
15:69
9

=

;

kg$m2
=s

or, in vector notation,
HG ¼ 0:8205^
i
0
0:1667^
j
0
þ 15:69^
k
0

kg$m2
=s

(h)
This is the absolute angular momentum of the panel about its own center of mass G, and it is used in Eqn (9.27) to
calculate the angular momentum HOrel
relative to the satellite’s center of mass O,
HOrel
¼ HG þ rG=O mvG=O (i)
rG/O is the position vector from O to G,
rG=O ¼

dO þ
‘
2

^
j
0
¼

1:5 þ
6
2

^
j
0
¼ 4:5^
j
0
ðmÞ (j)

The velocity of G relative to O, vG/O, is found from Eqn (9.2),
vG=O ¼ usatellite rG=O ¼ N^
k
0
rG=O ¼ 0:1^
k
0
4:5^
j
0
¼ 0:45^
i
0
ðm=sÞ (k)
Substituting Eqns (h), (j) and (k) into Eqn (i) finally yields
HOrel
¼

0:8205^
i
0
0:1667^
j
0
þ 15:69^
k
0

þ 4:5^
j
0

h
50

0:45^
i
0
i
¼ 0:8205^
i
0
0:1667^
j
0
þ 116:9^
k
0
kg$m2=s

(l)
Note that we were unable to use Eqn (9.21) to find the absolute angular momentum HO because that requires
knowing the absolute velocity vG, which in turn depends on the absolute velocity of O, which was not provided.
How can we find the direction cosine matrix [Q] such that Eqn (9.49) will yield a moment of inertia
matrix ½I0
that is diagonal, that is, of the form given by Eqn (9.41)? In other words, how do we find the
principal directions (“eigenvectors”) and the corresponding principal values (“eigenvalues”) of the
moment of inertia tensor?
Let the angular velocity vector u be parallel to the principal direction defined by the vector e, so
that u ¼ be, where b is a scalar. Since u points in the principal direction of the inertia tensor, so must
H, which means H is also parallel to e. Therefore, H ¼ ae, where a is a scalar. From Eqn (9.39), it
follows that
afeg ¼ ½IðbfegÞ
or
½Ifeg ¼ lfeg
where l ¼ a/b (a scalar). That is,
2
4
Ix Ixy Ixz
Ixy Iy Iyz
Ixz Iyz Iz
3
5
8

:
ex
ey
ez
9
=
;
¼ l
8

:
ex
ey
ez
9
=
;
This can be written
2
4
Ix l Ixy Ixz
Ixy Iy l Iyz
Ixz Iyz Iz l
3
5
8

:
ex
ey
ez
9
=
;
¼
8

:
0
0
0
9
=
;
(9.51)
The trivial solution of Eqn (9.51) is e ¼ 0, which is of no interest. The only way that Eqn (9.51) will
not yield the trivial solution is if the coefficient matrix on the left is singular. That will occur if its
determinant vanishes, that is, if
Ix l Ixy Ixz
Ixy Iy l Iyz
Ixz Iyz Iz l
¼ 0 (9.52)

Expanding the determinant, we find
Ix l Ixy Ixz
Ixy Iy l Iyz
Ixz Iyz Iz l
¼ l3
þ J1l2
J2l þ J3 (9.53)
where
J1 ¼ Ix þ Iy þ Iz
J2 ¼
Ix Ixy
Ixy Iy
þ
Ix Ixz
Ixz Iz
þ
Iy Iyz
Iyz Iz
J3 ¼
Ix Ixy Ixz
Ixy Iy Iyz
Ixz Iyz Iz
(9.54)
J1, J2, and J3 are invariants, that is, they have the same value in every Cartesian coordinate system.
Equations (9.52) and (9.53) yield the characteristic equation of the tensor [I],
l3
J1l2
þ J2l J3 ¼ 0 (9.55)
The three roots lp (p ¼ 1, 2, 3) of this cubic equation are real, since [I] is symmetric; furthermore, they
are all positive, since [I] is a positive-definite matrix. We substitute each root, or eigenvalue, lp back
into Eqn (9.51) to obtain
2
6
6
4
Ix lp Ixy Ixz
Ixy Iy lp Iyz
Ixz Iyz Iz lp
3
7
7
5
8

:
e
ðpÞ
x
e
ðpÞ
y
e
ðpÞ
z
9

=

;
¼
8

:
0
0
0
9

=

;
; p ¼ 1; 2; 3 (9.56)
Solving this system yields the three eigenvectors e(p)
corresponding to each of the three eigenvalues lp.
The three eigenvectors are orthogonal, also due to the symmetry of the matrix [I]. Each eigenvalue is a
principal moment of inertia, and its corresponding eigenvector is a principal direction.
EXAMPLE 9.9
Find the principal moments of inertia and the principal axes of inertia of the inertia tensor.
½I ¼
2
6
6
4
100 20 100
20 300 50
100 50 500
3
7
7
5 kg$m2

Solution
We seek the nontrivial solutions of the system [I]{e} ¼ l{e}, that is,
2
6
6
4
100 l 20 100
20 300 l 50
100 50 500 l
3
7
7
5
8

:
ex
ey
ez
9

=

;
¼
8

:
0
0
0
9

=

;
(a)
From Eqn (9.54),
J1 ¼ 100 þ 300 þ 500 ¼ 900
J2 ¼
100 20
20 300
þ
100 100
100 500
þ
300 50
50 500
¼ 217; 100
J3 ¼
100 20 100
20 300 50
100 50 500
¼ 11; 350; 000
(b)
Thus, the characteristic equation is
l3
900l2
þ 217;100l 11;350;000 ¼ 0 (c)
The three roots are the principal moments of inertia, which are found to be
l1 ¼ 532:052 l2 ¼ 295:840 l3 ¼ 72:1083 (d)
Each of these is substituted, in turn, back into Eqn (a) to find its corresponding principal direction.
Substituting l1 ¼ 532:052 kg$m2 into Eqn (a) we obtain
2
6
6
4
432:052 20:0000 100:0000
20:0000 232:052 50:0000
100:0000 50:0000 32:0519
3
7
7
5
8

:
e
ð1Þ
x
e
ð1Þ
y
e
ð1Þ
z
9

=

;
¼
8

:
0
0
0
9

=

;
(e)
Since the determinant of the coefficient matrix is zero, at most two of the three equations in Eqn (e) are inde-
pendent. Thus, at most, two of the three components of the vector e(1)
can be found in terms of the third. We can
therefore arbitrarily set e
ð1Þ
x ¼ 1 and solve for e
ð1Þ
y and e
ð1Þ
z using any two of the independent equations in Eqn (e).
With e
ð1Þ
x ¼ 1, the first two of Eqn (e) become
20:0000e
ð1Þ
y 100:000e
ð1Þ
z ¼ 432:052
232:052e
ð1Þ
y 50:000e
ð1Þ
z ¼ 20:0000
(f)
Solving these two equations for e
ð1Þ
y and e
ð1Þ
z yields, together with the assumption that e
ð1Þ
x ¼ 1,
e
ð1Þ
x ¼ 1:00000 e
ð1Þ
y ¼ 0:882793 e
ð1Þ
z ¼ 4:49708 (g)
The unit vector in the direction of e(1)
is
^
e1 ¼
eð1Þ

eð1Þ

¼
1:00000^
i þ 0:882793^
j 4:49708^
k
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:000002 þ 0:8827932 þ ð4:49708Þ2
q

or
^
e1 ¼ 0:213186^
i þ 0:188199^
j 0:958714^
k

l1 ¼ 532:052 kg$m2

(h)
Substituting l2 ¼ 295:840 kg$m2 into Eqn (a) and proceeding as above we find that
^
e2 ¼ 0:176732^
i 0:972512^
j 0:151609^
k

l2 ¼ 295:840 kg$m2

(i)
The two unit vectors ^
e1 and ^
e2 define two of the three principal directions of the inertia tensor. Observe that
^
e1$^
e2 ¼ 0, as must be the case for symmetric matrices.
To obtain the third principal direction ^
e3, we can substitute l3 ¼ 72:1083 kg$m2 into Eqn (a) and pro-
ceed as above. However, since the inertia tensor is symmetric, we know that the three principal directions
are mutually orthogonal, which means ^
e3 ¼ ^
e1 ^
e2. Substituting Eqns (h) and (i) into the crossproduct, we
find that
^
e3 ¼ 0:960894^
i 0:137114^
j 0:240587^
k

l3 ¼ 72:1083 kg$m2

(j)
We can check our work by substituting l3 and ^
e3 into Eqn (a) and verify that it is indeed satisfied:
2
6
6
4
100 72:1083 20 100
20 300 72:1083 50
100 50 500 72:1083
3
7
7
5
8

:
0:960894
0:137114
0:240587
9

=

;
¼
O
8

:
0
0
0
9

=

;
(k)
The components of the vectors ^
e1, ^
e2, and ^
e3 define the three rows of the orthogonal transformation [Q] from the
xyz system into the x0y0z0 system aligned along the three principal directions:
½Q ¼
2
6
6
4
0:213186 0:188199 0:958714
0:176732 0:972512 0:151609
0:960894 0:137114 0:240587
3
7
7
5 (l)
Indeed, if we apply the transformation in Eqn (9.49), ½I0
¼ ½Q½I½QT
, we find
½I0
¼
2
6
6
4
0:213186 0:188199 0:958714
0:176732 0:972512 0:151609
0:960894 0:137114 0:240587
3
7
7
5
2
6
6
4
100 20 100
20 300 50
100 50 500
3
7
7
5

2
6
6
4
0:213186 0:176732 0:960894
0:188199 0:972512 0:137114
0:958714 0:151609 0:240587
3
7
7
5
¼
2
6
6
4
532:052 0 0
0 295:840 0
0 0 72:1083
3
7
7
5

kg$m2


An alternative to the above hand calculations in Example 9.9 is to type the following lines in the
MATLAB Command Window:
I ¼ [ 100 -20 -100
-20 300 -50
-100 -50 500];
[eigenVectors, eigenValues] ¼ eig(I)
Hitting the Enter (or Return) key yields the following output to the Command Window:
eigenVectors ¼
0.9609 0.1767 -0.2132
0.1371 -0.9725 -0.1882
0.2406 -0.1516 0.9587
eigenValues ¼
72.1083 0 0
0 295.8398 0
0 0 532.0519
Two of the eigenvectors delivered by MATLAB are opposite in direction to those calculated in
Example 9.9. This illustrates the fact that we can determine an eigenvector only to within an arbitrary
scalar factor. That is, suppose e is an eigenvector of the tensor [I] so that [I]{e} ¼ l{e}. Multiplying
this equation through by an arbitrary scalar a yields [I]{e}a ¼ l{e}a, or [I]{ae} ¼ l{ae}, which means
that {ae} is an eigenvector corresponding to the same eigenvalue l.
Parallel axis theorem
Suppose the rigid body in Figure 9.14 is in pure rotation about point P. Then, according to Eqn (9.39),
fHPrel
g ¼ ½IPfug (9.57)
η
ζ
ξ
FIGURE 9.14
The moments of inertia are to be computed at P, given their values at G.

where [IP] is the moment of inertia tensor about P, given by Eqn (9.40) with
x ¼ xG=P þ x y ¼ yG=P þ h z ¼ zG=P þ z
On the other hand, we have from Eqn (9.27) that
HPrel
¼ HG þ rG=P mvG=P (9.58)
The vector rG/P mvG/P is the angular momentum about P of the concentrated mass m located at the
center of mass G. Using matrix notation, it is computed as follows:
n
rG=P mvG=P
o
hfH
ðmÞ
Prel
g ¼
h
I
ðmÞ
P
i
fug (9.59)
where
h
I
ðmÞ
P
i
, the moment of inertia of the point mass m about P, is obtained from Eqn (9.44), with
x ¼ xG/P, y ¼ yG/P, and z ¼ zG/P. That is,
½I
ðmÞ
P ¼
2
6
6
6
4
m

y2
G=P þ z2
G=P

mxG=P yG=P mxG=P zG=P
mxG=P yG=P m

x2
G=P þ z2
G=P

myG=P zG=P
mxG=P zG=P myG=P zG=P m

x2
G=P þ y2
G=P

3
7
7
7
5
(9.60)
Of course, Eqn (9.39) requires
fHGg ¼ ½IGfug
Substituting this together with Eqns (9.57) and (9.59) into Eqn (9.58) yields
½IPfug ¼ ½IGfug þ
h
I
ðmÞ
P
i
fug ¼

½IG þ
h
I
ðmÞ
P
i
fug
From this, we may infer the parallel axis theorem,
½IP ¼ ½IG þ
h
I
ðmÞ
P
i
(9.61)
The moment of inertia about P is the moment of inertia about the parallel axes through the center of
mass plus the moment of inertia of the center of mass about P. That is,
IPx
¼ IGx
þ m

y2
G=P þ z2
G=P

IPy
¼ IGy
þ m

y2
G=P þ x2
G=P

IPz
¼ IGz
þ m

x2
G=P þ y2
G=P

IPxy
¼ IGxy
mxG=P yG=P IPxz
¼ IGxz
mxG=P zG=P IPyz
¼ IGyz
myG=P zG=P
(9.62)

EXAMPLE 9.10
Find the moments of inertia of the rod in Example 9.5 (Figure 9.15) about its center of mass G.
Solution
From Example 9.5,
½IA ¼
2
6
6
6
6
6
6
6
6
6
4
1
3
m

b2
þ c2

1
3
mab
1
3
mac

1
3
mab
1
3
m

a2
þ c2

1
3
mbc

1
3
mac
1
3
mbc
1
3
m

a2
þ b2

3
7
7
7
7
7
7
7
7
7
5
Using Eqn (9.62)1, and noting the coordinates of the center of mass in Figure 9.15,
IGx
¼ IAx
m
h
ðyG 0Þ2
þ ðzG 0Þ2
i
¼
1
3
m

b2
þ c2

m

b
2
2
þ
c
2
2
¼
1
12
m

b2
þ c2

Equation (9.62)4 yields
IGxy
¼ IAxy
þ mðxG 0ÞðyG 0Þ ¼
1
3
mab þ m$
a
2
$
b
2
¼
1
12
mab
The remaining four moments of inertia are found in a similar fashion, so that
½IG ¼
2
6
6
6
6
6
6
6
6
6
6
4
1
12
m

b2
þ c2

1
12
mab
1
12
mac

1
12
mab
1
12
m

a2
þ c2

1
12
mbc

1
12
mac
1
12
mbc
1
12
m

a2
þ b2

3
7
7
7
7
7
7
7
7
7
7
5
(9.63)
FIGURE 9.15
A uniform slender rod.

EXAMPLE 9.11
Calculate the principal moments of inertia about the center of mass and the corresponding principal directions for
the bent rod in Figure 9.16. Its mass is uniformly distributed at 2 kg/m.
Solution
The mass of each of the rod segments is
m1 ¼ 2$0:4 ¼ 0:8 kg m2 ¼ 2$0:5 ¼ 1 kg m3 ¼ 2$0:3 ¼ 0:6 kg m4 ¼ 2$0:2 ¼ 0:4 kg (a)
The total mass of the system is
m ¼
X
4
i¼1
mi ¼ 2:8 kg (b)
The coordinates of each segment’s center of mass are
xG1
¼ 0 yG1
¼ 0 zG1
¼ 0:2 m
xG2
¼ 0 yG2
¼ 0:25 m zG2
¼ 0:2 m
xG3
¼ 0:15 m yG3
¼ 0:5 m zG3
¼ 0
xG4
¼ 0:3 m yG4
¼ 0:4 m zG4
¼ 0
(c)
If the slender rod in Figure 9.15 is aligned with, say, the x-axis, then a ¼ [ and b ¼ c ¼ 0, so that according to
Eqn (9.63),
½IG ¼
2
6
6
6
6
6
4
0 0 0
0
1
12
m‘2
0
0 0
1
12
m‘2
3
7
7
7
7
7
5
That is, the moment of inertia of a slender rod about the axes normal to the rod at its center of mass is 1
12 m‘2,
where m and [ are the mass and length of the rod, respectively. Since the mass of a slender bar is assumed to be
concentrated along the axis of the bar (its cross-sectional dimensions are infinitesimal), the moment of inertia
4
3
2
1
0.4 m
0.5 m
0.3 m
0.2 m
O
FIGURE 9.16
Bent rod for which the principal moments of inertia are to be determined.

about the centerline is zero. By symmetry, the products of inertia about the axes through the center of mass are all
zero. Using this information and the parallel axis theorem, we find the moments and products of inertia of each rod
segment about the origin O of the xyz system as follows:
Rod 1:
I
ð1Þ
x ¼ I
ð1Þ
G1

x
þ m1

y2
G1
þ z2
G1

¼

1
12
$0:8$0:42

þ
h
0:8

0 þ 0:22
i
¼ 0:04267 kg-m2
I
ð1Þ
y ¼ I
ð1Þ
G1

y
þ m1

x2
G1
þ z2
G1

¼

1
12
$0:8$0:42

þ
h
0:8

0 þ 0:22
i
¼ 0:04267 kg-m2
I
ð1Þ
z ¼ I
ð1Þ
G1

z
þ m1

x2
G1
þ y2
G1

¼ 0 þ ½0:8ð0 þ 0Þ ¼ 0
I
ð1Þ
xy ¼ I
ð1Þ
G1

xy
m1xG1
yG1
¼ 0 ½0:8ð0Þð0Þ ¼ 0
I
ð1Þ
xz ¼ I
ð1Þ
G1

xz
m1xG1
zG1
¼ 0 ½0:8ð0Þð0:2Þ ¼ 0
I
ð1Þ
yz ¼ I
ð1Þ
G1

yz
m1yG1
zG1
¼ 0 ½0:8ð0Þð0Þ ¼ 0
Rod 2:
I
ð2Þ
x ¼ I
ð2Þ
G2

x
þ m2

y2
G2
þ z2
G2

¼

1
12
$1:0$0:52

þ
h
1:0

0 þ 0:252
i
¼ 0:08333 kg-m2
I
ð2Þ
y ¼ I
ð2Þ
G2

y
þ m2

x2
G2
þ z2
G2

¼ 0 þ ½1:0ð0 þ 0Þ ¼ 0
I
ð2Þ
z ¼ I
ð2Þ
G2

z
þ m2

x2
G2
þ y2
G2

¼

1
12
$1:0$0:52

þ
h
1:0

0 þ 0:52
i
¼ 0:08333 kg-m2
I
ð2Þ
xy ¼ I
ð2Þ
G2

xy
m2xG2
yG2
¼ 0 ½1:0ð0Þð0:5Þ ¼ 0
I
ð2Þ
xz ¼ I
ð2Þ
G2

xz
m2xG2
zG2
¼ 0 ½1:0ð0Þð0Þ ¼ 0
I
ð2Þ
yz ¼ I
ð2Þ
G2

yz
m2yG2
zG2
¼ 0 ½1:0ð0:5Þð0Þ ¼ 0
Rod 3:
I
ð3Þ
x ¼ I
ð3Þ
G3

x
þ m3

y2
G3
þ z2
G3

¼ 0 þ 0:6

0:52 þ 0

¼ 0:15 kg-m2
I
ð3Þ
y ¼ I
ð3Þ
G3

y
þ m3

x2
G3
þ z2
G3

¼

1
12
$0:6$0:32

þ
h
0:6

0:152
þ 0
i
¼ 0:018 kg-m2
I
ð3Þ
z ¼ I
ð3Þ
G3

z
þ m3

x2
G3
þ y2
G3

¼

1
12
$0:6$0:32

þ
h
0:6

0:152
þ 0:52
i
¼ 0:1680 kg-m2
I
ð3Þ
xy ¼ I
ð3Þ
G3

xy
m3xG3
yG3
¼ 0 ½0:6ð0:15Þð0:5Þ ¼ 0:045 kg-m2
I
ð3Þ
xz ¼ I
ð3Þ
G3

xz
m3xG3
zG3
¼ 0 ½0:6ð0:15Þð0Þ ¼ 0
I
ð3Þ
yz ¼ I
ð3Þ
G3

yz
m3yG3
zG3
¼ 0 ½0:6ð0:5Þð0Þ ¼ 0

Rod 4:
I
ð4Þ
x ¼ I
ð4Þ
G4

x
þ m4

y2
G4
þ z2
G4

¼

1
12
$0:4$0:22

þ
h
0:4

0:42
þ 0
i
¼ 0:06533 kg-m2
I
ð4Þ
y ¼ I
ð4Þ
G4

y
þ m4

x2
G4
þ z2
G4

¼ 0 þ 0:4

0:32 þ 0

¼ 0:0360 kg-m2
I
ð4Þ
z ¼ I
ð4Þ
G4

z
þ m4

x2
G4
þ y2
G4

¼

1
12
$0:4$0:22

þ
h
0:4

0:32
þ 0:42
i
¼ 0:1013 kg-m2
I
ð4Þ
xy ¼ I
ð4Þ
G4

xy
m4xG4
yG4
¼ 0 ½0:4ð0:3Þð0:4Þ ¼ 0:0480 kg-m2
I
ð4Þ
xz ¼ I
ð4Þ
G4

xz
m4xG4
zG4
¼ 0 ½0:4ð0:3Þð0Þ ¼ 0
I
ð4Þ
yz ¼ I
ð4Þ
G4

yz
m4yG4
zG4
¼ 0 ½0:4ð0:4Þð0Þ ¼ 0
The total moments of inertia for all the four rods about O are
Ix ¼
P
4
i¼1
I
ðiÞ
x ¼ 0:3413 kg-m2
Iy ¼
P
4
i¼1
I
ðiÞ
y ¼ 0:09667 kg-m2
Iz ¼
P
4
i¼1
I
ðiÞ
z ¼ 0:3527 kg-m2
Ixy ¼
P
4
i¼1
I
ðiÞ
xy ¼ 0:0930 kg-m2
Ixz ¼
P
4
i¼1
I
ðiÞ
xz ¼ 0 Iyz ¼
P
4
i¼1
I
ðiÞ
yz ¼ 0
(d)
The coordinates of the center of mass of the system of four rods are, from Eqns (a)e(c),
xG ¼
1
m
X
4
i¼1
mixGi
¼
1
2:8
$0:21 ¼ 0:075 m
yG ¼
1
m
X
4
i¼1
miyGi
¼
1
2:8
$0:71 ¼ 0:2536 m
zG ¼
1
m
X
4
i¼1
mizGi
¼
1
2:8
$0:16 ¼ 0:05714 m
(e)
We use the parallel axis theorems to shift the moments of inertia in Eqn (d) to the center of mass G of the system.
IGx
¼ Ix m

y2
G þ z2
G

¼ 0:3413 0:1892 ¼ 0:1522 kg-m2
IGy
¼ Iy m

x2
G þ z2
G

¼ 0:09667 0:02489 ¼ 0:07177 kg-m2
IGz
¼ Iz m

x2
G þ y2
G

¼ 0:3527 0:1958 ¼ 0:1569 kg-m2
IGxy
¼ Ixy þ mxGyG ¼ 0:093 þ 0:05325 ¼ 0:03975 kg-m2
IGxz
¼ Ixz þ mxGzG ¼ 0 þ 0:012 ¼ 0:012 kg-m2
IGyz
¼ Iyz þ myGzG ¼ 0 þ 0:04057 ¼ 0:04057 kg-m2

Therefore, the inertia tensor, relative to the center of mass, is
½I ¼
2
6
6
4
IGx
IGxy
IGxz
IGxy
IGy
IGyz
IGxz
IGyz
IGz
3
7
7
5 ¼
2
6
6
4
0:1522 0:03975 0:012
0:03975 0:07177 0:04057
0:012 0:04057 0:1569
3
7
7
5

kg$m2

(f)
To find the three principal moments of inertia, we may proceed as in Example 9.9, or simply enter the following
lines in the MATLAB Command Window:
I ¼ [ 0.1522 -0.03975 0.012
-0.03975 0.07177 0.04057
0.012 0.04057 0.1569 ];
[eigenVectors, eigenValues] ¼ eig(IG)
to obtain
eigenVectors ¼
0.3469 -0.8482 -0.4003
0.8742 0.1378 0.4656
-0.3397 -0.5115 0.7893
eigenValues ¼
0.0402 0 0
0 0.1658 0
0 0 0.1747
Hence, the three principal moments of inertia and their principal directions are
l1 ¼ 0:04023 kg-m2
eð1Þ ¼ 0:3469^
i þ 0:8742^
j 0:3397^
k
l2 ¼ 0:1658 kg-m2
eð2Þ ¼ 0:8482^
i þ 0:1378^
j 0:5115^
k
l3 ¼ 0:1747 kg-m2
eð3Þ ¼ 0:4003^
i þ 0:4656^
j þ 0:7893^
k
9.6 Euler’s equations
For either the center of mass G or for a fixed point P about which the body is in pure rotation, we know
from Eqns (9.29) and (9.30) that
Mnet ¼ _
H (9.64)
Using a comoving coordinate system, with angular velocity U and its origin located at the point (G or
P), the angular momentum has the analytical expression
H ¼ Hx
^
i þ Hy
^
j þ Hz
^
k (9.65)
We shall henceforth assume, for simplicity, that
ðaÞ The moving xyz axes are the principal axes of inertia and (9.66a)
ðbÞ The moments of inertia relative to xyz are constant in time: (9.66b)

Equations (9.42) and (9.66a) imply that
H ¼ Aux
^
i þ Buy
^
j þ Cuz
^
k (9.67)
where A, B, and C are the principal moments of inertia.
According to Eqn (1.56), the time derivative of H is _
H ¼ _
HÞrel þ U H, so that Eqn (9.64) can be
written as
Mnet ¼ _
H

rel
þ U H (9.68)
Keep in mind that, whereas U (the angular velocity of the moving xyz coordinate system) and u (the
angular velocity of the rigid body itself) are both absolute kinematic quantities, Eqn (9.68) contains
their components as projected onto the axes of the noninertial xyz frame given by
u ¼ ux
^
i þ uy
^
j þ uz
^
k
U ¼ Ux
^
i þ Uy
^
j þ Uz
^
k
The absolute angular acceleration a is obtained using Eqn (1.56) as
a ¼ _
u ¼
dux
dt
^
i þ
duy
dt
^
j þ
duz
dt
^
k
zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{
arel
þ U u
that is,
a ¼

_
ux þ Uyuz Uzuy

^
i þ

_
uy þ Uzux Uxuz

^
j þ

_
uz þ Uxuy Uyux

^
k (9.69)
Clearly, it is generally true that
axs _
ux ays _
uy azs _
uz
From Eqn (1.57) and Eqn (9.67),
_
H

rel
¼
dðAuxÞ
dt
^
i þ
d

Buy

dt
^
j þ
dðCuzÞ
dt
^
k
Since A, B, and C are constant, this becomes
_
H

rel
¼ A _
ux
^
i þ B _
uy
^
j þ C _
uz
^
k (9.70)
Substituting Eqns (9.67) and (9.70) into Eqn (9.68) yields
Mnet ¼ A _
ux
^
i þ B _
uy
^
j þ C _
uz
^
k þ
^
i ^
j ^
k
Ux Uy Uz
Aux Buy Cuz
Expanding the crossproduct and collecting the terms leads to
Mxnet
¼ A _
ux þ CUyuz BUzuy
Mynet
¼ B _
uy þ AUzux CUxuz
Mznet
¼ C _
uz þ BUxuy AUyux
(9.71)
9.6 Euler’s equations 497

If the comoving frame is a rigidly attached body frame, then its angular velocity is the same as that
of the body, that is, U ¼ u. In that case, Eqn (9.68) reduces to the classical Euler’s equation of motion,
namely,
Mnet ¼ _
H

rel
þ u H (9.72a)
the three components of which are obtained from Eqn (9.71) as
Mxnet
¼ A _
ux þ ðC BÞuyuz
Mynet
¼ B _
uy þ ðA CÞuzux
Mznet
¼ C _
uz þ ðB AÞuxuy
(9.72b)
Equation (9.68) is sometimes referred to as the modified Euler equation.
When U ¼ u, it follows from Eqn (9.69) that
_
ux ¼ ax _
uy ¼ ay _
uz ¼ az (9.73)
That is, the relative angular acceleration equals the absolute angular acceleration when U ¼ u. Rather
than calculating the time derivatives _
ux, _
uy, and _
uz for use in Eqn (9.72), we may in this case first
compute a in the absolute XYZ frame
a ¼
du
dt
¼
duX
dt
^
I þ
duY
dt
^
J þ
duZ
dt
^
K
and then project these components onto the xyz body frame, so that
8

:
_
ux
_
uy
_
uz
9
=
;
¼ ½QXx
8

:
duX=dt
duY=dt
duZ=dt
9
=
;
(9.74)
where [Q]Xx is the time-dependent orthogonal transformation from the inertial XYZ frame to the
noninertial xyz frame.
EXAMPLE 9.12
Calculate the net moment on the solar panel of Examples 9.2 and 9.8 (Figure 9.17).
Solution
Since the comoving frame is rigidly attached to the panel, Euler’s equation (Eqn (9.72a)) applies to this problem,
that is
MGnet
¼ _
HG

rel
þ u HG (a)
where
HG ¼ Aux
^
i þ Buy
^
j þ Cuz
^
k (b)
and
_
HG

rel
¼ A _
ux
^
i þ B _
uy
^
j þ C _
uz
^
k (c)

In Example 9.2, the angular velocity of the panel in the satellite’s x0y0z0 frame was found to be
u ¼ _
q^
j
0
þ N^
k
0
(d)
In Example 9.8, we showed that the transformation from the panel’s xyz frame to that of the satellite is represented
by the matrix
½Q ¼
2
6
6
4
sin q 0 cos q
0 1 0
cos q 0 sin q
3
7
7
5 (e)
We use the transpose of [Q] to transform the components of u into the panel frame of reference,
fugxyz ¼ ½QT
fugx0y0z0 ¼
2
6
6
4
sin q 0 cos q
0 1 0
cos q 0 sin q
3
7
7
5
8

:
0
_
q
N
9

=

;
¼
8

:
N cos q
_
q
N sin q
9

=

;
or
ux ¼ N cos q uy ¼ _
q uz ¼ N sin q (f)
In Example 9.2, N and _
q were said to be constant. Therefore, the time derivatives of Eqn (f) are
_
ux ¼
dðN cos qÞ
dt
¼ N _
q sin q
_
uy ¼
d_
q
dt
¼ 0
_
uz ¼
dðN sin qÞ
dt
¼ N _
q cos q
(g)
In Example 9.8, the moments of inertia in the panel frame of reference were listed as
A ¼
1
12
m

‘2
þ t2

B ¼
1
12
m

w2
þ t2

C ¼
1
12
m

w2
þ ‘2

IGxy ¼ IGxz ¼ IGyz ¼ 0

(h)
θ
FIGURE 9.17
Free-body diagram of the solar panel in Examples 9.2 and 9.8.

Substituting Eqns (b), (c), (f), (g) and (h) into Eqn (a) yields
MGnet
¼
1
12
m

‘2
þ t2

N _
q sin q
^
i þ
1
12
m

w2
þ t2

$0$^
j þ
1
12
m

w2
þ ‘2

N _
q cos q

^
k
þ
^
i ^
j ^
k
N cos q _
q N sin q
1
12
m

‘2
þ t2

ðN cos qÞ
1
12
m

w2
þ t2

_
q
1
12
m

w2
þ ‘2

ðN sin qÞ
Upon expanding the cross product and collecting terms, this reduces to
MGnet
¼
1
6
mt2
N _
q sin q^
i þ
1
24
m

t2
w2

N2
sin 2q^
j þ
1
6
mw2
N _
q cos q^
k
Using the numerical data of Example 9.8 (m ¼ 50 kg, N ¼ 0.1 rad/s, q ¼ 40, _
q ¼ 0:01 rad=s, [ ¼ 6 m, w ¼ 2 m,
and t ¼ 0.025 m), we find
MGnet
¼ 3:348 106^
i 0:08205^
j þ 0:02554^
k ðN$mÞ
EXAMPLE 9.13
Calculate the net moment on the gyro rotor of Examples 9.3 and 9.6.
Solution
Figure 9.18 is a free-body diagram of the rotor. Since in this case the comoving frame is not rigidly attached to the
rotor, we must use Eqn (9.68) to find the net moment about G, that is
MGnet
¼ _
HG

rel
þ U HG (a)
where
HG ¼ Aux
^
i þ Buy
^
j þ Cuz
^
k (b)
and
_
HG

rel
¼ A _
ux
^
i þ B _
uy
^
j þ C _
uz
^
k (c)
From Eqn (i) of Example 9.3, we know that the components of the angular velocity of the rotor in the moving
reference frame are
ux ¼ _
q
uy ¼ N sin q
uz ¼ uspin þ N cos q
(d)

Since, as specified in Example 9.3, _
q, N, and uspin are all constant, it follows that
_
ux ¼
d_
q
dt
¼ 0
_
uy ¼
dðN sin qÞ
dt
¼ N _
q cos q
_
uz ¼
d

uspin þ N cos q

dt
¼ N _
q sin q
(e)
The angular velocity U of the comoving xyz frame is that of the gimbal ring, which equals the angular velocity of
the rotor minus its spin. Therefore,
Ux ¼ _
q
Uy ¼ N sin q
Uz ¼ N cos q
(f)
In Example 9.6, we found that
A ¼ B ¼
1
12
mt2
þ
1
4
mr2
C ¼
1
2
mr2
(g)
FIGURE 9.18
Free-body diagram of the gyro rotor of Examples 9.6 and 9.3.

Substituting Eqns (b) through (g) into Eqn (a), we get
MGnet
¼

1
12
mt2
þ
1
4
mr2

$0^
i þ

1
12
mt2
þ
1
4
mr2

N _
q cos q
^
j þ
1
2
mr2

N _
q sin q

^
k
þ
^
i ^
j ^
k
_
q N sin q N cos q

1
12
mt2
þ
1
4
mr2

_
q

1
12
mt2
þ
1
4
mr2

N sin q
1
2
mr2

uspin þ N cos q

Expanding the cross product, collecting terms, and simplifying leads to
MGnet
¼

1
2
uspin þ
1
12

3
t2
r2

N cos q mr2
N sin q^
i þ

1
6
t2
r2
N cos q
1
2
uspin

mr2 _
q^
j
1
2
N _
q sin qmr2^
k (h)
In Example 9.3, the following numerical data were provided: m ¼ 5 kg, r ¼ 0.08 m, t ¼ 0.025 m, N ¼ 2.1 rad/s,
q ¼ 60, _
q ¼ 4 rad=s, and uspin ¼ 105 rad/s. For this set of numbers, Eqn (h) becomes
MGnet
¼ 0:3203^
i 0:6698^
j 0:1164^
k ðN$mÞ
9.7 Kinetic energy
The kinetic energy T of a rigid body is the integral of the kinetic energy 1
2 v2dm of its individual mass
elements,
T ¼
Z
m
1
2
v2
dm ¼
Z
m
1
2
v$vdm (9.75)
where v is the absolute velocity _
R of the element of mass dm. From Figure 9.8, we infer that
_
R ¼ _
RG þ _
r. Furthermore, Eqn (1.52) requires that _
r ¼ u r. Thus, v ¼ vG þ u r, which means
v$v ¼ ½vG þ u r$½vG þ u r ¼ vG
2
þ 2vG$ðu rÞ þ ðu rÞ$ðu rÞ
We can apply the vector identity introduced in Eqn (1.21), namely
A$ðB CÞ ¼ B$ðC AÞ (9.76)
to the last term to get
v$v ¼ vG
2
þ 2vG$ðu rÞ þ u$½r ðu rÞ
Therefore, Eqn (9.75) becomes
T ¼
Z
m
1
2
vG
2
dm þ vG$
0
@u
Z
m
rdm
1
A þ
1
2
u$
Z
m
r ðu rÞdm

Since r is measured from the center of mass,
R
m
rdm ¼ 0. Recall that, according to Eqn (9.34),
Z
m
r ðu rÞdm ¼ HG
It follows that the kinetic energy may be written as
T ¼
1
2
mvG
2
þ
1
2
u$HG (9.77)
The second term is the rotational kinetic energy TR,
TR ¼
1
2
u$HG (9.78)
If the body is rotating about a point P that is at rest in inertial space, we have from Eqn (9.2) and
Figure 9.8 that
vG ¼ vP þ u rG=P ¼ 0 þ u rG=P ¼ u rG=P
It follows that
vG
2
¼ vG$vG ¼

u rG=P

$

u rG=P

Making use once again of the vector identity in Eqn (9.76), we find
vG
2
¼ u$
h
rG=P

u rG=P
i
¼ u$

rG=P vG

Substituting this into Eqn (9.77) yields
T ¼
1
2
u$
h
HG þ rG=P mvG
i
Equation (9.21) shows that this can be written as
T ¼
1
2
u$HP (9.79)
In this case, of course, all of the kinetic energy is rotational.
In terms of the components of u and H, whether it is HP or HG, the rotational kinetic energy
expression becomes, with the aid of Eqn (9.39),
TR ¼
1
2

uxHx þ uyHy þ uyHz

¼
1
2
P ux uy uz R
2
6
6
4
Ix Ixy Ixz
Ixy Iy Iyz
Ixz Iyz Iz
3
7
7
5
8

:
ux
uy
uz
9

=

;
Expanding, we obtain
TR ¼
1
2
Ixux
2
þ
1
2
Iyuy
2
þ
1
2
Izuz
2
þ Ixyuxuy þ Ixzuxuz þ Iyzuyuz (9.80)
9.7 Kinetic energy 503

Obviously, if the xyz axes are principal axes of inertia, then Eqn (9.80) simplifies considerably,
TR ¼
1
2
Aux
2
þ
1
2
Buy
2
þ
1
2
Cuz
2
(9.81)
EXAMPLE 9.14
A satellite in a circular geocentric orbit of 300-km altitude has a mass of 1500 kg, and the moments of inertia
relative to a body frame with origin at the center of mass G are
½I ¼
2
6
6
4
2000 1000 2500
1500 3000 1500
2500 1500 4000
3
7
7
5

kg$m2

If at a given instant the components of angular velocity in this frame of reference are
u ¼ 1^
i 0:9^
j þ 1:5^
k ðrad=sÞ
calculate the total kinetic energy of the satellite.
Solution
The speed of the satellite in its circular orbit is
v ¼
ffiffiffi
m
r
r
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
398; 600
6378 þ 300
r
¼ 7:7258 km=s
The angular momentum of the satellite is
fHGg ¼ ½IGfug ¼
2
6
6
4
2000 1000 2500
1500 3000 1500
2500 1500 4000
3
7
7
5
8

:
1
0:9
1:5
9

=

;
¼
8

:
6650
5950
9850
9

=

;

kg$m2
=s

Therefore, the total kinetic energy is
T ¼
1
2
mvG
2
þ
1
2
u$HG ¼
1
2
$1500$7725:82
þ
1
2
P 1 0:9 1:5 R
8

:
6650
5950
9850
9

=

;
¼ 44:766 106
þ 13; 390
T ¼ 44:766 MJ
Obviously, the kinetic energy is dominated by that due to the orbital motion.
9.8 The spinning top
Let us analyze the motion of the simple axisymmetric top in Figure 9.19. It is constrained to rotate
about point O.
The moving coordinate system is chosen to have its origin at O. The z-axis is aligned with the spin
axis of the top (the axis of rotational symmetry). The x-axis is the node line, which passes through O

and is perpendicular to the plane defined by the inertial Z-axis and the spin axis of the top. The y-axis
is then perpendicular to x and z, such that ^
j ¼ ^
k ^
i. By symmetry, the moment of inertia matrix of the
top relative to the xyz frame is diagonal, with Ix ¼ Iy ¼ A and Iz ¼ C. From Eqns (9.68) and (9.70),
we have
M0net
¼ A _
ux
^
i þ A _
uy
^
j þ C _
uz
^
k þ
^
i ^
j ^
k
Ux Uy Uz
Aux Auy Cuz
(9.82)
The angular velocity u of the top is the vector sum of the spin rate us and the rates of precession up and
nutation un, where
up ¼ _
f un ¼ _
q (9.83)
Thus,
u ¼ un
^
i þ up
^
K þ us
^
k
From the geometry, we see that
^
K ¼ sin q^
j þ cos q^
k (9.84)
Therefore, relative to the comoving system,
u ¼ un
^
i þ up sin q^
j þ

us þ up cos q

^
k (9.85)
FIGURE 9.19
Simple top rotating about the fixed point O.
9.8 The spinning top 505

From Eqn (9.85), we see that
ux ¼ un uy ¼ up sin q uz ¼ us þ up cos q (9.86)
Computing the time rates of these three expressions yields the components of angular acceleration
relative to the xyz frame, given by
_
ux ¼ _
un _
uy ¼ _
up sin q þ upun cos q _
uz ¼ _
us þ _
up cos q upun sin q (9.87)
The angular velocity U of the xyz system is U ¼ up
^
K þ un
^
i, so that, using Eqn (9.84),
U ¼ un
^
i þ up sin q^
j þ up cos q^
k (9.88)
From Eqn (9.88), we obtain
Ux ¼ un Uy ¼ up sin q Uz ¼ up cos q (9.89)
In Figure 9.19, the moment about O is that of the weight vector acting through the center of mass G:
M0net
¼

d^
k

mg^
K

¼ mgd^
k

sin q^
j þ cos q^
k

or
M0net
¼ mgd sin q^
i (9.90)
Substituting Eqns (9.86), (9.87), (9.89), and (9.90) into Eqn (9.82), we get
mgd sin q^
i ¼ A _
un
^
i þ A

_
up sin q þ upun cos q

^
j þ C

_
us þ _
up cos q upun sin q

^
k
þ
^
i ^
j ^
k
un up sin q up cos q
Aun Aup sin q C

us þ up cos q

(9.91)
Let us consider the special case in which q is constant, that is, there is no nutation, so that
un ¼ _
un ¼ 0. Then, Eqn (9.91) reduces to
mgd sin q^
i ¼ A _
up sin q^
j þ C

_
us þ _
up cos q

^
k þ
^
i ^
j ^
k
0 up sin q up cos q
0 Aup sin q C

us þ up cos q

(9.92)
Expanding the determinant yields
mgd sin q^
i ¼ A _
up sin q^
j þ C

_
us þ _
up cos q

^
k þ
h
Cupus sin q þ ðC AÞup
2
cos q sin q
i
^
i
Equating the coefficients of^
i,^
j, and ^
k on each side of the equation and assuming that 0 q 180
leads to
mgd ¼ Cupus þ ðC AÞup
2
cos q (9.93a)

A _
up ¼ 0 (9.93b)
C

_
us þ _
up cos q

¼ 0 (9.93c)
Equation (9.93b) implies _
up ¼ 0, and from Eqn (9.93c), it follows that _
us ¼ 0. Therefore, the rates
of spin and precession are both constant. From Eqn (9.93a), we find
ðA CÞcos qup
2
Cusup þ mgd ¼ 0 (9.94)
If the spin rate is zero, Eqn (9.94) yields
up

us¼0
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
mgd
ðC AÞcos q
s
if ðC AÞ cos q 0 (9.95)
In this case, the top rotates about O at this rate, without spinning. If A C (prolate), its symmetry axis
must make an angle between 90 and 180 to the vertical; otherwise up is imaginary. On the other
hand, if A C (oblate), the angle lies between 0 and 90. Thus, in steady rotation without spin, the
top’s axis sweeps out a cone that lies either below the horizontal plane (A C) or above the plane
(A C).
In the special case where (A C) cos q ¼ 0, Eqn (9.94) yields a steady precession rate that is
inversely proportional to the spin rate,
up ¼
mgd
Cus
if ðA CÞ cos q ¼ 0 (9.96)
If A ¼ C, this precession apparently occurs irrespective of tilt angle q. If A s C, this rate of precession
occurs at q ¼ 90, that is, the spin axis is perpendicular to the precession axis.
In general, Eqn (9.94) is a quadratic equation in up, so we can use the quadratic formula
to find
up ¼
C
2ðA CÞcos q
us
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
us
2
4mgdðA CÞcos q
C2
r !
(9.97)
Thus, for a given spin rate and tilt angle q (q s 90), there are two rates of precession _
f.
Observe that if (A C)cos q 0, then up is imaginary when us
2 4mgdðA CÞcos q=C2.
Therefore, the minimum spin rate required for steady precession at a constant inclination q is
usÞmin ¼
2
C
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
mgdðA CÞcos q
p
if ðA CÞcos q 0 (9.98)
If (A C) cos q 0, the radical in Eqn (9.97) is real for all us. In this case, as us / 0, up approaches
the value given above in Eqn (9.95).

EXAMPLE 9.15
Calculate the precession rate up for the top of Figure 9.19 if m ¼ 0.5 kg, A (¼ Ix ¼ Iy) ¼ 12 104
kg m2
, C
(¼ Iz) ¼ 4.5 104
kg m2
, and d ¼ 0.05 m.
Solution
For an inclination of, say, 60, (A C) cos q 0, so that Eqn (9.98) requires usÞmin ¼ 407:01 rpm. Let us choose
the spin rate to be us ¼ 1000 rpm ¼ 104.7 rad/sec. Then, from Eqn (9.97), the precession rate as a function of the
inclination q is given by either one of the following formulas:
up ¼ 31:42
1 þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 0:3312 cos q
p
cos q
and up ¼ 31:42
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 0:3312 cos q
p
cos q
(a)
These are plotted in Figure 9.20.
Figure 9.21 shows an axisymmetric rotor mounted so that its spin axis (z) remains perpendicular to the
precession axis (y). In that case, Eqn (9.85) with q ¼ 90 yields
u ¼ up
^
j þ us
^
k (9.99)
Likewise, from Eqn (9.88), the angular velocity of the comoving xyz system is U ¼ up
^
j. If we assume
that the spin rate and precession rate are constant ðdup=dt ¼ dus=dt ¼ 0Þ, then Eqn (9.68), written for
the center of mass G, becomes
MGnet
¼ U H ¼

up
^
j

Aup
^
j þ Cus
^
k

(9.100)
where A and C are the moments of inertia of the rotor about the x- and z-axes, respectively. Setting
Cus
^
k ¼ Hs, the spin angular momentum, and up
^
j ¼ up, we obtain
MGnet
¼ up Hs

Hs ¼ Cus
^
k

(9.101)
Since the center of mass is the reference point, there is no restriction on the motion G for which Eqn
(9.101) is valid. Observe that the net gyroscopic moment MGnet
exerted on the rotor by its supports is
90
ω
180
–6000
–3000
0
6000
45
50
55
θ (°)
180
90
0
0
3000
p (rpm) ωp (rpm)
θ (°)
45 135 45 135
(a) (b)
FIGURE 9.20
(a) High-energy precession rate (unlikely to be observed). (b) Low energy precession rate (the one most always
seen).

perpendicular to the plane of the spin and the precession vectors. If a spinning rotor is forced to
precess, the gyroscopic moment MGnet
develops. Or, if a moment is applied normal to the spin axis of a
rotor, it will precess so as to cause the spin axis to turn toward the moment axis.
EXAMPLE 9.16
A uniform cylinder of radius r, length L, and mass m spins at a constant angular velocity us. It rests on simple
supports (which cannot exert couples), mounted on a platform that rotates at an angular velocity of up. Find the
reactions at A and B. Neglect the weight (i.e., calculate the reactions due just to the gyroscopic effects).
Solution
The net vertical force on the cylinder is zero, so the reactions at each end must be equal and opposite in direction,
as shown on the free-body diagram insert in Figure 9.22. Noting that the moment of inertia of a uniform cylinder
about it axis of rotational symmetry is 1
2mr2, Eqn (9.101) yields
RL^
i ¼

up
^
j

1
2
mr2
us
^
k

¼
1
2
mr2
up us
^
i
so that
R ¼
mr2upus
2L
ω
ω
FIGURE 9.22
Illustration of the gyroscopic effect.
ω
ω
FIGURE 9.21
A spinning rotor on a rotating platform.

9.9 Euler angles
Three angles are required to specify the orientation of a rigid body relative to an inertial frame.
The choice is not unique, but there are two sets in common use: the Euler angles and the yaw,
pitch, and roll angles. We will discuss each of them in turn. The reader is urged to review Section
4.5 on orthogonal coordinate transformations and, in particular, the discussion of Euler angle
sequences.
The three Euler angles f, q, and j shown in Figure 9.23 give the orientation of a body-
fixed xyz frame of reference relative to the XYZ inertial frame of reference. The xyz frame is
obtained from the XYZ frame by a sequence of rotations through each of the Euler angles in
turn. The first rotation is around the Z (¼ z1) axis through the precession angle f. This takes
X into x1 and Y into y1. The second rotation is around the x2 (¼ x1) axis through the
nutation angle q. This carries y1 and z1 into y2 and z2, respectively. The third and final
rotation is around the z (¼ z2) axis through the spin angle j, which takes x2 into x and y2
into y.
The matrix [Q]Xx of the transformation from the inertial frame to the body-fixed frame is given by
the classical Euler angle sequence (Eqn (4.37)):
½QXx ¼ ½R3ðjÞ½R1ðqÞ½R3ðfÞ (9.102)
2
1
3
3
2
1
X
Y
x1 x2
,
Z z1
,
y1
x
y
y2
z2, z
φ
φ
φ
θ ψ
ψ
ψ
θ
θ
FIGURE 9.23
Classical Euler angle sequence. See also Figure 4.14.

From Eqns (4.32) and (4.34), we have
½R3ðjÞ ¼
2
6
6
6
4
cos j sin j 0
sin j cos j 0
0 0 1
3
7
7
7
5
½R1ðqÞ ¼
2
6
6
6
4
1 0 0
0 cos q sin q
0 sin q cos q
3
7
7
7
5
½R3ðfÞ ¼
2
6
6
6
4
cos f sin f 0
sin f cos f 0
0 0 1
3
7
7
7
5
(9.103)
According to Eqn (4.38), the direction cosine matrix is
½QXx ¼
2
6
6
4
sin f cos q sin j þ cos f cos j cos f cos q sin j þ sin f cos j sin q sin j
sin f cos q cos g cos f sin j cos f cos q cos j sin f sin j sin q cos j
sin f sin q cos f sin q cos q
3
7
7
5
(9.104)
Since this is an orthogonal matrix, the inverse transformation from xyz to XYZ is ½QxX ¼ ð½QXxÞT
,
½QxX ¼
2
6
6
4
sin f cos q sin j þ cos f cos j sin f cos q cos g cos f sin j sin f sin q
cos f cos q sin j þ sin f cos j cos f cos q cos j sin f sin j cos f sin q
sin q sin j sin q cos j cos q
3
7
7
5
(9.105)
Algorithm 4.3 is used to find the three Euler angles q, f, and j from a given direction cosine
matrix [Q]Xx.
EXAMPLE 9.17
The direction cosine matrix of an orthogonal transformation from XYZ to xyz is
½Q ¼
2
6
6
4
0:32175 0:89930 0:29620
0:57791 0:061275 0:81380
0:75000 0:43301 0:5000
3
7
7
5
Use Algorithm 4.3 to find the Euler angles f, q, and j for this transformation.
Solution
Step1 (precession angle):
f ¼ tan1

Q31
Q32

¼ tan1

0:75000
0:43301

ð0 f 360
Þ
9.9 Euler angles 511

Since the numerator is negative and the denominator is positive, the angle f lies in the fourth quadrant.
f ¼ tan1
ð1:7320Þ ¼ 300
Step 2 (nutation angle):
q ¼ cos1
Q33 ¼ cos1
ð0:5000Þ ð0 q 180
Þ
q ¼ 120
Step 3 (spin angle):
j ¼ tan1Q13
Q23
¼ tan1

0:29620
0:81380

ð0 j 360
Þ
Since both the numerator and denominator are negative, the angle j lies in the third quadrant.
j ¼ tan1
ð0:36397Þ ¼ 200
The time rates of change of the Euler angles f, q, and j are, respectively, the precession rate up, the
nutation rate un, and the spin us. That is,
up ¼ _
f un ¼ _
q us ¼ _
j (9.106)
The absolute angular velocity u of a rigid body can be resolved into components ux, uy, and uz along
the body-fixed xyz axes, so that
u ¼ ux
^
i þ uy
^
j þ uz
^
k (9.107)
Figure 9.23 shows that precession is measured around the inertial Z-axis (unit vector ^
K); nutation is
measured around the intermediate x1-axis (node line) with unit vector^
i1; and spin is measured around
the body-fixed z-axis (unit vector ^
k). Therefore, the absolute angular velocity can alternatively be
written in terms of the nonorthogonal Euler angle rates as
u ¼ up
^
K þ un
^
i1 þ us
^
k (9.108)
In order to find the relationship between the body rates ux, uy, and uz and the Euler angle rates up, un,
and us, we must express ^
K and ^
i1 in terms of the unit vectors ^
i^
j^
k of the body-fixed frame.
To accomplish that, we proceed as follows.
The first rotation ½R3ðfÞ in Eqn (9.102) rotates the unit vectors ^
I^
J^
K of the inertial frame into the
unit vectors^
i1
^
j1
^
k1 of the intermediate x1y1z1 axes in Figure 9.23. Hence^
i1
^
j1
^
k1 are rotated into ^
I^
J^
K by
the inverse transformation given by
8

:
^
I
^
J
^
K
9

=

;
¼ ½R3ðfÞT
8

:
^
i1
^
j1
^
k1
9

=

;
¼
2
6
6
4
cos f sin f 0
sin f cos f 0
0 0 1
3
7
7
5
8

:
^
i1
^
j1
^
k1
9

=

;
(9.109)

The second rotation [R1(q)] rotates^
i1
^
j1
^
k1 into the unit vectors^
i2
^
j2
^
k2 of the second intermediate frame
x2y2z2 in Figure 9.23. The inverse transformation rotates ^
i2
^
j2
^
k2 back into ^
i1
^
j1
^
k1:
8

:
^
i1
^
j1
^
k1
9

=

;
¼ ½R1ðqÞT
8

:
^
i2
^
j2
^
k2
9

=

;
¼
2
6
6
4
1 0 0
0 cos q sin q
0 sin q cos q
3
7
7
5
8

:
^
i2
^
j2
^
k2
9

=

;
(9.110)
Finally, the third rotation [R3(j)] rotates ^
i2
^
j2
^
k2 into ^
i^
j^
k, the target unit vectors of the body-fixed xyz
frame. ^
i2
^
j2
^
k2 are obtained from ^
i^
j^
k by the reverse rotation,
8

:
^
i2
^
j2
^
k2
9

=

;
¼ ½R3ðjÞT
8

:
^
i
^
j
^
k
9

=

;
¼
2
6
6
4
cos j sin j 0
sin j cos j 0
0 0 1
3
7
7
5
8

:
^
i
^
j
^
k
9

=

;
(9.111)
From Eqns (9.109), (9.110) and (9.111), we observe that
^
K
z}|{
9:109
¼ ^
k1
z}|{
9:110
¼ sinq^
j2 þ cosq^
k2
z}|{
9:111
¼ sinq

sinj^
i þ cosj^
j

þ cosq^
k
or
^
K ¼ sin q sin j^
i þ sin q cos j^
j þ cos q^
k (9.112)
Similarly, Eqns (9.110) and (9.111) imply that
^
i1 ¼ ^
i2 ¼ cos j^
i sin j^
j (9.113)
Substituting Eqns (9.112) and (9.113) into Eqn (9.108) yields
u ¼ up

sin q sin j^
i þ sin q cos j^
j þ cos q^
k

þ un

cos j^
i sin j^
j

þ us
^
k
or
u ¼

up sin q sin j þ un cos j

^
i þ

up sin q cos j un sin j

^
j þ

us þ up cos q

^
k (9.114)
Comparing Eqns (9.107) and (9.114), we see that
ux ¼ up sin q sin j þ un cos j
uy ¼ up sin q cos j un sin j
uz ¼ us þ up cos q
(9.115)

(Notice that the precession angle f does not appear.) We can solve these three equations to obtain the
Euler rates in terms of ux, uy, and uz:
up ¼ _
f ¼
1
sin q

ux sin j þ uy cos j

un ¼ _
q ¼ ux cos j uy sin j
us ¼ _
j ¼
1
tan q

ux sin j þ uy cos j

þ uz
(9.116)
Observe that if ux, uy, and uz are given functions of time, found by solving Euler’s equations of motion
(Eqn (9.72)), then Eqns (9.116) are three coupled differential equations that may be solved to obtain
the three time-dependent Euler angles, namely
f ¼ fðtÞ q ¼ qðtÞ j ¼ jðtÞ
With this solution, the orientation of the xyz frame, and hence the body to which it is attached, is known
for any given time t. Note, however, that Eqns (9.116) “blow up” when q ¼ 0, that is, when the xy plane
is parallel to the XY plane.
EXAMPLE 9.18
At a given instant, the unit vectors of a body frame are
^
i ¼ 0:40825^
I 0:40825^
J þ 0:81649^
K
^
j ¼ 0:10102^
I 0:90914^
J 0:40405^
K
^
k ¼ 0:90726^
I þ 0:082479^
J 0:41240^
K
(a)
and the angular velocity is
u ¼ 3:1^
I þ 2:5^
J þ 1:7^
K ðrad=sÞ (b)
Calculate up, un, and us (the precession, nutation, and spin rates) at this instant.
Solution
We will ultimately use Eqn (9.116) to find up, un, and us. To do so we must first obtain the Euler angles f, q, and j
as well as the components of the angular velocity in the body frame.
The three rows of the direction cosine matrix [Q]Xx comprise the components of the unit vectors ^
i, ^
j, and ^
k,
respectively,
½QXx ¼
2
6
6
4
0:40825 0:40825 0:81649
0:10102 0:90914 0:40405
0:90726 0:082479 0:41240
3
7
7
5 (c)
Therefore, the components of the angular velocity in the body frame are
fugx ¼ ½QXx fugX ¼
2
6
6
4
0:40825 0:40825 0:81649
0:10102 0:90914 0:40405
0:90726 0:082479 0:41240
3
7
7
5
8

:
3:1
2:5
1:7
9

=

;
¼
8

:
0:89817
2:6466
3:3074
9

=

;

or
ux ¼ 0:89817 rad=s uy ¼ 2:6466 rad=s uz ¼ 3:3074 rad=s (d)
To obtain the Euler angles f, q, and j from the direction cosine matrix in Eqn (c), we use Algorithm 4.3, as
illustrated in Example 9.17. That algorithm is implemented as the MATLAB function dcm_to_Euler.m
in Appendix D.20. Typing the following lines in the MATLAB Command Window:
Q ¼ [ .40825 -.40825 .81649
-.10102 -.90914 -.40405
.90726 .082479 -.41240];
[phi theta psi] ¼ dcm_to_euler(Q)
produces the following output:
phi ¼
95.1945
theta ¼
114.3557
psi ¼
116.3291
Substituting q ¼ 114.36 and j ¼ 116.33 together with the angular velocities of Eqn (d) into Eqn (9.116)
yields
up ¼
1
sin 114:36
½ 0:89817$sin 116:33
þ ð2:6466Þ$cos 116:33
¼ 0:40492 rad=s
un ¼ 0:89817$cos 116:33
ð2:6466Þ$sin 116:33
¼ 2:7704 rad=s
us ¼
1
tan 114:36
½ 0:89817$sin 116:33
þ ð2:6466Þ$cos 116:33
þ ð3:3074Þ ¼ 3:1404 rad=s
EXAMPLE 9.19
The mass moments of inertia of a body about the principal body frame axes with origin at the center of mass
G are
A ¼ 1000 kg$m2
B ¼ 2000 kg$m2
C ¼ 3000 kg$m2
(a)
The Euler angles in radians are given as functions of time in seconds as follows:
f ¼ 2te0:05t
q ¼ 0:02 þ 0:3 sin 0:25t
j ¼ 0:6t
(b)
At t ¼ 10 s, find
(a) The net moment about G and
(b) The components aX, aY, and aZ of the absolute angular acceleration in the inertial frame.

Solution
(a) We must use Euler’s equations (Eqns (9.72)) to calculate the net moment, which means we must first obtain
ux, uy, uz, _
ux , _
uy , and _
uz . Since we are given the Euler angles as a function of time, we can compute their time
derivatives and then use Eqn (9.115) to find the body frame angular velocity components and their derivatives.
Starting with Eqn (b), we get
up ¼
df
dt
¼
d
dt

2te0:05t

¼ 2e0:05t
0:1te0:05t
_
up ¼
dup
dt
¼
d
dt

2e0:05t
0:1e0:05t

¼ 0:2e0:05t
þ 0:005te0:05t
Proceeding to the remaining two Euler angles leads to
un ¼
dq
dt
¼
d
dt
ð0:02 þ 0:3 sin 0:25tÞ ¼ 0:075 cos 0:25t
_
un ¼
dun
dt
¼
d
dt
ð0:075 cos 0:25tÞ ¼ 0:01875 sin 0:25t
us ¼
dj
dt
¼
d
dt
ð0:6tÞ ¼ 0:6
_
us ¼
dus
dt
¼ 0
Evaluating all these quantities, including those in Eqn (b), at t ¼ 10 s yields
f ¼ 335:03 up ¼ 0:60653 rad=s _
up ¼ 0:09098 rad=s2
q ¼ 11:433 un ¼ 0:060086 rad=s _
un ¼ 0:011221 rad=s2
j ¼ 343:77 us ¼ 0:6 rad=s _
us ¼ 0
(c)
Equation (9.115) relates the Euler angle rates to the angular velocity components,
ux ¼ up sin q sin j þ un cos j
uy ¼ up sin q cos j un sin j
uz ¼ us þ up cos q
(d)
Taking the time derivative of each of these equations in turn leads to the following three equations:
_
ux ¼ upun cos q sin j þ upus sin q cos j unus sin j þ _
up sin q sin j þ _
un cos j
_
uy ¼ upun cos q cos j up us sin q sin j unus cos j þ _
up sin q cos j _
un sin j
_
uz ¼ upun sin q þ _
up cos q þ _
us
(e)
Substituting the data in Eqn (c) into Eqns (d) and (e) yields
ux ¼ 0:091286 rad=s uy ¼ 0:098649 rad=s uz ¼ 1:1945 rad=s
_
ux ¼ 0:063435 rad=s2
_
uy ¼ 2:2346 105 rad=s2
_
uz ¼ 0:08195 rad=s2
(f)
With Eqns (a) and (f) we have everything we need for Euler’s equations, namely,
Mxnet
¼ A _
ux þ ðC BÞuy uz
Mynet
¼ B _
uy þ ðA CÞuz ux
Mznet
¼ C _
uz þ ðB AÞux uy

from which we find
Mxnet
¼ 181:27 N$m
Mynet
¼ 218:12 N$m
Mznet
¼ 254:86 N$m
(b) Since the comoving xyz frame is a body frame, rigidly attached to the solid, we know from Eqn (9.74) that
8

:
aX
aY
aZ
9

=

;
¼ ½QxX
8

:
_
ux
_
uy
_
uz
9

=

;
(g)
In other words, the absolute angular acceleration and the relative angular acceleration of the body are the
same. All we have to do is project the components of relative acceleration in Eqn (f) onto the axes of the
inertial frame. The required orthogonal transformation matrix is given in Eqn (9.105),
½QxX ¼
2
6
6
4
sin f cos q sin j þ cos f cos j sin f cos q cos g cos f sin j sin f sin q
cos f cos q sin j þ sin f cos j cos f cos q cos j sin f sin j cos f sin q
sin q sin j sin q cos j cos q
3
7
7
5
Upon substituting the numerical values of the Euler angles from Eqn (c), this becomes
½QxX ¼
2
6
6
4
0:75484 0:65055 0:083668
0:65356 0:73523 0:17970
0:055386 0:19033 0:98016
3
7
7
5
Substituting this and the relative angular velocity rates from Eqn (f) into Eqn (g) yields
8

:
aX
aY
aZ
9

=

;
¼
2
6
6
4
0:75484 0:65055 0:083668
0:65356 0:73523 0:17970
0:055386 0:19033 0:98016
3
7
7
5
8

:
0:063435
2:2345 105
0:08195
9

=

;
¼
8

:
0:054755
0:026716
0:083833
9

=

;

rad=s2

EXAMPLE 9.20
Figure 9.24 shows a rotating platform on which is mounted a rectangular parallelepiped shaft (with dimensions
b, h, and [) spinning about the inclined axis DE. If the mass of the shaft is m, and the angular velocities up and
us are constant, calculate the bearing forces at D and E as a function of f and j. Neglect gravity, since we are
interested only in the gyroscopic forces. (The small extensions shown at each end of the parallelepiped are just for
clarity; the distance between the bearings at D and E is [.)
Solution
The inertial XYZ frame is centered at O on the platform, and it is right handed ð^
I ^
J ¼ ^
KÞ. The origin of the right-
handed comoving body frame xyz is at the shaft’s center of mass G, and it is aligned with the symmetry axes of the

parallelepiped. The three Euler angles f, q, and j are shown in Figure 9.24. Since q is constant, the nutation rate is
zero (un ¼ 0). Thus, Eqn (9.115) reduces to
ux ¼ up sin q sin j uy ¼ up sin q cos j uz ¼ up cos q þ us (a)
Since up, us, and q are constant, it follows (recalling Eqn (9.106)) that
_
ux ¼ up us sin q cos j _
uy ¼ up us sin q sin j _
uz ¼ 0 (b)
The principal moments of inertia of the parallelepiped are (Figure 9.10(c))
A ¼ ‘x ¼
1
12
m

h2
þ ‘2

B ¼ ‘y ¼
1
12
m

b2
þ ‘2

C ¼ ‘z ¼
1
12
m

b2
þ h2

(c)
Figure 9.25 is a free-body diagram of the shaft. Let us assume that the bearings at D and E are such as to exert just
the six body frame components of force shown. Thus, D is a thrust bearing to which the axial torque TD is applied
from, say, a motor of some kind. At E, there is a simple journal bearing.
D
X
Y
Z
y
z
x
O
E
(Measured in XY plane)
(Measured in Zz plane)
Shaft
Platform
˙
p
˙ s
x
x
(Measured in x'x plane,
perpendicular to shaft)
xyz axes attached
to the shaft
l/2
G
l/2
b h
φ
φ
θ
ψ ψ
ω
ω
=
=
′
′
FIGURE 9.24
Spinning block mounted on a rotating platform.
G
x
y
z
b
h
TD
Dx
Dz
Dy
Ey
Ex
l/2
l/2
FIGURE 9.25
Free-body diagram of the block in Figure 9.24.

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