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 2 nd 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
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]
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
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)));
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,
![Untitled-16 [pioneer.netserv.chula.ac.th]pioneer.netserv.chula.ac.th/~ltachai/210/solution... · PROBLEMS 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 Calculate AH and AS for the reaction](https://static.documents.pub/doc/80x56/5f5b483f1c7e9f287a26d0f1/untitled-16-ltachai210solution-problems-61-62-63-64-65-66-67-68.jpg)
