mechanics of rigid body

436- 353 MECHANICS 2 UNIT 2 MECHANICS OF A RIGID BODY J.M. KRODKIEWSKI 2008 THE UNIVERSITY OF MELBOURNE Department of ...

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436- 353 MECHANICS 2 UNIT 2

MECHANICS OF A RIGID BODY J.M. KRODKIEWSKI 2008

THE UNIVERSITY OF MELBOURNE Department of Mechanical and Manufacturing Engineering . 1

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

CONTENTS 1 THREE-DIMENSIONAL KINEMATICS OF A PARTICLE. 1.1 MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME. 1.1.1 Absolute linear velocity and absolute linear acceleration 1.2 MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Motion in terms of translating system of coordinates. . 1.2.2 Motion in terms of rotating system of coordinates. . . 1.2.3 Motion in terms of translating and rotating system of coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 6 6 8 8 10 22 24

2 THREE-DIMENSIONAL KINEMATICS OF A RIGID BODY 2.1 GENERAL MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 ROTATION ABOUT A POINT THAT IS FIXED IN THE INERTIAL SPACE (ROTATIONAL MOTION) . . . . . . . . . . . . . . . . . . . 2.3 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41 41

3 KINETICS OF SYSTEM OF PARTICLES. 3.1 MOTION OF CENTRE OF MASS - LINEAR MOMENTUM. . . . . 3.2 MOMENT OF MOMENTUM. . . . . . . . . . . . . . . . . . . . . . 3.2.1 Moment of momentum about a fixed point in the inertial space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Moment of momentum about a moving point in an inertial space. . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Moment of relative momentum. . . . . . . . . . . . . . . 3.3 EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS. . . 3.3.1 Conservation of momentum principle. . . . . . . . . . . . 3.3.2 Conservation of angular momentum principle. . . . . . 3.3.3 Impulse – momentum principle. . . . . . . . . . . . . . . 3.4 PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80 82 83

4 KINETICS OF RIGID BODY. 4.1 LINEAR AND ANGULAR MOMENTUM. . 4.2 PROPERTIES OF MATRIX OF INERTIA. 4.2.1 Parallel axis theorem. . . . . . . . 4.2.2 Principal axes. . . . . . . . . . . .

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43 44

83 85 87 88 88 89 90 91

96 . 96 . 99 . 100 . 102

CONTENTS

4

4.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 4.3 KINETIC ENERGY. . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 4.3.1 Rotational motion. . . . . . . . . . . . . . . . . . . . . . . . 127 4.3.2 General motion. . . . . . . . . . . . . . . . . . . . . . . . . 128 4.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 4.4 EQUATIONS OF MOTION . . . . . . . . . . . . . . . . . . . . . . . 146 4.4.1 Euler’s equations of motion . . . . . . . . . . . . . . . . . 146 4.4.2 Modified Euler’s equations of motion . . . . . . . . . . . 149 4.4.3 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 4.5 MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 4.5.1 Modelling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 4.5.2 Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 5 APPENDIXES 191 5.1 APPENDIX 1. REVISION OF THE VECTOR CALCULUS 191 5.2 APPENDIX 2. CENTRE OF GRAVITY, VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES. . . . . . . . . 194

CONTENTS

5

INTRODUCTION. The purpose of this text is to provide the students with the theoretical background and engineering applications of the three dimensional mechanics of a rigid body. It is divided into four chapters. The first one, Three-Dimensional Kinematics of A Particle, deals with the geometry of motion of an individual particle in terms of the inertial as well as in terms of the non-inertial system of coordinates. The introduced in this chapter translating, rotating as well as the translating and rotating system of coordinates allows motion of the rigid body with respect to the inertial frame to be determined and classified. The second chapter, entitled Three-Dimensional Kinematics of A Rigid Body, 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 and the motion about a fixed point is considered. The last two chapters are related with the relationships between motion and forces that act on bodies. The chapter Kinetics of A System of Particles oﬀers general principles that can be apply to any system of particles regardless of their number and internal forces acting between the individual particles. Because each continuum (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 expression for the kinetic energy of the moving rigid body as well as to derive its equations of 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 some of them is provided. Students should produce solution to the other problems during tutorials and in their own time.

Chapter 1 THREE-DIMENSIONAL KINEMATICS OF A PARTICLE. 1.1

MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME. Z Za

v O

Y

X Oa

Ya

Xa

Figure 1 To consider motion of a particle we assume the existence of so-called absolute (motionless) system of coordinates Xa Ya Za (see Fig. 1). DEFINITION: Inertial system of coordinated is one that does not rotate and which origin is fixed in the absolute space or moves along straight line at a constant velocity. Inertial systems of coordinates are usually denoted by upper characters, e.g. XY Z, to distinguish them from non-inertial systems of coordinates which are usually denoted by 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 equations 1.1 which determine the particle coordinates for any instant of time. rX = rX (t) rY = rY (t) rZ = rZ (t)

(1.1)

MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME.

7

Z

K I

r(t)

∆r

r(t+∆ t) J rZ (t) O Y rX (t) rY (t)

X

Figure 2 These coordinates represent scalar magnitude of components of so called absolute 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 the absolute velocity, as the first derivative of the absolute position vector r with respect to time, is given by the following formula. ∆r ∆rX ∆rY ∆rZ = I lim + J lim + K lim = Ir˙X + Jr˙Y + Kr˙Z ∆t→O ∆t ∆t→O ∆t ∆t→O ∆t ∆t→O ∆t (1.3) Similarly, vector of the absolute acceleration is defined as the second derivative of the position vector with respect to time. v = r˙ = lim

∆v = I¨ rX + J¨ rY + K¨ rZ ∆t→O ∆t

a=¨ r = lim

Scalar magnitude of velocity v (speed) can be expressed by formula q √ 2 v = |v| = v · v = r˙X + r˙Y2 + r˙Z2

(1.4)

(1.5)

Scalar magnitude of acceleration is a = |a| =

q √ 2 a · a = r¨X + r¨Y2 + r¨Z2

(1.6)

The distance done by the particle in a certain interval of time 0 < τ < t is given by the formula 1.7. Z tp Z t v dτ = r˙X (τ )2 + r˙Y (τ )2 + r˙Z (τ )2 dτ = s(t) (1.7) s= 0

0

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

1.2

8

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

Z

Z

z

Z

z

z o

o O

y

y

y

O

Y

O

o

Y

Y

x x X

x

a

c

b X

X

Figure 3 DEFINITION:System of coordinates which can not be classified as inertial is 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 (its axes 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 origin of 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 is called translating and rotating system of coordinates (Fig.3c) 1.2.1 Motion in terms of translating system of coordinates.

z

Z rP ro O

P rP,o

o

y Y

x X Figure 4

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

9

Let XY Z be the inertial system of coordinates and xyz be the translating system of coordinates. The relative motion of xyz system with respect to XY Z is usually defined 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 given by the position vector rP,o . The position vector of the particle P in the XY Z frame can 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 with respect to time is ˙ oX + Jr ˙ oY + Kr ˙ oZ r˙ P = Ir˙oX + Jr˙oY + Kr˙oZ + Ir ˙ P,oy + kr ˙ P,oz +ir˙P,ox + jr˙P,oy + kr˙P,oz + ˙irP,ox + jr

(1.9)

˙ = ˙i = j˙ = k˙ = 0 Since, I˙ = J˙ = K r˙ P = Ir˙oX + Jr˙oY + Kr˙oZ + ir˙P,ox + jr˙P,oy + kr˙P,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 it would 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 have if the non-inertial system of coordinates would be motionless (ro = const). According to the above definitions the velocity of transportation in the case considered is vT = r˙ o = Ir˙oX + Jr˙oY + Kr˙oZ (1.11) and the relative velocity is vR = ir˙P,ox + jr˙P,oy + kr˙P,oz

(1.12)

vP = r˙ P = vT + vR

(1.13)

Hence, the absolute velocity

Similarly, one can prove that the absolute acceleration of the particle P is rP = aT + aR aP = ¨

(1.14)

where aT and aR stand for the acceleration of transportation and the relative acceleration respectively. DEFINITION: Acceleration of transportation is the acceleration of the particle it would have if it would be motionless with respect to the noninertial frame (rP,o = const)

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

10

DEFINITION: Relative acceleration is the acceleration of the particle it would have if the non-inertial system of coordinates would be motionless (ro = const). In the case considered aT = ¨ ro = I¨ roX + J¨ roY + K¨ roZ

(1.15)

aR = i¨ rP,ox + j¨ rP,oy + k¨ rP,oz

(1.16)

1.2.2 Motion in terms of rotating system of coordinates. Z z K k

j O

I

o

y J

Y

i X

x

Figure 5 As it was mention before, the rotating system of coordinates has its origin coinciding the origin of an inertial system of coordinates (Fig.5). First of all, we have to establish matrix which transfers components of a vector from rotating system of coordinates xyz to the inertial XY Z. Matrix of direction cosines. Let the relative motion of a particle P be determined by a position vector r, which components 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 be obtained as scalar products of the vector r and unit vectors IJK. rX = = rY = = rZ = =

r · I = rx i · I + ry j · I + rz k · I rx cos ∠iI + ry cos ∠jI + rz cos ∠kI r · J = rx i · J + ry j · J + rz k · J rx cos ∠iJ + ry cos ∠jJ + rz cos ∠kJ r · K = rx i · K + ry j · K + rz k · K rx cos ∠iK + ry cos ∠jK + rz cos ∠kK

(1.18)

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

Z

z

11

P r rz

K

y

rZ

k

j

kI O I

rx

J o

Y

jI

i

rX

ry

iI rY

X

x Figure 6

The last relationship can be written in the a matrix form ⎡

⎤ ⎡ ⎤ ⎤ ⎡ ⎤⎡ rX rx cos ∠iI cos ∠jI cos ∠kI rx ⎣ rY ⎦=⎣ cos ∠iJ cos ∠jJ cos ∠kJ ⎦ ⎣ ry ⎦ = [Cr→i ] ⎣ ry ⎦ rZ rz rz cos ∠iK cos ∠jK cos ∠kK

(1.19)

The transfer matrix [Cr→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. [Cr→i ]−1 = [Cr→i ]T = [Ci→r ]

(1.20)

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

Jx iJ iY

J

Figure 7

cos ∠iJ =

Jx iY = iY = = Jx i J

(1.21)

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

12

Z i iZ O i Y

iX

Y

X

Figure 8 From Fig. 8 one can see that components iX, iY, iZ ,which are equal to corresponding direction cosines, fulfil the following relationship. i2X + i2Y + i2Z = cos2 ∠iI + cos2 ∠iJ + cos2 ∠iK = 1

(1.22)

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

Z ,z

O,o

Y, y

X, x Figure 9 Let us assume that the rotating system of coordinates xyz coincide the inertial one XY Z. as shown in Fig.9. Now, let us turn the system of coordinates xyz with respect to the inertial one XY Z about axis Z by an angle φ, so the system xyz takes position x1 y1 z1 (Fig. 10). The matrix of direction cosines between system x1 y1 z1 and XY Z is

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

13

Z ,z1 y1

φ

O,o

Y

φ x1

X

Figure 10 ⎡

⎤ cos φ − sin φ 0 [Cr1 →i ] = ⎣ sin 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 new position of the system xyz is shown in Fig. 11 as x2 y2 z2 . The matrix of direction cosines between system x2 y2 z2 and x1 y1 z1 has the following form.

θ

z2

Z ,z1 y2

θ φ

O,o

y1

Y

φ x1 ,x 2

X

Figure 11 ⎡

⎤ 1 0 0 [Cr2 →r1 ] = ⎣ 0 cos θ − sin θ ⎦ 0 sin θ cos θ

(1.24)

In the last step, the system xyz is turned by an angle ψ about axis z2 to its final position xyz (Fig. 12). The matrix of direction cosines between xyz and x2 y2 z2 is

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

θ

z2 ,z

14

y

Z ,z1

Ψ

y2

θ φ

O,o

y1

Y

φ Ψ

x

x2

X

Figure 12 ⎡

⎤ cos ψ − sin ψ 0 [Cr→r2 ] = ⎣ sin ψ cos ψ 0 ⎦ 0 0 1

According to Eq. 1.19, one can write the following relationships ⎤ ⎤ ⎡ ⎡ rx1 rX ⎣ rY ⎦ = [Cr1 →i ] ⎣ ry1 ⎦ rZ rz1 ⎤ ⎤ ⎡ ⎡ rx2 rx1 ⎣ ry1 ⎦ = [Cr2 →r1 ] ⎣ ry2 ⎦ rz1 rz2 ⎤ ⎤ ⎡ ⎡ rx rx2 ⎣ ry2 ⎦ = [Cr→r2 ] ⎣ ry ⎦ rz2 rz

(1.25)

(1.26)

(1.27)

(1.28)

Introducing Eq. 1.28 into Eq. 1.27) and than Eq. 1.27) into Eq. 1.26) one may obtained ⎤ ⎤ ⎤ ⎡ ⎡ ⎡ rx rx rX ⎣ rY ⎦ = [Cr1 →i ][Cr2 →r1 ][Cr→r2 ] ⎣ ry ⎦ = [Cr→i ] ⎣ ry ⎦ (1.29) rZ rz rz

Hence the matrix of direction cosines between rotating system xyz and the inertial XY Z is [Cr→i ] = [Cr1 →i ][Cr2 →r1 ][Cr→r2 ] (1.30) The last formula allows to express the direction cosines as function of three independent angles known as Euler’s angles. The angle φ is called angle of precession, angle θ is called angle of mutation and the angle ψ is called angle of spin.

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

15

Angular velocity and angular acceleration.

A1

Z

Z A2 r2

r1 r0

r2 A2

A0 O

Y

O

r0 r1

A0 A1

Y

b

a X

X

Figure 13 The introduced Euler angles can not be considered as vectorial values. To show it let us consider transformation of point Ao due to rotation about axes x and y by 90o . If we turn the vector ro by 90o first 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, angular displacement can not be considered as a vector because at least the commutative law would be violated.

dφ

φ

dφ

Figure 14 It is easy to show, but the proof is here omitted, that the infinitesimal angular displacement can be considered as vectorial magnitude. Vector of the infinitesimal angular displacement is perpendicular to the plane of rotation and its sense is determined by the law of right-handed screw (Fig. 14). Such vectors which have determined only direction and sense are called free vectors to distinguish them from linear vectors (sense and line of action is determined) and position vectors ( position of its tail, direction and sense is determined). According to the above rules the infinitesimal angular increments of Euler’s angles dφ, dψ and dθ may be drawn as shown in Fig. 15.

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

16

The instantaneous axis of rotation

z

Z dψ

dα dψ

dθ

dφ y O

Y dθ x

X

x1 Figure 15

The vectorial sum (1.31)

dα = dφ + dψ + dθ

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 by the angle dα about this axis is equivalent to rotation by angles dφ, dθ, dψ about axes Z x1 z respectively. The instantaneous axis go through origin of the rotating system of coordinates. Z

z

l dα dα h r

β

dr A

y

O Y x X

Figure 16 Now, let us consider point A fixed in the xyz system of coordinates determined by 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

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

17

angular displacement of the system of coordinates xyz. Hence, the point A at the instance considered moves along circle of radius (1.32)

h = r sinβ

The infinitesimal increment of vector r is tangential to the circle and is placed in plane 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 vector dα and r. dr = dα × r (1.34) The velocity of the point A is v= The vector

dα dt

dr dα = ×r dt dt

(1.35)

is called vector of angular velocity ω. ω=

dα dt

(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, its angular velocity, according to the above definition and Eq. ??, is ω=

dθ dα Kdφ + i1 dθ + kdψ dφ dψ = =K + i1 + k dt dt dt dt dt

The angular speed as well as its individual components are shown in Fig. 17. The instantaneous axis of rotation

z

Z ω

dφ dt

dψ dt

y O

Y

dθ dt X

x x1

Figure 17

(1.38)

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

18

The vector ω can be resolved along any system of coordinates. In particular it can be resolved along axes of the inertial system of coordinates XY Z (1.39)

ω = Iω X + JωY + KωZ

The scalar magnitude of the components ω X, ωY, ω Z can be expressed as functions of Euler’s angles and their first derivatives. dθ dφ dψ + I · i1 + I · k dt dt dt dθ dψ dφ = J · K + J · i1 + J · k dt dt dt dθ dψ dφ = K · K + K · i1 + K · k dt dt dt

ωX = I · K ωY ωZ

(1.40)

Having the components of angular speed as explicit function of time it is possible to find 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 called space cone (Fig. 18 a) space cone

instantaneous axis of rotation

body cone

Z

Z t0

ti

z

z

ωi

t0

Y

ti

ωi

t0

O

O

X

x

a

b

X

ωi

y Y

o

y

o

ti

x

c

Figure 18 The vector of angular speed ω can be as well resolved along axes of the system of coordinates xyz (1.41) ω = iωx + jωy + kωz and the components can be expressed as a function of time. dθ dψ dφ + i · i1 + i · k dt dt dt dψ dφ dθ = j · K + j · i1 + j · k dt dt dt dθ dφ dψ = k · K + k · i1 + k · k dt dt dt

ωx = i · K ωy ωz

(1.42)

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

19

The components ω x, ωy, ω z determine for any instant of time direction of the angular velocity in the rotating system of coordinates. The locus of these lines make up so called body cone (Fig. 18 b). The rotational motion of the system xyz with respect to the XY Z, can be hence considered as rolling without slipping of the body cone on the 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=

dω dt

(1.43)

= ω˙

Derivative of a vector expressed in terms of a rotating system of coordinates 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 defined by three independent angles (e.g. Euler angles) or, alternatively the rotational motion can 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 coordinates xyz (Fig. 19). z Z Az

A Ay

ω Ax

y

o

Y O

X

x

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 diﬀerentiate this vector with respect to time. ˙ = d (iAx + jAy + kAz ) A dt

(1.45)

˙ y + kA ˙ z ˙ = iA˙ x + jA˙ y + kA˙ z + ˙iAx + jA A

(1.46)

Hence The first three terms represent vector which can be obtained by direct diﬀerentiating of the components Ax , Ay , Az with respect to time. This vector will be denoted by A0 . A0 = iA˙ x + jA˙ y + kA˙ z (1.47)

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

20

Thus ˙ y + kA ˙ z ˙ = A0 + ˙iAx + jA A

(1.48)

According to definition of vector derivative, the first derivative of the unit vector i is the ratio of the infinitesimal vector increment di and dt (Fig. 20). z

ω

Z

y O Y

i vi X

x

Figure 20 ˙i = di = vi dt = vi dt dt were vi is a velocity of head of the vector i. But, according to Eq. 1.37

(1.49)

vi = ω × i

(1.50)

˙i = ω × i

(1.51)

Hence Similarly j˙ = ω × j

and

k˙ = ω × k

(1.52)

Introducing the above expressions into Eq. 1.48 we have ˙ = A0 + ω × iAx + ω × jAy + ω × kAz = A0 + ω × (iAx + jAy + kAz ) A

(1.53)

and eventually one may gain ˙ = A0 + ω × A A

(1.54)

where: A0 = iA˙ x + jA˙ y + kA˙ z ω - 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 diﬀerentiated vector. The last formula provides the rule for diﬀerentiation of a vector that it is resolved along a non-inertial system of coordinates. It can be applied to any vector (eg position vector, velocity, angular velocity etc.)

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

21

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 = irP x + jrP y + krP z

(1.55)

where rP x , rP y , rP z – 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 acceleration of this particle. z

P

Z ω

rP

y o

X

O

Y

x

Figure 21 The absolute velocity of the particle P is vP = r˙ P = r0P + ω × rP

(1.56)

In a similar manner, as it was done in case of translating system of coordinates, we introduce notions of the relative velocity and the velocity of transportation. For the case 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. 1.56 the relative velocity is (1.58) vR = r0P Hence vP = vR + vT

(1.59)

The absolute acceleration one may obtain diﬀerentiating equation 1.56 with respect to time. ¨ rP =

d 0 d rP + (ω × rP ) = r00P + ω × r0P + ω˙ × rP + ω × (r0P + ω × rP ) dt dt

(1.60)

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

22

Introducing Eq. 1.43 and developing the last term we have aP = ¨ rP = r00P + ε × rP + ω × (ω × rP ) + 2 ω × r0P

(1.61)

Assuming that the point P is motionless (rP = constant) one may obtain the following expression 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 term ω × (ω × rP ) is called normal acceleration of transportation. Assumption, that the system of coordinates is motionless yields expression for the 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 transportation, relative acceleration and Coriolis acceleration. (1.65)

aP = aT + aR + aC

1.2.3 Motion in terms of translating and rotating system of coordinates. Let us define motion of the translating and rotating system of coordinates xyz by the position vector ro and the vector of the angular velocity ω (see Fig..22). The relative 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 position vector rP in this case is rP = ro + rP,o (1.66) z P Z rP,o

rP ro

y

o

O

ω Y

x X

Figure 22

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

23

Let us assume that the relative position vector rP,o is determined by its components 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 r˙ P = r˙ o + r0P,o + ω × rP,o where r˙ o + ω × rP,o – velocity of transportation r0P,o – relative velocity The second derivative yields the absolute acceleration ro + r00P,o + ω × r0P ,o + ε × rP,o + ω × (r0P ,o + ω × rP,o ) ¨ rP = ¨ = ¨ ro + ε × rP,o + ω × ω × rP,o + r00P,o + 2 ω × r0P ,o Here: ¨ ro + ε × rP,o + ω × ω × rP,o = aT – acceleration of transportation r00P,o = aR – relative acceleration 2 ω × r0P ,o = aC – Coriolis acceleration.

(1.68)

PROBLEMS.

1.3

24

PROBLEMS.

Problem 1

Z z1 z2 2

1

ω1 ω21 y1 y2

X ω1 t x1

Y

ω2 1t x

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 relative angular velocity ω 21 . Calculate components of the absolute angular velocity ω2 of the disc 2 and its absolute acceleration ε2 . Solution. System of coordinates x1 y1 z1 is rigidly attached to the frame 1 and system x2 y2 z2 is body 2 system of coordinates. Absolute angular velocity of the body 2 is ω2 = k1 ω1 + j1 ω21

(1.69)

Its first vector derivative with respect to time yields absolute angular acceleration. ¯ ¯ ¯ i1 j1 k1 ¯ ¯ ¯ (1.70) ε2 = ω˙ 2 = ω02 + ω 1 × ω2 = k1 0 + j1 0 + ¯¯ 0 0 ω1 ¯¯ = −i1 ω 1 ω21 ¯ 0 ω21 ω1 ¯

PROBLEMS.

25

Problem 2

O o1 . α

ωo t z2

X

y1 Y

x1 x2

P

Z z1 y2

α

ωo y1

l

2 1

O

Figure 24 A radar antenna rotates about the vertical axis Z at the constant angular speed ω o . At the same time the angle α is being changed as follow α = a sin At Produce the components of the angular velocity and the angular acceleration of the antenna 1. along the inertial XY Z system of coordinates 2. the body 2 system of coordinates x2 y2 z2 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 .

PROBLEMS.

26

Solution. A. Matrices of direction cosines. Fig. 24 permits to produce matrices of direction cosines between the inertial system of coordinates XY Z, the rotating system of coordinates x1 y1 z1 and the rotating system of coordinates x2 y2 z2 . ⎤⎡ ⎤ ⎡ ⎡ ⎤ x1 X cos ω o t sin ω o t 0 X y1 = ⎣ − sin ω o t cos ω o t 0 ⎦ ⎣ Y ⎦ = [CI→1 ] ⎣ Y ⎦ (1.71) 0 0 1 Z z1 Z ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎤ ⎡ x1 1 0 0 x1 x2 ⎣ y2 ⎦ = ⎣ 0 cos α sin α ⎦ ⎣ y1 ⎦ = [C1→2 ] ⎣ y1 ⎦ (1.72) 0 − sin α cos α z2 z1 z1 ⎤ ⎡ ⎤ x2 X ⎣ y2 ⎦ = [C1→2 ][CI→1 ] ⎣ Y ⎦ z2 Z ⎡ ⎤⎡ 1 0 0 cos ω o t ⎣ ⎦ ⎣ − sin ωo t 0 cos α sin α = 0 0 − sin α cos α ⎡ sin ω o t cos ω o t ⎣ − sin ω o t cos α cos ω o t cos α = sin ωo t sin α − cos ω o t sin α ⎡

⎤⎡ ⎤ sin ω o t 0 X cos ωo t 0 ⎦ ⎣ Y ⎦ 0 1 Z ⎤⎡ ⎤ 0 X ⎦ ⎣ sin α Y ⎦ cos α Z

(1.73)

B. Angular velocities. The angular velocity of the body 2 is determined by the following vector equation. ω 2 = ω o k1 + αi ˙ 1 (1.74) Its components along the system of coordinates x2 y2 z2 may be calculated with help of Eq. 1.72. ⎤ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤⎡ ⎤⎡ ω2x2 1 0 0 ω 2x1 1 0 0 α˙ ⎣ ω 2y2 ⎦ = ⎣ 0 cos α sin α ⎦ ⎣ ω2y1 ⎦ = ⎣ 0 cos α sin α ⎦ ⎣ 0 ⎦ ω2z2 ω 2z1 0 − sin α cos α 0 − sin α cos α ωo ⎤ ⎡ α˙ ⎣ sin α ⎦ ω = (1.75) o ωo cos α

Since

α = asinAt

(1.76)

ω2x2 = aA cos At ω2y2 = ω o sin(a sin At) ω 2z2 = ω o cos(a sin At)

(1.77)

the above components are

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 angular velocity ω2 along inertial system of coordinates. These components may be obtained with help of the equation 1.71. ⎤⎡ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ α˙ ω2X ω2x1 cos ωo t − sin ω o t 0 ⎣ ω 2Y ⎦ = [CI→1 ]−1 ⎣ ω 2y1 ⎦ = ⎣ sin ωo t cos ω o t 0 ⎦ ⎣ 0 ⎦ 0 0 1 ωo ω 2Z ω2z1 ⎡ ⎤ α˙ cos ω o t = ⎣ α˙ sin ω o t ⎦ (1.78) ωo

Introduction of Eq. 1.76 into Eq. 1.78 yields the parametric equations of the space cone. ω2X = aA cos At cos ω o t ω 2Y = aA cos At sin ω o t ω 2Z = ω o

(1.79)

Both, the space cone and body cone, are presented in Fig. 25. They were computed for the following data a = 1[m],

ω o = 1[1/s],

A = 3[1/s] y2

Y 2

2

1

1

X

0

-3

-2

-1

-1

-2

-2

0

1

2

x2

0

-1

3

-3

Z

-2

-1

0

1

2

3

z2

x2

X

a)

b) Figure 25

Magnitude of the absolute angular velocity ω2 is q q q 2 2 2 2 2 2 ω2 = ω 2x2 + ω2y2 + ω2z2 = α˙ + (ω o sin α) + (ωo cos α) = α˙ 2 + ω2o C. Angular acceleration.

(1.80)

PROBLEMS.

28

The absolute angular acceleration of the antenna can be obtained by diﬀerentiation of the vector ω2 Eq. 1.75. ¨ + j2 ω o α˙ cos α − k2 ω o α˙ sin α ε2 = ω˙ 2 = ω02 + ω 2 × ω 2 = i2 α

(1.81)

Hence, its magnitude is q p ¨ 2 + (ωo α˙ cos α)2 + (ω o α˙ sin α)2 = α ¨ 2 + ωo α˙ 2 ε2 = α

(1.82)

where

α˙ = aA cos At α ¨ = −aA2 sin At

(1.83)

D. Velocity of the point P . Position of the point P is determined by the position vector l. (1.84)

l = k2 l

Its first derivative with respect to time yields the absolute velocity of the point P . ¯ ¯ ¯ i2 ¯ j2 k2 ¯ ¯ 0 (1.85) vP = ˙l = l + ω2 × l = ¯¯ α˙ ωo sin α ω o cos α ¯¯ = i2 ωo l sin α − j2 lα˙ ¯ 0 ¯ 0 1 Hence, magnitude of velocity of the point P is p ˙ 2 vP = (ω o l sin α)2 + (lα)

(1.86)

where α and α˙ 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 diﬀerentiation of the vector of its absolute velocity. aP = v˙ P = vP0 + ω 2 × vP

¯ ¯ i2 j2 k2 ¯ α˙ ω o sin α ωo cos α = i2 (ω o lα˙ cos α) − j2 (l¨ α) + ¯¯ ¯ ω o l sin α −lα˙ 0 = i2 (ω o lα˙ cos α + ω o lα˙ cos α) α + ω 2o l sin α cos α) +j2 (−l¨ +k2 (−lα˙ 2 − ω 2o l sin2 α) = i2 aPx2 + j2 aPy2 + k2 aPz2

Hence, its magnitude is aP =

q a2Px2 + a2Py2 + a2Pz2

¯ ¯ ¯ ¯ ¯ ¯ (1.87)

(1.88)

PROBLEMS.

29

Problem 3 3

4

ω

L v

β 2

1

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 with the constant angular speed ω = 1rad/s whereas the belt 3 moves with the constant linear velocity v = 2m/s in the direction shown. Calculate magnitude of velocity and acceleration of the particle 4 travelling without slipping for the position defined by the distance L = 5m.

PROBLEMS.

30

Solution.

O

X ω x t

Y

Z z

3

4

y v

ω L R O

β 2

1

Figure 27 Axes XY Z form the inertial system of coordinates. Axes xyz are fixed to the 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 particle 4. ¯ ¯ ¯ ¯ i j k ¯ ¯ 0 ˙ = R + ω 2 × R = jL˙ + ¯ 0 ω sin β ω cos β ¯ = −iLω cos β + jv (1.91) R ¯ ¯ ¯ ¯ 0 L 0

Hence, magnitude of the velocity is p p ˙ = (Lω cos β)2 + v2 = (5 · 1 · cos30o )2 + 22 = 4.76 m/s |R|

(1.92)

The first derivative of the absolute velocity yields the absolute acceleration of the particle. ¯ ¯ ¯ ¯ i j k ¯ ¯ 0 ¯ ˙ = −ivω cos β + ¯ ¨ = R ˙ + ω2 × R 0 ω sin β ω cos β R ¯ ¯ ¯ ¯ −Lω cos β v 0 = −i2vω cos β − jω 2 L cos2 β + kLω 2 sin β cos β

(1.93)

PROBLEMS.

31

Its magnitude may be calculated according to the following formula. p ¨ = (2ωv cos β)2 + (Lω 2 cos2 β)2 + (Lω 2 sin β cos β)2 |R| p = (2 · 1 · 2 · cos 30o )2 + (5 · 12 · cos2 30o )2 + (5 · 12 sin 30o cos30o )2 = 5.545 m/s2 (1.94)

PROBLEMS.

32

Problem 4

ω

2

1 L A

C

A

ρ

D

α

ρ

Figure 28 Wheel of radius ρ is free to rotate about axle CD which turns about the vertical axis 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.

PROBLEMS.

33

Solution. O

z1

ωt Z Y

y1 x1

X

1 L

ω2

ω 21

C O

2

A

ω1

β

x1

x2

F D L

β

ρ E

G

z1 z2

y1

ρ

α o

y2

Figure 29 Axes XYZ, in Fig. 29, form the inertial system of coordinates. Axis x1 y1 z1 are rigidly attached to the axle 1 and form the body 1 system of coordinates. Its axis x1 coincides axis X of the inertial system of coordinates. Therefore the angular speed of this system of coordinates is (1.95)

ω1 = Iω = i1 ω

Axes x2 y2 z2 are fixed to the wheel 2 and its axis x2 goes through the point A whereas its axis z2 coincides axis z1 . Its absolute angular velocity ω 2 is assembled of the absolute angular velocity ω 1 and the relative velocity ω21. (1.96)

ω 2 = ω1 + ω 21

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 the cone CEG may be considered as the space cone, the absolute angular velocity of the body 2 mast have direction of the common generating line EC. Therefore ω21 = ω 1 cot β = ω1

L ρ

(1.97)

Since the vector ω 21 has opposite direction to the positive direction of axis z1 , vector of the absolute angular velocity ω 2 is L ω 2 = i1 ω + k1 (−ω ) ρ

(1.98)

PROBLEMS.

34

Position vector of the point A is r = L + ρ = k1 L + i2 ρ

(1.99)

Components of the above position vector along system of coordinates x1 y1 z1 are as follows rx1 = r · i1 = k1 ·i1 L + i2 ·i1 ρ = ρ cos α ry1 = r · j1 = k1 ·j1 L + i2 ·j1 ρ = ρ sin α rz1 = r · k1 = k1 ·k1 L + i2 ·k1 ρ = L

(1.100)

Absolute velocity of the point A as the first derivative of the vector r is ¯ ¯ ¯ i1 j1 k1 ¯¯ ¯ 0 0 ¯¯ vA = r˙ = r0 + ω1 × r = −i1 ρα˙ sin α + j1 ρα˙ cos α + ¯¯ ω ¯ ρ cos α ρ sin α L ¯ = i1 (−ρα˙ sin α) + j1 (ρα˙ cos α − ωL) + k1 (ωρ sin α) (1.101) Magnitude of the absolute velocity is p |vA | = |˙r| = (−ρα˙ sin α)2 + (ρα˙ cos α − ωL)2 + (ωρ sin α)2

(1.102)

Similarly, the first derivative of the absolute velocity yields the absolute acceleration of the point A. ¯ ¯ ¯ i1 ¯ j k 1 1 ¯ ¯ 0 ¯ 0 0 ¯¯ aA = v˙ A = vA + ω 1 × vA = i1 v˙ Ax1 + j1 v˙ Ay1 + k1 v˙ Az1 + ¯ ω ¯ v Ax1 vAy1 v Az1 ¯ = i1 (v˙ Ax1 ) + j1 (v˙ Ay1 − ωvAz1 ) + k1 (v˙ Az1 + ωvAy1 ) (1.103) where, according to (1.101) vAx1 = −ρα˙ sin α vAy1 = ρα˙ cos α − ωL vAz1 = ωρ sin α Hence, the magnitude of absolute acceleration is q aA = (v˙ Ax1 )2 + (v˙ Ay1 − ωvAz1 )2 + (v˙ Az1 + ωvAy1 )2

(1.104)

(1.105)

PROBLEMS.

35

Problem 5

P

β ω

Ω l

Figure 30 A crane shown in Fig. 30 is revolving about vertical axis with the constant angular speed ω = 1rad/s in the direction shown. Simultaneously the boom is being lowered at the constant angular speed Ω = 0.5rad/s. Calculate the magnitude of the velocity and acceleration of the end P of the boom for the instant when it passes the position β = 30o . The boom has the length l = 10m. Answer: v = 7.07m/s

PROBLEMS.

36

Problem 6 0 y2 X

α

ωb

1 2 x1 z3

t

M 3

v

x2 x 3 z1 z2

ωt

Y y1

αb

o1

y3

y2 l

Figure 31 The turret on a tank (see Fig. 31) rotates about the vertical axis at angular speed ω t and the barrel is being raised at a constant angular speed ωb . The tank has constant forward speed v. When the barrel is in position defined by angles αt and αb a shell leaves the barrel with muzzle velocity vs and acceleration as . Determine the absolute velocity vm and acceleration am of the barrel muzzle as well as absolute velocity 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 the barrel rMbx2 = vt sin αt rMby2 = vt cos αt + l cos αb rMbz2 = l sin αb where l = constant The components of the absolute velocity of the point M that belong to the barrel vMbx2 = r˙Mbx2 −ωt l cos αb −ωt vt cos αt vMby2 = r˙Mby2 +ωt vt sin αt vMbz2 = r˙Mbz2 The components of the absolute acceleration of the point M that belong to the barrel aMbx2 = v˙ Mbx2 − ω t vMby2 aMby2 = v˙ Mby2 + ω t vMbx2 aMbz2 = v˙ Mbz2 The components of the absolute position vector of the point M that belong to the shell rMsx2 = vt sin αt rMsy2 = vt cos αt + l(t) cos αb rMsz2 = l(t) sin αb ˙ ¨ where l = vs l = as The components of the absolute velocity of the point M that belong to the shell vMsx2 = r˙Msx2 −ω t l cos αb −ω t vt cos αt vMsy2 = r˙Msy2 +ωt vt sin αt vMsz2 = r˙Msz2 The components of the absolute acceleration of the point M that belong to the barrel aMsz2 = v˙ Msz2 aMsx2 = v˙ Msx2 − ω t vMsy2 aMsy2 = v˙ Msy2 + ωt vMsx2

PROBLEMS.

37

Problem 7 O y

α X

z1

Y x

ω 21 t P

R O

Z z

ω 21

z1 l

y1

x1

γ y

O 2 1

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 the angle α. The link 2 rotates with respect to the link 1 with the constant relative angular velocity ω21 . Point P belongs to the body 2. Given are: l, R, ω 21 , γ, α(t) Produce the expression for: 1. the components of the linear absolute velocity of the point P along the body 1 system of coordinates x1 y1 z1 . Answer: vx1 = Rω 21 cos ω21 t + Rα˙ sin γ cos ω21 t − lα˙ cos γ vy1 = Rα˙ cos γ sin ω 21 t vz1 = −Rω 21 sin ω 21 t − Rα˙ sin γ sin ω 21 t 2. the components of the absolute acceleration of the point P along the body system of coordinates x1 y1 z1 . Answer: ax1 = v˙ x1 + vz1 α˙ sin γ − vy1 α˙ cos γ ay1 = v˙ y1 + vx1 α˙ cos γ az1 = v˙ z1 − vx1 α˙ sin γ

PROBLEMS.

38

Problem 8

Z,z 1

α

y1

X x1

Y

Z,z 1 1 R

2

β y1

Figure 33 The link 1, shown in Fig. 33, rotates about the vertical axis Z and its instantaneous angular position is determined by the angle α. The bead 2 moves along the circular slide of radius R and its relative angular position with respect to the link 1 is determined by the angle β. Produce 1. the expressions for the components of the absolute linear velocity along system of coordinates x1 y1 z1 Answer: v = i1 (−α(R ˙ + R cos β)) + j1 (−Rβ˙ sin β) + k1 (Rβ˙ cos β) 2. the expression for the absolute linear acceleration of the bead along system of coordinates x1 y1 z1 Answer: 2 a = i1 (−¨ α(R + R cos β) + 2Rα˙ β˙ sin β) + j1 (−Rβ¨ sin β − Rβ˙ cos β − α˙ 2 (R + R cos β)) + 2 k1 (Rβ¨ cos β − Rβ˙ sin β) .

PROBLEMS.

39

Problem 9

o1 y1

O

o2

α X

Y x2

x1

y2 P

Z z1

z2

L 2 1

β o1

O

o2

y1 D

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 angular displacement α. The system of coordinates x1 y1 z1 is attached to the base 1. At the same time the boom 2 is being raised. This relative motion about the axis x1 is determined by the angular displacement β. The system of coordinates x2 y2 z2 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 x2 y2 z2 system of coordinates Answer: ω2x2 = β˙ ω 2y2 = α˙ sin β ω2z2 = α˙ cos β 2. the components of the absolute angular acceleration of the boom 2 along the x2 y2 z2 system of coordinates Answer: ¨ ε2x2 = β ε2y2 = α ¨ sin β + α˙ β˙ cos β ε2z2 = α ¨ cos β − α˙ β˙ sin β

PROBLEMS.

40

3. the components of the absolute linear velocity of the point P along the x2 y2 z2 system of coordinates Answer: ˙ vP x2 = Dα˙ − Lα˙ cos β vP y2 = 0 vP z2 = βL 4. the components of the absolute linear acceleration of the point P along the x2 y2 z2 system of coordinates Answer: 2 aP x2 = D¨ α − L¨ α cos β + 2Lα˙ β˙ sin β aP y2 = Dα˙ 2 cos β − Lα˙ 2 cos2 β − Lβ˙ aP z2 = 2 2 ¨ Lβ − Dα˙ sin β + Lα˙ sin β cos β

Chapter 2 THREE-DIMENSIONAL KINEMATICS OF A RIGID BODY DEFINITION: A body which by assumption does not deform and therefore the distances between two of its points remains unchanged, regardless of forces acting on the body, is called rigid body.

z P Z G

k rP,o

rP

rG,o o

ro

j

i O

y

ω

Y x

X

Figure 1 To analyze motion of a rigid body usually we attach to the body a system of 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 be determined in the same manner as the motion of the translating and rotating system of 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 angular velocity 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

GENERAL MOTION

42

Fig.1) be defined by a position vector rP,o . Since components of the vector rP,o along the system of coordinates xyz are constant, the relative velocity of P with respect to the body system of coordinates, r0P ,o ,is always 0. Therefore its absolute velocity is vP = r˙ P = r˙ o + r˙ P,o = r˙ o + r0P ,o + ω × rP,o = r˙ o + ω × rP,o

(2.1)

Similarly, The absolute velocity of the point Q is vQ = r˙ o + ω × rQ,o

(2.2)

Hence, the relative velocity of the point Q with respect to the point P is vQP = vQ − vP = r˙ o + ω × rQ,o − (˙ro + ω × rP,o ) = ω × (rQ,o − rP,o ) = (2.3) ω × rQ,P The absolute acceleration of P is aP = ¨ rP = ¨ ro + ω˙ × 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 = ¨ ro + ω˙ × rQ,o + ω × (ω × rQ,o ) − ¨ ro − ω˙ × rP,o − ω × (ω × rP,o ) = (2.5) = +ω˙ × rQ,P + ω × (ω × rQ,P ) The first term atQP = ω˙ × rQ,P

(2.6)

anQP = ω × (ω × rQ,P )

(2.7)

is called the tangential relative acceleration and the second one

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 the centre of mass is defined by a vector rG,o , the components of its absolute velocity along the body axes are vGx = i · (˙ro + ω × rG,o ) vGy = j · (˙ro + ω × rG,o ) vGz = k · (˙ro + 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 = ω˙ = ω0 + ω × ω = ω0

(2.10)

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

2.2

ROTATION ABOUT A POINT THAT IS FIXED IN THE INERTIAL SPACE (ROTATIONAL MOTION)

DEFINITION: If one point of the body considered is motionless with respect 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 as the origin of the body system of coordinates (see Fig. 2).

ω

z

Z

G

P

rG

rP

O

y

Y

X x Figure 2 The linear velocity of an arbitrarily chosen point P of the rigid body as well as its acceleration is determined by the angular velocity of the body ω. Indeed vP = r˙ P = r0P + ω × rP = ω × rP

(2.11)

rP = ω˙ × rP + ω × (ω × rP ) aP = ¨

(2.12)

PROBLEMS.

2.3

44

PROBLEMS.

Problem 10 Y 1

X 2

y1

α

x11 1 Z z1 G µ

1

Ω

β 1

y1

L

Figure 3 Base 1 of the ventilator shown in Fig. 3, performs an oscillatory motion about the vertical axis Z of the inertial system of coordinates XY Z. This motion is determined by the following equation α = αo sinωt. The axis of relative rotation of the rotor 2 is fixed at the constant angle β with respect to the horizontal plane. The rotor 2 rotate with a constant angular velocity Ω in the direction shown. The centre of gravity G of the rotor is displaced from its axis of rotation by distance µ. Determine: 1. components of the absolute angular velocity of the rotor along a system of coordinates fixed to the rotor. 2. components of the absolute velocity of the centre of gravity of the rotor along the same system of coordinates. Given are: αo , ω, L, µ, β, Ω.

PROBLEMS.

45

Solution. Y

z 12 Ω t

1 y1

z2 X

α x11

µ y12 y2

z 12

z 11

y12

x12

L

β

1

x2

y1

x11 x12

Figure 4 In Fig. 4, system of coordinates x11 , y11 , z11 is rigidly attached to the base 1 and rotates about the vertical axis Z of the inertial system of coordinates XY Z. The angular displacement α determines uniquely its instantaneous position. System of coordinates x21 , y12 , z12 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 y12 and its instantaneous position is determined by the angular displacement Ωt. Absolute angular velocity of the rotor 2 is ω2 = ω1 + ω 2,1 = k11 α˙ + j2 Ω = (k21 cos β + j21 sin β)α˙ + j2 Ω = ((k2 cos Ωt − i2 sin Ωt) cos β + j2 sin β)α˙ + j2 Ω = i2 (−α˙ cos β sin Ωt) + j2 (α˙ sin β + Ω) + k2 (α˙ cos β cos Ωt)

(2.13)

Position vector of the centre of gravity G, according to Fig. 4, is RG = L + µ = j21 L + k2 µ = j2 L + k2 µ

(2.14)

Hence, its fist derivative with respect to time yields its absolute velocity ¯ ¯ i2 j2 k2 ¯ 0 ¯ ˙ RG = RG + ω2 × RG = ¯ −α˙ cos β sin Ωt α˙ sin β + Ω α˙ cos β cos Ωt ¯ 0 L µ = i2 (µ(α˙ sin β + Ω) − Lα˙ cos β cos Ωt) +j2 (−µα˙ cos β sin Ωt) +k2 (−Lα˙ cos β sin Ωt)

¯ ¯ ¯ ¯ ¯ ¯

(2.15)

PROBLEMS.

46

Problem 11

Z 3 B 1

lAB

O c

Y 2 A X Figure 5 Fig. 5 shows the kinematic diagram of a spatial mechanism. Its link 1 can move along axis X and is free to rotate about that axis. The link 2 is hinged to the link 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.

PROBLEMS.

47

Solution.

Z 3 z1 , z2

α

1

C rC O c

rA A X x1

α

β

rBC

B

y2

rBA 2

y1

Y

β x2 Figure 6

Let x1 y1 z1 be the body 1 system of coordinates. Its angular position with respect 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 relative position of the body 2 system of coordinates x2 y2 z2 is uniquely determined by angle β. A. Matrices of direction cosines. From Fig. 6 one can see that the matrix of direction cosines between system of coordinates x1 y1 z1 and inertial system of coordinates XY Z has the following form ⎤ ⎡ ⎤ ⎡ ⎤⎡ X 1 0 0 x1 ⎣ Y ⎦ = ⎣ 0 cos α − sin α ⎦ ⎣ y1 ⎦ (2.16) z1 Z 0 sin α cos α

In the same manner we can easily derive matrix of direction cosines between body 2 system of coordinates x2 y2 z2 and system of coordinates x1 y1 z1 . ⎤ ⎡ ⎤ ⎡ ⎤⎡ x1 cos β − sin β 0 x2 ⎣ y1 ⎦ = ⎣ sin β cos β 0 ⎦ ⎣ y2 ⎦ (2.17) z1 z2 0 0 1

Introduction of Eq. 2.17 into Eq. 2.16 yields matrix of direction cosines between the body 2 system of coordinates and the inertial one. ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤⎡ X 1 0 0 cos β − sin β 0 x2 ⎣ Y ⎦ = ⎣ 0 cos α − sin α ⎦ ⎣ sin β cos β 0 ⎦ ⎣ y2 ⎦ z2 Z 0 sin α cos α 0 0 1 ⎤ ⎡ ⎤⎡ cos β − sin β 0 x2 ⎣ ⎦ ⎣ y2 ⎦ cos α sin β cos α cos β − sin α (2.18) = z2 sin α sin β sin α cos β cos α

PROBLEMS.

48

B. Positions of individual links. To find positions of individual links let us consider the following vector equation rC + rBC = rA + rBA

(2.19)

Kc + JrBC = IrA + j2 lAB

(2.20)

or The above equation is equivalent to 3 scalar equations which may be obtained by subsequent multiplication its both sides by unit vectors I, J, K. 0 = rA + I · j2 lAB rBC = J · j2 lAB c = K · j2 lAB

(2.21)

The dot products, appearing in the above equations, can be taken directly from matrix 2.18. 0 = rA − lAB sin β rBC = lAB cos α cos β c = lAB sin α cos β

(2.22) (2.23) (2.24)

β = arcsin(rA /lAB ) α = arcsin(c/lAB cos β) rBC = lAB cos α cos β

(2.25) (2.26) (2.27)

Hence,

The three above equations determine position of all links for any instant of time. C. Angular velocities of individual links. Since the link 1 performs rotational motion about axis X, its absolute angular velocity ω 1 may be obtained by diﬀerentiation of 2.26 with respect to time. ω 1 = Iα˙

(2.28)

ω 2 = ω1 + ω 21

(2.29)

ω 21 = k2 β˙

(2.30)

Angular velocity of the link 2 is

Since the absolute angular velocity ω 2 takes the following form ω 2 = Iα˙ + k2 β˙

(2.31)

It components along body system of coordinates x2 y2 z2 are ω2x2 = I · i2 α˙ + k2 · i2 β˙ = α˙ cos β ω2y2 = I · j2 α˙ + k2 · j2 β˙ = −α˙ sin β = I · k2 α˙ + k2 · k2 β˙ = β˙ ω2 z2

(2.32)

PROBLEMS.

49

where α˙ and β˙ 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 diﬀerentiation of expression 2.27 with respect to time. r˙ B = vB = Jr˙BC = JlAB (−α˙ sin α cos β − β˙ cos α sin β)

(2.33)

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 XY Z. Motion of the point A is determined by the following function of time ZA = ZA (t) The point B of the body 1 can move along the horizontal slide located in the plane Y Z. This slide is by a apart from axis Y . The point C is in a constant contact with the plane XY . According to the described constraints,motion of the body 1 is uniquely determined by the function ZA . 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.

Z

B

a

A

90 o

a Z(t)

Y

O 1

a X

C

Figure 7

PROBLEMS.

51

Solution. A. direction cosines

Z

z

B

a rBA

x A

90 or

a

B

rA

Z(t) rCA O

Y

rCB

a

rC

X

C y Figure 8

Let us consider the following vector equation (see Fig. 8). rB = rA + rBA

(2.34)

JrBY + Ka = KZA (t) + ka

(2.35)

or Hence

ZA rBY +K(1 − ) (2.36) a a Multiplication of the above vector equation by the units vectors associated with the inertial system of coordinates I, J and K yields the direction cosines between the axis z and axes of the inertial system of coordinates XY Z. k=J

k · I = cos∠kI =0 rBY k · J = cos∠kJ = a k · K = cos∠kK = 1 − Since

ZA a

cos2 ∠kI + cos2 ∠kJ + cos2 ∠kK = 1

we have that 2

(0) +

³r

BY

a

´2

¶2 µ ZA + 1− =1 a

(2.37)

(2.38) (2.39)

PROBLEMS.

52

The last relationship permits the unknown component rBY to be determined as the explicit function of time q rBY = 2aZA − ZA2 ) (2.40)

To determine the direction cosines between the axis y and axes of the inertial system of coordinates, let us consider equation rC = rA + rCA

(2.41)

IXC + JYC = KZA + ja

(2.42)

or Since according to the above equation j=I

rCX rCY ZA +J −K a a a

(2.43)

the wanted direction cosines are rCX a rCY j · J = cos∠jJ = a ZA j · K = cos∠jK = − a j · I = cos∠jI =

(2.44)

The following property of the direction cosines cos2 ∠jI + cos2 ∠jJ + cos2 ∠jK = 1

(2.45)

2 2 + rCY + ZA2 = a2 rCX

(2.46)

yields The second equation for determination of the unknown rCX and rCY one may obtain by consideration of the following vector equation rC = rB + rCB

(2.47)

rCB = rB − rC

(2.48)

It follows that Multiplication of the above equation by unit vectors I, J and K respectively yields components of the vector rCB along the initial system of coordinates. rCBX = rB · I − rC · I = −rCX rCBY

= rB · J − rC · J =rBY −rCY =

rCBZ = rB · K − rC · K =a

q 2aZA − ZA2 −rCY

As one can see from Fig. 7 the length of the vector rCB is equal to 2 2 2 + rCBY + rCBZ = 2a2 rCBX

(2.49)

√ 2a. Hence (2.50)

PROBLEMS.

53

Therefore the second equation for determination of the components rCX and rCY takes form ¶2 µq 2 2 2aZA − ZA −rCY + a2 = 2a2 (2.51) rCX +

The above equation together with Eq. 2.46 form set of two equations that determine the components rCX and rCY as explicit functions of time. They are s 2(a − ZA ) rCX = a 2a − ZA ZA (a − ZA ) rCY = p ZA (2a − ZA ) rCZ = 0 (2.52) These equations allow the trajectory of the point C to be computed. This trajectory is shown in Fig. 9 for a = 1 0.35 0.3 0.25 r CY 0.2 0.15 0.1 0.05 0 0

0.5

1

r CX

Figure 9 The unit vector associated with axis x can be produced as a vector product of 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 Therefore

¯ ¯ ¯ ¯ I J K ¯ ¯ i = ¯¯ cos ∠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)

(2.54)

(2.55)

PROBLEMS.

54

Hence, the direction cosines between the axis x and axes XY Z are cos ∠iI = cos ∠jJ cos ∠kK− cos ∠jK cos ∠kJ cos ∠iJ = − cos ∠jI cos ∠kK cos ∠iK = cos ∠jI cos ∠kJ

(2.56)

B. Linear velocities The linear velocity of the point B can be obtained by diﬀerentiation of the position vector rB which according to Eq. 2.40 q (2.57) rB = J 2aZA − ZA2 + Ka Hence

a − ZA vB = J p Z˙ 2 A 2aZA − ZA

(2.58)

Similarly, diﬀerentiation of the vector rC yields velocity of the point C. The position vector rC according to Eq. 2.52 s 2(a − ZA ) ZA (a − ZA ) rC = Ia + Jp (2.59) 2a − ZA ZA (2a − ZA )

Hence

(2.60)

vC = IvCX + JvCY where vCX

√ 2 1 2 Z˙ A p = − a 2 2a − ZA (2a2 − 3aZA + ZA2 ) 2

vCY

2

a − 3aZA + ZA = Z˙ A √ ³p ´3 ZA (2a − ZA )

(2.61)

Angular velocities. The angular velocity of the link 1 may be obtained from the following vector relationship vB = vA + ω × rBA (2.62) or ω × 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 = 2aZA − ZA2 )

rBAZ = rBA · K =ak · K =acos∠kK =a − ZA

(2.64)

PROBLEMS.

55

Hence, the equation 2.63 yields ⎡ ⎤ I J K ⎣ ωx ω z ⎦ = I(0 − 0) + J(vBY − 0) + K(0 − Z˙ A ) p ωy 2 0 2aZA − ZA ) a − ZA = I(0) + J(vBY ) + K( − Z˙ A ) (2.65)

The above vector equation is equivalent to the following three scalar equation q ω y (a − ZA ) − ωz 2aZA − ZA2 ) = 0

Hence

−ω x (a − ZA ) = vBY q ω x 2aZA − ZA2 ) = −Z˙ A

(2.66)

−Z˙ A ωx = p (2.67) 2aZA − ZA2 ) To produce the remaining components of the angular velocity let as consider the vector relationship between the point C and A belonging to the same body 1. or

vC = vA + ω × rCA

(2.68)

(2.69) ω × rCA = vC − vA The developed earlier direction cosines (Eq. 2.44) permits the components of the relative position vector to be obtain as an explicit function of time. rCAX = rCA · I =aj · I =acos∠jI =rCX rCAY = rCA · J =aj · J =acos∠jJ =rCY rCAZ = rCA · K =aj · K =acos∠jK = − ZA

(2.70)

where rCX and rCY are given by equation 2.52.Introducing them into Eq. 2.69 one can get the following vector equation. ⎡ ⎤ I J K ⎣ ωx ωy ωz ⎦ = I(vCX − 0) + J(vCY − 0) + K(0 − Z˙ A ) rCX rCY −ZA (2.71) = I(vCX ) + J(vCY ) + K( − Z˙ A ) that is equivalent to three scalar equation of form 2.72

−ω y ZA − ωz rCY = vCX ω z rCX + ω x ZA = vCY ωx rCY − ω y rCX = −Z˙ A

Hence, the wanted components of the angular velocity are vCY − ωx ZA ωz = rCX Z˙ A + ωx rCY ωy = rCX

(2.72)

PROBLEMS.

56

Problem 13 Z z1 y1

α1 X

x1 Z

Y

P

z1

α32 a

l

s 21 O

1

2

3 y1

Figure 10 The base 1 of a robot shown in Fig. 10 rotates about the vertical Z and its 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 is hinged 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 given parameters, 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.

PROBLEMS.

57

Solution. Z z1 y1

α1 X

x1 Z

Y

P

z 1 z2 l

z3

r

α32

a

l y2

a s 21

y3

s21 O

1

2

3 y1

Figure 11 In Fig. 11, the following systems of coordinates were introduced: XY Z – inertial system of coordinates x1 y1 z1 – body 1 rotating system of coordinates x2 y2 z2 – body 2 rotating and translating system of coordinates x3 y3 z3 – body 3 rotating and translating system of coordinates. The angular velocity of the body 3 is ω3 = ω1 + ω 21 + ω32 Since ω21 = 0 one may obtain the following expression for the angular velocity of the link 3. ω3 = ω 1 + ω32 = k2 α˙ 1 + i3 α˙ 32 = (k3 cos α32 + j3 sin α32 )α˙ 1 + i3 α˙ 32 = i3 α˙ 32 + j3 α˙ 1 sin α32 + k3 α˙ 1 cos α32 (2.73) The angular acceleration of the body 3 can be obtain as vector derivative of ω3 with respect to time. ε3 = ω˙ 3 = ω03 + ω 3 × ω 3 = ω03 = i3 α ¨ 32 + j3 (¨ α1 sin α32 + α˙ 1 α˙ 32 cos α32 ) + k3 (¨ α1 cos α32 − α˙ 1 α˙ 32 sin α32(2.74) ) The position vector of the point P , according to Fig. 11 is r = s21 + a + l = k2 s21 + j2 a + j3 l = (k3 cos α32 + j3 sin α32 )s21 + (j3 cos α32 − k3 sin α32 )a + j3 l = j3 (s21 sin α32 + a cos α32 + l) + k3 (s21 cos α32 − a sin α32 )

(2.75)

PROBLEMS.

58

Its first derivative with respect to time represents the wanted velocity of the point P . r˙ = r0 + ω3 × r

(2.76)

r0 = j3 (s˙ 21 sin α32 + s21 α˙ 32 cos α32 − aα˙ 32 sin α32 ) +k3 (s˙ 21 cos α32 − s21 α˙ 32 sin α32 − aα˙ 32 cos α32 )

(2.77)

where

¯ ¯ ¯ ¯ i3 j k 3 3 ¯ ¯ ¯ α˙ 1 sin α32 α˙ 1 cos α32 ω3 × r = ¯¯ α˙ 32 ¯ ¯ 0 s21 sin α32 + a cos α32 + l s21 cos α32 − a sin α32 ¯ = i3 [(s21 cos α32 − a sin α32 )α˙ 1 sin α32 − (s21 sin α32 + a cos α32 + l)α˙ 1 cos α32 ] +j3 [−(s21 cos α32 − a sin α32 )α˙ 32 ] (2.78) +k3 [(s21 sin α32 + a cos α32 + l)α˙ 32 ] Upon adding the two above expression together, one may obtain components of velocity of the point P in the following form r˙ = i3 (−aα˙ 1 − lα˙ 1 cos α32 ) + j3 (s˙ 21 sin α32 ) + k3 (s˙ 21 cosα32 + lα˙ 32 )

(2.79)

PROBLEMS.

59

Problem 14 X X

1

2

3 P

P l r

x1

γx γz

Q

Z3

β

r O

l

γy

Z3

Q

z2

β

Z

O

Z z1

Y

a)

b)

y1

Figure 12 Fig. 12 shows the kinematic scheme of a mechanism. Its link 1 rotates with the constant angular velocity ω about the horizontal axis Z of the motionless system of coordinates XY Z. The link 3 is free to slide along and to rotate about axis Z3 . The axis Z3 is fixed in the plane XZ and its position is determined by angle β. The link 2 joins point P of the link 1 and point Q of the link 3 by means of kinematic constraints as is shown in Fig. 12. Derive the analytical expression for the linear velocity of the link 3. Given are: ω− angular velocity of the link 1. r− distance between points O and P . l− length of the link 2. β− angle between axis Z and Z3

PROBLEMS.

60

Solution. X rP

P

x1

γx

rQP

γz α =ω t r

l

γy

rQ

β

O

Q

Z3 z2 Z z1

Y y1

Figure 13 From Fig. 13 one can see that rQ = rP + rQP

(2.80)

The above vectors can be expressed as follows. rP = i1 r rQ = K rQ cos β + I rQ sin β rQP = k2 l = i1 l cos γ x + j1 l cos γ y + k1 l cos γ z

(2.81) (2.82) (2.83)

where γ x , γ y and γ z are angles between the axis z2 and axes of the system of coordinates x1 y1 z1 .Introduction of the above expressions into Eq. 2.80 yields KrQ cos β + IrQ sin β = i1 r + i1 l cos γ x + j1 l cos γ y + k1 l cos γ z

(2.84)

Multiplication of the equation Eq. 2.84 by the unit vectors i1 , j1 and k1 respectively oﬀers three scalar equations. rQ cos α sin β = r + l cos γ x −rQ sin α sin β = l cos γ y rQ cos β = l cos γ z

(2.85) (2.86) (2.87)

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

(2.88) (2.89) (2.90)

Hence

PROBLEMS.

61

The direction cosines have to fulfil the following relationship cos2 γ x + cos2 γ y + cos2 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. 2 + (−2r cos α sin β)rQ + (r2 − l2 ) = 0 rQ

(2.93)

Hence, the instantaneous position of the point Q is determined by the following expression. (2.94) rQ = r cos α sin β ± (r2 cos2 α sin2 β − r2 + l2 )1/2 Since axis Z3 is motionless, the fist derivative with respect to time of the above expression yields absolute velocity of the point Q. r˙Q = −rα˙ sin α sin β ± r2 α˙ 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 corresponds to sing ’-’. X

1

2

3

P l

Q+

r Q-

Z3

β O

Z

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

PROBLEMS.

62

Problem 15 y2

3

P

2 A

x3

ω 21 t

y1

y3 y2

β

O o2 r

z3

A o3

2

x2 x1 X

Z

z1

z2

P 3 s

z1 z2

o2

2 l

y1

1 O

o1

α

Y

Figure 15 A sketch of the Ferris Wheel is shown in Fig. 15. Its base 1 oscillates about the horizontal axis X of the XY Z inertial system of coordinates. The instantaneous position of this base is determined by the angular position α. The system of coordinates x1 y1 z1 is rigidly attached to the base 1. The relative angular velocity of the wheel 2 with respect to the base 1 is constant and is equal to ω 21 . The system of coordinates x2 y2 z2 is rigidly attached to the 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 displacement β. The system of coordinates x3 y3 z3 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 x3 y3 z3 2. the absolute angular acceleration of the seat 3 along the system of coordinates x3 y3 z3 3. the absolute linear velocity of the point P along the system of coordinates x3 y3 z3 Given are: r, l, s, ω 21 , α(t), β(t).

PROBLEMS.

63

Solution 3

P

y2

2 A

x3

ω21t

y1

y3

o2 r

z3

x1 X z1 z2

Z

A o3 3

2

x2

2

y2

β

O

z1

z2

P

s

o2

o2

l

y1

1 O

o1

α

Y

Figure 16 The absolute angular velocity of the system of coordinates x1 y1 z1 . (2.96)

ω1 = i1 α˙ The absolute angular velocity of the system of coordinates x2 y2 z2 . ω2 = ω1 + ω 21 = i1 α˙ + k2 ω 21 The absolute angular velocity of the system of coordinates x3 y3 z3 ω3 = ω 2 + ω32 = i1 α˙ + k2 ω 21 + i3 β˙

(2.97)

(2.98)

Since i1 = i2 cos ω21 t − j2 sin ω 21 t = i3 cos ω21 t − (j3 cos β − k3 sin β) sin ω21 t = (2.99) = i3 cos ω21 t + j3 (− cos β sin ω 21 t) + k3 (sin β sin ω 21 t) k2 = j3 sin β + k3 cos β the components of the angular velocity along the system of coordinates are x3 y3 z3 ω 3 = (i3 cos ω 21 t + j3 (− cos β sin ω 21 t) + k3 (sin β sin ω21 t)) α˙ + (j3 sin β + k3 cos β) ω 21 + i3 β˙ ´ ³ = i3 α˙ cos ω21 t + β˙ + j3 (−α˙ cos β sin ω21 t + ω 21 sin β)

+k3 (α˙ sin β sin ω21 t + ω 21 cos β) (2.100)

PROBLEMS.

64

The absolute angular velocity of the seat 3 is equal to ω 3 . The absolute angular acceleration of the seat is equal to the absolute angular acceleration of the system of coordinates x3 y3 z3 0

ε = ω˙ 3 = ω 3 + ω3 × ω3 = α cos ω 21 t − αω ˙ 21 sin ω 21 t + β¨ = i3 (¨ ´ ³ ˙ 21 cos β cos ω 21 t + ω21 β˙ cos β +j3 −¨ α cos β sin ω 21 t + α˙ β˙ cos β sin ω 21 t − αω ´ ³ ˙ ˙ +k3 α ˙ 21 sin β cos ω 21 t − ω 21 β sin(2.101) β ¨ sin β sin ω 21 t + α˙ β cos β sin ω 21 t + αω

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

RP = k2 l + j2 r + k3 (−s) = (j3 sin β + k3 cos β) l + (j3 cos β − k3 sin β) r + k3 (−s) = i3 (0) + j3 (l sin β + r cos β) + k3 (l cos β − r sin β − s)

(2.102)

The first derivative of this vector yields the absolute liner velocity of the point P. ˙ P = R0P + ω 3 × RP vP = R ³ ´ ³ ´ = j3 lβ˙ cos β − rβ˙ sin β) + k3 −lβ˙ sin β − rβ˙ cos β ¯ ¯ ¯ ¯ i3 j3 k3 ¯ ¯ + ¯¯ α˙ cos ω21 t + β˙ −α˙ cos β sin ω 21 t + ω21 sin β α˙ sin β sin ω21 t + ω 21 cos β ¯¯ ¯ ¯ 0 l sin β + r cos β l cos β − r sin β − s = i3 vP x3 + j3 vP y3 + k3 vP z3 (2.103)

PROBLEMS.

65

Problem 16 O

Z

α

x X

z1 Y y 1

1

y 2

ω

32 x2

P

β

o2

O

x 1 1 a

b 2

3

Figure 17 Fig. 17 shows a diagram of a wheel excavator. Its base 1 rotates about the vertical axis Y of the inertial system of coordinates XY Z. System of coordinates x1 y1 z1 is rigidly attached to the base 1. Its instantaneous position is determined by the angle α. The arm 2 rotates with respect to the base 1 about axis that is parallel to z1 . System of coordinates x2 y2 z2 is rigidly attached to the arm 2. Its relative angular position is determined by an angle β. The wheel 3 rotates with respect to the arm 2 about axis that is parallel to z1 with the angular velocity ω32 . Given are: α(t), β(t), ω 32 (t), a, b. Produce 1. the expressions for the components of the absolute angular velocity of the wheel 3 along system of coordinates x2 y2 z2 . Answer: ³ ´ ˙ ω3x2 = α˙ sin β ω 3y2 = α˙ cos β ω3z2 = β + ω 32 2. the expressions for the components of the absolute linear velocity of the point P along system of coordinates x2 y2 z2 . Answer: vP x2 = 0 vP y2 = −bβ˙ vP z2 = −aα˙ − bα˙ cos β 3. the expressions for components of the absolute acceleration of the point P along system of coordinates x2 y2 z2 . Answer: aP x2 = vP z2 α˙ cos β − vP y2 β˙ aP y2 = v˙ P y2 − vP z2 α˙ sin β aP z2 = v˙ P z2 + vP y2 α˙ sin β

PROBLEMS.

66

Problem 17

Z

z1

y3

ω1

4

z3

B 1 G

β y1

3

rA 2

l/2

A

l

Figure 18 Fig. 18 shows the kinematic scheme of a mechanism. Its link 1 rotates with a constant angular velocity ω 1 about the vertical axis Z. The link 2 can translate with respect to the link 1 and its motion is determined by position vector of the point A. rA = −k1 rA Given are: ω1 − angular velocity of the link 1 rA (t)− position of the point A as a function of time. l− length of the link 3 Produce expressions for 1. the components of the absolute angular velocity of the link 3 along the body 3 system of coordinates x3 y3 z3 . Answer: ω3x3 = β˙ ω 3y3 = ω 3z3 = ω1 cos β ¡ rAω¢1 sin β ˙ where β = arcsin l β = √ r2˙A 2 l −rA

2. the components of the absolute velocity of the point B along the body 1 system of coordinates x1 y1 z1 . Answer: p 2 vBx1 = −ω 1 l2 − rA vBy1 = √rA2r˙A 2 vBz1 = 0 l −rA

3. the components of the absolute velocity of the point G along the body 3 system of coordinates x3 y3 z3 . Answer: vGx1 = − 2l ω 1 cos β vGy1 = −r˙A sin β vGz1 = −r˙A cos β + 2l β˙

PROBLEMS.

67

Problem 18 y1

α Y X z2

x1 Z

z1 y3

z3

y2

γ L

β

3

P y1

2

R 1

Figure 19 The base 1 of the Ferris Wheel shown in Fig. 19 rotates about the vertical axis Z of the inertial system of coordinates XY Z. Its angular position is determined by the angular displacement α(t) which is a given function of time. The system of coordinates x1 y1 z1 is rigidly attached to this base. The wheel 2 of radius R performs rotational motion about axis x1 with respect to the base 1. This relative motion is determined by function β(t). The system of coordinates x2 y2 z2 is rigidly attached to the wheel 2. The seat 3 is hinged to the wheel 2. The relative angular position of the seat 3 with respect to the wheel 2 is determined by angle γ(t). The system of coordinates x3 y3 z3 is rigidly attached to the seat 3. 1. Derive expressions for the components of the absolute angular acceleration of the seat 3 along the system of coordinates x2 y2 z2 . 2. Derive expressions for the components of the absolute linear velocity of the point P along system of coordinates x2 y2 z2 . Given are: α(t), β(t), γ(t), R, L

PROBLEMS.

68

Problem 19 Z 1

z1

2 y1

x2

C

z2

α 21 R

α1

O

x1

P

Y

C X

L

x1

z1

O

y1 y2 C

O

Figure 20 The link 1 of the mechanical system shown in Fig. 20 is free to rotate about axis X of the inertial system of coordinates XY Z. Its instantaneous position is determined by the angle α1 . The system of coordinates is rigidly attached to the link 1. The link 2 can rotates with respect to the link 1 and its relative angular position is determined by the angle α21 . Derive expression for the components of 1. the absolute angular acceleration of the link 2 along the body 2 system of coordinates 2. the absolute linear velocity of the point P along the body 2 system of coordinates . Given are: α1 , α21 ,L, R

PROBLEMS.

69

Problem 20

1 y1

ω1 t X

P

x1 Z

Y

z1

z2

α

o2

O

y2

P 2 P

1

y1 a

c

y2

x2

o2 b

Figure 21 The helicopter blade 2 is hinged at to the helicopter rotor 1 as shown in Fig. 21. The distance between the hinge and the rotor axis is c. The helicopter body is stationary with respect to the inertial system of coordinates XY Z. Its rotor 1 rotates with a constant angular velocity ω 1 about the vertical axis Z. The system of coordinates x1 y1 z1 is rigidly attached to the rotor 1. The relative position of the blade 2 with respect to the rotor 1 is determined by the angular displacement α. This relative angular displacement α is a given function of time. The system of coordinates x2 y2 z2 is rigidly attached to the blade 2. The position of the point P which belong to the blade is determined by its coordinates a and b along the system of coordinates x2 y2 z2 . Produce: 1. the components of the absolute velocity and the absolute acceleration of the point P along the system of coordinates x2 y2 z2 2. the components of the absolute angular acceleration of the blade 2 along the system of coordinates x2 y2 z2

PROBLEMS.

70

Problem 21 0

α

R

E

F

vA

A a

Y

ω 2,1 y1

z1

β

b

X

x1

1

r

o1

o1

P

2 x1

Figure 22 The point A of the plane 1, shown in Fig. 22, follows the circular path of radius R with the constant linear velocity vA . Its longitudinal axis E − F is always tangential to the path of the point A. This path belongs to the horizontal plane of the inertial space XY Z. System of coordinates x1 , y1 , z1 is rigidly attached to the plane 1. Its propeller 2 rotates with the constant angular velocity ω2,1 with respect to the plane 1. Produce: 1. Components of the absolute velocity of the tip P of the propeller 2 along the system of coordinates x1 , y1 , z1 as a function of the angular displacements α and β. Answer: vP x1 = −rβ˙ sin β − bα˙ vP y1 = α(R ˙ + a + r cos β) vP z1 = −rβ˙ cos β 2. Components of the absolute acceleration of the tip P along the system of coordinates x1 , y1 , z1 as a function of the angular displacements α and β. Answer: aP x1 = v˙ P x1 − αv ˙ P y1 aP y1 = v˙ P y1 − αv ˙ P x1 aP z1 = v˙ P z1 3. Components of the angular acceleration of the propeller 2 along the system of coordinates x1 , y1 , z1 as a function of the angular displacements α and β. Answer: εx1 = −αω ˙ 2,1 εy1 = 0 εz1 = 0

PROBLEMS.

71

Problem 22 z1 z2 z 3

z1 3

2

z2

A

1 x1 x2 x3

z1

b o

x1

y2 y1 α 2

o

o

y3 y2

a

b)

a)

o

x3

α3 x2

Figure 23 Fig. 23a) shows the self-steering mechanism of the yacht 1. The system of coordinates x1 , y1 , z1 is attached to the yacht. The shaft 2 of this mechanism is free to rotate about the axis x1 . The system of coordinates x2 , y2 , z2 is attached to this shaft. Its instantaneous position is determined by the angle α2 (see Fig. 23b)). The link 3 is hinged to the link 2 at the point o. The instantaneous position of the body 3 system of coordinates x3 , y3 , z3 with respect to the system of coordinates x2 , y2 , z2 is determined by the angle α3 . The yacht is travelling along a straight line with velocity v. Given are: a, b, v(t), α2 (t), α3 (t). Produce the expression for 1. the components of the absolute angular velocity of the link 3 along the system of coordinates x3 , y3 , z3 Answer: ω3x3 = α˙ 2 cos α3 ω 3y3 = −α˙ 2 sin α3 ω3z3 = α˙ 3 2. the components of the absolute angular acceleration of the link 3 along the system of coordinates x3 , y3 , z3 Answer: ε3x3 = α ¨ 2 cos α3 − α˙ 2 α˙ 3 sin α3 ε3y3 = −¨ α2 sin α3 − α˙ 2 α˙ 3 cos α3 ε3z3 = α ¨3 3. the components of the absolute velocity of the point A shown in Fig. 23a) along the system of coordinates x3 , y3 , z3 . Answer: vAx3 = v cos α3 −bα˙ 2 sin α3 vAy3 = −v sin α3 −aα˙ 3 −bα˙ 2 cos α3 vAz3 = −aα˙ 2 sin α3

PROBLEMS.

72

Problem 23

Z

2

y2

z1

α

z2

Y y1

β

3 1

O

X

Ω a

P

x1 x2

b

Figure 24 The ship 1 (Fig. 24) rotates about the axis Y of the absolute system of coordinates XY Z. Its instantaneous position is determined by the angle α. System of coordinates x1 , y1 , z1 is rigidly attached to the ship. The housing 2 performs the rotational motion about axis x1 . The relative position of the housing with respect to the ship is given by the angular displacement β. System of coordinates x2 , y2 , z2 is fixed to the housing. The gyroscope 3 rotates about axis with respect to the housing. Its relative angular velocity is Ω. Produce: 1. components of the absolute angular velocity of the gyroscope along the system of coordinates x2 , y2 , z2 2. components of the absolute angular acceleration of the gyroscope along system of coordinates x2 , y2 , z2 3. components of the absolute linear velocity of the point P along the system of coordinates x2 , y2 , z2 4. components of the absolute linear velocity of the point P along the system of coordinates x1 , y1 , z1

PROBLEMS.

73

Problem 24 z1

z2 z3

3

γ

o3 y3

z1

α

x3

2

ω 21 x1

y1

o1 y1 o2 y2

2

β

x1 x2

o1 Z

R

1

L

O X Y

Figure 25 Fig. 25 shows the sketch of a Ferris Wheel. Its base 1 oscillates about the horizontal axis X of the inertial system of coordinates XY Z. Its instantaneous position is determined by the angular displacement α. The wheel 2 of radius R rotates with respect to the base 1 with the constant velocity ω21 about axis y1 of the body 1 system of coordinates x1 y1 z1 . The system of coordinates x2 y2 z2 is rigidly attached to the wheel 2. The seat 3 is free to rotate about axis y3 and its relative position is determined by the angular position γ. Produce the expressions for 1. the components of the absolute angular velocity of the seat 3 along the system of coordinates x2 y2 z2 Answer: ω3 = i2 (α˙ cos β) + j2 (ω 21 + γ) ˙ + k2 (α˙ sin β) 2. the components of the absolute angular acceleration of the seat 3 along the system of coordinates x2 y2 z2 Answer: ε3 = i2 (¨ α cos β − αω ˙ 21 sin β − α˙ γ˙ sin β) + j2 (¨ γ ) + k2 (¨ α sin β + αω ˙ 21 cos β − α˙ γ˙ cos β) 3. the components of the absolute linear velocity of the point o3 along the system of coordinates x2 y2 z2 Answer: vo3 = i2 (ω21 R) + j2 (−Lα˙ − Rα˙ cos β) + k2 (0)

PROBLEMS.

74

Problem 25 1

y1

2

Y

y1

A A

z1

α

f O

O

Z

x1 X

Y

y1 y2

1

α

f Z

2

z2 P

z1

A x2

L

β

β

O

β

α

x1 OA

Z z1 P z2

L

x1

a)

X

x2

b)

Figure 26 The link 1 of the mechanical system shown in Fig. 26 rotates about the horizontal axis Z of the inertial system of coordinates XY Z. Its instantaneous position is determined by the angular displacement α. The system of coordinates x1 y1 z is attached to the link 1. The link 2 is free to rotate about the axis y1 and its relative angular position is determined by the angle β. The system of coordinates x2 y2 z2 is attached to the link 2. The dimensions f and L locate position of the point P with respect to this system of coordinates. Produce 1. the components of the absolute angular velocity of the link 2 along the system of coordinates x2 y2 z2 . Answer: ˙ + k2 (α˙ cos β) ω2 = i2 (−α˙ sin β) + j2 (β) 2. the components of the absolute angular acceleration of the link 2 along the system of coordinates x2 y2 z2 .

PROBLEMS.

75

Answer: ¨ + k2 (¨ ε2 = i2 (−¨ α sin β − α˙ β˙ cos β) + j2 (β) α cos β − α˙ β˙ sin β) 3. the components of the absolute linear velocity of the point P along the system of coordinates x2 y2 z2 . Answer: ˙ − f α˙ cos β) + j2 (Lα˙ sin β) + k2 (−f α˙ sin β) vP = i2 (βL

PROBLEMS.

76

Problem 26

z1 Z

z1

0

α

1

y1 x1 O

o1

2

o1

Y z2

P

β

Y(t )

b

L

P

a y2

o2

o2 x2

Figure 27 Figure 27 shows the suspension of the casting ladle 2. The point o1 of the carriage 1 moves along the horizontal axis Y of the inertial system of coordinates XY Z. Its motion is determined by the displacement Y (t). This carriage is free to rotate about the axis x1 of the body 1 system of coordinates x1 y1 z1 . Its angular position is given by the function of time α(t). The ladle 2 rotates about the axis y2 of the body 2 system of coordinates x2 y2 z2 . Its relative angular position is determined by the angle β(t). Produce: 1. the expression for the components of the absolute angular velocity of the ladle 2 along the system of coordinates x2 y2 z2 . Answer: ω2 = i2 α˙ cos β + j2 β˙ + k2 α˙ sin β 2. the expression for the components of the absolute angular acceleration of the ladle 2 along the system of coordinates x2 y2 z2 . Answer: ε2 = i2 (¨ α cos β − α˙ β˙ sin β) + j2 β¨ + k2 (¨ α sin β + α˙ β˙ cos β) 3. the components of the absolute velocity of the point P along the system of coordinates x2 y2 z2 .

PROBLEMS.

Answer: vP = i2 (Y˙ sin α sin β + Y α˙ cos α sin β + Y β˙ sin α cos β + Lβ˙ cos β)+ +j2 (Y˙ cos α − Y α˙ sin α)+ +k¯ 2 (Y˙ sin α cos β − Y α˙ cos α cos β + Y β˙ sin α sin β + Lβ˙ sin β)+ ¯ i2 j2 k2 ¯ + ¯¯ α˙ cos β β˙ α˙ sin β ¯ Y sin α sin β + L sin β − a Y cos α −Y sin α cos β − L cos β + b

77

¯ ¯ ¯ ¯ ¯ ¯

PROBLEMS.

78

Problem 27

y2

2

Z z1

3

P

z2 G

y1

β

a O

A

b

c

Y

1

R

α

x2

a x1 X Figure 28

Fig. 28 shows the physical model of a mechanical system. The link 1 of this system rotates about the vertical axes Z of the inertial frame XYZ. Its instantaneous position is given by the absolute angular displacement α. The system of coordinates x 1 y 1 z 1 is rigidly attached to the link 1. The link 2 is hinged to the link 1 at the point A. The other end of this link P always stays in contact with the cylindrical surface 3 of radius R (b>R). The system of coordinates x 2 y 2 z 2 is attached to the body 2 and coincides with its principal axes. The link 2 possesses the mass m and its principal moments of inertia about the system of coordinates x 2 y 2 z 2 are I x2 , I y2 and I z2 . Its centre of gravity G is located by the distance c. The angular displacement β determines the relative position of the link 2 with respect to the system of coordinates x 1 y 1 z 1 . Produce: 1. The expression for the components of the absolute angular velocity of the link 2 along the system of coordinates x 2 y 2 z 2 in terms of α and β 2. The expression for the components of the absolute angular acceleration of the link 2 along the system of coordinates x 2 y 2 z 2 in terms of α and β 3. The expression for the components of the absolute velocity of the point P

PROBLEMS.

along the system of coordinates x 2 y 2 z 2 in terms of α and β 4. The kinetic energy of the link 2 as a function of α and β. 5. The expression for the angular displacement β as a function of α.

79

Chapter 3 KINETICS OF SYSTEM OF PARTICLES. Z

m r

F

O Y X Figure 1 Newton second law for a single particle (see Fig. 1) can be formulated in the following form d F = (m˙r) (3.1) dt where F - is the resultant force acting on the particle m - mass of the particle r˙ - is its absolute velocity. If the mass is constant, the second law can be rewritten in more simple way. F = m¨ r

(3.2)

If the position vector r and force F are determined by its components along the absolute rectangular system of coordinates (r = IrX + JrY + KrZ , F = IFX + JFY + KFZ ) the above equation is equivalent to three scalar equations. rX FX = m¨ FY = m¨ rY FZ = m¨ rZ

(3.3)

If number of particles n is relatively low, we are able to produce free body diagram for each particle separately and create 3n diﬀerential equations which permit each

KINETICS OF SYSTEM OF PARTICLES.

81

dynamic problem to be solved. But, if number of particles approaches infinity this way of solving dynamic problems fails. This chapter is concerned with formulation of equations of motion of any system of particles regardless of their number and internal forces acting between the individual particles. Because each continuum (fluid, gas, rigid or elastic body) can be considered as system of particles, the derived equations form a base for development of many branches of mechanics (fluid mechanics, solid mechanics, etc.).

Fi Z

Fij

mi r ij

ri

mj Fji

rj

Fj

O Y X Figure 2

Since the internal forces are to be eliminated from equations ofPmotion, the resultant force F have to be resolved into resultant of all internal forces N j=1 Fij and resultant of all external forces Fi acting on the i − th particle (see Fig. 2). F = Fi +

N X

Fij

(3.4)

j=1

Here N – represents number of particles involved Fi – represents resultant of all forces coming from sources external to the system considered (gravity force or any force explicitly determined in time). Fij - represents internal force acting on a particle i as result of interaction with a particle j. Hence, the second Newton law may be adopted in the following form r = Fi + mi¨

N X

Fij

(3.5)

j=1

According to the third Newton law we can assume that Fij = −Fji

and

Fii = Fjj = 0

(3.6)

MOTION OF CENTRE OF MASS - LINEAR MOMENTUM.

3.1

82

MOTION OF CENTRE OF MASS - LINEAR MOMENTUM.

Let us consider system of N particles in an inertial space. The motion of an individual particle mi is defined by the position vector ri (Fig. 3).

Z

mi G

ri rG O

Y X Figure 3

DEFINITION: The following vector P=

PN

i=1

(3.7)

mi r˙ i

is called inertial linear momentum of the system of particles. The first derivative of the linear momentum is ˙ = P

N X

(3.8)

mi¨ ri

i=1

According to Newton second law ˙ = P

N N N N N X X X X X (Fi + Fij ) = Fi + Fij i=1

j=1

i=1

(3.9)

i=1 j=1

Taking into account that according to Eq. 3.6 Fij = −Fji and Fii = Fjj = 0 the formula 3.9 can be rewritten in form ˙ = P

PN

i=1

Fi = Fex

(3.10)

The last formula permits to formulate the following statement. STATEMENT: The rate of change of the linear momentum of a system of particles is equal to the resultant of all external forces acting on the system of particles.

MOMENT OF MOMENTUM.

83

Let vector rG be the position vector of the centre of mass of a system of particles (Fig. 3 ). According to the definition of a centre of particles we have rG m =

N X

(3.11)

ri mi

i=1

where

P m= N i=1 mi - is the total mass of the system of particles. Diﬀerentiating the equation 3.11 with respect to time one can obtain r˙ G m =

N X

(3.12)

r˙ i mi = P

i=1

Second diﬀerentiation yields ˙ ¨ rG m = P

(3.13)

¨ rG m = Fex

(3.14)

Hence, according to Eq. 3.10 is The last result permits the following statement to be formulated. STATEMENT: The centre of mass of a system of particles moves as if the entire mass of the system was concentrated at that point and all external forces were applied there. 3.2

MOMENT OF MOMENTUM.

Considering system of particles, always three kinds of moment of momentum (angular momentum) are introduced. 1. Angular momentum about a fixed point in the inertial space. 2. Angular momentum about a moving point in the inertial space. 3. Relative angular momentum about a moving point in the inertial space. In this paragraph the three above types of angular momentum are defined and discussed. 3.2.1 Moment of momentum about a fixed point in the inertial space. Let mi be a particle which belongs to a system of N particles and ri be its position vector in the inertial space XY Z (Fig. 4). DEFINITION: The following vector HO =

PN

i=1 ri

× (mi r˙ i )

(3.15)

is called angular momentum about the fixed point O. The first derivative of angular momentum is ˙O= H

N N N X X X (˙ri × (mi r˙ i )) + (ri × (mi¨ ri )) = (ri × (mi¨ ri )) i=1

i=1

i=1

(3.16)

MOMENT OF MOMENTUM.

84

Z

mi ri

O Y X Figure 4 Applying the second Newton law to the equation 3.16 we have ˙O= H

N X i=1

ri × (Fi +

N X

Fij ) =

j=1

N X i=1

ri × Fi +

N X N X i=1 j=1

ri × Fij

(3.17)

To show that the last term is equal to zero, let us consider two particles mi and mj (Fig. 5)

Z

Fij

mi r ij

ri

mj

rj

O Y X Figure 5

From the above figure one can see that ri × Fij + rj × Fji = ri × Fij − rj × Fij = (ri − rj ) × Fij = rij × Fij

(3.18)

Since vectors rij and Fij are parallel then ri × Fij + rj × Fji = 0

(3.19)

Taking into account the above equation, one may say that N N X X i=1 j=1

ri × Fij = 0

(3.20)

MOMENT OF MOMENTUM.

85

and the equation 3.17 takes form ˙O= H

N X i=1

ri × Fi

(3.21)

The left hand side of equation 3.21 represents resultant moment MO of all external forces acting on the system of particles with respect to the fixed point O. Hence, we may conclude this paragraph with the following statement. STATEMENT: The rate of change of the angular momentum of a system of particles about a fixed point is equal to the sum of moments of all external forces acting on the system of particles about that point. ˙ O = MO H

(3.22)

3.2.2 Moment of momentum about a moving point in an inertial space. Let motion of the point C (6) be defined by the position vector rC and let us denote by mi an arbitrarily chosen particle of a system assembled out of N particles. Z

mi r i,C ri

C

rC

O Y X

Figure 6

If the motion of the particle mi is determined by vector ri , the relative motion of the particle mi with respect to the point C is determined by the formula (3.23). ri,C = ri − rC

(3.23)

DEFINITION: The following vector HC =

PN

i=1 ri,C

× (mi r˙ i )

(3.24)

is called angular momentum about the moving point C. Its first derivative is ˙C= H

N X i=1

r˙ i,C × (mi r˙ i ) +

N X i=1

ri,C × (mi¨ ri )

(3.25)

MOMENT OF MOMENTUM.

86

Introducing Eq. 3.23 into Eq. 3.25 and taking advantage from the second Newton law, the first derivative of the angular momentum can be written as follow. ˙C= H

N X i=1

r˙ i × (mi r˙ i ) −

N X i=1

r˙ C × (mi r˙ i ) +

N X i=1

ri,C × (Fi +

N X

Fij )

(3.26)

j=1

But

PN r˙ i × (mi r˙ i ) = 0 since r˙ i is parallel to mi r˙ i, PN Pi=1 N i=1 ri,C × j=1 Fij = 0 because ri,C × Fij + rj,C × Fji = (ri,C − rj,C ) × Fij = rij × Fij = 0 (see Fig. 7) mj ri,j

Z

mi

rj

Fi,j

r i,C

rj,C C

c

rC

ri O

Y X

Figure 7

Hence ˙C =− H where

N X i=1

r˙ C × (mi r˙ i ) +

N X i=1

ri,C × Fi = −˙rC × P + MC

(3.27)

P (m r˙ ) - is the linear momentum of the system considered P= N i=1 PN i i MC = i=1 ri,C × Fi - is resultant moment of all external forces about the moving point C. According to consideration curried out in chapter 2 section 1, P = r˙ G m. Hence ˙ C = −˙rC × r˙ G m + MC H

(3.28)

If the arbitrarily chosen point C coincides the gravity centre G (˙rC = r˙ G ), the above formula yields ˙ G = MG H (3.29) The last relationship allows to formulate the following statement. STATEMENT: The rate of change of angular momentum about centre of gravity of a system of particles is equal to moment about that centre of all external forces acting on the system of particles.

MOMENT OF MOMENTUM.

87

3.2.3 Moment of relative momentum. Let us once more consider a system of particles shown in Fig. 8 and the arbitrarily chosen point C. Z

G

mi ri

r i,C r G,C C

rC

O Y X

Figure 8

DEFINITION: The following vector hC =

PN

× (˙ri,C mi )

i=1 ri,C

(3.30)

is called moment of relative momentum. Its first derivative is h˙ C =

N X i=1

r˙ i,C × (˙ri,C mi ) +

N X i=1

ri,C × (¨ ri,C mi )

(3.31)

But, according to Fig 8, ri,C = ri −rC

(3.32)

Hence r˙ i,C = r˙ i − r˙ C

¨ ri,C = ¨ ri − ¨ rC

and

(3.33)

The first term in the right hand side of equation 3.31 is equal to 0. Hence, introduction of Eq. 3.33 into it yields h˙ C =

N X i=1

=

N X i=1

=

N X i=1

ri,C × (¨ ri mi ) − ri,C × (Fi +

N X i=1

N X j=1

Fij ) + ¨ rC ×

ri,C × Fi + ¨ rC ×

= MC + ¨ rC × mrG,C

ri,C × (¨ rC mi )

N X

N X (ri,C mi ) i=1

(ri,C mi )

i=1

(3.34)

EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS.

88

If C always coincides the centre of gravity G (rG,C = 0), moment of the relative angular momentum about the system’s centre of gravity is h˙ G = MG

(3.35)

The above considerations allows to formulate the following statement. STATEMENT: The rate of change of the relative angular momentum about the centre of gravity is equal to moment of all external forces acting on the system of particles about that centre. Taking into account that the right hand sides of equations 3.35 and 3.29 are equal, we arrive to conclusion that ˙ G = h˙ G H

and

HG = hG

(3.36)

The last expression is obvious because for motionless system of particles both moments of momentum are equal to 0. 3.3

EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS.

In the previous sections of this chapter four vectorial equations has been derived. Not all of them are independent. The independent pairs of equations are. ½ ½ ½ ˙ =F ˙ =F ˙ =F P P P (3.37) ˙ G = MG ˙ O = MO H H h˙ G = MG Depending on the dynamic problem to be solved, we can choose one of these three pairs of equations. Each of them is equivalent to six scalar equations. It means, that these problems can only be solved, without additional equations expressing interaction between individual particles, which have six scalar unknown only. Very small range of dynamic problems fulfil these requirements. 3.3.1 Conservation of momentum principle. Let us assumed that the component of the resultant of all external forces F along the fixed in the inertial space axis defined by the unit vector λ is equal to zero. Z

F

λ O Y X

Figure 9

EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS.

89

˙ = F by the unit vector λ yields the following scalar The scalar multiplication of P equation ˙ =λ·F=0 λ·P (3.38) Hence, λ · P = const

(3.39)

STATEMENT: If component of a resultant of all external forces along a fixed in an inertial space axis is equal to zero, its momentum is conserved along that axis (λ · P = const). 3.3.2 Conservation of angular momentum principle. Let us assumed that the component of the resultant moment MO of all external forces F along the fixed in the inertial space axis defined by the unit vector λ is equal to zero. Z

MO

λ O Y X

Figure 10

˙ O = MO by the unit vector λ, results in the The scalar multiplication of equation H following scalar relationship ˙ O = λ · MO = 0 λ·H

(3.40)

λ · HO = const

(3.41)

Hence,

STATEMENT: If component of a resultant moment MO of all external forces along a fixed in an inertial space axis is equal to zero, its angular momentum is conserved along that axis.

EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS.

90

3.3.3 Impulse – momentum principle. If the resultant force and moment of all external forces are explicit functions of time, the derived equations of motion ½ ½ ½ ˙ =F ˙ =F ˙ =F P P P (3.42) ˙ O = MO ˙ G = MG H H h˙ G = MG

can be integrated with respect to time. Z t Z t Z t Z t ˙ ˙ O dt = Pdt = H Fdt, MO dt, to to to to Z t Z t Z t Z t ˙ ˙ HG dt = MG dt, MG dt hG dt = to

to

to

(3.43)

to

The following expression are called respectively Z t ˙ = P(t) − P(to ) = ∆P Pdt

(3.44)

to

- increment of the linear momentum Z t ˙ O dt = HO (t) − HO (to ) = ∆HO H

(3.45)

to

- increment of the angular momentum about point O Z t ˙ G dt = HG (t) − HG (to ) = ∆HG H

(3.46)

to

- increment of the angular momentum about G Z t h˙ G dt = hG (t) − hG (to ) = ∆hG

(3.47)

to

- increment of the relative angular momentum about G Z t Fdt = UF

(3.48)

to

- the linear impulse

Z

t

MO dt = UMO

(3.49)

MG dt = UMG

(3.50)

to

-the angular impulse about point O Z

t

to

-the angular impulse about point G Introduction of the above notations into 3.43 yields ∆P = UF ,

∆HO = UMO ,

∆HG = ∆hG = UMG

(3.51)

The above formulae allows to formulate the following statement STATEMENT: For any system of particles an increment of the linear (angular) momentum is equal to the linear (angular) impulse of all external forces.

PROBLEMS.

3.4

91

PROBLEMS.

Problem 28

F

B

S

B

S

Figure 11 On a massless and free to rotate about the vertical axis Z table, two dogs take position at B and S as shown in Fig. 11. The dog B has a mass mB greater then that of the dog S (mB > mS ).At the instant t = 0, when a food was placed outside the table at F , the dogs as well as the table were motionless. Show, that regardless of relative velocities developed by the dogs with respect to the table, the small dog S will reach the food first.

PROBLEMS.

92

Solution . RB B

F . RS RB

RS

S

O

Y

X

Figure 12 According to the introduced definition (see Eq. 3.15), the angular momentum of the system considered about fixed in the inertial space point O (see Fig. 12) in arbitrarily chosen instance of time is ˙ B + RS × mS R ˙S HO = RB × mB R = K(−mB RB R˙ B + mS RS R˙ S )

(3.52)

This angular momentum has to be equal to zero since the system is conserved about the vertical axis Z and at the beginning the system was motionless. Taking into account that RS = RB = R, we have mB ˙ R˙ S = RB mS

(3.53)

B Since m > 1, the absolute velocity of the small dog R˙ S is always greater then the mS absolute velocity of the big dog R˙ B .

PROBLEMS.

93

Problem 29 1

2 D

A

ω

R

d

α

Figure 13 The sprinkler shown in Fig. 13 distributes water of density at the volume rate Q. Each of the two nozzles has the exit area equal to A. The friction moment between the rotating part 1 and the stationary one 2 is equal M. Produce expression for the steady state angular velocity ω of the rotating part 1. The diameter d of the nozzle as well as D are small as compare with the distance R shown in the drawing 13.

PROBLEMS.

94

Solution F

ω

F O

O

x

x

G p

E

Z,z

Z,z

A

α

B

B vA

X

R

M

B vR

M x

R

ω

X

α

O

y

C

x

y Y

Y

a)

B vT

b) Figure 14

Let us consider the system of particles limited by the boundary shown in Fig. 14a) by the dash-dot line. This system is assembled of the rotating part 1 and water that is shadowed in Fig. 14a). Axes XY Z forms the inertial system of coordinates and axes xyz are attached to the part 1. The forces F represents external forces acting on the system of particles due to its interaction with the stationary part 2. The resultant force due to gravitation is denoted by G. The friction between the rotating and stationary part of the sprinkler is represented by moment M. The pressure p is due to interaction of particles of water that belong to the system and the cut-oﬀ stream of water. All the above specified forces should be classified as external with respect to the system considered. The atmospheric pressure is evenly distributed over the entire outer surface of the system of particles. Since its resultant is always equal to zero, it is not shown in this diagram. Position of the particles shown in the Fig. 14a) corresponds to an arbitrarily chosen instant of time t. Fig. 14b) presents position of these particles after a small increment in time ∆t. The increment in the linear momentum associated with the increment of time ∆t is due to volume of water B and C as well as loss of water in the volume E. Let us produce expression for the linear momentum of particles in the volume B. If the diameter of the nozzle is small with respect to the distance R, one can assume B that all particles that belong to the volume B have the same absolute velocity vA . Therefore the increment in the linear momentum is B ∆PB = mB vA

(3.54)

PROBLEMS.

95

where mB stands for total mass of particles in the volume B and is equal to 1 mB = Q∆t (3.55) 2 B is equal to sum of the velocity of transportation vTB and the The absolute velocity vA B relative velocity vR . ¯ ¯ ¯ ¯ i j k ¯ Q Q ¯¯ B B B 0 0 ω ¯¯ + j(− ) = vA = vT + vR = ω × R + i = ¯ A ¯ A R cos α −R sin α 0 ¯ Q (3.56) = i(Rω sin α) + j(Rω cos α − ) + k(0) A Hence µ ¶ Q B B ∆P = m i(Rω sin α) + j(Rω cos α − ) (3.57) A Increment in the angular momentum about fixed point O is ¯ ¯ ¯ i j k ¯¯ ¯ B B¯ −R sin α 0 ¯¯ = ∆HB O = R × ∆P = m ¯ R cos α Q ¯ Rω sin α Rω cos α − 0 ¯ A µ ¶ Q = k mB (R2 ω − R cos α) (3.58) A In the same manner one may produce expressions for increment in the angular momentum of particles associated with the volume C and E. ¶ µ Q C C 2 (3.59) ∆HO = k m (R ω − R cos α) A

∆HE (3.60) O = 0 Hence the total increment in the angular momentum is ¶ µ Q B C E 2 (3.61) ∆HO =∆HO + ∆HO + ∆HO = k m(R ω − R cos α) A where m is m = mB + mC = Q ∆t (3.62) To solve the problem, one may take advantage of the angular momentum - angular impulse principle. ∆HO = UMO (3.63) where UMO = −kM∆t (3.64) Introduction of 3.61, 3.62 and 3.64 into 3.63 yields Q Q ∆t(R2 ω − R cos α)∆t = −M∆t (3.65) A Hence the wanted angular speed of the sprinkler is ω=

Q2 R cos α A

Q

−M

R2

(3.66)

Chapter 4 KINETICS OF RIGID BODY. 4.1

LINEAR AND ANGULAR MOMENTUM.

According to consideration presented in chapter 2, motion of a rigid body may be determined by its angular velocity ω and the position vector ro which defines motion of the origin o of the body system of coordinates xyz. (Fig. 1). i

z Z ω

ri,o G z

ri rG

rG,o o y

ro

O

dm

x

y

Y x X

Figure 1 The rigid body can be considered as a system of particles assembled of infinitesimal elements dm which positions with respect to the body frame is given by the position vector ri,o . The mass of an individual element dm is (4.1)

dm = ρ(xyz)dxdydz

where ρ is density. Hence, the total mass of the rigid body can be expressed as follow Z ZZZ ρ(xyz)dxdydz = dm (4.2) m= V

m

The centre of gravity of the rigid body, according to the above notations, with respect to the body system of coordinates is Z 1 rG,o = ri,o dm (4.3) m m

LINEAR AND ANGULAR MOMENTUM.

97

According to consideration in the previous chapter linear momentum P is P = r˙ G m

(4.4)

r˙ G = r˙ o + r˙ G,o = r˙ o + r0G,o + ω × rG,o = r˙ o + ω × rG,o

(4.5)

where Introducing Eq. 4.5 into Eq. 4.4 one can obtain P = (˙ro + ω × rG,o )m

(4.6)

Angular relative momentum of the element dm with respect to the origin o is dho = ri,o × r˙ i,o dm = ri,o × (r0i ,o + ω × ri,o )dm = ri,o × (+ω × ri,o )dm Hence, the total angular relative momentum has form Z ho = ri,o × (+ω × ri,o )dm

(4.7)

(4.8)

m

Let us calculate the triple cross product, ri,o × (ω × ri,o ) = ω · (ri,o · ri,o ) − ri,o · (ri,o · ω)

(4.9)

If both vectors involved are given by their components along the body system of coordinates, namely ω = iω x + jω y + kωz

and

ri,o = ix + jy + kz

(4.10)

the triple product is as follows. ri,o × (ω × ri,o ) = (iω x + jω y + kω z )(x2 + y 2 + z 2 ) − (ix + jy + kz)(xω x + yω y + zω z ) = i(ω x (y 2 + z 2 ) − xyωy − xzω z ) +j(ω y (x2 + z 2 ) − yxω x − yzω z ) (4.11) +k(ωz (x2 + y 2 ) − zxωx − zyω y ) Hence ho =

i(+ω x +j(−ω x

Z

Zm

m

+k(−ω x

Z

m

2

2

(y + z )dm − ω y yxdm + ω y

Z

Z

xydm − ω z

m

(x2 + z 2 )dm − ωz

m

zxdm − ω y

Z

m

zydm + ω z

Z

m

Z

Zm

xzdm) yzdm)

m

(x2 + y 2 )dm)

(4.12)

LINEAR AND ANGULAR MOMENTUM.

Introducing notations R Ix = (y 2 + z 2 )dm,

R Iy = (x2 + z 2 )dm,

m

Ixy =

R

98

m

xy dm,

Iyz =

m

R

yz dm,

R Iz = (x2 + y 2 )dm m

Ixz =

m

R

xz dm

(4.13)

m

the components of vector ho ,according to 4.12, are hox = Ix ω x − Ixy ωy − Ixz ωz hoy = −Iyx ω x + Iy ωy − Iyz ωz hoz = −Izx ωx − Izy ωy + Iz ω z

(4.14)

where: Ix , Iy , Iz −are called moments of inertia and Ixy , Iyz , Ixz −are called products of inertia. The above relations can be written in the following matrix form. ⎡ ⎤ ⎡ ⎤⎡ ⎤ hox ωx Ix −Ixy −Ixz ⎣ hoy ⎦ = ⎣ −Iyx Iy −Iyz ⎦ ⎣ ω y ⎦ (4.15) hoz −Izx −Izy Iz ωz or shorter

{ho } = [I]{ω}

(4.16)

Matrix [I] is called inertia matrix. The formula 4.16 produces components of vector ho along body system of coordinates. Matrix {ω} is assembled of components of angular velocity along body system of coordinates.

z

ω

Z i ri

dm

G rG z

O y

y

x

Y

X x Figure 2 For the particular case of rotational motion of the rigid body about the fixed in the inertial space point O (see Fig. 2) the angular momentum is Z Z Z (4.17) HO = dHO = ri × r˙ i dm = ri ×(ri ×ω)dm m

m

m

PROPERTIES OF MATRIX OF INERTIA.

99

or in the matrix form {HO } = [I]{ω} 4.2

(4.18)

PROPERTIES OF MATRIX OF INERTIA.

The introduced definitions of moment of inertia e.g. Z Ix = (y 2 + z 2 )dm

(4.19)

m

and product of inertia e.g. Iyz =

Z

(4.20)

yz dm

m

permit elements of the inertia matrix to be calculated for bodies of a simple geometrical shape like a cylinder, sphere, rectangular block etc. As an example let as calculate the moments and products of inertia for the rectangular block shown in Fig. 3.

z

c

dm o y

z x

y

a

b

x

Figure 3

Ix =

ZZZ

Zc Zb Za (y + z )ρdxdydz = ( ( (y 2 + z 2 )ρdx)dy)dz 2

2

0

0

0

Zc Z b Zc Z b 2 2 a = ρ ( (y + z )x |0 )dy)dz = ρa ( (y 2 + z 2 )dy)dz 0

= ρa

0

Zc 0

1 ( y 3 |b0 +z 2 y |b0 )dz = ρa 3

0

Zc

0

1 ( b3 + z 2 b)dz 3

0

1 1 1 = ρa( b3 z |c0 + z 3 b |c0 ) = ρa (b3 c + c3 b) 3 3 3 1 1 2 2 2 ρabc(b + c ) = m(b + c2 ) = 3 3

(4.21)

PROPERTIES OF MATRIX OF INERTIA.

100

In a similar manner the remaining elements may be calculated. The result of such calculations is collected in appendix B. For bodies having more complicated shape (see Fig. 4) a division into small elements have to be carried out.

z

∆ mi

zi yi

y xi

x Figure 4 Then, each element can be considered as a particle and the integration may be replaced by summation. N X (yi2 + zi2 )∆mi Ix ≈

(4.22)

i=1

Eventually, the inertia matrix with respect to the arbitrarily chosen system of coordinates may be analytically established. ⎡ ⎤ Ix −Ixy −Ixz [I] = ⎣ −Iyx Iy −Iyz ⎦ (4.23) −Izx −Izy Iz

The following considerations allow us to calculate inertia matrix with respect to any system of coordinates if once it has been established for a system of coordinates. 4.2.1 Parallel axis theorem. Let us assume, that the inertia matrix of a body is known about the body system of coordinates xyz (Fig. 5). ⎡ ⎤ Ix −Ixy −Ixz [I] = ⎣ −Iyx Iy −Iyz ⎦ (4.24) −Izx −Izy Iz

Let a, b, c be coordinates of the centre of gravity G of the body considered. Let xG, yG, zG be the body system of coordinates parallel to xyz having its origin at G.

PROPERTIES OF MATRIX OF INERTIA.

101

zG dm z xG

zG

G

yG yG z

xG

c O

y

a

x

b x y Figure 5

Moment of inertia of the body along axis x, according to the previously introduced definition, is Z Z 2 2 Ix = (y + z )dm = ((b + yG )2 + (c + zG )2 )dm = m

=

Z

m

2 2 (b2 + yG + 2byG + c2 + zG + 2czG )dm

m

=

Z

2 (yG

+

2 zG )dm

+

m

Z

2

b dm +

m

= IxG + b2 m + c2 m + 2b

Z

2

c dm +

m

Z

yG dm + 2c

1 m

Z

m

But, 1 m

Z

xG dm,

m

Z

Z

2byG dm +

m

Z

2czG dm

m

(4.25)

zG dm

m

zG dm,

1 m

m

Z

yG dm

(4.26)

m

represents components of the distance between origin G and the centre of gravity, which is actually 0. Hence, Ix = IxG + m(b2 + c2 ) Now, let us consider a product of inertia. Z Z yzdm = (yG + b)(zG + c)dm Iyz = m

=

Z

m

m

yG zG dm + c

Z

m

yG dm + b

Z

m

zG dm +

(4.27)

Z

m

bcdm = IyzG + bcm (4.28)

PROPERTIES OF MATRIX OF INERTIA.

102

Similarly, one may derive expressions for the remained products and moments of inertia. All these equations can be written in the following matrix form. ⎤ ⎡ 2 −ac b + c2 −ab −bc ⎦ [I] = [IG ] + m ⎣ −ba a2 + c2 (4.29) 2 −ca −cb a + b2

The above formula is known as the parallel axes theorem and allows for calculation of inertia matrix about any axes xyz parallel to xG, yG, zG or vise versa. ⎤ ⎡ 2 −ac b + c2 −ab −bc ⎦ [IG ]=[I]-m⎣ −ba a2 + c2 (4.30) 2 −ca −cb a + b2 4.2.2 Principal axes. Inertia matrices about axes having the same origin.

Let us assumed that the matrix of inertia of a body is known about the system of coordinates x1, y1, z1 (see Fig. 6). Introduce new body system of coordinates x2 , y2 , z2 which has the same origin O, but it is turned with respect to the first one. The relative position of the system x2 y2 z2 can be uniquely determined by the matrix of direction cosines [C1→2 ]. ω

z2

z1

dm

t 23 y2 o

y1

x1 x2

Figure 6 ⎡ ⎤ ⎤ ⎡ ⎤⎡ ⎤ x1 x2 t11 t12 t13 x1 ⎣ y2 ⎦ = [C1→2 ] ⎣ y1 ⎦ = ⎣ t21 t22 t23 ⎦ ⎣ y1 ⎦ z2 z1 t31 t32 t33 z1 ⎡

(4.31)

where e.g.t23 = cos(∠y2 z1 ) If the body rotates with an angular velocity ω, components of its angular momentum along axes x1 , y1 , z1 , according to Eq. 4.16, are ⎡ ⎤ ⎡ ⎤ hox1 ωx1 ⎣ hoy1 ⎦ = [I]1 ⎣ ωy1 ⎦ (4.32) hoz1 ωz1

PROPERTIES OF MATRIX OF INERTIA.

103

Or in shorter form, {ho }1 = [I]1 {ω}1

(4.33)

{ho }2 = [I]2 {ω}2

(4.34)

Similarly But {ho }2 = [C1→2 ]{h}1

and

{ω}2 = [C1→2 ]{ω}1

(4.35)

Hence, introducing Eq. 4.35 into 4.34 yields [C1→2 ]{h}1 = [I]2 [C1→2 ]{ω}1

(4.36)

Upon introducing 4.33 into the above formula we have [C1→2 ][I]1 {ω}1 = [I]2 [C1→2 ]{ω}1

(4.37)

and after a little manipulation one may obtain [I]2 = [C1→2 ][I]1 [C1→2 ]T

(4.38)

The above formula permits to calculate the matrix of inertia along the system of coordinates x2 , y2 , z2 when the matrix of inertia about the system of coordinates x1 , y1 , z1 is known. Principal axes. zp

z1

yp o

y1

x1 xp

Figure 7 In this paragraph will be proved existence of such a system of coordinates xp yp zp (see Fig. 7), turned with respect to the arbitrarily chosen system x1 y1 z1 , that all products of inertia about these axes are equal to 0. ⎤ ⎡ 0 Ixp 0 [I]p = ⎣ 0 Iyp 0 ⎦ (4.39) 0 0 Izp

PROPERTIES OF MATRIX OF INERTIA.

104

Axes xp yp zp are called principal axes. According to the previous considerations we have [I]p = [C1→p ][I]1 [C1→p ]T

(4.40)

[C1→p ]T [I]p = [I]1 [C1→p ]T

(4.41)

or or ⎫ ⎧ ⎫ ⎧ ⎫⎤ ⎡ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎤ ⎡⎧ ⎨ t11 Ixp ⎬ ⎨ t21 Iyp ⎬ ⎨ t31 Izp ⎬ ⎨ t11 ⎬ ⎨ t21 ⎬ ⎨ t31 ⎬ ⎣ t12 Ixp t12 , [I]1 t22 , [I]1 t32 ⎦ , t22 Iyp , t32 Izp ⎦ = ⎣[I]1 ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ t13 Ixp t23 Iyp t33 Izp t13 t23 t33 (4.42) The above relationship is fulfilled if and only if the corresponding columns are identical. Hence, ⎫ ⎧ ⎫ ⎧ ⎨ t11 ⎬ ⎨ t11 ⎬ = [I]1 t12 Ixp [1] t12 ⎭ ⎩ ⎭ ⎩ t13 t13 ⎧ ⎫ ⎧ ⎫ ⎨ t21 ⎬ ⎨ t21 ⎬ = [I]1 t22 Iyp [1] t22 ⎩ ⎭ ⎩ ⎭ t23 t23 ⎧ ⎫ ⎧ ⎫ ⎨ t31 ⎬ ⎨ t31 ⎬ Izp [1] t32 = [I]1 t32 (4.43) ⎩ ⎭ ⎩ ⎭ t33 t33

where

⎡

⎤ 1 0 0 [1] = ⎣ 0 1 0 ⎦ 0 0 1

The relationships 4.39 form three sets of linear ⎧ ⎨ t11 ([I]1 − Ixp [1]) t12 ⎩ t13 ⎧ ⎨ t21 ([I]1 − Iyp [1]) t22 ⎩ t23 ⎧ ⎨ t31 ([I]1 − Izp [1]) t32 ⎩ t33

homogeneous equations of form ⎫ ⎬ = 0 ⎭ ⎫ ⎬ = 0 ⎭ ⎫ ⎬ = 0 (4.44) ⎭

The above sets of equations have not trivial solutions if and only if their common characteristic determinant is equal to 0 |[I]1 − Ip [1]| = 0

(4.45)

PROPERTIES OF MATRIX OF INERTIA.

or

¯ ¯ Ix1 − Ip −Ix1 y1 −Ix1 z1 ¯ ¯ −Iy1 x1 Iy1 − Ip −Iy1 z1 ¯ ¯ −Iz1 x1 −Iz1 y1 Iz1 − Ip

105

¯ ¯ ¯ ¯=0 ¯ ¯

(4.46)

The last equation, if developed, forms an algebraic equation of third order. Hence, it has three roots. It may be proved that for any matrix of inertia [I]1 , the roots of equation 4.46 Ixp , Iyp , Izp are always real and positive. For each of these roots the sets of equations 4.44 becomes linearly dependent. Only two of them within each set are independent and may be used to identify the unknown direction cosines t11 , t12 , t13 . (Ix1 − Ixp )t11 + (−Ix1 y1 )t12 + (−Ix1 z1 )t13 = 0 (−Iy1 x1 )t11 + (Iy1 − Ixp )t12 + (−Iy1 z1 )t13 = 0

(4.47)

Since the direction cosines have to fulfill the following relationship t211 + t212 + t213 = 1

(4.48)

Equations 4.47 and 4.48 form set of equation which determines the direction cosines t11 , t12 , t13 . In similar manner one may obtain equations which allows the other direction cosines to be determined. (Ix1 − Iyp )t21 + (−Ix1 y1 )t22 + (−Ix1 z1 )t23 = 0 (−Iy1 x1 )t21 + (Iy1 − Iyp )t22 + (−Iy1 z1 )t23 = 0 t221 + t222 + t223 = 1

(4.49)

(Ix1 − Izp )t31 + (−Ix1 y1 )t32 + (−Ix1 z1 )t33 = 0 (−Iy1 x1 )t31 + (Iy1 − Izp )t32 + (−Iy1 z1 )t33 = 0 t231 + t232 + t233 = 1

(4.50)

Solutions of the above equations determine uniquely a position of the principal axes.

PROPERTIES OF MATRIX OF INERTIA.

106

4.2.3 Problems Problem 30

l

A

l

Figure 8 The figure above shows an element assembled of two identical thin and uniform bars. The bars were welded together at right angles at A. Each of the bars have a length l and mass m. Calculate the principal moments of inertia of the assembly along axes through its centre of gravity. Use the following numerical values; l = 1m, m = 12kg.

PROPERTIES OF MATRIX OF INERTIA.

107

Solution. y1

y2 x3

y3

l/4 G1 1

x2 G

l/4

A

G2

x1 2

Figure 9 Coordinates of the centre of gravity of the whole assembly are (see Fig. 9) l 4 Inertia matrix about axes x1 y1 z1 for the element 1. ⎡ ml2 ⎤ 0 0 3 I1,1 = ⎣ 0 0 0 ⎦ 2 0 0 ml3

(4.51)

xG = yG =

(4.52)

Inertia matrix about axes x1 y1 z1 for the element 2. ⎤ ⎡ 0 0 0 2 0 ⎦ I1,2 = ⎣ 0 ml3 ml2 0 0 3

(4.53)

Inertia matrix about axes x1 y1 z1 for the whole assembly. ⎤ ⎡ ml2 0 0 3 2 I1 = ⎣ 0 ml3 0 ⎦ 2ml2 0 0 3

(4.54)

Inertia matrix about axes x2 y2 z2 for the whole assembly. ⎡ 2 −ac b + c2 −ab 2 2 ⎣ −ba a + c −bc I2 = I1 − 2m 2 −ca −cb a + b2 ⎡ ml2 ⎤ ⎡ l2 l2 0 0 3 16 16 2 2 l2 = ⎣ 0 ml 0 ⎦ − 2m ⎣ l 3

0 ⎡

0

2ml2 3

⎤ 5 3 0 ml ⎣ 3 5 0 ⎦ = 24 0 0 10 2

16

16

0

0

⎤ ⎦

⎤ 0 0 ⎦

l2 8

(4.55)

PROPERTIES OF MATRIX OF INERTIA.

108

Since axis x3 is axis of symmetry of the body considered, x3 y3 z3 are principal axes. Transfer matrix C2→3 between axes x2 y2 z2 and x3 y3 z3 is. ⎡ ⎤ ⎤⎡ ⎤ ⎡ x3 x2 cos 45o cos 45o cos 90o ⎣ y3 ⎦ = ⎣ cos 135o cos 45o cos 90o ⎦ ⎣ y2 ⎦ z3 cos 90o cos 90o cos 0o z2 ⎤⎡ ⎤ ⎡ √2 √2 0 x2 2√ √2 2 2 ⎦ ⎣ ⎣ y2 ⎦ (4.56) = −2 0 2 z2 0 0 1

Hence, inertia matrix about the principal axes x3 y3 z3 ⎤⎡ ⎡ √2 √2 0 5 2 2 2 ml ⎣ √2 √2 T ⎦ ⎣ 3 I3 = C2→3 I2 C2→3 = −2 0 2 24 0 0 0 1 ⎡ ⎤ 4 0 0 = ⎣ 0 1 0 ⎦ 0 0 5

is

√ ⎤ ⎡ √2 ⎤ 2 − 0 3 0 2 2 √ √ 2 5 0 ⎦⎣ 2 0 ⎦ 2 2 0 10 0 0 1

(4.57)

PROPERTIES OF MATRIX OF INERTIA.

109

Problem 31 r1

R1

z

O1 O y l

l r2

R2 O2

Figure 10 The rigid body, shown in the figure 10, rotates about a fixed point O. To analyze its motion it is necessary to know the principal axes through the point O and principal moments of inertia. Calculate the angular position of the principal axes with respect to the given system of coordinates xyz as well as magnitudes of the principal moments of inertia. The balls 1 and 2 are homogeneous and have masses m1 and m2 respectively whereas the rods can be considered as rigid and massless. Assume the following numerical data: m1 = 1kg - mass of the ball 1 m2 =√2kg - mass of the ball 2 r1 = √5/2 m - radius of the ball 1 r2 = 5m- radius of the ball 2 R1 = 1m R2 = 2m l = 1m

PROPERTIES OF MATRIX OF INERTIA.

110

Solution. z1 r1

R1

O1

z

z2

y1 O y l

l

R2 O2

r2

y

2

Figure 11 Moment of inertia of the ball 1 about axes x1 y1 z1 (see Fig. 11) 2 Ix1 = Iy1 = Iz1 = m1 r12 = 0.5 kgm2 5

(4.58)

Moment of inertia of the ball 2 about axes x2 y2 z2 2 Ix2 = Iy2 = Iz2 = m2 r22 = 4 kgm2 5

(4.59)

Coordinates of the points O1 and O2 with respect to system of coordinates xyz are as follow O1 (0, −l, R1 )

O2 (0, l, −R2 )

(4.60)

Hence, according to parallel axes theorem the matrix of inertia of the ball 1 about system of coordinates xyz is ⎤ ⎡ 2 ⎤ ⎡ 0 0 0 (l + R12 ) Ix1 0 0 R12 −(−lR1 ) ⎦ [I1 ] = ⎣ 0 Iy1 0 ⎦ + m1 ⎣ 0 0 Iz1 0 −(R1 (−l)) l2 ⎡ ⎤ 2.5 0 0 = ⎣ 0 1.5 1 ⎦ kgm2 (4.61) 0 1 1.5 and the matrix of inertia of the ball 2 is ⎤ ⎡ 2 ⎤ ⎡ 0 0 0 (l + R22 ) Ix2 0 0 R22 −(−lR2 ) ⎦ [I2 ] = ⎣ 0 Iy2 0 ⎦ + m1 ⎣ l2 0 0 Iz2 0 −(R2 (−l)) ⎡ ⎤ 14 0 0 (4.62) = ⎣ 0 12 4 ⎦ kgm2 0 4 6

PROPERTIES OF MATRIX OF INERTIA.

111

Hence, matrix of inertia of the whole assembly about system of coordinates xyz is ⎡ ⎤ 16.5 0 0 [I] = [I1 ] + [I2 ] = ⎣ 0 13.5 5 ⎦ kgm2 (4.63) 0 5 7.5

Principal moments of inertia are roots of the following equation ¯ ¯ ¯ ¯ 16.5 − Ip 0 0 ¯ ¯ ¯=0 ¯ 0 13.5 − I 5 p ¯ ¯ ¯ 0 5 7.5 − Ip ¯

(4.64)

or in a more developed form

(16.5 − Ip )[(13.5 − Ip )(7.5 − Ip ) − 25] = 0

(4.65)

Root of the above equation are: Ipx = 16.5kgm2 ,

Ipy = 16.33kgm2 ,

Ipz = 4.67kgm2

(4.66)

If matrix of directional cosines between principal axes and the system of coordinates xyz is defined as follow ⎤ ⎡ ⎤⎡ ⎤ ⎡ x xp t11 t12 t13 ⎣ yp ⎦ = ⎣ t21 t22 t23 ⎦ ⎣ y ⎦ (4.67) zp t31 t32 t33 z

The corresponding equations for its elements take form

(16.5 − Ipx )t11 + (0)t12 + (0)t13 = 0 (0)t11 + (13.5 − Ipx )t12 + (5)t13 = 0 t211 + t212 + t213 = 1

(4.68)

(16.5 − Ipy )t21 + (0)t22 + (0)t23 = 0 (0)t21 + (13.5 − Ipy )t22 + (5)t23 = 0 t221 + t222 + t223 = 1

(4.69)

(16.5 − Ipz )t31 + (0)t32 + (0)t33 = 0 (0)t31 + (13.5 − Ipz )t32 + (5)t33 = 0 t231 + t232 + t233 = 1

(4.70)

They are fulfilled for t11 = 1 t12 = 0.00000 t21 = 0 t22 = 0.86700 t31 = 0 t32 = −0.4926

t13 = 0.0000 t23 = 0.4926 t33 = 0.8700

(4.71)

PROPERTIES OF MATRIX OF INERTIA.

112

Hence, matrix of directional cosines Cxyz→xp yp zp is ⎤ ⎡ ⎤ ⎡ cos 90o cos 90o 1 0 0 cos 0o Cxyz→xp yp zp = ⎣ 0 0.867 0.4926 ⎦ = ⎣ cos 90o cos 29.88o cos 60.48o ⎦ cos 90o cos 119.75o cos 29.88o 0 −0.4926 0.87 (4.72) Position of the principal axes with respect to the system of coordinates xyz is shown in Fig 12. zp

z 60.48 o yp 119.75 o

O

29.88 o y

x xp

Figure 12

PROPERTIES OF MATRIX OF INERTIA.

113

Problem 32 y

O

x

z

Figure 13 Matrix of inertia of a flat object shown in Fig. 13 about axes x, y, z is ⎤ ⎡ ⎡ ⎤ Ix −Ixy −Ixz 10 −5 0 ⎣ −Iyx Iy −Iyz ⎦ = ⎣ −5 3 0 ⎦ kgm2 −Izx −Izy Iz 0 0 13

Determine position of the principal axes of the object through the origin O.

PROPERTIES OF MATRIX OF INERTIA.

114

Solution. The principal moments of inertia can be obtained by solving the following equation. ¯ ¯ ¯ 10 − Ip −5 ¯ 0 ¯ ¯ ¯ −5 ¯=0 0 3 − Ip (4.73) ¯ ¯ ¯ ¯ 0 0 13 − Ip which can be developed as follow

(Ip2 − 3Ip − 10Ip + 30 − 25)(13 − Ip ) = 0

(4.74)

Hence Ipx Ipy Ipz

√ 169 − 20 = 12.6 = √2 13 − 169 − 20 = 0.4 = 2 = 13kgm2 13 +

(4.75)

Since the object is flat, the principal axis zp must coincide the axis z. The directional cosines between the principal axis xp and axes x and y may be obtained from the following set of equations. (10 − 12.6)t11 + (−5)t12 = 0 t211 + t212 = 0

(4.76) (4.77)

From Eq. 4.76 one can get t11 =

−5 t12 = −1.92t12 2.6

(4.78)

Introduction of Eq. 4.78 into Eq. 4.77 yields 3.7t212 + t212 = 1 Hence

r

1 = 0.46 =+ 4.7 r 1 t12 ” = − = −0.46 4.7 t012

Equation 4.78 yields

(4.79)

(4.80) (4.81)

t011 = −0.885

(4.82)

t11 ” = 0.885

(4.83)

∠(xp , x)0 = arccos(t011 ) = 152.5o ∠(xp , y)0 = arccos(t012 ) = 62.5o

(4.84)

Corresponding angles are

PROPERTIES OF MATRIX OF INERTIA.

115

∠(xp , x)” = arccos(t11 ”) = 27.5o ∠(xp , y)” = arccos(t12 ”) = 117.5o y

(4.85)

yp

117.5 o x

O z

27.5o xp

Figure 14 The solution 4.119 have to be neglected since those angles determine left handed system of coordinates and this is in disagreement with the adopted assumptions. The only possible position of axes xp and yp with respect to system of coordinates xy is shown in Fig. 14

PROPERTIES OF MATRIX OF INERTIA.

116

Problem 33

y b C y

a

x

α

x

Figure 15 The rectangular plate, shown in Fig. 15, rotates about axis x − x. Determine moment of inertia of the plate about this axis of rotation. Given are: a, b - length of sides of the plate. α - angle between side a and the axis of rotation x − x. m - mass of the plate. Moment of inertia of the rectangular plate about axis y − y is determine by the following formula 1 Iy = mb2 12

PROPERTIES OF MATRIX OF INERTIA.

117

Problem 34

2r

r

O

Figure 16 The cam shown in Fig. 16 can be considered as a thin, flat, uniform and rigid body of mass M. Compute the principal moments of inertia of the cam about axes through the point O. Given are: r = 1[m], M = 20[kg] Answer: Ipx = 16.375kgm2 Ipy = 21.625kgm2 Ipz = 38kgm2

PROPERTIES OF MATRIX OF INERTIA.

118

Problem 35 y (b)

(a)

l1

yG

l/2

xG 90 o

l2

G

α

l

x

Figure 17 Two uniform and rigid rods of length l1 and l2 are joined together to form a rigid body (see Fig. 17a). Their masses are m1 and m2 respectively. Determine matrix of inertia of this body about axis xyz. Moments of inertia of the uniform rod of length l and mass m, shown in Fig. 17b, about axis through its centre of gravity G are as follows. IxG = 0,

IyG = IzG =

1 2 ml 12

PROPERTIES OF MATRIX OF INERTIA.

119

Problem 36

yG

l1

G xG 1 2

l2

Figure 18 To find its matrix of inertia, the golf club shown in Fig. 18 was approximated by the slender rod 1 and the particle 2. Mass of the rod and the particle is m1 and m2 respectively. Produce 1. expression for the matrix of inertia of the golf club about axes xG yG zG through its centre of gravity G. Answer: ⎤ ⎡ 1 m l2 −b2 (m1 +m2 ) ab(m1 + m2 ) 0 3 1 1 ⎦ ab(m1 + m2 ) m2 l22 −a2 (m1 +m2 ) 0 IG = ⎣ 1 2 2 2 2 0 0 m l +m2 l2 −(a +b )(m1 +m2 ) 3 1 1

l1 2 l2 where a = mm1 +m b = 2(mm11+m 2 2) 2. expression for the principal moments of inertia about axes through the centre of gravity Answer: The principal moment of inertia are solution of the following equation ¯ £1 ¤ ¯ m1 l12 −b2 (m1 +m2 ) −Ip ab(m1 + m2 ) 0 ¯ 3 2 2 ¯ + m ) [m l −a (m +m )] −I 0 ab(m 1 2 2 2 1 2 p £ ¯ ¤ 1 2 2 2 2 ¯ m l +m l −(a +b )(m +m ) −Ip 0 0 1 2 1 2 1 2 3

¯ ¯ ¯ ¯ =0 ¯ ¯

PROPERTIES OF MATRIX OF INERTIA.

120

Problem 37

y

l1

1

2 G

l2

x

Figure 19 Two uniform bars of length l 1 and l 2 and mass m 1 and m 2 respectively were joined together to form the rigid body shown in Fig. B2. Produce: 1. The expression for the coordinates of the centre of gravity G of the rigid body. Answer: l2 l1 xG = 2(mm12+m ; yG = 2(mm11+m 2) 2) 2. The matrix of inertia of the rigid body about axes xyz. ⎤ ⎡ Answer: m1 l12 0 0 ⎥ ⎢ 3 m2 l22 I=⎣ 0 ⎦ 0 3 1 2 2 0 0 (m1 l1 + m2 l2 ) 3 3. The principal moments of inertia about the axes through the center of gravity G. Answer: IG ⎡= ⎤ m1 l12 2 − (m + m )y (m + m )x y 0 1 2 G 1 2 G G ⎢ 3 ⎥ m2 l22 2 = ⎣ (m1 + m2 )xG yG ⎦= − (m + m )x 0 1 2 G 3 1 2 2 2 2 0 0 (m1 l1 + m2 l2 ) − (m1 + m2 ) (xG + yG ) 3 ⎡ ⎤ −IxG yG 0 IxG IyG 0 ⎦ = ⎣ −IxG yG 0 0 IzG 4. The matrix of directional cosines between the system of coordinates xyz and the principal axes through the center of gravity G. Answer: √ (IxG +IyG )− (IxG +IyG )2 −4(IxG IyG −Ix2G yG ) Ixp = 2 √ (IxG +IyG )+ (IxG +IyG )2 −4(IxG IyG −Ix2G yG ) Iyp = 2 Izp = IzG

PROPERTIES OF MATRIX OF INERTIA.

121

The moment of inertia of a uniform bar of mass m and length l about the axis through its center of gravity is IG =

1 ml2 12

PROPERTIES OF MATRIX OF INERTIA.

122

Problem 38

Z

3 l 2 1

O Y l

l X Figure 20 Three uniform rods each of mass m and length l are joined together to form the rigid body shown in Fig. 20. Produce 1. The matrix of inertia of the body about axes XY Z 2. The principal moments of inertia about axes through the point O 3. The matrix of directional cosines that locate the principal axes through the point O in relation to the system of coordinates XY Z.

PROPERTIES OF MATRIX OF INERTIA.

123

Solution. Z 3

G3(a,b,c) 2 1

l c=l/2

O Y l

b=l X

Figure 21 The matrix of inertia of the rod 1 about axes XY Z. ⎤ ⎡ 0 0 0 I1O = ⎣ 0 13 0 ⎦ ml2 0 0 13

The matrix of inertia of the rod 2 about axes XY Z. ⎤ ⎡ 1 0 0 3 I2O = ⎣ 0 0 0 ⎦ ml2 0 0 13

The matrix of inertia of the rod 3 about axes XY Z. ⎤ ⎡ 2 −ac b + c2 −ab −bc ⎦ I3O = I1G3 + m ⎣ −ba a2 + c2 2 −ca −cb a + b2 ⎤ ⎤ ⎡ 2 ⎡ 1 l 2 + ( ) 0 0 l 0 0 2 12 2 1 0 ⎦ ml2 + m ⎣ 0 ( 2l )2 − l2 ⎦ = ⎣ 0 12 2 0 0 0 0 − l2 l2 ⎤ ⎡ 4 0 0 3 = ⎣ 0 13 − 12 ⎦ ml2 0 − 12 1

(4.86)

(4.87)

PROPERTIES OF MATRIX OF INERTIA.

124

The matrix of inertia of the body about axes XY Z. IO = I1O + I2O + I3O = ⎤ ⎤ ⎤ ⎡ 1 ⎡ 4 ⎡ 0 0 0 0 0 0 0 3 3 = ⎣ 0 13 0 ⎦ ml2 + ⎣ 0 0 0 ⎦ ml2 + ⎣ 0 13 − 12 ⎦ ml2 0 0 13 0 0 13 0 − 12 1 ⎤ ⎡ 5 0 0 3 = ⎣ 0 23 − 12 ml2 ⎦ 0 − 12 35

(4.88)

The principal moments of inertia about axes through the origin O are roots of the following equation ¯ ¯ 5 2 ¯ ¯ ml − Ip 0 0 ¯ ¯ 3 2 1 2 2 ¯=0 ¯ 0 ml − I − ml (4.89) p 3 2 ¯ ¯ 5 1 2 2 ¯ ¯ 0 − 2 ml ml − I p 3 They are

5 2 ml = 1.666ml2 : 1.873ml2 : 0.459ml2 3 Taking into consideration the matrix of inertia IO , one can see that Z Z XY dm = 0 and XZdm = 0 m

(4.90)

(4.91)

m

Therefore the axis X coincides with the principal axis Xp and the root associated with the first raw of the determinant 4.91 ( 53 ml2 = 1.666ml2 ) stands for the moment of inertia about the principal axis Xp . IXp = 1.666ml2

(4.92)

Indeed, if the matrix of directional cosines is of the following form ⎤ ⎡ ⎤ ⎡ t11 t12 t13 cosh(Xp X) cosh(Xp Y ) cosh(Xp Z) CXY Z→Xp Yp Zp = ⎣ t21 t22 t23 ⎦ = ⎣ cosh(Yp X) cosh(Yp Y ) cosh(Yp Z) ⎦ cosh(Zp X) cosh(Zp Y ) cosh(Zp Z) t31 t32 t33 (4.93) the equations for determination of its elements associated with the above root are 5 ( − 1.666)t11 + (0) t12 + (0) t13 = 0 3 2 (0) t11 + ( − 1.666)t12 − (0.5) t13 = 0 3 5 (0) t11 − (0.5) t12 + ( − 1.666)t13 = 0 3

(4.94)

The first equation is fulfilled for any magnitudes of t11 , t12 and t13 . The two last equations if and only if t12 = t13 = 0. Because t211 + t212 + t213 = 1

(4.95)

PROPERTIES OF MATRIX OF INERTIA.

125

the magnitude of t11 must be 1. Hence h(Xp X) = 0, h(Xp Y ) = 90o h(Xp Z) = 90o

(4.96)

Let us assign the remaining roots as follows IY p = 1.873ml2 and IZp = 0.459ml2

(4.97)

The axis Y is located by the directional cosines linked with second raw of 4.93. They are 5 ( − 1.873)t21 + (0)t22 + (0) t23 = 0 3 2 (0) t21 + ( − 1.873)t22 − (0.5) t23 = 0 3 t221 + t222 + t223 = 1

(4.98)

t21 = 0, t22 = ±0.3823, t23 = ∓0.9236

(4.99)

They are fulfilled for

Hence the angles between axis Yp and XY Z are h(Yp X) = 90o , h(Yp Y ) = 67.52o h(Yp Z) = 157.45

(4.100)

h(Yp X) = 90o , h(Yp Y ) = 112.47o h(Yp Z) = 22.54

(4.101)

The axis Z is located by the directional cosines linked with third raw of 4.93. They are 5 ( − 0.459)t31 + (0)t32 + (0) t33 = 0 3 2 (0) t31 + ( − 0.459)t32 − (0.5) t33 = 0 3 t231 + t232 + t233 = 1

(4.102)

These equations are fulfilled for t31 = 0, t32 = ±0.92360, t33 = ±0.3823

(4.103)

Hence the angles between axis Zp and XY Z are h(Zp X) = 90o , h(Zp Y ) = 22.54o h(Zp Z) = 67.52o

(4.104)

h(Zp X) = 90o , h(Zp Y ) = 157.45 h(Zp Z) = 112.47o

(4.105)

The analysis oﬀers two acceptable solutions shown in Fig. 22

PROPERTIES OF MATRIX OF INERTIA.

126 Z

Yp(2)

67.52o Zp(1)

X Y 22.54o Zp(2)

Yp(1)

Figure 22 To verify the result obtained let us transfer the inertia matrix from the system (1) (1) (1) of coordinates X, Y, Z into Xp , Yp , Zp .The matrix of directional cosines is ⎤ ⎡ ⎡ ⎤ t11 t12 t13 1 0 0 CXY Z→Xp(1) Yp(1) Zp(1) = ⎣ t21 t22 t23 ⎦ = ⎣ 0 0.3823 −0.9236 ⎦ (4.106) t31 t32 t33 0 0.92360 0.3823 Hence

= CXY Z→Xp(1) Yp(1) Zp(1) IO CTXY Z→X (1) Y (1) Z (1) I(1) p ⎤ ⎡ ⎤ ⎡ p5 p p 0 0 1 0 0 1 0 0 3 = ⎣ 0 0.3823 −0.9236 ⎦ ⎣ 0 23 − 12 ⎦ 0 0.382 3 0.923 6 0 − 12 35 0 −0.923 6 0.382 3 0 0.92360 0.3823 ⎡ ⎤ 1.6667 0.0 0.0 ⎣ 0.0 1.8723 0.0 ⎦ = (4.107) 0.0 0.0 0.45919

As one can see this transformation results in the diagonal matrix having elements equal to the principal moments of inertia IXp = 1.666ml2 , IY p = 1.873ml2 and IZp = 0.459ml2

(4.108)

KINETIC ENERGY.

4.3

127

KINETIC ENERGY.

4.3.1 Rotational motion. DEFINITION: It is said that a rigid body performs the rotational motion if one point of the body considered, due to its constraints, is motionless with respect to the inertial space. The kinetic energy of a particle of mass dm (see Fig. 23) is 1 1 dT = r˙ 2i dm = r˙ i · r˙ i dm (4.109) 2 2 where ri is the position vector of the particle. If the vector ri is given by its com-

ω

z

Z

dm

ri O

y

Y

X x Figure 23 ponents along the body system of coordinates rotating with angular velocity ω, the above expression may be developed as follow 1 1 1 dT = r˙ i · (r0i + ω × ri )dm = r˙ i · (ω × ri )dm = ω · (ri × r˙ i )dm 2 2 2 But, according to consideration in the previous chapter

(4.110)

(ri × r˙ i )dm = dHO Hence

1 dT = ω·dHO 2 Upon integrating of Eq. 4.111 over the entire body we are finally getting ⎤ ⎡ ωx T = 12 ω · HO = 12 [ωx , ω y , ω z ][I] ⎣ ω y ⎦ ωz

(4.111)

(4.112)

where

ωx , ωy , ω z -are components of the absolute angular velocity of the body along a body system of coordinates [I] – is the inertia matrix about that system of coordinates.

KINETIC ENERGY.

128

4.3.2 General motion. DEFINITION: It is said that a rigid body performs the general motion if its motion can not be classified as the rotational one. Kinetic energy of a particle dm of a rigid body (see Fig. 24) is 1 dT = r˙ 2i dm 2

(4.113)

In the general motion its position vector ri may be defined as follows

z

Z

ω

dm

i

ri

ri,G G

rG

y

O x

Y

X Figure 24 (4.114)

ri = rG +ri,G Introducing Eq. 4.114 into Eq. 4.113 one can obtain that 1 1 dT = (˙rG + r˙ i,G )2 dm = (˙r2G dm + r˙ 2i,G dm + 2˙rG · r˙ i,G dm) 2 2

(4.115)

Integration over the entire body yields the total kinetic energy in the fallowing form Z Z Z 1 2 2 T = (˙rG dm + r˙ i,G dm + 2˙rG · r˙ i,G dm) (4.116) 2 m

m

m

KINETIC ENERGY.

129

But

Z

m

Z

dm = m

m

r˙ 2i,G dm

=

Z

m

= ω· 2˙rG ·

Z

m

r˙ i,G · (ω × ri,G )dm = Z

m

r˙ i,G dm = 2˙rG ·

Z

ω · (ri,G × r˙ i,G )dm

m

ri,G × r˙ i,G dm = ω · hG Z

ω × ri,G dm

m

= 2˙rG · (ω×

Z

ri,G dm) = 2˙rG · (ω × rG,G m) = 0 (4.117)

m

The last relation becomes obvious if we notice that for the chosen system of coordinates rG,G = 0. Implementation of Eq. 4.117 into Eq. 4.116 gives the following formula for kinetic energy. ⎤ ωx T = 12 r˙ 2G m + 12 ω · hG = 12 r˙ 2G m + 12 [ω x , ωy , ωz ][I] ⎣ ωy ⎦ ωz ⎡

(4.118)

where: [I] - is inertia matrix about system of coordinates through centre of gravity of the body. ωx , ω y , ω z -are components of the angular velocity of the body along that system of coordinates. The last formula permits to formulate the following statement. STATEMENT: Kinetic energy of a rigid body is equal to the sum of its energy in the translational motion with velocity of its centre of gravity (energy of translation) and the energy in the rotational motion about its centre of gravity (energy of rotation).

KINETIC ENERGY.

130

4.3.3 Problems Problem 39 y1

Z, z1

X

α

Y

x1

z2

Z, z1 A

y1

x1

z1

A l

a

β

1 2

G

x2

Figure 25 Fig. 25 shows a mechanical system. Its link 1 is free to rotate about the vertical axis Z of the inertial system of coordinates XY Z. Moment of inertia of the link 1 about axis Z is I1Z . The link 2 of the system is hinged to the link 1 at the point A as it is shown in Fig. 25. Distance between the point A and axis of rotation Z 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 of inertia about these axes are I2x2 , I2y2 , I2z2 respectively. Mass of the link 2 is equal to m. Derive expression for the kinetic energy of the system as a function of angles α and β.

KINETIC ENERGY.

131

Solution y1

Z, z1

X

α

Y

x1

z2

Z, z1

rA

A

y1

x1

z1

A rGA

l

a

β

1 2

G

x2

Figure 26 The kinetic energy of the system T is sum of the kinetic energy possessed by the link 1 T1 and kinetic energy possessed by the link 2 T2 . T = T1 + T2

(4.119)

The link 1 performs the rotational motion about axis Z which is fixed in the inertial space. Therefore its kinetic energy is 1 T1 = I1Z ω 21 2

(4.120)

The link 2 performs the general motion in the inertial space. Therefore its kinetic energy is determined by the following equation. ⎤⎡ ⎤ ⎡ 0 ω 2x2 I2x2 0 1 1 2 T2 = m2 vG + [ω 2x2, ω2y2, ω2z2 ] ⎣ 0 I2y2 0 ⎦ ⎣ ω 2y2, ⎦ (4.121) 2 2 0 0 I2z2 ω 2z2 In the above equation vG stands for the absolute velocity of the centre of gravity of the link 2 and ω2x2, ω 2y2, ω 2z2 are components of its absolute angular velocity. Angular velocity of the link 1 is ω 1 = Kα˙ (4.122) The absolute angular velocity of the link 2 is ω2 = ω1 + ω 21 = Kα˙ + j1 β˙

(4.123)

KINETIC ENERGY.

132

Its components along the body system of coordinates x2 , y2 , z2 are ˙ · i2 = k1 · i2 α˙ + j2 · i2 β˙ = −α˙ sin β ω 2x2 = ω 2 · i2 = (Kα˙ + j1 β) ˙ · j2 = k1 · j2 α˙ + j2 · j2 β˙ = β˙ ω 2y2 = ω 2 · j2 = (Kα˙ + j1 β) ˙ · k2 = k1 · k2 α˙ + j2 · k2 β˙ = α˙ cos β ω 2z2 = ω 2 · k2 = (Kα˙ + j1 β)

(4.124)

The absolute velocity of the centre of gravity of the link 2 may be obtained by integration of its position vector. rG = rA + rGA = j2 a − k2 l

(4.125)

Hence, r˙ G

¯ ¯ ¯ i2 j2 k2 ¯¯ ¯ = vG = r0G + ω2 × rG = ¯¯ −α˙ sin β β˙ α˙ cos β ¯¯ ¯ 0 a −l ¯ = i2 (−lβ˙ − aα˙ cos β) + j2 (−lα˙ sin β) + k2 (−aα˙ sin β)

(4.126)

Taking into consideration Eq’s.. 4.122, 4.124 and 4.126 the wanted expression for the kinematic energy function is T =

1 I1Z α˙ 2 2

⎤ ˙ − aα˙ cos β −l β 1 ⎦ + m2 −lβ˙ − aα˙ cos β −lα˙ sin β −aα˙ sin β ⎣ −lα˙ sin β 2 −aα˙ sin β ⎡ ⎤⎡ ⎤ − α ˙ sin β I 0 0 2x2 ¤ 1£ ⎦ + −α˙ sin β β˙ −α˙ cos β ⎣ 0 I2y2 0 ⎦ ⎣ β˙ 2 0 0 I2z2 −α˙ cos β £

¤

⎡

KINETIC ENERGY.

133

Problem 40 x1

Ωt

y3

x3

O,G

1

z3

X

x3

y1

O

G

a

Z α

y3 Y

O

b

4 G

c 2

z3

ω

3

Figure 27 The mixing tank 1 and rotor of the electric motor 2, combined, are considered as rigid body (see Fig. 27). Its principal moments of inertia about axes through its centre of gravity are Ix1 , Iy1 , Iz1 and its mass is m. The tank rotates with the constant angular velocity Ω about axis z3 relatively to the housing 3. At the same time the housing 3 rotates about the axis X. Its motion is determined by the angular displacement α. Given are: a, b, c, Ω, Ix1 , Iy1 , Iz1 , m, α. Produce expressions for kinetic energy function of the mixing tank 1. Answer: ⎡ ⎤⎡ ⎤ 2 I + ma 0 0 α ˙ cos Ωt x1 £ ¤ 0 Iy1 + ma2 0 ⎦ ⎣ −α˙ sin Ωt ⎦ T = 12 α˙ cos Ωt −α˙ sin Ωt Ω ⎣ 0 0 Iz1 Ω

KINETIC ENERGY.

134

Problem 41 R L/2

α (t) X

y

L/2

x Y

z Z 1

2

b

3

4

o y

a

zG

b)

a) G

H(t)

xG

R yG L

O

Figure 28 To displace the cylinder 4 of mass m, the arm of the robot shown in Fig. 28a translates and rotates with respect to the inertial frame XY Z. The translation is determined by function H(t) and the rotation is determined by the angular displacement α(t). Upon assuming that the elements 2,3 and 4 forms one rigid body, derive expression for the kinetic energy of the cylinder 4. Given are: H(t), α(t), a, b, R, L, m The principal moments of inertia of a cylinder (see Fig. 28b) through centre of gravity G are IxG = T =

mR2 ; 2

IyG = IzG =

Answer: 1 + b2 α˙ 2 ) + 12 α˙ 2 12 m(3R2 + L2 )

1 m(H˙ 2

1 m(3R2 + L2 ) 12

KINETIC ENERGY.

135

Problem 42 y1

α X

x1 x2 Z z1

xG l

Y y2

a)

z2

r

y1

zG

A

β

B

yG

2

a 1 l

b)

G

m

r

Figure 29 The base 1 of the robot arm, shown in Fig. 29a), rotates about the vertical axis Z of the inertial system of coordinates XY Z. The instantaneous position of this base is determined by the angular displacement α.The link 2 is hinged to the base 1 at the point A. The relative instantaneous position of the link 2 with respect to the base 1 is determined by the angular displacement β. The link 2 can be considered as a rigid cylinder of length l, radius r and mass m attached rigidly to the massless element AB. The system of coordinates x1 y1 z1 and x2 y2 z2 are rigidly attached to the links 1 and 2 respectively. Produce kinetic energy function for the link 2 as a function of α(t) and β(t). The principal moments of inertia of the cylinder shown in Fig. 29b) about axes through its centre of gravity G are as follows 1 1 IxG = IyG = mr2 + ml2 4 12

1 IzG = mr2 2

KINETIC ENERGY.

136

Problem 43

X yG

Y

m r

G

X

xG

a)

α

yG

D

L/2

L/2 r

r G

2r 1

r

zG l

r

2

b)

m

2r

ω2,1

r r Z

Figure 30 The exciter shown in Fig. 30a) is design to produce the oscillatory motion of an object. Its rotor 2 can be treated as a rigid body assembled of two spheres each of mass ms and a cylinder of mass mc . These elements are join together by means of massless elements as it is shown in Fig. 30a). This rotor rotates with respect to its housing 1 with the constant angular velocity ω 2,1 . The oscillatory motion of the housing 1 is determined by the angular displacement α. Produce expression for the kinetic energy of the rotor of the exciter. The principal moments of inertia for the sphere shown in Fig. 30b) are 2 IxG = IyG = IzG = mr2 5 The principal moments of inertia for the cylinder shown in Fig. 30b) are 1 1 IxG = IyG = mr2 + ml2 4 12

1 IzG = mr2 2

KINETIC ENERGY.

137

Problem 44

0 y2 X

α

ωb

1 2 x1 z3

M

t

3

v

x2 x 3 z1 z2 ωt

Y y1

G

o1

αb

y3

y2 l

Figure 31 The turret 2 of the tank 1 shown in Fig. 31is rotating about the vertical axis with the angular velocity ω t and the barrel 3 is being raised with the constant angular velocity ω b . The tank has the constant forward linear velocity v. The system of coordinates x3 , y3 , z3 is rigidly attached to the barrel 3 and coincides with its principal axes. Moments of inertia about these axes are Ix3 , Iy3 and Iz3 respectively. Mass of the barrel is m and its centre of gravity is by l apart from o1 . Produce expression for the kinetic energy of the barrel when the barrel passes the position defined by angles αt and αb . Answer: ⎡ ⎤⎡ ⎤ 2 I − ml 0 0 ω x3 3x3 £ ¤ 2 ⎦ ⎣ ω3y3 ⎦ 0 Iy3 0 + 12 ω 3x3 ω 3y3 ω3z3 ⎣ T = 12 mvG 2 ω 3z3 0 0 Iz3 − ml where ω 3y3 = ω t sin αb ω 3z3 = ω t cos αb ω3x3 = ωb 0 vG = rG +ω 3 ×rG rG = i3 (vt sin αt )+j3 (vt cos αb cos αt +l)+k3 (−vt sin αb cos αt )

KINETIC ENERGY.

138

Problem 45

z2 Z z1

2

β 1

ω

G

o2 y2 x2 l

o1 y1

x1

Figure 32 The base 1 of the crane shown in Fig. 32 rotates with the constant angular velocity ω about the vertical axis Z of the inertial system of coordinates XY Z. The system of coordinates x1 y1 z1 is attached to the base. At the same time its boom 2 of mass m is being lowered. This relative motion about the axis is determined by the angular displacement β. The system of coordinates x2 y2 z2 is attached to the boom. Its origin coincides with the boom’s centre of gravity G. The distance l determines the position of the centre of gravity. The axes of the system of coordinates x2 y2 z2 coincide with principal axes of the boom. The principal moments of inertia about these axes are Ix , Iy , Iz respectively. Produce the expression for the kinetic energy of the boom. Answer: ⎡ ⎤⎡ ⎤ 2 −ω sin β I + ml 0 0 x £ ¤ ⎦ 0 Iy + ml2 0 ⎦ ⎣ T = 12 −ω sin β β˙ ω cos β ⎣ β˙ 0 0 Iz ω cos β

KINETIC ENERGY.

139

Problem 46

Y

y1

y2 x2

ω 21 t

o2 o1

1

x1

2 l

Z

α

z1 X

1

O

o2 o1

a

x1 Figure 33 The massless link 1 (see Fig. 33) is free to rotate about the horizontal axis Y of the inertial system of coordinates XY Z. The system of coordinates x1 y1 z1 is rigidly attached to the link 1. Its angular position is determined by the angle α. The link 2 that can be considered as a slender and uniform bar of mass m and length l rotates with respect to the link 1 with the constant velocity ω 21 . Produce the expression for the kinetic energy of the link 2. Answer: ⎤⎡ ⎤ ⎡ 1 2 2 ml + ma 0 0 t α ˙ sin ω 21 3 0 ma2 − 12 mal ⎦ ⎣ α˙ cos ω21 t ⎦ T = 12 [α˙ sin ω 21 t, α˙ cos ω21 t, ω21 ] ⎣ 0 − 12 mal 13 ml2 ω 21

KINETIC ENERGY.

140

Problem 47

z1 Z

z2

β

1

y1 y2

α

O

Y a b

2

α

l

0

β P

X

G

x1

x2 Figure 34 Figure 34 shows the physical model of a mechanism. The slide 0 is motionless with respect to the inertial system of coordinates XYZ . It is parallel to the axis Y and its position is determined by the dimensions a and b. The link 1 rotates about the vertical axis Z and its instantaneous position is given by the angular displacement α. The system of coordinates x 1 y 1 z 1 is rigidly attached to the link 1. The link 2 is hinged to the link 1 at the point O and it is supported by the slide 0. The system of coordinates x 2 y 2 z 2 is rigidly attached to the link 2 and coincides with its principal axis. The angular displacement β defines the relative position of the link 2 with respect to the link 1. The principal moments of inertia of the link 2 are I x2 , I y2 , I z2 and its mass is m. The distance l identifies the position of the center of gravity G of the link 2. Produce: 1. The expression for the components of the absolute angular velocity of the link 2 along the system of coordinates x 2 y 2 z 2 in terms of α and β. Answer: ω2 = i2 (−α˙ sin β) + j2 β˙ + k2 (α˙ cos β) 2. The expression for the components of the absolute angular acceleration of the link 2 along the system of coordinates x 2 y 2 z 2 in terms of α and β.

KINETIC ENERGY.

141

Answer: ε2 = i2 (−¨ α sin β − α˙ β˙ cos β) + j2 β¨ + k2 (¨ α cos β − α˙ β˙ sin β) 3. The expression for the components of the absolute velocity of the point G along the system of coordinates x 2 y 2 z 2 in terms of α and β. Answer: ˙ v2 = i2 (0) + j2 lα˙ cos β + k2 (−lβ) 4. The kinetic energy of the link 2 as a function of α and β. hAnswer: i 2 1 T = 2 Ix2 α˙ 2 sin2 β + Iy2 β˙ + Iz2 α˙ 2 cos2 β 5. The expression for the angular displacement β as a function of α. Answer: ¡ ¢ β = arctan ab cos α

KINETIC ENERGY.

142

Problem 48

y2

2

Z z1

3

P

z2 G

y1

β

a O

A

b

c

Y

1

R

α

x2

a x1 X

Figure 35 Fig. 35 shows the physical model of a mechanical system. The link 1 of this system rotates about the vertical axes Z of the inertial frame XYZ. Its instantaneous position is given by the absolute angular displacement α. The system of coordinates x 1 y 1 z 1 is rigidly attached to the link 1. The link 2 is hinged to the link 1 at the point A. The other end of this link P always stays in contact with the cylindrical surface 3 of radius R (b>R). The system of coordinates x 2 y 2 z 2 is attached to the body 2 and coincides with its principal axes. The link 2 possesses the mass m and its principal moments of inertia about the system of coordinates x 2 y 2 z 2 are I x2 , I y2 and I z2 . Its centre of gravity G is located by the distance c. The angular displacement β determines the relative position of the link 2 with respect to the system of coordinates x 1 y 1 z 1 . Produce: 1. The expression for the components of the absolute angular velocity of the link 2 along the system of coordinates x 2 y 2 z 2 in terms of α and β 2. The expression for the components of the absolute angular acceleration of the link 2 along the system of coordinates x 2 y 2 z 2 in terms of α and β 3. The expression for the components of the absolute velocity of the point P along the system of coordinates x 2 y 2 z 2 in terms of α and β 4. The kinetic energy of the link 2 as a function of α and β. 5. The expression for the angular displacement β as a function of α.

KINETIC ENERGY.

143

Solution 1. The absolute angular velocity of the link 2 is ω 2 = ω1 + ω21 = k1 α˙ + i2 β˙ Its components along the system coordinates x 2 y 2 z 2 are ´ ³ ω2x2 = i2 · k1 α˙ + i2 β˙ = β˙ ´ ³ ω2y2 = j2 · k1 α˙ + i2 β˙ = α˙ sin β ´ ³ ω 2k2 = k2 · k1 α˙ + i2 β˙ = α˙ cos β

2. The components of the absolute angular acceleration along the system of coordinates x 2 y 2 z 2 can be produced by diﬀerentiation of the vector of the absolute velocity ³ ´ ³ ´ ε2 = ω 02 + ω2 × ω2 = ω02 = i2 β¨ + j2 α ¨ sin β + α˙ β˙ cos β + k2 α ¨ cos β − α˙ β˙ sin β 3.

y2

2

Z z1

3

P

z2 rP

y1

β G

a O

A

b

c

Y

1

R

α

x2

a x1 X

Figure 36 The absolute position vector of the point P (see Fig. 36) is rP = j1 a + j2 b

(4.127)

KINETIC ENERGY.

144

Its components along the system of coordinates x 2 y 2 z 2 are as follows rP x2 = i2 · (j1 a + j2 b) = 0 rP y2 = j2 · (j1 a + j2 b) = a cos β + b rP z2 = k2 · (j1 a + j2 b) = −a sin β

(4.128)

The first time derivative of the position vector rP yields the absolute velocity of the point P 0

r˙ P = rP + ω2 × rP

¯ ¯ ¯ j2 k2 ³ ´ ³ ´ ¯¯ i2 ¯ = j2 −aβ˙ sin β + k2 −aβ˙ cos β + ¯¯ β˙ α˙ sin β α˙ cos β ¯¯ = ¯ 0 a cos β + b −a sin β ¯ ³ ´ ¢ ¡ 2 2 ˙ ˙ = i2 −aα˙ sin β − aα˙ cos β − bα˙ cos β + j2 −aβ sin β + aβ sin β ³ ´ + k2 −aβ˙ cos β + aβ˙ cos β + bβ˙ ³ ´ = i2 (−aα˙ − bα˙ cos β) + k2 bβ˙ (4.129) 4.

G

β

a

1

R

A

γ

3

P

z2

C

y2

2

Z z1

y1 D b

α c

O

Y

α

x2

a x1 X

Figure 37

KINETIC ENERGY.

145

The angular displacement β is not an independent variable. There is a relationship between the variable β and the independent coordinate α imposed by the kinematic constrains. To develop this relationship let us consider the vector equation shown in Fig. 37. −→ −−→ −−→ CO + OD = CD (4.130) or Ja+j1 (a + b cos β) = I (−R sin γ) + JR cos γ

(4.131)

Multiplication of the above equation by the unit vectors I and J results in the following set of scalar algebraic equations I · Ja+I · j1 (a + b cos β) = I · I (−R sin γ) + I · JR cos γ J · Ja+J · j1 (a + b cos β) = J · I (−R sin γ) + J · JR cos γ

(4.132)

or − sin α (a + b cos β) = −R sin γ a + cos α (a + b cos β) = R cos γ

(4.133)

Hence R2 = sin2 α (a + b cos β)2 + a2 + 2a cos α (a + b cos β) + cos2 α (a + b cos β)2 R2 = (a + b cos β)2 + a2 + 2a cos α (a + b cos β) (4.134) or (a + b cos β)2 + 2a cos α (a + b cos β) − (R2 − a2 ) = 0

(4.135)

Introducing the following substitution z = (a + b cos β)

(4.136)

z 2 + (2a cos α) z − (R2 − a2 ) = 0

(4.137)

µ ¶ q 1 2 2 2 (−2a cos α) ± (2a cos α) + 4(R − a ) z= 2

(4.138)

the above equation is Hence

Taking advantage of the substitution 4.136 the angular displacement β is specified by ¶ ¶ µ µ q 1 1 1 2 2 2 cos β = (z − a) = (−2a cos α) ± (2a cos α) + 4(R − a ) − a b b 2 µµ ¶ ¶ q 1 2 2 2 (−a cos α) ± (a cos α) + (R − a ) − a b ⎛ ⎞ s µ ¶2 R a⎝ − 1 − 1⎠ − cos α ± cos2 α + (4.139) b a

EQUATIONS OF MOTION

4.4

146

EQUATIONS OF MOTION

4.4.1 Euler’s equations of motion General case Each rigid body can be considered as a rigidly tied together system of particles. Hence, its motion is governed by the same equations which govern any system of particles. X ˙ = F= P Fi (4.140) X ri,G × Fi (4.141) h˙ G = MG =

Under influence of a set of external forces Fi (see Fig. 38) the considered body moves Fi

Z

ω

z

dm

i

ri

ri,G G

rG

y vG

O x

Y

X

Figure 38 in the inertial system of coordinates XY Z. Let vG be its absolute velocity of its centre of gravity G and ω its absolute angular velocity. To derive the formula for the linear momentum P let us introduce a body system of coordinates xyz. The body system of coordinates has its origin at the body centre of gravity G and its axes coincides principal axes of inertia. Such selection of the body system of coordinates, according to considerations in the previous section, is always possible. The linear momentum of the body considered is P = mvG

(4.142)

If components of velocity vG along the body system of coordinates are known vG = ivGx + jvGy + kvGz

(4.143)

the first derivative of the linear momentum is 0 ˙ = m(vG P + ω × vG )

(4.144)

EQUATIONS OF MOTION

147

where 0 vG = iv˙ Gx + jv˙ Gy + kv˙ Gz

and

(4.145)

ω = iωx + jωy + kω z

Introduction of Eq. 4.145 into Eq. 4.144)yields ⎛

¯ ¯ i j k ¯ ¯ ˙ ⎝ P = m (iv˙ Gx + jv˙ Gy + kv˙ Gz ) + ¯ ω x ω y ω z ¯ vGx vGy vGz = im(v˙ Gx + vGz ω y − vGy ω z ) +jm(v˙ Gy + vGx ω z − vGz ωx ) +km(v˙ Gz + vGy ωx − vGx ωy ) = iFx + jFy + kFz

¯⎞ ¯ ¯ ¯⎠ ¯ ¯ (4.146)

The above vector equation is equivalent to three following scalar equations m(v˙ Gx + vGz ωy − vGy ω z ) = Fx m(v˙ Gy + vGx ω z − vGz ωx ) = Fy m(v˙ Gz + vGy ωx − vGx ωy ) = Fz

(4.147)

Components of moment of the relative momentum, according to Eq. 4.15, in the case considered may be adopted in the following form. ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ 0 ωx hGx IGx 0 IGx ω x ⎣ hGy ⎦ = ⎣ 0 IGy 0 ⎦ ⎣ ωy ⎦ = ⎣ IGy ω y ⎦ (4.148) 0 0 IGz hGz ωz IGz ω z

Taking into account the above relationship, the equation 4.141 may be transformed to form 4.149. h˙ G = h0G +ω × hG

¯ ¯ i j k ¯ ωy ωz = iIGx ω˙ x + jIGy ω˙ y + kIGz ω˙ z + ¯¯ ω x ¯ IGx ω x IGy ω y IGz ωz = i(IGx ω˙ x + (IGz − IGy )ω z ω y ) +j(IGy ω˙ y + (IGx − IGz )ω x ω z ) +k(IGz ω˙ z + (IGy − IGx )ωy ω x ) = iMGx + jMGy + kMGz

¯ ¯ ¯ ¯ ¯ ¯ (4.149)

Now, one can formulate three additional scalar equations. IGx ω˙ x + (IGz − IGy )ωz ωy = MGz IGy ω˙ y + (IGx − IGz )ω x ω z = MGy IGz ω˙ z + (IGy − IGx )ω y ω x = MGz

(4.150)

Eventually, we obtained six independent equations which allow to obtain either kinematic parameters (ω and vG ) in case all the external forces are known, or resultant external force F and resultant moment MG one has to apply to the rigid body to keep it going according to the assumed motion.

EQUATIONS OF MOTION

148

Rotational motion In a case, when the rigid body performs the rotational motion about the point O which is fixed in the inertial space (see Fig. 39) it can be considered as system with three degree of freedom. Hence only three equations are necessary to describe its motion. The wanted equations may be obtained from equation 4.18. X ˙ O = MO = H ri ×Fi (4.151) ω

z

Z

dm

Fi G

rG O

ri

R y

Y

X x

Figure 39 Let us introduce through point of rotation O body system of coordinates which axes coincide principal axis of the body. Components of the angular momentum along these axes are ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ 0 ωx HOx IOx 0 IOx ωx ⎣ HOy ⎦ = ⎣ 0 IOy 0 ⎦ ⎣ ωy ⎦ = ⎣ IOy ωy ⎦ (4.152) 0 0 IOz HOz ωz IOz ωz

Hence, the equation 4.151 may be rewritten as follows

˙ O = H0O + ω × HO = MO H

(4.153)

In a similar way to that shown in the previous paragraph, one may transform the above equation into three equivalent equations known as Euler’s equations. IOx ω˙ x + (IOz − IOy )ωz ωy = MOz IOy ω˙ y + (IOx − IOz )ω x ω z = MOy IOz ω˙ z + (IOy − IOx )ω y ω x = MOz

(4.154)

The three unknown components of the reaction R in the pivot O are determined by Eq. 4.147 m(v˙ Gx + vGz ω y − vGy ω z ) = Fx m(v˙ Gy + vGx ω z − vGz ω x ) = Fy m(v˙ Gz + vGy ω x − vGx ω y ) = Fz

(4.155)

EQUATIONS OF MOTION

149

where vG = ω × rG

(4.156)

The external force F comprises both, the known driving forces Fd and the unknown reaction R. F = R + Fd (4.157) 4.4.2 Modified Euler’s equations of motion General case The modified equations of motion corresponds to the particular case of motion of a rigid body which fulfils the following requirements: 1. The body has an axis of symmetry (see Fig. 40). 2. The body rotates with a relative angular velocity Ω about its axis of symmetry z with respect to a translating and rotating system of coordinates xyz.

y

Ω

z

G

x Figure 40 For further consideration we will assume that the translating and rotating system of coordinates xyz has its origin at the centre of gravity of the symmetric body. As an example of possible application let us consider rotor of a turbo-compressor of a jet shown in Fig. 41.

y ω Ω z vG

x

Figure 41 The system of coordinates xyz is fixed to the jet and its origin is chosen at the centre of gravity G of the rotor. This system of coordinates possesses its own angular

EQUATIONS OF MOTION

150

speed ω. The rotor of the turbo-compressor itself rotates with angular velocity Ω about axis of symmetry z. Hence, all the above requirements are fulfilled. In such cases, equations of motion of a rigid body may be formulated in terms of kinematic parameters of the system xyz (vG , ω) and the relative angular velocity of the body Ω.

ω

y Fi

i

dm

yb β

y

ri,G

ωA

xb

Z Ω

rG

β

x

G

G

vG

x

Ω z zb

z O Y X

b)

a) Figure 42

The motion of the body ( Fig. 42) is govern by equations X ˙ = F= P Fi X ri,G ×Fi h˙ G = MG =

(4.158) (4.159)

Components of the angular momentum along the body system of coordinates xb , yb , zb (see Fig. 42b), according to Eq. 4.15(see page 98), are ⎡ ⎤ ⎡ ⎤ hGxb ω Ax b ¤ £ ⎣ hGyb ⎦ = IG ⎣ ωAyb ⎦ (4.160) xb, yb ,zb hGzb ωAzb where

[IG ]xb, yb ,zb is matrix of inertia of the body about the body system of coordinates. ωAxb , ω Ayb , ωAzb are components of the absolute angular velocity of the body along the body system of coordinates. Taking into consideration that ⎡ ⎤ ⎤ ⎡ hGxb hGx ⎣ hGyb ⎦ = [Cxyz→xb yb zb ] ⎣ hGy ⎦ (4.161) hGzb hGz

EQUATIONS OF MOTION

and

151

⎡

⎤ ⎤ ⎡ ω Axb ω Ax ⎣ ω Ayb ⎦ = [Cxyz→xb yb zb ] ⎣ ω Ay ⎦ ω Azb ω Az

(4.162)

the relationship 4.160 can be rewritten as follows ⎤ ⎤ ⎡ ⎡ ω Ax hGx £ ¤ ⎣ hGy ⎦ = ([Cxyz→xb yb zb ]T IG [Cxyz→xb yb zb ]) ⎣ ωAy ⎦ (4.163) xb, yb ,zb hGz ω Az £ ¤ Generally, the product [Cxyz→xb yb zb ]T IG x y ,z [Cxyz→xb yb zb ] is a function of time. b, b b But, in the case considered ⎡ ⎤ cos β(t) sin β(t) 0 [Cxyz→xb yb zb ] = ⎣ − sin β(t) cos β(t) 0 ⎦ (4.164) 0 0 1 and because of symmetry of the body £ Hence

IG

¤

xb, yb ,zb

⎤ 0 IG 0 = ⎣ 0 IG 0 ⎦ 0 0 IGz ⎡

(4.165)

£ ¤ [Cxyz→xb yb zb ]T IG x y ,z [Cxyz→xb yb zb ] b, b b ⎤⎡ ⎤ ⎡ ⎤⎡ 0 cos β(t) sin β(t) 0 cos β(t) − sin β(t) 0 IG 0 = ⎣ sin β(t) cos β(t) 0 ⎦ ⎣ 0 IG 0 ⎦ ⎣ − sin β(t) cos β(t) 0 ⎦ 0 0 IGz 0 0 1 0 0 1 ⎤ ⎡ 0 IG 0 ⎣ 0 IG 0 ⎦ = (4.166) 0 0 IGz

The Eq. 4.166 allows the relationship 4.163 to be simplified. ⎤ ⎡ ⎤⎡ ⎤ ⎡ 0 ω Ax hGx IG 0 ⎣ hGy ⎦ = ⎣ 0 IG 0 ⎦ ⎣ ωAy ⎦ 0 0 IGz hGz ω Az

(4.167)

ω A = ω + Ω = iω x + jω y + k(ω z + Ω)

(4.168)

It is easy to see from Fig. 42 a that the absolute angular velocity of the body ω A is

Introduction of Eq. 4.168 ⎤ ⎡ ⎡ hGx IG ⎣ hGy ⎦ = ⎣ 0 0 hGz

into Eq. 4.167 yields ⎤ ⎡ ⎤ ⎤⎡ 0 0 ωx IGx ω x ⎦ IG 0 ⎦ ⎣ ω y ⎦ = ⎣ IGy ω y 0 IGz ωz + Ω IGz (ω z + Ω)

(4.169)

EQUATIONS OF MOTION

152

Since the vector of the angular relative momentum hG is resolved along non-inertial system of coordinates its derivative takes form h˙ G = h0 +ω × hG

¯ ¯ i j k ¯ ˙ ¯ ωy ωz = iIG ω˙ x + jIG ω˙ y + kIGz (ω˙ z + Ω) + ¯ ω x ¯ IGx ω x IGy ω y IGz (ω z + Ω) = i(IG ω˙ x + (IGz − IG )ω z ω y + IGz ωy Ω) +j(IG ω˙ y − (IGz − IG )ω x ωz − IGz ω x Ω) ˙ +k(IGz (ω˙ z + Ω)) = iMGx + jMGy + kMGz

¯ ¯ ¯ ¯ ¯ ¯ (4.170)

The above vector equation may be rewritten in the scalar form IG ω˙ x + (IGz − IG )ω z ωy + IGz ω y Ω = MGx IG ω˙ y − (IGz − IG )ω x ωz − IGz ω x Ω = MGy ˙ = MGz IGz (ω˙ z + Ω)

(4.171)

The equation 4.158, treated in the same way as in the previous section, yields three additional equations m(v˙ Gx + vGz ω y − vGy ω z ) = Fx m(v˙ Gy + vGx ω z − vGz ω x ) = Fy m(v˙ Gz + vGy ω x − vGx ω y ) = Fz

(4.172)

Rotational motion. In case of rotational motion (see Fig. 43) the origin of the rotating system of coordinates x, y, z is always chosen at a point of rotation O. z Ω

Z R G

rG

ω

Fi

y ri

X

O

x

Figure 43

Y

EQUATIONS OF MOTION

153

Let ω be the angular velocity of the rotating system of coordinates xyz and the body rotates about its axis of symmetry z with the angular velocity Ω with respect to the rotating system of coordinates xyz. Hence the absolute angular velocity of the body considered is ω A = ω + Ω = i(ω x ) + j(ω y ) + k(ωz + Ω)

(4.173)

Similar consideration to this in the previous section leads to conclusion that the components of the angular momentum about the point O are ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 HOx ωx IO 0 IO ω x ⎣ HOy ⎦ = ⎣ 0 IO 0 ⎦ ⎣ ω y ⎦ = ⎣ ⎦ IO ω y (4.174) 0 0 IOz HOz ωz + Ω IOz (ωz + Ω)

Its first derivative with respect to time is ˙ O = H0O +ω × HO H

¯ ¯ i j k ¯ ˙ + ¯ ωx ω ω = iIO ω˙ x + jIO ω˙ y + kIOz (ω˙ z + Ω) y z ¯ ¯ IO ω x IO ω y IOz (ω z + Ω)

Implementation of Eq. 4.175 into the principle of moment of momentum ˙ O = MO H

¯ ¯ ¯ ¯(4.175) ¯ ¯ (4.176)

yields three scalar equations in the following form IO ω˙ x + (IOz − IO )ω z ω y + IOz ω y Ω = MOx IO ω˙ y − (IOz − IO )ω x ω z − IOz ωx Ω = MOy ˙ = MOz IOz (ω˙ z + Ω)

(4.177)

which permit the unknown vector ω to be obtained. The above equations are known as modified Euler’s equations. The equation 4.158, if treated in the same way as it has been done in the previous section, yields m(v˙ Gx + vGz ω y − vGy ω z ) = Fx m(v˙ Gy + vGx ω z − vGz ω x ) = Fy m(v˙ Gz + vGy ω x − vGx ω y ) = Fz ‘

(4.178)

Since velocity of the centre of gravity vG can be obtained from the formula below ¯ ¯ ¯ i ¯ j k ¯ ¯ (4.179) vG = ω × rG = ¯¯ ω x ω y ωz ¯¯ ¯ 0 0 rGz ¯

the equations 4.178 allow to the unknown components of the reactions R at the constrain O to be determined.

EQUATIONS OF MOTION

154

4.4.3 Problems. Problem 49

y1 z 3 z1

O,G

1

O A

Z

X

B

G

o1,o3

Ωt

α

z 3 z1

x1

a b

Y

O x3

4 G

c 2

x3

y3

ω

y3

3

Figure 44 ‘The mixing tank 1 and rotor of the electric motor 2, combined, are considered as rigid body. Its principal moments of inertia about axes through its centre of gravity are Ix , Iy , Iz and its mass is m. The tank rotates with the constant angular velocity Ω about axis z3 relatively to the housing 3. At the same time it rotates with the constant angular velocity ω about the axis X. Derive expressions for all components of reactions at A and B when the mixing tank passes an angular position α. The senses of the angular velocities and necessary dimensions are shown in Fig. 44. Given are: a, b, c, Ω, ω, Ix = Iy , Iz , m, α.

EQUATIONS OF MOTION

155

Solution. z3

Z

Ω ω

O R A =0 z3

Y

r

a G

R Ax3 A R Bz3 B R Bx3

R Ay3 b RB

y3

y3

c

Figure 45 In Fig. 45 the x3 y3 z3 system of coordinates is fixed to the housing 3. Its absolute angular velocity is ω 3 = −i3 ω (4.180) The relative angular velocity of the mixing tank 1 with respect to the x3 y3 z3 system of coordinates is Ω = k3 Ω (4.181) The absolute position vector of the centre of gravity G can be determine along the system x3 y3 z3 as follows rG = −k3 a (4.182) Hence, the absolute velocity of G is ¯ ¯ i3 j3 k3 ¯ vG = ω 3 ×rG = −ωi3 × (−ak3 ) = ¯¯ −ω 0 0 ¯ 0 0 −a

¯ ¯ ¯ ¯ = −j3 aω ¯ ¯

(4.183)

Since axis z3 is axis of symmetry for the tank, its motion is governed by the modified Euler equations. The tank performs the rotational motion about point O. But in the case considered it is more convenient to take advantage of the equations for the general case. m(v˙ Gx3 + vGz3 ωy3 − vGy3 ω z3 ) = RAx3 + RBx3 m(v˙ Gy3 + vGx3 ω z3 − vGz3 ωy3 ) = RAy3 + RBy3 + mgsinβ m(v˙ Gz3 + vGy3 ωx3 − vGx3 ωy3 ) = RBz3 − mgcosβ I ω˙ x3 + (Iz − I)ω y3 ω z3 + Iz Ωωy3 = RAy3 b + RBy3 (b + c) I ω˙ y3 − (Iz − I)ω x3 ω z3 − Iz Ωωx3 = −RAx3 b − RBx3 (b + c) ˙ = Mz3 Iz3 (ω˙ z3 + Ω)

(4.184)

(4.185)

EQUATIONS OF MOTION

156

where I and Iz are moments of inertia of the body about axes x1 y1 z1 through the centre of gravity G. According to Eq’s. 4.180 and 4.183 vGx3 = 0 ω x3 = −ω

vGy3 = −aω ω y3 = 0

vGz3 = 0 ω z3 = 0

(4.186) (4.187)

Introducing Eq’s. 4.57 and 4.187 into Eq’s. 4.184 and 4.185 one can obtain 0 0 2 m(aω ) 0 Iz Ωω 0

= = = = = =

RAx3 + RBx3 RAy3 + RBy3 + mg sin α RBz3 − mg cos α RAy3 b + RBy3 (b + c) −RAx3 b − RBx3 (b + c) Mz3

(4.188)

The first five equations allow for determination of the unknown reaction at A and B. RAx3 = RAy3 = RBx3 = RBy3 = RBz3 =

1 Iz Ωω c b+c mg sin α − c 1 − Iz Ωω c b mg sin α c ma2 ω 2 + mg cos α

(4.189)

The last equation oﬀers the driving moment of the tank. In the case considered it is equal to zero.

EQUATIONS OF MOTION

157

Problem 50

A

α

C G

A

l1

l2

Figure 46 Figure 46 shows a mobile concrete mixer. The line AC is the axis of symmetry of the mixing tank which is supported by a bearing at A and rollers at B. The bearing at A may react load in any direction but the rollers at B may only provide a reaction through C in a plane normal to AC. The mixing tank, which rotates about AC at 10RP M, clockwise when viewed from the rear, has a mass of 730kg; its centre of mass is at G. Its moment of inertia about AC is 9000kgm2 , and about any axis through G normal to AC is 11000kgm2 . The dimensions shown in the Fig. 46 are as follows: l1 = 1.2m, l2 = 1.8m α = 20o . The concrete mixer is driven on horizontal ground round a bend to the left of 30m radius at a steady speed 40km/h. Find the bearing reactions at A and C induced by motion of the vehicle.

EQUATIONS OF MOTION

158

Solution. According to the given data Ω = −10 RP M = −10 · π/30 rad/s = −1.05 rad/s v = 40 km/h = 40 · 1000/3600 m/s = 11.1 m/s ω = v/R = 11.1/30 = 0.37 rad/s l2 = 1.8 m l1 = 1.2 m Iz = 9000 kgm2 I = 11000 kgm2 G = 730 kg · 9.81 m/s = 7161 N α = 20o vx2 = v sin α = 11.1 sin 20o = 3.8 m/s vy2 = 0 vz2 = −v cos α = −11.1 cos 20o = −10.4 m/s ωx2 = ω cos α = 0.37 cos 20o = 0.348 rad/s ωy2 = 0 ωz2 = ω sin α = 0.37 sin 20o = 0.126 rad/s G

v

z1

R

y1 x1

x2

R Cy2 RA

R Cz2= 0

ω y2

RA z2

z2

α

C G

v A

Ω

z1 R Cx2

M A z2 G

RA x2 l1

Figure 47

l2

EQUATIONS OF MOTION

159

Euler’s modified equations applied to the tank yield m(v˙ x2 + vz2 ωy2 − vy2 ω z2 ) = RCx2 + RAx2 − G cos α m(v˙ y2 + vx2 ω z2 − vz2 ω x2 ) = RCy2 + RAy2 m(v˙ z2 + vy2 ωx2 − vx2 ω y2 ) = RAz2 − G sin α I ω˙ x2 + (Iz − I)ω y2 ω z2 + Iz ω y2 Ω = −RCy2 (l2 − l1 ) + RAy2 l1 I ω˙ y2 − (Iz − I)ωx2 ωz2 + Iz ω x2 Ω = +RCx2 (l2 − l1 ) − RAx2 l1 ˙ = MAz2 Iz (ω˙ z2 + Ω)

(4.190)

(4.191)

Introducing the given data into the above equations one may obtain RCx2 + RAx2 − 7161 cos 20o RCy2 + RAy2 RAz2 − 7161 sin 20o −RCy2 (1.8 − 1.2) + RAy2 1.2 +RCx2 (1.8 − 1.2) − RAx2 1.2 MAz2

= = = = = =

0 730(3.8 · 0.126 − (−10.4)0.348) 0 0 −(9000 − 11000)0.348 · 0.126 − 9000 · 0.348 · (−1.05) 0 (4.192)

Hence RCx2 + RAx2 RCy2 + RAy2 RAz2 −RCy2 + 2RAy2 +RCx2 − 2RAx2 MAz2

= = = = = =

6729 2990 2450 0 5627 0

Eventually RAx2 = 367 N RAy2 = 996 N RAz2 = 2450 N

RCx2 = 6362 N RCy2 = 1994 N Mz2 = 0

EQUATIONS OF MOTION

160

Problem 51

C

ω A l

B

Figure 48 Derive diﬀerential equations of motion of the system shown in Fig. 48. The link CA rotates with a constant angular speed ω about its vertical axis. The uniform rod AB of mass m and length l is hinged at A to the link CA.

EQUATIONS OF MOTION

161

Solution. y1

Y

x1

ωt

O

X

z1

Z

ω 1

z2 x1

A l/2 2

α G

mg

x2

Figure 49 Let us introduce the following systems of coordinates (see Fig. 49): XY Z – inertial system of coordinates. x1 y1 z1 – body 1 system of coordinates x2 y2 z2 – body 2 system of coordinates The absolute angular speed of the link 1 is ω1 = Kω = k1 ω

(4.193)

The relative angular speed of the link 2 with respect to the link 1 can be obtain by diﬀerentiation of the generalized coordinate α ˙ = j1 α˙ = j2 α˙ ω21 = α

(4.194)

The absolute angular velocity of the link 2 is ω2 = ω 1 + ω21 = j2 α˙ + k1 ω = −i2 ω sin α + j2 α˙ + k2 ω cos α

(4.195)

Since the link 2 performs rotational motion about the point A, its motion is governed by Euler’s equations Ix ω˙ 2x2 + (Iz − Iy )ω 2y2 ω2z2 = Mx2 Iy ω˙ 2y2 + (Ix − Iz )ω 2x2 ω2z2 = My2 Iz ω˙ 2z2 + (Iy − Ix )ω2x2 ω 2y2 = Mz2

(4.196)

EQUATIONS OF MOTION

162

z1

Z

ω

y2

R 2,1y2

1 M2,1y2 =0 R 2,1y2

z2 R 2,1z2 A

R 2,1x2 l/2

α

2

A R 2,1x2

M2,1z2 x1

M2,1x2

M2,1x2 G x2

mg

x2 Figure 50

were according to Fig. 50, Ix = 0

Iy = Iz = I =

ml2 3

Mx2 = M2,1x2 1 My2 = mgl cos α 2 Mz2 = M21z2

(4.197)

Upon introducing Eq’s. 4.195 and 4.197 into Eq. 4.196 one may obtain 0 = M2,1x2 1 Iα ¨ + Iω 2 sin α cos α = mgl cos α 2 −Iω α˙ sin α − I αω ˙ sin α = M21z2

(4.198) (4.199) (4.200)

Equation 4.199 presents the wanted equation of motion. 1 Iα ¨ + Iω2 sin α cos α − mgl cos α = 0 2

(4.201)

After solving it with respect to the unknown function α, the equation 4.200 allows the interaction moment M21z2 to be obtained.

EQUATIONS OF MOTION

163

Problem 52 1

y2 υ Y y1

2

c

3 O Z

α

υ

z1

Ω

k

α

z2

x1 x2 X

Figure 51 To record the angular velocity α˙ of the floating platform 1 about the vertical axis Y a gyroscope was installed as presented in Fig. 51. The shown spring, of a stiﬀness k, keeps the ring 2 in the horizontal position (ϑ = 0) if the platform does not rotate (α˙ = 0). The damper c produces a moment which can be approximated by a linear function Mc = −cϑ˙ (4.202) The ring 2 may be considered as massless. The gyroscope 3 is symmetrical with respect to its axis of relative rotation z2 and its moments of inertia are Ix2 = Iy2 = I, and Iz2 . The gyroscope rotates, relative to the ring 2, with the constant angular velocity Ω. Upon assuming that the platform can rotate about a vertical axis only, derive the relationship between its constant angular velocity α˙ = p and angular position of the gyroscope ϑ.

EQUATIONS OF MOTION

164

Solution. The system of coordinates x1 y1 z1 ,which is attached to the platform, rotates with respect to the inertial frame XY Z with angular velocity ω 1 . ω1 = j1 α˙ = j2 α˙ cos ϑ − k2 α˙ sin ϑ

(4.203)

The relative angular velocity of the ring 2 with respect to the system of coordinate x1 y1 z1 is ω21 = i2 ϑ˙

(4.204)

Therefore, the absolute angular velocity of the system of coordinates x2 y2 z2 is ω2 = ω1 + ω21 = i2 ϑ˙ + j2 α˙ cos ϑ + k2 (−α˙ sin ϑ)

(4.205)

Since the gyroscope 3 rotates about its axis of symmetry z2 with relative velocity Ω, the modified Euler’s equations may be used to derive its equations of motion. I ω˙ x2 + (Iz2 − I)ω y2 ω z2 + Iz2 Ωω y2 = Mx2 I ω˙ y2 − (Iz2 − I)ωx2 ωz2 − Iz2 Ωω x2 = My2 ˙ = Mz2 Iz2 (ω˙ z2 + Ω)

(4.206)

Since the ring 2 is massless its equilibrium conditions yields Mx2 = −kϑ − cϑ˙

(4.207)

Introduction of Eq’s. 4.205) and (4.207) into the first equation of the set 4.206 produces equation of motion of the gyroscope. I ϑ˙ + (Iz2 − I)(−α˙ 2 sin ϑ cos ϑ) + Iz2 Ωα˙ cos ϑ = −kϑ − cϑ˙

(4.208)

Because for a constant angular velocity α˙ = α˙ o the angle ϑo is supposed to be constant, the above equation yields (Iz2 − I)(−α˙ 2o sin ϑo cos ϑo ) + Iz2 Ωα˙ o cos ϑo = −kϑo

(4.209)

If angular velocity ϑo << Ω, the first term of Eq. 4.209 can be omitted and the required relationship has the following explicit form α˙ o = −

kϑo Iz2 Ω cos ϑo

(4.210)

EQUATIONS OF MOTION

165

Problem 53

A

1 2

β

Z l

Figure 52 Link 1 of the mechanical system shown in Fig. 52 performs rotational motion about the horizontal and motionless axis Z. Its motion is determined by angle α which is a known function of time. Link 2, which can be considered as a uniform and rigid rod of mass m and length l, is hinged to the link 1 at point A. Derive the equation of motion of the link 2. Given are: α(t)− equation of motion of the link 1 m− mass of the link 2 l− length of the link 2

EQUATIONS OF MOTION

166

Solution. z2

X A

ω21 x1 x2

Z z1

x1 x2 1

α

y1

ω1

A

β

rG

Z z1 l

G

2

y1

Y

y2

Figure 53 In Fig. 53 system of coordinates x1 y1 z1 is rigidly attached to the link 1 and rotates with respect to the inertial system of coordinates XY Z about the horizontal axis Z. Its instantaneous position is determined by the angle α. System of coordinates x2 y2 z2 is fixed to the link 2 and rotates about axis x1 of the system of coordinates x1 y1 z1 . Angle β determines its instantaneous position with respect to the system of coordinates x1 y1 z1 . Since the link 2 performs rotational motion about the point A, the following Euler equations of motion can be utilized. Ix ω˙ 2x2 + (Iz − Iy )ω2y2 ω 2z2 = M2x2 Iy ω˙ 2y2 + (Ix − Iz )ω2x2 ω 2z2 = M2y2 Iz ω˙ 2z2 + (Iy − Ix )ω2x2 ω 2y2 = M2z2

(4.211)

where Ix = 13 (ml2 )− moment of inertia of the body 2 about axis x2 Iy = 0− moment of inertia of the body 2 about axis y2 Iz = 13 (ml2 )− moment of inertia of the body 2 about axis z2 . Components of the absolute angular velocity of the link 2 along its body system of coordinates x2 y2 z2 may be determined as follows ω 2 = ω1 + ω21 = k1 α˙ + i2 β˙

(4.212)

Hence, ω2x2 = i2 · ω2 = β˙ ω 2y2 = j2 · ω 2 = α˙ sin β ω2z2 = k2 · ω2 = α˙ cos β

(4.213)

Moment M2 is due to the interaction force R21 and interaction moment M21 (see Fig. 54) in the constrain A and the gravity force G. M2 = M21 +M2G

(4.214)

EQUATIONS OF MOTION

167

M 2,1y2 R2,1y2

M 2,1y2

R2,1x2

x2

M 2,1x2 =0

z2

A M 2,1z2 R2,1z2 ω1 Z z1 A x1 x2 rG l 1 β G 2

R2,1y2

y2

y2

y1 Figure 54

Since the link 1 is free to rotate about axis x2 , component of the interaction moment along axis x is M21x2 = 0 (4.215) Moment produced by the gravity force G is a vector product of position vector rG and the gravity force G.Hence rG × G = j2 (l/2) × Jmg = (mlg/2)j2 × J (mlg/2)j2 × (j1 cos α + i2 sin α) (mlg/2)j2 × ((j2 cos β − k2 sin β) cos α + i2 sin α) (mlg/2)j2 × (i2 sin α + j2 cos α cos β − k2 cos α sin β) ¯ ¯ ¯ i2 ¯ j k 2 2 ¯ ¯ ¯ ¯ 1 0 = (mlg/2) ¯ 0 ¯ ¯ sin α cos α cos β − cos α sin β ¯ = (mlg/2)(i2 (−cosαsinβ) + k2 (−sinα))

M2G = = = =

(4.216)

Upon introducing Eq’s. 4.213 and 4.214 into first equation of the set 4.211 one can obtain equation of motion in the following form (4.217) β¨ + α˙ 2 sin β cos β + (3g/2l) cos α sin β = 0 The second and third equation allow the unknown components of interaction moment M21 to be obtained. Ix ω˙ 2x2 + (Iz − Iy )ω2y2 ω 2z2 = M2x2 Iy ω˙ 2y2 + (Ix − Iz )ω2x2 ω 2z2 = M2y2 Iz ω˙ 2z2 + (Iy − Ix )ω2x2 ω 2y2 = M2z2

(4.218)

EQUATIONS OF MOTION

M21y2 = 0 ˙ α cos β − α˙ β˙ sin β) − (1/3)ml2 α˙ βsinβ M21z2 = (1/3)ml2 (¨ +(1/2)mglsinα

168

(4.219)

EQUATIONS OF MOTION

169

Problem 54

ω

2

1 x1

R m

C D

z1 r

Figure 55 Wheel of radius r and mass m is free to rotate about axle CD which turns about the vertical axis with a constant angular speed ω. The wheel rolls without slipping on the horizontal plane. Determine the reaction between the wheel and the horizontal plane. Given are: r =0.5m R =2m m =100kg ω =1rad/s Iz1 =12.5kgm2 , Iy1 = Ix1 =10kgm2 – principal moments of inertia of the wheel

EQUATIONS OF MOTION

170

Solution z1

O,o

z2

ωt

Z

Y y2 x2 ω

2

1 x1 ω12 t x2

R m

C D

z1

O,o

y2

ω 12

z2 r

y1

Figure 56

m(v˙ Gx2 + vGz2 ω2y2 − vGy2 ω2z2 ) = F1x2 m(v˙ Gy2 + vGx2 ω 2z2 − vGz2 ω 2x2 ) = F1y2 m(v˙ Gz2 + vGy2 ω2x2 − vGx2 ω 2y2 ) = F1z2 Io ω˙ 2x2 + (Ioz − Io )ω 2y2 ω 2z2 + Ioz ω 12 ω 2y2 = M1x2 Io ω˙ 2y2 − (Ioz − Io )ω 2x2 ω2z2 − Ioz ω 12 ω2x2 = M1y2 Ioz (ω˙ 2z2 + ω˙ 12 ) = M1z2 Io = Ix1 + mR2 ;

Ioz = Iz1

ω2 = i2 ω ω 2x2 = ω

ω 2y2 = 0

ω2z2 = 0

R r rG = k2 R ¯ ¯ ¯ i2 j2 k2 ¯ ¯ ¯ vG = r˙ G = r0G + ω2 × rG = 0 + ¯¯ ω 0 0 ¯¯ = j2 (−ωR) ¯ 0 0 R ¯ ω12 = −ω

vGx2 = 0

vGy2 = −ωR

vGz2 = 0

EQUATIONS OF MOTION

171

0 = F1x2 0 = F1y2 −Rmω 2 = F1z2 0 = M1x2 R = M1y2 Ioz ω 2 r 0 = M1z2

0 = F1x2 = R12x2 − mg + N 0 = F1y2 = R12y2 + T −Rmω 2 = F1z2 = R12z2 0 = M1x2 = M12x2 − RT R = M1y2 = −Rmg + RN Ioz ω 2 r 0 = M1z2 = −rT T =0 N = mg + Ioz ω 2

R r

R12x2 = mg − N = −Ioz ω2 R12y2 = 0 R12z2 = −Rmω 2 M12x2 = RT = 0

R r

EQUATIONS OF MOTION

172

2 R12

1

M12

C D

Figure 57

x2 x2 2

R 12x2

1 x1

R

M12y2=0

m

M12x2 C

O,o M =0 12z2

D

M12x2 z1

y 2 R 12y2

z2

R 12z2 r

G

ω12 t R 12x2

G y1 T

N

Figure 58

N

EQUATIONS OF MOTION

173

Problem 55 X

x1

a

O

A

B

G

Ω

z1 1

b

b

Figure 59 On the rotating about the vertical axis X platform 1 (see Fig. 59) a turbine is installed. Rotor of the turbine has mass m and its centre of gravity G is at a distance a from the axis X. Axis z1 is axis of symmetry of the rotor and its principal moments of inertia I2x1 = I2y1 = I2 and I2z1 are known. If the platform rotates with the constant angular velocity ω and the rotor has the relative angular velocity Ω, determine components of reactions at the bearings A and B along system of coordinates x1 y1 z1 fixed to the platform 1. Given are: I2 , I2z1 , m, ω, Ω, a, b.

EQUATIONS OF MOTION

174

Problem 56 1

l

Z

B

β

A

a v

Y

Figure 60 The thin and uniform bar 1 has mass m and length l. Its end A is moving with constant velocity v along the horizontal axis Y of the inertial system of coordinates XY Z. Upon assuming that there is no friction in the constraints A and B and all the motion is in the vertical plane Y Z, produce the expression for the dynamic reactions in these constraints as a function of the bar angular position β. Answer: Solution of the folowing set of equations yields the wanted reactions. µ ¶ l¨ l ˙2 − β cos β + β sin β m = RAX + RB cos β 2 2 µ ¶ l¨ l ˙2 − β sin β − β cos β m = −mg + RB sin β + RAY 2 2 µ ¶ a ml2 l l l = RAY sin β + RAX cos β + RB − ε 12 2 2 2 cos β

EQUATIONS OF MOTION

175

Problem 57 B

X,x1 O

Y

T

y1 y2

X,x1 x2

1

3

2 l

ω y1 y2 O

z1

z2

G

α

B

N

z1

r z2 z 3

x3

X 2

1

ω

3 G y3

Figure 61 Body 3 (see Fig. 61) is driven by means of the massless links 1 and 2 that are hinged to each other at the point O. Link 1 rotates about the vertical axis X with the constant angular velocity ω and the body 3 rolls over the horizontal motionless ring 4 without slipping. Axis z2 is the axis of symmetry of body 3 and its moments of inertia about the body system of coordinates x3 y3 z3 are Ix3 = Iy3 = I and Iz respectively. Derive the expression for the components of the reaction N and T between the body 3 and the ring 4. Given are: m - mass of the body 3 I, Iz - moments of inertia of the body 3 l - distance between the point of rotation O and centre of gravity G of the body 3 r - radius of the ring 4 ω - angular velocity of the link 1 α - angle between axis z2 and the horizontal plane Hint: The body 3 performs rotational motion about the point O.

EQUATIONS OF MOTION

176

Problem 58 z2

Z

Z

α Ω

3 Y G3

y2

2 1

X x2

Figure 62 Fig. 62 shows the physical model of a shaker frequently used to excite the rolling motion of a ship. The housing 2 of the gyroscope 3 performs oscillatory motion about axis Y of the inertial system of coordinates XY Z. This motion is determined by the following equation α = α0 sin ωt The system of coordinates x2 y2 z2 is rigidly attached to the housing. The gyroscope 3 of mass m and moments of inertia I3x2 = I3y2 = I3 and I3z about axes x2 y2 z2 respectively, rotates with the constant angular velocity Ω. Upon assuming that the mass of the housing is negligible, derive an expression for the components of the moment transmitted to the board 1 of the ship.

EQUATIONS OF MOTION

177

Problem 59 Y

Y

a)

ω 3

4

1

ω

2

1

c

Ω

Ω

G3 G1

2

3

G3 G1

b

a O

X

Z

x2

b)

G3 y2

x1

z2

ωt z1

y1

Y

G3 G1

O

α

Z z1

Figure 63 Fig. 63a) shows a gyrostabilizer for stabilization of the monorail car 1. To test this gyrostabilizer, the car 1 was forced to move along the coordinate α (see Fig. 63b)) according to the following equation α = A sin ft Simultaneously the housing 2 was rotated with respect to the car 1 with constant angular velocity ω. The relative angular velocity of the gyroscope 3 with respect to

EQUATIONS OF MOTION

178

the housing 2 was Ω. XY Z is the inertial system of coordinates. x1 y1 z1 is the body 1 system of coordinates. x2 y2 z2 is the body 2 system of coordinates. Given are: A, f - amplitude and frequency of the oscillatory motion of the car 1 ω - angular velocity of the housing 2 with respect to the car 1 Ω- angular velocity of the gyroscope 3 with respect to the housing 2 a, b, c, dimensions shown in Fig.63a) m- mass of the gyroscope 3 Ix = Iy = I, Iz - principal moments of inertia of the gyroscope 3 about axes through its centre of gravity. Produce expressions for interaction forces between the gyroscope 3 and its housing 2. Answer: m(−(a + b)¨ α sin ωt) − mg sin α sin ωt = R32x2 m(−(a + b)α˙ 2 ) + mg cos α = R32y2 m((a + b)¨ α cos ωt) − mg sin α cos ωt = R32z2 I(¨ α cos ωt − αω ˙ sin ωt) + (Iz − I)αω ˙ sin ωt + Iz Ωω = M32x2 2 (Iz − I)α˙ sin ωt cos ωt − Iz Ωα˙ cos ωt = M32y2 Iz (¨ α sin ωt + αω ˙ cos ωt) = Md

EQUATIONS OF MOTION

179

Problem 60 0

x1 A

α z1

Ω

a 1 G

2 l

l b

Figure 64 The aircraft landing gear shown in Fig. 64 is being retracted while the aircraft 0 moves with constant velocity along a horizontal straight line. The instantaneous position of the arm 1 is determined by the angular displacement α. The wheel 2 rotates with the constant relative angular velocity Ω with respect to the arm 1. Given are: a, b, l, - dimensions shown m - mass of the wheel 2 Iz - moment of inertia of the wheel 2 about its axis of relative rotation I = Ix = Iy - moments of inertia of the wheel 2 about axes perpendicular to the axis of relative rotation α(t) - the angular displacement of the arm 1 with respect to the plane. Ω - the angular velocity of the wheel 2 with respect to the arm Produce the expression for the components of the interaction forces between the wheel 2 and the arm 1 along the body 1 system of coordinates x1 y1 z1 .

EQUATIONS OF MOTION

180

Problem 61

1

a O

ωt o1

b

x1 z1 , z2

Z

X

2 2

Y

y1

y2

x2

1

ω 21 t O o1

x1

Figure 65 Fig. 65 presents the physical model of a ventilator. Its base 1 rotates with the constant angular velocity ω about the vertical axis Y of the inertial system of coordinates XY Z. The system of coordinates x1 y1 z1 is attached to this base. The rotor 2 is free to rotate about the axis z1 . The relative angular velocity of the rotor 2 with respect to the base 1 is constant and its magnitude is ω 21 . The system of coordinates x2 y2 z2 is attached to the rotor 2. The axis z2 is the axis of symmetry of the rotor and its moments of inertia about the system of coordinates x2 y2 z2 are Ix2 = Iy2 = I2 , Iz2 respectively. The mass of the rotor is m. Produce the expression for the components of the interaction force between the rotor 2 and its base 1. Answer: Rx1 = −maω 2 Ry1 = mg Rz1 = −bmω 2 Mx1 = Iz2 ωω21 My1 = 0 Mz1 = Md = 0

EQUATIONS OF MOTION

181

Problem 62

Z z1 z2

α a/2 A

2

a/2

y1 y2

B

ωt

O 1

Y

ωt α

a x1 x2

X b b

Figure 66 The link 1 of the mechanical system shown in Fig.66 rotates about the vertical axis Z of the inertial system of coordinates XYZ with a constant angular velocity ω. The system of coordinates x 1 y 1 z 1 is rigidly attached to this link. Two uniform bars, each of length a and mass m, were joined together to form the link 2 of this system. The system of coordinates x 2 y 2 z 2 is rigidly attached to the link 2. Its instantaneous relative position with respect to the system of coordinates x 1 y 1 z 1 is determined by the angular displacement α. Produce: 1. The free body diagram for the link 2. 2. The diﬀerential equation of motion of the link 2 in terms of the variable α. 3. The equations for the determination of the interaction forces between the link 2 and 1 in the kinematic constraints A and B. The moment of inertia of a uniform bar of mass m and length l about the axis through its centre of gravity is

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

1 ml2 12

IG = 4.5

182

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

4.5.1 Modelling. Physical model. Let us consider motion of a rigid body about a fixed in an inertial space point O (Fig. 67). z

Z z1

Θ

Ω

r

G . Φ

Θ

G

Φ

y1

Φ O

X

y

. Θ

Y 1

x1 x

Figure 67 Axis x of the rotating system of coordinates xyz stays always in plane XY of the absolute system of coordinates XY Z. The angle Φ, between those two axes is called angle of precession and together with angle of mutation Θ determines uniquely position of axes the rotating system of coordinates xyz. For further consideration, we shell assume that axis z is axis of symmetry of the body. Since the body is free to rotate about axis z it has three degree of freedom. Let Ω be the angular velocity of the body with respect to xyz. The body has mass m. Its centre of gravity is located on axis z at the point G. Its position is given by a distance r. IOx = IOy = IO and IOz represent principal moment of inertia of the body about axis xyz. Kinematic analysis Angular velocity of the system of coordinates xyz is ˙ ˙ +Θ ˙ = KΦ˙ + iΘ ω=Φ

(4.220)

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

183

Scalar multiplication of the above equation by unit vectors i, j, and k respectively yields its components along the rotating system of coordinates xyz. ˙ Θ) ˙ =Θ ˙ ω x = i · (KΦ+i ˙ Θ) ˙ = Φ˙ sin Θ ω y = j · (KΦ+i ˙ Θ) ˙ = Φ˙ cos Θ ω z = k · (KΦ+i

(4.221)

Equations of motion. The introduced assumptions allow to take advantage from the modified Euler’s equations. IO ω˙ x + (IOz − IO )ω z ω y + IOz ω y Ω = MOx IO ω˙ y − (IOz − IO )ωx ω z − IOz ωx Ω = MOy ˙ = MOz IOz (ω˙ z + Ω)

(4.222)

Components of moment MO = kr × K(−mg)

(4.223)

are Mx

My

Mz

¯ ¯ ¯ ¯ 1 0 0 ¯ ¯ ¯ 0 1 ¯¯ = rmgsinθ = −(i · (k × K))rmg = −rmg ¯ 0 ¯ 0 sin Θ cos Θ ¯ ¯ ¯ ¯ ¯ 0 1 0 ¯ ¯ ¯ 0 1 ¯¯ = 0 = −(j · (k × K))rmg = −rmg ¯ 0 ¯ 0 sin Θ cos Θ ¯ ¯ ¯ ¯ 0 0 1 ¯¯ ¯ 0 1 ¯¯ = 0 (4.224) = −(k · (k × K))rmg = −rmg ¯¯ 0 ¯ 0 sin Θ cos Θ ¯

Upon introducing Eq. 4.224 and 4.221 into Eq. 4.222 one can obtain equations of motion in the following form. ¨ + (IOz − IO )Φ˙ 2 sin Θ cos Θ + IOz ΩΦ˙ sin Θ − rmg sin Θ = 0 IO Θ ¨ sin Θ + IO Φ˙ Θ ˙ cos Θ − (IOz − IO )Φ˙ Θ ˙ cos Θ − IOz ΩΘ ˙ = 0 IO Φ ¨ ˙ sin Θ + Ω) ˙ = 0 − Φ˙ Θ IOz (Φcosθ

(4.225)

4.5.2 Analysis. General solutions of the mathematical model 4.225 can not be obtain by means of any analytical methods. But, very often, we can procure a number of particular solutions by guessing their form.

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

184

Particular solutions. Let us predict the particular solution of the set of equations 4.225 in the following form Φ = pt Θ = Θo Ω = Ωo

(4.226)

where p, Θo , Ωo are constant values. Hence Φ˙ = p

¨ =0 Φ

˙ =0 Θ

¨ =0 Θ

Ω˙ = 0

(4.227)

Upon introducing Eq. 4.226 into Eq. 4.225 one can see that second and third equation is fulfilled for any instant of time. The first equation yields (IOz − IO )p2 sin Θo cos Θo + IOz Ωo p sin Θo − rmg sin Θo = 0

(4.228)

The predicted solution 4.226 may be considered as an solution, if the equation 4.228 is fulfilled. It is fulfilled for Θo = 0 Θo = π

(4.229)

regardless of magnitude p, Ωo or for any set of parameters p, θo , Ωo satisfying the following relationship p −IOz Ωo ± (IOz Ωo )2 + 4(IOz − IO )rmg cos Θo (4.230) p1,2 = 2(IOz − IO ) cos Θo The precession p1 , which corresponds to sign + is called slow precession to distinguish it from fast precession p2 corresponding to sign −. The relationship 4.230 permits to compute so called static characteristic of a gyroscope. Let us adopt, for further analysis, dimensions of the gyroscope shown in Fig. 68

R b

G

R=0.04m

O

b=0.015m m=0.6kg 0.025m r= 0.015m

r

{

Figure 68

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

185

For r = 0.025m, IO = 0.000629kgm2 > IOz = 0.000482kgm2 the static characteristic of the gyroscope for a few values of angular velocity (Ωo = 0, 5, 10, 15, 20 and 25s−1 ) is shown in Fig. 69.For high angular velocity Ωo = 25s−1 (see Fig. 70) the gyroscope can perform the fast and slow regular precession p for any mutation angle Θo . If angular velocity Ωo is slower e.g. Ωo = 15s−1 , the gyroscope can perform regular precession only for certain range of angle of mutation Θo . Fig. 71 present static characteristics for r = 0.015m. In this case IO = 0.000388kgm2 < IOz = 0.000482kgm2 . The presented characteristics determine regular precession which may be obtained only for certain set of initial conditions which strictly correspond to the assumed form of particular solution 4.226. More particular solution can be obtained by numerical integration of the mathematical model 4.225.∂ Numerical solution of equations of motion. Fig. 73. presents a few numerical solutions carried out for r = .025m. The diagrams in this figure show instantaneous positions yG of the point G of the gyroscope versus its angle of precession Φ. For all cases the initial speed of the gyroscope was assumed to be Ωi = 15s−1 . Fig 73A presents solution for initial conditions chosen in vicinity of upper ˙ i = 0, Φi = 0, Φ˙ i = 0, Ωi = 15s−1 ). equilibrium position Θo = 0(Θi = 0.01s−1 , Θ Since the gyroscope can not perform regular precession (see Fig. 70), it execute large oscillations. Fig. 73B corresponds to situation when the gyroscope, performing regular ˙ o = 0, Φ˙ o = p1 = 25s−1 , Ωo = 15s−1 see Fig. 70) was pushed precession (Θo = 1.26, Θ ˙ i = 8s−1 . Due to the following initial out of its motion at t = 0 with initial velocity Θ ˙ i = 8s−1 , Φi = 0, Φ˙ i = 0, Ωi = 15s−1 the gyroscope conditions Θi = 1.26rad, Θ performs small oscillation around regular precession. Similar behavior can be observed if it is pushed out of its regular fast precession ˙ o = 0, Φ˙ o = p2 = 136, Ωo = 15 see Fig. 70). This situation is (Θo = 1.26rad, Θ presented in Fig. 73C. Figures 73D and 73E show solution of disturbed motion of the gyroscope about ˙ slow (Φo = 16.17s−1 ) and fast (Φ˙ o = −77.1s−1 ) precession which correspond to angle of mutation Θo = 2.51rad (see Fig. 70). Those solution were obtained for initial ˙ i = 8s−1 , Φi = 0, Φ˙ i = 16.170s−1 , Ωi = 15s−1 and conditions Θi = 2.51rad, Θ −1 ˙ i = 8s , Φi = 0, Φ˙ i = −77.1s−1 , Ωi = 15s−1 respectively. Θi = 2.51rad, Θ Fig. 73F presents motion of the gyroscope in vicinity of the lower equilibrium ˙ i = 20s−1 , position (Θo = π) caused by the following initial conditions Θi = 2.51, Θ −1 Φi = 0, Φ˙ i = 0, Ωi = 15s .

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

186

Slow precession p1 versus angle of nutation Θ ο r=0.025 [m] 60 50 40 30

p

1

[rad/s] 20 10 0 0

0.63

Ωο = 0

Ωο = 5

1.26

Ωο = 10

Θ o [rad]

1.88

Ωο = 15

2.51

Ωο = 20

3.14

Ωο = 25 [rad]

Fast precession p versus angle of nutation Θ ο 2 r=0.025 [m] 400 300 200 p

100

2

[rad/s]

0 -100 -200 -300 -400 0

0.63

1.26

1.88

2.51

3.14

Θ o [rad] Ωο = 0

Ωο = 5

Ωο = 10

Ωο = 15

Figure 69

Ωο = 20

Ωο = 25 [rad]

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

Ω ο = 25 [rad/s]

Slow & fast precession for

187

Θο

versus angle of nutation

r=0.025 [m] 200 150 100 50 p [rad/s]

0 -50 -100 -150 -200 0

0.63

1.26

1.88

Θ slow precession

o

2.51

p

fast precession

1

p 2

Ω = 15 [rad/s] versus angle of nutation ο

Slow & fast precession for

3.14

[rad]

Θο

r=0.025 [m] 200 150 100 50 p [rad/s]

0 -50

-100 -150 -200 0

0.63

1.26

slow precession

Θ

1.88 o

p

1

Figure 70

2.51

3.14

[rad] fast precession

p 2

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

Slow precession p1 versus angle of nutation

188

Θο

r = 0.015 [m] 60 50 40 p

30

1

[rad/s] 20 10 0 0

0.63

Ωο = 0

Ωο = 5

1.26

Ωο = 10

Θ o [rad]

1.88

Ωο = 15

2.51

Ωο = 20

3.14

Ωο = 25 [rad]

Fast precession p versus angle of nutation Θ ο 2

r = 0.015 [m] 400 300 200 100 p 2

[rad/s]

0 -100 -200 -300 -400 0

0.63

1.26

1.88

2.51

3.14

Θ o [rad] Ωο = 0

Ωο = 5

Ωο = 10

Ωο = 15

Figure 71

Ωο = 20

Ωο = 25 [rad]

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

Ω ο = 20 [rad/s]

Slow & fast precession for

versus angle of nutation

189

Θο

r = 0.015 [m] 200 150 100 50 p [rad/s]

0 -50 -100 -150 -200 0

0.63

1.26

1.88

Θ slow precession

2.51

o

p

fast precession

1

p 2

Ω = 10 [rad/s] versus angle of nutation ο

Slow & fast precession for

3.14

[rad]

Θο

r = 0.015 [m] 200 150 100 50 p [rad/s]

0 -50 -100 -150 -200 0

0.63

1.26

Θ

slow precession

1.88 o

p

1

Figure 72

2.51

3.14

[rad] fast precession

p 2

MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.

4

Φ [rad]

F

0 0

E

Φ [rad] -20 20

Φ [rad]

D

0 20

Φ [rad]

C

0 20

Φ [rad]

B

0 20

A

Φ [rad] yG [m]

0 G

Z

yG

O Figure 73

0 -0.04

190

Chapter 5 APPENDIXES 5.1

APPENDIX 1. REVISION OF THE VECTOR CALCULUS

NOTATION. z

V k i

j

o

Vz

y Vx

x

Vy

The vector quantities are printed in boldface type V or, in handwriting, should always be indicated by symbol V to distinguish them from the scalar quantities V. Vector quantities are usually defined in the right-handed system of coordinates by its scalar components Vx , Vy , Vz . (5.1)

V = iVx + jVy + kVz were i, j, k are unit vectors of the system of coordinates xyz.

SCALAR MAGNITUDE of a vector V is V =

q Vx2 + Vy2 + Vz2

(5.2)

DIRECTION COSINES l, m, n are the cosines of angles between a vector V and axes xyz.. Thus l = cos ]Vi =

Vx V

n = cos ]Vj =

Vy V

m = cos ]Vk =

Vz V

(5.3)

Useful relations l2 + n2 + m2 = 1

(5.4)

DOT OR SCALAR PRODUCT of two vectors P and Q is defined as scalar

APPENDIX 1. REVISION OF THE VECTOR CALCULUS

192

magnitude P · Q =P Q cos α

(5.5)

where α stands for the angle between the two vectors.

P

α

Q

Useful relations

P·Q=

£

Px

P·Q=Q·P

(5.6)

i·i=j·j=k·k=1

(5.7)

i·j=j·k=i·j=0 ⎡ ⎤ Q x ¤ Py Pz ⎣ Qy ⎦ = Px Qx + Py Qy + Pz Qz Qz

Px = i · P

Py = j · P

Pz = k · P

(5.8) (5.9) (5.10)

VECTOR OR CROSS PRODUCT of two vectors P and Q is defined as a vector with the magnitude P Q sin α and direction specified by the right-hand rule as shown. Px Q

Q

α P

Useful relations P× (Q + R) = P × Q + P × R

(5.11)

Q × P = −P × Q

(5.12)

i×j=k

j×k=i

k×i=j

(5.13)

i×i=j×j=k×k=0 ¯ ¯ ¯ i j k ¯¯ ¯ P × Q = ¯¯ Px Py Pz ¯¯ ¯ Qx Qy Qz ¯

(5.14)

(P × Q) · R

(5.16)

(5.15)

TRIPLE CROSS-SCALAR PRODUCT is defined as

APPENDIX 1. REVISION OF THE VECTOR CALCULUS

193

Useful relations (P × Q) · R = (R × P) · Q = (Q × R) · P ¯ ¯ ¯ Px Py Pz ¯ ¯ ¯ (P × Q) · R = ¯¯ Qx Qy Qz ¯¯ ¯ Rx Ry Rz ¯

(5.17)

P× (Q × R)

(5.19)

P× (Q × R) = Q· (R · P) − R· (P · Q)

(5.20)

µ ¶ dP ∆P ˙ = lim =P ∆t→0 dt ∆t

(5.21)

(5.18)

TRIPLE VECTOR PRODUCT is defined as

Useful relations

DERIVATIVE OF A VECTOR

Useful relations d(Pf ) ˙ + Pf˙ = Pf dt d(P × Q) ˙ ×Q+P×Q ˙ = P dt d(P · Q) ˙ ·Q+P·Q ˙ = P dt

(5.22) (5.23) (5.24)

APPENDIX 2. CENTRE OF GRAVITY, VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES. 194

5.2

APPENDIX 2. CENTRE OF GRAVITY, VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES.

Sphere z

R G

y

x

4 V = πR3 3

2 Ixx = Iyy = Izz = mR2 5

Hemisphere z 3R 8 y

G R x

2 V = πR3 3

Ixx = Iyy = 0.259mR2

2 Izz = mR2 5

Cone z 1h 4 y

h G R

x

1 V = πR2 h 3

Ixx = Iyy =

3 m(4R2 + h2 ) 80

Cylinder z

1h 2 1h 2 x

y G R

Izz =

3 mR2 10

APPENDIX 2. CENTRE OF GRAVITY, VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES. 195

V = πR2 h

Ixx = Iyy =

1 m(3R2 + h2 ) 12

1 Izz = mR2 2

Rectangular block z

G

y c

x

a b

V = abc

Ixx =

1 m(b2 + c2 ) 12

Iyy =

1 m(a2 + c2 ) 12

Slender rod z R l G

y

x

V =0

Ixx = Iyy =

1 2 ml 12

Izz = 0

Izz =

1 m(a2 + b2 ) 12