Adv Dyna Book

2103-617 Advanced Dynamics

Thitima Jintanawan

2

c©Copyright 2005Thitima Jintanawan

Contents

1 Kinematics 51.1 Evolution of Kinematics . . . . . . . . . . . . . . . . . . . . . 51.2 Position Vector, Velocity, and Acceleration . . . . . . . . . . . 51.3 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Rate of Change of a Constant-Length Vector . . . . . . . . . 81.5 Moving Coordinate Systems . . . . . . . . . . . . . . . . . . . 101.6 Coordinate Transformation . . . . . . . . . . . . . . . . . . . 17

1.6.1 First set of Euler angles–precession-nutation-spin (φθψ) 201.6.2 Second set of Euler angles–yaw-pitch-row (ψθφ) . . . . 20

1.7 Angular velocity related to Euler angles . . . . . . . . . . . . 231.8 A Finite Motion . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.8.1 Transformation matrices for a finite rotation . . . . . 261.8.2 Transformation matrices for a general motion . . . . . 30

2 Linear and Angular Momentums 352.1 Dynamics of a System of Particles: a Review . . . . . . . . . 35

2.1.1 Total mass . . . . . . . . . . . . . . . . . . . . . . . . 352.1.2 First moment of mass . . . . . . . . . . . . . . . . . . 352.1.3 Linear momentum . . . . . . . . . . . . . . . . . . . . 372.1.4 Angular momentum . . . . . . . . . . . . . . . . . . . 372.1.5 Moment of force . . . . . . . . . . . . . . . . . . . . . 382.1.6 Laws of linear and angular momentum . . . . . . . . . 38

2.2 Angular Momentum of a Rigid Body . . . . . . . . . . . . . . 402.3 Mass Moment of Inertia . . . . . . . . . . . . . . . . . . . . . 42

3 Dynamics of a Rigid Body 453.1 Newton-Euler Equations of a rigid body . . . . . . . . . . . . 453.2 Modified Euler’s equations . . . . . . . . . . . . . . . . . . . . 493.3 Introduction to stability of a spin body . . . . . . . . . . . . 53

3

4 CONTENTS

4 Multi-Body Mechanical System 574.1 Degrees of Freedom (DOF) . . . . . . . . . . . . . . . . . . . 574.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3 Constraint Equations . . . . . . . . . . . . . . . . . . . . . . . 574.4 Classification of Constraints . . . . . . . . . . . . . . . . . . . 604.5 Number of DOF vs. Driving Forces . . . . . . . . . . . . . . . 614.6 Dynamic Analysis of Multi-Body Mechanical Systems . . . . 614.7 Example Problem: Dynamics of Two-Link Arms . . . . . . . 62

5 Principle of Virtual work 695.1 Virtual Displacement and Virtual Work . . . . . . . . . . . . 695.2 Holonomic and Nonholonomic Constraints . . . . . . . . . . . 695.3 Generalized Coordinates and Jacobian . . . . . . . . . . . . . 715.4 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . 755.5 D’Alembert principle . . . . . . . . . . . . . . . . . . . . . . . 76

6 Lagrange Mechanics 816.1 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 816.2 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . 826.3 Remarks on Properties of Generalized Coordinates for the

System with Holonomic constraints . . . . . . . . . . . . . . . 846.4 Derivation of Lagrange’s equations . . . . . . . . . . . . . . . 856.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.6 Lagrange Multiplier . . . . . . . . . . . . . . . . . . . . . . . 94

7 Stability Analysis 997.1 Equilibrium, Quasi-Equilibrium, and Steady States . . . . . . 997.2 Stability of Equilibrium or Steady State . . . . . . . . . . . . 99

Chapter 1

Kinematics

In this chapter various coordinate systems, such as cartesian and cylindri-cal coordinates, are introduced. Position vector, velocity and accelerationof particles and rigid bodies are formulated using different reference coordi-nates. Each coordinate system is related to the other through the coordinatetransformation. For the 3-D transformation, two different sets of Euler an-gles: precession-nutation-spin and yaw-pitch-roll, are conventionally used.Finally the transformation matrix used to describe a finite motion of rigidbodies is revealed.

1.1 Evolution of Kinematics

Prior to 1950s: Express velocity v and acceleration a in terms of scalarcomponents and use graphical method to determine total magnitudeand direction

1950s and later: Express velocity v and acceleration a using vector ap-proach

Recent years: Express the rotation with a matrix and utilize the matrixoperation for calculating the cross product. The matrix approach canbe simply implement in a computer simulation program.

1.2 Position Vector, Velocity, and Acceleration

Fig. 1.1 shows a particle moving in a 3-dimensional (3-D) space. Let’sintroduce a cartesian or rectangular coordinate system XY Z as shown in

5

6 CHAPTER 1. KINEMATICS

X

Y

Z

ij

k

rx

ry

r

rz

particle

moving path

O

Figure 1.1: A cartesian coordinate system

Fig. 1.1 in which all coordinates are orthogonal to each other and its axesdo not change in direction. If we choose XY Z in Fig. 1.1 as an inertial orfixed reference frame1, the absolute motion of the particle in Fig. 1.1 can bedescribed 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 scalarcomponents of r in X, Y , and Z coordinates. The position vector r can bealternatively 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 thechosen inertial frame.

Figure 1.2 shows another set of coordinate system so called the cylindri-cal coordinates ρθz, with their unit vectors eρeθez. In Fig. 1.2, the positionvector 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 vectorr 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

1.3. ANGULAR VELOCITY 7

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 vgiven 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 abso-lute 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 arbitraryrectangular coordinate system with unit vectors e1, e2, and e3. The angularvelocity can be expressed in a matrix form as

ω =

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

(1.7)


v

n

r

θ(t)A e1

e2

e3

Figure 1.3: Angular velocity

The velocity at point A in Fig. 1.3 is then

v = ω × r (1.8)≡ ωr

Equation (1.9) indicates that the cross product can be represented by thematrix 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 rmust 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 ofa fixed length vector c is given by the cross product of its rotation rate ωand 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 withrate of φ and simultaneously possesses a yaw maneuver (turn to left) witha rate of ψ. Determine a relative velocity of point C on the horizontalstabilizer at coordinates (b, a, 0), observed from the C.G. of the plane.

1.4. RATE OF CHANGE OF A CONSTANT-LENGTH VECTOR 9

X

Y

Z

G

C

ω

ez

ey

ex

Figure 1.4: A defense jet plane

SolutionFrom Fig. 1.4, the position vector of point C relative to the C.G. is a

fixed length vector 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 thereference coordinates moving with the body. ω can be written in a matrixform 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

a0

= [ −aψ bψ −bφ]T

As another example, the unit vectors ijk for any rotating system ofcoordinates xyz is also the fixed length vector. Hence the rate of change ofthese 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 rotatingcoordinate system xyz.


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 Moving Coordinate Systems

Any moving coordinate system xyz used to describe the motion can bedivided into 3 types depending on its motion with respect to the inertialframe 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 movingbody, is normally used as a reference frame to describe kinematics of the

1.5. MOVING COORDINATE SYSTEMS 11

X

Y

Z

x

z

y

o

path of origin o

Figure 1.7: Translating and rotating coordinate systems

body. Specifically, such reference coordinate system is arranged such thatits origin o is fixed to and translate with the body’s C.G. and its axessynchronously rotate with the body.

Fig. 1.8 shows a particle moving in 3-D space. XY Z is an inertial framewith unit vectors ijk. Also xyz is the moving reference frame with unitvectors e1e2e3. If the angular velocity of xyz is ω and the position vector ris

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 (1.14), 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 ve-locity observed by the observer moving (both translating and rotating) withxyz, whereas dρ

dt is the relative velocity observed by the observer who onlytranslates but not rotates with xyz. ω × ρ in (1.14) is hence the differencebetween these two relative velocities. If ρ is

ρ = ρ1e1 + ρ2e2 + ρ3e3 (1.15)


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

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 (1.14)dρ

dt= vr + ω × ρ (1.18)

Plug (1.18) into (1.17) 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.

• ar is the relative acceleration of the particle as observed in the movingreference frame xyz.


• ω × ω × ρ is a centripetal acceleration, or the correction term for thelocal position vector ρ considering that the observer rotates with themoving reference frame.

• ω×ρ is another correction term for the angular acceleration vector ωof the moving reference frame.

• 2ω × vr is the Coriolis acceleration which is the other correction termfrom two sources, both of which measure the rotation of the basis(unit) vectors of the moving reference frame and associate with aninteraction of motion along more than one coordinate curve.

Example 1.2 :The satellite shown in Fig. 1.9 has a steady spin Ω about the body fixed axise3 The solar panel arm rotates about the e2-axis with a rate θ, and angularacceleration θ = 0. The panel arm also moving along the radial direction erwith a steady rate s = α. Determine an absolute acceleration of the pointP at the end of the solar panel.Solution:

Let [e1e2e3] be the coordinate system that rotates with the body. Henceωe1e2e3 = Ωe3. Choose [ereθe2] in Fig. 1.9 as a rotating reference frame.For this case we obtain the terms in (1.13) and (1.14) 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

ω = Ωe3 × ω = −Ωθe1

Components in (1.20) are now expressed in terms of two different coordinatesystems [e1e2e3] and [ereθe2]. To express these terms in only one coordinatesystem, i.e. [e1e2e3], we need the coordinate transformation.


b

s(t) + c

P

e2

θe2

e3

eθ

er

e1

θ

Ω

Ω

P

Figure 1.9: A satellite

e3

e1

er

u

eθ

θ

θe2

u3

u1

ur

uθ

Figure 1.10: Coordinate systems [e1e2e3] and [ereθe2]


From Fig. 1.10, we obtain the transformation relation of an arbitraryvector u as

u =

(ur

uθ

)=

[cosθ sinθ−sinθ cosθ

](u3

u1

)= T

(u3

u1

)

The reader can prove that T is an orthogonal matrix or T−1 = TT . Asa 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) interms 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 systemof 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

0cosθ

+(s(t) + c)

0 0 0

0 0 Ωθ

0 −Ωθ 0

sinθ

0cosθ

+2α

0 −Ω θ

Ω 0 0−θ 0 0

sinθ

0cosθ

(1.21)


L

µG

y1

z1

α.

β

Ω

y2

z2

Figure 1.11: A ventilator

Now your task is to follow the previous procedure and express ap in termsof the coordinate system [ereθe2].Example 1.3 :Figure 1.11 shows a ventilator mounted on a rotating base. The base hasan oscillatory motion with α = α0sinωt. The rotor spins with a constantangular velocity Ω in the direction shown. Its center of gravity G is offsetby µ from the axis of rotation. Determine:

1. components of the absolute angular velocity of the rotor along thesystem of coordinates fixed to the rotor.

2. components of the absolute velocity of the center of gravity of the rotoralong the same system of coordinates.

Solution: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

)

(xz

)=

[cosαt −sinαtsinαt cosαt

](x2

z2

)

1.6. COORDINATE TRANSFORMATION 17

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

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

Lµ

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 transformcomponents of a vector in one system of coordinates to another system of


X

Y

Z

x

z

y

iI

r

Figure 1.13: A vector r and two sets of coordinate systems XY Z and xyz

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 inFig. 1.13. An arbitrary vector r in Fig. 1.13 can be expressed as

r = rXI + rY J + rZK

= rxi + ryj + rzk (1.22)

Components of r in XY Z coordinates are

rX = r · I = (rxi + ryj + rzk) · IrY = r · J = (rxi + ryj + rzk) · JrZ = 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

rYrZ

= C

rx

ryrz

(1.25)


J

i

iJ

iY

Ji

Figure 1.14: components of the unit vectors

where C is a matrix of directional cosines so called a coordinate transfor-mation matrix given by

C =

cos iI cos jI cos kI

cos iJ cos jJ cos kJcos iK cos jK cos kK

(1.26)

Note that C is the orthogonal matrix where C−1 = CT . This yields rx

ryrz

= CT

rX

rYrZ

(1.27)

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 following6 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 = 3independent components of C. Specifically, only three independent angular


transformation terms are needed to describe the coordinate transformation.There exist many possible sets of angular transformation, but two popularsets called Euler angles are normally used. Each set consists of three anglesdescribing 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 systemssuch as rotordynamics. 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′′

• Spin: rotation about z′′ axis by ψ(t) to get xyz or xyz′′

The coordinate transformations are then rX

rYrZ

=

cosφ −sinφ 0

sinφ cosφ 00 0 1

rx′

ry′

rz′

= C1

rx′

ry′

rz′

(1.29)

rx′

ry′

rz′

=

1 0 0

0 cosθ −sinθ0 sinθ cosθ

rx′′

ry′′

rz′′

= C2

rx′′

ry′′

rz′′

(1.30)

rx′′

ry′′

rz′′

=

cosψ −sinψ 0

sinψ cosψ 00 0 1

rx

ryrz

= C3

rx

ryrz

(1.31)

Combine (1.29)-(1.31), therefore rX

rYrZ

= C1C2C3

rx

ryrz

(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


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


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

1.7. ANGULAR VELOCITY RELATED TO EULER ANGLES 23

• 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

rYrZ

=

cosψ −sinψ 0

sinψ cosψ 00 0 1

rx′

ry′

rz′

= [Rψ]

rx′

ry′

rz′

(1.33)

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 −sinφ0 sinφ cosφ

rx

ryrz

= [Rφ]

rx

ryrz

(1.35)

Therefore rX

rYrZ

= [Rψ] [Rθ] [Rφ]

rx

ryrz

(1.36)

1.7 Angular velocity related to Euler angles

For the first set of Euler angles, the absolute angular velocity ω of xyzcoordinates is given by

ω = φk + θex′ + ψez≡ ωxex + ωyey + ωzez (1.37)

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 0cosθ 0 1

φ

θ

ψ

(1.38)

Similarly, for the second set of Euler angles, the absolute angular velocityω of xyz coordinates is given by

ω = ψk + θey′ + φex= ωxex + ωyey + ωzez (1.39)


x

y’

Z

θ

ψ

φ

Figure 1.17: Yaw, pitch, and roll axes of vehicle dynamics

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 cosθsinφ0 −sinφ cosθcosφ

φ

θ

ψ

(1.40)

(1.38) and (1.40) relate the Euler angles, the rotation that measured in realapplications, with the components of the angular velocity, ωx, ωy and ωz, inthe reference coordinate system.Example 1.4 :A submarine shown in Fig. 1.17 undergoes a yaw rate ψ = AcosΩt and apitch 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 ofmass.Solution:

From Fig. 1.17, let xyz with their unit vectors ex, ey, and ez be the body-fixed rotating system of coordinates. The velocity of the bow observed fromthe 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 xyzcoordinates given by

ω = ψk + θey′ (1.43)

1.8. A FINITE MOTION 25

φ

u

A

A'

Figure 1.18: A finite motion of a rigid body

ω in terms of ex, ey, and ez can be obtained from (1.40) as

ω =

ωx

ωy

ωz

=

1 0 −sinθ

0 cosφ cosθsinθ0 −sinφ cosθcosφ

0

θ

ψ

(1.44)

Note that, in this case, the submarine performs only pitch and yaw rotationsbut 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, thenψ(t) = A

ΩsinΩt 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 uof 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 afinite rotation. Then the transformation matrix for a general finite motion,possessing both translation and rotation, is considered.

1.8.1 Transformation matrices for a finite rotation

Define a position vector of any point P on the body before and after therotation as rp and r′p, respectively. A transformation matrix T relating rpand 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 pointp 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

HenceTTT = I (1.53)

(1.53) indicates that T−1 = TT or T is the orthogonal matrix.

The transformation T for a rotation about Z-, Y -, and X-axes can be de-termined subsequently as follows.


φ

X

Y

(x, y, z)

(x', y', z')

Figure 1.19: Finite rotation about Z-axis

1. Rotation about Z-axis with φ

If the previous coordinates of any point p is r = [ x y z ]T andthe new coordinates of this point is r′ = [ x′ y′ z′ ]T as seen inFigure 1.19, then

x′

y′

z′

=

cosφ −sinφ 0

sinφ cosφ 00 0 1

x

yz

= T1

x

yz

(1.54)

The transformation matrix T1 in this case is

T1 =

cosφ −sinφ 0

sinφ cosφ 00 0 1

(1.55)

2. Rotation about Y -axis with θ

From Figure 1.20 we obtain the transformation matrix T2 for a rota-tion 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 aboutX-axis is

T3 =

1 0 0

0 cosψ −sinψ0 sinψ cosψ

(1.57)


θ

Z

X

Figure 1.20: Finite rotation about Y -axis

ψ

Z

Y

Figure 1.21: Finite rotation about X-axis


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 andso on. Also φ ≈ ωz∆t, θ ≈ ωy∆t, and ψ ≈ ωx∆t. In this case T1, T2, andT3 do commute. If the rigid body has an infinitesimal rotation about anarbitrary axis during time ∆t, the new position vector r′ related to theprevious 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 representedin a matrix form as previously introduced in (1.7).

The transformation matrix for a finite rotation is useful for a com-puter graphic programming simulating the dynamics of rigid-body motionas shown in the following example.Example 1.5 :


a

bφ

θ

R

mgO

L

Cθ.

Figure 1.23: A free body diagram of the falling box

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.Solution:

First we need to derive the equation governing the motion of this box. Afree body diagram (FBD) of the box is shown in Figure 1.23. The dynamicsof 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 atthe 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 writtenin state form as (

x1

x2

)=

(x2

esin(x1 + φ0) − cx2

)(1.62)

where e = mgLIo

and c = CIo

. (1.62) together with the transformation matrixfor the finite rotation in (1.55) are used in the MatLab program to determinethe new position of the falling box. The detail of this program is presentedin 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 intotwo parts: a translation u and a finite rotation θ. The position vector r


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


-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


describing the finite rotation of the rigid body is then

r = u + ρ′ (1.63)= u + Aρ

where A is the transformation matrix of the rotation relating ρ and ρ′. Inaddition r = [ x y z ]T and u = [ ux uy uz ]T expressed in the inertialreference coordinates xyz, and ρ = [ ρx ρy ρz ]T expressed in the localcoordinate system exeyez. Substituting the component vectors into (1.64),the finite motion in Fig. 1.26 is governed by

xyz

=

ux

uy

uz

+

cosθ −sinθ 0

sinθ cosθ 00 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 =

| 0A3×3 | 0

| 0−− −− −− −|− −−0 0 0 | 1

Note that T = T1T2.


Chapter 2

Linear and AngularMomentums

This chapter is organized into three parts: 1) dynamics of a system of par-ticles; 2) an angular momentum of a rigid body; and 3) a mass moment ofinertia. These topics are used as fundamentals for a study of dynamics of arigid body and a multi-body mechanical system in Chapter 3 and Chapter4, 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-thparticle is subjected to the applied force Fi. Also c is the center of mass orcenter 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.

2.1.2 First moment of mass

The first moment of the total mass about its C.G. is the sum of the firstmoment of each mass given by:

rcM =∑

rimi (2.2)

35

36 CHAPTER 2. LINEAR AND ANGULAR MOMENTUMS

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

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 aboutthe system’s C.G. is zero.

2.1. DYNAMICS OF A SYSTEM OF PARTICLES: A REVIEW 37

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 momen-tum. The angular momentum of the system of particles about the origin ois 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).


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 andmoments 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 reactionforce that the j-th particle acts on the the i-th particle. Also note thatfij = −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 thatthe rate of change of linear momentum of the system is equal to the sum ofall 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)

2.1. DYNAMICS OF A SYSTEM OF PARTICLES: A REVIEW 39

Now consider (2.17) term by term. The term on the left is rewritten as

∑i

(ri × mivi) =∑i

ri × midvi

dt

=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 anyarbitrary 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 ofchange of angular momentum about o is equal to the moment of all externalforces about o.

Alternatively, we could formulate the law of angular momentum aboutthe 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)


X

Y

Z

A

ω

rA

e1

e2e3

r

ρ dm

o

Figure 2.2: A rigid body

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)


∑ρi × Fi =

d

dtHc (2.25)

OrHc =

∑ρi × Fi = Mc (2.26)

From equation (2.26), if proper coordinates are used to described ρi thenwe can define a set of geometric quantities so called moments of inertia ofa rigid body. The moments of inertia measure the angular momentum perunit 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 thebody be ω. Consider a rigid body as a continuous media of particles with

2.2. ANGULAR MOMENTUM OF A RIGID BODY 41

no deformation, the angular momentum of this rigid body is the integralform of (2.7) or

Ho =∫r× vdm (2.27)

where r = rA+ρ is a position vector from the fixed origin o to the differentialmass dm of the body. Let e1e2e3 in Figure 2.2 be the rotating referenceframe with its origin located at an arbitrary point A on the body. If thereference coordinate system and the body have the same angular velocity,i.e. ω, then

v = r = vA + ω × ρ (2.28)

Note that the velocity of the differential mass dm relative to e1e2e3 is zerobecause: 1) the rigid body has no deformation and 2) e1e2e3 rotates syn-chronously 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 differentcases: pure rotation and general motion (combined rotation and transla-tion).

1. Pure rotation about fixed point o

In this case, if we choose point A in Figure 2.2 fixed at o. HencerA = 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)

HenceHo = rc × mvc +Hc (2.33)

2.3 Mass Moment of Inertia

Due to a constant geometric property of the rigid body, the angular mo-mentum can be more simplified as follows. First it is noted that the angularmomentum about o (2.30), in case of pure rotation, and the angular momen-tum about c (2.32), in case of general motion, have the same form. Thereforewe will drop out the subscripts in (2.30) and (2.32) for convenience and gen-erally 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

]TNext 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 00 0 ρ2

ωdm

=

∫

ρ2 0 0

0 ρ2 00 0 ρ2

dm

ω

(2.36)

2.3. MASS MOMENT OF INERTIA 43

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

yz

dm

= −∫

(xωx + yωy + zωz)

x

yz

dm

= −∫ x2ωx + xyωy + xzωz

xyωx + y2ωy + yzωz

xyωx + yzωy + z2ωz

dm

= −∫ x2 xy xz

xy y2 yzxz yz z2

ωx

ωy

ωz

dm

= −∫ x2 xy xz

xy y2 yzxz yz z2

dm(ω)

(2.37)


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 Ialong 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 givenby

I12 = I21 = −∫

xydm

I13 = I31 = −∫

xzdm

I23 = I32 = −∫

yzdm

(2.40)

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 canbe determined from the eigenvalues of the matrix I. With the principalinertias, 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 orthogonal eigenvectors which represent the principalaxes e′1e

′2e

′3.

3. For a basis with at least two symmetry planes, the off-diagonal termsor 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.

Chapter 3

Dynamics of a Rigid Body

In this chapter, the dynamics of a rigid body for two different cases: 1) apure rotation; and 2) a general motion consisting of both translation androtation, are analyzed using the Newton-Euler approach. According to thelaws of angular momentum stated in the previous chapter, we can formulatethe 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 thiscase, we choose the reference coordinate system e1e2e3, with its originfixed 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 com-ponents are along e1e2e3 axes.

For a general motion of a rigid body, the equations governing both trans-lation and rotation are ∑

F = mvc (3.3)

and ∑Mc = Hc (3.4)

where point c is the C.G. of the rigid body.

45

46 CHAPTER 3. DYNAMICS OF A RIGID BODY

Equation (3.3) is the law of linear momentum for a rigid body or so calledthe Newton’s equation, and equation (3.4) is the law of angular momentum.For the general motion, we normally choose the reference coordinate systemsuch that its origin is fixed at the C.G. of the body and its coordinates rotatewith 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 com-ponents are along the reference coordinates.

For both cases of motion, if the reference coordinates, i.e. e1e2e3, arein 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)

andHc = 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)

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.

3.1. NEWTON-EULER EQUATIONS OF A RIGID BODY 47

m, L

g θex

eymassless cart

u(t)

Figure 3.1: A cart-pendulum system

Example 3.1 : Dynamics of a pendulum-cart systemThe cart with a negligible weight moves along a frictionless floor as shown inFigure 3.1. The pendulum with mass m and length L is hinged to the cartat one end. If the cart motion is prescribed by u(t), derive the equation ofmotion of the system.Solution:

First, consider the pendulum or the uniform rod which has a generalplane (2-D) motion. The degree of freedom used to describe the motion ofthis rod is θ(t).Kinematics: With the coordinate systems shown in Figure 3.2, the velocityvc 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

[∑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)


θ

er

eθ

θ

θex

ey

er

Fr

eθ

Fθ

c

mg

Figure 3.2: Coordinate systems and FBD

Note that Ic in (3.15) is the moment of inertia about the C.G. of the rodalong z-axis. We now have three unknowns: Fr, Fθ, and θ, and three scalarequations, two from (3.14) and one from (3.15). To derive the equation ofmotion, we need to eliminate all unknown forces which are Fr and Fθ andreduce the Newton-Euler equations to only one differential equation.

Figure 3.2 shows the two coordinate systems with the coordinate trans-formation given by

ex = sinθer + cosθeθey = −cosθer + sinθeθ

(3.16)

To eliminate Fθ, we substitute (3.15) into (3.14) and transform all coordi-nates to ereθ using (3.16). Then (3.14) can be expressed in scalar formas:

r-component:

mθ2L

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 nonlinearequation. The solution of (3.18) can be obtained from a numerical integra-tion using Matlab. Otherwise if only small oscillation is interested, we can

3.2. MODIFIED EULER’S EQUATIONS 49

C.G. e1

e2

e3

ω0

XY

Zdm

ρ

Ω

Figure 3.3: A rigid body with symmetric shape

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 = mθ2 L

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, twoconditions are defined.

1. The rigid body spins about the symmetry axis with a constant speedωo.

2. The reference coordinate system e1e2e3 is chosen such that one of theaxes, i.e. e3, is the symmetry axis. In addition, e1e2e3 only precessesbut does not spin with the body. Also the origin of e1e2e3 is fixed atthe 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 areprincipal axes and I1 = I2 ≡ I. If the angular velocity of e1e2e3 is

Ω = Ω1e1 + Ω2e2 + Ω3e3

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 generalmotion, respectively. The derivation of the modified Euler’s equations isshown for the case of general motion as follows. From Figure 3.3, the angularmomentum 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 00 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)

3.2. MODIFIED EULER’S EQUATIONS 51

o

e1

k

e3

φ

ψ

θ

.

.

L

mg

Figure 3.4: A gyro top

Substitution of (3.24) into (3.4) hence results in (3.22).Example 3.2 : Steady precession of a gyro top

Derive the dynamic equation governing steady precession of a gyro topshown in Figure 3.4. With the steady precession, the top has constantprecession rate ψ and constant spin rate φ, and the nutation angle θ is alsoconstant.

Method1: Direct approach In Figure 3.4, the angular velocity of thereference 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 thesymmetric axis, therefore I1 = I2 and all cross products of inertia arezero. As the previous proof in (3.23), it can be similarly shown thatthe 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)

Note that the angular momentum of the steady gyro in (3.28) has aconstant magnitude, and the direction of the angular momentum is onthe plane of rotations, i.e. k− e3 plane. Moreover, the rate of changeof 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, givenby ∑

Mo = Le3 ×−mgk = −mgLsinθe2 (3.31)

Note that moment of the resultant force about o shown in (3.31) isalways perpendicular to both rotation axes (k and e3), resulting inthe precession. For the steady precession, this moment has a constantmagnitude 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 thetop as

mgL = I3ψφ + (I3 − I1) ψ2cosθ (3.33)

From (3.33), with a given θ we can determine the relation between theprecession 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 equationsare presented again as follows

M1o = IoΩ1 + (I3o − Io) Ω2Ω3 + I3oωoΩ2

M2o = IoΩ2 + (Io − I3o) Ω3Ω1 − I3oωoΩ1

M3o = I3oΩ3

(3.35)

3.3. INTRODUCTION TO STABILITY OF A SPIN BODY 53

y

x

z

1

Figure 3.5: Stability of a spin plate

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.

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 principalaxis.

By stability of rotation, we ask the question: during a steady spin abouteach axis, if the initial rotation is applied so close to the principal axes (isperturbed a bit in every directions), will the rotation remain close to theprincipal axes (does the perturbation die out), or will the body begin to seeincreasing rotation about one of the other axes (does the perturbation growwith 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 theaxis ‘1’ with a constant speed ω0. (Note that axis-1 can be any arbitraryprincipal axis, i.e. x-, y-, or z-axis in Figure 3.5.) Then the plate is perturbedwith small angular velocities η1(t), η2(t), and η3(t), respectively, about allprincipal axes. Hence 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, suchas η1η2, η2η3, etc., yield

I1cη1 = 0I2cη2 + (I1c − I3c) ω0η3 = 0I3cη3 + (I2c − I1c) ω0η2 = 0

(3.39)

The first row of (3.39) implies that η1(t) is constant. In addition, the lasttwo 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)

)=

(00

)(3.40)

orη(t) +Kη(t) = 0 (3.41)

To solve (3.40), assume the solution as the following form

η(t) =

(η2(t)η3(t)

)=

(ab

)eλt (3.42)

Substitution (3.42) into (3.41) yields

[λI + K]

(ab

)eλt =

(00

)(3.43)

3.3. INTRODUCTION TO STABILITY OF A SPIN BODY 55

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 casesas follows

Case I: (λ2 ≤ 0) In this case, λ1,2 are positive and negative imaginary partsand the rotation are marginally stable. Specifically, the perturbationcauses the oscillatory motion about the steady state. To satisfy thisstable condition, I1c > I2c > I3c or I1c < I2c < I3c. In other words, themoment of inertia about the spin axis I1c should be either maximumor minimum.

Case II: (λ2 > 0) In this case, one of the root is positive real and the otheris negative real. With the positive real root, the solution (3.46) showsthat the rotation is about to increase exponentially with time andhence the rotation of the plate is unstable.

From this analysis together with a real demonstration, the students shouldbe able to figure out that in which directions the rotation of the spin plateare stable.


Chapter 4

Multi-Body MechanicalSystem

4.1 Degrees of Freedom (DOF)

Degrees of freedom are a complete set of independent coordinates that usedto describe the motion. For example, a rigid body performing free motion(without any constraints) in 3-D space needs six degrees of freedom (coor-dinates) to describe its motion, i.e. three for translations and another threefor 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 con-necting the bodies is called constraint. The constraint imposes additionalrelative motion of one body with respect to anothers. With constraints, themotion of each rigid body in all six coordinates are not independent, hencethe number of DOF for each body is reduced to less than six.

4.3 Constraint Equations

The constraint equations describe the relative motions of any two connectedbodies. We can learn to construct these constraint equations by the followingexamples.

57

58 CHAPTER 4. MULTI-BODY MECHANICAL SYSTEM

x

y

ry

zφ

Rx

RzMz

Mz

Aθx

θz

Figure 4.1: A slider

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 fourconstraint equations, i.e. rx = 0, rz = 0, θx = 0 and θz = 0. Withoutfriction, the slider translates free along y-direction and also rotate freeabout y-axis. In this case, two coordinates such as ry and φ as seenin Figure 4.1 can be chosen as the DOFs to describe such translationand rotation.

Example 2: spherical joint Three constraint forces Rx, Ry, and Rz inthe spherical joint as shown in Fig. 4.2 result in three constraint equa-tions, i.e. rx = 0, ry = 0, and rz = 0. In Figure 4.2, link B thatconnected to the stationary link A through the joint can rotate freeabout its center, assuming no friction. In this case, three sphericalcoordinates or the conventional Euler angles θ, ψ, and φ are the DOFsused to describe the rotation.

Example 3: rolling sphere Consider the spherical ball rolls without slip-ping as shown in Fig. 4.3. The first geometric constraint relation, i.e.rcz = a, can be simply observed. Another two relations are derivedfrom the fact that the contact point A on the sphere is motionless withrespect to the contact point A′ on the surface. Hence

(vAx)rel = rcx − θya = 0

4.3. CONSTRAINT EQUATIONS 59

y

x

z

Rx

Ry

Rz

θψ

φ

A

B

Figure 4.2: Ball and socket

yx

z

a

c

rcz

rcy

rcxA

A’

θy

θx

Figure 4.3: A rolling sphere


(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 equalto [6 − number of constraint equations (or constraint forces)]. The chosenDOFs in each case are called generalized coordinates.

Now let’s consider the multi-body linkages in Fig. 4.4. From the previousexamples, we can conclude that the total constraint equations is equal to 4(from the slider) + 3 (from the spherical joint) = 7. The number of DOFsis 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-bodysystem 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 generalizedcoordinates and time, e.g. examples 1 and 2 in section 4.3, these constraintsare classified as holonomic constraints. In addition, the holonomic con-straints can be divided into two classes: scleronomic and rheonomic. Theconstraint equation for the scleronomic constraint is an implicit function oftime whereas the equation for rheonomic constraint is an explicit functionof time.

If one of the constraint equation is a function of both the generalizedcoordinates and their time derivatives, such constraint is classified as non-holonomic constraint, e.g. example 3 in Section 4.3: rolling sphere.

4.5. NUMBER OF DOF VS. DRIVING FORCES 61

X

Y

φ yx

Z

z

θx θy

θz

rY

Figure 4.4: Combined constraints

4.5 Number of DOF vs. Driving Forces

If the motion along L coordinates can be prescribed as functions of time, socalled the prescribed motions, the number of DOF is then reduced by L. Inorder to have the mechanical system perform such prescribed motions, thecorresponding 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 therotor. With the prescribed motion in the system, the number of DOF ofmulti-body system is

M = 6 × N −∑

C − L

where L is number of the prescribed motions.

4.6 Dynamic Analysis of Multi-Body MechanicalSystems

Dynamic analysis of a multi-body mechanical system can be separated intotwo main parts: kinematics and kinetics. Detailed analysis of each part isdescribed 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 ofthe generalized coordinates along the reference coordinate system

Kinetic analysis :

1. Express Newton-Euler’s equations governing dynamics of eachrigid body

2. With free body diagram (FBD), determine components of forcesand moments corresponding to the reference coordinates

3. Substitute forces and kinematic relations into Newton-Euler’sequations

4. Eliminate all unknown forces to obtain equations of motion (num-ber of equations of motion is equal to number of DOF.)

5. Solve the equations of motion to determine the time responsesand then use them to obtain all unknown forces

4.7 Example Problem: Dynamics of Two-Link Arms

The two-link arms are connected by the hinge support A as shown in Fig-ure 4.5. Link 1 is approximately massless and is driven by a motor whichis excluded from the system. The driving torque Md provided by the motoris 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 equationgoverning the motion of link 2 and solve for time response, given the initialconditions: β(0) = 0, β(0) = 0, ω(0) = 0, and ω(0) = 0.Kinematic analysisNumber 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. Inthis case we choose α and β as the generalized coordinates. Fig. 4.5 alsoshows the coordinate systems and their unit vectors.

The angular velocities of link 1 and 2 and the velocity at the C.G. (pointc) of link 2 are, respectively,

Ω1 = αk1 (4.1)

4.7. EXAMPLE PROBLEM: DYNAMICS OF TWO-LINK ARMS 63

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


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, andx2y2z2, 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 β2 + aα2sinβ + l

2 α2sin2β

)k2

(4.4)

Kinetic analysisFigure 4.5 shows the free body diagram of both links. First let’s considerlink 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 sixscalar equations for link 2:

m

(−aα − l

2αsinβ − l

2αβcosβ

)= RAx (4.11)

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 theresultant force and resultant couple are then zero. From FBD of link 1 inFigure 4.5, consider only the Euler’s equation in z1-direction which is

z1 : Md − MAzcosβ − MAysinβ + RAxa = 0 (4.17)

OrMAy =

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 equationas get

ma2α + ml2

3 αsin2β + malαsinβ + 512ml2αβsinβcosβ

+mal2 αβcosβ − (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, andx3 = β. By substituting the state variables into (4.20) and (4.21), theequations 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)


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

where

f1 =M0 − ∆Mx1 − (mal/2)x1x3cosx2 − (5ml2/12)x1x3sinx2cosx2

ma2 + (ml2/3)sin2x2 + malsinx2

f2 = x3

f3 = x21sinx2cosx2 +

3a2l

x21cosx2 −

3g2l

sinx2

Then the state equation (4.22) is numerically solved using Matlab, wherethe time response plots is shown in Figure 4.6. Note that the Matlab m-fileis described in Fig. 4.7


%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


Chapter 5

Principle of Virtual work

5.1 Virtual Displacement and Virtual Work

A virtual displacement δr is defined as an infinitesimal and instantaneousdisplacement in an arbitrary direction that does not oppose or violate con-straints. Fig. 5.1 and Fig. 5.2 show examples of a particle under constrainedmotion. 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 constraintforce F(c) can be expressed as

F(c) = Fcn

where n is a unit vector normal to the path of motion. According thedefinition of δr, the virtual displacement is always tangent to the path ofmotion 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)

5.2 Holonomic and Nonholonomic Constraints

For the holonomic constraint, the constraint equation can be derived as afunction of the generalized coordinates and time. Therefore the positionvector can be put in following form

r(t) = f (q1(t), q2(t), . . . , qM (t)) (5.2)

69

70 CHAPTER 5. PRINCIPLE OF VIRTUAL WORK

X

YnF(a)

F(c)

δr

r

Path of motion

Figure 5.1: Virtual displacement of a particle moving along a constrainedpath

X

Z

n

g

F(a)F(c)

δrθ

Figure 5.2: Virtual displacement of a bead moving in a circular ring

5.3. GENERALIZED COORDINATES AND JACOBIAN 71

orr(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 numberof DOFs. Equation (5.2) is for the case of scleronomic constraint wherer(t) is an implicit function of time, and (5.3) is for the case of rheonomicconstraint where r(t) is an explicit function of time.

For the nonholonomic constraint, the constraint equations are functionsof both the generalized coordinates and generalized velocities. Hence thegeneral 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 con-straint.

5.3 Generalized Coordinates and Jacobian

Generalized coordinates denoted by qi, where i = 1, 2, . . . ,M1, are the setof independent coordinates used to describe the motion of the system. For amotion under geometric (holonomic) constraints, qi can be any generalizedvariable natural to the constraints. In other words, they are not necessarilythe spatial position variables. The following examples can elaborate thisconcept.The bead moving along the 3-D constrained pathFrom 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

where x, y, and z are spatial position variables. From (5.2), the positionvector 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)

1M is the number of degrees of freedom.


X

Z

Y

path

v(t)s(t)

r

Figure 5.3: Moving particle along the constrained path

where j is defined a Jacobian vector, representing a unit vector that tangentto 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 coordi-nate through this Jacobian. In addition (5.5) can be expressed in a matrixform as

δr =[

dxds

dyds

dzds

]Tδs ≡ Jδs (5.6)

where J is a Jacobian matrix. We can see that the generalized coordinates(t) is chosen such that it is tangent to the path or natural to the constraint.Moving cart with the coin on its inclined planeLet q1 and q2 in Fig. 5.4 be the generalized coordinates of the coin C. Theposition vector r of the coin is given by

r = [ x y z ]T = r (q1, q2, t) (5.7)

Note that the position vector r in (5.7) is an explicit function of time becauseof the prescribed motion u(t) of the cart which is an explicit function of time.Virtual displacement δr of the coin is

δr = ∂r∂q1

δq1 + ∂r∂q2

δq2 + ∂r∂t δt

=2∑

i=1

∂r∂qi

δqi(5.8)

5.3. GENERALIZED COORDINATES AND JACOBIAN 73

X

Y

Z

q2

ru(t)

q1

path

C

Figure 5.4: Moving coin on the inclined plane of the moving cart

Note that δt in (5.8) is zero because of the instantaneous virtual displace-ment. 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)

whereδq =

[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 particlesFig. 5.5 shows a system of N -particles with K-geometric constraints. Num-ber of degree-of-freedom of this system is M = 3N−K. Let q1, q2, . . . , qM beall M generalized coordinates. The position vector r to describe the motionof all particles is

r =[rT1 , rT2 , . . . , rTN

]T= [x1, y1, z1, x2, y2, z2 . . . , xN , yN , zN ]T3N×1


X

Y

Z

m1 m2

mj

m4mN

m3

r1

rN

Figure 5.5: A constrained system of particles

For a system with holonomic constraints, the position vector is also a func-tion 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)

5.4. PRINCIPLE OF VIRTUAL WORK 75

and J is the Jacobian matrix given by

J =

J11 J12 . . . J1M

J21 J22 . . . J2M...

.... . .

...JN1 JN2 . . . JNM

3N×M

(5.15)

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 thephysical or spatial variables with respect to the generalized coordinates.Moreover, the virtual work is related to the generalized coordinates throughthis Jacobian. Note that the Jacobians derived in the previous examples arefor 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 thesystem is in equilibrium then

N∑i=1

Fi = 0 (5.16)

where Fi is the total forces acting on the i-th particle. Fi can be divided intothree types of forces: 1) applied or external forces F(a)

i ; 2) internal spring ordamping 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 constraintforces is always zero, i.e.

N∑i=1

F(c)i · δri = 0 (5.18)

2Constraints in this case excludes the springs and dampers


If the system of particles is in equilibrium, the virtual work δW done by allforces 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 therelation (5.18), (5.19) is simply

δW (ac) =N∑i=1

F(ac)i · δri = 0 (5.20)

Equation (5.20) is the principle of virtual work stating that if the systemis in equilibrium then virtual work done by all active forces in the system iszero.

5.5 D’Alembert principle

Let’s consider a system of N -particles and M -degrees of freedom. The New-ton’s second law applied to such system can be rewritten as the followingalternative 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 thei-th particle, and ri is the acceleration of the i-th particle. The term −mirirepresents an inertia force resisting the motion of the system. From (5.21),virtual work δW done by all forces, including the inertia force −miri, isthen 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) issimply

δW ′ =N∑i=1

(F(ac)

i − miri)· δri = 0 (5.23)

5.5. D’ALEMBERT PRINCIPLE 77

stating that the virtual work δW ′ done by all active forces and inertia forcesis zero. For the holonomic constraints, we can substitute the Jacobian rela-tion (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) as

M∑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), eachk-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. Equa-

tions (5.26) are the D’Alembert principle that automatically yield the equa-tions of motion. Again the D’Alembert principle can only be applied todynamics of the systems with holonomic constraints.Example: Direct application of D’Alembert principleThe sliders of masses m1 and m2 are constrained by springs and move alongthe frictionless disk slot as shown in Figure 5.6. The disk also rotates aboutits center with angular displacement θ. If the unstretched length of thesprings is a, derive the equations of motion.

From the system shown in Figure 5.6, let’s choose ρ1, ρ2, and θ as thegeneralized coordinates. Hence

q =[

ρ1 ρ2 θ]T

(5.27)

The system has three degrees of freedom or M = 3. The system consists oftwo particles or N = 2. The position vectors of the sliders 1 and 2 are

r1 = f (ρ1, ρ2, θ) = ρ1er (5.28)

andr2 = f (ρ1, ρ2, θ) = −ρ2er (5.29)


X

Y

er

eθ

k

k

m2θθ

m1

ρ1

ρ2

Figure 5.6: Example

The accelerations of both sliders are

r1 (ρ1, ρ2, θ) =(ρ1 − θ2ρ1

)er +

(ρ1θ + 2ρ1θ

)eθ (5.30)

andr2 (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)

andF(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

5.5. D’ALEMBERT PRINCIPLE 79

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)

orm1

(ρ1 − θ2ρ1

)+ K (ρ1 − a) = 0 (5.35)

For k = 2, (F(ac)

1 − m1r1

)· β12 +

(F(ac)

2 − m2r2

)· β22 = 0 (5.36)

orm2

(ρ2 − θ2ρ2

)+ K (ρ2 − a) = 0 (5.37)

For k = 3, (F(ac)

1 − m1r1

)· β13 +

(F(ac)

2 − m2r2

)· β23 = 0 (5.38)

orm1

(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 as

d

dt

(m1ρ

21θ + m2ρ

22θ)

= 0 (5.40)

which indicates the conservation of angular momentum of the system orHo = 0.


Chapter 6

Lagrange Mechanics

6.1 Kinetic Energy

Kinetic energy of a system of particles is the total sum of kinetic energy ofeach 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 velocityof the i-th particle. Kinetic energy of a rigid body as shown in Figure 6.1 istherefore

T =12

∫v · vdm (6.2)

where v is the velocity of element dm. From Figure 6.1, the velocity v ofmass 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)

81

82 CHAPTER 6. LAGRANGE MECHANICS

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 =12mvT

c vc +12ωT Icω (6.6)

6.2 Potential Energy

Some active forces such as gravitational forces and internal forces due toelastic deformation can be represented by a gradient operator of a scalarpotential energy function V as

F = −∇V (r) (6.7)

where ∇ is the gradient operator given (in cartesian coordinates) by

∇ =∂

∂xi +

∂

∂yj +

∂

∂zk

6.2. POTENTIAL ENERGY 83

Such forces in form of (6.7) are called conservative forces and the functionV is called potential energy. Consequently work done by the conservativeforces is is independent to the path of motion and equal to the change of thepotential 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)

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 potentialenergy 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 Coor-

dinates for the System with Holonomic con-straints

1. For a multi-degree-of-freedom dynamical system, all generalized coor-dinates q1, q2, . . . , qM (where M is the number of degrees of freedom)are independent.

2. For a system of particles with holonomics (geometric) constraints, theposition vectors ri; i = 1, 2, . . . , N (N is number of particles) can beexpressed in terms of the generalized coordinates and time t as

ri = ri (q1, q2, . . . , qM , t)

3. For a system of particles with holonomics (geometric) constraints, thevirtual displacement can be expressed in terms of the generalized co-ordinates 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 instanta-

neous displacement, and∂ri∂qj

= f (q1, q2, . . . , qM , t).

4. With only the holonomic constraints in the system, the actual velocityof the i-th particle can be derived in terms of generalized coordinatesas

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

6.4. DERIVATION OF LAGRANGE’S EQUATIONS 85

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

Let’s consider a system of N -particles with M -degrees of freedom. From theD’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)

With the remarkable observation 1 and 2, (6.11) can be rearranged as

N∑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 asdescribed in the following details.


1. From (6.12), Qk is defined as the component of all active forces thatprojected 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 workdone by enforcing virtual displacement in the particular j-coordinatefor one unit (δqj = 1), and enforcing zero virtual displacements in theother coordinates (δqi = 0, i = j).

In addition, the generalized forces can be separated into two cases: conser-vative 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 po-

tential 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

Substitution of the conservative generalized forces Q(c)k into (6.13) yields the

Lagrange’s equations

d

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 87

θ

m

lg

Figure 6.2: A pendulum

6.5 Examples

Example 6.1 :Derive the equation of motion of the pendulum in Figure 6.2 using theLagrange’s equation.SolutionThe pendulum shown in Fig. 6.2 has one degree of freedom. Let’s chooseθ as the generalized coordinate. The kinetic energy T of the pendulum isthen

T =12mv2 =

12ml2θ2 (6.17)

Also the potential energy V of the pendulum (with respect to the datum atthe hinge level) is

V = −mglcosθ (6.18)

There is no external forces applied to the pendulum, hence the generalizedforce 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


m1

m2

k

g θ

x

er

eθ

Figure 6.3: A cart and pendulum

The cart-pendulum system has two degrees of freedom. Let the generalizedcoordinates 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)

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)

6.5. EXAMPLES 89

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 theequations of motion:

(m1 + m2) x + m2Lθcosθ − m2Lθ2sinθ + kx = 0 (6.36)

andm2L

2θ + m2Lxcosθ + m2gLsinθ = 0 (6.37)

Lagrange’s equations can be used to derive equations of motion of therigid body or multi-body system. In this case, derivation of the Lagrange’sequation is similar to that for the system of particles and will be omittedhere. Also the Lagrange equations for the rigid body or multi-body systemare identical to (6.15) and (6.16).Example 6.3 :


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

The two-link arm robot as shown in Figure 6.4 is operated in a horizontalplane. The motion of the arms is controlled by two motors installed at thejoints. The motors generate moments τ1 and τ2 as shown in Figure 6.4.Derive equations of motion for the two-link arm robot.SolutionKinematics: There are two degrees of freedom in this case. Let’s define α1

and α2 as the generalized coordinates. The position vector and the velocityof 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)

6.5. EXAMPLES 91

τ1

τ2

τ2g

α1

α2

Figure 6.5: Reaction torques

Since the whole system operates in the horizontal plane and there is norestoring 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)

HenceQα1 = τ1 − τ2 (6.43)

andQα2 = 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)

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 equa-tions of motion:(

I1 + m1a21 + m2l

21

)α1 + m2a2l1α2cos (α1 − α2)

+m2a2l1α22sin (α1 − α2) = τ1 − τ2

(6.53)

(m2a

22 + I2

)α2 + m2a2l1α1 cos (α1 − α2)

−m2a2l1α21sin (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 link1 which is driven to spin with a constant speed ω. Derive the differentialequations governing the motion of the thin bar using Lagrange’s equations.SolutionWith the prescribed motion ω is constant, the number of degrees of freedomis M = 6 × N −∑C − L = 6 × 2 − (5 + 5) − 1 = 1. Let’s choose β as thegeneralized 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 00 I 00 0 0

β

ωsinβωcosβ

= 12IZω2 + 1

2

(β2 + ω2sin2β

)(6.57)

6.5. EXAMPLES 93

β

ω

g

k1

k2 j2

l

Figure 6.6: A spinning pendulum

Potential energy is V = −mg(l/2)cosβ. The Lagrange’s equation can beformulated as

d

dt

(∂T

∂β

)− ∂T

∂β+

∂V

∂β= Q (6.58)

where∂T

∂β= Iβ (6.59)

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, itcan 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, eachbracket in (6.64) is zero and we obtain the Lagrange equations as shownin (6.15). If additional p constraints are introduced later, and result in thedependency of some qk, what happens? Let’s consider the equation (6.64).The introduction of new constraints will lead to the following conditions:

1. There exist new constraints which can be expressed by the generalform of p constraint equations

M∑k=1

aikdqk = 0; i = 1, 2, . . . , p (6.65)

where p is the number of additional constraints. (6.65) is also a generalform of nonholonomic 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 rear-ranged 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. Ifthe virtual generalized coordinates are arranged such that δq1, δq2, . . . , δqM−p

are independent, and δqM−p+1, δqM−p+2, . . . , δqM are dependent. Then thefollowing procedure is performed to derive the equations of motion. First

6.6. LAGRANGE MULTIPLIER 95

we choose λ1, λ2, . . . , λp so that each coefficient in the bracket in (6.67) cor-responding 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 inthe 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 asthe generalized forces as seen from the second term on the right of (6.69).Moreover the Lagrange equation with Lagrange Multiplier can deal withdynamics with nonholonomic constraints.

Then we solve (6.69) together with the revised constraint relations (6.65),putting in the form of

M∑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)

Example 6.5 :Derive equation of motion of a pendulum shown in Figure 6.2 using theLagrange multiplier.SolutionFirst if we assume that the pendulum is not constrained in the radial direc-tion, i.e. l is not fixed, this system will have two degrees of freedom. Letr and θ be the generalized coordinates as shown in Figure 6.7. The kineticand potential energies are

T =12m(r2 + r2θ2

)

V = −mgr cos θ


θ

m

rg

Figure 6.7: A pendulum without constraint

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)

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)

6.6. LAGRANGE MULTIPLIER 97

In (6.76) and (6.77), there are three unknowns: r, θ and λ. Therefore tosolve these equations for r, θ and λ, we need another one equation which isthe constraint equation:

r = l (6.78)

Plugging (6.78) into (6.76) and (6.77) yields

mlθ2 + mg cos θ = λ (6.79)

andml2θ + mgl sin θ = 0 (6.80)

Note that (6.80) is equivalent to the equation of motion that we obtain inExample 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.


Chapter 7

Stability Analysis

7.1 Equilibrium, Quasi-Equilibrium, and SteadyStates

Equilibrium is the state in which a system is at stationary; i.e. q = 0 andq = 0, where q is the vector of generalized coordinates.

Quasi-equilibrium or steady state is the state that some coordinates of asystem are in equilibrium, meanwhile the system has steady rotations (withconstant speed) about some axes, e.g. steady precession of the top.

7.2 Stability of Equilibrium or Steady State

To determine if the equilibrium or the steady state is stable, we initiallyperturb the system from each state with a small perturbation, and theninvestigate how the perturbation changes with time. If the perturbationdies out or possesses a small oscillation, the perturbed state is stable. If theperturbation grows with time, that perturbed state is unstable.

From the previous chapters, the equations of motion can be put in ageneral 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 variablesX1, X2, . . . , Xk, Xk+1, Xk+2, . . . , X2k as

X1 = q1,X2 = q2, . . . ,Xk = qk (7.2)

andXk+1 = q1,Xk+2 = q2, . . . ,X2k = qk (7.3)

99

100 CHAPTER 7. STABILITY ANALYSIS

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 equi-

librium or the steady state with a small perturbation ∆X(0) in every coor-dinates. 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)

7.2. STABILITY OF EQUILIBRIUM OR STEADY STATE 101

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 pertur-bation 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 statewill 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 positivereal 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. De-termine the steady states and analyze if each state is stable or unstable.SolutionKinematics:

The system has only one degree of freedom. From Fig. 7.1, let α be thegeneralized coordinate. Hence the absolute velocity of the bead is

v = rαeθ + (rsinα) ωez (7.19)


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 then

d

dt

(∂T

∂α

)− ∂T

∂α+

∂V

∂α= 0 (7.22)

Formulate each term in (7.22) and substitute into the equation to get equa-tion 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)

7.2. STABILITY OF EQUILIBRIUM OR STEADY STATE 103

For the steady state, X = 0. Therefore, (7.25) becomes

(00

)=

(X2

ω2sinX1cosX1 −g

rsinX1

)(7.26)

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) =

(00

), 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 analysisTo analyze the stability of each steady state, the system is initially perturbedwith ∆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

= 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 thecharacteristic equation: |A− λI| = 0, or∣∣∣∣∣ −λ 1

ω2cos2X1 − gr cosX1 −λ

∣∣∣∣∣ (7.32)

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.

Bibliography

[1] J. H. Ginsberg, Advanced Engineering Dynamics, Cambridge, 1998.

[2] D. T. Greenwood, Priciples of Dynamics, Prentice Hall, 1988.

[3] F. C. Moon, Applied Dynamics: With Applications to Multibody andMechatronic Systems, John Wiley & Sons, 1998.

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