dynamics of mechanical systems

436-354 MECHANICS 3UNIT 2

DYNAMICSOF

MECHANICAL SYSTEMSJ. M. KRODKIEWSKI

2008

THE UNIVERSITY OF MELBOURNEDepartment of Mechanical and Manufacturing Engineering

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1

2

DYNAMICS OF MECHANICAL SYSTEMS

Copyright C 2008 by J.M. Krodkiewski

ISBN 0-7325-1536-X

The University of MelbourneDepartment of Mechanical and Manufacturing Engineering

CONTENTS

I MODELLING 8

1 MODELLINGOFMECHANICAL SYSTEMSBYMEANSOFEULEREQUATIONS 101.1 CONSTRAINTS - CLASSIFICATION OF MECHANICAL SYSTEMS. 10

1.1.1 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.2 Classification of mechanical systems. . . . . . . . . . . . 14

1.2 MOBILITY - GENERALIZED COORDINATES. . . . . . . . . . . . 151.3 NUMBER OF DEGREE OF FREEDOM - DRIVING FORCES. . . . 161.4 EQUATIONS OF MOTION. . . . . . . . . . . . . . . . . . . . . . . 171.5 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 MODELLINGOFMECHANICAL SYSTEMS BYMEANS OF LA-GRANGE EQUATIONS. 362.1 VIRTUAL DISPLACEMENT. . . . . . . . . . . . . . . . . . . . . . . 362.2 VIRTUAL WORK - GENERALIZED FORCE . . . . . . . . . . . . . 402.3 IMPRESSED AND CONSTRAINT FORCES. . . . . . . . . . . . . . 422.4 PRINCIPLES OF THE VIRTUAL WORK. . . . . . . . . . . . . . . 46

2.4.1 Principle of virtual work for a system in equilibrium. . 462.4.2 Principle of virtual work for a system in motion. . . . . 47

2.5 VIRTUAL WORK PERFORMED BY GRAVITY FORCES. . . . . . 472.6 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.7 LAGRANGE’S EQUATIONS OF MOTION. . . . . . . . . . . . . . . 75

2.7.1 Properties of a position vector partial derivatives. . . . 752.7.2 Lagrange’s equations – general case. . . . . . . . . . . . 752.7.3 Lagrange’s equations for conservative forces. . . . . . . 78

2.8 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

3 MODELLING AND ANALYSIS OF COLLISIONS. 1093.1 COLLISION OF TWO UNCONSTRAINED BODIES. . . . . . . . . 1093.2 COLLISION OF CONSTRAINED BODIES. . . . . . . . . . . . . . . 1123.3 CENTRE OF PERCUSSION. . . . . . . . . . . . . . . . . . . . . . . 1143.4 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

CONTENTS 4

II ANALYSIS. 149

4 ANALYTICAL SOLUTIONS AND THEIR STABILITY. 1514.1 ANALYTICAL SOLUTION OF EQUATIONS OF MOTION. . . . . 1514.2 STATE - SPACE FORMULATION OF EQUATIONS OF MOTION. 1524.3 EQUATIONS OF PERTURBATIONS. . . . . . . . . . . . . . . . . . 1534.4 DEFINITIONS OF STABILITY IN LAPUNOV’S SENSE. . . . . . . 1544.5 CRITERIA OF STABILITY OF EQUILIBRIUM POSITION. . . . . 1574.6 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

5 MODELLING AND ANALYSIS OF A CENTRIFUGE. 1615.1 MODELLING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

5.1.1 Description of the centrifuge. . . . . . . . . . . . . . . . . 1615.1.2 Physical model . . . . . . . . . . . . . . . . . . . . . . . . . 1625.1.3 Mathematical model. . . . . . . . . . . . . . . . . . . . . 163

5.2 ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1645.2.1 Space state formulation of equations of motion. . . . . 1645.2.2 Equilibrium positions. . . . . . . . . . . . . . . . . . . . . . 1655.2.3 Equations of perturbation - stability of the equilibrium

positions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

III EXPERIMENTAL INVESTIGATIONS. 170

6 INVESTIGATION OF THE EFFECT OF A GYROSTABILIZERON MOTION OF A SHIP. 1726.1 DESCRIPTION OF THE GYROSTABILIZER. . . . . . . . . . . . . 1726.2 MODELLING. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

6.2.1 Physical model. . . . . . . . . . . . . . . . . . . . . . . . . . 1736.2.2 Mathematical model. . . . . . . . . . . . . . . . . . . . . . 174

6.3 ANALYSIS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1786.4 EXPERIMENTAL INVESTIGATIONS. . . . . . . . . . . . . . . . . 180

6.4.1 Description of the laboratory installation . . . . . . . . 1806.4.2 Mathematical model of the laboratory installation. . . 1816.4.3 Identification of parameters. . . . . . . . . . . . . . . . . . 1826.4.4 Verification of the mathematical model. . . . . . . . . . 185

7 INVESTIGATION OF AN INDICATOR OF ANGULAR VELOC-ITY. 1867.1 DESCRIPTION OF THE INDICATOR. . . . . . . . . . . . . . . . . 1867.2 MODELLING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

7.2.1 Physical model. . . . . . . . . . . . . . . . . . . . . . . . . . 1877.2.2 Mathematical model. . . . . . . . . . . . . . . . . . . . . . 188

7.3 ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1907.3.1 Particular solutions. (equilibrium positions). . . . . . . 1907.3.2 Stability analysis. . . . . . . . . . . . . . . . . . . . . . . . 191

CONTENTS 5

7.3.3 Stable and unstable equilibrium positions. . . . . . . . 1927.4 EXPERIMENTAL INVESTIGATION. . . . . . . . . . . . . . . . . . 197

7.4.1 Description of the laboratory installation. . . . . . . . 1977.4.2 Identification of the system’s parameters. . . . . . . . . 1977.4.3 Collection of experimental data. . . . . . . . . . . . . . . 199

CONTENTS 6

INTRODUCTION.The purpose of this text is to provide the students with the theoretical back-

ground of the three dimensional mechanics of rigid body and its applications to engi-neering problems existing in mechanical systems. As most of the engineering subjectsthis part of mechanics is presented in three parts: Modelling, Analysis and Experi-mental Investigations (see Fig. 1).

EXPERIENCE AND KNOWLEDGE

MACHINE

ASSUMPTIONS

PHYSICAL MODEL

MATHEMATICAL MODEL

DYNAMIC PROBLEM

ANALYTICAL ANALYSIS

NUMERICAL ANALYSIS

SOLUTION

STABILITY ANALYSIS

SOLUTION OF THE DYNAMIC PROBLEM

ASSESSEMENT OF ASSUMPTIONS

EXPERIMENTAL INVESTIGATION

MACHINE DYNAMIC PROBLEM

D

Y

A

M

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C

S

O

F

M

A

C

H

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M O D E L L I N G

A N A L Y S I S

E X P E R I

M E N T A L

I N V E S T I G A T I O N

PHYSICS

MECHANICS

MATHEMATICS

COMPUTATIONAL MECHANICS

THEORY OF

STABILITY

DESIGN

TECHNOLOGY

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Figure 1

Development of the physical and the mathematical model of the dynamicproblem is the main aim of the first part. In the scope of this text it is assumed thatthe dynamic problems exist in machines which can be approximated by a chain of rigidbodies connecting to each other by means of kinematic constraints. This kinematicconstraints restrict the relative motion of the connecting links. This chain of rigidbodies (the mechanical system) performs general motion in the three-dimensionalinertial space. The process of creation of the mathematical model (a set of differentialequation) is based upon the Euler and Lagrange Equations.

Solution of the mathematical model and its analysis is the aim of part two.It allows to predict motion of the mechanical system if the forces acting on thesystem are known. If the motion of the system can be assumed known, analysis ofthe mathematical model yields driving forces which are necessary to maintain theassumed motion. The stability analysis provides informations about feasibility of thesolution obtained.

CONTENTS 7

Experimental verification of the developed mathematical models is the mainaim of the last part. Usually it can be accomplished by comparison of the obtainedsolution with results of measurements taken from the real objects.

The interaction between the Dynamics of Machines and the other engineeringsubjects is shown in Fig. 1.

This text is divided into six chapters. When relevant, chapters are ended withseveral engineering problems. Solution to some of them are provided. Solution tothe other problems should be produced by students during tutorials and in their owntime.

Part I

MODELLING

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9

INTRODUCTION.

In many applications we deal with a number of rigid bodies connected to eachother in some manner.. These connections, called constrains, impose additional con-ditions on the relative motion of one body with respect to another. Such a constrainedset of rigid bodies forms a mechanical system.

This part is concerned with creation of mathematical models of mechanicalsystems.

Chapter 1

MODELLING OF MECHANICAL SYSTEMS BY MEANS OFEULER EQUATIONS

1.1 CONSTRAINTS - CLASSIFICATIONOFMECHANICAL SYSTEMS.

1.1.1 ConstraintsThe unconstrained rigid body j has six degrees of freedom. Therefore, its relative po-sition with respect to another body i can be uniquely determined by six independentcoordinates. Usually the six coordinates xi, yi, zi, αx, αy, αz are chosen as shown inFig. 1

zj

yxj

j

j

Oj

α x

yα

zα

zi

i

zi

iO

y x i

x i

iyi

Figure 1

CONSTRAINTS - CLASSIFICATION OF MECHANICAL SYSTEMS. 11

yj

x j

j

α x

zαz

iO

x i

iyi

Oj

iy

izj

R xijMxij

R zij

Mzij

Figure 2

If the body is connected to another, the six coordinates are not independentand we can produce a number of analytical relationships between them. These an-alytical relationships are called constraint equations. For example, if the two bodiesi and j are connected as shown in Fig. 2, one may produce the following constraintequations

xi = 0

zi = 0

αz = αx

αy = 0 (1.1)

Since there is NCE = 4 constraint equations, only 6 − NCE = 2 coordinates maybe considered as independent.. In the case considered, yi and αx may be chosen asindependent coordinates.

DEFINITION: The number of independent coordinates c which uniquelydetermine the relative position of two constrained bodies is called class ofconstraint.

Hence the class of a constraint is determined by formula

c = 6−NCE (1.2)

Figures 3 to 6 provide more examples of possible constraints, their constraint equa-tions and class. Figures 2 to 6 show the scalar components of the resultant force Rij

and moment Mij of interaction between those two bodies. By inspection of thosediagrams one can notice that their number r is

r = 6− c = 6− (6−NCE) = NCE (1.3)


x j

j

α x

zαz

iO

x i

yj

iyi

Oj

iy

izj

R xijMxij

R zij

Mzij

=aR yij

Figure 3 Constraint equations: xi = 0, yi = a, zi = 0, αx = αz, αy = 0; Number ofconstraint equations: NCE = 5; Class: c = 6− 5 = 1.

xj

j

α x

zα

z

iO

xi

iy

i

Oj

iy

i zj

R yij

R zij

α y

R xij

Figure 4 Constraint equations: xi = 0, yi = 0, zi = 0; Number of constraint equations:NCE = 3; Class: c = 6− 3 = 3.


xj

j

αx

zα

z

iO

xi

iy

i

jy

i zj

R zij

α y

R xijjy

Oi Oj

Figure 5 Constraint equations: xi = 0, zi = 0, Number of constraint equations:NCE = 2; Class: c = 6− 2 = 4.

j

i

x j α x

zα

Oj

zj

α y

jyiy

xi

zi

Oi

iy

zi

xi

R zij

Figure 6 Constraint equations: zi = a Number of constraint equations: NCE = 1;Class: c = 6− 1 = 5.


1.1.2 Classification of mechanical systems.All constraint equations corresponding to the constraints considered in the previousparagraph were not dependent on derivatives of coordinates involved.

DEFINITION: If all constraint equations of a mechanical system are notdependent on derivative of coordinates involved, the mechanical system iscalled holonomic system.

DEFINITION: If at least one constraint equation of a mechanical systemis dependent on derivative of a coordinate involved, the mechanical systemis called nonholonomic system.

j

i

xj α x

zα

Oj

zj

α y

jyiy

xi

zi

Oi

iy

zi

xi

Rzij

a

ω yi

ω xi

Ryij

Rxij

A

Figure 7

To show difference between holonomic and nonholonomic system let us con-sider constraint shown in Fig. 7. If we assume that there is no friction between surfaceof the plane and sphere, we can produce only one constraint equation, namely

zi = a (1.4)

Hence, there are five independent coordinates which may be chosen as xi, yi, αx, αy,αz. But, if we assume that the ball rolls over the plane without slipping, the fivecoordinates are not independent, and one has to produce constraint equations whichreflect the fact that the relative velocity of the contact point Aj with respect to thepoint Ai is equal to 0.

vAj,Ai = iixi + jiyi +ωj,i × rAj,Oj

= iixi + jiyi+

¯¯ ii ji kiωj,ixi ωj,iyi ωj,izi

0 0 −a

¯¯ = 0 (1.5)

MOBILITY - GENERALIZED COORDINATES. 15

The above vector equation yields two scalar constraint equations

xi − ωj,iyi(αx, αy, αz, αx, αy, αz) · a = 0

yi + ωj,ixi(αx, αy, αz, αx, αy, αz) · a = 0 (1.6)

Therefore, in this case, class of this constrain is c = 3. The constraint equations1.6 depend on time derivatives of the coordinates involved, therefore the systemcomprising such a constraint is nonholonomic.

1.2 MOBILITY - GENERALIZED COORDINATES.

Let us consider a mechanical system which comprises n rigid bodies connected toeach other by p constraints. Among the p constraints there is p1 constraints of classc = 1, p2 constraints of class c = 2, p3 of class c = 3, p4 of class c = 4 and p5 of classc = 5. Evidently,

p = p1 + p2 + p3 + p4 + p5 (1.7)

Unconnected bodies of the mechanical system would have 6n degree of freedom.Since each constraint of class c takes away from the system considered 6− c degreeof freedom, the number of degree of freedom which is left after imposition of the pconstraints is

MO = 6n− (6− 1)p1 − (6− 2)p2 − (6− 3)p3 − (6− 4)p4 − (6− 5)p5MO = 6n− 5p1 − 4p2 − 3p3 − 2p4 − 1p5 (1.8)

DEFINITION: The number MO determined by the formula 1.8 is calledmobility.

As an example let us consider mechanical system presented in Fig. 8.

NUMBER OF DEGREE OF FREEDOM - DRIVING FORCES. 16

Z

X

Yy

x1

11

q1

Oq2

o

o

2

z1

z2

y

x2

2

q4

q5

q3

ro

r r1

r2

Figure 8

The system is assembled of n = 2moving links connected by p = 2 constraints.There is one constraint of class 2 (p2 = 1) and one constraint of class 3 (p3 = 1).Hence, its mobility according to Eq. 1.8 is

MO = 6 · 2− 4 · 1− 3 · 1 = 5

The mobility MO is equal to number of independent coordinates which must be in-troduced to determine uniquely a position of the system considered in the inertialspace. These independent coordinates are called generalized coordinates.

DEFINITION: Independent coordinates which uniquely determine posi-tion of a mechanical system with respect to the inertial space are calledgeneralized coordinates.

Figure 8 shows one of the possible introductions of the five generalized coordinates q1,q2, q3, q4, q5. Kinematics provides methods which allow position vector of any pointof mechanical system to be expressed in terms of the MO generalized coordinates.

r = r(q1, q2, .....qMO) (1.9)

1.3 NUMBER OF DEGREE OF FREEDOM - DRIVING FORCES.

In some practical applications, often we need to assume that motion along some ofthe generalized coordinates is given as an explicit function of time. Let L < MO benumber of coordinates along which motion of the system is known. In this case thenumber of degree of freedom of the system is reduced by L. Hence, the actual numberof degree of freedom is

M =MO − L (1.10)

EQUATIONS OF MOTION. 17

The above assumption requires introduction of L independent forces which assure theassumed motion along the L coordinates. These unknown forces are called drivingforces. In the considered case, each position vector can be expressed byM generalizedcoordinates and time t.

r = r(q1, q2, qM , t) (1.11)

DEFINITION: If all possible points of a mechanical system have positionsvector of form 1.9, the system is called scleronomic.

DEFINITION: If at least one point of a mechanical system has positionvector of form 1.11, the system is called rheonomic.

1.4 EQUATIONS OF MOTION.

The carried out in the previous section discussion leads to conclusion that for anymechanical system assembled of n rigid bodies we deal with

M – unknown functions representing a motion along M generalized coordi-nates qm,

L – unknown driving forces Fd,r =

P5i=1 pi(6− i) – number of unknown scalar components at p constraints.

Hence, the total number of all unknowns is

TNU = M + L+5X

i=1

pi(6− i)

= MO + 5p1 + 4p2 + 3p3 + 2p4 + 1p5 (1.12)

Upon introducing Eq. 1.8 into Eq. 1.12) we have

TNU = 6n− 5p1 − 4p2 − 3p3 − 2p4 − 1p5 + 5p1 + 4p2 + 3p3 + 2p4 + 1p5= 6n (1.13)

On the other hand, we may produce n free body diagrams for each body separately.An example of such a free body diagram ,corresponding to body 1 (see Fig. 8), isgiven in Fig. 9.

EQUATIONS OF MOTION. 18

z y

x

1

1

1

v

ω

G

R

M

10z

10z R10y

M10y

G

G

R12z

R12y

R12x

m g1

1

Figure 9

Always we are able to produce for each body involved six equations havingthe following form

m(vGx + vGzωy − vGyωz) = Fx

m(vGy + vGxωz − vGzωx) = Fy

m(vGz + vGyωx − vGxωy) = Fz (1.14)

IGxωx + (IGz − IGy)ωzωy = MGz

IGyωy + (IGx − IGz)ωxωz = MGy

IGzωz + (IGy − IGx)ωyωx = MGz (1.15)

Kinematics (Chap. 2) provides methods which permit the kinematic parameters vGand ωG to be expressed as functions of the M unknown generalized coordinates qmand time t.

vG = vG(q1...qm....qM , q1...qm....qM , t)

ω = ω(q1...qm....qM , q1...qm....qM , t) (1.16)

Since the system considered is assembled of n bodies we are able to generate 6nequations with 6n unknown. These equations are linear with respect to the 6n−Munknown components of reactions and driving forces. Therefore always it is possibleto eliminate them and produceM differential equations known as differential equationof motion of a mechanical system. Their solution represents motion of the systemconsidered. Upon introducing this solution into the remaining equations one canobtain expressions for all reactions and driving forces.

PROBLEMS. 19

1.5 PROBLEMS.

Problem 1

Z z1

ω

Md

a β

Az2

y2

G

y1

l

1

2

Figure 10

A uniform and rigid rod 2 of length l and mass m is hinged at A to the link 1as shown in the figure 10. The link 1, which can be considered as massless is drivenby a motor. Its driving moment depend on angular speed of the link 1 ω and can beapproximated by the following function

Md =Mo −∆Mω (1.17)

Derive equations of motion of the system and solve them for the following initialconditions

β(0) = 0, β(0) = 0, ω(0) = 0 (1.18)

PROBLEMS. 20

Solution.

KINEMATIC ANALYSIS.

Z z1

ω

β

Az2

y2

G

y1

l

1 2

j1 a

- k2l2

rG

y1

x1X

α

O

Y

Figure 11

The system considered has two degrees of freedom. Its instantaneous positioncan be uniquely determined by two independent coordinates α and β ( see Fig. 11).Angular velocity of the system of coordinates x1y1z1

ω1 = k1α (1.19)

Angular velocity of the system of coordinates x2y2z2

ω2 = ω1 +ω21 = k1α+ i2β (1.20)

Its components along system of coordinates x2y2z2 are

ω2x2 = ω2 · i2 = k1 · i2α+ i2 · i2β = β

ω2y2 = ω2·j2 = k1·j2α+ i2·j2β = α sinβ

ω2z2 = ω2·k2 = k1·k2α+ i2·k2β = α cosβ (1.21)

Position vector of the centre of gravity of the link 2 is

rG = j1a− k2(12 l) (1.22)

PROBLEMS. 21


rGx2 = rG·i2= j1·i2a− k2·i2(12 l) = 0rGy2 = rG·j2 = j1·j2a− k2·j2(12 l) = a cos β

rGz2 = rG·k2 = j1·k2a− k2·k2(12 l) = −a sinβ − 12l (1.23)

Absolute velocity of the centre of gravity

vG = rG = r0G +ω2×rG

= j2(−aβ sinβ) + k2(−aβ cosβ) +¯¯ i2 j2 k2β α sinβ α cosβ0 a cosβ −a sinβ − 1

2l

¯¯

= i2(−aα− l2α sinβ) + j2(

l2β) + k2(0) (1.24)

Hence, its components along x2y2z2 are

vGx2 = −aα− l2α sinβ

vGy2 = l2β

vGz2 = 0 (1.25)

KINETIC ANALYSIS.

The link 2 performs a general motion. Hence, its motion is governed byunmodified Euler’s equations.

m(vGx2 + vGz2ωy2 − vGy2ωz2) = F2x2

m(vGy2 + vGx2ωz2 − vGz2ωx2) = F2y2

m(vGz2 + vGy2ωx2 − vGx2ωy2) = F2z2 (1.26)

Ix2ωx2 + (Iz2 − Iy2)ωy2ωz2 = M2x2

Iy2ωy2 + (Ix2 − Iz2)ωx2ωz2 = M2y2

Iz2ωz2 + (Iy2 − Ix2)ωx2ωy2 = M2z2 (1.27)

Since,Ix2 = Iy2 = I = ml2/12 and Iz2 = 0 (1.28)

the last three equations may be rewritten in form 1.29

I(ωx2 − ωy2ωz2) = M2x2

I(ωy2 + ωx2ωz2) = M2y2

0 = M2z2 (1.29)

Kinematic parameters as function of the generalized coordinates are given by formulae1.21 and 1.25.

PROBLEMS. 22

Components of the resultant force and the resultant moment can be obtainedfrom the free body diagram of the body 2 shown in Fig. 12

β

A

2

y2G

2

z

MAz2

MAy2

R Az2

R Ay2

mg

R Ax2MAx2

=0

Figure 12

According to Fig. 12 the components of resultant force are

F2x2 = RAx2 (1.30)

F2y2 = RAy2 −mg sinβ (1.31)

F2z2 = RAz2 −mg cosβ (1.32)

and the components of resultant moment are

M2x2 = −RAy2(12l) (1.33)

M2y2 = RAx2(12l) +MAy2 (1.34)

M2z2 = MAz2 (1.35)

The link 1 (see Fig. 13) is considered as massless (I1z = 0), therefore the componentof resultant moment along axis z1 M1z1 must be equal to zero (see free body diagramin Fig. 1d).

M1z1 =Md −MAz2 cosβ −MAy2 sinβ +RAx2a = 0 (1.36)

The last equation yields an expression for MAy2.

MAy2 =Md/ sinβ −MAz2 cotβ +RAx2a/ sinβ (1.37)

Upon introducing Eq’s. 1.30, 1.35 and 1.37 into Eq. 1.34 one can obtain

M2y2 = F2x2(l/2 + a/ sinβ)−M2z2 cotβ +Md/ sinβ (1.38)

PROBLEMS. 23

Z z1

β

Az2

y2

y1

1

MAz2

MAy2

R Cz1

R Ay2

R Ax2MAx2

=0O

C

MCz1=0

R Az2

Md

R Cy1MCy1R Cx1MCx1

Figure 13

Introduction of Eq. 1.31 into Eq. 1.33 yields

M2x2 = −F2y2(l/2)−mg(l/2) sinβ (1.39)

Introduction of Eq’s. 1.21 and 1.25 into Eq’s. 1.26 and 1.29 yields

F2x2 = m(−aα− (l/2)α sinβ − lαβ cosβ)

F2y2 = m((l/2)β − a(α)2 cos β − (l/2)(α2) sinβ cosβ)F2z2 = m((l/2)(β)2 + a(α)2 sinβ + (l/2)(α)2 sin2 β) (1.40)

M2x2 = I(β − (α)2 sinβ cosβ)M2y2 = I(α sinβ + 2αβ cos β)

M2z2 = 0 (1.41)

Introduction of Eq’s. 1.40, 1.41 and 1.17 into Eq’s. 1.38 and 1.39 yields the wantedequations of motion in the following form.

0 = (ma2)α+ (I +m(l/2)2)α sin2 β +malα sinβ

+(2I +m(l2/2))αβ sinβ cosβ +mlaαβ cosβ − (Mo −∆Mα)

0 = (I +m(l/2)2)β − (I +m(l/2)2)α2 sinβ cosβ

−ma(l/2)α2 cosβ +mg(l/2) sinβ (1.42)

Since the above equations do not depend on α it is easy to lower their order byintroduction of ω and ω instead of α and α respectively.

0 = (ma2)ω + (I +m(l/2)2)ω sin2 β +malω sinβ

+(2I +m(l2/2))ωβ sinβ cosβ +mlaωβ cosβ − (Mo −∆Mω)

0 = (I +m(l/2)2)β − (I +m(l/2)2)ω2 sinβ cosβ

−ma(l/2)ω2 cosβ +mg(l/2) sinβ (1.43)

PROBLEMS. 24

Solution of equations 1.43 yields motion of the system. The equations 1.43 werenumerically integrated for m = 5kg, a = 0.2m, l = 0.4m, Mo = 1Nm and ∆M =0.2Nms. The time diagrams of the angular displacement β of the arm 2 and thecentrifuge angular velocity ω is shown in Fig. 14 and Fig. 15 respectively. The systemtends to an equilibrium position (βo = 0.75rad and ωo = 5rad/s). This equilibriumposition can be alternatively obtained by solution of Eq. 1.43 for ω = ωo =constantand β = βo =constant.

(Mo −∆Mωo) = 0

−(I +m(l/2)2)ω2o sinβo cosβo −ma(l/2)ω2o cosβo +mg(l/2) sinβo = 0

(1.44)

β[rad]

-0.4

0.4

0.8

1.2

1.6

0.0

0.0

4 8 12 16 20t[sec]

Figure 14

0.0 4 8 12 16 20t[sec]

ω[rad/s]

0.0

2.0

4.0

6.0

8.0

10

Figure 15

PROBLEMS. 25

Problem 2

zz

y

y

βω

Ωr

l

b

1

2

3

B

A

G

1 2

1

2

Figure 16

The wheel 3 of mass m rotates with a constant relative angular velocity Ωabout axis z2. The massless axle BG is hinged at B to a vertical shaft 1 whichrotates with a constant angular speed ω. Determine equilibrium position βo of theaxle BG.

Given are: Ω, ω, b, l, m, r

PROBLEMS. 26

Solution.

zz

y βω

Ωr

l

b

1

23

B

A

G

1 2

y1

2

mgB3

Z

ω ty1

x1

O

XY

Figure 17

It is easy to notice that the point B3 which belong to the link 3 is motionlesswith respect to the inertial frame XY Z. Hence, motion of the link 3 may be con-sidered as the rotational one about point B (see Fig. 17). Because z2 is the axis ofsymmetry of this link, the modified Euler’s equation can be applied.

I.ω2x2 + (Iz2 − I)ω2y2ωz2 + Iz2Ωω2y2 = Mx2

I.ω2y2 − (Iz2 − I)ω2x2ωz2 − Iz2Ωω2x2 = My2

Iz2(.ω2z2 +

.

Ω) = Mz2 (1.45)

In the case considered the relative angular velocity of the body 3 with respect to therotating system of coordinates x2y2z2 is

Ω = k2Ω (1.46)

Angular velocity of the system of coordinates x2y2z2 is

ω2= ω1+ω21= k1ω + i1β = i2β + j2ω sinβ + k2ω cosβ (1.47)

Moments of inertia of the body 3 about axis x2z2y2 are

Iz2 =mr2

2

I = Ix2 = Iy2 = IG +ml2 = m(r2

4+1

12b2) +ml2

= m(r2

4+1

12b2 + l2) (1.48)

PROBLEMS. 27

β

l

G

z1z2

y1

y2

mgB3

R 32y2

M32y2 M32z2R32z2

M32x2R32x2

z1

x1

x2B

Figure 18

β

z1z2

y1

y2

BR 32y2

M32y2M32z2

R32z2

M32x2R32x2

M21z1

z1

M21z1

x1

x2

R 21z1

M21y1 R 21y1R 21x1

B

Figure 19

PROBLEMS. 28

Components of the resultant moment due to external forces acting on the link 3,according to the free body diagram shown in Fig. 18 are

Mx2 = M32x2 +mgl sinβ

My2 = M32y2

Mz2 = M32z2 (1.49)

The free body diagram of the link 2, shown in Fig. 19, yields the following relation-ships.

M32x2 = 0

M32y2 = M21y1 cosβ +M21z1 sinβ

M32z2 = −M21y1 sinβ +M21z1 cosβ (1.50)


Mx2 = mgl sinβ

My2 = M21y1 cosβ +M21z1 sinβ

Mz2 = −M21y1 sinβ +M21z1 cosβ (1.51)

Introduction of Eq’s. 1.47, 1.48 and 1.51 into the Euler’s equation 1.45 yields

Iβ + (Iz2 − I)ω2 sinβ cosβ + Iz2Ωω sinβ = mgl sinβ (1.52)

Iωβ cosβ − (Iz2 − I)βω cosβ − Iz2Ωβ = M21y1 cosβ +M21z1 sinβ

(1.53)

Iz2(−βω sinβ) = −M21y1 sinβ +M21z1 cosβ

(1.54)

In the case of given ω and Ω the equation 1.52 allow to obtain motion β as a functionof time. It is called equation of motion. The particular solution of Eq. 1.52 whichcorresponds to β = β = 0 determines equilibrium position βo.

((Iz2 − I)ω2 cos βo + Iz2Ωω −mgl) sinβo = 0 (1.55)

Roots of the above equation are as follow

βo = 0, βo = 180o, βo = arccos

mgl − Iz2Ωω

(Iz2 − I)ω2(1.56)

Equations 1.53 and 1.54 determine the interaction moments (M21y1, M21z1) betweenthe link 2 and 1 as functions of time. Then, the equations 1.50 allow the interactionmoments (M32y1, M32z1) between link 3 and 2 to be determined. Moments M21z1

and M32z1 are the driving moments which one has to apply to the link 1 and 3 tokeep them going with the assumed velocities ω and Ω respectively.

PROBLEMS. 29

Problem 3

AA

b

a

1

23

ω

z

Ω

yGG

α

R

Figure 20

To test the gyroscope which is commonly used for the stabilization of theoperating theatre of a hospital ship, it was mounted on a rigid foundation as shownin Fig. 20. The gyroscope 1 rotates with a constant rotating speed Ω relative tothe housing 2, and the housing 2 is driven by the electric motor 3. Determine thereaction at bearing A if the motor turns the housing 2 with the constant angularspeed ω. The centre of gravity the gyroscope are at their axes of rotation as shownin the Fig. 20. The housing 2 may be considered as massless.

Given are:Ω = 1000rad/s,ω = 1rad/s,a = 1m,b = 2m,R = 0.8mIz = 100kgm2, Ix = Iy = 70kgm2 – principal moments of inertia of the

gyroscope about axes through its centre of gravity,α = 45o.

PROBLEMS. 30

Problem 4

A

B

l

C

ω21

ε

β

1

2

Figure 21

A uniform and thin bar 2 of mass m and length l is hinged to link 1 at A.The link 1 rotates with a constant angular acceleration ε. The relative angular speedof the link 2 with respect to the link 1 is constant and is equal to ω21. Determinereactions at the hinge A and C as well as the necessary driving moments.

PROBLEMS. 31

Problem 5

Y

X

y1Z, z1

x1

Z, z

α

Ay1

12

z1

1z2

x1

x2

G

A

β

la

Figure 22

Fig. 22 shows a mechanical system. Its link 1 is free to rotate about thevertical axis Z of the inertial system of coordinates XY Z. Moment of inertia of thelink 1 about axis Z is I1Z . The link 2 of the system is hinged to the link 1 at thepoint A as shown in Fig. 22. Distance between the point A and axis of rotationZ is a. Distance between the centre of gravity G and the point A is equal to l.Axes x2, y2, z2 are principal axis of inertia of the link 2 and the principal moments ofinertia about these axes are respectively I2x2, I2y2, I2z2. Mass of the link 2 is equal tom. Derive equations of motion of the system assuming that the angles α and β arethe generalized coordinates.

PROBLEMS. 32

Problem 6

Y αv

Z

y

z

a b

c

d

l β

1 2

3

4

A

B G

Figure 23

The anti-pollution bus 3, shown in Fig. 23, moves in the vertical plane Y Z ofthe inertial system of coordinates XY Z. The rear wheels 2 of the bus 3 are drivenwith the constant linear velocity v by the flywheel 1. When this bus is moving overthe hump 4 its angular velocity about axis x is

β =v

l

sinα

cos(α− β)

The angular velocity of the gyroscope 1 about axis z is variable. At the position βits magnitude is equal to Ω and its time derivative is Ω. Mass of the gyroscope ism and its principal moments of inertia about system of coordinates xyz through itscentre of gravity are Ix = Iy = I and Iz respectively.

1. Determine interaction forces between the gyroscope 1 and the bus at theconstraints A and B as a function of β .

2. Prove that the angular velocity β of the bus is determined by the aboveformula.

PROBLEMS. 33

Problem 7

Z

12

GOz L

3

Fω

4

α

x

y y

R

G

X

Y

L Sz

x

a b

A B

Figure 24

The arm 1 of the crane shown in Fig. 24 rotates with the angular velocity ωabout the inertial axis Z. In the same time the carriage 2 moves along the arm 1and its relative motion is determined by the position vectors S and L. The electricalmotor 4 drives the drum 3. The driving force produced by this motor is F. The drummay be considered as a symmetric rigid body of mass m and the principal momentsof inertia along the system of coordinates xyz are respectively Ix = Iy = I. Iz.

Produce:1. equation of motion of the drum2. expressions for the interaction forces between the drum and the carriageGiven are: S, L(t), ω(t), m, I, , F, a, R, a, b

PROBLEMS. 34

Problem 8

A

Z

X

12

z

l

α 1

x1

G2

z2

Figure 25

The massless link 1 of the mechanical system shown in Fig. 25 can rotateabout the horizontal axis Y of the inertial system of coordinates XY Z only. Itsmotion is determined by the following function

α = a sinωt

The system of coordinates x1y1z1 is attached to the link 1. The link 2 possesses massm and its principal moments of inertia about the axes through its centre of gravityG2 are Ix2 = Iyy2 = I and Iz2. The relative angular velocity of the link 2 with respectto the link 1 is determined by the following function

Ω = Ω0 sinWt

Produce:1. the expression for the interaction forces at the constraint A.

Answer: Forces and moments acting on the base.RAx1 = −aω2lm sinωt−mg sinαRAy1 = 0RAz1 = −a2ω2lm cos2 ωt+mg sinαMAx1 = Iz2Ω0aω sinWt cosωtMAz1 = Iz2(−ΩW cosWt)2. the expression for the driving moment that must be applied the link 1MAy1 =MAY = (I +ml2)aω2 sinωt+mgl sinα

PROBLEMS. 35

Problem 9

Y

X x

a

β

1

2

11

z 11

z 11

y 11

z 12

x 12

Z

α

O

Ω

M d

G 2

O

Figure 26

The housing of the ventilator shown in Fig. 26 is fixed to its base at theconstant angle β to form one rigid body 1. The instantaneous position of the body 1with respect to the inertial system of coordinates XY Z is determined by the angulardisplacement α.

α = A sinωt (1.57)

The rotor 2 of this ventilator rotates with respect to its housing with the constantangular velocity Ω.Given are:

I1 - moment of inertia of the body 1 about axis X.Ix = Iy = I, Iz- the principal moments of inertia of the rotor 2m - mass of the rotor 2a, A, Ω, ω, β - given constant parameters

Produce the expression for the driving moment Md that must be applied to the base1 to assure its motion according to the equation 1.57.

Answer:Md = I1(−Aω2 sinωt)− IAω2 sinωt cos2 β − IzAω

2 sinωt sin2 β

Chapter 2

MODELLING OF MECHANICAL SYSTEMS BY MEANS OFLAGRANGE EQUATIONS.

In the previous chapter it has been shown that dynamic problems can be solvedby decomposition of a mechanical system into individual rigid bodies and then byapplication of equations of motion corresponding to each body, equations of motionof the whole system may be formulated. Such a treatment of mechanical systems iscalled Euler’s approach.

Equations of motion of mechanical system, in many cases, can be formulatedwithout the need of decomposition of the system considered. They can be derivedfrom expressions for the kinetic energy function of the system, its potential energyfunction and the virtual work performed by external forces acting on it. Such anapproach is called Lagrange’s approach.

2.1 VIRTUAL DISPLACEMENT.

From previous consideration we remember that in case of holonomic and scleronomicsystems a position vector rn of a particle n (see Fig. 1) can be expressed as a functionof M generalized coordinates q1, ....qm, .....qM .

rn = rn(q1, q2, ....qM) (2.1)

Infinitesimal displacement drn of the particle n, corresponding to infinitesimal incre-ment of time dt can be calculated according to the total differential formula.

drn =MX

m=1

∂rn∂qm

dqm (2.2)

If qm would be determined in time, its infinitesimal displacement dqm would haveform

dqm =∂qm∂t

dt (2.3)

and the formula 2.2, in that case, determines the ’true’ infinitesimal displacement ofthe particle n. But, the generalized coordinates, at the present stage, are consideredas independent coordinates and do not have to be obeyed to the formula 2.3. Theymay be chosen arbitrarily but must be consistent with constraints of the system.

VIRTUAL DISPLACEMENT. 37

Z

X

Yy

x1

11

q1

Oq2

o

o

2

z1

z2

y

x2

2

q4

q5

q3

ro

rr1

r2

n

n

Figure 1

Such infinitesimal increments of generalized coordinates are called virtual dis-placements and are denoted by δqm to distinguish them from the ’true’ ones dqm.Hence, the virtual displacement of the particle n may be expressed as follows

δrn =MX

m=1

∂rn∂qm

δqm (2.4)

In case of holonomic and rheonomic systems the infinitesimal displacement of a par-ticle n is

drn =MX

m=1

∂rn∂qm

dqm +∂rn∂t

dt (2.5)

The virtual displacement is calculated under additional assumption that at the instantconsidered motion determined as explicit function of time along coordinates ql(t) isceased. Hence,

∂rn∂t

dt = 0 (2.6)

and the virtual displacement is determined by formula (2.4)

δrn =PM

m=1∂rn∂qm

δqm (2.7)

Figure 2 presents a simple system which configuration is determined by two gen-eralized coordinates q1 and q2. If both coordinates are independent, the system isscleronomic and virtual displacement of point P is shown in Fig. 3.If, for example, q1is determined in time (q1 = q1(t)) the system is rheonomic. Figure 4. gives graphicalinterpretation of virtual displacement for that case.


O

X

Z

Y

x

o

y

z

q1

r

q2

n

Figure 2

O

X

Z

Y

x

o

y

z

q1

r

q2

n

δ q1

δ q2

δ rn

rnq1

δ q1

rnq2

δ q2

Figure 3


O

X

Z

Y

x

o

y

z

q1

r

q2

n

δ q2

δ rnrnq2

δ q2 =

(t)

Figure 4

VIRTUAL WORK - GENERALIZED FORCE 40

2.2 VIRTUAL WORK - GENERALIZED FORCE

Consider a mechanical system shown in Fig. 5. Let us assume that the system hasM degree of freedom and its motions is caused by a set of forces F1,F2, ..Fn..FN . Letrn be the position vector of the point of application of the force Fn in the inertialsystem of coordinates XY Z.

Z

X

Yy

x1

1

1

q1

Oq2

o

o

2

z1

z2

y

x2

2

q4

q5

q3

ro

rr1

n

n

rnX

rnY

rnZ

r2

Fn

FnZ

Fn YFn X

Figure 5

DEFINITION: The expression

δWn = Fn · δrn (2.8)

is called virtual work performed by a force Fn on the virtual displacementδrn.

The virtual work performed by all forces is

δW =NXn=1

δWn =NXn=1

Fn · δrn (2.9)

If the system has M degree of freedom, there exists M generalized coordinatesq1, q2, ..qm, ..qM and each of the position vectors rn can be expressed as a func-tion of these coordinates.

rn = rn(q1, q2..qM , t) (2.10)

VIRTUAL WORK - GENERALIZED FORCE 41

Taking into account that the virtual displacement is

δrn =MX

m=1

∂rn∂qm

δqm (2.11)

the expression for the virtual work 2.9 may by rewritten in form

δW =NXn=1

Fn · (MX

m=1

∂rn∂qm

δqm)

=NXn=1

MXm=1

Fn · ∂rn∂qm

δqm

=MX

m=1

NXn=1

Fn · ∂rn∂qm

δqm

=MX

m=1

δqm(NXn=1

Fn · ∂rn∂qm

)

=MX

m=1

(NXn=1

Fn · ∂rn∂qm

)δqm (2.12)

DEFINITION: The expressionPN

n=1Fn· ∂rn∂qm( in the equation 2.12 it is inside

the bracket ) is called generalized force.

The generalised force is usually denoted by Qm.

Qm =PN

n=1Fn · ∂rn∂qm(2.13)

Decomposition of the force Fn and the position vector rn along axes of the system ofcoordinates XY Z yields

rn = IrnX + JrnY +KrnZ (2.14)

Fn = IFnX + JFnY +KFnZ (2.15)

Introducing Eq’s. 2.14) and 2.15 into Eq. 2.13 one may obtain

Qm =PN

n=1(FnX∂rnX∂qm

+ FnY∂rnY∂qm

+ FnZ∂rnZ∂qm

) (2.16)

As can be seen from the above expression, the generalized force is a scalar magnitude.The productQmδqm represents the virtual work done by all forces acting on the systemon displacements corresponding to δqm 6= 0 on assumption that all the others virtualdisplacements are equal to 0.)

IMPRESSED AND CONSTRAINT FORCES. 42

2.3 IMPRESSED AND CONSTRAINT FORCES.

For further analysis it is convenient to distinguish forces that produce the virtualwork equal to zero from forces that produce the virtual force not equal to zero.

DEFINITION: Forces that produce the virtual work equal to zero arecalled constraint forces and will be denoted by R.

DEFINITION: Forces that, in general, produce non-zero virtual work arecalled impressed forces and will be denoted by F.

It will be shown that the following forces produce the virtual work equal tozero.

1. interaction forces between the particles the individual links are made of2. interaction forces between individual links3. driving forces1. The virtual work produced by the interaction forces between the

particles the individual links are made of

i

j

Z

X

YOri

rj

δ rij

Fijδ ri

rij

Figure 6

Let us calculate the virtual work produced by the interaction forces betweenparticles which belong to the same link. Particle i (see Fig. 6) interacts with all theother particles the body is made of. Therefore the virtual work done by forces actingon the particle i is

δWi =NXj=1

Fij · δri (2.17)

The virtual work done by forces acting on all N particles the body is made of can beexpressed by the following formula.

δW =NXi=1

NXj=1

Fij · δri (2.18)


To show that the virtual work δW is equal to 0, let us consider the virtual work doneby interaction forces between the particle i and the particle j.

δWij = Fij · δri + Fji · δrj= Fij · δri − Fij · δrj= Fij · (δri − δrj)

= Fij · δrij = 0 (2.19)

Since rij has a constant length, the vector δrij is always perpendicular to the vectorFij. Hence, because the expression 2.18 is assembled of terms 2.19, one may concludethat

δW = 0 (2.20)

2.The virtual force produced by the interaction forces between individuallinks.

Z

X

YO

A

rA RAj

i

jδ rAj

RAi

δ rAi

δ rAij

Figure 7

Now let us calculate the virtual work performed by constraint forces betweentwo links i and j shown in Fig. 7. If friction between those two links is neglected,direction of a constraint force Ri

A, representing reaction of the link j on the link i,is always perpendicular to the slide i. Let δriA be virtual displacement of the link iat point of application A of the force Ri

A. Hence the virtual work performed by thisforce is

δW i = RiA · δriA (2.21)

Similarly, the virtual work performed by RjA is

δW j = RjA · δrjA (2.22)


The total virtual work done by both forces is

δW = δW i + δW j

= RiA · δriA +Rj

A · δrjA= Ri

A · (δriA − δrjA)

= RiA · δrijA (2.23)

But, the relative virtual displacement δrijA is always tangential to the slide i, hencethe total virtual work δW must be equal to zero.

δW = 0 (2.24)

Z

X

YO

A

RrA Aj

ij

δ rAj

Figure 8

When rheonomic system is considered, some of its links has motion assumed tobe known. Let us assume that link i has motion determined by an explicit function oftime (see Fig. 8). Because the virtual displacements are calculated under assumptionthat motion given by explicit functions of time is ceased, the virtual displacement ofthe point of application of reaction Rj

A is always tangential to the slide i. Hence thevirtual work performed by that reaction must be equal to zero.

δW j = RjA · δrjA = 0 (2.25)

3. The virtual work produced by the driving forces


Z

X

YO

A

r A

ij

z

z

δ r Aj=δ r A

i

F dj F d

i

Figure 9

Let us assume that the relative motion between the link i and the link j alongthe axis z is by assumption known. Hence we have to apply to the link i and the linkj two driving forces Fi

d and Fjd (see Fig 9). The virtual work produced by these two

forces isδW = Fi

d · δriA + Fjd · δrjA (2.26)

Since the virtual displacements correspond to the ’frozen’ links,

δriA = δrjA (2.27)

ThereforeδW =

¡Fid + F

jd

¢ · δriA = 0 · δriA = 0The above results can be obtained for any possible constraints.

PRINCIPLES OF THE VIRTUAL WORK. 46

2.4 PRINCIPLES OF THE VIRTUAL WORK.

2.4.1 Principle of virtual work for a system in equilibrium.

Z

X

YO

n

rn

mn

Fn

Rn

Figure 10

If a mechanical system is in an equilibrium, each of particle the system is made upfulfills its equilibrium conditions (see Fig. 10).

Rn + Fn = 0 n = 1, 2, ......N (2.28)

In the above equation Rn is resultant of all constraint forces acting on a particlen, Fn stands for resultant of all impressed forces acting on a particle n and N isnumber of particles the system is made up. Multiplying the above equation by virtualdisplacement of the particle n one may obtain expression for virtual work.

δWn = Rn · δrn + Fn · δrn = 0 (2.29)

Hence, virtual work performed by all forces acting on the system must be 0 as well.

δW =NXn=1

Rn · δrn +NXn=1

Fn · δrn = 0 (2.30)

But, the virtual work performed the constraint forces is equal to zero

NXn=1

Rn · δrn = 0 (2.31)

Therefore the virtual work is produced by the impressed forces only.

δW =NXn=1

Fn · δrn = 0 (2.32)

VIRTUAL WORK PERFORMED BY GRAVITY FORCES. 47

The last formula permits to formulate the following statement.

STATEMENT: A holonomic and scleronomic system with perfect con-straints is in its equilibrium if and only if the virtual work produced byall impressed forces is equal to zero.

If there is a set of L impressed forces F1,F2, .....Fl, ...FL acting on a system with Mdegree of freedom, the virtual work can be calculated according to formula 2.12.

δW =MX

m=1

Qm · δqm = 0 (2.33)

Since δqm, as independent variables, can not be equal to 0, the generalized forces Qm

must disappear in an equilibrium position.

Qm = 0 m = 1, 2, ......M (2.34)

Upon solving the above equations with respect toM unknown generalized coordinatesqm one may always obtain all possible system’s equilibrium positions.

2.4.2 Principle of virtual work for a system in motion.If a mechanical system is not stationary, each of its particles has to be obey toNewton’s law.

mnrn = Rn + Fn n = 1, 2, N (2.35)

Repetition of consideration carried out in the previous paragraph lead to equation

δW =NXn=1

(Fn −mnrn) · δrn = 0 (2.36)

which permit to formulate the following statement.

STATEMENT: For any holonomic system with perfect constraints the vir-tual work produced by all impressed forces and D’Alembert forces is alwaysequal to zero.

The above statement has no direct applications but forms a base for derivation ofequations of motion of any holonomic system.

2.5 VIRTUAL WORK PERFORMED BY GRAVITY FORCES.

The derived expression for virtual work (see Eq. 2.32) δW =PN

n=1Fn · δrn containsall impressed forces Fn acting on a particle n. One of its component is always gravityforce Gn (see Fig. 11).


Z

X

YO

rG

rnGn

G

G

n mn

Figure 11

Let us calculate the virtual work performed by gravity forces acting on a linkof a mechanical system. According to the above formula, the virtual work performedby the gravity forces acting on all particles the body is made of, may be adopted inthe following form.

δWG =NXn=1

Gn · δrn (2.37)

Taking into consideration that δrn =PM

m=1∂rn∂qm

δqm

δWG =NXn=1

Gn ·MXm=1

∂rn∂qm

δqm

=NXn=1

MXm=1

Gn · ∂rn∂qm

δqm

=MXm=1

NXn=1

Gn · ∂rn∂qm

δqm (2.38)

On the other hand, the virtual displacement of the centre of gravity of the bodyconsidered is

δrG =MX

m=1

∂rG∂qm

δqm (2.39)

Implementation to the above formula definition of position of the centre of gravityyields

δrG =MX

m=1

∂

∂qm(1

m

NXn=1

rnmn)δqm (2.40)


Since, qm does not depends on index of summation n, the formula 2.40 may betransformed as follow

δrGm =MXm=1

NXn=1

∂rn∂qm

mnδqm (2.41)

After multiplication of both sides of the above equation by Kg one can obtain

G · δrG =MXm=1

NXn=1

Gn · ∂rn∂qm

δqm (2.42)

Since right hand sides of equations 2.42 and 2.38 are the same, their left hand sidesmast be equal. Hence

δWG = G · δrG (2.43)

PROBLEMS 50

2.6 PROBLEMS

Problem 10

l

A

kC1 C2

l

2l

2l

B C

α

Figure 12

Two thin and uniform bars, each of mass m and length 2l, are hinged at Aand supported at B and C as shown in Fig. 12. The centres of gravity of these barsC1 and C2 are connected by a spring of length l and stiffness k. Upon assuming thatall constraints have no friction and the whole assembly can move only in the verticalplane, determine the stiffness k to ensure an equilibrium position of the system atα = 45o. Use the principle of virtual work.Given are:

α = 45o

m = 10kgl = 1m.AC1 = BC1AC2 = CC2

PROBLEMS 51

Solution.

A

C1 C2

2l

2l

B C

α

y

x

l

lr1 r2

F1

G1

F2

G2

l

Figure 13

The only impressed forces acting on the system considered are.

F1 = ik(2l cosα− l) = ilk(2 cosα− 1) (2.44)

F2 = −ilk(2 cosα− 1) (2.45)

G1 = −jmg (2.46)

G2 = −jmg (2.47)

Corresponding position vectors

r1 = il cosα+ jl sinα (2.48)

r2 = i(2l + l) cosα+ jl sinα = i3l cosα+ jl sinα (2.49)

Virtual displacement of the points C1 and C2.

δr1 =∂r1∂α

δα = (−il sinα+ jl cosα)δα (2.50)

δr2 =∂r2∂α

δα = (−i3l sinα+ jl cosα)δα (2.51)

The virtual work is.

δW = F1 · δr1 +G1 · δr1 + F2 · δr2+G2·δr2= (−l2k sinα(2 cosα− 1)−mgl cosα+ 3kl2 sinα(2 cosα− 1)− lmg cosα)δα

= (2kl2 sinα(2 cosα− 1)− 2mgl cosα)δα = 0 (2.52)

Since the generalized coordinate α can not be equal to 0.

2kl2 sinα(2 cosα− 1)− 2mgl cosα = 0 (2.53)

Hence

k =mg cosα

l sinα(2 cosα− 1) =mg

l(2 cosα− 1) tanα =10 · 9.81

1(2 cos 45o − 1) tan 45o = 236.8N/m

(2.54)

PROBLEMS 52

Problem 11

y

r

α

l l

12

3

l

xO

A

Figure 14

Three uniform and rigid bars 1, 2, and 3 (see Fig. 14), are hinged together atpoint A. Each of them has the same length l and mass m. The bar 1 can slide alonga vertical axis y. The bars 2 and 3 are resting on a cylinder of radius r. Find theangle α when the system is in an equilibrium. Use the principle of virtual work.

PROBLEMS 53

Solution.

y

αl l

l

xO

A

C 1

C 2 C 3G1

G2 G 3

r1

r2 r3

l/2

Figure 15

The system has one degree of freedom and α can be considered as the gener-alized coordinate. Impressed forces G1, G2 and G3 are shown in Fig. 15.

G1= G2= G3= −jmg (2.55)

Corresponding position vectors of the points of application C1, C2, C3 of the aboveimpressed forces as function of the generalized coordinate α are as follow.

r1 = j(r/ sinα+ l/2) (2.56)

r2 = i(−(l/2) sinα) + j(r/ sinα− (l/2) cosα) (2.57)

r3 = i((l/2) sinα) + j(r/ sinα− (l/2) cosα) (2.58)

Hence, virtual displacements of the points C1, C2 and C3 are

δr1 =∂r1∂α

δα = j(−r · cosαsin2 α

)δα (2.59)

δr2 =∂r2∂α

δα = i(−(l/2) cosα)δα+ j(−r · cosαsin2 α

+ (l/2) sinα)δα (2.60)

δr3 =∂r3∂α

δα = i((l/2) cosα)δα+ j(−r · cosαsin2 α

+ (l/2) sinα)δα (2.61)

For the system in an equilibrium position the virtual work have to be equal to O.

δW = δr1 ·G1 + δr2 ·G2 + δr3 ·G3

= (−mg(−r cosαsin2 α

)− 2mg(−r cosαsin2 α

+ (l/2) sinα))δα = 0 (2.62)

PROBLEMS 54

Since δα can not be equal to zero, the equation 2.62 yields

rcosα

sin2 α− 2(−r cosα

sin2 α+ (l/2) sinα) = 0 (2.63)

The equation 2.63 can be easily simplified to the following form

3r cosα− l sin3 α = 0 (2.64)

The above equation can be solved numerically and its roots represents the equilibriumpositions of the system considered.

PROBLEMS 55

Problem 12

x

y

l1

l2P

O

A

B

Figure 16

Two uniform bars of the length l1, l2 and mass m1, m2 respectively are jointtogether to form a double pendulum in the vertical plane. There is force P appliedat the point B. Determine the system equilibrium position.

PROBLEMS 56

Solution.

x

y

P

0l1_2

l2_2

C1

C2

A

B

G1

G 2

rC1

rC2

rB

q1

q2

Figure 17

The system has two degree of freedom. Its position can be uniquely determinedby two generalized coordinates q1 and q2 (see Fig. 17). There are three impressedforces acting on the system.

G1 = jm1g = jG1

G2 = jm2g = jG2

P = iP (2.65)

Position vectors of the points of application of the above forces as functions of thegeneralized coordinates

rC1 = i(rC1x) + j(rC1y) = i((l1/2) sin q1) + j((l1/2) cos q1)

rC2 = i(rC2x) + j(rC2y) = i(l1 sin q1 + (l2/2) sin q2) + j(l1 cos q1 + (l2/2) cos q2)

rB = i(rBx) + j(rBy) = i(l1 sin q1 + l2 sin q2) + j(l1 cos q1 + l2 cos q2) (2.66)

PROBLEMS 57

Virtual work done by the impressed forces is

δW = G1 · δrC1 +G2 · δrC2 +P · δrB= G1 · (∂rC1

∂q1δq1 +

∂rC1∂q2

δq2) +

G2 · (∂rC2∂q1

δq1 +∂rC2∂q2

δq2) +

P · (∂rB∂q1

δq1 +∂rB∂q2

δq2)

= jG1 · (∂(i(rC1x) + j(rC1y))∂q1

δq1 +∂(i(rC1x) + j(rC1y))

∂q2δq2) +

jG2 · (∂(i(rC2x) + j(rC2y))∂q1

δq1 +∂(i(rC2x) + j(rC2y))

∂q2δq2) +

iP · (∂(i(rBx) + j(rBy))∂q1

δq1 +∂(i(rBx) + j(rBy))

∂q2δq2)

= G1∂(rC1y)

∂q1δq1 +G1

∂(rC1y)

∂q2δq2 +

G2∂(rC2y)

∂q1δq1 +G2

∂(rC2y)

∂q2δq2 +

P∂(rBx)

∂q1δq1 + P

∂(rBx)

∂q2δq2 (2.67)

The partial derivatives according to (2.66) are respectively

∂rC1y∂q1

= −(l1/2) sin q1∂rC1y∂q2

= 0

∂rC2y∂q1

= −l1 sin q1∂rC2y∂q2

= −(l2/2) sin q2∂rBx∂q1

= l1 cos q1

∂rBx∂q2

= l2 cos q2 (2.68)


δW = (−G1(l1/2) sin q1 −G2l1 sin q1 + Pl1 cos q1)δq1 +

(−G2(l2/2) sin q2 + Pl2 cos q2)δq2 (2.69)

Since virtual work for system in an equilibrium has to be equal 0 and the virtualdisplacements along the generalized coordinates can not be equal 0, the last equation

PROBLEMS 58

yields two algebraic equations for the unknown coordinates q1 and q2.

(−G1(l1/2) sin q1 −G2l1 sin q1 + Pl1 cos q1 = 0

−G2(l2/2) sin q2 + Pl2 cos q2 = 0 (2.70)

Their solutions are

q11 = q01q12 = q02q21 = q01 + 180

o

q22 = q02 + 180o (2.71)

where

q01 = arccot2P

G1 + 2G2

q02 = arccot2P

G2(2.72)

The physical interpretation of the above solutions is given in Fig. 18.

x

y

P

x

y

Px

y

P

x

y

P

a) b) c) d)

Figure 18 a) q1 = q11, q2 = q12 b) q1 = q11, q2 = q22 c) q1 = q21, q2 = q12 d)q1 = q21, q2 = q22

PROBLEMS 59

Problem 13

k k

k k

q1

q2 l

l

A1

A2

Figure 19

Two identical uniform rods, each of mass m and length l, are joined togetherto form an inverse double pendulum (see Fig. 19). The pendulum is supported byfour springs, all of stiffness k, in such way that its vertical position (q1 = 0 and q2 = 0)is its equilibrium position. Using the principle of virtual work derive equations whichdetermine the other possible equilibrium positions.Given are: l, m, k,

PROBLEMS 60

Solution

y

x

q1

q2

A1

A2

G2

G1

F2

F1

rG1

rG2

rF2

rF1

Figure 20

The interaction forces between the springs and the individual links of thesystem shown in Fig. 20 are represented by the vectors F1 and F2.

F1 = −j(2kl sin q1) = −jF1F2 = −j(2kl sin q1 + 2kl sin q2) = −jF2 (2.73)

The vectors G1 and G2 represent the gravity forces acting on the link 1 and 2 re-spectively.

G1 = −img = −iG1

G2 = −img = −iG2 (2.74)

Points of application of the impressed forces F1, F2, G1 and G2 as a function of thegeneralized coordinates q1and q2 are determined by the following position vectors.

rF1 = i(l cos q1) + j(l sin q1) = ixF1 + jyF1

rF2 = i(l cos q1l cos q2) + j(l sin q1l sin q2) = ixF2 + jyF2

rG1 = i(1

2l cos q1) + j(

1

2l sin q1) = ixG1 + jyG1

rG2 = i(l cos q11

2l cos q2) + j(l sin q1

1

2l sin q2) = ixG2 + jyG2 (2.75)

PROBLEMS 61

The virtual work performed by all the impressed forces acting on the system is

δW = F1 · δrF1 + F2 · δrF2 +G1 · δrG1 +G2 · rG2 == −jF1

µ∂(ixF1 + jyF1)

∂q1δq1 +

∂(ixF1 + jyF1)

∂q2δq2

¶+

−jF2 ·µ∂(ixF2 + jyF2)

∂q1δq1 +

∂(ixF2 + jyF2)

∂q2δq2

¶+

−iG1 ·µ∂(ixG1 + jyG1)

∂q1δq1 +

∂(ixG1 + jyG1)

∂q2δq2

¶+

−iG2 ·µ∂(ixG2 + jyG2)

∂q1δq1 +

∂(ixG2 + jyG2)

∂q2δq2

¶(2.76)

Hence, simplifying the above expression one can get the following scalar expressionfor the virtual work.

δW = −F1µ∂yF1∂q1

δq1 +∂yF1∂q2

δq2

¶− F2

µ∂yF2∂q1

δq1 +∂yF2∂q2

δq2

¶+

−G1

µ∂xG1∂q1

δq1 +∂xG1∂q2

δq2

¶−G2

µ∂xG2∂q1

δq1 +∂xG2∂q2

δq2

¶(2.77)

Collecting terms in front of δq1 and δq2 we have

δW =

µ−F1∂yF1

∂q1− F2

∂yF2∂q1−G1

∂xG1∂q1

−G2∂xG2∂q1

¶δq1 +

+

µ−F1∂yF1

∂q2− F2

∂yF2∂q2−G1

∂xG1∂q2

−G2∂xG2∂q2

¶δq2 (2.78)

Since δq1 and δq2 can not be equal to zero, the virtual work become zero if

−F1∂yF1∂q1− F2

∂yF2∂q1

−G1∂xG1∂q1

−G2∂xG2∂q1

= 0

−F1∂yF1∂q2− F2

∂yF2∂q2

−G1∂xG1∂q2

−G2∂xG2∂q2

= 0 (2.79)

According to expressions 2.75 the partial derivatives are

∂yF1∂q1

= l cos q1∂yF2∂q1

= l cos q1

∂xG1∂q1

= −12l sin q1

∂xG2∂q1

= −l sin q1∂yF1∂q2

= 0∂yF2∂q2

= l cos q2

∂xG1∂q2

= 0∂xG2∂q2

= −12l sin q2 (2.80)

PROBLEMS 62

Introduction of equations 2.73, 2.74 and 2.80 into equation 2.79 yields

2kl2 sin q1 cos q1 + (2kl sin q1 + 2kl sin q2)(l cos q1)− 12mgl sin q1 −mgl sin q1 = 0

(2kl sin q1 + 2kl sin q2)(l cos q2)− 12mgl sin q2 = 0

(2.81)

or after simplification

8kl sin q1 cos q1 + 4kl sin q2 cos q1 − 3mg sin q1 = 0

4kl sin q1 cos q2 + 4kl sin q2 cos q2 −mg sin q2 = 0 (2.82)

One can see from the above equations that if q1 = q2 = 0 or q1 = q2 = ±π thesystem is in equilibrium. These trivial solutions of the equation 2.82 are interpretedin Fig. 23a). In order to obtain the non-trivial equilibrium positions one have tosolve numerically the non-linear set of the algebraic equations 2.82. To this end letus introduce the following functions.

f1(q1, q2) = 8kl sin q1 cos q1 + 4kl sin q2 cos q1 − 3mg sin q1

f2(q1, q2) = 4kl sin q1 cos q2 + 4kl sin q2 cos q2 −mg sin q2 (2.83)

These functions are shown in Fig. 21. Their zero point are presented in Fig. 22. Thecontinuous line represents zero points of the function f1(q1, q2) and the dotted onezero points of the function f2(q1, q2). The intersection points of those two lines offersthe wanted solutions of the set of equations 2.82. The geometrical interpretation ofthese solution is given in Fig. 23b),c),d) and e).

PROBLEMS 63

0

10

20

-10

-20

0

10

20

-10

-20

q1

2q

2q

q1

π

−ππ

−π

π

−π

−π

π

f (q ,q )1 21

f (q ,q )2 21

Figure 21

zero points of f (q ,q )1 21zero points of f (q ,q )2 21

q1

q2π

π

−π

−π

0

Figure 22

PROBLEMS 64

(0,0)

π( ,0)

π( 0, )−π( 0, )

−π( ,0) −ππ ( , )ππ( , )

−π π ( , )

−π −π ( , )

( 85 ,87 )o ( 40 ,-60 )

( 68 ,-126 ) ( 153 ,-75 )

a)

b) c)

d) e)

o o

o

o

oo o

Figure 23

PROBLEMS 65

Problem 14

l

A

B

l

a

X

Yq

Figure 24

Two uniform bars, each of mass m and length l, are joint as is shown in Fig.24 to form a planar system. The point A is stationary whereas the point B can movesalong the vertical plane which is apart from A by distance a < 2l. Find angle q whichdetermines the equilibrium position of the system. Use principle of the virtual work.

AnswerSolution of the following equation−32sin q ± 1

2cos q

al−sin q

1−(al )2+2a

lsin q−sin2 q

= 0

PROBLEMS 66

Problem 15

x

yll l l

l

Figure 25

Five uniform and rigid bars each of mass m and length l are joined togetheras shown in Fig. 25. Find equilibrium position by means of the principle of virtualwork.

PROBLEMS 67

Problem 16

l

l

l α

F A B

C

G

x

y

Figure 26

Fig 26 shows the kinematic scheme of a mechanism. The link AB can beconsidered as an uniform rod of mass m and length l. The link BC is massless andits length is l. Both links can move only in the vertical plane xy. At point A thehorizontal force F is applied. The distance between point C and the line of action ofthe force F is equal to l. By means of the virtual work principle determine the angleα which corresponds to the system equilibrium position .Given are: l, F, m

Answer:Solution of the following equation-12Gl sinα+ Fl

³cosα+ (1−cosα) sinα√

2 cosα−cos2 α

´= 0

PROBLEMS 68

Problem 17

l

l/2

G

c

Θ

Figure 27

A uniform and rigid bar of mass m and length l is supported as shown in Fig.27. Upon neglecting friction at the supports, determine the equilibrium position θ bymeans of principle of the virtual work.Given are: l, c, m

Answer:Solution of the following equationcos3Θ = 2 c

l

PROBLEMS 69

Problem 18

G

mg

l

y

x

R

Figure 28

A thin and uniform bar of the length l and mass m shown in Fig. 28 is placedinto a smooth bowl of radius R. Applying the principle of the virtual work determinethe equilibrium position of the bar.

PROBLEMS 70

Problem 19

2l

2l

Figure 29

x

z

o

α

β

A

Figure 30

Two uniform bars each of mass m and length 2l were welded together to formthe rigid body shown in Fig. 29). Two such rigid bodies were joined by means ofhinge A and suspended in the vertical plane xz as shown in Fig. 30). The general-ized coordinates α and β determine uniquely position of this system. Upon takingadvantage of the principle of the virtual work, produce magnitudes of the generalizedcoordinates corresponding to the system equilibrium position.

PROBLEMS 71

Problem 20

B

A

C

D

Y

X

1

23

4

H

lL

α

r

k

G

Figure 31

The mechanism shown in Fig. 31 can move in the vertical plane XY of aninertial system of coordinates. The links 2 and 3 are massless. The link 1 can beconsidered as a uniform rod of mass m and length l. The uncompressed length ofthe uncompressed spring 4 is lo and its stiffness is k. By means of the virtual workprinciple, determine the angle α corresponding to the equilibrium position of thismechanism.

Answer:Solution of the following equationk(lo −H + x sinα) ∂

∂α(L sin2 α± sinα√r2 − L2 cos2 α) + (−mg) ∂

∂α( l2sinα) = 0

where x = L sinα±√r2 − L2 cos2 α

PROBLEMS 72

Problem 21

Y

XO

A

b

a

α Br=0

1 23

Figure 32

The uniform beam 1 (see Fig. 39) of mass m and length a is hinged at thepoint O to the ground. Its end A is suspended on the massless rope 3. The rope isloaded by the block 2 of mass M . By means of the virtual work principle determinethe angle α which corresponds to the system equilibrium position.

PROBLEMS 73

Problem 22

1

2R

l1

2R

α

3

a) b)

Figure 33

The arm 1 of the balance shown in Fig. 33 can roll over the cylinder 2 withoutslipping. If this balance is not loaded, its arm remains in the horizontal position as isshown in Fig. 33a). By means of the virtual work principle determine the relationshipbetween the weight 3 and the angular position α of the arm 1 (Fig. 33b)).

Given are:R, l - distances shown in Fig. 33M - mass of the unloaded arm 1m - mass of the weight 3

Answer:α = ml

R(M+m)

PROBLEMS 74

Problem 23

X

Y

O

1

3

z

α a

a

A

B

2

90 o

Figure 34

Two massless rods, each of length a, were joint together at the point A toform the rigid body 1 (see Fig. 34). This body is free to rotate about the horizontalaxis Z of the inertial system of coordinates XY Z. The body 2 which can be treatedas a particle of mass m can slide along the arm AB. It is supported by the spring 3of stiffness k. The uncompressed length of this spring is equal to a/2. The systemhas two degrees of freedom and the two generalized coordinates α and z determineits position with respect to the inertial system of coordinates XY Z.

By means of the virtual work principle produce equations that determine thecoordinates α and z corresponding to the system equilibrium position.

Answer:a sinα− z cosα = 0Mg sinα+ ka

2− kz = 0

LAGRANGE’S EQUATIONS OF MOTION. 75

2.7 LAGRANGE’S EQUATIONS OF MOTION.

2.7.1 Properties of a position vector partial derivatives.Any position vector associated with a holonomic system has the following form.

rn = rn(q1, ..qm, ..qM , t) (2.84)

Since the generalized coordinates qm are themselves functions of time, the first deriv-ative of the position vector with respect to time is

rn =∂rn∂q1

q1 + ...∂rn∂qm

qm + ...∂rn∂qM

qM +∂rn∂t

(2.85)

The functions qm are called generalized velocities. Since all position vectors donot depend on qm, the partial derivative ∂rn

∂qmdo not depend on qm either. Hence,

∂

∂qm(∂rn∂qm

) = 0 (2.86)

Therefore, differentiation of Eq. 2.85 with respect to qm yields

∂rn∂qm= ∂rn

∂qm(2.87)

It is one of important properties of a position vector rn. Another one we obtaindifferentiating ∂rn

∂qmwith respect to time

d

dt(∂rn∂qm

) =∂2rn

∂q1∂qmq1 + ...

∂2rn∂qm∂qm

qm + ...∂2rn

∂qM∂qmqM +

∂2rn∂t∂qm

(2.88)

Now, let us differentiated both sides of equation 2.85 with respect to qm

∂rn∂qm

=∂2rn

∂q1∂qmq1 + ...

∂2rn∂qm∂qm

qm + ...∂2rn

∂qM∂qmqM +

∂2rn∂t∂qm

(2.89)

The right hand sides of equation 2.88 and 2.89 are the same. Therefore we canconclude that

ddt( ∂rn∂qm)= ∂rn

∂qm(2.90)

The equalities 2.87 and 2.90 are to be used in the next paragraph.2.7.2 Lagrange’s equations – general case.According to the principle of the virtual work, for any particle of a holonomic systemwith perfect constraints, we have.

NXn=1

(Fn −mnrn) · δrn = 0 n = 1, 2, .....N (2.91)

where Fn is resultant of impressed forces, mn is mass of the particle and rn is positionvector of the particle


If the system considered has M degree of freedom, the virtual displacementδrn is

δrn =MX

m=1

∂rn∂qm

δqm (2.92)


NXn=1

(Fn −mnrn) ·MX

m=1

∂rn∂qm

δqm = 0 (2.93)

Since generalized coordinates are independent, the above equation has to be fulfilledfor any combination of the virtual displacements along these generalized coordinates.In particular, it has to be fulfilled for

δq1 = 0, δq2 = 0, ......δqm 6= 0, ........δqM−1 = 0, δqM = 0 (2.94)

HenceNXn=1

(Fn −mnrn) · ∂rn∂qm

δqm = 0 (2.95)

orNXn=1

mnrn · ∂rn∂qm

=NXn=1

Fn · ∂rn∂qm

(2.96)

Right hand side of the above equation represent generalized force along m− th coor-dinate.

Qm =NXn=1

Fn · ∂rn∂qm

(2.97)

Now, let us manipulate on the left hand side of equation 2.96.

L =NXn=1

mnrn · ∂rn∂qm

=NXn=1

mn(d

dtrn) · ( ∂rn

∂qm) (2.98)

After adding and subtracting the same expression

L =NXn=1

mn

∙(d

dtrn) · ( ∂rn

∂qm) +

d

dt(rn · ∂rn

∂qm)− d

dt(rn · ∂rn

∂qm)

¸(2.99)

Upon developing the last term we have

L =NXn=1

mn

∙(d

dtrn) · ( ∂rn

∂qm) +

d

dt(rn · ∂rn

∂qm)− ( d

dtrn) · ( ∂rn

∂qm)− rn · ( d

dt

∂rn∂qm

)

¸(2.100)

Reduction of the same terms yields

L =NXn=1

mn

∙d

dt(rn · ∂rn

∂qm)− rn · ( d

dt

∂rn∂qm

)

¸(2.101)


Implementation of Eq. 2.87 and Eq. 2.90 produces

L =NXn=1

mn

∙d

dt(rn · ∂rn

∂qm)− rn · ( ∂rn

∂qm)

¸(2.102)

Now, it will be shown that the first term in Eq. 2.102 may by replaced by ddt( ∂∂qm

T )

and the second one by ∂∂qm

T where T is the kinetic energy of the system considered

T =NXn=1

mn(rn)2

2(2.103)

Indeed

d

dt(∂

∂qmT ) =

d

dt(∂

∂qm

NXn=1

mn(rn)2

2)

=NXn=1

d

dt(∂

∂qm

mn(rn)2

2)

=NXn=1

mnd

dt(∂

∂qm

(rn)2

2)

=NXn=1

mnd

dt(1

22rn · ∂rn

∂qm)

=NXn=1

mnd

dt(rn · ∂rn

∂qm) (2.104)

and

∂

∂qmT =

∂

∂qm

NXn=1

mn(rn)2

2

=NXn=1

∂

∂qm

mn(rn)2

2

=NXn=1

mn∂

∂qm

(rn)2

2

=NXn=1

mn(1

22rn · ∂rn

∂qm)

=NXn=1

mn(rn · ∂rn∂qm

) (2.105)

Introduction of Eq’s. 2.104 and 2.105 into Eq. 2.102 yields

L =d

dt(∂

∂qmT )− ∂

∂qmT (2.106)


Introduction of Eq’s. 2.106 and 2.97 into Eq. 2.96 gives the following final form ofLagrange’s equations.

ddt( ∂∂qm

T )− ∂∂qm

T = Qm m = 1, 2, ....M (2.107)

According to the presented derivation, T refers to kinetic energy of the whole system(links which motion is determined as explicit function of time do not have to beincluded). In a general case of a system with n links, its kinetic energy is determinedby formula

T =nXi=1

⎛⎝12mi[vGxi, vGyi, vGzi]

⎡⎣ vGxivGyivGzi

⎤⎦+ 12[ωxi, ωyi, ωzi]

⎡⎣ Ixi Ixiyi IxiziIyixi Iyi IyiziIzixi Iziyi Izi

⎤⎦⎡⎣ ωxi

ωyi

ωzi

⎤⎦⎞⎠(2.108)

wherevGxi, vGyi, vGzi−are components of absolute velocity of centre of gravity of the i− thlink vGi along a body system of coordinates xiyizi through the centre of gravity.ωxi, ωyi, ωzi−are components of absolute angular velocity ωi along the xiyizi systemof coordinates.[I ] – matrix of inertia about the xiyizi system of coordinates.mi – mass of the body.

If on the system considered acts a set of L impressed forces Fl, the generalizedforce Qm may be obtained from the formula 2.16.

Qm =LXl=1

(FlX∂rnX∂qm

+ FlY∂rnY∂qm

+ FlZ∂rnZ∂qm

) (2.109)

whereFlX , FlY , FlZ−are components of l − th force along an inertial system of coordinatesXY Z.rlX , rlY , rlZ−are components of the absolute position vector of a point of applicationof that force along the inertial system of coordinates XY Z.

2.7.3 Lagrange’s equations for conservative forces.All impressed forces can be divided into two categories: conservative forces C andnon-conservative forces F.

DEFINITION: If there exists such a function V , called potential energy func-tion, that its partial derivatives with respect to the coordinates of the pointof application of an impressed force taken with sign ’-’ are equal to thecomponent of the impressed force along these coordinates, the impressedforce is called conservative.

DEFINITION: Forces for which the potential energy function does notexists are called non-conservative.

Hence, if V is a potential energy function corresponding to a set of the conservative


forces C1,C2, ...Ck, ...CK, and rkX , rkY , rkZ are coordinates of point of application ofa conservative force Ck, its components are

CkX = − ∂V

∂rkX

CkY = − ∂V

∂rkY

CkZ = − ∂V

∂rkZ(2.110)

The generalized force corresponding to all conservative forces is

Qm =KXk=1

CkX∂rkX∂qm

+ CkY∂rkY∂qm

+ CkZ∂rkZ∂qm

=KXk=1

− ∂V

∂rkX

∂rkX∂qm

− ∂V

∂rkY

∂rkY∂qm

− ∂V

∂rkZ

∂rkZ∂qm

= −∂V (q1, q2, ...qm, ...qM)∂qm

(2.111)

Introduction of Eq. 2.111) into Eq. 2.107 yields

ddt( ∂∂qm

T )− ∂∂qm

T + ∂V∂qm

= Qm m = 1, 2, ....M (2.112)

where Qm represents a generalized force of all non-conservative forces only.If all impressed forces of a system are conservative equation 2.112 may be

simplify to the following form.

d

dt(∂

∂qmT )− ∂

∂qmT +

∂V

∂qm= 0 m = 1, 2, ....M (2.113)

Introducing expression known as Lagrangian

L = T − V (2.114)

and taking into account that potential energy function does not depend on qm, theequation 2.113 can be rewritten in form 2.115.

d

dt(∂

∂qmL)− ∂

∂qmL = 0 m = 1, 2, ....M (2.115)

If apart of conservative forces there are non-conservative forces involved, the Lagrangeequations takes form

ddt( ∂∂qmL)− ∂

∂qmL = Qm m = 1, 2, ....M (2.116)

where Qm, similarly to Eq. 2.112, represents generalized force of the non-conservativeforces only.


Z

X

YO

rG

i

Gi

i

rGiZ

Figure 35

Z

XYO

A

Bs

rArB

Figure 36

The only conservative forces which can be encountered in a mechanical systemare caused by gravity and springs connecting its links. Potential energy function forthe gravity force of the link i shown in Fig. 35 is

Vi = migrGiZ (2.117)

Potential energy for forces at A and B caused by a spring s of stiffness ks and lengthls (see Fig. 36) is

Vs =1

2ks(|rA − rB|− ls)

2 (2.118)

Potential energy function for all conservative forces acting on the system is

V =IX

i=1

Vi +SXs=1

Vs (2.119)

PROBLEMS 81

2.8 PROBLEMS

Problem 24

A

B

l

ω

β

1

2

Figure 37

An uniform and thin bar 2 of mass m and length l is hinged to link 1 whichrotates with a constant angular speed ω. Derive the differential equation of motion oflink 2 by means of Lagrange equations. Assume that β is the generalized coordinate.

PROBLEMS 82

B

l/2

ω

β

z1z2

y1

y2

o

G

Z

Figure 38

Solution.Angular velocity of the link 2 is a sum of the absolute angular velocity of the

link 1 ω and the relative velocity of the link 2 with respect to the link 1.

ω2 = ω + i2β = k1ω + i2β (2.120)

Sincek1 = j2 sinβ + k2 cosβ (2.121)

the absolute angular velocity of the link 2 is

ω2 = i2β + j2 sinβω + k2ω cosβ

Its components are

ω2x = β

ω2y = ω sinβ

ω2z = ω cos β (2.122)

The link 2 performs rotational motion about point O. Hence, its total kinetic energyis.

T =1

2[ω2x, ω2y, ω2z]

⎡⎣ I 0 00 I 00 0 0

⎤⎦⎡⎣ ω2xω2yω2z

⎤⎦=

1

2I(ω22x + ω22y) =

1

2I(β

2+ ω2 sin2 β) (2.123)

The potential energy function may be expressed as follow.

V = − l

2mgl cosβ (2.124)

PROBLEMS 83

Since all impressed forces are conservative ,motion of the system considered is gov-erned by Lagrange equations of the following form.

d

dt

µ∂T

∂β

¶− ∂T

∂β+

∂V

∂β= 0 (2.125)

where

d

dt

µ∂T

∂β

¶= Iβ

∂T

∂β= Iω2 sinβ cosβ

∂V

∂β=

1

2mgl sinβ (2.126)

Introducing Eq’s. 2.126 into Eq. 2.125 one can obtained

Iβ − Iω2 sinβ cosβ +1

2mgl sinβ = 0 (2.127)

Since I = 13ml2, the final form of equation of motion is

β − ω2 sinβ cos β + 32lg sinβ = 0 (2.128)

PROBLEMS 84

Problem 25

α

Rω

Z

1

2

Figure 39

The bead 1 which can be considered as a particle of mass m, may slideswithout any friction along the slide 2. The slide 2 rotates with the constant angularvelocity ω about the vertical axis Z ( see Fig. 39). The angle α can be consideredas the generalized coordinate. By means of Lagrange’s equations derive equation ofmotion of the bead 1 and determine all possible equilibrium positions.Given are : m = 1 kg, R = .25 m, ω = 10 rad/s.

PROBLEMS 85

Solution.

α

ω

Z

R

,z

xO,o

O,o

Y

X

x

y

ω t

Figure 40

In Fig. 40 the inertial system of coordinates is denoted by XY Z. System ofcoordinates xyz is rigidly attached to the slide and rotates with the angular velocityω about axis Z. Vector of the absolute position vector of the bead 1 is

R = iR sinα+ k(−R cosα) (2.129)

Its first derivative produces the absolute linear velocity v.

v = R = iRα cosα+kRα sinα+

¯¯ i j k

0 0 ωR sinα 0 −R cosα

¯¯

= iRα cosα+ jRω sinα+ kRα sinα (2.130)

Hencev2 = α2R2 + ω2R2 sin2 α (2.131)

All forces acting on the system considered are conservative, therefore Lagrange’sequations may be taken in the following form.

d

dt

µ∂T

∂α

¶− ∂T

∂α+

∂V

∂α= 0 (2.132)

PROBLEMS 86

where

T =mv2

2=1

2m(α2R2 + ω2R2 sin2 α) (2.133)

V = −mgRcosα (2.134)

Hence

∂T

∂α= mR2α

d

dt

µ∂T

∂α

¶= mR2α

∂T

∂α= mR2ω2 sinα cosα

∂V

∂α= mgR sinα (2.135)

Introduction of Eq. 2.135 into Eq. 2.132 yields equation of motion.

mR2α−mω2R2 sinα cosα+mgR sinα = 0 (2.136)

Since for the static equilibrium position α = α = 0 they, according to (2.136), haveto satisfy the following equation.

ω2R sinα cosα = g sinα (2.137)

Hence the possible equilibrium position are

αo = 0

αo = π

αo = arccosg

ω2R= arccos

9.81

102 · 0.25 = ±66.89o

PROBLEMS 87

Problem 26

1

2

C

Z

Y

r

y2

z2

ω21

Figure 41

A block 2 of mass m has the following matrix of inertia about axes x2, y, z2through the body centre of gravity C as shown in Fig. 41.

[I2] =

⎡⎣ Ix2 0 00 Iy2 −Iy2z20 −Iz2y2 Iz2

⎤⎦ where Iy2z2 = Iz2y2

The block rotates about the axle 1 with a constant relative angular velocity ω21. Theaxle 1 is free to rotate about a horizontal axis Y which is fixed in the inertial spaceXY Z. Centre of gravity C of the block 2 is on the axis of relative rotation at theknown distance r from axis Y . Use Lagrange’s approach to derive equation of motionof the block 2.

PROBLEMS 88

Solution.

Zz2

X

z1

x1

C,o 2

α

O,o1

x 1

x2y1

y2

o1 o2

ω 21 t

ω21

ω1

Y

mg

Figure 42

The kinetic energy of the body considered is

T = TT + TR

=1

2mv2C +

1

2[ωx2, ωy2, ωz2]

⎡⎣ Ix2 0 00 Iy2 −Iy2z20 −Iz2y2 Iz2

⎤⎦⎡⎣ ωx2

ωy2

ωz2

⎤⎦ (2.138)

where vC is the absolute velocity of a centre of gravity of the body and ωx2, ωy2, ωz2

are components of its absolute angular velocity. The angular velocity of the systemof coordinates x1y1z1, according to Fig. 42, is

ω1 = j1α = (i2 sinω21t+ j2 cosω21t)α (2.139)

The body considered rotates with respect to the system of coordinates x1y1z1 withangular velocity ω21

ω21 = k2ω21 (2.140)

Hence,its absolute angular velocity is

ω2 = ω1 +ω21 = i2α sinω21t+ j2α cosω21t+ k2ω21 (2.141)

PROBLEMS 89

The kinetic energy in rotational motion about its centre of gravity is

TR =1

2[ α sinω21t α cosω21t ω21 ]

⎡⎣ Ix2 0 00 Iy2 −Iy2z20 −Iz2y2 Iz2

⎤⎦⎡⎣ α sinω21tα cosω21t

ω21

⎤⎦=

1

2(Ix2α

2 sin2 ω21t− 2Iy2z2ω21α cosω21t+ Iy2α2 cos2 ω21t+ Iz2ω

221) (2.142)

The position vector of centre of gravity C is

r = −k1r (2.143)

Since its derivative with respect to time is

r = vC = r0 + ω1 × r =

¯¯ i1 j1 k10 α 00 0 −r

¯¯ = −i1αr (2.144)

the kinetic energy in translating motion is

TT =1

2mα2r2 (2.145)

It follows that expression for the total kinetic energy has form.

T =1

2(mα2r2+Ix2α

2 sin2 ω21t−2Iy2z2ω21α cosω21t+Iy2α2 cos2 ω21t+Iz2ω221) (2.146)

The potential energy function for the considered case is

V = −mgr cosα) (2.147)

Since the conservative forces are involved only, the following form of Lagrange’s equa-tion may be used.

d

dt

µ∂T

∂α

¶− ∂T

∂α+

∂V

∂α= 0 (2.148)

whereµ∂T

∂α

¶=

1

2(2mr2α+ 2Ix2α sin

2 ω21t− 2Iy2z2ω cosω21t+ 2Iy2α cos2 ω21t)= mr2α+ Ix2α sin

2 ω21t− Iy2z2ω21 cosω21t+ Iy2α cos2 ω21t (2.149)

d

dt

µ∂T

∂α

¶= mr2α+ Ix2α sin

2 ω21t+ Ix2α2 sinω21t(ω21 cosω21t) + Iy2z2ω221 sinω21t

+Iy2α cos2 ω21t− Iy2α2 cosω21t(ω21 sinω21t)

= α(mr2 + Ix2 sin2 ω21t+ Iy2 cos

2 ω21t)

+α(Ix2ω21 sin 2ω21t− Iy2ω21 sin 2ω21t) + Iy2z2ω221 sinω21t (2.150)

PROBLEMS 90

∂T

∂α= 0 (2.151)

∂V

∂α= mgr sinα (2.152)

Introduction of expressions. 2.149, 2.150, 2.151 and 2.152 into Eq. 2.148 yields theequation of motion in form

α(mr2 + Ix2 sin2 ω21t+ Iy2 cos

2 ω21t) + αω21 sin 2ω21t(Ix2 − Iy2) +mgr sinα+Iy2z2ω

221 sinω21t = 0

(2.153)

PROBLEMS 91

Problem 27

x

y

l1

l2

O1

O2

G1

G 2

a1

α 1

M 1

a2 α 2

Figure 43

Fig. 43 shows an arm of a robot operating in the horizontal plane. Motion ofthe arm is controlled by two actuators installed at joints O1 and O2. The actuatorsproduce moments which are functions of the angular position of the link 1 and 2. Themoment M1 is a function of the angular displacement α1 and the moment M2 is afunction of the relative angular displacement of the link 2 with respect to 1 (α2−α1).Derive differential equations of motion of the robot’s arm.

Given are:I1 - moment of inertia of the link 1 about a vertical axis through its centre of gravityG1.I2 - moment of inertia of the link 2 about a vertical axis through its centre of gravityG2.m1,m2 - masses of the link 1 and 2 respectively.a1, a2, l1, l2 - dimensions shown in Fig. 43

PROBLEMS 92

Solution.

x

y

l1

l2

O1

O2

G1

G2

a1

α 1

a2 α 2

rG2

Figure 44

The system considered has 2 degree of freedom and the angles α1 and α2 maybe considered as the generalized coordinates. Hence, Lagrange equations for this casecan be adopted in the following form.

d

dt

µ∂T

∂α1

¶−µ∂T

∂α1

¶= Q1

d

dt

µ∂T

∂α2

¶−µ∂T

∂α2

¶= Q2 (2.154)

The kinetic energy T is a sum of kinetic energy of the link 1 (T1) and kinetic energyof the link 2 (T2).

T1 =1

2(I1 +m1a

21)α

21 (2.155)

T2 =1

2(m2r

2G2 + I2α

22) (2.156)

whererG2 - is the absolute velocity of the centre of gravity G2.

The position vector of centre of gravity G2 is

rG2 = i(l1 cosα1 + a2 cosα2) + j(l1 sinα1 + a2 sinα2) (2.157)

Its first derivative yields the velocity of the centre of gravity G2.

rG2 = i(−l1α1 sinα1 − a2α2 sinα2) + j(l1α1 cosα1 + a2α2 cosα2) (2.158)

PROBLEMS 93

Squared magnitude of the velocity is

r2G2 = (−l1α1 sinα1 − a2α2 sinα2)2 + (l1α1 cosα1 + a2α2 cosα2)

2

= l21α21 + a22α

22 + 2l1a2α1α2 cos(α1 − α2) (2.159)

Upon introducing Eq. 2.159) into Eq. 2.156) one can obtain

T2 =1

2m2(l

21α

21 + a22α

22 + 2l1a2α1α2 cos(α1 − α2)) +

1

2I2α

22 (2.160)

Hence the total kinetic energy is.

T = T1 + T2 =1

2(I1 +m1a

21)α

21

+1

2m2(l

21α

21 + a22α

22 + 2l1a2α1α2 cos(α1 − α2)) +

1

2I2α

22 (2.161)

Since the robot operates in the horizontal plane, the only non-conservative forcesacting on the system are the driving moments shown in Fig. 45.

x

yO1

O2

α 1

M1

α 2

1

2

M12

M21

Figure 45

The virtual work produced by these forces is

∂W = kM1 · kδα1 + kM12 · kδα1 + kM21 · kδα2 (2.162)

SinceM12 =M2 and M21 = −M2 (2.163)

PROBLEMS 94

the expression for the virtual force takes the following form

∂W = (M1 +M2)δα1 −M2δα2 (2.164)

Therefore, the generalized forces in the equations 2.154 are

Q1 = M1 +M2

Q2 = −M2 (2.165)

PROBLEMS 95

Problem 28

1

2x

y y

z

RL

O

A

Β

α=ω tβ

Figure 46

Link 1 of the mechanical system shown in Fig. 46 rotates about the horizontalaxis z with the constant angular speed ω. The link 2 is hinged to the link 1 at pointA. The system has one degree of freedom and the generalized coordinate β determinesthe absolute angular position of the link 2. The link 2 comprises the massless rodAB and a particle of mass m attached at its end B. Apply the Lagrange approach toderive the differential equation of the system motion.

Given are:m - mass of the particleL - length of the link 2R - distance between the axis of rotation z and the hinge Aω - angular speed of the link 1.

PROBLEMS 96

Problem 29

ZZ

Y

1 2

X

Y,y

x

z

z

R

O O

O

α

β

Figure 47

The circular slide 1 of radius R is free to rotate about the horizontal axis Yof the inertial system of coordinates XY Z. Its moment of inertia about that axis isI. The body 2, which can be considered as a particle of mass m, can move alongthe slide without friction. System of coordinates xyz, shown in Fig. 47, is rigidlyattached to the slide 1.By means of Lagrange equations derive the differential equations of motion of thesystem along the generalized coordinates α and β.

Answer:¡I +mR2 sin2 β

¢α+ 2mR2αβ sinβ cos β −mgR sinα sinβ = 0

mR2β −mR2α2 sinβ cosβ +mgR cosα cosβ = 0

PROBLEMS 97

Problem 30

zz

y

y

βω

Ωr

l

b

1

2

3

B

A

C

1 2

1

2

Figure 48

The wheel 3 of mass m (see Fig. 48)) rotates with a constant angular velocityΩ about axis of the body 2 system of coordinates . The massless axle BC is hingedto the vertical shaft 1 which rotates with a constant angular velocity ω. The systemhas one degree of freedom, therefore the angle β determines uniquely its position.Given are: Ω, ω, m, b, l, rTake advantage of Lagrange’s approach to produce the equation of motion of thesystem.

Answer:IBx3β + (IBz3 − IBy3)ω

2 sinβ cosβ + IBz3ωΩ sinβ −mgl sinβ = 0

where IBx3 = IBy3 =mr2

4+ mb2

12+ml2 IBz3 =

mr2

2

PROBLEMS 98

Problem 31

1

2

k k Rr34

A

G

Figure 49

The link 1 of a mass m1, shown in Fig. 49), can move along the horizontalslide and is supported by two springs 3 each of stiffness k. The ball 2 of mass m2

and a radius r is hinged to the link 1 at the point A by means of the massless andrigid rod 4. All motion is in the vertical plane. Use Lagrange’s approach to deriveequations of motion of the system.

Given are:m1, m2, r, R, k, I = 2

5m2r

2 - moment of inertia of the ball about axes through itscentre of gravity G2.

Answer:(m1 +m2)x+m2Rϕ cosϕ−m2Rϕ

2 sinϕ+ 2kx = 0(m2R

2 + I)ϕ+m2Rx cosϕ+m2gR sinϕ = 0x - linear displacement of the block 1, ϕ - angular displacement of the link 2-4

PROBLEMS 99

Problem 32

B

D

OY

X

xE

C

αll

l

l l

l

A

12 34 56 7

Figure 50

Fig. 50 shows a mechanical system. Link 1 of the system moves along thevertical axis X and its motion is governed by the following equation

x = xosinωt

The links 2 and 3 are hinged to the link 1 at the point D. The links 4 and 5 join thelinks 2 and 3 with the collar 6 as is shown in Fig. 50. The spring 7 has a stiffness kand its length (when the spring is not compressed) is equal to 2l. The system has onedegree of freedom and its position may be determined by one generalized coordinateα. The links 4 5 and 6 are assumed to be massless. The links 2 and 3 can be treatedas thin and uniform bars each of length 2l and mass m.

Derive equations of motion of the system.

PROBLEMS 100

Problem 33

Gl

L 12

a

Figure 51

Fig. 51 shows the physical model of a centrifugal hammer. Its member 1rotates in the vertical plane with a constant angular velocity ω. The member 2 hasmass m and moment of inertia I about its centre of gravity G. Applying Lagrangeapproach derive differential equations governing motion of the member 2.

Given are: ω, L, l, a, m, I.

PROBLEMS 101

Problem 34

A

a

a

r

r

B

G

y

C

Z,z1 23

4

5

ω

1

1

kl q

Figure 52

Fig. 52 presents a physical model of Watt’s regulator. The link 1 performsrotational motion about the vertical axis Z with a constant velocity ω. The link 2 ishinged at A to the link 1. The link 2 is composed of a massless and rigid rod anda ball of mass m which may be considered as a particle. The spring 5 has stiffnessk. Length of the not compressed spring is equal to 2r. Links 3 and 4 are rigid andmassless. Due to the imposed constraints, point A, B,C and G stay always in planey1z1 which rotates with the angular speed ω. Derive equation of motion of the systemby means of Lagrange’s equations. Choose angular displacement q as the generalizedcoordinate.

Given are:m− mass of the link 2 concentrated at the point G,ω− angular speed of the link 1,lAB = lBC = r,lAG = l,a - distance of point A and C from axis Z,k− stiffness of the spring 5,lo = 2r− length of the not compressed spring 5

PROBLEMS 102

Problem 35

O

l , m 1 1 l , m 2 2

l 0

k 1 k 2 q 1

q 2

Figure 53

Produce equations of motion of the system shown in Fig. 53. Each link may beconsidered as a uniform rod. Their mass and length is m1, l1 and m2, l2 respectively.The links can move in a vertical plane of the inertial space only. Length of the notcompressed springs is lo and their stiffness is k1and k2. In Fig. 53, q1 and q2 standfor the generalized coordinates. Take advantage of the Lagrange equations.

PROBLEMS 103

Problem 36

R

r

O

oα

1 2

Z

Figure 54

The cylinder 2 may rolls over the stationary cylindrical surface 1 withoutslipping. Taking advantage of the Lagrange’s equations produce equation of thisrolling motion. The angular displacement α may be considered as the generalizedcoordinate. R and r stand for radius of the cylindrical surface and radius of thecylinder respectively. Axis Z is the vertical axis an inertial system of coordinates.Mass of the cylinder 2 is m and its moment of inertia about its axis of symmetry isI.

Answer:α (R− r)2

¡m+ I

r2

¢+mg sinα(R− r) = 0

PROBLEMS 104

Problem 37

Z

X

x

z

G

C

R

4R3π

O

α

Figure 55

The semi-cylinder of mass m and radius R shown in Fig. 55 is free to roll overthe horizontal plane XY without slipping. The instantaneous angular position of thissemi-cylinder is determined by the angular displacement α. Produce the equation ofoscillations of the semi-cylinder.

Answer:¡IG +mR2

¡1 + 16

9π2− 8

3πcosα

¢¢α+

³43mR2

πsinα

´α2 + 4

3Rπmg sinα = 0

where IG = 12mR2 −m

¡43Rπ

¢2

PROBLEMS 105

Problem 38

G

x1

y1

Y

x1

z1 X Z

ωt

l

1

5

4

3

2

α A

B

C

O o1

o1 O

y2

x2

Figure 56

The slide 1 of the mechanical system shown in Fig.56 rotates about the verticalaxis Y of the inertial system of coordinates XY Z with a constant angular velocity ω.The system of coordinates x1y1z1 is rigidly attached to the slide 1. The two sliders3 and 4 are massless. They are joined together by the link 2. The link 2 can beconsidered as a uniform rod of mass m and length l. The system is of one degreeof freedom and the only one generalized coordinate is denoted by α. The spring 5possesses stiffness k. The length of the uncompressed spring is lo. Its left hand sideend is attached to the axle of the joint A and its right hand side end is attached tothe slide 1 at the point C. Produce:

1. the kinetic energy function of the systemAnswer:

T = 16ml2(ω2 cos2 α+ α2)2. the potential energy function of the system

V = 12lmg sinα+ 1

2k(lo − l cosα)2

3. the differential equation of motion of the system13ml2α+ 1

3ml2ω2 cosα sinα+ 1

2mgl cosα+ kllo sinα− kl2 cosα sinα = 0

4. the equation for the angle α that defines the steady-state motion of thesystem (α=constant)(13ml2ω2 − kl2) cosα sinα+ 1

2mgl cosα+ kllo sinα = 0

PROBLEMS 106

Problem 39

Y

X

O q1

A

G

q2 c

k

l

12

3

4

α

a

Figure 57

The angle α locates the angular position of the stationary slide 4 with respectto the vertical plane XY of the inertial system of coordinates XY Z. The masslesslink 1 is free to move along this slide and is supported by the spring 3 of stiffness k.The length of the uncompressed spring is l. The link 2 is hinged to the link 1 at thepoint A. The distance c locates the position of the centre of gravity G of the link 2.The link 2 possesses mass m and its moment of inertia about axis through the centreof gravity G is I. This system possesses two degrees of freedom and q1 and q2 standfor the generalized independent coordinates.

1. Produce the kinetic energy function of the system2. Produce the potential energy function of the system3. Produce the equations of motion of the system (take advantage of the

Lagrange’s equations of motion)4. Produce the expressions for the generalized coordinates corresponding

to the possible equilibrium positions of the system.

PROBLEMS 107

Solution1. Since the link 1 is massless the total kinetic energy of the system is asso-

ciated with the link 2 only. The link 2 performs a general plane motion hence itskinetic energy is.

T =1

2mv2G +

1

2Iq22 (2.166)

where vG stands for the absolute linear velocity of the centre of gravity G ofthe link 2 and q2is its absolute angular velocity.The velocity vG can be produced by differentiation of the following absolute positionvector

rG = I(q1 cosα+ a sinα+ c cos q2) + J(q1 sinα− a cosα+ c sin q2) (2.167)

Hence the wanted velocity is

vG = rG = I(q1 cosα− cq22 sin q2) + J(q1 sinα+ cq22 cos q2) (2.168)

Introduction of the expression 2.167 and 2.168 into 2.168 yields the wanted kineticenergy function

T =1

2m¡(q1 cosα− cq22 sin q2)

2 + (q1 sinα− a cosα+ c sin q2)2¢+1

2Iq22 =

=1

2mq21 +mcq1q2 sin(α− q2) +

1

2mc2q22 +

1

2Iq22 (2.169)

2. The potential energy function due to gravitation is

Vg = −mgrGX = mg(q1 cosα+ a sinα+ c cos q2) (2.170)

The potential energy of the spring is

Vs =1

2kq21 (2.171)

Therefore the potential energy function is

V = Vg + Vs = mg(q1 cosα+ a sinα+ c cos q2)1

2kq21 (2.172)

3. Since the system is of two degree of freedom and the generalized coordinatesare q1and q2and all the impressed forces are conservative one can take advantage ofthe following Lagrange’s equations

d

dt

∂T

∂q1− ∂T

∂q1+

∂V

∂q1= 0

d

dt

∂T

∂q2− ∂T

∂q2+

∂V

∂q2= 0 (2.173)

According to 2.169 and 2.172 the equations of motion are

mq1 +mc sin(α− q2)q2 −mc cos(α− q2)q22 −mg cosα+ kq1 = 0

(I +mc2)q2 +mc sin(α− q2)q1 −mc cos(α− q2)q21+

+mc cos(α− q2)q1q2 −mgc sin q2 = 0

(2.174)

PROBLEMS 108

4. The above set of equations allows the equilibrium position of the system tobe determined. If q1and q2 are constant

−mg cosα+ kq1 = 0

−mgc sin q2 = 0 (2.175)

Henceq1 =

mg cosα

kq2 = 0 (2.176)

Chapter 3

MODELLING AND ANALYSIS OF COLLISIONS.

3.1 COLLISION OF TWO UNCONSTRAINED BODIES.

Let us assume, that two bodies 1 and 2 (see Fig. 1a) collide each other at a instantta.

Z

Y

X

O

1

2

G

G

1

2

rG1a

rG2a

ω1a

ω2a

v aG1

v aG2C

t=ta

Z

Y

X

O

1

2G

G1

2

C

t=t

Z

Y

X

O

1

2

G

G

1

2

ω1s

ω2s

v sG1

v sG2C

s

z1

y1

x1

y2z2

x2

rC2G2rC1G1

F

UF

UF

F

t <t<ta s

rG2a rG2

s=

rG1a rG1

s=

a b c

1

2

Figure 1

The instant ta is called instant of approach. Formulated in previous paragraphsequations of motion allow us to assume that the linear and angular velocities of thetwo bodies are known at the instant ta. Let vaG1,v

aG2,ω

a1,ω

a2 be the linear velocities

of their centres of gravity and angular velocities respectively at the instant ta. Forfurther considerations we assume that the surface in the vicinity of the contact pointC is perfectly smooth. Hence, the contact force F is normal to the surface of thebody 1 and 2 at the contact point C (see Fig. 1b). If F represents reaction of thebody 2 on body 1, reaction of the body 1 on 2 can be represented by reaction −F.

Let us introduce the body system of coordinates x1y1z1 and x2y2z2 in such away that axis x1 and x2 are parallel to the line of action of the contact force F. Aftera short period of time ∆t, at the instant ts = ta + ∆t called instant of separation,the two bodies are separated. Then, they continue their motion due to externalforces and the initial conditions corresponding to the instant ts (see Fig. 1c). theinitial velocities vsG1,v

sG2,ω

s1,ω

s2, called velocities of separation, are to be obtained on

bases of the following analysis. To simplify this analysis we introduce the followingadditional assumptions:

COLLISION OF TWO UNCONSTRAINED BODIES. 110

1. All external forces are small with respect to contact force F.2. Position of the body 1 and 2 at the instant ts is the same as it was at the instantta

3. The line of action of the contact force is unchanged during period of time ta < t <ts.

Each body involved in a collision may be considered as a separate system ofparticles. Hence, the impulse – momentum principle can be applied to each of them.

∆P = UF ∆hG = UMG(3.1)

Application of this principle for ta < t < ts to the body 1 yields

m1(vsG1 − vaG1) = UF1 (3.2)

∆hG = rC1G1 ×UF1 (3.3)

whereUF1 = i1UF UF2 = −UF1 = −i2UF (3.4)

rC1G1 ×UF =

¯¯ i1 j1 k1rC1G1x rC1G1y rC1G1zUF 0 0

¯¯

= j1(rC1G1zUF ) + k1(−rC1G1yUF ) (3.5)

Introduction of Eq’s. 3.4 and 3.5 into Eq’s. 3.2 and 3.3 produces six scalar equations.

m1(vsG1x − vaG1x) = UF

m1(vsG1y − vaG1y) = 0

m1(vsG1z − vaG1z) = 0 (3.6)⎡⎣ I1x −I1xy −I1xz

−I1yx I1y −I1yz−I1zx −I1zy I1z

⎤⎦⎡⎣ ωs1x − ωa

1x

ωs1y − ωa

1y

ωs1z − ωa

1z

⎤⎦ =⎡⎣ 0

rC1G1zUF

−rC1G1yUF

⎤⎦ (3.7)

Similar set of equation may be obtained for body 2.

m2(vsG2x − vaG2x) = −UF

m2(vsG2y − vaG2y) = 0

m2(vsG2z − vaG2z) = 0 (3.8)⎡⎣ I2x −I2xy −I2xz

−I2yx I2y −I2yz−I2zx −I2zy I2z

⎤⎦⎡⎣ ωs2x − ωa

2x

ωs2y − ωa

2y

ωs2z − ωa

2z

⎤⎦ =⎡⎣ 0−rC2G2zUF

rC2G2yUF

⎤⎦ (3.9)

So far we have produced 12 equations with 13 unknown.For perfectly elastic bodies there is no dissipation of energy during the contact

time∆t. Hence the additional equation may be derived from principle of conservationof energy. Since position of the two bodies at ta and ts has not been changed,

COLLISION OF TWO UNCONSTRAINED BODIES. 111

according to this principle, kinetic energy of the two bodies at instant ta and ts mustbe the same.

T a = T s (3.10)

It is easy to show (proof is enclosed in Appendix 3), that the equation 3.10 is equiv-alent to

−vsC1x − vsC2xvaC1x − vaC2x

= 1 (3.11)

where vsC1x − vsC2x, vaC1x − vaC2x are normal components of relative velocity of body 1

and 2 at contact point C at the instant of separation and approach respectively (seeFig. 2).

1

2G

G1

2

C

z1 y1

x1

y2z2

x2

vC2a

vC1a

vC2a-

C1va

C1vs vC2s

-

C1vs

vC2s

C1xvs

vC2xs

-

vC1xa

vC2xa-

Figure 2

On the other hand if the bodies involved in the collision are perfectly plasticthey stick to each other and travel after the collision together. Hence their relativevelocity of separation is equal to zero (vsC1x = vsC2x). Therefore


= 0 (3.12)

For the real materials we assume that the ratio of normal components of relativevelocity −vsC1x−vsC2x

vaC1x−vaC2x takes value within range between 0 and 1, and it depends only onnormal component of relative velocity of approach vaC1x − vaC2x. This ratio is calledcoefficient of restitution and is denoted by e.


= e 0 < e < 1 (3.13)

The above formula expresses so called Newton’s hypothesis.

COLLISION OF CONSTRAINED BODIES. 112

vC1xa

vC2xa- ||

e elastic materials

plastic materials

real materials

0

1

1

Figure 3

Magnitude of the coefficient of restitution for different material may be foundin technical literature. Fig. 3. presents an experimentally obtained relationshipbetween the coefficient of restitution and the relative velocity of approach.

The equations 3.6, 3.7, 3.8, 3.9 and 3.13 form a closed set of the algebraic equa-tions which permits for computation of all velocities of separation and the magnitudeof impulse at the contact point.

3.2 COLLISION OF CONSTRAINED BODIES.

Let us consider system shown in Fig. 4 and let us assume that its body i and j collideeach other at C. The system, before and after collision, has mobility MO Analysis sofar, allows to assume that all velocities of approach are known.

COLLISION OF CONSTRAINED BODIES. 113

i

j

C

Figure 4

During collision ,similarly to consideration in the previous section, we willassume that all the driving forces are small as compare with the forces of interactionin constraints. Hence, all driving forces during the time of collision can be neglected.It follows that during the very short time of collision assumption that we know motionof system along some of the generalized coordinates is not acceptable. Therefore,number of degree of freedom during collision is always equal to its mobility.

M =MO (3.14)

Hence, during the collision, there is M unknown parameters which determine veloci-ties of separation,r unknown components of impulses at constraints and 1 unknownimpulse at collision point C. Therefore, a total number of unknown TNU is

TNU =M + r + 1 (3.15)

where according to Eq. 1.8

M =MO = 6n− 5p1 − 4p2 − 3p3 − 2p4 − p5 (3.16)

r = 5p1 + 4p2 + 3p3 + 2p4 + p5 (3.17)

Introduction of Eq. 3.16 and 3.17 to Eq. 3.18 yields

TNU = 6n+ 1 (3.18)

On the other hand, for each body involved, we can produce 6 equations reflectingimpulse – momentum principle of form

mi(vsGix − vaGix) = UFix

m1(vsGiy − vaGiy) = UFiy

m1(vsGiz − vaGiz) = UFiz (3.19)

CENTRE OF PERCUSSION. 114

C

G

O

rGrC

v

2

1

1

Figure 5

⎡⎣ Iix −Iixy −Iixz−Iiyx Iiy −Iiyz−Iizx −Iizy Iiz

⎤⎦⎡⎣ ωsix − ωa

ix

ωsiy − ωa

iy

ωsiz − ωa

iz

⎤⎦ =⎡⎣ UMGix

UMGiy

UMGiz

⎤⎦ (3.20)

wherei = 1, 2, ....nUFi, UMGi− are resultant impulse and angular impulse of all impulses acting on i− thbody respectively.The lacking equation is offered by Newton’s hypothesis


= e 0 < e < 1 (3.21)

Hence, always we are able to produce 6n+ 1 equations with 6n+ 1 unknown whichallows for determination of all parameters of separation.

After collision motion of the system is controlled again by driving forces andinitial conditions (position at instance of approach and velocities of separation).

3.3 CENTRE OF PERCUSSION.

To introduce notion of centre of percussion let us consider the system shown in Fig.5 The stationary pendulum 1 is hit at C by the bullet 2 travelling horizontally withthe velocity of approach v.


C

G

O

rGrC

2

1

x

y1

1

x2

y2

UC UC

UOx

UOy

1

Figure 6

The system has three degree of freedom hence its position can determined bythree independent coordinates ϕ, x2 and y2 (see Fig. 6). At the instant of approach

ϕa = 0, ϕa = ωa1 = 0, xa2 = 0, xa2 = va2 = v, ya2 = 0 ya2 = 0 (3.22)

Impulse – momentum principle applied to the pendulum yields

m1(vsG1y1 − vaG1y1) = UOy1 (3.23)

m1(vsG1x1 − vaG1x1) = UC + UOx1 (3.24)

IO(ωs1 − ωa

1) = UCrC (3.25)

wherem1 – mass of the pendulumIO – moment of inertia of the pendulum about point OUC –-Impulse at contact point CUOx1, UOy1– Components of impulse at constraint O

Impulse – momentum principle applied to the bullet yields

m2(vs2y2 − va2y2) = 0 (3.26)

m2(vs2x2 − va2x2) = −UC (3.27)

wherem2 – mass of the bullet

Newton’s hypothesis for the case considered can be written as follows

−vsC1x1 − vsC2x2vaC1x1 − vaC2x2

= e (3.28)


Since vsG1y1 = 0 and vaG1y1 = 0, equation 3.23 yields UOy1 = 0. Since va2y2 = 0,according to 3.26 the component y of the bullet velocity of separation vs2y2 = 0.

Taking into consideration thatvsG1x1 = ϕsrGvaG1x1 = 0ωs1 = ϕs

ωa1 = 0

vs2 = xs2va2 = vvsC1x1 = ϕsrCvsC2x2 = xs2vaC1x1 = 0vaC2x2 = vthe equations 3.24, 3.25, 3.27 and 3.28 take form

m1(ϕsrG) = UC + UOx (3.29)

IO(ϕs) = UCrC (3.30)

m2(xs2 − v) = −UC (3.31)

e = −ϕsrC − xs2−v (3.32)

The above equations are linear with respect to unknown ϕs, xs2, UOx, and UC. Theirsolution is

ϕs =m2vrC(1 + e)

IOx +m2r2C(3.33)

xs2 =v(m2r

2C − eIO)

IOx +m2r2C(3.34)

UC =m2vIO(1 + e)

IOx +m2r2C(3.35)

UOx =m2v(1 + e)

IOx +m2r2C(m1rCrG − IO) (3.36)

The last equation allows to find position of the point of contact C having such apeculiarity that any impact at that point produce no impulse at O. Indeed, for

rC =IOxm1rG

(3.37)

UOx is always zero. Point of contact, fulfilling the formula 3.37 is called centre ofpercussion.

Motion of the system after the instance of separation will be controlled by


gravity forces and the following initial conditions

ϕo = ϕa = 0

ϕo = ϕs

x2o = xa2x2o = xs2y2o = ya2 = 0

y2o = ys2 = 0

PROBLEMS 118

3.4 PROBLEMS

Problem 40

L l

C

A

21

ω

Figure 7

The motion of the shuttle 1 of a loom is initiated by an impact given byhammer 2. The hammer 2 may be considered as a rigid and uniform bar of lengthL and a mass M . The shuttle 1 has mass m and may be considered as a particle.The point of contact C is at distance l from hinge A. At the instant of approach theangular velocity of the hammer 2 is ω and the shuttle 1 is motionless. Derive anexpression for velocity of the shuttle at the instant of separation and the impulse ofreaction at A if the coefficient of restitution is e.

PROBLEMS 119

Solution.

L l

C

A

21

ω

z

x

z

xUA

U U

ωa=

Figure 8

Angular impulse – momentum principle applied to the hammer 2 yields.

I(ωs − ω) = −Ul (3.38)

Impulse – momentum principle applied to the shuttle 1 yields.

m(vs − 0) = U (3.39)

By means of Newton hypothesis one can gain.

e = −vs − ωsl

0− ωl=

vs − ωsl

ωl(3.40)

Upon introducing of Eq. 3.39 into Eq. 3.38 one may obtain.

Iωs − Iω = −mlvs (3.41)

From Eq. 3.40vs = eωl + ωsl (3.42)

Incorporation of Eq. 3.42 into Eq. 3.41 yields.

Iωs − Iω = −ml2(eω + ωs) (3.43)

The above formula allows to determine the angular velocity of separation.

ωs =Iω −ml2eω

I +ml2(3.44)

Introduction of Eq. 3.44 into Eq. 3.42 yields velocity of separation of the shuttle.

vs =Ilω −ml3eω + eωl(I +ml2)

I +ml2=

Ilω(1 + e)

I +ml2(3.45)

PROBLEMS 120

Impulse – momentum principle applied to the hammer yields

M(ωsL

2− ω

L

2) = −U + UA (3.46)

Hence, the impulse of reaction at A is.

UA = U +ML

2(ωs − ω) = mvs +M

L

2ωs −M

L

2ω (3.47)

Upon introducing Eq’s. 3.44 and 3.45 into Eq. 3.47 one may obtain the impulse UA

in its final form.

UA =mlω(e+ 1)(I −M L

2l)

I +ml2(3.48)

PROBLEMS 121

Problem 41

v

l

l

C

O

Figure 9

Two thin uniform bars, each of mass m and length l, are connected by apin joint to form a double pendulum. Initially the bars are hanging vertically andare motionless. Then, a particle of mass mo traveling horizontally with a constantvelocity v strikes the joint between the bars. Upon assumption that the collision isplastic one ( the coefficient of restitution e = 0) solve this problem for the angularvelocity of each bar immediately after the impact.

PROBLEMS 122

Solution.

v

l

C

O

x

y

1

3

2

4

Figure 10

Collision of two mechanical systems is considered. First system is assembledof the particle 1 ( see Fig. 10). Second system is assembled of three bodies denotedin Fig. 10 by 2, 3, 4. The collision, between the body 1 and 2, takes place at thepoint C. Upon taking into account Newton’s hypothesis one may write the followingrelationship.

e = −vsC1x − vsC2xvaC1x − vaC2x

= 0 (3.49)

Hence,

vsC1x = vsC2x = ωs3l (3.50)

Impulse - momentum principle applied to the particle 1 (Fig. 11) yields

mo(vsC1x − v) = −U (3.51)

Angular impulse - momentum principle about the point of rotation O applied to thelink 3 (Fig. 12) yields.

ml2

3(ωs

3 − 0) = U3l (3.52)

Link 4 performs a general motion (Fig. 13.). It is possible to formulate two equation.First, reflected the impulse - momentum principle

m(vsG4x − vaG4x) = m(ωs3l + ωs

4(l/2)− 0) = U4 (3.53)

PROBLEMS 123

y

x

C

U

1 2

Figure 11

ω 3

l

O

x

y

3

U

UO

3

Figure 12

ω 4

l/2

Cx

y

U4

G

4

Figure 13

y

xC

U

2

U

U3

4

Figure 14

PROBLEMS 124

and second, reflected the angular impulse - momentum principle about the centre ofgravity G.

ml2

12(ωs

4 − 0) = −U4(l

2) (3.54)

Since the link 2 is considered as massless, it has to fulfill equilibrium conditions (Fig.14.)

U = U3 + U4 (3.55)

So far, we have produce 6 independent equations 3.50 to 3.55 with 6 unknown (U ,U3, U4, ωs

3, ωs4, v

sC1x). Since this is a set of linear equations, it can be easily solved

for the unknown angular velocities ωs3 and ω24.

ωs3 =

v

1(1 + 7m12mo

)

ωs4 = − 3v

1(1 + 7m12mo

)(3.56)

PROBLEMS 125

Problem 42

G

C

r

rG

C

2 1

O

α

Figure 15

To assess muzzle velocity of a bullet, it was fired to strike a stationary pendu-lum at a point C (Fig. 15). After the collision the maximum swing of the pendulum2 was observed to be α. Upon assuming that velocity of the bullet at the instant ofapproach is horizontal and coefficient of restitution is equal to zero, derive a formulafor the muzzle velocity of the bullet.

Given are:m - mass of the bulletIO - moment of inertia of the pendulum about the point OM - mass of the pendulum.rG - distance between axis of rotation of the pendulum and its centre of gravity.rC - distance between axis of rotation of the pendulum and the point of collision C.α - maximum swing of the pendulum.

PROBLEMS 126

Solution

G

C

rG

2 1

O

αy2

x2

x1

y1

U U

Uox2

Uoy2

Figure 16

Angular momentum – impulse principle applied to the pendulum yields.

IOωs2 = UrC (3.57)

Since the collision is plastic, from Newton’s hypothesis one can see that the velocityof separation of the bullet vs1x1 is

vs1x1 = ωs2rC (3.58)

The momentum – impulse principle applied to the bullet can be written as follows

m(ωs2rC − v) = −U (3.59)

Implementation of Eq. 3.59 into 3.57 allows the velocity of the bullet to be expressedas function of angular velocity of separation.

IOωs2 = −mrC(ω

s2rC − v) (3.60)

Hence

v =Io +mr2CmrC

ωs2 (3.61)

Principle of energy conservation

1

2IO(ω

s2)2 =Mg(rG − rGcosα) (3.62)

PROBLEMS 127

yields the angular velocity of separation in form.

ωs2 =

s2MgrG(1− cosα)

Io(3.63)

Upon introducing Eq. 3.63 into Eq. 3.61 one may obtain

v =Io +mr2CmrC

s2MgrG(1− cosα)

Io(3.64)

PROBLEMS 128

Problem 43

G

C

xyp

p

h

r

v

α

Figure 17

A football of mass m and principal moments of inertia Ixp, Iyp, Izp falls downvertically and hits the ground at the angle α with velocity v and the angular velocityω = 0. Find the equations of motion of the ball after the impact. Dimensions h andr locate the position of the centre of gravity G at the instant of impact.

Answer:

PROBLEMS 129

Problem 44

mv

x

ya

b

G

Figure 18

A rectangular block, resting vertically as shown in Fig. 18, is struck at pointC by a bullet of mass m traveling with a constant velocity v. Assuming that theimpact is inelastic (the coefficient of restitution e = 0) determine the smallest velocityv which is needed to overturn the block.

Given are:m – mass of the bulletM – mass of the blockIx, Iy, Iz- principal moments of inertia of the block about axes through its centre ofgravity Ga, b, c, – dimensions of the block

PROBLEMS 130

Problem 45

a

b

2

l

1

A

Figure 19

The mast 1, which can by considered as an uniform beam of mass m andlength l, is released from its vertical position without any initial velocity. The mastpivots about point A to strike the wall 2. Assuming that the coefficient of restitutionis e determine the angular velocity of the mast immediately after the impact with thewall.Given are:

m = 200kgl = 10ma = 2mb = 3me = 0.5

PROBLEMS 131

Problem 46

1

2v

k

α =45o

C

p

3

Figure 20

The body 1, which can be considered as a particle of mass m, is travellingwith the constant velocity v along path p to strike the block 2 of mass M at thepoint of collision C. The block 2 is supported by spring 3 of stiffness k and beforethe collision is motionless. The coefficient of restitution is e.

Produce the expression for:1. velocity of separation of the particle 1

Answer:vs1 = v

q12(m−Me)2

(M+m)2+ 1

2

2. velocity of separation of the block 2Answer:

vs2 =√22v mM+m

(1 + e)3. impulse of force during collision at the point C

Answer:U =

√22v mMM+m

(1 + e)4. amplitude of oscillation of the block 2 after collision.

A =√22

qMkv mM+m

(1 + e)

PROBLEMS 132

Problem 47

O

G

R

r

CG

rα

1

2 3

Figure 21

The pendulum, shown in Fig. 21, is assembled of the ball 1 of radius r and themassless and rigid rod 2. Mass of the pendulum is m and its moment of inertia aboutits centre of gravity is G is I. The pendulum was release from position determinedby angle α with initial velocity equal to zero to collide the vertical wall 3. Determinethe angular velocity of separation of the pendulum, impulse of force at the point ofcollision C and impulse of force at the constraint O.

PROBLEMS 133

Problem 48

1

2

3

α

GR

Figure 22

The cylinder 1 of mass m and radius R rolls without slipping over the hori-zontal floor 2 to eventually collide with the wall 3 (see Fig. 22). The angular velocityof approach of the cylinder is ω. Produce expression for the angular velocity of thecylinder and the linear velocity of its centre of gravity G at the time of separation ifthe coefficient of restitution is e.

Answer:vsGx1 = Rωe sinα vsGy1 = Rω cosα ωs = ω

PROBLEMS 134

Problem 49

x

y

G

ba

γ

p

z

p

p

c G

v

Figure 23

A rectangular block which sides are equal to a, b and c translates in thehorizontal plane with the constant velocity v to strike the vertical wall (see Fig. 23).Direction of motion of the block with respect to the wall is determined by the angleγ. The principal moments of inertia of the block about its principal axis xp, yp, zpthrough the centre of gravity G are Ixp, Iyp, Izp respectively and its mass is equal tom.Derive expression for the linear and angular velocity of the block immediately afterthe collision.

PROBLEMS 135

Problem 50

G1

G2

a

v1

2v

1 2

C

Figure 24

The ship 1 and 2 are traveling with constant velocities v1 and v2 to collideeach other at point C as shown in Fig. 24. Upon assuming that the coefficient ofrestitution is equal to 0, produce equations of motion of these ships after collisionand formulate the initial conditions. Friction between water and the ships can beneglected.Given are:

m1,m2 - mass of the ship 1 and 2 respectivelyI1, I2 - moment of inertia of the ship 1 and 2 about the vertical axis through

their centers of gravity G1, G2 respectivelyv1, v2 - velocity of the ship 1 and 2 respectivelye = 0 - coefficient of restitutiona - the distance shown in Fig. 24

PROBLEMS 136

Problem 51

h

α

1

2

H

L

ba

CG

O

ω

A

B

Figure 25

The tennis racket 2 (see Fig. 25rotates about the point O with the constantangular velocity ω. The tennis ball 1, that can be considered as a particle of massm, is released from the position A with the initial velocity equal to zero. The ballcollides with the racket at the point C. At the instant of approach the racket makesangle α with horizon. The coefficient of restitution is e. After the collision the ballhits the ground at the point B.Given are:

m - mass of the ball 1M, I - mass and the moment of inertia of the racket 2 about its centre of

gravity G respectivelye - coefficient of restitutionh,H, a, b - given distancesα - angular position of the racket at the instance of approachω - angular velocity of the racket at the instance of approach.

Produce the expression for the horizontal distance L between the point C and thepoint B.

Answer:L = vY

vXg

³1 +

q1 + gH

v2X

´wherevX = vsx1 cosα− vsy1 sinα vY = vsx1 sinα+ vsy1 cosα

vsx1 =1

1+mb2

Io

³√2gh cosα

³mb2

Io+ e´+ ωb (1 + e)

´vsy1 =

√2gh sinα

PROBLEMS 137

Problem 52

1 r

a

b

c

α

2

C

position 1

position 2

position 3

Figure 26

A spherical object 1 of massm and the principal moment of inertia I is releasedfrom position 1 with the initial linear and angular velocity equal to zero (see Fig. 26).This object collides with the wall 2 at position 2. The slope of the wall at the pointof collision C is equal to α. After separation, this object collides with the horizontalwall at position 3.

Determine coefficient of restitution e at the point of collision C.Given are:a, b, c - dimensions shown in Fig. 26m, I - mass an the principal moment of inertia of the object 1 respectivelyα - slope of the wall

PROBLEMS 138

Problem 53

G1

G2 a2

a2

b2 b2

v1

v2

a1

a1

b1

b1

1

2

C

α

Figure 27

The two cars shown in Fig. 27, 1 and 2, collide each other at the point ofcollision C. Their masses are m1 and m2 respectively. The moments of inertia ofthe cars about the axes through the centre of gravity G1 and G2 are I1 and I2. Atthe instant of approach the linear velocity of their centers of gravity were v1 and v2respectively and their angular velocities were equal to zero. Upon assuming that thecoefficient of restitution is equal to zero, produce:1. the equations which allow the kinematic parameters of separation to be determined

Answer:Solution of the following set of equationsm1(v

sG1x1 − v1 sinα) = −U

m1(vsG1y1 − v1 cosα) = 0

I1(ωs1 − 0) = U(b1 cosα− a1 sinα)

m2(vsG2x2 − 0) = U

m2(vsG2y2 − v2) = 0

I2(ωs2 − 0) = 0

vsG1x1 − ωs1(+b1 cosα− a1 sinα)− vsG2x2 = 0

2. the expression for the energy dissipated during the collisionAnswer

E =¡12m1v

21 +

12m2v

22

¢−³12m1

³(vsG1x1)

2 +¡vsG1y1

¢2´+ 1

2I1 (ω

s1)2 + 1

2m2

³(vsG2x2)

2 +¡vsG2y2

¢2´+ 1

2I2 (ω

s2)2´

PROBLEMS 139

Problem 54

G1 v

1

G2 a

3 2

C

A

G3

Figure 28

The two cars shown in Fig.28, 1 and 2, collide with each other at the point ofcollision C. At the instant of approach the car 1 was translating with velocity v. Thesecond car at the instant of approach was stationary. The passenger 3 of the secondcar is modeled as a rigid body hinged to the car at the point A. Its mass is m3 andits moment of inertia about the center of gravity G3 is I. The distance a determinesthe position of the center of gravity G3 at the instant of approach. The mass of thecar 1 is m1 and the mass of the car 2 without the passenger is m2.

Assuming that the coefficient of restitution is equal to e, produce:1. the relative angular velocity of the passenger 3 at the instant of separation2. the expression for the energy dissipated during the collision

PROBLEMS 140

Answer:Question 1

G1 v

1

G2

a

3

2

C1

A

G3

A

C2x1

y1

UC UC

x2

UA y2

x3

UAx

y3

UK L N M

UL UN UM

UAy

UAx

UAy

Figure 29

The relative angular velocity of the passenger ωs3 is the solulution of the fol-

lowing set of equationsm1(v

sG1x1 − v) = −UC

m2(vsG2x2 − 0) = +UC + UAx

m3(vsG2x2 − ωs

3a− 0) = −UAx

I(ωs3 − 0) = −aUAx

e = −vsG1x1−vsG2x2v−0

Question 2∆T = T a−T s=1

2m1v

2−(12m1 (v

sG1x1)

2+12m2 (v

sG2x2)

2+12m3 (v

sG2x2 − ωs

3a)2+1

2I (ωs

3)2 )

PROBLEMS 141

Problem 55

v

1 2 3

Figure 30

The carriage 1 (see Fig.30) ) moves with a constant velocity v to collide withthe two stationary carriages 2 and 3. The carriages 2 and 3 are unconstrained (thereis a negligible gap between them). Mass of each carriage is identical and it is equalto m. The coefficient of restitution e is the same for each point of collision.

1. Produce the equations for the velocity of separation of individual car-riages and the impulses between the carriages.

2. Solve the equations for the coefficient of restitution e = 1 and e < 13. Produce the expression for the energy lost due to the collision if e < 1.

PROBLEMS 142

Solution.

Collision 1.

3 1 2 v 1a1=v

v2a1=0

C1 C2

U11 U1

1

y 1 x 1

y2

x2

Figure 31

At the time of approach ta1 the carriage 1 collides with the carriage 2 at thepoint of collision C1(see Fig. 31). Since the carriage 2 and 3 are unconstrained (thereis a gap between them) and during the collision carriages 1 and 2 are stationary, theimpulse at the point C2 is equal to zero. Hence one can write the following equationsreflecting the linear impulse-linear momentum principle applied to the carriage 1 andthe carriage 2.

m(vs11 − v) = −U11 (3.65)

m(vs12 − 0) = +U11 (3.66)

The Newton’s hypothesis yields

e = −vs11 − vs12va11 − va12

= −vs11 − vs12

v(3.67)

Solving these equations one can get

vs11 =(1− e)v

2(3.68)

vs12 =(1 + e)v

2(3.69)

U11 = m

(1 + e)v

2(3.70)

Since the carriage 2 after collision possesses positive velocity of separation it willcollide with the carriage 3. Taking into account that the gap between the carriage2 and 3 is negligible, one can assume that the velocity of separation vs12 is equal tovelocity of approach va22 for the second collision

va22 = vs12 (3.71)

Collision 2

PROBLEMS 143

In Fig.32 the free body diagram for the collision 2 is presented.

3 1 2 v 1s1

v2a2

C1 C2

U2

2

y 3

x 3

y2

x2

v3a2= 0

U22

Figure 32

Application of the linear impulse-linear momentum principle to the carriage 2and 3 yields

m(vs22 − vs12 ) = −U22 (3.72)

m(vs23 − 0) = U22 (3.73)

The Newton’s hypothesis yields

e = −vs22 − vs23va22 − va23

= −vs22 − vs23vs12

(3.74)

where according to 3.69

vs12 =(1 + e)v

2

Solution of these equations gives:

vs22 =(1− e)vs12

2=(1− e)

2

(1 + e)v

2=(1− e2)v

4(3.75)

vs23 =(1 + e)vs12

2=(1 + e)

2

(1 + e)v

2=(1 + e)2v

4(3.76)

U22 = m

(1 + e)vs122

= m(1 + e)

2

(1 + e)v

2= m

(1 + e)2v

4(3.77)

According to the consideration so far after the second collision:the first carriage possesses velocity vs11 =

(1−e)v2

the second carriage possesses velocity vs22 =(1−e2)v

4

the third carriage possesses velocity vs23 =(1+e)2v

4

If the coefficient of restitution is equal to one (perfectly elastic collision) the situationis clear:

the first carriage possesses velocity vs11 = 0the second carriage possesses velocity vs22 = 0the third carriage possesses velocity vs23 = v

It means that the carriage 1 and 2 are stationary after collision and the carriage 3moves with the velocity v. Such a behaviour of the unconstrained bodies is oftenreferred to as "Newton’s cradle".

PROBLEMS 144

The situation is more complicated if e 6= 1. It is easy to see that in this case,after the second collision, the velocity of the first carriage is greater then the velocityof the second one regardless the coefficient of restitution e.

vs11 =(1− e)v

2> vs22 =

(1− e2)v

4

2 > (1 + e)

Therefore immediately after the second collision mast taken place third collision be-tween the first and second carriage.

The velocity of approach of the first carriage is

va31 = vs11 =(1− e)v

2(3.78)

and the velocity of approach of the second carriage is

va32 = vs22 =(1− e2)v

4(3.79)

Collision 3.

3 1 2v 1a3

v2a3

C1 C2

U31 U3

1

y 1 x 1

y2

x2

v3s2

Figure 33

The free body diagram shown in Fig.33 allows the following set of equationsto be formulated.

m(vs31 − va31 ) = −U31 (3.80)

m(vs32 − va32 ) = U31 (3.81)

e = −vs31 − vs32va31 − va32

(3.82)

Its solution yields the wanted velocities of separation after the collision 3.

vs31 =

µ1

2(1− e) va31 + (1 + e) va32

¶= (1− e)

µ1

4(1− e) +

1

8(1 + e)2

¶v (3.83)

vs32 =

µ1

2(1 + e) va31 + (1− e) va32

¶= (1 + e)

µ1

4(1− e) +

1

8(1− e)2

¶v (3.84)

PROBLEMS 145

Since the third carriage does not take part in this collision its velocity is

vs33 = vs23 =1

4(1 + e)2v (3.85)

The results shown below was computed for the coefficient of restitution e = 0.5After the first collision the velocities of the individual carriages are:vs11 =

(1−e)2

v = (1−0.5)2

v = 0.25v

vs12 =(1+e)2

v = (1+0.5)2

v = 0.75vvs13 = 0The energy lost due to the first collision∆T = T0−T1 = 1

2mv2− 1

2m (0.25v)2− 1

2m (0.75v)2 = 1

2mv2(1−(0.25)2−(0.75)2)

= 12mv2 (0.375)

After the second collision the velocities of the individual carriages arevs21 = vs11 = 0.25v

vs22 =(1−e2)4

v = (1−0.52)4

v = 0.187 5

vs23 =(1+e)2

4v = (1+0.5)2

4v = 0.562 5

The energy lost due to both collisions is∆T = T0 − T12 =

12mv2(1− (0.25)2 − (0.1875)2 − (0.562 5)2)

= 12mv2 (0.585 94)

After the third collision the velocities of the individual carriages arevs31 = (1− 0.5)

¡14(1− 0.5) + 1

8(1 + 0.5)2

¢v = 0.203 13v

vs32 = (1 + 0.5)¡14(1− 0.5) + 1

8(1− 0.5)2¢ v = 0.234 38v

vs33 =14(1 + 0.5)2v = 0.562 5v

The energy lost due to all three collisions is∆T = T0 − T123 =

12mv2(1− (0.203 13)2 − (0.234 38)2 − (0.562 5)2)

= 12mv2 (0.587 40)

It is necessary to notice that our analysis is true for arbitrarily small gap betweenthe carriage 2 and 3. Therefore it is true in the extreme case of zero gap.

PROBLEMS 146

Problem 56

v

1 3

2

Figure 34

The carriage 1 (see Fig.34) ) moves with a constant velocity v to collide withthe two stationary carriages 2 and 3. The carriages 2 and 3 are constrained by meansof a hinge. Mass of each carriage is identical and it is equal to m. The coefficient ofrestitution at the point of collision is e.

1. Produce the equations for the velocity of separation of individual car-riages and the impulses between the carriages.

2. Solve the equations3. Produce the expression for the energy lost due to the collision for e = 1,

e = 0.5 and e = 0.

PROBLEMS 147

Solution

C1 C2

v2a= v3

a =0v1a=v

1

U1

y 1

x 1

2

U1 U2

y2

x2

3

U2

y 3

x 3

Figure 35

If the carriges 2 and 3 are constrained the collision at the point C1 causes theinpulse U1 at the point of collision and the impulse U2 at the constrain C2. Thereforeone may produce the following three equations reflected the linear impulse linearmomentum principle.

m(vs1 − va1) = −U1m(vs2 − va2) = +U1 − U2 (3.86)

m(vs3 − va3) = +U2

In the above equationsva1 = v va2 = 0 va3 = 0 (3.87)

and due to the constraint at C2

vs2 = vs3 = vs2−3 (3.88)

Introducing the equations 3.87 and 3.88 in equations 3.86 we have

m(vs1 − v) = −U1mvs2−3 = +U1 − U2 (3.89)

mvs2−3 = +U2

The fourth equation is offered by the Newton’s hypothesis

e = −vs1 − vs2va1 − va2

= −vs1 − vs2−3

v(3.90)

Solution of the equations 3.89 and 3.90 yields

vs1 =1− 2e3

v

vs2−3 =1 + e

3v (3.91)

and

U1 =2(1 + e)

3mv

U2 =1 + e

3mv (3.92)

PROBLEMS 148

If the coefficient of restitution is equal to 1vs1 = −13vvs2−3 =

23v

the energy lost is∆T = T a − T s = 1

2mv2 − 1

2m¡13v¢2 − 1

22m

¡23v¢2= 0

If the coefficient of restitution is equal to 0.5vs1 =

1−2e3

v = 0vs2−3 =

1+e3v = 1

2v


2mv2 − 1

22m

¡12v¢2= 1

4mv2

If the coefficient of restitution is equal to 0vs1 =

13v

vs2−3 =13v


2mv2 − 1

23m

¡13v¢2= 1

3mv2

Part II

ANALYSIS.

149

150

INTRODUCTION.

This part provides bases for the analytical solution and analysis of the math-ematical model developed in the previous part. It allows to predict motion of themechanical system if the forces acting on the system are known. If the motion ofthe system can be assumed known, analysis of the mathematical model yields drivingforces which are necessary to maintain the assumed motion. The stability analysisprovides informations about feasibility of the solution obtained.

Chapter 4

ANALYTICAL SOLUTIONS AND THEIR STABILITY.

4.1 ANALYTICAL SOLUTION OF EQUATIONS OF MOTION.

Consideration carried out in the previous part leads to conclusion that motion ofany dynamic system of N degree of freedom, can be described by a set of N ordinarydifferential equations of second order. These equations may be written in the followingmatrix form.

[M ]q = g(q1, q2..qn...qN , q1, q2..qn..qN , t) n = 1, 2.....N (4.1)

Since the equations 4.1 are always linear with respect to the second derivatives of thegeneralize coordinates qn, it follows that

q = [M ]−1g =

⎧⎪⎪⎨⎪⎪⎩ϕ1(q1, q2..qn..qN , q1, q2..qn..qN , t)

.

.ϕN(q1, q2..qn..qN , q1, q2..qn..qN , t)

⎫⎪⎪⎬⎪⎪⎭ (4.2)

Hence, the equation of motion can be always transferred to the form 4.3

qn = ϕn(q1, q2..qn..qN , q1, q2..qn..qN , t) n = 1, 2..N (4.3)

Functions ϕn are nonlinear and because of that, a general solution can not be ef-fectively obtained by means of analytical manipulations. However, same particularsolutions, which represent usually so called steady state motion, may by predicted.Each set of function qn(t) which fulfill the equation 4.3 may be considered as itsparticular solution.

Relatively easy is to obtain particular solution if the equation 4.3 does notcontain time in explicit form. Such a systems are called autonomous systems andcorresponding equations take form.

qn = ϕn(q1, q2..qn..qN , q1, q2..qn..qN) n = 1, 2..N (4.4)

Successful prediction of a particular solution is demonstrated in part 1,chapter IIsection 6. Always, for autonomous systems, it is possible to predict particular solu-tions which presents possible equilibrium positions of the system considered. In thiscase, velocities qn as well as accelerations qn may be assumed to be 0. Hence, theparticular solutions of mathematical model 4.4 can be obtained from the followingset of algebraic nonlinear equations.

ϕn(q1, q2..qn..qN) = 0 n = 1, 2..N (4.5)

STATE - SPACE FORMULATION OF EQUATIONS OF MOTION. 152

Each set of solution qon of equations 4.5 is a particular solution of mathematicalmodel (4.4) and represents an equilibrium position of the system considered.

Obtained in this way analytical solutions can by divided into two categories:stable solutions and unstable solutions. To classify a particular solution to one ofthe two categories it is necessary to consider effect of small perturbations in initialconditions of the particular solution. If the initially small perturbation remain smallfor any instant of time the solution is stable. If the perturbations grows in timethe particular solution is called unstable. It means that only stable solutions arerealizable in practice.

Strict definition of stable solution and unstable solution as well as analysiswhich allows the particular solutions to be classified will be given in the followingchapters.

4.2 STATE - SPACE FORMULATION OF EQUATIONS OF MOTION.

For further consideration is convenient to introduce the following substitutions.

qn = xn (4.6)

qn = xN+n (4.7)

Differentiation of Eq. 4.6 with respect to time yields

qn = xn (4.8)

According to 4.8, the Eq. 4.7 may be rewritten as follow

xn = xN+n (4.9)

Introduction of Eq’s. 4.7 and 4.8 into Eq. 4.3 yields

xN+n = ϕn(x1, x2..xn..xN , xN+1, xN+2..xN+n..x2N , t) (4.10)

The equations 4.9 and 4.10 form one set of differential equation of first order, whichmay be written in the following form.

xk = fk(x1, x2..xk..xK , t) (4.11)

whereK = 2N (4.12)

The coordinates xk are called state - space coordinates and the vector

x = x1, x2..xk..xKT (4.13)

is called state - space vector.The first K

2state - space coordinates represent displacement of the system

along generalized coordinates

xk = qk for k ≤ K

2(4.14)

EQUATIONS OF PERTURBATIONS. 153

The others represent velocities along generalized coordinates

xk = qk−K2

for k >K

2(4.15)

The equations 4.11 represents equations of motion of the system considered in 2N -dimensional Euclidean space called state - space.

4.3 EQUATIONS OF PERTURBATIONS.

Let us assume that

xk k = 1, 2, ..K (4.16)

is a particular solution of the set of equations 4.11

xk = fk(x1, x2..xk..xK, t). (4.17)

The solution 4.16 is called nominal motion or undisturbed motion. All other solutionsare called perturbed motion or disturbed motion.

Let us consider the perturbed motion

xk k = 1, 2, ..K (4.18)

caused by perturbation in initial conditions

∆xok k = 1, 2, ..K (4.19)

The perturbed motion 4.18 always can be determined as follows.

xk = xk +∆xk k = 1, 2, ..K (4.20)

where ∆xk is called perturbation.

∆ kx

okx

∆ okx kx~

kx

t

kx

~

Figure 1

The geometrical interpretation of the introduced motions is given in Fig. 1

DEFINITIONS OF STABILITY IN LAPUNOV’S SENSE. 154

To obtain differential equations for the perturbations, let us introduce thesolution 4.20 into differential equations 4.17.

.˜xk +∆xk = fk(x1 +∆x1, x2 +∆x2, ..xk +∆xk..xK +∆xK , t) (4.21)

Since xk is a particular solution it fulfills the equations of motion 4.17. Hence,

.˜xk= fk(x1, x2..xk..xK , t). (4.22)

Incorporation of Eq. 4.22 into Eq. 4.21 yields

∆xk = −fk(x1, x2..xk..xK , t)+fk(x1 +∆x1, x2 +∆x2, ..xk +∆xk..xK +∆xK , t) (4.23)

The right hand side of equation 4.21 may be developed into Taylor’s series in a vicinityof the nominal motion xk

fk(x1 +∆x1, x2 +∆x2, ..xk +∆xk..xK +∆xK , t)

= fk(x1, x2..xk..xK , t) +

Ãi=KXi=1

∂fk∂xi

(x1, x2, .., xK, t) ·∆xi

!+ h (4.24)

Since the perturbations ∆xi are small, h which contains term of second order andhigher may be neglected. Introduction of Eq. 4.24 into Eq. 4.23 yields equations ofperturbations.

∆xk =i=KXi=1

∂fk∂xi

(x1, x2, .., xK , t) ·∆xi (4.25)

Upon adopting notation

Ak,i =∂fk∂xi

(x1, x2, .., xK, t) (4.26)

the equation of perturbations takes form

∆xk =i=KXi=1

Ak,i ·∆xi k = 1, 2, .....,K (4.27)

From Eq. 4.27 one can see that perturbations are determined by a set of lineardifferential equations of first order.

4.4 DEFINITIONS OF STABILITY IN LAPUNOV’S SENSE.

Let1. xk = fk(x1, x2..xk..xK , t) k = 1, 2, ..K be equations of motion of a mechanicalsystem,2. xk be its nominal solution,3. ∆xk = Ak,i∆xi be equations of perturbations corresponding to the nominal solu-tion xk,


4. k∆xk = p∆x21 +∆x22 + ..+∆x2k + ..+∆x2K be a norm in the K− dimensionalEuclidean space.

DEFINITION: The nominal motion xk is stable in the Lapunov’s sense if forevery ε > 0, there exists a δ > 0, where δ depends on ε and to only, such thatk∆x(to)k ≤ δ implies kx(t)k ≤ ε for t > to.

DEFINITION: The nominal motion xk is asymptotically stable if it is stableand limt→∞ k∆xk = 0DEFINITION: The nominal motion xk is unstable if there exists ε > 0 suchthat for any δ > 0, where δ depends on ε and to only, there exists such atime t > to that k∆x(to)k ≤ δ implies kx(t)k > ε.

The following figures give geometrical interpretation of stable, asymptotically stableand unstable solutions respectively.


tδε

x =q1

2x =q

x (t )∆ 1 o

x (t )∆ 2 o

2|| x(t )||=∆ o x (t )∆ 2 o x (t )∆ 1 o + <δ2

~x x

t>toto

2|| x(t)||=∆ x (t)∆ 2x (t)∆ 1 + <ε2

x (t)∆ 2

x (t)∆ 1

Figure 2

tδε

x =q1

2x =q

x (t )∆ 1 o

x (t )∆ 2 o

2|| x(t )||=∆ o x (t )∆ 2 o x (t )∆ 1 o + <δ2

~x x

t>toto

2|| x(t)||=∆ x (t)∆ 2x (t)∆ 1 + <ε2

x (t)∆ 2

x (t)∆ 1

lim|| x(t)||=0∆t

Figure 3

tδε

x =q1

2x =q

x (t )∆ 1 o

x (t )∆ 2 o

2|| x(t )||=∆ o x (t )∆ 2 o x (t )∆ 1 o + <δ2

~x x

t>toto

2|| x(t)||=∆ x (t)∆ 2x (t)∆ 1 + >ε2

x (t)∆ 2

x (t)∆ 1

Figure 4

CRITERIA OF STABILITY OF EQUILIBRIUM POSITION. 157

4.5 CRITERIA OF STABILITY OF EQUILIBRIUM POSITION.

In a case of autonomous system always exists a particular solution which does notdepends time

qon (4.28)

Such a solution presents equilibrium position of the system considered. Equations ofperturbations for equilibrium position has form 4.25, but the elements Ak,i are timeindependent.

∆xk =KXi=1

Ak,i∆xk k = 1, 2, .....,K (4.29)

General solution of equation of perturbations 4.29 as the linear combination of itsparticular solution is.

∆xk =KXi=1

Ck,ierit (4.30)

where ri are roots of the following characteristic equation¯¯¯A11 − r A12 A13 ..... A1KA21 A22 − r A23 ..... A24..... ..... ..... ..... .......... ..... ..... ..... .....AK1 AK2 AK3 ..... AKK − r

¯¯¯ = 0 (4.31)

The following criteria, which permit any equilibrium position qon to be classified asstable or unstable, are formulated in terms of roots ri.

STATEMENT: If all roots of the characteristic equation 4.31 has negativereal part, the nominal motion qon is asymptotically stable.STATEMENT: If at least one of roots of the characteristic equation 4.31has positive real part, the nominal motion is unstable.

STATEMENT: If one root of the characteristic equation 4.31 has real partequal 0 and all the other have negative real part, the nominal motion isstable.

STATEMENT: If more then one root of the characteristic equation 4.31has real part equal to 0 and all other have negative real part, stability of thenominal motion can not be determined in terms of roots of characteristicequation only.

PROBLEMS. 158

4.6 PROBLEMS.

Problem 57

Motion of a mechanical system of one degree of freedom is governed by thefollowing equation.

q + (1 + cos2 q)q − π2q + q3 = 0 (4.32)

In the above equation q is the generalized coordinate. Find possible equilibriumpositions of the system and determine their stability.

Solution.

STATE - SPACE FORMULATION OF EQUATIONS OF MOTION AND THESYSTEM EQUILIBRIUM POSITIONS.

Let us denote the generalised displacement q by x1 and the generalised velocityq by x2.

q = x1

q = x2 (4.33)

Introduction of the above notations into the equation 4.32 results in the followingstate-space equation of motion of the system considered.

x1 = x2 = f1

x2 = −(1 + cos2 x1)x2 + π2x1 − x31 = f2 (4.34)

For x1 = 0, and x2 = 0 the above equations yields equations which allow the equilib-rium positions of the system to be determined.

0 = x2

0 = −(1 + cos2 x1)x2 + π2x1 − x31 (4.35)

The equation 4.35 has the following solutions.

x11 = 0

x12 = 0 (4.36)

x21 = π

x22 = 0 (4.37)

x31 = −πx32 = 0 (4.38)

ANALYSIS OF STABILITY OF THE EQUILIBRIUM POSITIONS.

PROBLEMS. 159

To formulate the characteristic equation, one has to produce the followingpartial derivatives.

A11 =∂f1∂x1

= 0

A12 =∂f1∂x2

= 1

A21 =∂f2∂x1

= 2x2 cosx1 sinx1 + π2 − 3x21

A22 =∂f2∂x2

= −(1 + cos2 x1) (4.39)

Stability analysis of solution 4.36For x11 = 0 and x

12 = 0 the above partial derivatives have the following magni-

tudesA111 = 0 A112 = 1 A121 = π2 A122 = −2 (4.40)

Therefore the characteristic equation is¯ −r 1π2 −2− r

¯= 0

r2 + 2r − π2 = 0 (4.41)

Its roots are

r1 =−2 +√4 + 4π2

2= −1 +√1 + π2 > 0

r2 =−2−√4 + 4π2

2= −1−√1 + π2 < 0 (4.42)

Since the root r1 > 0 this equilibrium position is unstable.Stability analysis of solution 4.37For x11 = π and x12 = 0 the above partial derivatives have the following mag-

nitudesA111 = 0 A112 = 1 A121 = −2π2 A122 = −2 (4.43)

Therefore the characteristic equation is¯ −r 1−2π2 −2− r

¯= 0

r2 + 2r + 2π2 = 0 (4.44)

Its roots are

r1 =−2 +√4− 8π2

2= −1 +√1− 2π2 = −1 + i

√2π2 − 1

r2 =−2−√4 + 4π2

2= −1−√1 + π2 = −1− i

√2π2 − 1 (4.45)

PROBLEMS. 160

Since both roots have negative real parts this equilibrium position is stable.Stability analysis of solution 4.38For x11 = −π and x12 = 0 the above partial derivatives have the following

magnitudesA111 = 0 A112 = 1 A121 = −2π2 A122 = −2 (4.46)

Therefore the characteristic equation is¯ −r 1−2π2 −2− r

¯= 0

r2 + 2r + 2π2 = 0 (4.47)

Its roots are

r1 =−2 +√4− 8π2

2= −1 +√1− 2π2 = −1 + i

√2π2 − 1

r2 =−2−√4 + 4π2

2= −1−√1 + π2 = −1− i

√2π2 − 1 (4.48)

Since both roots have negative real parts this equilibrium position is stable.

Chapter 5

MODELLING AND ANALYSIS OF A CENTRIFUGE.

5.1 MODELLING

5.1.1 Description of the centrifuge.

1

2

Figure 1

To separate heavier fractions from lighter ones, the fluid containing them is subjectedto high acceleration. The high acceleration is produced by installations called cen-trifuges. The kinematic scheme of a centrifuge is shown in Fig. 1. Container withthe fluid is attached to the whirling arm 2 which is hinged to the rotating with highspeed column 1. Due to centrifugal acceleration the heavier fractions concentrates atthe bottom of the container whereas the lighter ones are still inside the fluid.

MODELLING 162

5.1.2 Physical model

Z z1

ω

β

Az2

y2

G

y1l

1 2

j1 a

- k2 lrG

y1

x1X

α

O

Y

Figure 2

Let XY Z be the inertial system of coordinates. The column 1 is considered as arigid body of moment of inertia I1 about axis Z. System of coordinates x1y1z1 isfixed to the body 1. Its instantaneous position with respect to the inertial system ofcoordinates XY Z is determined by angle α. The column 1 is driven by a DC electricmotor which characteristic can be approximated by the following relationship.

M =Mo +∆Mω (5.1)

M− is the driving momentMo, ∆M− are parametersω = α− is the instantaneous angular velocity of the column.

If all arms are identical and can by considered as rigid bodies, they can bereplaced by one, marked in Fig. 2 by number 2, which possesses a mass m equal tomass of all arms. Its centre of gravity G is chosen as the origin of its body systemof coordinates. The angular position of this body system of coordinates x2y2z2 isdetermined by the angular displacement β. Axes x2y2z2 are principal axis of thebody 2 and moments of inertia, which represent moments of inertia of all arms, arerespectively I2x, I2y, I2z. The system has two degree of freedom. The angles α and βare chosen as the generalized coordinates.

MODELLING 163

5.1.3 Mathematical model.KINEMATICS.


ω1 = k1α (5.2)


ω2 = ω1+ω21 = k1α+ i2β (5.3)


ω2x2 = ω2 · i2 = k1·i2α+ i2·i2β = β

ω2y2 = ω2·j2 = k1·j2α+ i2·j2β = α sinβ

ω2z2 = ω2·k2 = k1·k2α+ i2·k2β = α cosβ (5.4)

Position vector of the centre of gravity of the link is

rG= j1a−k2l (5.5)


rGx2 = rG·i2 = j1·i2a− k2·i2l = 0rGy2 = rG·j2 = j1·j2a− k2·j2l = a cosβ

rGz2 = rG·k2 = j1·k2a− k2·k2l = −a sinβ − l (5.6)

Absolute velocity of the centre of gravity is

vG = rG = r0G +ω2 × rG

= j2(−aβ sinβ) + k2(−aβ cosβ) +¯¯ i2 j2 k2β α sinβ α cosβ0 a cosβ −a sinβ − l

¯¯

= i2(−aα− lα sinβ) + j2lβ + k2(0) (5.7)

Hence, its components along x2y2z2 are

vGx2 = −aα− lα sinβ

vGy2 = lβ

vGz2 = 0 (5.8)

KINETICS.

Motion of the system considered is governed by Lagrange’s equations.

d

dt(∂T

∂α)− ∂T

∂α+

∂V

∂α= Qα

d

dt(∂T

∂β)− ∂T

∂β+

∂V

∂β= Qβ (5.9)

ANALYSIS 164

Kinetic energy function T is assembled of kinetic energy of the link 1 (T1) and kineticenergy of the link 2 (T2).

T = T1 + T2 (5.10)

whereT1 = 0.5I1α

2 (5.11)

T2 = 0.5m(v2Gx2 + v2Gy2 + v2Gz2) + 0.5[ω2x2, ω2y2, ω2z2]

⎡⎣ I2x 0 00 I2y 00 0 I2z

⎤⎦⎡⎣ ω2x2ω2y2ω2z2

⎤⎦= 0.5m(a2α2 + l2α2 sin2 β + 2alα2 sinβ + l2β

2)

+0.5(I2xβ2+ I2y sin

2 β + I2zα2 cos2 β) (5.12)

There is only one conservative force G. The corresponding potential energy functionis

V = mgrGz1 = mgk1·rG = mgk1 · (i2rGx2 + j2rGy2 + k2rGz2)= mgk1·(i20 + j2a cosβ + k2(−a sinβ − l)) = −mgl sinβ (5.13)

The non-conservative driving moment M produces virtual work.

δW = δα M (5.14)

Hence, the generalized forces are

Qα = M

Qβ = 0 (5.15)

Introduction of Eq’s. 5.12, 5.13, 5.15 and 5.1 into Eq. 5.9 yields the following equationof motion.

[(I1 +ma2) + (I2y +ml2) sin2 β + 2mal sinβ + I2z cos2 β]α+

+[2(I2y − I2z +ml2) sinβ cosβ + 2mal cosβ]αβ −Mo +∆Mα = 0

(I2x +ml2)β − (I2y − I2z +ml2)α2 sinβ cosβ −malα2 cosβ +mgl sinβ = 0 (5.16)

5.2 ANALYSIS

5.2.1 Space state formulation of equations of motion.Since the equations of motion do not depend on α, it is easy to lower the order of thefirst equation. Upon introducing

α = ω (5.17)

the equations 5.16 can be rewritten in the following form

[(I1 +ma2) + (I2y +ml2) sin2 β + 2mal sinβ + I2z cos2 β]ω+

ANALYSIS 165

+[2(I2y − I2z +ml2) sinβ cosβ + 2mal cosβ]ωβ −Mo +∆Mω = 0

(I2x +ml2)β − (I2y − I2z +ml2)ω2 sinβ cosβ −malω2 cosβ +mgl sinβ = 0 (5.18)

Introduction of the following notation

β = Ω (5.19)

yields equation of motion in terms of space state coordinates β, ω, and Ω

β = Ω

ω =1

A(BΩω sinβ cos β + CΩω cosβ +Mo −∆Mω)

Ω =1

D(Eω2 sinβ cos β + Fω2 cos β +G sinβ) (5.20)

where

A = (I1 +ma2) + (I2y +ml2) sin2 β + 2mal sinβ + I2z cos2 β

B = −2(I2y − I2z +ml2)

C = −2mal

D = I2x +ml2

E = I2y − I2z +ml2

F = mal

G = −mgl (5.21)

5.2.2 Equilibrium positions.Upon assuming the particular solution of equations 5.20 in the following form

β = βs = constant

ω = ωs = constant (5.22)

one can obtain the following equilibrium conditions

Ωs = 0

Mo −∆Mωs = 0

Eω2s sinβs cosβs + Fω2s cosβs +G sinβs = 0 (5.23)

Solutions of the above equations for different magnitude of ωs are presented in Fig.3 The following set of data was adopted for numerical computation.I1 = 2.5 [kgm

2]m = 8 [kg]I2x = 2 [kgm

2]I2y = 2 [kgm

2]I2z = 0.2 [kgm

2]a = 0.3 [m]

ANALYSIS 166

l = 0.5 [m]

0

5

10

-3 -2 -1 0 1

ωrad/s

β rad

βs2 βs3β s1 βs4

ω s

s

s

Figure 3

β s1 βs2

βs3

βs4

Figure 4

From Fig. 3 one can see that for 0 < ωs < 5.1[rad/s] there exists two equilib-rium position. For ωs > 5.1[rad/s] there exists four possible equilibrium position. Thefour equilibrium positions βs1, βs2, βs3 and βs4 which correspond to ωs = 7[rad/s]are presented in Fig. 4. Not all of them have to be stable. Hence, stability analysisis required to distinguish between stable and unstable equilibrium positions.5.2.3 Equations of perturbation - stability of the equilibrium positions.The already found particular solution forms the nominal motion.

β = βs ω = ωs Ω = Ωs = 0 (5.24)

Equations of motion 5.20 may by rewritten as follow.

β = ϕ1(β, ω,Ω)

ω = ϕ2(β, ω,Ω)

Ω = ϕ3(β, ω,Ω) (5.25)

ANALYSIS 167

where

ϕ1 = Ω

ϕ2 =1

A(BΩω sinβ cos β + CΩω cosβ +Mo −∆Mω)

ϕ3 =1

D(Eω2 sinβ cosβ + Fω2 cosβ +G sinβ) (5.26)

Equation of perturbations has the following form.⎡⎣ ∆β∆ω

∆Ω

⎤⎦ =⎡⎣ A11 A12 A13

A21 A22 A23A31 A32 A33

⎤⎦⎡⎣ ∆β∆ω∆Ω

⎤⎦ (5.27)

where

A11 =∂ϕ1∂β

(βs, ωs,Ωs)= 0

A12 =∂ϕ1∂ω

(βs, ωs,Ωs) = 0

A13 =∂ϕ1∂Ω

(βs, ωs,Ωs) = 1

A21 =∂ϕ2∂β

(βs, ωs,Ωs)

=(2(I2y +ml2) sinβs cosβs + 2mal cosβs − 2I2z sinβs cosβs)(Mo −∆Mωs)

A(βs)2

Since according to Eq. 5.23 Mo −∆Mωs = 0

A21 = 0

A22 =∂ϕ2∂ω

(βs, ωs,Ωs) =−∆M

A(βs)

A23 =∂ϕ2∂Ω

(βs, ωs,Ωs) =B sinβs cosβsωs + C cosβsωs

A(βs)

A31 =∂ϕ3∂β

(βs, ωs,Ωs) =Eω2s cos

2 βs −Eω2s sin2 βs − Fω2s sinβs +G cosβsD

A32 =∂ϕ3∂ω

(βs, ωs,Ωs) =2Eωs sinβs cosβs + 2Fωs cosβs

D

A33 =∂ϕ3∂Ω

(βs, ωs,Ωs) = 0 (5.28)

The characteristic equation is¯¯ A11 − r A12 A13

A21 A22 − r A23A31 A32 A33 − r

¯¯ = 0 (5.29)

If all roots of the above equation have negative real parts, solution βs, ωs,Ωs is con-sidered as stable. Application of this criterion to the system considered leads toconclusion that only solution shown in Fig. 5 are stable.

ANALYSIS 168

0

5

10

-3 -2 -1 0 1

ωrad/s

β rad

βs2 βs4

ω s

s

s

Figure 5

For ωs = 7[s−1] there exists two stable equilibrium positions. They are ωs =

7[s−1], βs2 = −1.25[rad] and ωs = 7[s−1], βs4 = 1.399[rad]. Roots of the equation5.29 corresponding to these equilibrium positions were computed to be

r21 = −0.1352 + i5.733 r22 = −0.1352− i5.733 r23 = −1.2765 + i0

r41 = −0.0200 + i7.811 r42 = −0.0200− i7.881 r43 = −0.7215 + i0

(5.30)

respectively. Their real parts proves stability of the equilibrium positions. Their imag-inary conjugated parts represents frequencies of the decaying oscillations in vicinityof the equilibrium positions ( see Fig. 6 and 7).

Which stable position the system finally obtains depend on initial conditions.Numerical simulation carried out forMo = 50[Nm] and∆M =Mo/7[Nms], indicatesthat for initial conditions

β(0) = 0 ω(0) = 0 Ω(0) = −2.5 (5.31)

the system tends to equilibrium position βs2 = −1.25[rad] (see Fig. 6). But, forinitial conditions

β(0) = 0 ω(0) = 0 Ω(0) = 0 (5.32)

the system tends to position βs4 = 1.399[rad] (see Fig. 7).

ANALYSIS 169

ω

β rad

rad/s

0 10 20 30 40time s

-2

0

2

4

6

8

Figure 6

0

2

4

6

8

0 10 20 30 40

ω

β rad

rad/s

time s

Figure 7

Part III

EXPERIMENTALINVESTIGATIONS.

170

171

INTRODUCTION.

The results obtained by modeling and analyzing a dynamic problem corre-spond strictly to the physical model we have created. Assumptions adopted for thephysical model always simplify the problem considered. Hence, the obtained results(e.g.. dynamic behavior of the physical model) are different then a true dynamicbehavior of the real object. Experimental investigation allows to assess these differ-ences. If they are negligible from practical point of view, one can adopt the developedmathematical model as well as the methods used for its solution for further design.If experimental investigation shows unacceptable differences the physical model ormethods of its analysis must be improved.

If the object is still on design stage, experimental investigation can be car-ried out on especially design models supplied with measuring equipment (laboratoryinstallations). Models for experimental investigation should possesses all features ofthe real object which has an qualitative influence on the problem considered. Resultsof the experimental investigation can not be transfer directly to the real object, butform a base for verification of mathematical model and methods used for its analysis.After verification the mathematical model can be used for qualitative description ofthe problem existing in the real object.

In some cases it is possible to design model which reflect the dynamic prob-lem both qualitatively and quantitatively. Theory of dynamic similarity permits thescaling factors between the real object and the model to be calculated. Upon usingthese scaling factors, experimental results can be transfer to the real object withoutany mathematical model.

The following chapters provide examples of modeling analysis and experimen-tal investigation of two dynamic problems.

Chapter 6

INVESTIGATION OF THE EFFECT OF A GYROSTABILIZERON MOTION OF A SHIP.

6.1 DESCRIPTION OF THE GYROSTABILIZER.

Y

X

Zz

x

y1

1

1

x 2

y2

z 2 α

β

OΩ

1

23

Figure 1

Housing 2 of the gyroscope 3 is free to rotate about axis x1 as shown in Fig. 1. Thegyroscope 3, driven by a motor, rotates with respect to the housing about axis z2.Such device attenuates greatly the rolling motion of the ship 1. caused by impacts ofwaves and blows of wind. Fig. 2 shows the stabilizer installed on the board of a ship.

MODELLING. 173

Figure 2

6.2 MODELLING.

6.2.1 Physical model.The physical model of the system considered is shown in Fig. 3. In this figure XY Zare axis of the inertial system of coordinates. System of coordinates x1y1z1 is rigidlyattached to the ship whereas x2y2z2 makes body 2 ( housing) system of coordinates.Axis x2y2z2 are principal axis of the body 2 through centre of rotation O.

Z z

X

x

1

1

O

H1 a1

123

G3G2

a2

a3

z1

y1

βz2

y2

o

Ω

α

h

G1

G1

M = G hr 1-

Mb

Figure 3

The ship 1 is considered as a rigid body which is free to rotate about axis Y .Its angular position is determined by the angle α. All linear displacements as well as

MODELLING. 174

pitching and yawing of the ship are neglected. Moment of inertia of the ship aboutits axis of rotation will be denoted by I1y and its centre of gravity is determined bythe distance a1. Due to the gravity force G1 and the buoyancy force H1 there existsthe righting moment which in this physical model is denoted byMr.

Housing 2 can to rotate about axis x2 and the angle β uniquely determinesits relative angular position with respect to the ship. Its principal moments of inertiaalong axes x2y2z2 will be denoted by I2x, I2y, I2z respectively. Centre of gravity G2 isdetermined by distance a2 and its mass will be denoted by m2. Between the housingand the ship there is a brake installed. It produces the braking moment about theaxis of relative rotation x2. This braking moment is denoted in Fig. 3byMb.

The gyroscope 3 rotates with the constant angular speed Ω about axis z2relatively to its housing. Axis z2 is an axis of symmetry of the gyroscope. Theprincipal moments of inertia of the gyroscope are denoted by I3 = I3x = I3y, I3zand its mass by m3. The centre of gravity of the gyroscope G3 is determined by thedistance a3

If one adopts the above assumptions, the system can be considered as systemof two degree of freedom. The two generalized coordinates are denoted by α and β.

6.2.2 Mathematical model.Kinematics.

Angular velocity of the ship 1. along body 1 system of coordinates x1y1z1.

ω1 = j1α (6.1)

Angular velocity of the housing 2 along body 2 system of coordinates x2y2z2

ω2 = ω1 +ω21 = j1α+ i2β = i2β + j2α cosβ − k2α sinβ (6.2)

Angular velocity of the gyroscope 3 along body 2 system of coordinates x2y2z2.

ω3 = ω2 +ω32 = i2β + j2α cosβ + k2(−α sinβ + Ω) (6.3)

Velocity of the centre of gravity G1 along system of coordinates x1y1z1 may be ob-tained by differentiation of its position vector rG1 = k1(−a1).

vG1 = ω1 × k1(−a1) =¯¯ i1 j1 k10 α 00 0 −a1

¯¯ = −i1αa1 (6.4)

Velocity of the centre of gravity G2 along system of coordinates z2y2z2 may be ob-tained by differentiation of its position vector rG2 = k2(−a2).

vG2 = ω2 × k2(−a2) =¯¯ i2 j2 k2β α cosβ −α sinβ0 0 −a2

¯¯ = −i2a2α cosβ + j2a2β (6.5)

MODELLING. 175

Equations of motion of the gyroscope (body 3).

3

G3a3

z1

y1

βy2

Ωz2

o2

G3

x2

Ω

z2

o1

M32x2R 32x2

M32z2

R 32z2

R 32z2

M32z2M32y2

R 32y2

Figure 4

The free body diagram of the gyroscope 3 is shown in Fig. 4. The origin o2 ischosen as the point of reduction of the interaction forces (M32 and R32) betweenthe gyroscope and its housing. The component M32z2 stands for the driving momentthat has to apply to the gyroscope to keep it going with the assumed velocity Ω. Thegyroscope performs rotational motion about point O and its axis of relative rotationis axis of symmetry. Hence, the modified Euler’s equations may be used.

I3ω2x + (I3z − I3)ω2yω2z + I3zω2yΩ = M32x2 −m3a3g cosα sinβ (6.6)

I3ω2y − (I3z − I3)ω2xω2z − I3zω2xΩ = M32y2 −m3a3g sinα (6.7)

I3z(ω2z + Ω) = M32z2 (6.8)

Introduction of Eq. 6.2 into Eqs. 6.6, 6.7 and 6.8 yields.

I3β + (I3z − I3)(−α2 sinβ cosβ) + I3zΩα cos β =

=M32x2 −m3a3g cosα sinβ (6.9)

I3(α cosβ − αβ sinβ)− (I3z − I3)(−αβsinβ)− I3zΩβ =M32y2 −m3a3g sinα (6.10)

I3z(−α sinβ − αβ cos β) =M32z2 (6.11)

MODELLING. 176

Equations of motion of the housing (body 2).

2

G2

a2

z1

y1

β

z2y2

o

Mb R 23z2

M23z2M23y2

R 23y2

R21z2

M21z2M21y2

R 21y2

Figure 5

The free body diagram of the housing is shown in Fig. 5. The housing is interactingwith both the gyroscope 3 and the ship body 1. M23 and R23 represent the momentand force due to interaction with the gyroscope whereas M21 and R21 stand for themoment and force due to interaction with the ship. Mb represents the breakingmoment produced by a break installed between the body of ship and the housing.The housing performs rotational motion about the point O. Hence Euler equationsmay be used.

I2xω2x + (I2z − I2y)ω2yω2z = M23x2 −m2a2g cosα sinβ +Mb (6.12)

I2yω2y + (I2x − I2z)ω2xω2z = M23y2 +M21y2 −m2a2g sinα (6.13)

I2zω2z + (I2y − I2x)ω2xω2y = M23z2 +M21z2 (6.14)

Introducing Eq. 6.2 into Eqs. 6.12, 6.13 and 6.14 one can obtain

I2xβ + (I2z − I2y)(−α2 sinβ cosβ) =

=M23x2 −m2a2g sinβ +Mb (6.15)

I2y(α cosβ − αβ sinβ) + (I2x − I2z)(−αβ sinβ) ==M23y2 +M21y2 −m2a2g sinα (6.16)

I2z(−α sinβ − αβ cosβ) + (I2y − I2x)αβ cosβ =M23z2 +M21z2 (6.17)

MODELLING. 177

Equations of motion of the ship (body 1).

Z z

X

x

1

1

O

H1 a1

1

α

h

G1

G1

M = G hr 1-

2

z1

y1

βz2

y2

o

M12z212y2M

M rY

Figure 6

The ship performs rotational motion about axis y1, which is fixed in the inertial space.Hence its motion may be described by Newton’s generalized equation.

I1yα =M12y1 +Mr (6.18)

Since according to Fig. 6

M12y1 =M12y2 cosβ −M12z2 sinβ (6.19)

the equation (6.18) can be rewritten in the following form.

I1yα = (M12y2 cosβ −M12z2 sinβ) +Mr (6.20)

Equations of motion.

Taking into consideration, that

M23x2 = −M32x2 M23y2 = −M32y2 M23z2 = −M32z2

M12y2 = −M21y2 M12z2 = −M21z2 (6.21)

the derived equations may be rewritten in the following form.

I3β + (I3z − I3)(−α2sinβcosβ) + I3zΩα cos β = M32x2 −m3a3g cosα sinβ

I3(α cosβ − αβ sinβ)− (I3z − I3)(−αβsinβ)− I3zΩβ = M32y2 −m3a3 sinα

I3z(−α sinβ − αβ cosβ) = M32z2

ANALYSIS. 178

I2xβ + (I2z − I2y)(−α2 sinβ cos β) = −M32x2 −m2a2g sinβ +Mb

I2y(α cosβ − αβ sinβ) + (I2x − I2z)(−αβ sinβ) = −M32y2 −M12y2 −m2a2g sinα

I2z(−α sinβ − αβ cosβ) + (I2y − I2x)αβ cos β = −M32z2 −M12z2

I1yα = (M12y2 cosβ −M12z2 sinβ) +Mr (6.22)

The above 7 equations comprises 7 unknown: α, β,M12y2,M12z2,M32x2,M32y2,M32z2.Upon eliminating the five unknown components of reaction moments, one may obtaintwo differential equations which govern motion of the system considered.

[I1y + (I2y + I3) cos2 β + (I2z + I3z) sin

2 β]α+ (I2z − I2y + I3z − I3)αβ sin 2β

−I3zΩβ cosβ + (m2a2 +m3a3)g cosβ sinα−Mr = 0

(I2x + I3)β +1

2(I3 − I3z + I2y − I2z)α

2 sin 2β + I3zΩα cosβ

+(m2a2 +m3a3)g cosα sinβ −Mb = 0 (6.23)

6.3 ANALYSIS.

For analysis of the mathematical model derived in the previous section, it is assumedthat the righting moment may by approximated by function

Mr = −Mro sinα (6.24)

and the breaking moment is considered as a linear function of the angular displace-ment β and its first derivative β

Mb = −kβ − cβ (6.25)

Introducing the above relationships into equations 6.23, the equations of motion takethe following form


2 β]α+ (I2z − I2y + I3z − I3)αβ sin 2β

−I3zΩβ cosβ + (m2a2 +m3a3)g cosβ sinα+Mro sinα = 0

(I2x + I3)β +1

2(I3 − I3z + I2y − I2z)α

2 sin 2β + I3zΩα cosβ

+(m2a2 +m3a3)g cosα sinβ + kβ + cβ = 0 (6.26)

The above equations has been solved by means of Runge Kutta method for numericaldata corresponding to a liner of length 110m. and total mass 5.1 × 106kg. Figure7 presents proportions between the ship and gyroscope. To see the influence of thegyroscope on the dynamic behavior of the ship let us compute motion of the shipcaused by an impact of wave for rotating and stationary gyroscope. Let us assumethat the impact initiates motion of the previously motionless ship with initial angularvelocity αo = 0.1rad/s. Hence, the initial conditions may be chosen as follow.

αo = 0 αo = .1 βo = 0 βo = 0 (6.27)

ANALYSIS. 179

4m15m

0.4m

110m

Figure 7 Parameters of the ship and gyroscope., I3 = 0.1 × 106kgm2, I3z = .05 ×106kgm2, I2x = 0.2 × 106kgm2, I2y = 0.1 × 106kgm2, I2z = 0.2 × 106kgm2, I1y =198.5× 106kgm2, m1 = 5.1× 106kg, Mro = 9.× 106Nm, m2a2+m3a3 = 1.× 103Nm,k = 1.× 104Nm/rad, c = 1.× 106Nms/rad, Ω = 300 rad/s

0

0.5

-0.5

-1.0

-1.50 50 100 150 200

time [s]

βα &[rad]

α β

Figure 8 Motion for the stationary gyroscope.

0

0.5

-0.5

-1.0

-1.50 50 100 150 200

time [s]

βα &[rad]

α β

Figure 9 Motion for the rotating gyroscope.

EXPERIMENTAL INVESTIGATIONS. 180

Solution of equations with the above initial conditions for stationary gyroscope (Ω =0) is presented in Fig. 8. The ship, in this case, performs an oscillatory motion withperiod T ≈ 30s whereas the housing is motionless with respect to the ship.

Solution of the equations 6.26 with the same initial conditions for the gyroscoperotating with angular velocity Ω = 300rad/s is presented in Fig. 9 In this casethe energy of the impact is adopted not only by the ship, but by the housing withgyroscope as well. Due to large relative motion of the housing the energy is dissipatedby the brake. It results in three times smaller swing of the ship and fast return to itsequilibrium position.

6.4 EXPERIMENTAL INVESTIGATIONS.

Results of the above analysis are more reliable if they are supported by an experimen-tal investigation which usually is aimed to assess assumptions done during creation ofphysical model, evaluate mathematical methods adopted for solution of mathematicalmodel as well as to make sure that there is no mistakes in derivation of mathematicalmodel or programming.

The experimental investigation can be curried out on real objects (if theyexist and are available for experiment), dynamically similar models (if feasible), oron especially design laboratory installations which contains all important features ofthe real system but only qualitatively reflect its dynamic properties.

This section presents the last approach to experimental investigations aimedto verify mathematical model 6.23.

6.4.1 Description of the laboratory installationSchematic of the laboratory installation is shown in Fig. 10.

S

H

B1011768

A123A514

Figure 10


Frame 1, which represents the ship, is free to rotate about the horizontal axiswith respect to the motionless base 4. Housing 2 of the gyroscope 3 is free to rotatewith respect to the frame 1. the hydraulic damper 8 produces a damping momentwhich depend on relative motion of the housing with respect to the frame 1. Thegyroscope 3 is driven by electric motor 5 with a constant speed. Protractors 6 and7 allows to measure the angular position of the ship α and housing β. They are usedfor scaling of transducers A and B. The transducers A and B produce the analogsignals which are proportional to the angular displacements α and β. These signals,in a digitized form, are send to a memory of the computer 11.

The laboratory installation was design to fulfil some of the adopted assump-tions. Similarly to the physical model the body which represents the ship may onlyrotates about one axis Y. Of course, those assumption can not be evaluated by meansof the following experimental investigation.

6.4.2 Mathematical model of the laboratory installation.In the case of laboratory installation the righting momentMr is produced by gravityforces and friction which is here approximated by a linear damping with coefficientcα. Hence

Mr = −m1a1gsinα− cαα (6.28)

Assume that the real friction produced by the break is a linear friction with coefficientcβ.

Mb = −cββ (6.29)

Introduction of Eqs.6.28 and 6.29 into Eq. 6.23 yields


2 β]α+ (I2z − I2y + I3z − I3)αβ sin 2β+

−I3zΩβ cosβ + (m2a2 +m3a3)g cosβ sinα+m1a1g sinα+ cαα = 0

(I2x + I3)β +1

2(I3 − I3z + I2y − I2z)α

2 sin 2β + I3zΩα cosβ+

+(m2a2 +m3a3)g cosα sinβ + cββ = 0 (6.30)

The mathematical model above has the following parameters:m2a2 +m3a3− static moment of the housing and gyroscopem1a1− static moment of the shipI1y− moment of inertia of the ship about axis y1I2x− moment of inertia of the housing about axis x2I2y− moment of inertia of the housing about axis y2I2z− moment of inertia of the housing about axis z2I3− moment of inertia of the gyroscope about axis x2 and y2I3z− moment of inertia of the gyroscope about axis z2cα− damping coefficient corresponding to velocity αcβ− damping coefficient corresponding to velocity βΩ− angular speed of the gyroscopeAll of the above parameters have to be identified.


6.4.3 Identification of parameters.Identification of static moments (m2a2 +m3a3 and m1a1).

Static moment m2a2 + m3a3 may be obtained by measuring angle βH caused by aweight of the known mass mH suspended from the point H. (see Fig. 10).

m3 gm2 g

m g

H

a2 a3

β H

O

l

G2

G3

32

H

H

Figure 11

From Fig. 11 one can see that

(m2a2 +m3a3)g sinβH = mHglH cosβH (6.31)

Hencem2a2 +m3a3 = mHlH cotβH (6.32)

In a similar manner, upon suspending a weight mS at point S, (see Fig. 10), one canidentify static moment m1a1 +m2a2 +m3a3

m1a1 +m2a2 +m3a3 = mSlS cotαS (6.33)

Hencem1a1 = mSlS cotαS −mH lH cotβH (6.34)

Identification of damping coefficients and moments of inertia of the wholeassembly.

Estimation by means of the linear theory of vibrations.

Damping coefficient and moment of inertia of the whole assembly (I1y+I2y+I3)about axis y1 can be obtained by analysis of the free motion of the whole assembly


about axis y1 for stationary gyroscope and motionless housing with respect to theship (Ω = 0, β = 0, β = 0, β = 0).

According to Eq. 6.30 the free motion of the ship is governed by the followingequation

(I1y + I2y + I3)α+ (m1a1 +m2a2 +m3a3)g sinα+ cαα = 0 (6.35)

The estimation of moment of inertia I1y + I2y + I3 as well as damping coefficient cαmay be obtained with help of the linearized equation 6.35. It is as follows.

Iα+ kα+ cαα = 0 (6.36)

whereIα = I1y + I2y + I3kα = (m1a1 +m2a2 +m3a3)g

or

α+ ω2oα+ 2hα = 0 (6.37)

whereωα =

qkαIα

2hα =cαIα

The moment of inertia Iα may be assessed from relationship between the periodof free oscillation Tα and the natural frequency ωα.

Tα =2π

ωα= 2π

rIαkα

(6.38)

The period Tα can be obtained from the recorded free motion of the ship. An exampleof the recorded free motion of the ship αex(t) is presented in Fig. 12.

Tt

t

t+T

α

α (t) α (t+T )ex

ex

ex

α

α

α

Figure 12

Since kα has been already identified the formula 6.38 yields approximate mag-nitude of the moment of inertia Iα.

Iα =T 2αkα4π2

=T 2α(m1a1 +m2a2 +m3a3)g

4π2(6.39)


The damping coefficient may be assessed from the following relationship.

ln

µαex(t)

αex(t+ Tα)

¶= Tαhα = Tα

cα2Iα

(6.40)

Hence,

cα = ln

µαex(t)

αex(t+ Tα)

¶2IαTα

(6.41)

Damping coefficient cβ and moment of inertia Iβ = I2x + I3 of the gyroscopeand housing about axis x1 can be obtained by means of analysis of the free motion ofthe gyroscope and housing about axis x1 for the stationary gyroscope and motionlessship 1 (see Fig. 10) (Ω = 0, α = 0, α = 0, α = 0). In the same manner as previouslyone can obtained the following formulae for estimation of parameters Iβ = I2x + I3and cβ.

Iβ =T 2βkβ

4π2=

T 2β (m2a2 +m3a3)g

4π2(6.42)

cβ = ln

µαex(t)

αex(t+ Tβ)

¶2IβTβ

(6.43)

Magnitude of these parameters can be improved by fitting solution of the non-linearmodel 6.35 into the recorded motion by means of the curve fitting technique. Com-puted in this way parameters may be considered as the ’identified’ ones.

The curve fitting technique.

Very often, to identify parameters of a mathematical model, the curve fittingtechnique is used. To explain it, let us assume that xex(t) is the recorded motion of asystem. This motion is approximated by solution of x(t) of the following mathematicalmodel.

ϕ(x, x, x, t, a, b) = 0 (6.44)

where a and b are parameters to be identify.The recorded motion always allows to assess the initial conditions.

xo = xex(0)

xo =xex(∆t)− xex(0)

∆t(6.45)

where ∆t is the sampling interval.At this stage one can solve equation 6.44 for estimated magnitudes of para-

meters a, b and initial conditions 6.45. Let x(t) be this solution.Accuracy of approximation of the recorded motion xex(t) by the solution x(t)

may by determined by error function as follow.

ER =

Ãn=NXn=1

(xex(tn)− x(tn))2

!1/2(6.46)


Now, upon varying the parameters a and b one can minimize the error ER. Parame-ters ai and bi for which error has the smallest magnitude can be considered as theidentified ones.

Parameters I3 and I3z may be assessed with sufficient accuracy analyticallyand parameter I2y may be assumed as equal to I2x. Therefore

I2x = Iβ − I3

I2y = I2x

I1y = Iα − I2y − I3 (6.47)

The gyroscope is driven by one phase AC electrical motor, hence its angular speedmay by assumed to be 3000RPM .

6.4.4 Verification of the mathematical model.Having all parameters of the mathematical model identified, one may verify it. Thisverification can be obtained by comparison of the recorded motion of the laboratoryinstallation with solution of its mathematical model. The necessary initial conditionshave to coincide initial condition of the recorded motion.

Chapter 7

INVESTIGATION OF AN INDICATOR OF ANGULARVELOCITY.

7.1 DESCRIPTION OF THE INDICATOR.

β

α

Ω

Y

X

Z,z 1

1x

1 y

2 y

2z

ω 1

1 2 3 4

Figure 1

It is easy to measure the absolute angular velocity of an object if it performs itsrotational motion in an immediate neighborhood of another which can be consideredas motionless. For example, relative angular velocity of a rotor measured with respectto its housing may be considered as absolute one if the housing is founded on theEarth. Much more difficult is to measure or control the absolute velocity of an objectwhich performs its motion far away from the Earth (planes, ships, floating platforms,etc.) or are installed on vehicle performing its own rotational motion (a radar antennainstalled on a track). In such cases the principle of work of indicators of the absoluteangular velocity are based on gyroscopic effect.

Fig. 1. presents an indicator which may be used to measure and control

MODELLING 187

absolute angular velocity of a floating platform.The gyroscope 3 is driven by the motor 4 which is fixed to the housing 4 of

the gyroscope. Since the housing is free to rotate about the horizontal axis x1, its axisz2, due to gravity forces, is vertical if the platform 1 is motionless. If the platformrotates, the housing changes its angular position β.

The following analysis is aimed to establish a relationship between the an-gular velocity of the platform and angular position of the housing. Experimentalinvestigations allows to verify the derived mathematical model.

7.2 MODELLING

7.2.1 Physical model.

X x Y

y1

1

Oαβ

1zZ

y1O

ω1

β

2G3G

2

l3

l

y2z2

2

3

Figure 2

The physical model of the described indicator is shown in Fig. 2. Axes XY Z formsthe inertial system of coordinates. The system of coordinates x1y1z1 is rigidly con-nected to the platform 1 and rotates with the constant angular velocity ω1. Housing,with all attached accessories, form the rigid body 2, and x2y2z2 is the body 2 sys-tem of coordinates. The centre of gravity G2 of the body 2 is located at distance l2from the centre of rotation C. Its angular position with respect to the platform 1 isdetermined by an angle β.

The gyroscope 3, together with rotor of the motor 4, is regarded as rigid body3. Axis z2 is its axis of symmetry and the dimension l3 locates its centre of gravity G3.The body 3 rotates with the constant angular rate Ω about axis z2. Axes x2y2z2 are

MODELLING 188

principal axes of body 2 and 3. Moments of inertia of the two bodies are I2x2, I2y2, I2z2and I3x2 = I3y2 = I3, I3z2 respectively.

Between the housing 2 and platform 1 there exists damping which can beapproximated as viscous damping Md = −cβ.7.2.2 Mathematical model.Equations of motion of the gyroscope.

X x

Y

y1

1

Oα

1zZ

y1

β

3G

l3

y2z2

O

x2

M32x2

M32y2M32z2

m g3

Figure 3

The gyroscope 3, whose the free body diagram is shown in Fig. 3, performs rotationalmotion about the origin O, and axis z2 is its axis of symmetry. Hence, the modifiedEuler equations may be used to describe its motion. The gyroscope 3 rotates with theangular velocity Ω with respect to system of coordinates x2y2z2 whereas the systemof coordinates has its own angular velocity ω2.

ω2 = ω1 +ω21 = k1ω1 + i2β (7.1)

Its components along the system coordinates x2y2z2 are

ω2x2 = i2 · ω2 = i2 · (k1ω1 + i2β) = β

ω2y2 = j2 · ω2 = j2 · (k1ω1 + i2β) = ω1 sinβ

ω2z2 = k2 · ω2 = k2 · (k1ω1 + i2β) = ω1 cosβ (7.2)

MODELLING 189

Introduction of equations 7.2 into the modified Euler equations

Iωx + (Iz − I)ωyωz + IzΩωy = Mx

Iωy − (Iz − I)ωxωz − IzΩωx = My

Iz(ωz + Ω) = Mz (7.3)

yields

I3β + (I3z2 − I3)ω21 sinβ cosβ + I3z2Ωω1 sinβ = M32x2 −m3gl3 sinβ

I3(ω1 sinβ + ω1β cosβ)− (I3z2 − I3)ω1β cosβ − I3z2Ωβ = M32y2

I3z2(ω1 cosβ − ω1β sinβ + Ω) = M32z2 (7.4)

whereM32x2, M32y2, M32z2 are components of the resultant moment of reactions of the body2 on the gyroscope 3.

Equations of motion of the housing 2.

1zZ

y1O

β

2G

2l

y2z2

X x

Y

y1

1

Oα

x2

M23x2

M23z2

gm2

M21z2M21y2

M23y2

M dM21x2 =

Figure 4

ANALYSIS 190

The housing 2 whose the physical model is shown in Fig. 4 performs rotationalmotion about point O. Hence Euler unmodified equations may be used.

Ixωx + (Iz − Iy)ωyωz = Mx

Iyωy + (Ix − Iz)ωxωz = My

Izωz + (Iy − Ix)ωyωx = Mz (7.5)

Upon introducing the relationship 7.2, the equations of motion takes the followingform.

I2x2β + (I2z2 − I2y2)ω21 sinβ cosβ = M23x2 −m2gl2 sinβ − cβ

I2y2(ω1 sinβ + ω1β cosβ) + (I2x2 − I2z2)ω1β cosβ = M23y2 +M21y2

I2z2(ω1 cosβ − ω1β sinβ) + (I2y2 − I2x2)ω1β sinβ = M23z2 +M21z2 (7.6)

Equations of motion of the system.

Taking into consideration that

M23x2 = −M32x2 (7.7)

the first of equation 7.4 and first of equation 7.6 yield the following equation of motion.

Ixβ + (Iz − Iy)ω21 sinβ cos β + I3z2Ωω1 sinβ +MG sinβ + cβ = 0 (7.8)

whereIx = I2x2 + I3 – moment of inertia of the whole assembly about axis x2Iy = I2y2 + I3 – moment of inertia of the whole assembly about axis y2Iz = I2z2 + I3z2 – moment of inertia of the whole assembly about axis z2MG = m2gl2 +m3gl3 – static moment with respect to point of rotation O.

7.3 ANALYSIS

7.3.1 Particular solutions. (equilibrium positions).Let us find a particular solution which can be predicted in the following form.

β = βo = constant (7.9)

Introduction of Eq. 7.9 into Eq. 7.8) yields.

((Iz − Iy)ω21 cosβo + I3z2Ωω1 +MG) sinβo = 0 (7.10)

Hence,

βo = arccos

µ−I3z2Ωω1 +MG

(Iz − Iy)ω21

¶, βo = 0 and βo = π (7.11)

For the following set of numerical data

Ix = 0.5kgm2 Iy = 0.6kgm2 Iz = 0.25kgm

2 I3z2 = 0.01kgm2

Ω = 150.rad/s MG = 5.Nm c = 0.2 Nms2/rad.

ANALYSIS 191

the particular solutions 7.11, representing possible equilibrium positions, are pre-sented in Fig. 5. as a function of angular speed ω1.

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]

ω 1 [rad/s]

Figure 5

Not all these solution are stable. To assess stability of these solutions let usanalyze motion of the system in vicinity of an equilibrium position βo.7.3.2 Stability analysis.Motion in vicinity of the equilibrium position βo always can be assumed in the fol-lowing form.

β = βo +∆β (7.12)

where ∆β ,called perturbation, is an unknown function of time. To determine thisfunction let us introduce 7.12 into the equation of motion 7.8.

Ix∆β + (Iz − Iy)ω21 sin(βo +∆β) cos(βo +∆β) + I3z2Ωω1 sin(βo +∆β)

+MG sin(βo +∆β) + c∆β = 0 (7.13)

Since the perturbations ∆β can be treated as a small magnitudes, it is possible tolinearize the equation 7.13 in vicinity of the equilibrium position βo.

(Iz − Iy)ω21 sin(βo) cos(βo) + I3z2Ωω1 sin(βo) +MG sin(βo)

+Ix∆β + ((Iz − Iy)ω21 cos 2βo + I3z2Ωω1 cosβo +MG cosβo)∆β + c∆β = 0 (7.14)

According to 7.10 the first three terms are equal to zero. Therefore the equation ofperturbations takes the following form.

Ix∆β + ((Iz − Iy)ω21 cos 2βo + I3z2Ωω1 cosβo +MG cos βo)∆β + c∆β = 0 (7.15)

or∆β +B∆β + C∆β = 0 (7.16)

whereC =

1

Ix((Iz − Iy)ω

21 cos 2βo + I3z2Ωω1 cosβo +MG cosβo) (7.17)

ANALYSIS 192

B =c

Ix(7.18)

Solution of the equation of perturbations depends on roots of the characteristic equa-tion associated with the equation 7.16.

r2 +Br + C = 0 (7.19)

Roots of the characteristic equation of equation are

r1,2 =−B ±√B2 − 4C

2(7.20)

Hence, the solution of equation 7.16 is as follows.

∆β = C1er1t + C2e

r2t (7.21)

It can be seen from Eq. 7.21 that if the real part of both roots is negative, theperturbations decay to zero and the equilibrium considered is stable. Since Ix and care always positive, roots 7.20 have negative real parts if and only if

C > 0 (7.22)

Hence, only those solutions 7.11 can be considered as stable which additionally fulfillstability condition 7.22.

7.3.3 Stable and unstable equilibrium positions.In Figures 6 to 11 the equilibrium positions which fulfil the condition 7.22 (the stablesolutions) are marked by solid line. The broken line corresponds to the unstableequilibrium positions. Stable and unstable equilibrium positions for Iy > Iz areshown in Fig. 6 to Fig. 8 for Ω = 0 Ω = 150 and Ω = 300 rad/s respectively.

Different behavior may be predicted in case when Iy < Iz. From solutionspresented in Fig. 9to Fig. 11 one can see that the only possible equilibrium positions(solid line) are βo = 0, βo = π or both of them. It means, that for certain magnitudeof angular velocity of the platform (e.g.. ω1 = −10, Fig. 9.) the housing consideredmay perform oscillations about equilibrium position βo = 0, as well as βo = π. Thisoscillatory motion depends on initial conditions. Fig. 12 presents solution of equationof motion 7.8 for two sets of initial conditions very close to each other. One of them(βin = 1, βin = 0) corresponds to the point A shown in Fig. 10. The other (βin = 1.1,βin = 0) corresponds to the point B in the same figure. Despite of the fact that theabove initial conditions differs very little, the corresponding solutions (see Fig. 12)are totally different. One of them tends to equilibrium position βo = 0, whereas theother tends to equilibrium position βo = π. In the case considered, the unstablesolution (broken line) divides all possible initial positions βo corresponding to initialvelocity βin = 0 into two categories. One of them produces solutions which tendto equilibrium position βo = 0. Solutions belonging to the other category of initialconditions tend to equilibrium βo = π.

ANALYSIS 193

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]

ω 1 [rad/s]stable equilibrium positions unstable equilibrium positions

Figure 6 Stable and unstable solution for : Ω = 0 and 0.6 = Iy > Iz = 0.25 kgm2

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]

ω 1 [rad/s]

Figure 7 Stable and unstable solution for : Ω = 150rad/s and 0.6 = Iy > Iz = 0.25kgm2

ANALYSIS 194

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]


Figure 8 Stable and unstable solution for : Ω = 300rad/s and 0.6 = Iy > Iz = 0.25kgm2

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]

ω 1 [rad/s]

stable equilibrium positions unstable equilibrium positions

Figure 9 Stable and unstable solution for : Ω = 0 rad/s and 0.6 = Iy < Iz = 0.8 kgm2

ANALYSIS 195

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]

ω 1 [rad/s]

B

A

stable equilibrium positions unstable equilibrium positions

Figure 10 Stable and unstable solution for : Ω = 150 rad/s and 0.6 = Iy < Iz = 0.8kgm2

0

0.5

1

1.5

2

2.5

3

3.5

-30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30

βo

[rad]


Figure 11 Stable and unstable solution for : Ω = 300 rad/s and 0.6 = Iy < Iz = 0.8kgm2

ANALYSIS 196

0

-1

1

2

3

4

5

0 25t [s]

β

[rad]BA

Figure 12 Solution of equation of motion for : Ω = 150 rad/s, ω1 = −10 rad/s and0.6 = Iy < Iz = 0.8 kgm

2; A) initial magnitude of βin = 1rad; B) initial magnitude ofβin = 1.1rad

EXPERIMENTAL INVESTIGATION. 197

7.4 EXPERIMENTAL INVESTIGATION.

7.4.1 Description of the laboratory installation.

6

7

10

2

2

5

4

3

1

9

FM

11

8

R

z

x

1

Figure 13

The gyroscope 3 is driven by the electric motor 4. with a constant angular speedΩ. The housing 2 of the gyroscope is free to rotate about axis x2 with respect tothe frame 1, which represent the platform. The transducer 5 allows for recordingthe actual relative angular position of the housing as a function of time. The frame1 is driven by the electric motor 6 via gearbox 7. Regulator 8 permits variation ofangular speed of the frame 1. Transducer 9 allows to measure this angular speed.The indicator of angular position 10 permits to scale the transducer 5.7.4.2 Identification of the system’s parameters.The mathematical model of the laboratory installation has form 7.8 and it is repeatedhere.

Ixβ + (Iz − Iy)ω21 sinβ cosβ + I3z2Ωω1 sinβ +MG sinβ + cβ = 0 (7.23)

The following parameters of the mathematical model above have to be identified:I3z2 – moment of inertia of the gyroscope about axis z2MG = m2gl2 +m3gl3 – static moment with respect to point of rotation O.Ix = I2x2 + I3 – moment of inertia of the whole assembly about axis x2Iy = I2y2 + I3 – moment of inertia of the whole assembly about axis y2Iz = I2z2 + I3z2 – moment of inertia of the whole assembly about axis z2Ω – angular speed of the gyroscope.


Identification of static moment MG.

1zZ

y1O

β

2G3G

2

l3

l

y2z2

2

3

s

L

F

m2 gm3 g

Figure 14

The arrangement for identification of the static moment due to gravity forces m2gand m3g is shown in Fig. 14. Its magnitude may be obtained from the followingformula.

MG = m2l2g +m3l3g = FL (7.24)

whereF - is the reading from spring balance (see Fig. 14)L = - is the distance between point of application of the force F and point of rotationO

Identification of moments of inertia.

The moments of inertia which appear in the mathematical model 7.23 may be assessedin the following manner.

Moment of inertia I3z2 is to be assessed analytically.Moment of inertia Ix is to be obtained experimentally by means of analysis

of a small oscillations of the housing for stationary frame (ω1 = 0) and stationarygyroscope (Ω = 0). Period of small oscillations of the assembly is

T =2π

α(7.25)


where

α2 =MG

Ix(7.26)

Hence

Ix =MG

α2=

MGT2

4π2(7.27)

whereT - is the measured period of small free oscillations of housingMG - is the static moment measured in the previous experiment.

Moment of inertia Iy, due to symmetry of the housing, is to be assumed tobe equal to Ix.

Moment of inertia Iz is to be obtained from the equilibrium position of thesystem for stationary gyroscope (Ω = 0) and rotating frame 1. According to equation7.11, for the case considered, one can obtain

Iz = Iy − MG

cosβsω21

(7.28)

whereω1 - is the angular velocity of the frame.βs - is the corresponding equilibrium position of the housing.

7.4.3 Collection of experimental data.The transducer 9 produces 500 impulses per one revolution of the frame 1. Frequencyof these impulses can be read from the frequency meter 11 (see Fig. 13). Thereforethe angular velocity of the frame can be computed from the following formula.

ω1 =2π

500 1f

rad/s

wheref - is the frequency measured in Hz.

For any angular velocities of the frame 1 within the range −10 < ω1 <+10 rad/s,the indicator of the angular position 10 (see Fig. 13) allows the equilibriumposition βo to be assessed. This data forms so called experimental static character-istics of the laboratory installation. The developed software permits to compute, forthe identified parameters, static characteristic of the physical model. Comparison ofthe analytical and experimental static characteristic allows to assess correctness andaccuracy of the derived mathematical model as well as the applied stability analysis.

dynamics of mechanical systems

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