fundamentals of lagrange's mechanics....
TRANSCRIPT
Preface
I write this book, shortly because I love to do it. With this book, I would like to
share my own experience on Dynamics from education and researches for over
ten years with students and the people in the same field. The book is organized
and written from my viewpoints of dynamics, and it is appropriate for the one
who study dynamics in the intermediate level. Why written in English? It is about
the right time and right situation. When Chulalongkorn University started to
promote the faculties to carry out the class in English in the academic year of
2001, I joined the program and have an opportunity to teach the Advanced
Dynamics class in English. I started to write the first draft in English for the
class's lecture notes and continuingly improve it since then. It is the right situation
when we have the ME graduate foreign student attending the class in the Year
2005. Language is not the obstacle for communication. Instead good writing
communication needs a well-organized manuscript that indeed my book still has a
room for improvement.
My first experience in dynamics during the undergraduate years is not different
from everyone's experience in that we simply start with the Newton's 2nd Law, and
laws of energy and momentum. Taking the motion of a particle as an example,
both the Newton's Law and the principle of energy in dynamics have the same
root from the law of linear momentum. Later, when I was doing my Master
Degree, I learnt the whole new aspect of dynamics, namely, 3-D Dynamics,
Dynamic Model and Analysis, Derivation of Equations of Motion, Lagrange's
Mechanics, Stability and Rotordynamics. During the time for PhD, I got the first
lesson of dynamics there from my advisor. Not as a coursework requirement, he
kindly gave me the intensive lectures on dynamics and vibration of deformable
bodies such as plate and shell, so that I could have a necessary background to start
the research. Next I began to learn the Halmiton’s Principle and the Variational
Principle from several courseworks and from self-study. These two principles are
fundamentals of Lagrange's mechanics. Truly, My PhD research is the best lesson
of dynamics that I have learnt. At that time, I started to use Matlab as a program
tool for dynamic simulation and continue writing the Matlab codes nowadays.
Also, one of a good memory for Dynamics during those years is the opportunity
to attend the seminar "a New Paradigm of Dynamics" by a world famous
dynamist who develops the Kane’s method, Prof. Thomas Kane of Stanford
University.
There are two premium dynamists who are my role model. The first person is my
supervisor at the University of Melbourne, Dr. Januzt Krodkiewski. He is an icon
of the discipline and logic. The second one is my PhD advisor, Prof. Steve Shen.
He is an icon of making things simple (no matter how complicate they are). Both
of them similarly have an excellent background in Mathematics. I wish I could be
a half of the people that I admire. Therefore, it is to them that I dedicate this book.
Thitima Jintanawan
Chapter 1
Kinematics
In this chapter various coordinate systems, such as cartesian and cylindrical coordinates, are
introduced. Position vector, velocity and acceleration of particles and rigid bodies are for-
mulated using different reference coordinates. Each coordinate system is related to the other
through the coordinate transformation. For the 3-D transformation, two different sets of Eu-
ler angles: precession-nutation-spin and yaw-pitch-roll, are conventionally used. Finally the
transformation matrix used to describe a finite motion of rigid bodies is revealed.
1.1 Evolution of Kinematics
Prior to 1950s: Express velocity v and acceleration a in terms of scalar components and use
graphical method to determine total magnitude and direction
1950s and later: Express velocity v and acceleration a using vector approach
Recent years: Express the rotation with a matrix and utilize the matrix operation for calcu-
lating the cross product. The matrix approach can be simply implement in a computer
simulation program.
1
X
Y
Z
ij
k
rx
ry
r
rz
particle
moving path
O
Figure 1.1: A cartesian coordinate system
1.2 Position Vector, Velocity, and Acceleration
Fig. 1.1 shows a particle moving in a 3-dimensional (3-D) space. Let’s introduce a cartesian or
rectangular coordinate system XY Z as shown in Fig. 1.1 in which all coordinates are orthogonal
to each other and its axes do not change in direction. If we choose XY Z in Fig. 1.1 as an inertial
or fixed reference frame1, the absolute motion of the particle in Fig. 1.1 can be described by a
position vector r as follows
r = rxi+ ryj+ rzk (1.1)
where i, j, and k are the unit vectors of XY Z and rx, ry, and rz are scalar components of r
in X, Y , and Z coordinates. The position vector r can be alternatively presented in a matrix
form as a 3 × 1 column matrix given by
r = [ rx ry rz ]T (1.2)
Note that the position vector r must be measured from the origin O of the chosen inertial frame.
Figure 1.2 shows another set of coordinate system so called the cylindrical coordinates
ρθz, with their unit vectors eρeθez. In Fig. 1.2, the position vector r expressed in terms of
eρeθez is
r = ρeρ + zez (1.3)
The absolute velocity v is defined as a time derivative of the position vector r given by
v =drdt
(1.4)
= rxi+ ryj+ rzk
= ρeρ + ρθeθ + zez1An inertial or fixed reference frame is the coordinate system whose origin O is fixed in space
2
X
Y
Z
ρ
θ
z
eθ
eρ
r
ez
ρ
θ
z
Figure 1.2: A cylindrical coordinate system
The absolute acceleration a is defined as a time derivative of the velocity v given by
a =dvdt
(1.5)
= rxi+ ryj+ rzk
=(ρ − ρθ2
)eρ +
(ρθ + 2ρθ
)eθ + zez
1.3 Angular Velocity
Figure 1.3 shows a rigid cylinder having a rotation about n axis. The absolute angular velocity
ω of the rigid body is defined as
ω =dθ
dtn (1.6)
= ω1e1 + ω2e2 + ω3e3
where ω1, ω2, and ω3 are components of the angular velocity in an arbitrary rectangular co-
ordinate system with unit vectors e1, e2, and e3. The angular velocity can be expressed in a
matrix form as
ω =
0 −ω3 ω2
ω3 0 −ω1
−ω2 ω1 0
(1.7)
The velocity at point A in Fig. 1.3 is then
v = ω × r (1.8)
≡ ωr
3
v
n
r
θ(t)A e1
e2
e3
Figure 1.3: Angular velocity
Equation (1.9) indicates that the cross product can be represented by the matrix multiplication
or
v =
v1
v2
v3
=
0 −ω3 ω2
ω3 0 −ω1
−ω2 ω1 0
r1
r2
r3
(1.9)
Note that for the matrix multiplication in (1.9), components of ω and r must be expressed in
the same coordinate system.
1.4 Rate of Change of a Constant-Length Vector
The Theorem in the Vector of Calculus states that “The time derivative of a fixed length vector
c is given by the cross product of its rotation rate ω and the vector c itself.”
dcdt
= ω × c (1.10)
Example 1.1:
The defense jet plane as shown in Fig. 1.4 operates in a roll maneuver with rate of φ and
simultaneously possesses a yaw maneuver (turn to left) with a rate of ψ. Determine a relative
velocity of point C on the horizontal stabilizer at coordinates (b, a, 0), observed from the C.G.
of the plane.
Solution
From Fig. 1.4, the position vector of point C relative to the C.G. is a fixed length vector
4
X
Y
Z
G
C
ω
ez
ey
ex
Figure 1.4: A defense jet plane
given in terms of the body coordinate system as
r(rel)c = bex + aey = [ b a 0 ]T (1.11)
Hence the relative velocity of C is
v(rel)c = ω × r(rel)c ≡ ωr(rel)c (1.12)
where ω = φey + ψez is the rotation rate or the angular velocity of the reference coordinates
moving with the body. ω can be written in a matrix form as
ω =
0 −ψ φ
ψ 0 0
−φ 0 0
Therefore
v(rel)c =
0 −ψ φ
ψ 0 0
−φ 0 0
b
a
0
= [ −aψ bψ −bφ]T
As another example, the unit vectors ijk for any rotating system of coordinates xyz is
also the fixed length vector. Hence the rate of change of these ijk vectors can be determined
from the same theorem as i = ω × i, j = ω × j, and k = ω × k, where ω is the angular velocity
of such rotating coordinate system xyz.
5
X
Y
Z
x
z
yo
path of origin o
Figure 1.5: Translating coordinate systems
X
Y
Z
x
z
y
o
Figure 1.6: Rotating coordinate systems
1.5 Kinematics Relative To Moving Coordinate Systems
Any moving coordinate system xyz used to describe the motion can be divided into 3 types
depending on its motion with respect to the inertial frame XY Z. They are
1. Translating coordinate systems (Fig. 1.5)
2. Rotating coordinate systems (Fig. 1.6)
3. Translating and rotating coordinate systems (Fig. 1.7)
A moving coordinate system, chosen such that it is attached to a moving body, is nor-
mally used as a reference frame to describe kinematics of the body. Specifically, such reference
coordinate system is arranged such that its origin o is fixed to and translate with the body’s
C.G. and its axes synchronously rotate with the body.
6
X
Y
Z
x
z
y
o
path of origin o
Figure 1.7: Translating and rotating coordinate systems
X
Y
path of particle
path of moving reference frame
Z
ij
kr
ρ
ω
R
e1
e2
e3
x
y
z
O
O'
Figure 1.8: Moving coordinate systems
7
Fig. 1.8 shows a particle moving in 3-D space. XY Z is an inertial frame with unit vectors
ijk. Also xyz is the moving reference frame with unit vectors e1e2e3. If the angular velocity
of xyz is ω and the position vector r is
r = R+ ρ (1.13)
Then the velocity v of the particle is given by
v =drdt
(1.14)
=dRdt
+dρ
dt
= R+ vr + ω × ρ
In (1.15), R = dRdt is the velocity of the origin o′ of the reference frame xyz and ω is its angular
velocity. In addition vr, sometimes denoted by(dρdt
)rel
, is a relative velocity of the particle
with respect to xyz or the relative velocity observed by the observer moving (bothe translating
and rotating) with xyz, whereas dρdt is the relative velocity observed by the observer who only
translates but not rotates with xyz. ω × ρ in (1.15) is hence the difference between these two
relative velocities. If ρ is
ρ = ρ1e1 + ρ2e2 + ρ3e3 (1.15)
Then
vr ≡(
dρ
dt
)rel
= ρ1e1 + ρ2e2 + ρ3e3 (1.16)
The absolute acceleration a of the particle is then
a =dvdt
(1.17)
= R+ ar + ω × vr + ω × dρ
dt+
dω
dt× ρ
From (1.15)dρ
dt= vr + ω × ρ (1.18)
Plug (1.18) into (1.18) yields
a = R+ ar + ω × ω × ρ + ω × ρ + 2ω × vr (1.19)
where ar = ρ1e1 + ρ2e2 + ρ3e3.
We can describe the physical meaning of each term in (1.19) as follows.
• R is the acceleration of the origin o of the moving reference frame xyz.
8
• ar is the relative acceleration of the particle as observed in the moving reference frame
xyz.
• ω×ω×ρ is a centripetal acceleration, or the correction term for the local position vector
ρ considering that the observer rotates with the moving reference frame.
• ω × ρ is another correction term for the angular acceleration vector ω of the moving
reference frame.
• 2ω × vr is the Coriolis acceleration which is the other correction term from two sources,
both of which measure the rotation of the basis (unit) vectors of the moving reference
frame and associate with an interaction of motion along more than one coordinate curve.
Example 1.2:
The Hubble Space satellite shown in Fig. 1.9 has a steady spin Ω about the body fixed axis e3
The solar panel arm rotates about the e2-axis with a rate θ, and angular acceleration θ = 0.
The panel arm also moving along the radial direction er with a steady rate s = α. Determine
an absolute acceleration of the point P at the end of the solar panel.
Solution:
Let [e1e2e3] be the coordinate system that rotates with the body. Hence ωe1e2e3 = Ωe3.
Choose [ereθe2] in Fig. 1.9 as a rotating reference frame. For this case we obtain the terms in
(1.13) and (1.15) as
R = be1, ρ = ρrer + ρθeθ + ρ2e2 = (s(t) + c)er
ω = Ωe3 + θe2
Note that R and ω are the fixed-length vectors with constant magnitudes. From (1.19), the
absolute acceleration of point P is
ap = R+ ar + ω × ω × ρ + ω × ρ + 2ω × vr (1.20)
where
R = be1, R = Ωe3 ×R = bΩe2, R = Ωe3 × R = −bΩ2e1
vr = ρrer + ρθeθ + ρ2e2 = s(t)er = αer
ar = ρrer + ρθeθ + ρ2e2 = s(t)er = 0
9
b
s(t) + c
P
e2
θe2
e3
eθ
er
e1
θ
Ω
Ω
P
Figure 1.9: A Hubble Space satellite
ω = Ωe3 × ω = −Ωθe1
Components in (1.20) are now expressed in terms of two different coordinate systems [e1e2e3]
and [ereθe2]. To express these terms in only one coordinate system, i.e. [e1e2e3], we need the
coordinate transformation.
From Fig. 1.10, we obtain the transformation relation of an arbitrary vector u as
u =
ur
uθ
=
cosθ sinθ
−sinθ cosθ
u3
u1
= T
u3
u1
e3
e1
er
u
eθ
θ
θe2
u3
u1
ur
uθ
Figure 1.10: Coordinate systems [e1e2e3] and [ereθe2]
10
The reader can prove that T is an orthogonal matrix or T−1 = TT . As a result u3
u1
=
cosθ sinθ
−sinθ cosθ
−1 ur
uθ
= T−1
ur
uθ
= TT
ur
uθ
With the coordinate transformation, we can express all terms in (1.20) in terms of [e1e2e3]
components as
vr = αer = α (cosθe3 + sinθe1) = [ αsinθ, 0, αcosθ ]T
ω = [ 0 θ Ω ]T
ω =
0 −Ω θ
Ω 0 0
−θ 0 0
ω = [ −Ωθ 0 0 ]T
˙ω =
0 0 0
0 0 Ωθ
0 −Ωθ 0
ρ = (s(t) + c)er = (s(t) + c) (cosθe3 + sinθe1) = (s(t) + c)[ sinθ, 0, cosθ ]T
R = [ −bΩ2 0 0 ]T
Plug these terms into (1.20), we obtain ap in terms of the rotating system of coordinates [e1e2e3]
as
ap = [ −bΩ2 0 0 ]T + 0+ (s(t) + c)
0 −Ω θ
Ω 0 0
−θ 0 0
0 −Ω θ
Ω 0 0
−θ 0 0
sinθ
0
cosθ
+(s(t) + c)
0 0 0
0 0 Ωθ
0 −Ωθ 0
sinθ
0
cosθ
+ 2α
0 −Ω θ
Ω 0 0
−θ 0 0
sinθ
0
cosθ
(1.21)
Now your task is to follow the previous procedure and express ap in terms of the coordinate
system [ereθe2].
Example 1.3:
11
L
µG
y1
z1
α.
β
Ω
y2
z2
Figure 1.11: A ventilator
X
Y
x1
y1Z, z1
x1, x2y1
z1
y2
z2
x2
z2
x
z
y2, y
α
α
β
β
Ωt
Ωt
Figure 1.12: Coordinate Systems
Figure 1.11 shows a ventilator mounted on a rotating base. The base has an oscillatory motion
with α = α0sinωt. The rotor spins with a constant angular velocity Ω in the direction shown.
Its center of gravity G is offset by µ from the axis of rotation. Determine:
1. components of the absolute angular velocity of the rotor along the system of coordinates
fixed to the rotor.
2. components of the absolute velocity of the center of gravity of the rotor along the same
system of coordinates.
Solution:
12
Figure 1.12 shows the coordinate systems XY Z, x1y1z1, x2y2z2, and xyz. From Fig. 1.12
coordinate transformations are: y2
z2
=
cosβ sinβ
−sinβ cosβ
y1
z1
x
z
=
cosαt −sinαt
sinαt cosαt
x2
z2
Absolute angular velocity of the rotor is then
ω = αez1 + Ωey2
= α(sinβey2 + cosβez2) + Ωey2
= (αsinβ + Ω)ey2 + αcosβez2
= (αsinβ + Ω)Ωey + αcosβ(−sinΩtex + cosΩtez)
= −αcosβsinΩtex + (Ω + αsinβ)ey + αcosβcosΩtez
The position vector of G is
rG = Ley2 + µez
= Ley + µez
= [ 0 L µ ]T
Absolute velocity of G is
rG = ω × rG
= ω
0
L
µ
1.6 Coordinate Transformation
In Section 1.5, we simply transform the coordinates in two dimensional (2-D) space. Now
let’s consider a general 3-D coordinate transformation. Specifically, we want to establish a
transformation matrix C that transform components of a vector in one system of coordinates
to another system of coordinates. Let XY Z be an inertial reference frame with unit vectors
IJK, and xyz be a rotating coordinate system with unit vectors ijk as shown in Fig. 1.13. An
arbitrary vector r in Fig. 1.13 can be expressed as
r = rXI+ rY J+ rZK
= rxi+ ryj+ rzk (1.22)
13
X
Y
Z
x
z
y
iI
r
Figure 1.13: A vector r and two sets of coordinate systems XY Z and xyz
Components of r in XY Z coordinates are
rX = r · I = (rxi+ ryj+ rzk) · I
rY = r · J = (rxi+ ryj+ rzk) · J
rZ = r ·K = (rxi+ ryj+ rzk) ·K (1.23)
Or
rX = rxcos iI+ rycos jI+ rzcos kI
rY = rxcos iJ+ rycos jJ+ rzcos kJ
rZ = rxcos iK+ rycos jK+ rzcos kK (1.24)
(1.24) can be put in a matrix form as
rX
rY
rZ
= C
rx
ry
rz
(1.25)
where C is a matrix of directional cosines so called a coordinate transformation matrix given
by
C =
cos iI cos jI cos kI
cos iJ cos jJ cos kJ
cos iK cos jK cos kK
(1.26)
Note that C is the orthogonal matrix where C−1 = CT . This yields
rx
ry
rz
= CT
rX
rY
rZ
(1.27)
14
J
i
iJ
iY
Ji
Figure 1.14: components of the unit vectors
Q: Are all nine components of C independent?
To answer this question, let’s consider Fig. 1.14, showing the following relations.
cos iJ =|iY ||i| = |iY | (1.28)
With similar expressions for the other axes, we come up with the following 6 relationships
|iX |2 + |iY |2 + |iZ |2 = cos2 iI+ cos2 iJ+ cos2 iK = 1
|jX |2 + |jY |2 + |jZ |2 = cos2 jI+ cos2 jJ+ cos2 jK = 1
|kX |2 + |kY |2 + |kZ |2 = cos2 kI+ cos2 kJ+ cos2 kK = 1
|Ix|2 + |Iy|2 + |Iz|2 = cos2 Ii+ cos2 Ij+ cos2 Ik = 1
|Jx|2 + |Jy|2 + |Jz|2 = cos2 Ji+ cos2 Jj+ cos2 Jk = 1
|Kx|2 + |Ky|2 + |Kz|2 = cos2 Ki+ cos2 Kj+ cos2 Kk = 1
With these 6 relations of the directional cosines, there are only 9 − 6 = 3 independent
components of C. Specifically, only three independent angular transformation terms are needed
to describe the coordinate transformation. There exist many possible sets of angular transfor-
mation, but two popular sets called Euler angles are normally used. Each set consists of three
angles describing the sequence of rotations as described in the following subsections.
1.6.1 First set of Euler angles–precession-nutation-spin (φθψ)
This set of Euler angles is normally used to describe the gyroscopic systems such as rotordy-
namics. The sequence of rotations as shown in Fig. 1.15 is
• Precession: rotation about Z axis by φ(t) to get x′y′z′ or x′y′Z
• Nutation: rotation about x′ axis by θ(t) to get x′′y′′z′′ or x′y′′z′′
15
• Spin: rotation about z′′ axis by ψ(t) to get xyz or xyz′′
The coordinate transformations are then
rX
rY
rZ
=
cosφ −sinφ 0
sinφ cosφ 0
0 0 1
rx′
ry′
rz′
= C1
rx′
ry′
rz′
(1.29)
rx′
ry′
rz′
=
1 0 0
0 cosθ −sinθ
0 sinθ cosθ
rx′′
ry′′
rz′′
= C2
rx′′
ry′′
rz′′
(1.30)
rx′′
ry′′
rz′′
=
cosψ −sinψ 0
sinψ cosψ 0
0 0 1
rx
ry
rz
= C3
rx
ry
rz
(1.31)
Combine (1.29)-(1.31), therefore
rX
rY
rZ
= C1C2C3
rx
ry
rz
(1.32)
1.6.2 Second set of Euler angles–yaw-pitch-row (ψθφ)
This set of Euler angles is normally used to describe the dynamics of vehicles. The sequence of
rotations as shown in Fig. 1.16 is
• Yaw: rotation about Z axis by ψ(t) to get x′y′z′ or x′y′Z
• Pitch: rotation about y′ axis by θ(t) to get x′′y′′z′′ or x′′y′z′′
• Roll: rotation about x′′ axis by φ(t) to get xyz or x′′yz
The coordinate transformations are then
rX
rY
rZ
=
cosψ −sinψ 0
sinψ cosψ 0
0 0 1
rx′
ry′
rz′
= [Rψ]
rx′
ry′
rz′
(1.33)
16
X
Y
Z
x’
z’
y’φ
φ
φ
x’ x’’
z’z’’
y’’
y’
θ
θ
θ
x’’
z’’
y’’
y
x
ψ
ψ
ψ
z
Figure 1.15: First set of Euler’s angles and sequence of rotation
17
XY
Z
θ
ψ
φ
XY
Z
x’
z’
y’
φφ
φ
x’’x’
z’z’’
y’’y’
θ
θ
θ
x’’
z’’
y’’
y
x
ψ
ψ
ψ
z
Figure 1.16: Yaw, pitch, and roll axes of vehicle dynamics
rx′
ry′
rz′
=
cosθ 0 sinθ
0 1 0
−sinθ 0 cosθ
rx′′
ry′′
rz′′
= [Rθ]
rx′′
ry′′
rz′′
(1.34)
rx′′
ry′′
rz′′
=
1 0 0
0 cosφ −sinφ
0 sinφ cosφ
rx
ry
rz
= [Rφ]
rx
ry
rz
(1.35)
Therefore
rX
rY
rZ
= [Rψ] [Rθ] [Rφ]
rx
ry
rz
(1.36)
1.7 Angular velocity related to Euler angles
For the first set of Euler angles, the absolute angular velocity ω of xyz coordinates is given by
ω = φk+ θex′ + ψez
≡ ωxex + ωyey + ωzez (1.37)
18
x
y’
Z
θ
ψ
φ
Figure 1.17: Yaw, pitch, and roll axes of vehicle dynamics
Rewrite ex′ and k in terms of ex, ey and ez, using (1.29)-(1.31), we get
ωx
ωy
ωz
=
sinθsinψ cosψ 0
sinθcosψ −sinφ 0
cosθ 0 1
φ
θ
ψ
(1.38)
Similarly, for the second set of Euler angles, the absolute angular velocity ω of xyz coor-
dinates is given by
ω = ψk+ θey′ + φex
= ωxex + ωyey + ωzez (1.39)
Rewrite ey′ and k in terms of ex, ey and ez, using (1.33)-(1.35), we get
ωx
ωy
ωz
=
1 0 −sinθ
0 cosφ cosθsinφ
0 −sinφ cosθcosφ
φ
θ
ψ
(1.40)
(1.38) and (1.40) relate the Euler angles, the rotation that measured in real applications, with
the components of the angular velocity, ωx, ωy and ωz, in the reference coordinate system.
Example 1.4:
A submarine shown in Fig. 1.17 undergoes a yaw rate ψ = AcosΩt and a pitch rate θ = BsinΩt.
If the local x-axis is in the long-body direction, describe the velocity of the bow of the submarine
relative to its center of mass.
Solution:
From Fig. 1.17, let xyz with their unit vectors ex, ey, and ez be the body-fixed rotating
19
system of coordinates. The velocity of the bow observed from the submarine C.G. is then
v = ω × ρ (1.41)
where
ρ = Lex (1.42)
and ω is the angular velocity of the body or the angular velocity of the xyz coordinates given
by
ω = ψk+ θey′ (1.43)
ω in terms of ex, ey, and ez can be obtained from (1.40) as
ω =
ωx
ωy
ωz
=
1 0 −sinθ
0 cosφ cosθsinθ
0 −sinφ cosθcosφ
0
θ
ψ
(1.44)
Note that, in this case, the submarine performs only pitch and yaw rotations but no row.
Neglecting the higher order terms, ω is therefore
ω = −ψsinθex + θey + ψcosθez (1.45)
Substitution of (1.42) and (1.45) into (1.41) yields
v = Lψcosθey − Lθez (1.46)
From given ψ = AcosΩt and θ = BsinΩt, and if ψ(0) = θ(0) = 0, then ψ(t) = AΩsinΩt and
θ(t) = −BΩ cosΩt. Substitution of these conditions into (1.46) yields
v = −ALcosΩtcos
(B
ΩcosΩt
)ey − LBsinΩtez (1.47)
For the small value of BΩ , cos
(BΩ cosΩt
)≈ 1. In addition if A = B, the velocity vector v =
−AL(cos Ωtey + sin Ωtez) performs a circular path.
1.8 A Finite Motion
A general motion of any rigid body can be resolved into the translation u of an arbitrary point
on the body and a finite rotation φ about this point as shown in Figure 1.18. First we consider
the transformation matrix for a finite rotation. Then the transformation matrix for a general
finite motion, possessing both translation and rotation, is considered.
20
φ
u
A
A'
Figure 1.18: A finite motion of a rigid body
1.8.1 Transformation matrices for a finite rotation
Define a position vector of any point P on the body before and after the rotation as rp and r′p,
respectively. A transformation matrix T relating rp and r′p is given by
r′p = Trp (1.48)
Properties of the transformation matrix T are described as follows:
1. Because of no deformation of a rigid body, T is the same for any point p in the body.
Hence the subscript p in (1.48) can be drop out.
r′ = Tr (1.49)
2. The rotation should be invertible.
r = T−1r′ (1.50)
3. The length of r is unchanged, hence
r · r = rT r = r′ · r′ =(r′)T r′ (1.51)
Or
rT r =(r′)T r′ (1.52)
= (Tr)T Tr
= rTTTTr
Hence
TTT = I (1.53)
(1.53) indicates that T−1 = TT or T is the orthogonal matrix.
21
φ
X
Y
(x, y, z)
(x', y', z')
Figure 1.19: Finite rotation about Z-axis
The transformation T for a rotation about Z-, Y -, and X-axes can be determined subsequently
as follows.
1. Rotation about Z-axis with φ
If the previous coordinates of any point p is r = [ x y z ]T and the new coordinates of
this point is r′ = [ x′ y′ z′ ]T as seen in Figure 1.19, then
x′
y′
z′
=
cosφ −sinφ 0
sinφ cosφ 0
0 0 1
x
y
z
= T1
x
y
z
(1.54)
The transformation matrix T1 in this case is
T1 =
cosφ −sinφ 0
sinφ cosφ 0
0 0 1
(1.55)
2. Rotation about Y -axis with θ
From Figure 1.20 we obtain the transformation matrix T2 for a rotation about Y -axis as
T2 =
cosθ 0 sinθ
0 1 0
−sinθ 0 cosθ
(1.56)
3. Rotation about X-axis with ψ
From Figure 1.21, the transformation matrix T3 for a rotation about X-axis is
T3 =
1 0 0
0 cosψ −sinψ
0 sinψ cosψ
(1.57)
22
θ
Z
X
Figure 1.20: Finite rotation about Y -axis
ψ
Z
Y
Figure 1.21: Finite rotation about X-axis
23
a
bφ
θL
Figure 1.22: A falling box
In general the transformation matrices do not commute; i.e., T1T2 = T2T1. However, for the
infinitesimal angular displacement, cosφ ≈ 1, sinφ ≈ φ and so on. Also φ ≈ ωz∆t, θ ≈ ωy∆t,
and ψ ≈ ωx∆t. In this case T1, T2, and T3 do commute. If the rigid body has an infinitesimal
rotation about an arbitrary axis during time ∆t, the new position vector r′ related to the
previous position vector r, according to (1.48), is then
r′ = r(t + ∆t) = T1T2T3r(t) (1.58)
Substitution of (1.55), (1.56), and (1.57) into (1.58) yields
r(t + ∆t) =
1 −ωz∆t ωy∆t
ωz∆t 1 −ωx∆t
−ωy∆t ωx∆t 1
x(t)
y(t)
z(t)
(1.59)
Therefore the velocity v is given by
v = lim∆t→0
r(t + ∆t) − r(t)∆t
=
0 −ωz ωy
ωz 0 −ωx
−ωy ωx 0
x(t)
y(t)
z(t)
≡ ωr (1.60)
From (1.60), it is proved that the angular velocity ω can be represented in a matrix form as
previously introduced in (1.7).
The transformation matrix for a finite rotation is useful for a computer graphic program-
ming simulating the dynamics of rigid-body motion as shown in the following example.
Example 1.5:
A box, considered as the planar problem, is hinged as shown in Figure 1.22. Construct the
Matlab m-file to simulate the dynamics of this falling box.
24
a
bφ
θ
R
mgO
L
Cθ.
Figure 1.23: A free body diagram of the falling box
Solution:
First we need to derive the equation governing the motion of this box. A free body diagram
(FBD) of the box is shown in Figure 1.23. The dynamics of the falling box is governed by the
law of angular momentum given by
[∑
Mo = Ho]; mgLsin(θ + φ0) − Cθ = Ioθ (1.61)
where C is the torsional damping coefficient used to model the friction at the hinge and Io is
the mass moment of inertia about o. To solve (1.61), let’s define state variables as x1 = θ and
x2 = θ Then (1.61) can be written in state form as x1
x2
=
x2
esin(x1 + φ0) − cx2
(1.62)
where e = mgLIo
and c = CIo
. (1.62) together with the transformation matrix for the finite rotation
in (1.55) are used in the MatLab program to determine the new position of the falling box. The
detail of this program is presented in Figure 1.24 and the result is shown in Figure 1.25.
1.8.2 Transformation matrices for a general motion
A general motion of a rigid body as shown in Fig. 1.26 can be divided into two parts: a
translation u and a finite rotation θ. The position vector r describing the finite rotation of the
rigid body is then
r = u+ ρ′ (1.63)
= u+Aρ
25
clear all% MATLAB Animation Program for Falling Box%===Define the vertices of the boxa=0.1;b=0.2;x=[0 a a 0 0];y=[0 0 b b 0];%===Define a matrix whose column vectors are the box verticesr=[x; y];%===Draw the box in the initial positionfigure(1), clfaxis([-0.3 0.3 -0.3 0.3])line(x, y,'linestyle','--');grid on%===Define parameters m=1; g=9.81;C=0.001;L=0.5*sqrt(a^2+b^2);I=m*(a^2+b^2)/12;e=m*g*L/I;c=C/I;%===Define initial conditionstheta = 0;omega = 0;phi_0 = atan(a/b);%===stepsdt = 0.001; % time step for simulationn=10; % # of animation M=moviein(n); % # define a matrix M for movie in%========= Finish data input ============================
%===Numerically integrate the equations of motion using Newton methodfor j = 1:n; % Do loop for new box graphic
for n =1:20; % Do loop for elapsed time integration omega = omega+dt*e*sin(theta+phi_0)-dt*c*omega; theta=theta+dt*omega;end%===Rotate box graphic using finite rotation matrixA=[cos(theta) sin(theta); -sin(theta) cos(theta)];r1=A*r;x1=r1(1,:);y1=r1(2,:);patch(x1,y1,'r');axis('equal')M(:,j)=getframe;end%===Show movie%figure(2), clf%movie(M,1,2);
Figure 1.24: Matlab program for animation of box falling
26
-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
Figure 1.25: Simulation of the falling box
x
y
z
A
A
i
j
k
r
’
u
θ
ez
ex
Figure 1.26: Finite motion
27
where A is the transformation matrix of the rotation relating ρ and ρ′. In addition r =
[ x y z ]T and u = [ ux uy uz ]T expressed in the inertial reference coordinates xyz,
and ρ = [ ρx ρy ρz ]T expressed in the local coordinate system exeyez. Substituting the
component vectors into (1.64), the finite motion in Fig. 1.26 is governed by
x
y
z
=
ux
uy
uz
+
cosθ −sinθ 0
sinθ cosθ 0
0 0 1
ρx
ρy
ρz
(1.64)
If r and ρ are expanded as r = [ x y z 1 ]T and ρ = [ ρx ρy ρz 1 ]T , (1.64) can be
rewritten as
r4×1 = T4×4ρ4×1 (1.65)
where T is the transformation matrix for a general finite motion given by
T =
| ux
A3×3 | uy
| uz
−− −− −− −|− −−
0 0 0 | 1
For a finite translation, the transformation matrix is simply
T1 =
1 0 0 | ux
0 1 0 | uy
0 0 1 | uz
−− −− −− −|− −−
0 0 0 | 1
For a finite rotation, the transformation matrix is simply
T2 =
| 0
A3×3 | 0
| 0
−− −− −− −|− −−
0 0 0 | 1
Note that T = T1T2.
28
Chapter 2
Introduction to Linear and Angular
Momentums
This chapter is organized into three parts: 1) dynamics of a system of particles; 2) an angu-
lar momentum of a rigid body; and 3) a mass moment of inertia. These topics are used as
fundamentals for a study of dynamics of a rigid body and a multi-body mechanical system in
Chapter 3 and Chapter 4, respectively.
2.1 Dynamics of a System of Particles: a Review
Figure 2.1 shows a system consisting of n-particles in 3-D where the i-th particle is subjected
to the applied force Fi. Also c is the center of mass or center of gravity (C.G.) of the system.
2.1.1 Total mass
A total mass of the system shown in Figure 2.1 is
M =∑
mi (2.1)
where mi is a mass of the i-th particle and∑
is the sum over i, for i = 1, 2, . . . , n.
29
X
Y
Z
m1
m2
mi
m
mN
m3
rc
ri
vi
ρi
Fi
FN
F1
F3
F2
c
o
Figure 2.1: A system of particle
2.1.2 First moment of mass
The first moment of the total mass about its C.G. is the sum of the first moment of each mass
given by:
rcM =∑rimi (2.2)
From (2.2), the center of mass rc can be obtained as
rc =1M
∑rimi (2.3)
From Figure 2.1, the displacement of the i-th particle relative to the C.G. is
ρi = ri − rc (2.4)
In addition, sum of the first moment of each mass about C.G. is given by
∑ρimi =
∑[ri − rc] mi
=∑rimi − rc
∑mi
= rcM − rcM
= 0
(2.5)
Equation (2.5) indicates that sum of the first moment of each mass about the system’s C.G. is
zero.
30
2.1.3 Linear momentum
A linear momentum P of the system of particles is defined as follows:
P ≡ ∑mivi
=∑
midridt
=d
dt
[∑miri
]=
d
dt(Mrc)
= Mvc
(2.6)
2.1.4 Angular momentum
An angular momentum is defined as the first moment of the linear momentum. The angular
momentum of the system of particles about the origin o is then given by
Ho ≡∑ri × mivi (2.7)
The angular momentum of the system of particles about the system’s C.G. is defined as
Hc ≡∑
ρi × miρi (2.8)
Ho and Hc are related through the following equation
Ho = Hc + rc × Mvc (2.9)
To prove the relation (2.9), we rewrite (2.7) as follows
Ho =∑
[(ρi + rc) × mivi]
=∑
ρi × mivi + rc ×∑
mivi(2.10)
The second term on the right of (2.10) is then
rc ×∑
mivi = rc × Mvc (2.11)
The first term on the right of (2.10) can be rewritten as
∑ρi × mivi =
∑ρi × mi (rc + ρi)
=∑
ρi × mirc +∑
ρi × miρi
=∑
(miρi) × rc +Hc
= 0+Hc
(2.12)
Substitution of (2.11) and (2.12) into (2.10), therefore, yields (2.9).
31
2.1.5 Moment of force
The moment due to all applied forces about o is
Mo =∑ri × Fi (2.13)
2.1.6 Laws of linear and angular momentum
The laws of linear and angular momentum relate the applied forces and moments to the linear
and angular momentums of the system.
Law of linear momentum
Applying the Newton’s 2nd law to each i-th particle, we obtain
mivi = Fi +∑j
fij ; i = j (2.14)
where Fi are external forces applied to the mass mi, and fij is a reaction force that the j-th
particle acts on the the i-th particle. Also note that fij = −fji. Summation of (2.14) for all
particles then yields
∑i
mivi =∑i
Fi +∑i
∑j
fij; i = j (2.15)
Since fij = −fji, therefore∑i
∑j
fij = 0. Hence (2.15) becomes
dPdt
= M vc =∑i
Fi (2.16)
(2.16) is the law of linear momentum for the system of particles, stating that the rate of change
of linear momentum of the system is equal to the sum of all external forces applied to the
system.
Law of angular momentum
Let’s take the first moment of (2.14) about o and sum over all particles:
∑i
(ri × mivi) =∑i
(ri × Fi) +∑i
ri ×∑
j
fij
; i = j (2.17)
32
Now consider (2.17) term by term. The term on the left is rewritten as
∑i
(ri × mivi) =∑i
ri × midvidt
=d
dt
[∑i
ri × mivi
]
= Ho
(2.18)
The first term on the right of (2.17) is∑i
(ri ×Fi) ≡Mo, and the second term on the right is
zero as shown in the following proof.
Let’s consider any two particles m and n. Since fmn = −fnm and rm−rn is approximately
colinear with fmn, then the action-reaction pair of any arbitrary internal moments are zero, or
rm × fmn + rn × fnm = (rm − rn) × fmn = 0 (2.19)
According to (2.19), therefore ∑i
ri ×∑
j
fij
= 0 (2.20)
Substituting (2.18) to (2.20) into (2.17), we get
dHo
dt= Ho =Mo (2.21)
Equation (2.21) is the law of angular momentum, stating that the rate of change of angular
momentum about o is equal to the moment of all external forces about o.
Alternatively, we could formulate the law of angular momentum about the system’s C.G..
First, differentiate (2.9) with time:
d
dtHo =
d
dtHc +
d
dt(rc × Mvc) (2.22)
From (2.22), the first term in (2.22) is rewritten as
d
dtHo = Mo
=∑ri × Fi
=∑
(rc + ρi) × Fi
=∑rc × Fi +
∑ρi × Fi
(2.23)
In addition, the third term in (2.22) is then
d
dt(rc × Mvc) =
d
dtrc × Mvc + rc ×
d
dt(Mvc)
= vc × Mvc + rc ×∑Fi
= 0+ rc ×∑Fi
(2.24)
33
X
Y
Z
A
ω
rA
e1
e2e3
r
ρ dm
o
Figure 2.2: A rigid body
Substitution of (2.23) and (2.24) into (2.22) yields
∑ρi × Fi =
d
dtHc (2.25)
Or
Hc =∑
ρi × Fi =Mc (2.26)
From equation (2.26), if proper coordinates are used to described ρi then we can define a set
of geometric quantities so called moments of inertia of a rigid body. The moments of inertia
measure the angular momentum per unit rate of rotation. They will be derived in detail in
Section 2.3.
2.2 Angular Momentum of a Rigid Body
Figure 2.2 shows a rigid body moving in 3D. Let the angular velocity of the body be ω. Consider
a rigid body as a continuous media of particles with no deformation, the angular momentum
of this rigid body is the integral form of (2.7) or
Ho =∫r× vdm (2.27)
where r = rA + ρ is a position vector from the fixed origin o to the differential mass dm of the
body. Let e1e2e3 in Figure 2.2 be the rotating reference frame with its origin located at an
arbitrary point A on the body. If the reference coordinate system and the body have the same
angular velocity, i.e. ω, then
v = r = vA + ω × ρ (2.28)
34
Note that the velocity of the differential mass dm relative to e1e2e3 is zero because: 1) the rigid
body has no deformation and 2) e1e2e3 rotates synchronously with the body. Substitution of
(2.28) into (2.27) yields
Ho =∫
[(rA + ρ) × (vA + ω × ρ)] dm
= rA × vA∫
dm +[∫
ρdm
]× vA
+rA ×[ω ×
∫ρdm
]+∫
ρ × (ω × ρ) dm
= rA × mvA + mρc × vA + rA × [ω × mρc] +∫
ρ × (ω × ρ) dm
(2.29)
where m =∫
dm is the mass of the rigid body and ρc is the position of the body’s C.G. measured
with respect to A and given by ρc = 1m
∫ρdm. Furthermore, the rotation of a rigid body can
be considered as two different cases: pure rotation and general motion (combined rotation and
translation).
1. Pure rotation about fixed point o
In this case, if we choose point A in Figure 2.2 fixed at o. Hence rA = vA = 0, and (2.29)
becomes
Ho =∫
ρ × (ω × ρ) dm (2.30)
2. General motion
In this case if point A in Figure 2.2 is fixed at the body’s C.G, i.e. point c. Hence rA = rc
and ρc = 0. (2.29) then becomes
Ho = rc × mvc +∫
ρ × (ω × ρ) dm (2.31)
For a rigid body, the angular momentum about c is defined as
Hc ≡∫
ρ × (ω × ρ) dm (2.32)
Hence
Ho = rc × mvc +Hc (2.33)
2.3 Mass Moment of Inertia
Due to a constant geometric property of the rigid body, the angular momentum can be more
simplified as follows. First it is noted that the angular momentum about o (2.30), in case of
35
pure rotation, and the angular momentum about c (2.32), in case of general motion, have the
same form. Therefore we will drop out the subscripts in (2.30) and (2.32) for convenience and
generally rewrite both equations as
H =∫
ρ × (ω × ρ) dm (2.34)
Since a triple cross product can be rewritten as A × (B × C) = (A · C)B − (A · B)C, then
(2.34) becomes
H =∫
(ρ · ρ)ω − (ρ · ω)ρdm (2.35)
ρ and ω can be expressed in terms of components with respect to e1e2e3 coordinate system as
follows
ρ =[
x y z
]T, ω =
[ωx ωy ωz
]T
Next we will rewrite (2.35) in terms of its components. Let’s consider (2.35) term by term. The
first term is ∫(ρ · ρ)ωdm =
∫(ρ · ρ) [δ] ωdm
=∫
ρ2 0 0
0 ρ2 0
0 0 ρ2
ωdm
=
∫
ρ2 0 0
0 ρ2 0
0 0 ρ2
dm
ω
(2.36)
36
where ρ2 = x2 + y2 + z2 and [δ] is the identity matrix. The second term of (2.35) is
−∫
(ρ · ω)ρdm = −∫[
x y z
]
ωx
ωy
ωz
x
y
z
dm
= −∫
(xωx + yωy + zωz)
x
y
z
dm
= −∫
x2ωx + xyωy + xzωz
xyωx + y2ωy + yzωz
xyωx + yzωy + z2ωz
dm
= −∫
x2 xy xz
xy y2 yz
xz yz z2
ωx
ωy
ωz
dm
= −∫
x2 xy xz
xy y2 yz
xz yz z2
dm(ω)
(2.37)
Substitution of (2.36) and (2.37) into (2.35) yields
H =
∫
y2 + z2 −xy −xz
−xy x2 + z2 −yz
−xz −yz x2 + y2
dm
ω
=
I11 I12 I13
I21 I22 I23
I31 I32 I33
ω = Iω
(2.38)
where I is the matrix of (second) moments of inertia. The components of I along diagonal are
called moments of inertia given by
I11 =∫ (
y2 + z2)
dm
I22 =∫ (
x2 + z2)
dm
I33 =∫ (
x2 + y2)
dm
(2.39)
and the off-diagonal components so called cross product of inertia are given by
I12 = I21 = −∫
xydm
I13 = I31 = −∫
xzdm
I23 = I32 = −∫
yzdm
(2.40)
37
Any three orthogonal axes e′1e′2e
′3 that yields all zero cross product of inertia, i.e. I12 = I13 =
I23 = 0, are called principal axes. In this case I11 = I1, I22 = I2, and I33 = I3 are called the
principal inertias. I1, I2, and I3 can be determined from the eigenvalues of the matrix I. With
the principal inertias, the angular momentum of a rigid body can be simplified as
H = I1ω1e′1 + I2ω2e′2 + I3ω3e′3 (2.41)
Properties of I
1. I is a symmetric matrix
2. I has positive eigenvalues which are principal inertias I1, I2, and I3, and has three or-
thogonal eigenvectors which represent the principal axes e′1e′2e
′3.
3. For a basis with at least two symmetry planes, the off-diagonal terms or the cross-product
of inertia are zero.
4. The parallel axes theorem states that
Ikk = I(c)kk + m∆2
k, k = 1, 2, 3 (2.42)
and
Iij = I(c)ij − mdidj , i, j = 1, 2, 3, i = j (2.43)
where ∆k is the distance between the two parallel axes, and di and dj are the relative
displacements along i and j coordinates, respectively.
5. The inertia matrix calculation is an additive operator.
Practical methods used to determine the inertia matrix are: 1) look-up table, 2) computer
calculation, and 3) experiment.
38
Chapter 3
Dynamics of a Rigid Body:
Newton-Euler Approach
In this chapter, the dynamics of a rigid body for two different cases: 1) a pure rotation; and 2)
a general motion consisting of both translation and rotation, are studied. According to the laws
of angular momentum stated in the previous chapter, we can formulate the dynamics equations
governing the motion of a rigid body.
3.1 Newton-Euler Equations of a rigid body
For a rigid body having pure rotation about o with the angular velocity ω, the governing
equation is ∑Mo = Ho (3.1)
Equation (3.1) is the law of angular momentum for a rigid body. In this case, we choose the
reference coordinate system e1e2e3, with its origin fixed at o, that rotates with the body with
the same angular velocity ω. Hence, the angular momentum about o can be simplified as
Ho = Ioω (3.2)
where Io is the constant matrix of moments of inertia about o whose components are along
e1e2e3 axes.
For a general motion of a rigid body, the equations governing both translation and rotation
39
are ∑F = mvc (3.3)
and ∑Mc = Hc (3.4)
where point c is the C.G. of the rigid body.
Equation (3.3) is the law of linear momentum for a rigid body or so called the Newton’s
equation, and equation (3.4) is the law of angular momentum. For the general motion, we
normally choose the reference coordinate system such that its origin is fixed at the C.G. of the
body and its coordinates rotate with the body. If the rigid body has the angular velocity ω,
then
Hc = Icω (3.5)
where Ic is the constant matrix of moments of inertia about c whose components are along the
reference coordinates.
For both cases of motion, if the reference coordinates, i.e. e1e2e3, are in the directions
such that they are the principal axes, then Ho and Hc in (3.2) and (3.5) are simply
Ho = Ioω = I1oω1e1 + I2oω2e2 + I3oω3e3 (3.6)
and
Hc = Icω = I1cω1e1 + I2cω2e2 + I3cω3e3 (3.7)
Hence (3.1) and (3.4) can be more simplified as
∑Mo =
M1o
M2o
M3o
=
I1oω1
I2oω2
I3oω3
+
0 −ω3 ω2
ω3 0 −ω1
−ω2 ω1 0
I1oω1
I2oω2
I3oω3
(3.8)
and
∑Mc =
M1c
M2c
M3c
=
I1cω1
I2cω2
I3cω3
+
0 −ω3 ω2
ω3 0 −ω1
−ω2 ω1 0
I1cω1
I2cω2
I3cω3
(3.9)
Or they can be written in a scalar form as
M1o = I1oω1 + (I3o − I2o) ω2ω3
M2o = I2oω2 + (I1o − I3o) ω1ω3
M3o = I3oω3 + (I2o − I1o) ω1ω2
(3.10)
40
m, L
g θex
eymassless cart
u(t)
Figure 3.1: A cart-pendulum system
andM1c = I1cω1 + (I3c − I2c) ω2ω3
M2c = I2cω2 + (I1c − I3c) ω1ω3
M3c = I3cω3 + (I2c − I1c) ω1ω2
(3.11)
Equation sets (3.10) and (3.11) are called Euler’s equations.
Example 1: Dynamics of a pendulum-cart system
The cart with a negligible weight moves along a frictionless floor as shown in Figure 3.1.
The pendulum with mass m and length L is hinged to the cart at one end. If the cart motion
is prescribed by u(t), derive the equation of motion of the system.
Solution:
First, consider the pendulum or the uniform rod which has a general plane (2-D) motion.
The degree of freedom used to describe the motion of this rod is θ(t).
Kinematics: With the coordinate systems shown in Figure 3.2, the velocity vc and acceleration
ac at C.G. of the rod are
vc = u(t)ex + θL
2eθ (3.12)
ac = vc = u(t)ex + θL
2eθ − θ2L
2er (3.13)
With the free body diagram (FBD) shown in Figure 3.2, we set Newton-Euler equations as
41
θ
er
eθ
θ
θex
ey
er
Fr
eθ
Fθ
c
mg
Figure 3.2: Coordinate systems and FBD
[∑F = mvc];
Frer + Fθeθ − mgey = m
(u(t)ex + θ
L
2eθ − θ2 L
2er)
(3.14)
and [∑
Mc = Icω];
FθL
2= Icθ (3.15)
Note that Ic in (3.15) is the moment of inertia about the C.G. of the rod along z-axis. We now
have three unknowns: Fr, Fθ, and θ, and three scalar equations, two from (3.14) and one from
(3.15). To derive the equation of motion, we need to eliminate all unknown forces which are Fr
and Fθ and reduce the Newton-Euler equations to only one differential equation.
Figure 3.2 shows the two coordinate systems with the coordinate transformation given by
ex = sinθer + cosθeθ
ey = −cosθer + sinθeθ(3.16)
To eliminate Fθ, we substitute (3.15) into (3.14) and transform all coordinates to ereθ using
(3.16). Then (3.14) can be expressed in scalar form as:
r-component:
mθ2 L
2+ mgcosθ = Fr + musinθ (3.17)
θ-component: (Ic +
mL2
4
)θ +
mgL
2sinθ =
mL
2ucosθ (3.18)
Equation (3.18) is the equation of motion. Note that (3.18) is a nonlinear equation. The
solution of (3.18) can be obtained from a numerical integration using Matlab. Otherwise if only
42
C.G. e1
e2
e3
ω0
XY
Zdm
ρ
Ω
Figure 3.3: A rigid body with symmetric shape
small oscillation is interested, we can linearize (3.18) to get a closed-form solution. With the
solution of (3.18), the dynamic forces Fr and Fθ are obtained from (3.15) and (3.17) as
Fr = mθ2L
2+ mgcosθ − musinθ (3.19)
and
Fθ = −2IcL
θ (3.20)
3.2 Modified Euler’s equations
A set of modified Euler’s equations is used in the case of the symmetric-shape rigid body which
spins about its symmetry axis with a constant speed, as shown in Figure 3.3. To formulate the
Modified Euler’s equations, two conditions are defined.
1. The rigid body spins about the symmetry axis with a constant speed ωo.
2. The reference coordinate system e1e2e3 is chosen such that one of the axes, i.e. e3, is
the symmetry axis. In addition, e1e2e3 only precesses but does not spin with the body.
Also the origin of e1e2e3 is fixed at the point of rotation for the case of pure rotation,
and is fixed at the body’s C.G. for the case of general motion. In these cases, e1e2e3 are
principal axes and I1 = I2 ≡ I. If the angular velocity of e1e2e3 is
Ω = Ω1e1 + Ω2e2 + Ω3e3
43
The angular velocity of the rigid body is then
ωb = Ω+ ωoe3
According to the conditions above, the modified Euler’s equations become
M1o = IoΩ1 + (I3o − Io) Ω2Ω3 + I3oωoΩ2
M2o = IoΩ2 + (Io − I3o) Ω3Ω1 − I3oωoΩ1
M3o = I3oΩ3
(3.21)
andM1c = IcΩ1 + (I3c − Ic) Ω2Ω3 + I3cωoΩ2
M2c = IcΩ2 + (Ic − I3c) Ω3Ω1 − I3cωoΩ1
M3c = I3cΩ3
(3.22)
Equation sets (3.21) and (3.22) are for the cases of pure rotation and general motion, respec-
tively. The derivation of the modified Euler’s equations is shown for the case of general motion
as follows. From Figure 3.3, the angular momentum of the rigid body about its C.G. is
Hc =∫
(ρ × ρ) dm
=∫ (
ρ ×[(
dρdt
)rel
+Ω× ρ])
dm
=∫
(ρ × [(ωoe3 × ρ) +Ω× ρ]) dm
=∫
(ρ × (ωoe3 +Ω) × ρ) dm
=∫
(ρ × ωb × ρ) dm
= Iωb =
Ic 0 0
0 Ic 0
0 0 I3c
Ω1
Ω2
Ω3 + ωo
(3.23)
Then
Hc = I
Ω1
Ω2
Ω3
+Ω×Hc
=
IcΩ1
IcΩ2
I3cΩ3
+
0 −Ω3 Ω2
Ω3 0 −Ω1
−Ω2 Ω1 0
IcΩ1
IcΩ2
I3c (Ω3 + ωo)
(3.24)
Substitution of (3.24) into (3.4) hence results in (3.22).
Example 2: Steady precession of a gyro top
44
o
e1
k
e3
φ
ψ
θ
.
.
L
mg
Figure 3.4: A gyro top
Derive the dynamic equation governing steady precession of a gyro top shown in Figure 3.4.
With the steady precession, the top has constant precession rate ψ and constant spin rate φ,
and the nutation angle θ is also constant.
Method1: Direct approach In Figure 3.4, the angular velocity of the reference coordinate
system e1e2e3 is
ωe1e2e3 = ψk+ θe2 (3.25)
Also the angular velocity of the body is
ωb ≡ ω1e1 + ω2e2 + ω3e3
= ψk+ θe2 + φe3
= ψ (cosθe3 + sinθe1) + θe2 + φe3
= ψsinθe1 + θe2 +(φ + ψcosθ
)e3
(3.26)
Hence ω1 = ψsinθ, ω2 = θ, and ω3 = φ+ ψcosθ. Since e3 is the symmetric axis, therefore
I1 = I2 and all cross products of inertia are zero. As the previous proof in (3.23), it can
be similarly shown that the angular momentum of the body about o is Ho = Ioωb. Or
Ho = I1ω1e1 + I1ω2e2 + I3ω3e3
= I1ψsinθe1 + I1θe2 + I3
(φ + ψcosθ
)e3
(3.27)
For a steady motion, θ is constant or θ = 0, and ω1 = ω2 = ω3 = 0. The angular
momentum is then
Ho = I1ψsinθe1 + I3
(φ + ψcosθ
)e3 (3.28)
45
Note that the angular momentum of the steady gyro in (3.28) has a constant magnitude,
and the direction of the angular momentum is on the plane of rotations, i.e. k−e3 plane.
Moreover, the rate of change of angular momentum is
Ho = ψk×Ho (3.29)
Substituting (3.28) into (3.29) and performing a matrix operation yield
Ho =[(I1 − I3) ψ2sinθcosθ − I3ψφsinθ
]e2 (3.30)
From the law of angular momentum[∑Mo = Ho
], the moment sum
∑Mo about o, in
this case, is due to only the gravitation force, given by
∑Mo = Le3 ×−mgk = −mgLsinθe2 (3.31)
Note that moment of the resultant force about o shown in (3.31) is always perpendicular
to both rotation axes (k and e3), resulting in the precession. For the steady precession,
this moment has a constant magnitude due to the constant nutation angle θ. Substitution
of (3.30) and (3.31) into the law of angular momentum yields
−mgLsinθ = (I1 − I3) ψ2sinθcosθ − I3ψφsinθ (3.32)
For sinθ = 0, we get the equation governing a steady precession of the top as
mgL = I3ψφ + (I3 − I1) ψ2cosθ (3.33)
From (3.33), with a given θ we can determine the relation between the precession rate ψ
and the spin rate φ. For example, if θ =π
2equation (3.33) becomes
ψφ =mgL
I3(3.34)
Method 2: modified Euler’s equations The modified Euler’s equations are presented again
as followsM1o = IoΩ1 + (I3o − Io) Ω2Ω3 + I3oωoΩ2
M2o = IoΩ2 + (Io − I3o) Ω3Ω1 − I3oωoΩ1
M3o = I3oΩ3
(3.35)
In (3.35), Ω1 = ψsinθ, Ω2 = 0, and Ω3 = ψcosθ. Also Ω1 = Ω2 = 0 because of a steady
motion. Furthermore, the spin rate ωo = φ. Substitution of these terms into (3.35) yields
M2o = −mgLsinθ = (I1 − I3) ψ2sinθcosθ − I3ψφsinθ (3.36)
Equation (3.36) is equivalent to (3.32) from Method 1.
46
y
x
z
1
Figure 3.5: Stability of a spin plate
3.3 Introduction to stability of a spin body
Stability analysis of a spin plate:
Let xyz be principal axes of a spinning rectangular plate as shown in Figure 3.5. We want
to analyze the stability of rotation about each principal axis.
By stability of rotation, we ask the question: during a steady spin about each axis, if the
initial rotation is applied so close to the principal axes (is perturbed a bit in every directions),
will the rotation remain close to the principal axes (does the perturbation die out), or will the
body begin to see increasing rotation about one of the other axes (does the perturbation grow
with time)?
To analyze this problem, let’s first formulate the Euler’s equations for the spin plate as
follows:M1c = I1cω1 + (I3c − I2c) ω2ω3
M2c = I2cω2 + (I1c − I3c) ω1ω3
M3c = I3cω3 + (I2c − I1c) ω1ω2
(3.37)
where subscripts 1, 2, and 3 in (3.37) denote the principal axes of the plate. Due to a steady
spin, the system is moment-free. Hence in (3.37) M1c = M2c = M3c = 0. Let’s assume that the
plate has a steady spin about the axis ‘1’ with a constant speed ω0. (Note that axis-1 can be
any arbitrary principal axis, i.e. x-, y-, or z-axis in Figure 3.5.) Then the plate is perturbed
with small angular velocities η1(t), η2(t), and η3(t), respectively, about all principal axes. Hence
47
the angular velocities in each direction are
ω1(t) = ω0 + η1(t)
ω2(t) = η2(t)
ω3(t) = η3(t)
(3.38)
Substitution (3.38) into (3.37) and neglecting the higher order terms, such as η1η2, η2η3, etc.,
yield
I1cη1 = 0
I2cη2 + (I1c − I3c) ω0η3 = 0
I3cη3 + (I2c − I1c) ω0η2 = 0
(3.39)
The first row of (3.39) implies that η1(t) is constant. In addition, the last two rows of (3.39)
can be written in a matrix form as η2(t)
η3(t)
+
0 (I1c−I3c)ω0
I2c
(I2c−I1c)ω0
I3c0
η2(t)
η3(t)
=
0
0
(3.40)
or
η(t) +Kη(t) = 0 (3.41)
To solve (3.40), assume the solution as the following form
η(t) =
η2(t)
η3(t)
=
a
b
eλt (3.42)
Substitution (3.42) into (3.41) yields
[λI+K]
a
b
eλt =
0
0
(3.43)
For a nontrivial solution, we get the characteristic equation: |λI+K| = 0. The characteristic
roots λ can be solved as
λ2 =(I1c − I3c) (I2c − I1c) ω2
0
I2cI3c(3.44)
There are two roots of λ which are
λ1,2 = ±[
(I1c − I3c) (I2c − I1c) ω20
I2cI3c
] 12
(3.45)
With two roots, the solution (3.42) is then η2
η3
=
a1
b1
eλ1t +
a2
b2
eλ2t (3.46)
To analyze the stability from the values of λ, we can divide λ2 into two cases as follows
48
Case I: (λ2 ≤ 0) In this case, λ1,2 are positive and negative imaginary parts and the rotation
are marginally stable. Specifically, the perturbation causes the oscillatory motion about
the steady state. To satisfy this stable condition, I1c > I2c > I3c or I1c < I2c < I3c. In
other words, the moment of inertia about the spin axis I1c should be either maximum or
minimum.
Case II: (λ2 > 0) In this case, one of the root is positive real and the other is negative real.
With the positive real root, the solution (3.46) shows that the rotation is about to increase
exponentially with time and hence the rotation of the plate is unstable.
From this analysis together with a real demonstration, the students should be able to figure
out that in which directions the rotation of the spin plate are stable.
49
Chapter 4
Dynamics of a Multi-Body
Mechanical System: Newton-Euler
Approach
4.1 Degrees of Freedom (DOF)
Degrees of freedom are a complete set of independent coordinates that used to describe the
motion. For example, a rigid body performing free motion (without any constraints) in 3-D
space needs six degrees of freedom (coordinates) to describe its motion, i.e. three for translations
and another three for rotations. For a system of N -rigid bodies having the 3-D free motion,
the number of DOFs is 6 × N .
4.2 Constraints
If any two rigid bodies are connected to each other, the mechanism connecting the bodies is
called constraint. The constraint imposes additional relative motion of one body with respect
to anothers. With constraints, the motion of each rigid body in all six coordinates are not
independent, hence the number of DOF for each body is reduced to less than six.
50
x
y
ry
zφ
Rx
RzMz
Mz
Aθx
θz
Figure 4.1: A slider
y
x
z
Rx
Ry
Rz
θψ
φ
A
B
Figure 4.2: Ball and socket
4.3 Constraint Equations
The constraint equations describe the relative motions of any two connected bodies. We can
learn to construct these constraint equations by the following examples.
Example 1: slider Four constraint forces and couples Rx, Rz ,Mx andMz in the frictionless
slider A as shown in Figure 4.1 result in four constraint equations, i.e. rx = 0, rz = 0,
θx = 0 and θz = 0. Without friction, the slider translates free along y-direction and
also rotate free about y-axis. In this case, two coordinates such as ry and φ as seen in
Figure 4.1 can be chosen as the DOFs to describe such translation and rotation.
Example 2: spherical joint Three constraint forces Rx, Ry, and Rz in the spherical joint
51
yx
z
a
c
rcz
rcy
rcxA
A’
θy
θx
Figure 4.3: A rolling sphere
as shown in Fig. 4.2 result in three constraint equations, i.e. rx = 0, ry = 0, and rz = 0.
In Figure 4.2, link B that connected to the stationary link A through the joint can rotate
free about its center, assuming no friction. In this case, three spherical coordinates or the
conventional Euler angles θ, ψ, and φ are the DOFs used to describe the rotation.
Example 3: rolling sphere Consider the spherical ball rolls without slipping as shown in
Fig. 4.3. The first geometric constraint relation, i.e. rcz = a, can be simply observed.
Another two relations are derived from the fact that the contact point A on the sphere is
motionless with respect to the contact point A′ on the surface. Hence
(vAx)rel = rcx − θya = 0
(vAy)rel = rcy + θxa = 0
Or the velocities of the C.G. are then
vcx = rcx = θya
vcy = rcy = −θxa
Note that there exist three unknown constraint forces Rx, Ry, and Rz for this case.
From these previous examples, the number of DOFs of each body is equal to [6 − number of
constraint equations (or constraint forces)]. The chosen DOFs in each case are called generalized
coordinates.
Now let’s consider the multi-body linkages in Fig. 4.4. From the previous examples, we
can conclude that the total constraint equations is equal to 4 (from the slider) + 3 (from the
52
X
Y
φ yx
Z
z
θx θy
θz
rY
Figure 4.4: Combined constraints
spherical joint) = 7. The number of DOFs is therefore equal to 2 × 6 − 7 = 5. The generalized
coordinates, in this case, are ry, φ, and the other three spherical coordinates at the spherical
joint.
Generally speaking, the number of degrees of freedom of a multi-body system is
M = 6 × N −∑
C
where M is the number of degrees of freedom, N is number of rigid bodies,∑
C is number of
all constraint equations.
4.4 Classification of Constraints
If the constraint equation can be derived as a function of only generalized coordinates and time,
e.g. examples 1 and 2 in section 4.3, these constraints are classified as holonomic constraints. In
addition, the holonomic constraints can be divided into two classes: scleronomic and rheonomic.
The constraint equation for the scleronomic constraint is an implicit function of time whereas
the equation for rheonomic constraint is an explicit function of time.
If one of the constraint equation is a function of both the generalized coordinates and
their time derivatives, such constraint is classified as nonholonomic constraint, e.g. example 3
in Section 4.3: rolling sphere.
53
4.5 Number of DOF vs. Driving Forces
If the motion along L coordinates can be prescribed as functions of time, so called the prescribed
motions, the number of DOF is then reduced by L. In order to have the mechanical system
perform such prescribed motions, the corresponding driving forces need to be applied to the
system. For instant, the driving torque is applied to the motor to assure a constant speed of
the rotor. With the prescribed motion in the system, the number of DOF of multi-body system
is
M = 6 × N −∑
C − L
where L is number of the prescribed motions.
4.6 Dynamic Analysis of Multi-Body Mechanical Systems
Dynamic analysis of a multi-body mechanical system can be separated into two main parts:
kinematics and kinetics. Detailed analysis of each part is described as follows.
Kinematic analysis :
1. Choose reference coordinate system for each body
2. Define generalized coordinates
3. Formulate components of velocity and angular velocity in terms of the generalized
coordinates along the reference coordinate system
Kinetic analysis :
1. Express Newton-Euler’s equations governing dynamics of each rigid body
2. With free body diagram (FBD), determine components of forces and moments cor-
responding to the reference coordinates
3. Substitute forces and kinematic relations into Newton-Euler’s equations
4. Eliminate all unknown forces to obtain equations of motion (number of equations of
motion is equal to number of DOF.)
5. Solve the equations of motion to determine the time responses and then use them to
obtain all unknown forces
54
yy
m, l y2
z2
z2
Z, z1
Z, z1
y1
a
y1
x1X
Yα
α
β
β
Link 1
Link 2
y2
c
c
Md
Md
rG
RAy
RAy
MAz
MAz
MCY
MCX
MAy
MAy
RAz
RAz
RCZ
RCY
RCX
RAx
RAx
mg
A
A
A
C
FBD of link 1
FBD of link 2
Figure 4.5: Two-link arms
4.7 Example Problem: Dynamics of Two-Link Arms
The two-link arms are connected by the hinge support A as shown in Figure 4.5. Link 1 is
approximately massless and is driven by a motor which is excluded from the system. The driving
torque Md provided by the motor is related to the speed ω (in rad/s) as Md = M0 − ∆Mω,
where M0 and ∆M are constant parameters. Link 2 has mass m and length l. In addition, the
rest dimensions and coordinates are shown in Fig. 4.5. Derive equation governing the motion
of link 2 and solve for time response, given the initial conditions: β(0) = 0, β(0) = 0, ω(0) = 0,
and ω(0) = 0.
Kinematic analysis
Number of DOF = (2 × 6) - number of constraint equations = 2 × 6 − (5 + 5) = 2
Therefore we need two DOFs to describe the motion of this system. In this case we choose
α and β as the generalized coordinates. Fig. 4.5 also shows the coordinate systems and their
unit vectors.
55
The angular velocities of link 1 and 2 and the velocity at the C.G. (point c) of link 2 are,
respectively,
Ω1 = αk1 (4.1)
Ω2 = αk1 + βi2
= βi2 + αsinβj2 + αcosβk2
(4.2)
vG2 =(−aα − l
2αsinβ
)i2 +
l
2βj2 (4.3)
where IJK, i1j1k1, and i2j2k2 are the unit vectors of XY Z, x1y1z1, and x2y2z2, respectively.
The acceleration at CG of link 2 is then
vG2 =(−aα − l
2 αsinβ − l2 αβcosβ
)i2
+(
l2 β − aα2cosβ − l
2 α2sinβcosβ
)j2
+(
l2 β
2 + aα2sinβ + l2 α
2sin2β)k2
(4.4)
Kinetic analysis
Figure 4.5 shows the free body diagram of both links. First let’s consider link 2. The Newton’s
equation governing the translation of link 2 is
mvG2 = Fx2i2 + Fy2j2 + Fz2k2 (4.5)
where the resultant forces are determined from the free body diagram as
Fx2 = RAx, Fy2 = RAy − mgsinβ, Fz2 = RAz − mgcosβ (4.6)
Euler’s equations governing the rotation of link 2 are
x2 : M1c = I1cω1 + (I3c − I2c) ω2ω3
y2 : M2c = I2cω2 + (I1c − I3c) ω1ω3
z2 : M3c = I3cω3 + (I2c − I1c) ω1ω2
(4.7)
where
I1c = I2c =ml2
12, I3c = 0 (4.8)
ω1 = β, ω2 = αsinβ, ω3 = αcosβ (4.9)
M1c = −RAyl
2, M2c = RAx
l
2+ MAy, M3c = MAz (4.10)
Substitution of (4.4), (4.6), and (4.8)-(4.10) into (4.5) and (4.7) yields six scalar equations for
link 2:
m
(−aα − l
2αsinβ − l
2αβcosβ
)= RAx (4.11)
56
m
(l
2β − aα2cosβ − l
2α2sinβcosβ
)= RAy − mgsinβ (4.12)
m
(l
2β2 + aα2sinβ +
l
2α2sin2β
)= RAz − mgcosβ (4.13)
−RAyl
2=
ml2
12β − ml2
12α2sinβcosβ (4.14)
RAxl
2+ MAy =
ml2
12
(αsinβ + 2αβcosβ
)(4.15)
MAz = 0 (4.16)
Now let’s consider link 1. Since link 1 is massless, all components of the resultant force and
resultant couple are then zero. From FBD of link 1 in Figure 4.5, consider only the Euler’s
equation in z1-direction which is
z1 : Md − MAzcosβ − MAysinβ + RAxa = 0 (4.17)
Or
MAy =Md
sinβ− MAzcotβ +
RAxa
sinβ(4.18)
Plug (4.18) and (4.16) into (4.15) to obtain
RAxl
2+
Md
sinβ+
RAxa
sinβ=
ml2
12
(αsinβ + 2αβcosβ
)(4.19)
Then plug (4.11) into (4.19) to eliminate RAx and rearrange the equation as get
ma2α+ml2
3αsin2β+malαsinβ+
512
ml2αβsinβcosβ+mal
2αβcosβ−(M0 − ∆Mα) = 0 (4.20)
To eliminate RAy, plug (4.14) into (4.12) and rearrange the equation as
23lβ − 2
3lα2sinβcosβ − aα2cosβ + gsinβ = 0 (4.21)
Note that (4.20) and (4.21) are the set of equations of motion.
To solve the equations of motion numerically, we rewrite (4.20) and (4.21) in state form.
First, let’s define the state variables x1 = α, x2 = β, and x3 = β. By substituting the state
variables into (4.20) and (4.21), the equations of motion can be put into in the state form as
follows.
x =
x1
x2
x3
= f(x1, x2, x3) =
f1
f2
f3
(4.22)
where
f1 =M0 − ∆Mx1 − (mal/2)x1x3cosx2 − (5ml2/12)x1x3sinx2cosx2
ma2 + (ml2/3)sin2x2 + malsinx2
57
0 1 2 3 4 5 6 7 8 9 10-0.5
0
0.5
1
1.5
2
2.5
time (s)
α (s
olid
) an
d β
(das
h); r
ad/s
Time reponses with zero initial conditions
0 1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
6
time (s)
β (d
egre
e)
.
.
Figure 4.6: Time responses of the two link arms
f2 = x3
f3 = x21sinx2cosx2 +
3a2l
x21cosx2 −
3g2l
sinx2
Then the state equation (4.22) is numerically solved using Matlab, where the time response
plots is shown in Figure 4.6. Note that the Matlab m-file is described in Fig. 4.7
58
%Simulation of two-link arms%x(:,1) is omega%x(:,2) is beta%x(:,3) is beta_dotclear allt=[0 10]; %initial and final timex0=zeros(3,1); %initial conditions[t,x]=ode45('link_eqns',t,x0); %solve nonlinear ode of 2-link armsfigure(1), clfsubplot(2,1,1) plot(t,x(:,1), 'r', t,x(:,3), 'b--') xlabel('time (s)') ylabel('omega (solid) and d(beta)/dt (dash); rad/s') %grid on title('Time reponses with zero initial conditions')subplot(2,1,2) plot(t,x(:,2)*180/pi) xlabel('time (s)') ylabel('beta (degree)') grid on%===============================================================function xdot=link_eqns(t,x)m= 2; % in kga= 0.1; % in ml= 0.5; % in mdelta_M= 0.5; %in (Nm)sec/radM_0= 1; % in Nmxdot=zeros(3,1);kk=m*(a^2+l^2/3*sin(x(2))^2+a*l*sin(x(2)));
xdot(1)= (-5/12*m*l^2*x(1)*x(3)*sin(x(2))*cos(x(2)) ... -m*a*l/2*x(1)*x(3)*cos(x(2)) ... -delta_M*x(1)+M_0)/kk;xdot(2)=x(3);xdot(3)=x(1)^2*sin(x(2))*cos(x(2))+3/2/l*a*x(1)^2*cos(x(2)) ... -3/2*9.81/l*sin(x(2));
Figure 4.7: Matlab m-file
59
Chapter 5
Principle of Virtual work and
D’Alembert Principle
5.1 Virtual Displacement and Virtual Work
A virtual displacement δr is defined as an infinitesimal and instantaneous displacement in
an arbitrary direction that does not oppose or violate constraints. Fig. 5.1 and Fig. 5.2 show
examples of a particle under constrained motion. In both figures, F(a) is the applied force,
F(c) is the constraint force, and δr is the virtual displacement. In Fig. 5.1 and Fig. 5.2, the
constraint force F(c) can be expressed as
F(c) = Fcn
where n is a unit vector normal to the path of motion. According the definition of δr, the
virtual displacement is always tangent to the path of motion and orthogonal to n. Hence
δr · n = 0
As a result, the virtual work δW (c) done by a constraint force is zero, i.e.,
δW (c) ≡ F(c) · δr = Fcn · δr = 0 (5.1)
60
X
YnF(a)
F(c)
δr
r
Path of motion
Figure 5.1: Virtual displacement of a particle moving along a constrained path
X
Z
n
g
F(a)F(c)
δrθ
Figure 5.2: Virtual displacement of a bead moving in a circular ring
61
5.2 Holonomic and Nonholonomic Constraints
For the holonomic constraint, the constraint equation can be derived as a function of the
generalized coordinates and time. Therefore the position vector can be put in following form
r(t) = f (q1(t), q2(t), . . . , qM (t)) (5.2)
or
r(t) = f (q1(t), q2(t), . . . , qM (t), t) (5.3)
where qi(t), i = 1, 2, . . . ,M are generalized coordinates and M is the number of DOFs. Equation
(5.2) is for the case of scleronomic constraint where r(t) is an implicit function of time, and
(5.3) is for the case of rheonomic constraint where r(t) is an explicit function of time.
For the nonholonomic constraint, the constraint equations are functions of both the gen-
eralized coordinates and generalized velocities. Hence the general form of position vector is
given by
r(t) = f (q1(t), q2(t), . . . , qM (t), q1(t), q2(t), . . . , qM (t), t) (5.4)
Note that most of the following contents, we deal with the holonomic constraint.
5.3 Spatial Coordinates, Generalized Coordinates, and Jaco-
bian
Generalized coordinates denoted by qi, where i = 1, 2, . . . ,M1, are the set of independent coor-
dinates used to describe the motion of the system. For a motion under geometric (holonomic)
constraints, qi can be any generalized variable natural to the constraints. In other words, they
are not necessarily the spatial position variables. The following examples can elaborate this
concept.
The bead moving along the 3-D constrained path
From Fig. 5.3, let’s choose q = s(t) as a generalized coordinate of the bead. The position vector
r of the bead in terms of spatial coordinates is
r = [ x y z ]T
1M is the number of degrees of freedom.
62
X
Z
Y
path
v(t)s(t)
r
Figure 5.3: Moving particle along the constrained path
where x, y, and z are spatial position variables. From (5.2), the position vector r also can be
written as a function of the generalized coordinate as
r = r(s)
The virtual displacement δr of the bead is then
δr =drds
δs ≡ jδs (5.5)
where j is defined a Jacobian vector, representing a unit vector that tangent to the path of
motion, i.e.,
j =dr(x, y, z)
ds=
dx
dsi+
dy
dsj+
dz
dsk
Specifically, the virtual displacement δr is related to the generalized coordinate through this
Jacobian. In addition (5.5) can be expressed in a matrix form as
δr =[
dxds
dyds
dzds
]Tδs ≡ Jδs (5.6)
where J is a Jacobian matrix. We can see that the generalized coordinate s(t) is chosen such
that it is tangent to the path or natural to the constraint.
Moving cart with the coin on its inclined plane
Let q1 and q2 in Fig. 5.4 be the generalized coordinates of the coin C. The position vector r of
the coin is given by
r = [ x y z ]T = r (q1, q2, t) (5.7)
63
X
Y
Z
q2
ru(t)
q1
path
C
Figure 5.4: Moving coin on the inclined plane of the moving cart
Note that the position vector r in (5.7) is an explicit function of time because of the prescribed
motion u(t) of the cart which is an explicit function of time. Virtual displacement δr of the
coin isδr = ∂r
∂q1δq1 + ∂r
∂q2δq2 + ∂r
∂t δt
=2∑
i=1
∂r∂qi
δqi(5.8)
Note that δt in (5.8) is zero because of the instantaneous virtual displacement. Rewrite δr in
terms of spatial coordinates xyz, therefore
δr =2∑
i=1
[∂x
∂qii+
∂y
∂qij+
∂z
∂qik]δqi (5.9)
Alternatively, the virtual displacement can be put in a matrix form as
δr = Jδq (5.10)
where
δq =[
δq1 δq1
]Tand J is the Jacobian matrix given by
J =
∂x∂q1
∂x∂q2
∂y∂q1
∂y∂q2
∂z∂q1
∂z∂q2
3×2
A system of particles
Fig. 5.5 shows a system of N -particles with K-geometric constraints. Number of degree-of-
freedom of this system is M = 3N − K. Let q1, q2, . . . , qM be all M generalized coordinates.
64
X
Y
Z
m1 m2
mj
m4mN
m3
r1
rN
Figure 5.5: A constrained system of particles
The position vector r to describe the motion of all particles is
r =[rT1 , rT2 , . . . , rTN
]T= [x1, y1, z1, x2, y2, z2 . . . , xN , yN , zN ]T3N×1
For a system with holonomic constraints, the position vector is also a function of generalized
coordinates:
r = r (q1, q2, . . . , qM )
The virtual displacement δr is then
δr =∂r∂q1
δq1 +∂r∂q2
δq2 + . . . +∂r
∂qMδqM
=M∑j=1
∂r∂qj
δqj(5.11)
or
δri =M∑j=1
∂ri∂qj
δqj ; i = 1, 2, . . . , N (5.12)
Equation (5.11) can be put in a matrix form as
δr = Jδq (5.13)
where
δq =[
δq1 δq2 . . . δqM
]TM×1
(5.14)
and J is the Jacobian matrix given by
J =
J11 J12 . . . J1M
J21 J22 . . . J2M...
.... . .
...
JN1 JN2 . . . JNM
3N×M
(5.15)
65
In (5.15), components of the Jacobian matrix Jkl =∂rk∂ql
for k = 1, 2, . . . , N and l = 1, 2, . . . ,M .
In summary, Jacobian embodies the information of unit vectors tangent to the geometric
constraints, which is determined by differentiation of the physical or spatial variables with
respect to the generalized coordinates. Moreover, the virtual work is related to the generalized
coordinates through this Jacobian. Note that the Jacobians derived in the previous examples
are for the holonomic constraints that satisfy (5.2) or (5.3).
5.4 Principle of Virtual Work
Let’s consider a system of N -particles and M -degrees of freedom. If the system is in equilibrium
thenN∑i=1
Fi = 0 (5.16)
where Fi is the total forces acting on the i-th particle. Fi can be divided into three types of
forces: 1) applied or external forces F(a)i ; 2) internal spring or damping forces between ith and
jth particles fij; and 3) constraint forces2 F(c)i . Therefore
Fi = F(a)i +
N∑j=1
fij +F(c)i , i = j (5.17)
The applied forces F(a)i and the internal spring or damping forces fij can be grouped as working
forces so called active forces F(ac)i . In cases of no friction and plastic deformation at constraints,
the constraint forces F(c)i are workless forces. According to (5.1), the virtual work done by all
constraint forces is always zero, i.e.
N∑i=1
F(c)i · δri = 0 (5.18)
If the system of particles is in equilibrium, the virtual work δW done by all forces given by
δW =N∑i=1
Fi · δri = 0
=N∑i=1
(F(ac)
i + F(c)i
)· δri = 0, i = j
(5.19)
The virtual work δW is zero because of the zero sum of all forces. With the relation (5.18),
(5.19) is simply
δW (ac) =N∑i=1
F(ac)i · δri = 0 (5.20)
2Constraints in this case excludes the springs and dampers
66
Equation (5.20) is the principle of virtual work stating that if the system is in equilibrium then
virtual work done by all active forces in the system is zero.
5.5 D’Alembert principle (Dynamic Principle of Virtual Work)
Let’s consider a system of N -particles and M -degrees of freedom. The Newton’s second law
applied to such system can be rewritten as the following alternative form
N∑i=1
(Fi − miri) = 0; (5.21)
where Fi is the total forces acting on the i-th particle, mi is the mass of the i-th particle, and ri
is the acceleration of the i-th particle. The term −miri represents an inertia force resisting the
motion of the system. From (5.21), virtual work δW done by all forces, including the inertia
force −miri, is then zero. Or
δW =N∑i=1
(Fi − miri) · δri = 0 (5.22)
Since the virtual work done by the constraint forces is always zero, (5.22) is simply
δW ′ =N∑i=1
(F(ac)
i − miri)· δri = 0 (5.23)
stating that the virtual work δW ′ done by all active forces and inertia forces is zero. For the
holonomic constraints, we can substitute the Jacobian relation (5.12) into (5.23) to get
N∑i=1
(F(ac)
i − miri)·
M∑k=1
∂ri∂qk
δqk = 0 (5.24)
Rewrite(5.24) asM∑k=1
N∑i=1
(F(ac)i − miri) ·
∂ri∂qk
δqk = 0 (5.25)
Since each δqk is independent and nonzero, therefore to satisfy (5.25), each k-term in the bracket
must be zero, i.e.
N∑i=1
(F(ac)
i − miri)· ∂ri∂qk
= 0; k = 1, 2, . . . ,M (5.26)
For a short notation, we define βik ≡ ∂ri∂qk
for the rest of the chapter. Equations (5.26) are the
D’Alembert principle that automatically yield the equations of motion. Again the D’Alembert
principle can only be applied to dynamics of the systems with holonomic constraints.
67
X
Y
er
eθ
k
k
m2θθ
m1
ρ1
ρ2
Figure 5.6: Example
Example: Direct application of D’Alembert principle
The sliders of masses m1 and m2 are constrained by springs and move along the frictionless disk
slot as shown in Figure 5.6. The disk also rotates about its center with angular displacement
θ. If the unstretched length of the springs is a, derive the equations of motion.
From the system shown in Figure 5.6, let’s choose ρ1, ρ2, and θ as the generalized coordi-
nates. Hence
q =[
ρ1 ρ2 θ
]T(5.27)
The system has three degrees of freedom or M = 3. The system consists of two particles or
N = 2. The position vectors of the sliders 1 and 2 are
r1 = f (ρ1, ρ2, θ) = ρ1er (5.28)
and
r2 = f (ρ1, ρ2, θ) = −ρ2er (5.29)
The accelerations of both sliders are
r1 (ρ1, ρ2, θ) =(ρ1 − θ2ρ1
)er +
(ρ1θ + 2ρ1θ
)eθ (5.30)
and
r2 (ρ1, ρ2, θ) =(−ρ2 + θ2ρ2
)er −
(ρ2θ + 2ρ2θ
)eθ (5.31)
The active spring forces acting on both sliders are
F(ac)1 = −k (ρ1 − a) er (5.32)
68
and
F(ac)2 = k (ρ2 − a) er (5.33)
The components of the Jacobian matrix can be formulated as follows
β11 ≡ ∂r1∂ρ1
= er
β12 ≡ ∂r1∂ρ2
= 0
β13 ≡ ∂r1∂θ
= ρ1eθ
β21 ≡ ∂r2∂ρ1
= 0
β22 ≡ ∂r2∂ρ2
= −er
β23 ≡ ∂r2∂θ
= −ρ2eθ
Substitution of (5.30) to (5.33) into (5.26) yields three sets of equations:
For k = 1, (F(ac)
1 − m1r1)· β11 +
(F(ac)
2 − m2r2)· β21 = 0 (5.34)
or
m1
(ρ1 − θ2ρ1
)+ K (ρ1 − a) = 0 (5.35)
For k = 2, (F(ac)
1 − m1r1)· β12 +
(F(ac)
2 − m2r2)· β22 = 0 (5.36)
or
m2
(ρ2 − θ2ρ2
)+ K (ρ2 − a) = 0 (5.37)
For k = 3, (F(ac)
1 − m1r1)· β13 +
(F(ac)
2 − m2r2)· β23 = 0 (5.38)
or
m1
(ρ1θ + 2ρ1θ
)ρ1 + m2
(ρ2θ + 2ρ2θ
)ρ2 = 0 (5.39)
Equations (5.35) and (5.37) are the equations of motions. In addition, (5.39) can be arranged
asd
dt
(m1ρ
21θ + m2ρ
22θ)
= 0 (5.40)
which indicates the conservation of angular momentum of the system or Ho = 0.
69
Chapter 6
Dynamics Analysis through
Lagrange Mechanics
6.1 Kinetic Energy
Kinetic energy of a system of particles is the total sum of kinetic energy of each particle given
by
T =12
N∑i=1
mivi · vi (6.1)
where N is number of particles and mi and vi are the mass and the velocity of the i-th particle.
Kinetic energy of a rigid body as shown in Figure 6.1 is therefore
T =12
∫v · vdm (6.2)
where v is the velocity of element dm. From Figure 6.1, the velocity v of mass dm is
v = vc + ω × ρ (6.3)
Substitution of (6.3) into (6.2) yields1
T = 12
∫(vc + ω × ρ) · (vc + ω × ρ) dm
= 12
∫vc · vcdm +
∫vc · ω × ρdm + 1
2
∫(ω × ρ) · (ω × ρ) dm
= 12vc · vc
∫dm + (vc × ω) ·
∫ρdm + 1
2
∫ω · (ρ × ω × ρ) dm
(6.4)
The terms∫
ρdm = 0 and∫
dm = m. Also
12
∫ω · (ρ × ω × ρ) dm =
12ω ·
∫ρ × (ω × ρ) dm
1Note that (A × B) · C = A · (B × C)
70
X
Y
Z
Cρ
ω
rc
vc
Figure 6.1: Kinetic energy of a rigid body
where∫
ρ × (ω × ρ) dm = Hc is the angular momentum about the C.G. Hence kinetic energy
of a rigid body (6.4) is simply
T =12mvc · vc +
12Hc · ω (6.5)
Equation (6.5) can be put in the matrix form as
T =12mvTc vc +
12ωT Icω (6.6)
6.2 Potential Energy
Some active forces such as gravitational forces and internal forces due to elastic deformation
can be represented by a gradient operator of a scalar potential energy function V as
F = −∇V (r) (6.7)
where ∇ is the gradient operator given (in cartesian coordinates) by
∇ =∂
∂xi+
∂
∂yj+
∂
∂zk
Such forces in form of (6.7) are called conservative forces and the function V is called potential
energy. Consequently work done by the conservative forces is is independent to the path of
motion and equal to the change of the potential energy ∆V . Therefore the work done by the
conservative force F, resulting in any path of motion from A to B, is
W =∫ B
AF · δr ≡ V (rA) − V (rB) (6.8)
71
Or virtual work δW done by the conservative force is
δW = F · δr = −dV (6.9)
Examples of the conservative forces F and their corresponding potential energy V are as follows.
1. Gravity near the Earth’s surface
F = −mgez, V = mgz
2. Gravitational force
F = −GMm
r2er, V = −GMm
r
3. Elastic spring force
F = −kx, V =12kx2
4. Magnetic force between two wires
F =µ0I1I2
2πr, V = −µ0I1I2logr
2π
where µ0 = 4π × 10−7 in mks units.
5. Electric force between two charges
F =Q1Q2
4πε0r2, V = −Q1Q2
4πε0r
where1
4πε0= 8.99 × 109 in mks units.
6.3 Remarks on Properties of Generalized Coordinates for the
System with Holonomic constraints
1. For a multi-degree-of-freedom dynamical system, all generalized coordinates q1, q2, . . . , qM
(where M is the number of degrees of freedom) are independent.
2. For a system of particles with holonomics (geometric) constraints, the position vectors
ri; i = 1, 2, . . . , N (N is number of particles) can be expressed in terms of the generalized
coordinates and time t as
ri = ri (q1, q2, . . . , qM , t)
72
3. For a system of particles with holonomics (geometric) constraints, the virtual displacement
can be expressed in terms of the generalized coordinates and time as
δri =∂ri∂q1
δq1 +∂ri∂q2
δq2 + . . . +∂ri∂qM
δqM +∂ri∂t
δt
=M∑j=1
∂ri∂qj
δqj
Note that δt is zero because the virtual displacement is an instantaneous displacement,
and∂ri∂qj
= f (q1, q2, . . . , qM , t).
4. With only the holonomic constraints in the system, the actual velocity of the i-th particle
can be derived in terms of generalized coordinates as
ri =∂ri∂q1
dq1
dt+
∂ri∂q2
dq2
dt+ . . . +
∂ri∂qM
dqMdt
+∂ri∂t
=M∑j=1
∂ri∂qj
qj +∂ri∂t
≡ f (q1, q2, . . . , qM , q1, q2, . . . , qM , t)
where qj ≡dqjdt
is the generalized velocity.
5. Remarkable observation 1
∂ri∂qk
=∂ri∂qk
, k = 1, 2, . . . ,M ; i = 1, 2, . . . , N
6. Remarkable observation 2
∂ri∂qk
=d
dt
(∂ri∂qk
), k = 1, 2, . . . ,M ; i = 1, 2, . . . , N
6.4 Derivation of Lagrange’s equations through D’Alembert Prin-
ciple
Let’s consider a system of N -particles with M -degrees of freedom. From the D’Alembert
principle:N∑i=1
(miri − F(ac)
i
)· ∂ri∂qk
= 0; k = 1, 2, . . . ,M (6.10)
Consider (6.10) term by term. The first term on the left can be written as
N∑i=1
miri ·∂ri∂qk
=N∑i=1
[d
dt
(miri ·
∂ri∂qk
)− miri ·
d
dt
(∂ri∂qk
)](6.11)
73
With the remarkable observation 1 and 2, (6.11) can be rearranged asN∑i=1
miri ·∂ri∂qk
=N∑i=1
[d
dt
(miri ·
∂ri∂qk
)− miri ·
∂ri∂qk
]
=N∑i=1
(d
dt
[∂
∂qk
(12miri · ri
)]− ∂
∂qk
[12miri · ri
])
=N∑i=1
(d
dt
[∂Ti
∂qk
]− ∂Ti
∂qk
)
=d
dt
(∂T
∂qk
)− ∂T
∂qk
where Ti is the kinetic energy of i-particle, and T is the total kinetic energy.
The second term of (6.10) is defined as generalized forces Qk. Or
Qk ≡N∑i=1
F(ac)i · ∂ri
∂qk; k = 1, 2, . . . ,M (6.12)
Consequently (6.10) becomes
d
dt
(∂T
∂qk
)− ∂T
∂qk= Qk; k = 1, 2, . . . ,M (6.13)
There are two alternative ways to determine the generalized forces Qk as described in the
following details.
1. From (6.12), Qk is defined as the component of all active forces that projected on the
direction of k-generalized coordinate.
2. Since virtual work done by all active forces is
δW (ac) =M∑k=1
[N∑i=1
F(ac)i · ∂ri
∂qk
]δqk =
M∑k=1
Qkδqk (6.14)
Any generalized force Qj is therefore determined from the virtual work done by enforcing
virtual displacement in the particular j-coordinate for one unit (δqj = 1), and enforcing
zero virtual displacements in the other coordinates (δqi = 0, i = j).
In addition, the generalized forces can be separated into two cases: conservative generalized
forces Q(c)k and nonconservative generalized forces Q
(nc)k . The conservative generalized force
Q(c)k can be expressed in term of the potential energy as
Q(c)k =
N∑i=1
F(c)i · ∂ri
∂qk
=N∑i=1
[−∂V
∂ri(q1, q2, . . . , qM ) · ∂ri
∂qk
]
= − ∂V
∂qk
74
θ
m
lg
Figure 6.2: A pendulum
Substitution of the conservative generalized forces Q(c)k into (6.13) yields the Lagrange’s equa-
tionsd
dt
(∂T
∂qk
)− ∂T
∂qk+
∂V
∂qk= Q
(nc)k ; k = 1, 2, . . . ,M (6.15)
Let’s define a Lagrange function L such that
L(q, q) ≡ T (q, q) − V (q)
An alternative form of the Lagrange’s equations is then
d
dt
(∂L∂qk
)− ∂L
∂qk= Q
(nc)k ; k = 1, 2, . . . ,M (6.16)
6.5 Examples
Example 6.1:
Derive the equation of motion of the pendulum in Figure 6.2 using the Lagrange’s equation.
Solution
The pendulum shown in Fig. 6.2 has one degree of freedom. Let’s choose θ as the generalized
coordinate. The kinetic energy T of the pendulum is then
T =12mv2 =
12ml2θ2 (6.17)
Also the potential energy V of the pendulum (with respect to the datum at the hinge level) is
V = −mglcosθ (6.18)
75
m1
m2
k
g θ
x
er
eθ
Figure 6.3: A cart and pendulum
There is no external forces applied to the pendulum, hence the generalized force is zero. The
Lagrange’s equation for the pendulum is
d
dt
(∂T
∂θ
)− ∂T
∂θ+
∂V
∂θ= 0 (6.19)
Substitution of (6.17) and (6.18) into (6.19) yields the equation of motion:
ml2θ + mglsinθ = 0 (6.20)
Example 6.2:
Derive the equations of motion for the cart-pendulum as shown in Fig. 6.3.
Solution
The cart-pendulum system has two degrees of freedom. Let the generalized coordinates be
q1 = x, q2 = θ. Absolute velocity v2 of the pendulum is
v2 = xex + Lθeθ
= xsinθer +(xcosθ + Lθ
)eθ
(6.21)
where L is the length of the pendulum. Also
v2 · v2 = x2sin2θ +(xcosθ + Lθ
)2
= x2 + 2Lxθcosθ + L2θ2(6.22)
The total kinetic energy T of the system is
T =12m1x
2 +12m2v2 · v2 (6.23)
76
Substitute (6.22) into (6.23) to get
T =12
(m1 + m2) x2 + m2Lxθcosθ +12m2L
2θ2 (6.24)
The potential energy of the system is
V =12kx2 − m2gL cos θ (6.25)
The Lagrange’s equations are
d
dt
(∂T
∂x
)− ∂T
∂x+
∂V
∂x= 0 (6.26)
andd
dt
(∂T
∂θ
)− ∂T
∂θ+
∂V
∂θ= 0 (6.27)
Formulate each term in (6.26) and (6.27) as follows:
∂T
∂x= (m1 + m2) x + m2Lθcosθ (6.28)
d
dt
(∂T
∂x
)= (m1 + m2) x + m2Lθcosθ − m2Lθ2sinθ (6.29)
∂T
∂x= 0 (6.30)
∂T
∂θ= m2Lxcosθ + m2L
2θ (6.31)
d
dt
(∂T
∂θ
)= m2Lxcosθ − m2Lxθsinθ + m2L
2θ (6.32)
∂T
∂θ= −m2Lxθsinθ (6.33)
∂V
∂x= kx (6.34)
∂V
∂θ= m2gLsinθ (6.35)
Plug (6.29), (6.30) and (6.32) to (6.35) into (6.26) and (6.27) to obtain the equations of motion:
(m1 + m2) x + m2Lθcosθ − m2Lθ2sinθ + kx = 0 (6.36)
and
m2L2θ + m2Lxcosθ + m2gLsinθ = 0 (6.37)
Lagrange’s equations can be used to derive equations of motion of the rigid body or multi-
body system. In this case, derivation of the Lagrange’s equation is similar to that for the system
of particles and will be omitted here. Also the Lagrange equations for the rigid body or multi-body
system are identical to (6.15) and (6.16).
77
l1
τ1
G1
l2
τ2G2
g
α1
α2
a1
a2
m1,
I1(about G)
m2,
I2(about G)
Figure 6.4: A rigid two link arm system
Example 6.3:
The two-link arm robot as shown in Figure 6.4 is operated in a horizontal plane. The motion
of the arms is controlled by two motors installed at the joints. The motors generate moments
τ1 and τ2 as shown in Figure 6.4. Derive equations of motion for the two-link arm robot.
Solution
Kinematics: There are two degrees of freedom in this case. Let’s define α1 and α2 as the
generalized coordinates. The position vector and the velocity of G2 can be written in terms of
both generalized coordinates as
rG2 = (l1cosα1 + a2cosα2) i+ (l1sinα1 + a2sinα2) j (6.38)
rG2 = (−l1α1sinα1 − a2α2sinα2) i+ (l1α1cosα1 + a2α2cosα2) j (6.39)
The dot product of the velocity rG2is then
rG2 · rG2 = (−l1α1sinα1 − a2α2sinα2)2 + (l1α1cosα1 + a2α2cosα2)2
= l21α21 + a2
2α22 + 2l1a2α1α2cos (α1 − α2)
(6.40)
Kinetic and potential energy:
The total kinetic energy T of the two-link arms is
T = T1 + T2
= 12Io1α
21 +
(12m2rG2 · rG2 + 1
2IG2α22
)= 1
2
(I1 + m1a
21
)α2
1 + 12m2
[l21α
21 + a2
2α22 + 2l1a2α1α2cos (α1 − α2)
]+ 1
2I2α22
(6.41)
78
τ1
τ2
τ2g
α1
α2
Figure 6.5: Reaction torques
Since the whole system operates in the horizontal plane and there is no restoring forces, the
potential energy is zero, i.e. V = 0.
Generalized forces Qα1 and Qα2 :
From Figure 6.5, the virtual work done by all nonconservative torques is
δW = τ1δα1 − τ2δα1 + τ2δα2 (6.42)
Hence
Qα1 = τ1 − τ2 (6.43)
and
Qα2 = τ2 (6.44)
Formulate the Lagrange’s equations as follows:
d
dt
(∂T
∂α1
)− ∂T
∂α1+
∂V
∂α1= Qα1 (6.45)
d
dt
(∂T
∂α2
)− ∂T
∂α2+
∂V
∂α2= Qα2 (6.46)
where∂T
∂α1=(I1 + m1a
21
)α1 + m2l
21α1 + m2a2l1α2cos (α1 − α2) (6.47)
d
dt
(∂T
∂α1
)=
(I1 + m1a
21 + m2l
21
)α1 + m2a2l1α2cos (α1 − α2)
−m2a2l1α2 (α1 − α2) sin (α1 − α2)(6.48)
∂T
∂α1= −m2a2l1α1α2sin (α1 − α2) (6.49)
∂T
∂α2= m2a
22α2 + m2a2l1α1cos (α1 − α2) + I2α2 (6.50)
79
d
dt
(∂T
∂α2
)=
(m2a
22 + I2
)α2 + m2a2l1α1cos (α1 − α2)
−m2a2l1α1 (α1 − α2) sin (α1 − α2)(6.51)
∂T
∂α2= m2a2l1α1α2sin (α1 − α2) (6.52)
Substitution of (6.47)-(6.52) into (6.45) and (6.46) yields the following equations of motion:
(I1 + m1a
21 + m2l
21
)α1 + m2a2l1α2cos (α1 − α2) + m2a2l1α
22sin (α1 − α2) = τ1 − τ2 (6.53)
(m2a
22 + I2
)α2 + m2a2l1α1cos (α1 − α2) − m2a2l1α
21sin (α1 − α2) = τ2 (6.54)
Example 6.4:
Fig. 6.6 shows a uniform and thin bar of mass m and length l hinged to link 1 which is driven
to spin with a constant speed ω. Derive the differential equations governing the motion of the
thin bar using Lagrange’s equations.
Solution
With the prescribed motion ω is constant, the number of degrees of freedom is M = 6 × N −∑C − L = 6 × 2 − (5 + 5) − 1 = 1. Let’s choose β as the generalized coordinate. The angular
velocity of the link 2 is
ω2 = ωk1 + βi2
= βi2 + ωsinβj2 + ωcosβk2
≡[
β ωsinβ ωcosβ
]T (6.55)
Kinetic energy T is
T =12IZω2 +
12ωT
2 Iω2 (6.56)
Substituting (6.55) into (6.56) yields
T = 12IZω2 + 1
2
[β ωsinβ ωcosβ
]
I 0 0
0 I 0
0 0 0
β
ωsinβ
ωcosβ
= 12IZω2 + 1
2
(β2 + ω2sin2β
)(6.57)
Potential energy is V = −mg(l/2)cosβ. The Lagrange’s equation can be formulated as
d
dt
(∂T
∂β
)− ∂T
∂β+
∂V
∂β= Q (6.58)
where∂T
∂β= Iβ (6.59)
80
β
ω
g
k1
k2 j2
l
Figure 6.6: A spinning pendulum
d
dt
(∂T
∂β
)= Iβ (6.60)
∂T
∂β= Iω2sinβcosβ (6.61)
∂V
∂β= mg
l
2sinβ (6.62)
and Q = 0. Plug (6.59)-(6.62) into (6.58) to get the equation of motion:
Iβ − Iω2sinβcosβ + mgl
2sinβ = 0 (6.63)
6.6 Lagrange Multiplier
Lagrange equation is derived from the D’Alembert principle. Originally, it can be put in the
variational form as
M∑k=1
d
dt
(∂T
∂qk
)− ∂T
∂qk+
∂V
∂qk− Q
(nc)k
δqk = 0; k = 1, 2, . . . ,M (6.64)
where M is number of degrees of freedom. If all δqk are independent, each bracket in (6.64) is
zero and we obtain the Lagrange equations as shown in (6.15). If additional p constraints are
introduced later, and result in the dependency of some qk, what happens? Let’s consider the
equation (6.64). The introduction of new constraints will lead to the following conditions:
81
1. There exist new constraints which can be expressed by the general form of p constraint
equationsM∑k=1
aikdqk = 0; i = 1, 2, . . . , p (6.65)
where p is the number of additional constraints. (6.65) is also a general form of nonholo-
nomic constraints.
2. With conditions (6.65), Equation (6.64) can be rewritten as
M∑k=1
d
dt
(∂T
∂qk
)− ∂T
∂qk+
∂V
∂qk− Q
(nc)k
δqk +
p∑i=1
λi
M∑k=1
aikδqk
= 0 (6.66)
where λi is called Lagrange Multiplier. Equation (6.66) can be rearranged as
M∑k=1
d
dt
(∂T
∂qk
)− ∂T
∂qk+
∂V
∂qk− Q
(nc)k +
p∑i=1
λiaik
δqk = 0 (6.67)
Now we have M equations from (6.67) plus p constraint equations from (6.65) and have M + p
unknowns which are q1, q2, . . . , qM , λ1, λ2, . . . , λp. If the virtual generalized coordinates are
arranged such that δq1, δq2, . . . , δqM−p are independent, and δqM−p+1, δqM−p+2, . . . , δqM are
dependent. Then the following procedure is performed to derive the equations of motion.
First we choose λ1, λ2, . . . , λp so that each coefficient in the bracket in (6.67) corresponding to
δqM−p+1, δqM−p+2, . . . , δqM is zero. Or
d
dt
(∂T
∂qk
)− ∂T
∂qk+
∂V
∂qk− Q
(nc)k +
p∑i=1
λiaik = 0; k = M − p + 1,M − p + 2, . . . ,M (6.68)
With the chosen λi and independent q1, q2, . . . , qM−p, the rest coefficients in the the bracket of
(6.67) corresponding to δq1, δq2, . . . , δqM−p are all zero. In summary, we obtain the following
relation:d
dt
(∂T
∂qk
)− ∂T
∂qk+
∂V
∂qk= Q
(nc)k −
p∑i=1
λiaik; k = 1, 2, . . . ,M (6.69)
Note that the new constraints are introduced to the Lagrange equations as the generalized
forces as seen from the second term on the right of (6.69). Moreover the Lagrange equation
with Lagrange Multiplier can deal with dynamics with nonholonomic constraints.
Then we solve (6.69) together with the revised constraint relations (6.65), putting in the
form ofM∑k=1
aikqk = 0; i = 1, 2, . . . , p (6.70)
or in the integral form
fi (q1, q2, . . . , qM ) = 0; i = 1, 2, . . . , p (6.71)
82
θ
m
rg
Figure 6.7: A pendulum without constraint
Example 6.5:
Derive equation of motion of a pendulum shown in Figure 6.2 using the Lagrange multiplier.
Solution
First if we assume that the pendulum is not constrained in the radial direction, i.e. l is not
fixed, this system will have two degrees of freedom. Let r and θ be the generalized coordinates
as shown in Figure 6.7. The kinetic and potential energies are
T =12m(r2 + r2θ2
)
V = −mgr cos θ
The Lagrange equation in r-coordinate is
ddt
(∂T∂r
)− ∂T
∂r + ∂V∂r = 0
ddt (mr) − mrθ2 − mg cos θ = 0
(6.72)
The Lagrange equation in θ-coordinate is
ddt
(∂T∂θ
)− ∂T
∂θ + ∂V∂θ = 0
ddt
(mr2θ
)+ mgr sin θ = 0
(6.73)
Then we impose the constraint equation, i.e. r = l or δr = 0. Combine (6.72) and (6.73)
together with the imposed constraint, we get
[mr − mrθ2 − mg cos θ
]δr
[d
dt
(mr2θ
)+ mgr sin θ
]δθ + λδr = 0 (6.74)
or [mr − mrθ2 − mg cos θ + λ
]δr
[d
dt
(mr2θ
)+ mgr sin θ
]δθ = 0 (6.75)
83
where λ is the Lagrange multiplier. Choose λ such that
mr − mrθ2 − mg cos θ + λ = 0 (6.76)
andd
dt
(mr2θ
)+ mgr sin θ = 0 (6.77)
In (6.76) and (6.77), there are three unknowns: r, θ and λ. Therefore to solve these equations
for r, θ and λ, we need another one equation which is the constraint equation:
r = l (6.78)
Plugging (6.78) into (6.76) and (6.77) yields
mlθ2 + mg cos θ = λ (6.79)
and
ml2θ + mgl sin θ = 0 (6.80)
Note that (6.80) is equivalent to the equation of motion that we obtain in Example 6.1. In
addition (6.79) gives us the Lagrange multiplier λ which is, in this case, the tension or the
constraint force in the string.
84
Chapter 7
Stability Analysis
7.1 Equilibrium, Quasi-Equilibrium, and Steady States
Equilibrium is the state in which a system is at stationary; i.e. q = 0 and q = 0, where q is
the vector of generalized coordinates.
Quasi-equilibrium or steady state is the state that some coordinates of a system are in
equilibrium, meanwhile the system has steady rotations (with constant speed) about some axes,
e.g. steady precession of the top.
7.2 Stability of Equilibrium States or Steady States
To determine if the equilibrium or the steady state is stable, we initially perturb the system
from each state with a small perturbation, and then investigate how the perturbation changes
with time. If the perturbation dies out or possesses a small oscillation, the perturbed state is
stable. If the perturbation grows with time, that perturbed state is unstable.
From the previous chapters, the equations of motion can be put in a general form as
qi = fi (q1, q2, . . . , qk, q1, q2, . . . , qk, t) ; i = 1, 2, . . . , k (7.1)
where k is the number of degrees of freedom. Let’s define the state variables X1, X2, . . . , Xk,
Xk+1, Xk+2, . . . , X2k as
X1 = q1,X2 = q2, . . . ,Xk = qk (7.2)
85
and
Xk+1 = q1,Xk+2 = q2, . . . ,X2k = qk (7.3)
Also define a state vector X as
X = [X1,X2, . . . ,Xk,Xk+1,Xk+2, . . . ,X2k]T (7.4)
Then equation (7.1) can be written in terms of the state variables given by:
X = f (X1,X2, . . . ,Xk,Xk+1,Xk+2, . . . ,X2k, t)
= [f1, f2, . . . , f2k]T(7.5)
Equation (7.5) is called state equation. For equilibrium, X = 0 and (7.5) becomes
0 = f(X1, X2, . . . , Xk, Xk+1, Xk+2, . . . , X2k, t
)(7.6)
where X1, X2, . . . , X2k are the set of equilibrium state determined from (7.6).
To analyze the stability, the system is initially perturbed from the equilibrium or the
steady state with a small perturbation ∆X(0) in every coordinates. Thus after the initial time,
the state vector X(t) is then
X(t) = X+ ∆X(t) (7.7)
where
∆X(t) = [∆X1(t), ∆X2(t), . . . , ∆Xk(t), ∆Xk+1(t), ∆Xk+2(t), . . . , ∆X2k(t)]T (7.8)
In (7.8), ∆X1, ∆X2, . . . , ∆X2k are small perturbation. Substitution of (7.7) into the equations
of motion (7.5) yields
˙X+ ∆X(t) = f(X1 + ∆X1, X2 + ∆X2, . . . , X2k + ∆X2k, t
)(7.9)
Equation (7.9) can be expanded using the Taylor’s series as
˙X+ ∆X(t) ≈ f(X1, X2, . . . , X2k, t
)+[
∂f∂X
]X=X
∆X(t) (7.10)
Since ˙X = 0 and with the equilibrium condition (7.6), Equation (7.10) becomes
∆X(t) ≈[
∂f∂X
]X=X
∆X(t) = A∆X(t) (7.11)
where
A ≡[
∂f∂X
]X=X
=
A11 A12 . . . A1,2k
A21 A22 . . . A2,2k
......
. . ....
A2k,1 A2k,2 . . . A2k,2k
(7.12)
86
The component Aij in (7.12) is given by
Aij =
[∂fi∂Xj
]X=X
; i = 1, 2, . . . , 2k; j = 1, 2, . . . , 2k (7.13)
For a short notation, replace ∆X(t) in (7.11) with Y(t). Hence the perturbation equation
(7.11) becomes
Y(t) = AY(t) (7.14)
To analyze the stability, we then solve for Y(t). First let the solution be
Y(t) = Ceλt (7.15)
Substituting (7.15) into (7.14) yields
λCeλt = ACeλt (7.16)
Or
(A− λI)Ceλt = 0 (7.17)
Thus the nontrivial solution of (7.17) is
|A− λI| = 0 (7.18)
From (7.18), there are 2k values of λ. The equilibrium or the steady state will be unstable if
either one of the following conditions is satisfied.
1. There exist real roots of λ and at least one of them is positive.
2. There exist complex roots of λ and at least one of them has a positive real part.
Example 7.1:
The bead is constrained to move along the circular ring as shown in Fig. 7.1. If the ring is
rotated about the vertical axis with a constant speed ω. Determine the steady states and
analyze if each state is stable or unstable.
Solution
Kinematics:
The system has only one degree of freedom. From Fig. 7.1, let α be the generalized
coordinate. Hence the absolute velocity of the bead is
v = rαeθ + (rsinα) ωez (7.19)
87
er
eθr
g
ω constant
smooth
Figure 7.1: Stability analysis of a bead steady motion
Kinetic energy T and potential energy V can be formulated as
T =12mv · v =
12m(r2α2 + ω2r2sin2α
)(7.20)
V = −mgrcosθ (7.21)
Lagrange equation is thend
dt
(∂T
∂α
)− ∂T
∂α+
∂V
∂α= 0 (7.22)
Formulate each term in (7.22) and substitute into the equation to get equation of motion:
mr2α − mω2r2sinαcosα + mgrsinα = 0 (7.23)
Determine steady states:
Let’s define the state variables: X1 = α and X2 = α.
Hence the state equations are
X1 = X2
X2 = ω2sinX1cosX1 −g
rsinX1
(7.24)
(7.24) can be put in the matrix form as
X =
X1
X2
=
X2
ω2sinX1cosX1 −g
rsinX1
= f (X1,X2) (7.25)
For the steady state, X = 0. Therefore, (7.25) becomes 0
0
=
X2
ω2sinX1cosX1 −g
rsinX1
(7.26)
88
From (7.26), X2 = 0 and
sinX1
(ω2cosX1 −
g
r
)= 0 (7.27)
There are two possible solutions for (7.27): X1 = 0 or X1 = cos−1(
g
ω2r
). Therefore the system
has two steady states which are
X(1) =
0
0
, X(2) =
cos−1
(g
ω2r
)0
(7.28)
Note that the second steady state X(2) exists if and only ifg
r≤ ω2.
Stability analysis
To analyze the stability of each steady state, the system is initially perturbed with ∆X(0) from
the steady state. After t > 0, the motion is described by
X(t) = X+
∆X1(t)
∆X2(t)
(7.29)
Substitute (7.29) into (7.25), and then linearize the equation. Let Y(t) = ∆X(t) for a short
notation, the perturbation equation is therefore
Y(t) = AY(t) (7.30)
where
A11 =[
∂f1
∂X1
]X=X
=[∂X2
∂X1
]X=X
= 0
A12 =[
∂f1
∂X2
]X=X
=[∂X2
∂X2
]X=X
= 1
A21 =[
∂f2
∂X1
]X=X
=
[∂
∂X1
(ω2
2sin2X1 −
g
rsinX1
)]X=X
= ω2cos2X1 −g
rcosX1
A22 =[
∂f2
∂X2
]X=X
= 0
Or the matrix A is
A =
0 1
ω2cos2X1 − gr cosX1 0
(7.31)
Let the solution of (7.30) be Y(t) = Ceλt, λ can be obtained from the characteristic equation:
|A− λI| = 0, or ∣∣∣∣∣∣∣−λ 1
ω2cos2X1 − gr cosX1 −λ
∣∣∣∣∣∣∣ (7.32)
89
Equation (7.32) yields
λ1,2 = ±(ω2cos2X1 −
g
rcosX1
) 12
(7.33)
For the 1st steady state, substitute X1 = 0 and X2 = 0 into (7.33) to get
λ1,2 = ±(ω2 − g
r
) 12
(7.34)
In (7.34), if ω2 >g
rthen one of λ will be a positive real resulting in unstable steady state. On
the other hand if ω2 ≤ g
rthen both λ will be conjugate imaginary resulting in stable steady
state.
For the 2nd steady state, X1 = cos−1(
g
ω2r
)and X2 = 0. Equation (7.33) then becomes
λ1,2 = ±[ω2cos2X1 − g
r cosX1] 1
2
= ±(ω2(cos2X1 − sin2X1
)− g
r cosX1) 1
2
= ±[g2−ω4r2
ω2r2
] 12 , g
r ≤ ω2
(7.35)
Since the 2nd steady state exists if and only ifg
r≤ ω2, both λ in (7.35) are conjugate imaginary.
Thus this steady state is always stable.
90