Mechanics of Rigid Body by Janusz Krodkiewski Melbourne Text Book

8/20/2019 Mechanics of Rigid Body by Janusz Krodkiewski Melbourne Text Book

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436- 353 MECHANICS 2UNIT 2

MECHANICSOF

A RIGID BODY

J.M. KRODKIEWSKI

2008

THE UNIVERSITY OF MELBOURNEDepartment of Mechanical and Manufacturing Engineering

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1


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2

MECHANICS OF A RIGID BODY

Copyright C 2008 by J.M. Krodkiewski

ISBN 0-7325-1535-1

The University of Melbourne

Department of Mechanical and Manufacturing Engineering


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CONTENTS

1 THREE-DIMENSIONAL KINEMATICS OF A PARTICLE. 61.1 MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME. 6

1.1.1 Absolute linear velocity and absolute linear acceleration 61.2 MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELA-

TIVE MOTION). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.1 Motion in terms of translating system of coordinates. . 81.2.2 Motion in terms of rotating system of coordinates. . . 101.2.3 Motion in terms of translating and rotating system of

coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.3 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 THREE-DIMENSIONAL KINEMATICS OF A RIGID BODY 412.1 GENERAL MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2 ROTATION ABOUT A POINT THAT IS FIXED IN THE INERTIAL

SPACE (ROTATIONAL MOTION) . . . . . . . . . . . . . . . . . . . 432.3 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3 KINETICS OF SYSTEM OF PARTICLES. 803.1 MOTION OF CENTRE OF MASS - LINEAR MOMENTUM. . . . . 823.2 MOMENT OF MOMENTUM. . . . . . . . . . . . . . . . . . . . . . 83

3.2.1 Moment of momentum about a fixed point in the iner-tial space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3.2.2 Moment of momentum about a moving point in an in-ertial space. . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

3.2.3 Moment of relative momentum. . . . . . . . . . . . . . . 87

3.3 EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS. . . 883.3.1 Conservation of momentum principle. . . . . . . . . . . . 883.3.2 Conservation of angular momentum principle. . . . . . 893.3.3 Impulse – momentum principle. . . . . . . . . . . . . . . 90

3.4 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4 KINETICS OF RIGID BODY. 964.1 LINEAR AND ANGULAR MOMENTUM. . . . . . . . . . . . . . . . 964.2 PROPERTIES OF MATRIX OF INERTIA. . . . . . . . . . . . . . . 99

4.2.1 Parallel axis theorem. . . . . . . . . . . . . . . . . . . . . . 100

4.2.2 Principal axes. . . . . . . . . . . . . . . . . . . . . . . . . . 102


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CONTENTS 4

4.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1064.3 KINETIC ENERGY. . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

4.3.1 Rotational motion. . . . . . . . . . . . . . . . . . . . . . . . 127

4.3.2 General motion. . . . . . . . . . . . . . . . . . . . . . . . . 1284.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

4.4 EQUATIONS OF MOTION . . . . . . . . . . . . . . . . . . . . . . . 1464.4.1 Euler’s equations of motion . . . . . . . . . . . . . . . . . 1464.4.2 Modified Euler’s equations of motion . . . . . . . . . . . 1494.4.3 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

4.5 MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREE-DOM. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1824.5.1 Modelling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1824.5.2 Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

5 APPENDIXES 1915.1 APPENDIX 1. REVISION OF THE VECTOR CALCULUS 191

5.2 APPENDIX 2. CENTRE OF GRAVITY, VOLUME ANDMOMENTS OF INERTIA OF RIGID BODIES. . . . . . . . . 194


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CONTENTS 5

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

ground and engineering applications of the three dimensional mechanics of a rigidbody. It is divided into four chapters.

The first one, Three-Dimensional Kinematics of A Particle, deals withthe geometry of motion of an individual particle in terms of the inertial as well asin terms of the non-inertial system of coordinates. The introduced in this chaptertranslating, rotating as well as the translating and rotating system of coordinatesallows motion of the rigid body with respect to the inertial frame to be determinedand classified.

The second chapter, entitled Three-Dimensional Kinematics of A RigidBody, provides procedures for determination of the absolute velocity and the absolute

acceleration of any point that belong to the rigid body. Both, the general motion andthe motion about a fixed point is considered.

The last two chapters are related with the relationships between motion andforces that act on bodies. The chapter Kinetics of A System of Particles off ersgeneral principles that can be apply to any system of particles regardless of theirnumber and internal forces acting between the individual particles. Because eachcontinuum (fluid, gas, rigid or elastic body) can be considered as a system of particles,the derived equations form a base for development of many branches of mechanics(fluid mechanics, solid mechanics, rigid body mechanics etc.).

The developed principles are widely utilized in the chapter entitled Kinetics

of A Rigid Body. This chapter gives procedures for determination of matrix of inertia of a rigid body and its principal axes. This makes possible to produce expres-sion for the kinetic energy of the moving rigid body as well as to derive its equationsof motion. Both, the general motion as well as rotation of a rigid body is considered.

Each chapter is ended with several engineering problems. Solution to someof them is provided. Students should produce solution to the other problems duringtutorials and in their own time.


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Chapter 1

THREE-DIMENSIONAL KINEMATICS OF A PARTICLE.

1.1 MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME.

X

Y

Z

O

Oa Y a

Z a

X a

v

Figure 1

To consider motion of a particle we assume the existence of so-called absolute (mo-tionless) system of coordinates X aY aZ a(see Fig. 1).

DEFINITION: Inertial system of coordinated is one that does not rotate andwhich origin is fixed in the absolute space or moves along straight line ata constant velocity.

Inertial systems of coordinates are usually denoted by upper characters, e.g. XY Z , todistinguish them from non -inertial systems of coordinates which are usually denotedby lower characters, e.g. xyz.

1.1.1 Absolute linear velocity and absolute linear acceleration

Let us assume that a motion of a particle is given by a set of parametric equations1.1 which determine the particle coordinates for any instant of time.

rX = rX (t)

rY = rY (t)

rZ = rZ (t) (1.1)


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MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME. 7

X

Y

Z

O

K

I

Jr(t)

∆ r

(t+ t)∆rr Z

(t)

r Y (t) r X (t)

Figure 2

These coordinates represent scalar magnitude of components of so called ab-solute position vector r along the inertial system of coordinates XY Z .

r = IrX (t) + JrY (t) + KrZ (t) (1.2)

were I, J, K are unit vectors of the inertial system of coordinates XY Z . Vector of theabsolute velocity, as the first derivative of the absolute position vector r with respectto time, is given by the following formula.

v = ṙ = lim∆t→O

∆r

∆t = I lim

∆t→O

∆rX ∆t

+ J lim∆t→O

∆rY ∆t

+ K lim∆t→O

∆rZ ∆t

= IṙX + JṙY + KṙZ

(1.3)

Similarly, vector of the absolute acceleration is defined as the second derivative of theposition vector with respect to time.

a = r̈ = lim∆t→O

∆v

∆t = Ir̈X + Jr̈Y + Kr̈Z (1.4)

Scalar magnitude of velocity v (speed) can be expressed by formula

v = |v| =√

v · v =q ̇

r2X + ṙ2Y + ṙ

2Z (1.5)

Scalar magnitude of acceleration is

a = |a| =√

a · a =q ̈

r2X + r̈2Y + r̈

2Z (1.6)

The distance done by the particle in a certain interval of time 0 < τ < t is given bythe formula 1.7 .

s =

Z t0

v dτ =

Z t0

p ̇rX (τ )2 + ṙY (τ )2 + ṙZ (τ )2dτ = s(t) (1.7)


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MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). 8

1.2 MOTION IN TERMS OF THE NON-INERTIAL FRAMES (REL-ATIVE MOTION).

Z

Y

X

Oo

x

y

z

X

Y

Z

O

o

z

x

y

Z

Y

X

O

o

x

y

z

a b c

Figure 3

DEFINITION:System of coordinates which can not be classified as inertialis called non -inertial system of coordinates.

The non-inertial systems of coordinates are denoted by lower characters (e.g.xyz) to distinguish it from inertial one.

DEFINITION: If a non-inertial system of coordinates does not rotate (itsaxes xyz are always parallel to an inertial system) the system is called translating system of coordinates (Fig. 3 a).

DEFINITION: If a non-inertial system of coordinates rotates about originof an inertial system of coordinates, the system is called rotating system of coordinates (Fig. 3b)

In a general case a non-inertial system of coordinates may translate and rotate.

DEFINITION: System of coordinates which can translate and rotate iscalled translating and rotating system of coordinates (Fig.3c)

1.2.1 Motion in terms of translating system of coordinates.

X

Y

Z

Oo

z

x

y

P

ro

r P,or P

Figure 4


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Let XY Z be the inertial system of coordinates and xyz be the translating systemof coordinates. The relative motion of xyz system with respect to XY Z is usuallydefined by a position vector ro (Fig.4). Let us consider a particle P which moves

with respect to the translating system of coordinates and its relative motion is givenby the position vector rP,o. The position vector of the particle P in the X Y Z framecan be composed from vectors ro and rP,o.

rP = ro + rP,o = IroX + JroY + KroZ + irP,ox + jrP,oy + krP,oz (1.8)

Hence, the absolute velocity of the particle P , as the first derivative of vector rP withrespect to time is

ṙP = IṙoX + JṙoY + KṙoZ + İroX + J̇roY + K̇roZ+iṙP,ox + jṙP,oy + kṙP,oz + i̇rP,ox + ˙ jrP,oy + k̇rP,oz (1.9)

Since, İ = J̇ = K̇ = i̇ = ˙ j = k̇ = 0

ṙP = IṙoX + JṙoY + KṙoZ + iṙP,ox + jṙP,oy + kṙP,oz (1.10)

In the expression for the absolute velocity one may distinguish two parts called velocity of transportation and relative velocity .

DEFINITION: Velocity of transportation is the velocity of the particle itwould have if it would be motionless with respect to the non-inertial frame(rP,o = const)

DEFINITION: Relative velocity is the velocity of the particle it would haveif the non-inertial system of coordinates would be motionless (ro = const).

According to the above definitions the velocity of transportation in the case consideredis

vT = ṙo = IṙoX + JṙoY + KṙoZ (1.11)

and the relative velocity is

vR = iṙP,ox + jṙP,oy + kṙP,oz (1.12)

Hence, the absolute velocity

vP = ṙP = vT + vR (1.13)

Similarly, one can prove that the absolute acceleration of the particle P is

aP = r̈P = aT + aR (1.14)

where aT and aR stand for the acceleration of transportation and the relative accel-eration respectively.

DEFINITION: Acceleration of transportation is the acceleration of theparticle it would have if it would be motionless with respect to the non-inertial frame (rP,o = const)


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DEFINITION: Relative acceleration is the acceleration of the particle itwould have if the non-inertial system of coordinates would be motionless(ro = const).

In the case consideredaT = r̈o = Ir̈oX + Jr̈oY + Kr̈oZ (1.15)

aR = ir̈P,ox + jr̈P,oy + kr̈P,oz (1.16)

1.2.2 Motion in terms of rotating system of coordinates.

Z

Y

X

Oo

x

y

z

j

i

I J

k K

Figure 5

As it was mention before, the rotating system of coordinates has its origin coincidingthe origin of an inertial system of coordinates (Fig.5). First of all, we have to establishmatrix which transfers components of a vector from rotating system of coordinatesxyz to the inertial XY Z .

Matrix of direction cosines.

Let the relative motion of a particle P be determined by a position vector r, whichcomponents in the rotating system of coordinates xyz (Fig.6) are

r = irx + jry + krz (1.17)

Components of the vector r along the inertial system of coordinates XY Z may beobtained as scalar products of the vector r and unit vectors IJK.

rX = r · I = rxi · I + ry j · I + rzk · I

= rx cos∠iI + ry cos∠ jI + rz cos∠kI

rY = r · J = rxi · J + ry j · J + rzk · J

= rx cos∠iJ + ry cos∠ jJ + rz cos∠kJ

rZ = r · K = rxi · K + ry j · K + rzk · K

= rx cos∠iK + ry cos∠ jK + rz cos∠kK (1.18)


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Z

Y

X

Oo

x

y

z

j

iI

J

k

K

P

r

r x

r y

r z

r X

r Y

r Z

kI

iI

jI

Figure 6

The last relationship can be written in the a matrix form

⎡⎣

rX

rY rZ

⎤⎦

=

⎡⎣

cos∠iI cos∠ jI cos∠kI

cos∠iJ cos∠ jJ cos∠kJcos∠iK cos∠ jK cos∠kK

⎤⎦⎡⎣

rx

ryrz

⎤⎦

= [C r→i]

⎡⎣

rx

ryrz

⎤⎦

(1.19)

The transfer matrix [C r→i] is called matrix of direction cosines.From the manner we have developed the matrix of direction cosines it is easy

to notice that the inverse matrix equal the transpose one.

[C r→i]−1 = [C r→i]T = [C i→r] (1.20)

Another useful relationship should be noticed from Fig. 7. Cosine of angle betweentwo unit vectors e.g. i and J is equal to the component of one on the other.

iJ

J

i J

iY

x

Figure 7

cos∠iJ =

iY i = iY =

J xJ = J x (1.21)


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X

Y

Z

O

i

i X iY

i Z

Figure 8

From Fig. 8 one can see that components iX,iY,iZ ,which are equal to corre-

sponding direction cosines, fulfil the following relationship.

i2X + i2Y + i

2Z = cos

2∠iI + cos2∠iJ + cos2∠iK = 1 (1.22)

There exists six such relationships. Hence, only three direction angles can be chosenindependently. The three independent angles, which uniquely determined the positionof the rotating system of coordinates with respect to the inertial one, are called Euler’s angles.

Euler angles.

X

Y

Z

O ,o

x

y

z ,

,

,

Figure 9

Let us assume that the rotating system of coordinates xyz coincide the inertial oneXY Z . as shown in Fig.9. Now, let us turn the system of coordinates xyz with respectto the inertial one XY Z about axis Z by an angle φ, so the system xyz takes positionx1y1z1 (Fig. 10). The matrix of direction cosines between system x1y1z1 and XY Z is


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X

Y

Z

O ,o

z ,

x1

y1

1

φ

φ

Figure 10

[C r1→i] =

⎡⎣

cos φ − sin φ 0sin φ cos φ 0

0 0 1

⎤⎦ (1.23)

In the next step, let us turn the system xyz about axis x1 by an angle θ. The newposition of the system xyz is shown in Fig. 11 as x2y2z2. The matrix of directioncosines between system x2y2z2 and x1y1z1 has the following form.

X

Y

Z

O ,o

z ,

x1

y1

1

φ

φ θ

θ

y

2

2

,x2

Figure 11

[C r2→r1] =

⎡⎣

1 0 00 cos θ − sin θ0 sin θ cos θ

⎤⎦ (1.24)

In the last step, the system xyz is turned by an angle ψ about axis z2 to its finalposition xyz (Fig. 12). The matrix of direction cosines between xyz and x2y2z2 is


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X

Y

Z

O ,o

z ,

x

y1

1

φ

φ θ

θ

y2

2

y2 ,z

x

Ψ

Ψ

Figure 12

[C r→r2 ] =

⎡⎣

cos ψ − sin ψ 0sin ψ cos ψ 0

0 0 1

⎤⎦ (1.25)

According to Eq. 1.19, one can write the following relationships

⎡⎣

rX rY rZ

⎤⎦ = [C r1→i]

⎡⎣

rx1ry1rz1

⎤⎦ (1.26)

⎡⎣

rx1ry1rz1

⎤⎦ = [C r2→r1]

⎡⎣

rx2ry2rz2

⎤⎦ (1.27)

⎡⎣

rx2ry2rz2

⎤⎦ = [C r→r2]

⎡⎣

rxryrz

⎤⎦ (1.28)

Introducing Eq. 1.28 into Eq. 1.27) and than Eq. 1.27) into Eq. 1.26) one mayobtained

⎡⎣rX

rY rZ

⎤⎦= [C r1→i][C r2→r1][C r→r2]

⎡⎣rx

ryrz

⎤⎦= [C r→i]

⎡⎣rx

ryrz

⎤⎦ (1.29)

Hence the matrix of direction cosines between rotating system xyz and the inertialXY Z is

[C r→i] = [C r1→i][C r2→r1][C r→r2 ] (1.30)

The last formula allows to express the direction cosines as function of three indepen-dent angles known as Euler’s angles. The angle φ is called angle of precession, angleθ is called angle of mutation and the angle ψ is called angle of spin.


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Angular velocity and angular acceleration.

X

Y

Z

O

A1

A2

A00r1r

2r

X

Y

Z

O A1

A2

A00r

1r

2r

a b

Figure 13

The introduced Euler angles can not be considered as vectorial values. To show it letus consider transformation of point Ao due to rotation about axes x and y by 90o.If we turn the vector ro by 90

ofirst about axis x and then about axis y, the final

position of the point Ao is represented by vector r2 (Fig.13a). Let us do the same,but now the vector ro is turned about axis y first and then about axis x (Fig. 13 b).We can see that the final position depend on the order of rotation. Hence, angulardisplacement can not be considered as a vector because at least the commutative lawwould be violated.

φ d φ

d

Figure 14

It is easy to show, but the proof is here omitted, that the infi

nitesimal angulardisplacement can be considered as vectorial magnitude. Vector of the infinitesimalangular displacement is perpendicular to the plane of rotation and its sense is de-termined by the law of right-handed screw (Fig. 14). Such vectors which havedetermined only direction and sense are called free vectors to distinguish them fromlinear vectors (sense and line of action is determined) and position vectors ( positionof its tail, direction and sense is determined). According to the above rules the infin-itesimal angular increments of Euler’s angles dφ, dψ and dθ may be drawn as shownin Fig. 15.


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X

Y

Z

O

x

x

y

1

d

d

d

d ψ

d ψ

The instantaneous axis of rotation

d α

Figure 15

The vectorial sumdα = dφ + dψ + dθ (1.31)

of the three infinitesimal angular increments of Euler’s angles determines direction of so called instantaneous axis of rotation, having such a property that the rotation bythe angle dα about this axis is equivalent to rotation by angles dφ, dθ, dψ about axes

Z x1z respectively. The instantaneous axis go through origin of the rotating systemof coordinates.

X

Y

Z

O

β

d α

d α

rd r

A

h

l z

y

x

Figure 16

Now, let us consider point A fixed in the xyz system of coordinates determinedby a position vector r (Fig. 16). And let the axis l be the instantaneous axis of rotation of the system xyz with respect to the inertial one. Let dα be an infinitesimal


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angular displacement of the system of coordinates xyz. Hence, the point A at theinstance considered moves along circle of radius

h = r sinβ (1.32)

The infinitesimal increment of vector r is tangential to the circle and is placed inplane perpendicular to the axis l. Its scalar magnitude is

dr = h dα = r dα sinβ (1.33)

Hence, the infinitesimal increment dr can be considered as a vector product of vectordα and r.

dr = dα× r (1.34)

The velocity of the point A is

v = dr

dt =

dα

dt × r (1.35)

The vector dαdt

is called vector of angular velocity ω.

ω = dαdt

(1.36)

and the velocity of A can be expressed as follow.

v = ω × r (1.37)

If motion of the rotating system of coordinates is determined by Euler’s angles, itsangular velocity, according to the above definition and Eq. ??, is

ω =dα

dt =

Kdφ + i1dθ + kdψ

dt =K

dφ

dt + i1

dθ

dt + k

dψ

dt (1.38)

The angular speed as well as its individual components are shown in Fig. 17.

X

Y

Z

O

z

x

x

y

1

The instantaneous axis of rotation

d

dt

d

dt

d ψ

dt

ω

Figure 17


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The vector ω can be resolved along any system of coordinates. In particularit can be resolved along axes of the inertial system of coordinates X Y Z

ω = IωX + JωY + KωZ (1.39)

The scalar magnitude of the components ωX,ωY,ωZ can be expressed as functions of Euler’s angles and their first derivatives.

ωX = I · Kdφ

dt + I · i1

dθ

dt + I · k

dψ

dt

ωY = J · Kdφ

dt + J · i1

dθ

dt + J · k

dψ

dt (1.40)

ωZ = K · Kdφ

dt + K · i1

dθ

dt + K · k

dψ

dt

Having the components of angular speed as explicit function of time it is possible tofind the vector of angular velocity for any instant of time. The locus of the lines of action of the vector of angular speed in the inertial system of coordinates is calledspace cone (Fig. 18 a)

z

y

x

o

t 0t i

ωi

body cone

X

Y

Z

O

z

y

x

o

t 0t i

ωi

instantaneous axis of rotation

Y

Z

O

t 0t i

ωi

space cone

a b c

Figure 18

The vector of angular speed ω can be as well resolved along axes of the systemof coordinates xyz

ω = iωx + jωy + kωz (1.41)

and the components can be expressed as a function of time.

ωx = i · Kdφ

dt + i · i1

dθ

dt + i · k

dψ

dt

ωy = j · Kdφ

dt + j · i1

dθ

dt + j · k

dψ

dt (1.42)

ωz = k · K

dφ

dt + k · i1dθ

dt + k · k

dψ

dt


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The components ωx,ωy,ωz determine for any instant of time direction of the angularvelocity in the rotating system of coordinates. The locus of these lines make up socalled body cone (Fig. 18 b). The rotational motion of the system xyz with respect

to the X Y Z , can be hence considered as rolling without slipping of the body cone onthe space cone (Fig. 18 c).

The angular acceleration is defined as the first derivative of the vector of angular velocity with respect to time.

ε = dωdt

= ω̇ (1.43)

Derivative of a vector expressed in terms of a rotating system of coordi-nates

According to the consideration carried out in the previous paragraphs the rotational

motion of a system of coordinates xyz with respect to the inertial one can be definedby three independent angles (e.g. Euler angles) or, alternatively the rotational motioncan be determined by its initial position and the vector of its angular velocity ω.

Let ω be the absolute angular velocity of the rotating system of coordinatesxyz (Fig. 19).

X

Y

Z

O

z

y

x

o

ω

A

A x

A y

A z

Figure 19

Consider a vector A which is given by its components along the rotating system

of coordinates xyz. A = iAx + jAy + kAz (1.44)

Let us diff erentiate this vector with respect to time.

Ȧ = d

dt(iAx + jAy + kAz) (1.45)

HenceȦ = i Ȧx + j Ȧy + k Ȧz + i̇Ax + ˙ jAy + k̇Az (1.46)

The first three terms represent vector which can be obtained by direct diff erentiatingof the components Ax, Ay, Az with respect to time. This vector will be denoted byA0.

A0 = i Ȧx + j Ȧy + k Ȧz (1.47)


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ThusȦ = A0 + i̇Ax + ˙ jAy + k̇Az (1.48)

According to defi

nition of vector derivative, the fi

rst derivative of the unit vector i

isthe ratio of the infinitesimal vector increment di and dt (Fig. 20).

Y

Z

O

i vi

ω

x

z

y

Figure 20

i̇ = di

dt =

vidt

dt = vi (1.49)

were vi is a velocity of head of the vector i. But, according to Eq. 1.37

vi = ω × i (1.50)

Hencei̇ = ω × i (1.51)

Similarly˙ j = ω × j and k̇ = ω × k (1.52)

Introducing the above expressions into Eq. 1.48 we have

Ȧ = A0 + ω × iAx + ω × jAy + ω × kAz = A0 + ω × (iAx + jAy + kAz) (1.53)

and eventually one may gain

Ȧ = A0 + ω × A (1.54)

where:A0 = i Ȧx + j Ȧy + k Ȧzω - is the absolute angular velocity of the rotating system of coordinates xyz

along which the vector A was resolved to produce vector A0

A - is the diff erentiated vector.The last formula provides the rule for diff erentiation of a vector that it is resolvedalong a non-inertial system of coordinates. It can be applied to any vector (eg position

vector, velocity, angular velocity etc.)


21/195


Motion in terms of rotating system of coordinates.

Let relative motion of a particle P (see Fig. 21) with respect to the rotating system

of coordinates xyz be determined by position vector rP .

rP = irPx + jrPy + krPz (1.55)

whererPx, rPy, rPz – components of the vector rP along system of coordinates xyz.The system xyz itself has its own rotational motion determined by absolute

angular velocity ω. We are interested in the absolute velocity and absolute accelera-tion of this particle.

X

Y

Z

O

z

y

x

o

ω

P

r P

Figure 21

The absolute velocity of the particle P is

vP = ṙP = r0

P + ω × rP (1.56)

In a similar manner, as it was done in case of translating system of coordinates, weintroduce notions of the relative velocity and the velocity of transportation. For thecase of a particle motionless in the rotating system of coordinates, according to Eq.1.56 the velocity of transportation is

vT = ω × rP (1.57)

For the case of motionless system of coordinates xyz ( ω = 0), according to Eq. 1.56the relative velocity is

vR = r0

P (1.58)

HencevP = vR + vT (1.59)

The absolute acceleration one may obtain diff erentiating equation 1.56 with respectto time.

r̈P = ddt

r0P + ddt(ω × rP ) = r00P + ω × r

0

P + ω̇ × rP + ω × (r0

P + ω × rP ) (1.60)


22/195


Introducing Eq. 1.43 and developing the last term we have

aP = r̈P = r00

P + ε× rP + ω × (ω × rP ) + 2 ω × r0

P (1.61)

Assuming that the point P is motionless (rP = constant) one may obtain the followingexpression for acceleration of transportation

aT = ε × rP + ω × (ω × rP ) (1.62)

Here, the term ε× rP is called tangential acceleration of transportation and the termω × (ω × rP ) is called normal acceleration of transportation .

Assumption, that the system of coordinates is motionless yields expression forthe relative acceleration .

aR = r00 (1.63)

The last term in equation 1.61 is called Coriolis acceleration aC .

aC = 2 ω × r0 (1.64)

Now, we can state that the absolute acceleration is composed of acceleration of trans-portation, relative acceleration and Coriolis acceleration.

aP = aT + aR + aC (1.65)

1.2.3 Motion in terms of translating and rotating system of coordinates.

Let us defi

ne motion of the translating and rotating system of coordinates xyz bythe position vector ro and the vector of the angular velocity ω (see Fig..22). Therelative motion of a point P with respect to the rotating and translating system of coordinates xyz is determined by a position vector rP,o. Hence, the absolute positionvector rP in this case is

rP = ro + rP,o (1.66)

X

Y

Z

O

z

y

x

o

P

r P

ro

r P,o

ω

Figure 22


23/195


Let us assume that the relative position vector rP,o is determined by its com-ponents along the translating and rotating system of coordinates xyz.

rP,o = irP,ox + jrP,oy + krP,oz (1.67)

According to the previously developed rules, the absolute velocity of the point P is

ṙP = ṙo + r0

P,o + ω × rP,o

whereṙo + ω × rP,o – velocity of transportation r0P,o– relative velocity

The second derivative yields the absolute acceleration

r̈P = r̈o + r00

P,o + ω × r0

P ,o + ε× rP,o + ω × (r0

P ,o + ω × rP,o)= r̈o + ε× rP,o + ω ×ω × rP,o + r

00

P,o + 2 ω × r0

P ,o (1.68)

Here:r̈o + ε× rP,o + ω ×ω × rP,o = aT – acceleration of transportation r00P,o = aR – relative acceleration 2 ω × r0P ,o = aC – Coriolis acceleration .


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PROBLEMS. 24

1.3 PROBLEMS.

Problem 1

X Y

y

Z z 1

x1

1 y

x

z 2

2

2

ω

ω

1

12

ω 1 t

ω 1t 2

1 2

Figure 23

The frame 1 of the system shown in Fig. 23 rotates about the vertical axis Z with a constant angular velocity ω1 whereas the disc 2 has its own constant relativeangular velocity ω21. Calculate components of the absolute angular velocity ω2 of

the disc 2 and its absolute acceleration ε2.Solution.

System of coordinates x1y1z1 is rigidly attached to the frame 1 and systemx2y2z2 is body 2 system of coordinates. Absolute angular velocity of the body 2 is

ω2 = k1ω1 + j1ω21 (1.69)

Its first vector derivative with respect to time yields absolute angular acceleration.

ε2 = ω̇2 = ω0

2 + ω1 × ω2 = k10 + j10 + ¯̄̄̄̄̄i1 j1 k10 0 ω1

0 ω21 ω1¯̄̄̄̄̄ = −i1ω1ω21 (1.70)


25/195

PROBLEMS. 25

Problem 2

X

Y

y

x1

1O o1

2 x

ω o t α

.

Z z 1

1

2 y

ωoα

y

O12

P

l

2 z

Figure 24

A radar antenna rotates about the vertical axis Z at the constant angularspeed ωo. At the same time the angle α is being changed as follow

α = a sin At

Producethe components of the angular velocity and the angular acceleration of the

antenna1. along the inertial X Y Z system of coordinates2. the body 2 system of coordinates x2y2z2

the magnitude of the angular velocity and acceleration of the antenna..the components of velocity and acceleration of the probe P along the body 2

system of coordinates.the magnitude of velocity and acceleration of the probe P .


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PROBLEMS. 26

Solution.A. Matrices of direction cosines.Fig. 24 permits to produce matrices of direction cosines between the iner-

tial system of coordinates XY Z , the rotating system of coordinates x1y1z1 and therotating system of coordinates x2y2z2.

x1y1z1

=

⎡⎣

cos ωot sin ωot 0− sin ωot cos ωot 0

0 0 1

⎤⎦⎡⎣

X Y Z

⎤⎦ = [C I →1]

⎡⎣

X Y Z

⎤⎦ (1.71)

⎡⎣

x2y2z2

⎤⎦ =

⎡⎣

1 0 00 cos α sin α0 − sin α cos α

⎤⎦⎡⎣

x1y1z1

⎤⎦ = [C 1→2]

⎡⎣

x1y1z1

⎤⎦ (1.72)

⎡⎣

x2y2z2

⎤⎦ = [C 1→2][C I →1]

⎡⎣

X Y Z

⎤⎦

=

⎡⎣

1 0 00 cos α sin α0 − sin α cos α

⎤⎦⎡⎣

cos ωot sin ωot 0− sin ωot cos ωot 0

0 0 1

⎤⎦⎡⎣

X Y Z

⎤⎦

=

⎡⎣

cos ωot sin ωot 0− sin ωot cos α cos ωot cos α sin α

sin ωot sin α − cos ωot sin α cos α

⎤⎦⎡⎣

X Y Z

⎤⎦ (1.73)

B. Angular velocities.The angular velocity of the body 2 is determined by the following vector

equation.ω2 = ωok1 + α̇i1 (1.74)

Its components along the system of coordinates x2y2z2 may be calculated with helpof Eq. 1.72.⎡⎣

ω2x2ω2y2ω2z2

⎤⎦ =

⎡⎣

1 0 00 cos α sin α0 −

sin α cos α

⎤⎦⎡⎣

ω2x1ω2y1ω2z1

⎤⎦ =

⎡⎣

1 0 00 cos α sin α0 −

sin α cos α

⎤⎦⎡⎣

α̇0

ωo

⎤⎦

=

⎡⎣

α̇ωo sin αωo cos α

⎤⎦ (1.75)

Sinceα = asinAt (1.76)

the above components are

ω2x2 = aA cos At

ω2y2 = ωo sin(a sin At) (1.77)

ω2z2 = ωo cos(a sin At)


27/195

PROBLEMS. 27

The above equations may be considered as parametric equations of the body cone.Parametric equations of the space cone are determined by components of the angularvelocity ω2 along inertial system of coordinates. These components may be obtained

with help of the equation 1.71.⎡⎣

ω2X ω2Y ω2Z

⎤⎦ = [C I →1]−1

⎡⎣

ω2x1ω2y1ω2z1

⎤⎦ =

⎡⎣

cos ωot − sin ωot 0sin ωot cos ωot 0

0 0 1

⎤⎦⎡⎣

α̇0

ωo

⎤⎦

=

⎡⎣

α̇ cos ωotα̇ sin ωot

ωo

⎤⎦ (1.78)

Introduction of Eq. 1.76 into Eq. 1.78 yields the parametric equations of the spacecone.

ω2X = aA cos At cos ωot

ω2Y = aA cos At sin ωot (1.79)

ω2Z = ωo

Both, the space cone and body cone, are presented in Fig. 25. They were computedfor the following data

a = 1[m], ωo = 1[1/s], A = 3[1/s]

-2

-1

0

1

2

-3 -2 -1 0 1 2 3

-2

-1

0

1

2

-3 -2 -1 0 1 2 3

y2

x2

Y

X

Z

X

z 2

x2

a) b)

Figure 25

Magnitude of the absolute angular velocity ω2 is

ω2 = q ω22x2 + ω

22y2 + ω

22z2 = q α̇

2 + (ωo sin α)2 + (ωo cos α)2 = q α̇2 + ω2o (1.80)

C. Angular acceleration.


28/195

PROBLEMS. 28

The absolute angular acceleration of the antenna can be obtained by diff eren-tiation of the vector ω2 Eq. 1.75.

ε2 = ω̇2 = ω0

2 + ω2 × ω2 = i2α̈ + j2ωo α̇ cos α− k2ωo α̇ sin α (1.81)

Hence, its magnitude is

ε2 =q ̈

α2 + (ωo α̇ cos α)2 + (ωo α̇ sin α)2 =p ̈

α2 + ωo α̇2 (1.82)

where

α̇ = aA cos At

α̈ = −aA2 sin At (1.83)

D. Velocity of the point P .Position of the point P is determined by the position vector l.

l = k2l (1.84)

Its first derivative with respect to time yields the absolute velocity of the point P .

vP = l̇ = l0 + ω2 × l =

¯̄̄¯¯̄

i2 j2 k2α̇ ωo sin α ωo cos α0 0 1

¯̄̄¯¯̄

= i2ωol sin α− j2lα̇ (1.85)

Hence, magnitude of velocity of the point P is

vP =p

(ωol sin α)2 + (lα̇)2 (1.86)

where α and α̇ are determined by formulae 1.76 and 1.83.E. Acceleration of the point P .The absolute acceleration of point P may be obtained by diff erentiation of the

vector of its absolute velocity.

aP = v̇P = v0

P + ω2 × vP

= i2(ωolα̇ cos α)− j2(lα̈) +¯̄̄̄̄¯ i2 j2 k2α̇ ωo sin α ωo cos αωol sin α −lα̇ 0

¯̄̄̄̄¯

= i2(ωolα̇ cos α + ωolα̇ cos α)

+ j2(−lα̈ + ω2ol sin α cos α)+k2(−lα̇2 − ω2ol sin2 α)

= i2aP x2 + j2aP y2 + k2aP z2 (1.87)

Hence, its magnitude is

aP = q a2P x2 + a2P y2 + a2P z2 (1.88)


29/195

PROBLEMS. 29

Problem 3

ω L

12

34

v

β

Figure 26

The belt conveyor 1 shown in Fig. 26 is mounted at the constant angle β = 30o

with respect to the horizontal plane on the rotating table 2 . The table rotates withthe constant angular speed ω = 1rad/s whereas the belt 3 moves with the constantlinear velocity v = 2m/s in the direction shown. Calculate magnitude of velocity andacceleration of the particle 4 travelling without slipping for the position defined bythe distance L = 5m.


30/195

PROBLEMS. 30

Solution.

L

12

34

v

β

Z

R

O

O

Y ω t

Figure 27

Axes XY Z form the inertial system of coordinates. Axes xyz are fixed tothe rotating table 2 and have its origin at O. Angular velocity of the system of coordinates xyz is vertical and its components along xyz system of coordinates are

ω2 = jω sin β + kω cos β (1.89)

Position vector which determines a position of the particle 4 is

R = jL (1.90)

Its first derivative with respect to time produces the absolute velocity of the particle4.

˙

R = R

0

+ ω2 × R = j˙

L +¯̄̄̄̄̄ i j k

0 ω sin β ω cos β 0 L 0¯̄̄̄̄̄

= −iLω cos β + jv (1.91)

Hence, magnitude of the velocity is

| Ṙ| =p

(Lω cos β )2 + v2 =p

(5 · 1 · cos30o)2 + 22 = 4.76 m/s (1.92)

The first derivative of the absolute velocity yields the absolute acceleration of theparticle.

R̈ = Ṙ0 + ω2 × Ṙ = −ivω cos β +

¯̄

¯̄̄̄i j k

0 ω sin β ω cos β

−Lω cos β v 0

¯̄

¯̄̄̄= −i2vω cos β − jω2L cos2 β + kLω2 sin β cos β (1.93)


31/195

PROBLEMS. 31

Its magnitude may be calculated according to the following formula.

|R̈| = p (2ωv cos β )2 + (Lω2 cos2 β )2 + (Lω2 sin β cos β )2=

p (2 · 1 · 2 · cos 30o)2 + (5 · 12 · cos2 30o)2 + (5 · 12 sin30ocos30o)2

= 5.545 m/s2 (1.94)


32/195

PROBLEMS. 32

Problem 4

ρ

L

C D ρ α

ω

A A

1 2

Figure 28

Wheel of radius ρ is free to rotate about axle CD which turns about the verticalaxis with a constant speed ω. The wheel rolls without slipping on the horizontal plane.Determine, as a function of its angular position α, the magnitudes of velocity and

acceleration of the shown in the Fig. 28 point A .Given are: ρ, l, ω.


33/195

PROBLEMS. 33

Solution.

ρ

L

C D ρα

ω

z

z

x

y

x

y

12

1

2

1

2

x1 X

z

Z

Y y1

1O

o

A

ω t

2

β

β

ω

ω

1

2

2

1 O

L

1

F

G E

Figure 29

Axes XYZ, in Fig. 29, form the inertial system of coordinates. Axis x1y1z1are rigidly attached to the axle 1 and form the body 1 system of coordinates. Itsaxis x1coincides axis X of the inertial system of coordinates. Therefore the angular

speed of this system of coordinates is

ω1 = Iω = i1ω (1.95)

Axes x2y2z2 are fixed to the wheel 2 and its axis x2 goes through the point Awhereas its axis z2 coincides axis z1. Its absolute angular velocity ω2 is assembled of the absolute angular velocity ω1 and the relative velocity ω21.

ω2 = ω1 + ω21 (1.96)

Direction of the relative angular velocity ω21, according to the imposed constraints,

coincide axis z2. Since the cone CEF may by considered as the body 2 cone and thecone CEG may be considered as the space cone, the absolute angular velocity of thebody 2 mast have direction of the common generating line EC. Therefore

ω21 = ω1 cot β = ω1L

ρ (1.97)

Since the vector ω21 has opposite direction to the positive direction of axis z1, vectorof the absolute angular velocity ω2 is

ω2 = i1ω + k1(−

ωL

ρ

) (1.98)


34/195

PROBLEMS. 34

Position vector of the point A is

r = L + ρ = k1L + i2ρ (1.99)

Components of the above position vector along system of coordinates x1y1z1 are asfollows

rx1 = r · i1 = k1·i1L + i2·i1ρ = ρ cos α

ry1 = r · j1 = k1· j1L + i2· j1ρ = ρ sin α

rz1 = r · k1 = k1·k1L + i2·k1ρ = L (1.100)

Absolute velocity of the point A as the first derivative of the vector r is

vA = ṙ = r0 + ω1 × r = −i1ρα̇ sin α + j1ρα̇ cos α +¯̄̄̄̄¯

i1 j1 k1ω 0 0

ρ cos α ρ sin α L

¯̄̄̄̄¯

= i1(−ρα̇ sin α) + j1(ρα̇ cos α− ωL) + k1(ωρ sin α) (1.101)

Magnitude of the absolute velocity is

|vA| = |ṙ| =p

(−ρα̇ sin α)2 + (ρα̇ cos α− ωL)2 + (ωρ sin α)2 (1.102)

Similarly, the first derivative of the absolute velocity yields the absolute accelerationof the point A.

aA = v̇A = v0

A + ω1 × vA = i1 v̇Ax1 + j1 v̇Ay1 + k1 v̇Az1 +

¯̄̄¯̄̄ i1 j1 k1ω 0 0

vAx1 vAy1 vAz1

¯̄̄¯̄̄

= i1(v̇Ax1) + j1(v̇Ay1 − ωvAz1) + k1(v̇Az1 + ωvAy1) (1.103)

where, according to (1.101)

vAx1 = −ρα̇ sin αvAy1 = ρα̇ cos α− ωLvAz1 = ωρ sin α (1.104)

Hence, the magnitude of absolute acceleration is

aA =q

(v̇Ax1)2 + (v̇Ay1 − ωvAz1)2 + (v̇Az1 + ωvAy1)2 (1.105)


35/195

PROBLEMS. 35

Problem 5

P

l

β Ω

ω

Figure 30

A crane shown in Fig. 30 is revolving about vertical axis with the constantangular speed ω = 1rad/s in the direction shown. Simultaneously the boom is beinglowered at the constant angular speed Ω = 0.5rad/s. Calculate the magnitude of thevelocity and acceleration of the end P of the boom for the instant when it passes theposition β = 30o. The boom has the length l = 10m.

Answer:v = 7.07m/s


36/195

PROBLEMS. 36

Problem 6

1 23

Y y x x

y

x

M

z 1

12 3

1 2 z

2

0

z

y

y

3

2

3

l

o1

α t

bα

v

ωt

ωb

Figure 31

The turret on a tank (see Fig. 31) rotates about the vertical axis at angularspeed ω t and the barrel is being raised at a constant angular speed ωb. The tank hasconstant forward speed v. When the barrel is in position defined by angles αt andαb a shell leaves the barrel with muzzle velocity vs and acceleration as. Determinethe absolute velocity vm and acceleration am of the barrel muzzle as well as absolutevelocity v and acceleration a of the shell when it leaves the barrel.

Answer:The components of the absolute position vector of the point M that belong to thebarrelrMbx2 = vt sin αt rMby2 = vt cos αt + l cos αb rMbz2 = l sin αbwhere l = constantThe components of the absolute velocity of the point M that belong to the barrelvMbx2 = ṙMbx2−ωtl cos αb−ωtvt cos αt vMby2 = ṙMby2+ωtvt sin αt vMbz2 = ṙMbz2The components of the absolute acceleration of the point M that belong to the barrel

aMbx2 = v̇Mbx2 −ωtvMby2 aMby2 = v̇Mby2 + ωtvMbx2 aMbz2 = v̇Mbz2The components of the absolute position vector of the point M that belong to theshellrMsx2 = vt sin αt rMsy2 = vt cos αt + l(t)cos αb rMsz2 = l(t)sin αbwhere l̇ = vs l̈ = asThe components of the absolute velocity of the point M that belong to the shellvMsx2 = ṙMsx2−ωtl cos αb−ωtvt cos αt vMsy2 = ṙMsy2+ωtvt sin αt vMsz2 = ṙMsz2The components of the absolute acceleration of the point M that belong to the barrelaMsx2 = v̇Msx2 −ωtvMsy2 aMsy2 = v̇Msy2 + ωtvMsx2 aMsz2 = v̇Msz2


37/195

PROBLEMS. 37

Problem 7

X Y

O

x

y

Z z

z y

y

α

x

z

RO

P

11

1

1

O

21

1

2

ω 21

t

γ

l

Figure 32

The link 1 of the mechanical system shown in Fig. 32 performs the rotational

motion about the absolute axis Z . Its instantaneous position is determined by theangle α. The link 2 rotates with respect to the link 1 with the constant relativeangular velocity ω21 . Point P belongs to the body 2 .

Given are: l ,R, ω21, γ , α(t)Produce the expression for:

1. the components of the linear absolute velocity of the point P along the body 1system of coordinates x1y1z1.

Answer:vx1 = Rω21 cos ω21t + Rα̇ sin γ cos ω21t− lα̇ cos γ vy1 = Rα̇ cos γ sin ω21t

vz1 = −Rω21 sin ω21t−Rα̇ sin γ sin ω21t2. the components of the absolute acceleration of the point P along the body systemof coordinates x1y1z1.

Answer:ax1 = v̇x1 + vz1 α̇ sin γ − vy1 α̇ cos γ ay1 = v̇y1 + vx1 α̇ cos γ az1 = v̇z1 − vx1 α̇ sin γ


38/195


39/195

PROBLEMS. 39

Problem 9

Y

Z

O

X

P

1 z

2 y 2 x

2 z

2 o

1 x

1 y

1 o 1 y

β

α

D

1

2

1 o

2 o

O

L

Figure 34

The base 1 of the crane shown in Fig. 34 rotates about the vertical axis Z of the inertial system of coordinates XY Z . Its motion is determined by the angulardisplacement α. The system of coordinates x1y1z1 is attached to the base 1. Atthe same time the boom 2 is being raised. This relative motion about the axis x1is determined by the angular displacement β . The system of coordinates x2y2z2 is

attached to the boom.Given are: L, D, α(t), β (t)Produce the expressions for

1. the components of the absolute angular velocity of the boom 2 along the x2y2z2system of coordinates

Answer:ω2x2 = β̇ ω2y2 = α̇ sin β ω2z2 = α̇ cos β 2. the components of the absolute angular acceleration of the boom 2 along the x2y2z2system of coordinates

Answer:ε2x2 = β̈ ε2y2 = α̈ sin β + α̇β̇ cos β ε2z2 = α̈ cos β

− α̇β̇ sin β


40/195

PROBLEMS. 40

3. the components of the absolute linear velocity of the point P along the x2y2z2system of coordinates

Answer:

vPx2 = Dα̇− Lα̇ cos β vPy2 = 0 vPz2 = β̇L4. the components of the absolute linear acceleration of the point P along the x2y2z2system of coordinates

Answer:aPx2 = Dα̈−Lα̈ cos β + 2Lα̇β̇ sin β aPy2 = Dα̇2 cos β −Lα̇2 cos2 β −Lβ̇

2aPz2 =

Lβ̈ −Dα̇2 sin β + Lα̇2 sin β cos β


41/195

Chapter 2

THREE-DIMENSIONAL KINEMATICS OF A RIGID BODY

DEFINITION: A body which by assumption does not deform and there-fore the distances between two of its points remains unchanged, regardlessof forces acting on the body, is called rigid body.

X

Y

Z

O

z

y

x

o

P

r P

ro

r P,o

ω

G

rG,o

i

j

k

Figure 1

To analyze motion of a rigid body usually we attach to the body a systemof coordinates xyz at an arbitrarily chosen point o (see Fig.1). Such a system of coordinates is called body system of coordinates. The body system of coordinates, in

a general case, may translate and rotate. Hence, motion of the rigid body may bedetermined in the same manner as the motion of the translating and rotating systemof coordinates. As we remember, motion of the translating and rotating system of coordinates can be determined by a position vector ro and a vector of the angularvelocity ω. The angular velocity ω of the body system of coordinates is called angular velocity of the rigid body.

2.1 GENERAL MOTION

DEFINITION: If the body system of coordinates translates and rotates,it is said that the body performs the general motion .

Let the position of a point P with respect to the body system of coordinates (see


42/195

GENERAL MOTION 42

Fig.1) be defined by a position vector rP,o. Since components of the vector rP,o alongthe system of coordinates xyz are constant, the relative velocity of P with respect tothe body system of coordinates, r0P ,o,is always 0. Therefore its absolute velocity is

vP = ṙP = ṙo + ṙP,o = ṙo + r0

P ,o + ω × rP,o = ṙo + ω × rP,o (2.1)

Similarly, The absolute velocity of the point Q is

vQ = ṙo + ω × rQ,o (2.2)

Hence, the relative velocity of the point Q with respect to the point P is

vQP = vQ − vP = ṙo + ω × rQ,o − (ṙo + ω × rP,o) = ω × (rQ,o − rP,o) =ω × rQ,P (2.3)

The absolute acceleration of P is

aP = r̈P = r̈o + ω̇ × rP,o + ω × (ω × rP,o) (2.4)

The relative acceleration of the point Q with respect to the point P is

aQP = aQ − aP = r̈o + ω̇ × rQ,o + ω × (ω × rQ,o)− r̈o − ω̇ × rP,o −ω × (ω × rP,o) == + ω̇ × rQ,P + ω × (ω × rQ,P ) (2.5)

The first termatQP = ω̇ × rQ,P (2.6)

is called the tangential relative acceleration and the second one

anQP = ω × (ω × rQ,P ) (2.7)

is called the normal relative acceleration.For the kinetics purposes we are often interested in components of the absolute

velocity of the centre of mass G along the body coordinates xyz. If position of thecentre of mass is defined by a vector rG,o, the components of its absolute velocityalong the body axes are

vGx = i · (ṙo + ω × rG,o)

vGy = j · (ṙo + ω × rG,o)vGz = k · (ṙo + ω × rG,o (2.8)

The components of angular velocity along the body coordinates are

ωx = i ·ω

ωy = j · ω

ωz = k · ω (2.9)

Absolute angular acceleration ε may be obtained as follow

ε = ω̇ = ω 0 + ω ×ω = ω 0 (2.10)


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ROTATION ABOUT A POINT THAT IS FIXED IN THE INERTIAL SPACE (ROTATIONALMOTION) 43

2.2 ROTATION ABOUT A POINT THAT IS FIXED IN THE INER-TIAL SPACE (ROTATIONAL MOTION)

DEFINITION: If one point of the body considered is motionless withrespect to the inertial frame, it is said that the body performs rotational motion .

This motionless point is called centre of rotation and usually this centre is chosen asthe origin of the body system of coordinates (see Fig. 2).

Z

Y O

x

y

z

r

ω

rG

G P

P

Figure 2

The linear velocity of an arbitrarily chosen point P of the rigid body as wellas its acceleration is determined by the angular velocity of the body ω. Indeed

vP = ṙP = r0

P + ω × rP = ω × rP (2.11)

aP = r̈P = ω̇ × rP + ω × (ω × rP ) (2.12)


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45/195

PROBLEMS. 45

Solution.

X

Y

α

x11

y11

z 11

y11

x11

β

1

2 y12

1

2 x

x2

Ω t z 12

2 z

µ

L

y12

y2

1

2 x

Figure 4

In Fig. 4, system of coordinates x11, y11, z

11 is rigidly attached to the base 1 and

rotates about the vertical axis Z of the inertial system of coordinates XY Z . Theangular displacement α determines uniquely its instantaneous position. System of coordinates x21, y

21, z

21 is rigidly attached to the base 1 and it is turned by angle β about

axis x11. System of coordinates x2, y2, z2 is fixed to the rotor 2 . Its axis y2 coincides

axis y

2

1 and its instantaneous position is determined by the angular displacement Ω

t.Absolute angular velocity of the rotor 2 is

ω2 = ω1 + ω2,1 = k11 α̇ + j2Ω = (k

21 cos β + j

21 sin β )α̇ + j2Ω

= ((k2 cosΩt− i2 sinΩt)cos β + j2 sin β )α̇ + j2Ω= i2(−α̇ cos β sinΩt) + j2(α̇ sin β + Ω) + k2(α̇ cos β cosΩt) (2.13)

Position vector of the centre of gravity G, according to Fig. 4, is

RG = L +µ = j21L + k2µ = j2L + k2µ (2.14)

Hence, its fist derivative with respect to time yields its absolute velocity

ṘG = R0

G + ω2 × RG =

¯̄̄¯̄̄ i2 j2 k2−α̇ cos β sinΩt α̇ sin β + Ω α̇ cos β cosΩt

0 L µ

¯̄̄¯̄̄

= i2(µ(α̇ sin β + Ω)− Lα̇ cos β cosΩt)+ j2(−µα̇ cos β sinΩt)+k2(−Lα̇ cos β sinΩt)

(2.15)


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PROBLEMS. 46

Problem 11

B

Z

O

X

1

2

3

c

Y

l AB

Figure 5

Fig. 5 shows the kinematic diagram of a spatial mechanism. Its link 1 canmove along axis X and is free to rotate about that axis. The link 2 is hinged to thelink 1 at the point A and at the point B is connected to the link 3 through a ball joint. The link 3 can move in plane Y Z along axis which is parallel to Y .

Given are:1. Motion of the point A (its position vector rA).

2. Distance c between the point B and axis Y .3. Length of the link 2 lAB.Determine:1. Positions of individual links.2. Linear velocity of the link 3 .3. Angular velocity of the link 1.4. Angular velocity of the link 2 .


47/195

PROBLEMS. 47

Solution.

B

Z

O

A

1

2

3

c z z

12

1 2

1

x x

,

y

α

β α

β

2 y

C

r

r

r BC

r BAC

A Y

Figure 6

Let x1y1z1 be the body 1 system of coordinates. Its angular position withrespect to the inertial system of coordinates XY Z , may be determined by angle α.Since the link 2 can rotate with respect to the link 1 about axis z1 only, the relativeposition of the body 2 system of coordinates x2y2z2 is uniquely determined by angleβ .

A. Matrices of direction cosines.From Fig. 6 one can see that the matrix of direction cosines between systemof coordinates x1y1z1 and inertial system of coordinates X Y Z has the following form

⎡⎣

X Y Z

⎤⎦ =

⎡⎣

1 0 00 cos α − sin α0 sin α cos α

⎤⎦⎡⎣

x1y1z1

⎤⎦ (2.16)

In the same manner we can easily derive matrix of direction cosines between body 2 system of coordinates x2y2z2 and system of coordinates x1y1z1.

⎡⎣x1y1z1

⎤⎦ = ⎡⎣cos β

−sin β 0

sin β cos β 00 0 1

⎤⎦⎡⎣x2y2z2

⎤⎦ (2.17)

Introduction of Eq. 2.17 into Eq. 2.16 yields matrix of direction cosines between thebody 2 system of coordinates and the inertial one.

⎡⎣

X Y Z

⎤⎦ =

⎡⎣

1 0 00 cos α − sin α0 sin α cos α

⎤⎦⎡⎣

cos β − sin β 0sin β cos β 0

0 0 1

⎤⎦⎡⎣

x2y2z2

⎤⎦

=

⎡⎣

cos β − sin β 0

cos α sin β cos α cos β − sin αsin α sin β sin α cos β cos α

⎤⎦⎡⎣

x2

y2z2

⎤⎦

(2.18)


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49/195

PROBLEMS. 49

where α̇ and β̇ are derivatives of expressions 2.25 and 2.26 respectively.D. Velocity of the point B .Velocity of the point B can be obtained by direct diff erentiation of expression

2.27 with respect to time.

ṙB = vB = JṙBC = JlAB(−α̇ sin α cos β − β̇ cos α sin β ) (2.33)


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PROBLEMS. 50

Problem 12

The point A of the body 1 shown in Fig. 7 can move along the vertical slide 2

which is located along the axis Z of the inertial system of coordinates X Y Z . Motionof the point A is determined by the following function of time

Z A = Z A(t)

The point B of the body 1 can move along the horizontal slide located in the planeY Z . This slide is by a apart from axis Y . The point C is in a constant contactwith the plane XY . According to the described constraints,motion of the body 1 isuniquely determined by the function Z A.

Produce:1. Components of the linear velocity of the point B and the point C .2. Components of the absolute angular velocity of the body 1.

X

Z

O

a

a

a

Y

90o

A

B

C

Z(t)

1

Figure 7


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PROBLEMS. 51

Solution.A. direction cosines

X

Z

O

a

a

a

Y

90o

A

B

C

Z(t)

z

x

y

r A

r B

r BA

rC

rCA

rCB

Figure 8

Let us consider the following vector equation (see Fig. 8).

rB = rA + rBA (2.34)

orJrBY + Ka = KZ A(t) + ka (2.35)

Hence

k = JrBY

a +K(1− Z A

a ) (2.36)

Multiplication of the above vector equation by the units vectors associated with theinertial system of coordinates I, J and K yields the direction cosines between the axisz and axes of the inertial system of coordinates X Y Z .

k · I = cos∠kI =0k · J = cos∠kJ =

rBY a

k · K = cos∠kK = 1− Z Aa

(2.37)

Sincecos2∠kI + cos2∠kJ + cos2∠kK = 1 (2.38)

we have that

(0)2 + ³rBY

a ´2

+µ1−Z Aa ¶

2

= 1 (2.39)


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PROBLEMS. 52

The last relationship permits the unknown component rBY to be determined as theexplicit function of time

rBY = q 2aZ A − Z 2A) (2.40)To determine the direction cosines between the axis y and axes of the inertial systemof coordinates, let us consider equation

rC = rA + rCA (2.41)

orIX C + JY C = KZ A + ja (2.42)

Since according to the above equation

j = I

rCX

a + J

rCY

a −KZ A

a (2.43)

the wanted direction cosines are

j · I = cos∠ jI =rCX

a

j · J = cos∠ jJ = rCY

a

j · K = cos∠ jK = −Z Aa

(2.44)

The following property of the direction cosines

cos2∠ jI + cos2∠ jJ + cos2∠ jK = 1 (2.45)

yieldsr2CX + r

2CY + Z

2A = a

2 (2.46)

The second equation for determination of the unknown rCX and rCY one may obtainby consideration of the following vector equation

rC = rB + rCB (2.47)

It follows that

rCB = rB − rC (2.48)Multiplication of the above equation by unit vectors I, J and K respectively yieldscomponents of the vector rCB along the initial system of coordinates.

rCBX = rB · I− rC · I = −rCX rCBY = rB · J− rC · J =rBY −rCY =

q 2aZ A − Z 2A−rCY

rCBZ = rB · K− rC · K =a (2.49)

As one can see from Fig. 7 the length of the vector rCB is equal to√

2a. Hence

r2CBX + r2CBY + r

2CBZ = 2a

2 (2.50)


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PROBLEMS. 53

Therefore the second equation for determination of the components rCX and rCY takes form

r2CX

+µq 2aZ A −

Z 2A−

rCY ¶2 + a2 = 2a2 (2.51)

The above equation together with Eq. 2.46 form set of two equations that determinethe components rCX and rCY as explicit functions of time. They are

rCX = a

s 2(a− Z A)2a− Z A

rCY = Z A(a− Z A)p

Z A(2a− Z A)rCZ = 0 (2.52)

These equations allow the trajectory of the point C to be computed. This trajectoryis shown in Fig. 9 for a = 1

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.5 1

r CX

r CY

Figure 9

The unit vector associated with axis x can be produced as a vector productof unit vectors j and k.

i = j × k (2.53)

where

j = I cos∠ jI + J cos∠ jJ + K cos∠ jK

k = I cos∠kI + J cos∠kJ + K cos∠kK (2.54)

Therefore

i =

¯̄̄¯̄̄ I J Kcos∠ jI cos∠ jJ cos∠ jK

0 cos∠kJ cos∠kK

¯̄̄¯̄̄

= I(cos∠ jJ cos∠kK− cos∠ jK cos∠kJ)+J(

−cos∠ jI cos∠kK)

+K(cos∠ jI cos∠kJ)

(2.55)


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PROBLEMS. 54

Hence, the direction cosines between the axis x and axes X Y Z are

cos∠iI = cos∠ jJ cos∠kK

−cos∠ jK cos∠kJ

cos∠iJ = − cos∠ jI cos∠kKcos∠iK = cos∠ jI cos∠kJ (2.56)

B. Linear velocitiesThe linear velocity of the point B can be obtained by diff erentiation of the

position vector rB which according to Eq. 2.40

rB = Jq

2aZ A − Z 2A + Ka (2.57)

Hence

vB = J a− Z Ap 2aZ A − Z 2A

Ż A (2.58)

Similarly, diff erentiation of the vector rC yields velocity of the point C . The positionvector rC according to Eq. 2.52

rC = Ia

s 2(a− Z A)2a− Z A

+ J Z A(a− Z A)p

Z A(2a− Z A)(2.59)

HencevC = IvCX + JvCY (2.60)

where

vCX = −1

2a2

Ż A2a− Z A

√ 2p

(2a2 − 3aZ A + Z 2A)

vCY = Ż Aa2 − 3aZ A + Z 2A

√ Z A

³p (2a− Z A)

´3 (2.61)Angular velocities.

The angular velocity of the link 1 may be obtained from the following vector

relationshipvB = vA + ω × rBA (2.62)

orω × rBA = vB − vA (2.63)

The components of the vector rBA along the inertial system of coordinates are

rBAX = rBA · I =ak · I =acos∠kI =0

rBAY = rBA · J =ak · J =acos∠kJ =rBY =q

2aZ A − Z 2A)rBAZ = rBA · K =ak · K =acos∠kK =a

−Z A (2.64)


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56/195

PROBLEMS. 56

Problem 13

Z z

a

s

1 2 3

l

α

α

Z z y

X Y x

y

O

1

1

1

1

1

32

21

1

P

Figure 10

The base 1 of a robot shown in Fig. 10 rotates about the vertical Z andits angular position is determined by the angle α1. The link 2 can move along the

vertical slide of the base and its relative position is determined by s21. The link 3 ishinged to the link 2 and the angle α32 determines its relative position.

Upon assuming that α1, s21, α32 are given functions of time and a, l, are givenparameters, derive expressions for components of:

1. absolute angular velocity and acceleration of the link 3 along a body 3 system of coordinates.

2. linear velocity of the point P along the same system of coordinates.


57/195

PROBLEMS. 57

Solution.

Z z

a

s

1 2 3

l α

α

Z z

y

X Y x

y

O

1

1

1

1

1

32

21

1

a

l

s 21

r

z

z

y 2 3

3

2 y

P

Figure 11

In Fig. 11, the following systems of coordinates were introduced:XY Z – inertial system of coordinatesx1y1z1 – body 1 rotating system of coordinatesx2y2z2 – body 2 rotating and translating system of coordinatesx3y3z3 – body 3 rotating and translating system of coordinates.The angular velocity of the body 3 is

ω3 = ω1 + ω21 + ω32

Since ω21 = 0 one may obtain the following expression for the angular velocity of thelink 3 .

ω3 = ω1 + ω32 = k2 α̇1 + i3 α̇32 = (k3 cos α32 + j3 sin α32)α̇1 + i3 α̇32

= i3 α̇32 + j3 α̇1 sin α32 + k3 α̇1 cos α32 (2.73)

The angular acceleration of the body 3 can be obtain as vector derivative of ω3 withrespect to time.

ε3 = ω̇3 = ω0

3 + ω3 × ω3 = ω0

3

= i3α̈32 + j3(α̈1 sin α32 + α̇1 α̇32 cos α32) + k3(α̈1 cos α32 − α̇1 α̇32 sin α32)(2.74)

The position vector of the point P , according to Fig. 11 is

r = s21 + a + l = k2s21 + j2a + j3l

= (k3 cos α32 + j3 sin α32)s21 + ( j3 cos α32−

k3 sin α32)a + j3l

= j3(s21 sin α32 + a cos α32 + l) + k3(s21 cos α32 − a sin α32) (2.75)


58/195

PROBLEMS. 58

Its first derivative with respect to time represents the wanted velocity of the point P .

ṙ = r0 + ω3 × r (2.76)

where

r0 = j3(ṡ21 sin α32 + s21 α̇32 cos α32 − aα̇32 sin α32)+k3(ṡ21 cos α32 − s21 α̇32 sin α32 − aα̇32 cos α32) (2.77)

ω3 × r =

¯̄̄¯¯̄

i3 j3 k3α̇32 α̇1 sin α32 α̇1 cos α32

0 s21 sin α32 + a cos α32 + l s21 cos α32 − a sin α32

¯̄̄¯¯̄= i3[(s21 cos α32 − a sin α32)α̇1 sin α32 − (s21 sin α32 + a cos α32 + l)α̇1 cos α32]+ j3[−(s21 cos α32 − a sin α32)α̇32]

+k3[(s21 sin α32 + a cos α32 + l)α̇32] (2.78)

Upon adding the two above expression together, one may obtain components of ve-locity of the point P in the following form

ṙ = i3(−aα̇1 − lα̇1 cos α32) + j3(ṡ21 sin α32) + k3(ṡ21cosα32 + lα̇32) (2.79)


59/195


60/195

PROBLEMS. 60

Solution.

X x

Z Q

P γ

γ

γ

z O

y

Y

β

r

l

1

x

z

y

2

3

1

Z z 1

α =ω t

r P

rQ

rQP

Figure 13

From Fig. 13 one can see that

rQ = rP + rQP (2.80)

The above vectors can be expressed as follows.

rP = i1r (2.81)

rQ = K rQ cos β + I rQ sin β (2.82)

rQP = k2l = i1l cos γ x + j1l cos γ y + k1l cos γ z (2.83)

where γ x, γ y and γ z are angles between the axis z2 and axes of the system of coordi-nates x1y1z1.Introduction of the above expressions into Eq. 2.80 yields

KrQ cos β + IrQ sin β = i1r + i1l cos γ x + j1l cos γ y + k1l cos γ z (2.84)

Multiplication of the equation Eq. 2.84 by the unit vectors i1, j1 and k1 respectively

off ers three scalar equations.

rQ cos α sin β = r + l cos γ x (2.85)

−rQ sin α sin β = l cos γ y (2.86)rQ cos β = l cos γ z (2.87)

Hence

cos γ x = (rQ cos α sin β − r)/l (2.88)cos γ y = (−rQ sin α sin β )/l (2.89)cos γ z = (rQ cos β )/l (2.90)


61/195

PROBLEMS. 61

The direction cosines have to fulfil the following relationship

cos2 γ x + cos2 γ y + cos

2 γ z = 1 (2.91)

Introduction of Eq’s.. 2.88, 2.89 and 2.90 into Eq. (2.91) yields

(rQ cos α sin β − r)2 + (−rQ sin α sin β )2 + (rQ cos β )2 = l2 (2.92)

After simple manipulation, the equation 2.15 takes form 2.93.

r2Q + (−2r cos α sin β )rQ + (r2 − l2) = 0 (2.93)

Hence, the instantaneous position of the point Q is determined by the followingexpression.

rQ = r cos α sin β ± (r2

cos2

α sin2

β − r2

+ l2

)1/2

(2.94)Since axis Z 3 is motionless, the fist derivative with respect to time of the aboveexpression yields absolute velocity of the point Q.

ṙQ = −rα̇ sin α sin β ± r2α̇ cos α sin α sin2 β/(r2 cos2 α sin2 β − r2 + l2)1/2 (2.95)

There are two possible solution. One corresponds to sign ’+’ and the other corre-sponds to sing ’-’.

X

P

Q Z

Z

1 2

O

β

r

l

3

3

Q-

+

Figure 14

The physical interpretation of those two solutions is given in Fig. 14.


62/195

PROBLEMS. 62

Problem 15

Y

Z

z 1

y1

y1

X x1

1

y2

x2

z 1 z 2

y2

2

2

2

3

3

z

x3

y3

P

P

3

ω21t

α

β

A

A

O

O

o1

o2

o3

o2

l

s

r

z 2

Figure 15

A sketch of the Ferris Wheel is shown in Fig. 15. Its base 1 oscillates about thehorizontal axis X of the X Y Z inertial system of coordinates. The instantaneous po-sition of this base is determined by the angular position α. The system of coordinatesx1y1z1 is rigidly attached to the base 1.

The relative angular velocity of the wheel 2 with respect to the base 1 isconstant and is equal to ω21. The system of coordinates x2y2z2 is rigidly attached tothe wheel 2 .

The seat 3 is hinged to the wheel at the point A. The instantaneous position

of the seat 3 with respect to the wheel 2 is determined by the angular displacementβ . The system of coordinates x3y3z3 is rigidly attached to the seat 3 .Produce expression for components of:1. the absolute angular velocity of the seat 3 along the system of coordinates

x3y3z32. the absolute angular acceleration of the seat 3 along the system of coordi-

nates x3y3z33. the absolute linear velocity of the point P along the system of coordinates

x3y3z3Given are: r,l,s,ω21, α(t), β (t).


63/195

PROBLEMS. 63

Solution

Y

Z z 1

y1

y1

X x1

1

y2

x2

z 1

z 2

y 2

2

2

2

3

3

z

x3

y3

P

P

3

ω21t

α

β

A

A

O

O

o1

o2

o3

o2

l

s

r

z 2

o2

Figure 16

The absolute angular velocity of the system of coordinates x1y1z1.

ω1 = i1 α̇ (2.96)

The absolute angular velocity of the system of coordinates x2y2z2.

ω2 = ω1 + ω21 = i1 α̇ + k2ω21 (2.97)

The absolute angular velocity of the system of coordinates x3y3z3

ω3 = ω2 + ω32 = i1 α̇ + k2ω21 + i3 β̇ (2.98)

Since

i1 = i2 cos ω21t− j2 sin ω21t = i3 cos ω21t− ( j3 cos β − k3 sin β )sin ω21t == i3 cos ω21t + j3(− cos β sin ω21t) + k3(sin β sin ω21t) (2.99)

k2 = j3 sin β + k3 cos β

the components of the angular velocity along the system of coordinates are x3y3z3

ω3 = (i3 cos ω21t + j3(− cos β sin ω21t) + k3(sin β sin ω21t)) α̇+ ( j3 sin β + k3 cos β ) ω21 + i3 β̇

= i3

³α̇ cos ω21t + β̇ ́ + j3 (−α̇ cos β sin ω21t + ω21 sin β )

+k3(α̇ sin β sin ω21t + ω21 cos β )

(2.100)


64/195

PROBLEMS. 64

The absolute angular velocity of the seat 3 is equal to ω3.The absolute angular acceleration of the seat is equal to the absolute angular

acceleration of the system of coordinates x3y3z3

ε = ω̇3 = ω0

3 + ω3 × ω3 =

= i3(α̈ cos ω21t− α̇ω21 sin ω21t + β̈ + j3

³−α̈ cos β sin ω21t + α̇β̇ cos β sin ω21t− α̇ω21 cos β cos ω21t + ω21 β̇ cos β ́

+k3

³α̈ sin β sin ω21t + α̇β̇ cos β sin ω21t + α̇ω21 sin β cos ω21t− ω21 β̇ sin β ́(2.101)

According to Fig. 16 the absolute position vector of the point P is

RP = k2l + j2r + k3(−s)= ( j3 sin β + k3 cos β ) l + ( j3 cos β − k3 sin β ) r + k3(−s) (2.102)= i3 (0) + j3 (l sin β + r cos β ) + k3 (l cos β − r sin β − s)

The first derivative of this vector yields the absolute liner velocity of the point P.

vP = ṘP = R0

P + ω3 × RP

= j3

³lβ̇ cos β ́ − rβ̇ sin β ) + k3

³−lβ̇ sin β − rβ̇ cos β ́

+¯̄̄̄̄̄ i3 j3 k3˙

α cos ω21t + ˙

β −˙α cos β sin ω21t + ω21 sin β

˙α sin β sin ω21t + ω21 cos β 0 l sin β + r cos β l cos β − r sin β − s

¯̄̄̄̄̄= i3vPx3 + j3vPy3 + k3vPz3

(2.103)


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PROBLEMS. 65

Problem 16

Y

Z X z

x

y

y

x

x

1 2 3

1

1

2

2

1

1

P

Oo2

O α

β

ω 32

a

b

Figure 17

Fig. 17 shows a diagram of a wheel excavator. Its base 1 rotates about thevertical axis Y of the inertial system of coordinates XY Z . System of coordinatesx1y1z1 is rigidly attached to the base 1. Its instantaneous position is determined bythe angle α. The arm 2 rotates with respect to the base 1 about axis that is parallelto z1. System of coordinates x2y2z2 is rigidly attached to the arm 2. Its relativeangular position is determined by an angle β .

The wheel 3 rotates with respect to the arm 2 about axis that is parallel toz1 with the angular velocity ω32.

Given are: α(t), β (t), ω32(t), a, b.Produce

1. the expressions for the components of the absolute angular velocity of the wheel 3 along system of coordinates x2y2z2.

Answer:

ω3x2 = α̇ sin β ω3y2 = α̇ cos β ω3z2 =³

β̇ + ω32´

2. the expressions for the components of the absolute linear velocity of the point P along system of coordinates x2y2z2.

Answer:vPx2 = 0 vPy2 = −bβ̇ vPz2 = −aα̇− bα̇ cos β 3. the expressions for components of the absolute acceleration of the point P alongsystem of coordinates x2y2z2.

Answer:aPx2 = vPz2 α̇ cos β − vPy2 β̇ aPy2 = v̇Py2 − vPz2 α̇ sin β aPz2 = v̇Pz2 + vPy2 α̇ sin β


66/195

PROBLEMS. 66

Problem 17

1

2

3

4 Z z

r

ω

B β

y

y

l l/2

G

A

3

1 3

1

A

1

Figure 18

Fig. 18 shows the kinematic scheme of a mechanism. Its link 1 rotates with aconstant angular velocity ω

Documents

Mechanics of Rigid Body by Janusz Krodkiewski Melbourne Text Book