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Engineering Dynamics: A
Comprehensive Introduction—Errata
N. Jeremy Kasdin & Derek A. Paley
Last updated November 14, 2014
PRINCETON UNIVERSITY PRESS
PRINCETON AND OXFORD
1
Chapter 1:
Chapter 2:
• In tutorial 2.4, the solution for the critically damped case (starting at
the bottom of page 33) solves for λ1 , λ2 = −ζω0 . This is propagated
through to the solution (Eq. (2.26)), which makes it appear as if ζ is
a variable in the solution. This is misleading as ζ can only equal 1 in
this case. The equation should be rewritten without ζ.
• In problem 2.16, the downward acceleration should be g −
c 2
m y˙ .
Chapter 3:
• In Section 3.4.1, the final equation defining the transformation matrix
should be in terms of the scalar component magnitudes, not the unit
vectors. That is, the elements of the two matrices should not be bold.
It should thus read:
b1
cos θ sin θ
a1
=
b2 B
− sin θ cos θ
a2 A
• In Section 3.8.4 on page 93 describing the vector Taylor series the
vector (and components) were mistakenly written at t+a rather than t.
Thus, the sentence before Eq. (3.68), that equation, and the following
two equations should be written:
First, write the vector r(t) as components in a frame A = (O, a1 , a2 , a3 ):
r(t) = r1 (t)a1 + r2 (t)a2 + r3 (t)a3 .
(1)
Then take the Taylor series of the scalar magnitude of each component.
For i = 1, 2, 3, we have
1 d2 d (t − a) +
ri
(t − a)2 + . . .
ri (t) = ri (a) + ri dt t=a
2! dt2 t=a
Expanding and rearranging each coefficient in Eq. (3.68) yields
r(t) = r1 (a)a1 + r2 (a)a2 + r3 (a)a3
d d d + r1 (t − a)a1 + r2 (t − a)a2 + r3 (t − a)a3
dt t=a
dt t=a
dt t=a
1 d2 1 d2 1 d2 2
2
+
r1
(t − a) a1 +
r2
(t − a) a2 +
r3
2! dt2 2! dt2 2! dt2 t=a
t=a
t=a
+...
• In Problem 3.2,
√ the magnitudes of the ey components of F1 and F3
must √sum to 96 for the problem to have a solution. Thus, F3 =
15−12 6
ey N.
3
(t − a)2 a3
2
• Figure 3.45 is missing a gravity vector. (Gravity acts down.)
• Problem 3.30. Assume that the hill’s center of curvature is 50 m below
the surrounding ground level.
Chapter 4:
• In Example 4.1 on pgs. 115-116, the expressions for x(t) and x(t)
˙
should be written relative to the initial time t2 ,
−F¯P (t1 , t2 )
sin [ω0 (t − t2 )]
mP ω0
−F¯P (t1 , t2 )
x(t)
˙ =
cos [ω0 (t − t2 )]
mP
x(t) =
• On page 138, the first unnumbered equation should not have a minus
sign.
• In tutorial 4.3, the moment expressions on page 141 should be divided
by the satellite mass to correspond to the change in specific angular
momentum.
• In Problem 4.11 on page 146, the feedback equation for the base motion
should be for the acceleration of the base, u
¨(t):
u
¨(t) = −
g sin θ + kθ + bθ˙
cos θ
Chapter 5:
• In Problem 5.14, the value given for C in part (b) is actually the square
root of the correct value. C should equal 0.33.
• Top of page 155 should not be gray box.
• On page 163, Example 5.8, the second equation has a typo; the sign
is wrong on the last term. It should read:
T b1 − µc N b1 + N b2 − mP g sin θb1 − mP g cos θb2 = 0
• Figure 5.19, the height of the particle should be same in both figures.
Chapter 6:
(ext)
(ext)
(t1 , t2 ) instead of F
. The follow• Equation 6.10 should use F
(ext)
ing line should read “where F
(t1 , t2 ) is the total external impulse
acting on the system.”
3
• Equation 6.23 on page 203 should be:
B d2
dt2
r1/2 = B a1/G − B a2/G
=
−k(m1 + m2 )
kr1/2 k − l0 ˆr1/2 .
m1 m2
(ext)
• On page 235, the second bullet of the Key Ideas should use F
(t1 , t2 )
(ext)
(ext)
instead of F
. The following line should read “where F
(t1 , t2 )
is the total external impulse.”
• In Problem 6.7 on p. 239, the summation index should be i not j.
• In Problem 6.23 on p. 244, the values for the mass flow rate and
exhaust velocity should be m
˙ = 0.5 kg/s and ue = 2.75 km/s.
• In Problem 6.24 on p. 244, the value for the spring rest length, l0
should be 1 m, and the value for the damper constant, b should be
25 N·s/m. The integration should be initialized with the spring at its
rest length.
• In Problem 6.25 on p. 244, the numerical values should be as follows:
the rocket weighs 2 kg and carries 3 kg of fuel, the mass flow rate is
0.1 kg/s, and the exhaust velocity is 500 m/s.
Chapter 7:
• In Example 7.8 on p. 265 there is a typo in the expression for TG/O .
The square should be on y˙ G not outside the parenthesis,
1
1
2
).
TG/O = mG kI vG/O k2 = mG (x˙ 2G + y˙ G
2
2
• In Example 7.11 on p. 273, the unit vectors of the polar frame have
been defined as er and eθ so the b1 and b2 are incorrect. The equations
for rP/G and I vP/G should be written
r
rP/G = − er
2
r˙
r˙
I
vP/G = − er − θe
θ.
2
2
Likewise, in the sentence that follows the unit vector b1 should be
replaced by er .
In this same example, the final conditions in the last paragraph are not
correct. The rotation rate at infinity is θ˙ = 2hG /r02 . The expression
for the nonconservative work is thus
mh2G m
4h2G
1
(nc,int)
2
W
= 2 −
r(0)
˙
+
− k(r(0) − r0 ).
2
4
r(0)
2
r0
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
4
I
m2
I
e2
G
eθ
r2/O
r1/O
G θ
m1
e1
e3
B
r1/2
rG/O
O
275
r
(a)
er
(b)
Figure 7.10 Two bodies of masses m1 and m2 in orbit about their common center of
mass in the plane orthogonal to e3. (a) Inertial frame I = (O, e1, e2 , e3). (b) Polar frame
B = (G, er , eθ , e3).
Figure 1 Annotated correction to Figure 7.10(b)
•
•
also perhaps different from our intuitive image of the earth orbiting the sun. In fact,
the earth and sun orbit their common center of mass; because the sun is so much
larger than the earth, the center of mass of the two bodies is actually inside the sun.
3 (The same
Consequently,
the sun’s
motion isthe
very small.
of the clear.
earth’s motion
On p.
275, Figure
7.10(b),
definition
of isθtrue
is not
It is defined
when
it
is
orbited
by
a
small
satellite.)
Nevertheless,
it
is
very
common
and
useful
as the rotation angle of the line between the particles relative to the
to find a description of the earth’s motion relative to the sun (or, to be more general,
inertial
should
body frame.
1 relative toThe
body figure
2) given by
r1/2(t). be corrected as annotated here.
Using the inertial frame I = (O, e1, e2, e3), we write the dynamics of body 1
relative
to body
We start
with the definition
of the
center of mass,
On p.
277
the 2.final
equation
for the
angular
momentum relative to
the center of mass is incorrect.
Theremshould
not be a 2 multiplying
m1
2
r2/O ,
1/O +
G/O =correct requation
the reduced mass. rThe
is:
m1 + m 2
m1 + m 2
! "# $
! "# $
!
m1 m2=µ
2
˙ 3,
hG =
r2 θe
m
+
m
!
1 masses
2 µ =
where we have introduced the dimensionless
m /(m
I
!
=µ1
!
and µ2 =
m2/(m1 + m2). Note µ1 + µ2 = 1 and m1 + m2 = mG. Rearranging and using the
vector
r2/1 + r1/O , we find
p.
277triad
inr2/O
the= description
following the above equation for I h
!
1
1
1 + m2 )
• On
G the
specific angularµmomentum
from
Tutorial
4.2
should
be
h
rather
than
O
1r1/O = rG/O − µ2 r2/O = rG/O − µ2 (r2/1 + r1/O )
hP .
=r
+µ r −µ r ,
G/O
2 1/2
2 1/O
• On p.
287,
the last bullet should use EO instead of E (four times).
which
means
r1/2. problem should read “speed
rG/O +ofµ2the
1/O =text
• On p. 290, Problem 7.7, rthe
v0 ” rather than velocity.
3 Scientists searching for a planet about another star infer the presence of a planet by measuring the (small)
motion of the
Chapter
8:star.
• on p. 298, Example 8.2 the equation at the bottom of the page should
be expressed in terms of inertial-frame unit vectors rather than bodyframe unit vectors:
Kasdin third pages 2010/12/9 6:33
p. 275 (chap07)
I
Princeton Editorial Assoc., PCA
ZzTEX 13.9
vO0 /O = −ωB e3 × (−Re2 ) = −RωB e1
Likewise, all the b unit vectors in the preceeding paragraph should be
e unit vectors.
5
• p. 313 The second equation is correct, but is perhaps more insightful
if written in terms of the fixed inertial velocity, I vP/O . That can be
done by adding an extra term to the equation:
B
aP/O = −2I ω B × B vP/O − I ω B × I ω B × rP/O = −2I ω B × I vP/O
• p. 315 In the second equation the velocity should be with respect to
frame B rather than I. The equation should thus be:
B
aP/L = −
fl
ˆr
− 2ωr b3 × B vP/L − ωr2 b3 × (b3 × rP/L ).
mP P/L
• Kindle Edition Only Problem 8.1(a), I vP/O0 should be I vP/O .
• p. 330, Problem 8.3, I ω A should be negative to be consistent with the
handedness of the rotation as drawn. Thus, I ω A = −2 rpm.
• p. 333, Problem 8.10, disk radius should be R = 1 m.
Chapter 9:
• On p. 365, the phrase “implying that I aP/O = 0 in Eq. (9.30)” should
read “. Although I aP/O 6= 0, it is perpendicular to the ramp, which
implies the cross product rP/G × I aP/O in Eq. (9.30) is zero.”
• On p. 384, the sentence before Eq. (9.61) should read “...two unknowns
θ˙ and I ω B .”
• On p. 389, all four vector expressions for the moment about the center
of mass, MG , are missing the unit vector in the b3 direction.
• On p. 399, Problem 9.5. In figure, the distance l should go to center
of bar, not the edge. The arrow on θ should be reversed.
• On p. 401, Problem 9.7, the area moment of inertia should be J =
a4 /12.
• On p. 405, Problem 9.19 the equation of motion for the motor has an
incorrect sign on the motor torque. It should read:
IG θ¨ = −bθ˙ + ki.
Also, in text for part b it should read “the inductance L to 0.75 H”
rather than h.
Chapter 10:
• On p. 418, the second-to-last equation should start “β¨ − θ˙2 ...”
6
• On
p p. 422 the third unnumbered equation from the top should have
x2 + y 2 in the denominator (two times).
• On p. 439, in Eq. (10.50) and in the following unnumbered set of three
equations, csc θ should be sec θ.
• On p. 430, the description of the elements of the direction cosine matrix
are transposed and the lack of superscripts creates an ambiguity. The
penultimate sentence in the paragraph following Eq. (10.37) should
thus read:
B =
The elements of the direction cosine matrix I C B are given by I Cij
ei ·bj , i, j = 1, 2, 3 (the transpose of the elements of the transformation
table in Eq. (10.33)).
• On pp. 448-449, as with the previous errata, the elements of the direction cosine matrix are transposed. The second and third paragraphs,
including equations, should therefore read:
We can now use Eq. (10.60) to find expressions for the elements of the
direction cosine matrix in terms of the component magnitudes of the
unit vector k = k1 e1 +k2 e2 +k3 e3 and the rotation angle θ. Recall from
Section 10.4.1 that the elements of the direction cosine matrix, I C B ,
between reference frame I = (O, e1 , e2 , e3 ) and B = (O, b1 , b2 , b3 ) are
given by
I
B
Cij
= ei · bj
i, j = 1, 2, 3.
Using the fact that bj = ej prior to the rotation, we can substitute
for bj from Eq. (10.60) to obtain
I
B
Cij
= ei · (ej cos θ − (ej × k) sin θ + (ej · k)k(1 − cos θ)) .
Using the component magnitudes of k in I (or B) allows us to compute the following expressions for each element of the transformation
matrix:
I
B
C11
= cos θ + k12 (1 − cos θ)
I B
C12 = −k3 sin θ + k1 k2 (1 − cos θ)
I B
C13 = k2 sin θ + k1 k3 (1 − cos θ)
I B
C21 = k3 sin θ + k1 k2 (1 − cos θ)
I B
C22 = cos θ + k22 (1 − cos θ)
I B
C23 = −k1 sin θ + k2 k3 (1 − cos θ)
I B
C31 = −k2 sin θ + k3 k1 (1 − cos θ)
I B
C32 = k1 sin θ + k3 k2 (1 − cos θ)
I B
C33 = cos θ + k32 (1 − cos θ).
7
As we expected, we can write the transformation matrix either in terms
of the three Euler angles describing the orientation of B in I (as in
Eq. (10.37)) or in terms of the components of the unit vector k along
the Euler axis of rotation and the rotation angle θ. Either set is a
valid description of the orientation of B in I.
There is something even more important about these relationships.
We can use them to prove Euler’s theorem. Suppose that the rigid
body (reference frame B) has some arbitrary orientation in I. This
orientation could be described by the Euler angles, (ψ, θ, φ)IB , as in
the previous section. And as in Section 10.4.1, we can write the transformation matrix in terms of these angles. The inverse of the above
relationships then provides expressions for the components of the unit
vector k along the Euler axis of rotation and the rotation angle in
terms of the components of this matrix:
1
B
B
B 1/2
1 + I C11
+ I C22
+ I C33
2
I CB − I CB
32
23
k1 =
2 sin θ
I CB − I CB
13
31
k2 =
2 sin θ
I CB − I CB
21
12
k3 =
.
2 sin θ
θ = 2 cos−1
(2)
(3)
(4)
(5)
Eqs. (10.63)-(10.66) constitute a proof of Euler’s theorem! In other
words, given any arbitrary orientation and the transformation matrix
(or Euler angles) describing it, we can always find a single axis of
rotation and angle about that axis corresponding to the orientation
by using Eqs. (2)–(5).
• On p. 449, Section 10.5.2 contains a derivation of the angular velocity
vector as the instantaneous Euler axis. While technically correct, it
lacks rigor and can be difficult to understand. Below is an alternative
derivation.
Finally, we turn our attention back to the angular velocity. Suppose
now that the rigid body (frame B) is changing orientation with time
in I. We have already shown that the rate of change of vectors with
respect to I can be found using using the transport equation; that
is, the angular velocity acts as an operator to provide derivatives.
But what does the angular velocity physically represent? For simple
rotations, we showed in Chapters 3 and 8 that the angular velocity
is directed along the axis of rotation and its magnitude is the rate of
rotation. Here we show, using Euler’s theorem and the corollary in
8
Eq. (10.60), that in three dimensions the angular velocity is directed
along the instantaneous axis of rotation.
If the orientation of B is changing with time, then over any small
time interval δt that change could be represented by an Euler axis
rotation of an amount δθ from the orientation of B at time t to a new
orientation at time t + δt. We can thus use Eq. (10.60) to represent
the change in each unit vector of B. That is, consider one of the unit
vectors defining frame B at time t, bi (t). At time t + δt we can write
bi (t) using Eq. (10.60) as
bi (t + δt) = bi (t) cos δθ − (bi (t) × k) sin δθ + (bi · k)k(1 − cos δθ)
where k is the unit vector directed along the Euler axis of the small
rotation. Subtracting bi (t) from both sides leaves
bi (t+δt)−bi (t) = −bi (t)(1−cos δθ)−(bi (t)×k) sin δθ+(bi ·k)k(1−cos δθ).
(6)
We showed in section 3.3.2, using the definition of the vector derivative,
that the derivative of a vector is given by the limiting operation
Id
dt
bi = lim
δt→0
bi (t + δt) − bi (t)
δt
where it is understand that by this limit we mean the limit taken
individually on each of the components of bi in the inertial frame I.
Thus, from Eq. (6)
Id
dt
bi = lim
δt→0
1
[−bi (t)(1 − cos δθ) − (bi (t) × k) sin δθ + (bi · k)k(1 − cos δθ)] .
δt
We can take this limit by using a Taylor series expansion on sin δθ and
cos δθ or by using L’Hopital’s rule. In either case we find
Id
δθ
˙ × bi
bi = lim
k × bi = θk
δt→0 δt
dt
where we used the definition of the scalar derivative (see Appendix A).
We know from earlier in the chapter, however, that
Id
dt
bi = I ω B × bi
for the unit vectors of B. These two equations tell us that the angular
velocity is given by
I
˙
ω B = θk.
The angular velocity is thus directed along the instantaneous Euler
axis and is equal in magnitude to the instantaneous rate of rotation
9
about that axis. The angular velocity represents the rotation rate about
an instantaneous axis of rotation parallel to it.
• On p. 462, Problem 10.7, change “when it reaches the bottom of the
funnel” to “when it reaches a height of h/2.”
• On p. 463, Problem 10.12 is worded incorrectly. The C130 measures
and telemeters to the ground the relative position and velocity of the
vehicle, rP/O0 and I vP/O0 . The proper wording is as follows:
Consider an experimental high-altitude vehicle dropped from a large
C-130 transport plane. The plane and vehicle are beyond the visible
range of the ground system. Sensors on the plane measure the position and velocity of the released vehicle relative to the C-130, rP/O0
and B vP/O0 . The C-130 telemeters this information to the ground observers as well as its own inertial velocity I vO0 /O , its angular velocity
in inertial space I ω B , and its inertial position rO0 /O , obtained from
its inertial navigation system and GPS sensors. The ground observers
must point their cameras and other equipment to pick up the experimental vehicle; thus they must know the vehicle’s velocity in inertial
space. Write down the vector expression they use to find the vehicle’s
velocity from the telemetered data.
Chapter 11:
• On
R p. 467, in the penultimateR line of Example 11.1, the equation m =
p B dV . . . should be m = ρ B dV . . .
• On p. 471, the first equation at the top of the page has a sign error.
It should read:
0 = −2mP l2 θ˙φ˙ cos φ − mP l2 θ¨ sin φ + hφ˙
This means that Eq. (11.5) also has a sign error. It should read:
θ¨ + 2θ˙φ˙ cot φ −
hφ˙
=0
mP l2 sin φ
• On p. 494, Eq. (11.45), csc θ should be sec θ.
• On p. 501, Example 11.13, equation at the bottom of the page the
leading I2 in the (2, 1) element of the first matrix should be deleted:


I10 θ¨
 d (I2 ψ˙ sin θ + h) 
dt
I10 (ψ¨ cos θ − ψ˙ θ˙ sin θ) B
10
12
ζ = 0.05
ζ = 0.1
ζ = 0.3
ζ = 0.5
ζ = 0.8
10
8
Stretch (m)
6
4
2
0
−2
−4
−6
−8
0
2
4
6
8
10
t (s)
12
14
16
18
20
Figure 2 Figure 12.4 Stretch in a 50 m bungee cord for five different damping ratios and
a natural frequency of 0.5 Hz.
• On p. 533, in Problem 11.19, the value given for D is actually the
value of D/IT , so that the problem should read D/IT = 0.5 . . .
Chapter 12:
• On p. 541, the solution given to the differential equation of motion
is not consistent with the initial conditions in the problem statement.
With x the distance from the unstretched length of the bungee cord,
the solution in the last equation on p. 541 is
!
p
2 − gζ
2glω
g
g
p 0
x(t) = 2 + e−ζω0 t
sin ωd t − 2 cos ωd t .
ω
ω0
ω02 1 − ζ 2
As a result, Fig. 12.4 needs to be corrected as shown below.
Alternatively, x could be measured from the platform (which is perhaps more consistent with Fig. 12.3), which would change the equation
of motion on p. 541 to
x
¨ + 2ζω0 x˙ + ω 2 (x − l) = g
where here the stretch in the cord is x − l. The solution then becomes
!
p
2 − gζ
2glω
g
g
p 0
sin ωd t − 2 cos ωd t .
x(t) = 2 + l + e−ζω0 t
ω
ω0
ω02 1 − ζ 2
11
25
ζ = 0.3
ζ = 0.4
ζ = 0.5
ζ = 0.6
ζ = 0.7
Stretch (m)
20
15
10
5
0
0
5
10
15
t (s)
Figure 3 Figure 12.4a Stretch in a 50 m bungee cord for five different damping ratios
and a natural frequency of 0.2 Hz.
The plot of cord stretch in Fig. 12.4 is the same for both interpretations.
Also, it is perhaps an unphysical model of a bungee cord to allow it to
have compression as shown in Fig. 12.4 using the parameters in the
book. Bungee cords only stretch, they do not compress. This can be
easily fixed by, for example, only examining slightly higher damping
ratios and a softer bungee cord with a natural frequency of 0.2 Hz.
The result is shown in Figure 12.4a.
• On p. 553, first of the two unnumbered equations before Eq. (12.19)
should be “y˙ 1 = y2 ”.
• On p. 558, the last unnumbered equation and the preceding sentence
should be “... from the initial conditions y(0) and y(0):
˙
y(0) = c1 v1 + c2 v2
y(0)
˙
= λ1 c1 v1 + λ2 c2 v2 .”
Chapter 13:
• On p. 583, bottom of the page, the expression for the holonomic constraint should be indexed from 1 to K rather than 1 to N :
fk (r1/O , r2/O , . . . , rN/O , t) = 0,
k = 1, . . . , K.
12
• On p. 591, Section 13.3.2, the indices in the double sum (i and j should
be reversed, i spans 1 to NC and j spans 1 to N :
N
X
j=1
(act)
Fj
I
· δrj/O =
NC X
N
X
i=1 j=1
|
(act)
Fj
·
{z
=Qi
I ∂r
j/O
∂qi
δqi ,
}
Appendix A:
Appendix B:
Appendix C:
• On p. 658, fifth line of text, “TSPAN” should be “TSPAN”
• On p. 658, seven lines from the bottom lowercase “l” should be uppercase “L”, two times.
Appendix D: