436- 353 MECHANICS 2UNIT 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 44 83 85 87 88 88 89 90 91 96 . 96 . 99 . 100 . 102 . . .2 Modified Euler’s equations of motion . . . . . . . . . . . . . . . . .4 EQUATIONS OF MOTION .4. . . 194 . . . . . . . . . . .3. . . .4. . . .5. .1 Euler’s equations of motion . . . . . . . . . . . . . . . . . . . . . . . . .2 APPENDIX 2. . . . . . 183 5 APPENDIXES 191 5. . . . . . CENTRE OF GRAVITY. . . . . . . . . . 146 4. . . . . . . . . . . . . . . 127 4. . . . . . . . . .1 Modelling. . . . . . . . . . . .CONTENTS 4 4. . . . . . . . 130 4. . .3 Problems . . . . .4. . . . . . . . . . . . . . . . . . . .5 MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. . . . . .2 Analysis. . . . . . . . . . . . . . . . . 182 4. . 154 4. .2. . . . . . . 182 4. . . . . . . . . . . . . . . .3. . . . . . . . . . . 127 4. .1 Rotational motion. . . . . . . . . . . . . . . . . . . . . 106 4. . . . .2 General motion. . . . . 128 4. . . . . . . . . . . . . . . . . .3.1 APPENDIX 1. . . . . .3 Problems. .3 KINETIC ENERGY. . . . . . . . . 149 4. . . . . . . . . . . . .3 Problems . . REVISION OF THE VECTOR CALCULUS 191 5. . . . . . . . . . . . . . . . . . . . . . . . . . . 146 4. . . .5. . VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES. . the derived equations form a base for development of many branches of mechanics (fluid mechanics. gas. Because each continuum (fluid. The developed principles are widely utilized in the chapter entitled Kinetics of A Rigid Body. Each chapter is ended with several engineering problems. This chapter gives procedures for determination of matrix of inertia of a rigid body and its principal axes.). The last two chapters are related with the relationships between motion and forces that act on bodies. . Both. solid mechanics. provides procedures for determination of the absolute velocity and the absolute acceleration of any point that belong to the rigid body. the general motion as well as rotation of a rigid body is considered. 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. Both. The introduced in this chapter translating. rigid or elastic body) can be considered as a system of particles. It is divided into four chapters. rigid body mechanics etc. entitled Three-Dimensional Kinematics of A Rigid Body. Students should produce solution to the other problems during tutorials and in their own time. 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. Three-Dimensional Kinematics of A Particle. The first one. This makes possible to produce expression for the kinetic energy of the moving rigid body as well as to derive its equations of motion. The second chapter.CONTENTS 5 INTRODUCTION. the general motion and the motion about a fixed point is considered. Solution to some of them is provided. 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 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. e.1) . XY Z.Chapter 1 THREE-DIMENSIONAL KINEMATICS OF A PARTICLE. 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.g. 1. e. rX = rX (t) rY = rY (t) rZ = rZ (t) (1. Inertial systems of coordinates are usually denoted by upper characters.g. xyz.1.1 which determine the particle coordinates for any instant of time. to distinguish them from non-inertial systems of coordinates which are usually denoted by lower characters.1 MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME. 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. 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). 7.6) The distance done by the particle in a certain interval of time 0 < τ < t is given by the formula 1. is given by the following formula.7) s= 0 0 . r = IrX (t) + JrY (t) + KrZ (t) (1.5) Scalar magnitude of acceleration is a = |a| = q √ 2 a · a = r¨X + r¨Y2 + r¨Z2 (1.MOTION OF A PARTICLE IN TERMS OF THE INERTIAL FRAME. Z tp Z t v dτ = r˙X (τ )2 + r˙Y (τ )2 + r˙Z (τ )2 dτ = s(t) (1.4) (1. Vector of the absolute velocity. 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 (1.2) were I. as the first derivative of the absolute position vector r with respect to time. J. 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. K are unit vectors of the inertial system of coordinates XY Z. ∆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. DEFINITION: System of coordinates which can translate and rotate is called translating and rotating system of coordinates (Fig.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). xyz) to distinguish it from inertial one.1 Motion in terms of translating system of coordinates. 3 a). DEFINITION: If a non-inertial system of coordinates rotates about origin of an inertial system of coordinates.g. the system is called rotating system of coordinates (Fig.3c) 1.o o y Y x X Figure 4 .2 8 MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). The non-inertial systems of coordinates are denoted by lower characters (e. 1. 3b) In a general case a non-inertial system of coordinates may translate and rotate.2. 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. z Z rP ro O P rP. 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. 12) vP = r˙ P = vT + vR (1.oy + kr ˙ P. 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.ox + jrP. 9 Let XY Z be the inertial system of coordinates and xyz be the translating system of coordinates.11) and the relative velocity is vR = ir˙P.oy + kr˙P.oy + kr˙P. one can prove that the absolute acceleration of the particle P is rP = aT + aR aP = ¨ (1. The position vector of the particle P in the XY Z frame can be composed from vectors ro and rP. 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 . 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.o .MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).ox + jr˙P.oy + kr˙P.9) ˙ = ˙i = j˙ = k˙ = 0 Since. the absolute velocity of the particle P .oz (1.oz + ˙irP. The relative motion of xyz system with respect to XY Z is usually defined by a position vector ro (Fig.13) Hence.oz (1. According to the above definitions the velocity of transportation in the case considered is vT = r˙ o = Ir˙oX + Jr˙oY + Kr˙oZ (1.4). the absolute velocity Similarly.o = const) . I˙ = J˙ = K r˙ P = Ir˙oX + Jr˙oY + Kr˙oZ + ir˙P.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 = IroX + JroY + KroZ + irP.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).oy + krP.8) Hence.oz (1. rP = ro + rP.14) where aT and aR stand for the acceleration of transportation and the relative acceleration respectively.ox + jr˙P.ox + jr (1.oz +ir˙P.ox + jr˙P. MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).15) aR = i¨ rP. we have to establish matrix which transfers components of a vector from rotating system of coordinates xyz to the inertial XY Z. which components in the rotating system of coordinates xyz (Fig. 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).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. the rotating system of coordinates has its origin coinciding the origin of an inertial system of coordinates (Fig. Z z K k j O I o y J Y i X x Figure 5 As it was mention before. Matrix of direction cosines.oz (1. In the case considered aT = ¨ ro = I¨ roX + J¨ roY + K¨ roZ (1. Let the relative motion of a particle P be determined by a position vector r.5).2.2 Motion in terms of rotating system of coordinates.6) are r = irx + jry + krz (1.16) 1.ox + j¨ rP.18) . First of all. 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.oy + k¨ rP. 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. i Jx iJ iY J Figure 7 cos ∠iJ = Jx iY = iY = = Jx i J (1. Cosine of angle between two unit vectors e. i and J is equal to the component of one on the other.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). From the manner we have developed the matrix of direction cosines it is easy to notice that the inverse matrix equal the transpose one.19) The transfer matrix [Cr→i ] is called matrix of direction cosines.g.20) Another useful relationship should be noticed from Fig. [Cr→i ]−1 = [Cr→i ]T = [Ci→r ] (1. 7.21) . 10). iZ . Z .which are equal to corresponding direction cosines. i2X + i2Y + i2Z = cos2 ∠iI + cos2 ∠iJ + cos2 ∠iK = 1 (1.o Y. let us turn the system of coordinates xyz with respect to the inertial one XY Z about axis Z by an angle φ.22) There exists six such relationships. Euler angles. which uniquely determined the position of the rotating system of coordinates with respect to the inertial one. are called Euler’s angles. 8 one can see that components iX. fulfil the following relationship. The matrix of direction cosines between system x1 y1 z1 and XY Z is . The three independent angles. only three direction angles can be chosen independently. Now. x Figure 9 Let us assume that the rotating system of coordinates xyz coincide the inertial one XY Z. iY. 12 Z i iZ O i Y iX Y X Figure 8 From Fig.z O. as shown in Fig. so the system xyz takes position x1 y1 z1 (Fig.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).9. Hence. y X. z1 y2 θ 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. The matrix of direction cosines between system x2 y2 z2 and x1 y1 z1 has the following form. 12). 13 Z . the system xyz is turned by an angle ψ about axis z2 to its final position xyz (Fig. θ z2 Z . 11 as x2 y2 z2 .23) In the next step. The matrix of direction cosines between xyz and x2 y2 z2 is .o y1 Y φ x1 .MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).x 2 X Figure 11 ⎡ ⎤ 1 0 0 [Cr2 →r1 ] = ⎣ 0 cos θ − sin θ ⎦ 0 sin θ cos θ (1.z1 y1 φ O. let us turn the system xyz about axis x1 by an angle θ.24) In the last step. The new position of the system xyz is shown in Fig. 27) into Eq. θ z2 .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.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 ] (1.27) (1. 1. 1.26) (1.z1 Ψ y2 θ φ O. 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. 1. 1.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).z 14 y Z .30) The last formula allows to express the direction cosines as function of three independent angles known as Euler’s angles.19. The angle φ is called angle of precession.28) Introducing Eq.28 into Eq. 1. angle θ is called angle of mutation and the angle ψ is called angle of spin.25) (1. .27) and than Eq.26) one may obtained ⎤ ⎤ ⎤ ⎡ ⎡ ⎡ rx rx rX ⎣ rY ⎦ = [Cr1 →i ][Cr2 →r1 ][Cr→r2 ] ⎣ ry ⎦ = [Cr→i ] ⎣ ry ⎦ (1. angular displacement can not be considered as a vector because at least the commutative law would be violated. Hence. 15. . dψ and dθ may be drawn as shown in Fig. direction and sense is determined). but now the vector ro is turned about axis y first and then about axis x (Fig. 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. To show it let us consider transformation of point Ao due to rotation about axes x and y by 90o . 14).13a). dφ φ dφ Figure 14 It is easy to show. but the proof is here omitted. Let us do the same. If we turn the vector ro by 90o first about axis x and then about axis y. 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. 15 Angular velocity and angular acceleration. We can see that the final position depend on the order of rotation. 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. the final position of the point Ao is represented by vector r2 (Fig. 13 b). According to the above rules the infinitesimal angular increments of Euler’s angles dφ. that the infinitesimal angular displacement can be considered as vectorial magnitude.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). let us consider point A fixed in the xyz system of coordinates determined by a position vector r (Fig. dψ about axes Z x1 z respectively. Z z l dα dα h r β dr A y O Y x X Figure 16 Now.31) dα = dφ + dψ + dθ of the three infinitesimal angular increments of Euler’s angles determines direction of so called instantaneous axis of rotation.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). Let dα be an infinitesimal . 16). The instantaneous axis go through origin of the rotating system of coordinates. And let the axis l be the instantaneous axis of rotation of the system xyz with respect to the inertial one. having such a property that the rotation by the angle dα about this axis is equivalent to rotation by angles dφ. 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. dθ. 38) .MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). ??. Its scalar magnitude is dr = h dα = r dα sinβ (1.37) If motion of the rotating system of coordinates is determined by Euler’s angles.35) is called vector of angular velocity ω.34) The velocity of the point A is v= The vector dα dt dr dα = ×r dt dt (1. the infinitesimal increment dr can be considered as a vector product of vector dα and r. dr = dα × r (1.36) and the velocity of A can be expressed as follow. the point A at the instance considered moves along circle of radius (1. v =ω×r (1. 17.32) h = r sinβ The infinitesimal increment of vector r is tangential to the circle and is placed in plane perpendicular to the axis l. ω= dα dt (1. 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. according to the above definition and Eq. its angular velocity. 17 angular displacement of the system of coordinates xyz. Hence.33) Hence. The instantaneous axis of rotation z Z ω dφ dt dψ dt y O Y dθ dt X x x1 Figure 17 (1. 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 (1.39) ω = Iω X + JωY + KωZ The scalar magnitude of the components ω X.41) ω = iωx + jωy + kωz and the components can be expressed as a function of time.42) . In particular it can be resolved along axes of the inertial system of coordinates XY Z (1. 18 The vector ω can be resolved along any system of coordinates. The locus of the lines of action of the vector of angular speed in the inertial system of coordinates is called space cone (Fig. ωY.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. 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.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). 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. ω Z can be expressed as functions of Euler’s angles and their first derivatives. Let ω be the absolute angular velocity of the rotating system of coordinates xyz (Fig. 18 c). 19 The components ω x. The angular acceleration is defined as the first derivative of the vector of angular velocity with respect to time. The locus of these lines make up so called body cone (Fig. Euler angles) or. ε= dω dt (1. ω z determine for any instant of time direction of the angular velocity in the rotating system of coordinates.47) . This vector will be denoted by A0 . alternatively the rotational motion can be determined by its initial position and the vector of its angular velocity ω.44) Let us differentiate this vector with respect to time. 18 b). 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. can be hence considered as rolling without slipping of the body cone on the space cone (Fig.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). A0 = iA˙ x + jA˙ y + kA˙ z (1. The rotational motion of the system xyz with respect to the XY Z.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. Ay . ωy.g.46) Hence The first three terms represent vector which can be obtained by direct differentiating of the components Ax . A = iAx + jAy + kAz (1. 19).45) ˙ y + kA ˙ z ˙ = iA˙ x + jA˙ y + kA˙ z + ˙iAx + jA A (1. Az with respect to time. ˙ = d (iAx + jAy + kAz ) A dt (1. is the absolute angular velocity of the rotating system of coordinates xyz along which the vector A was resolved to produce vector A0 A .50) ˙i = ω × i (1.37 (1. It can be applied to any vector (eg position vector.is the differentiated vector. the first derivative of the unit vector i is the ratio of the infinitesimal vector increment di and dt (Fig.53) and eventually one may gain ˙ = A0 + ω × A A (1.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).) .49) vi = ω × i (1. But.52) Introducing the above expressions into Eq. according to Eq.51) Hence Similarly j˙ = ω × j and k˙ = ω × k (1. 20 Thus ˙ y + kA ˙ z ˙ = A0 + ˙iAx + jA A (1. 1.48 we have ˙ = A0 + ω × iAx + ω × jAy + ω × kAz = A0 + ω × (iAx + jAy + kAz ) A (1. The last formula provides the rule for differentiation of a vector that it is resolved along a non-inertial system of coordinates.54) where: A0 = iA˙ x + jA˙ y + kA˙ z ω . 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. 20). 1. angular velocity etc.48) According to definition of vector derivative. velocity. 56 with respect to time.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION). The system xyz itself has its own rotational motion determined by absolute angular velocity ω. 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 (1.58) vR = r0P Hence vP = vR + vT (1. For the case of a particle motionless in the rotating system of coordinates. We are interested in the absolute velocity and absolute acceleration of this particle. ¨ rP = d 0 d rP + (ω × rP ) = r00P + ω × r0P + ω˙ × rP + ω × (r0P + ω × rP ) dt dt (1. rP z – components of the vector rP along system of coordinates xyz. 21) with respect to the rotating system of coordinates xyz be determined by position vector rP . 21 Motion in terms of rotating system of coordinates.55) where rP x .56 the velocity of transportation is vT = ω × rP (1.60) . rP = irP x + jrP y + krP z (1. Let relative motion of a particle P (see Fig. according to Eq. 1. we introduce notions of the relative velocity and the velocity of transportation. as it was done in case of translating system of coordinates. 1.59) The absolute acceleration one may obtain differentiating equation 1. rP y .56) In a similar manner. according to Eq.56 the relative velocity is (1.57) For the case of motionless system of coordinates xyz ( ω = 0). 2.61 is called Coriolis acceleration aC .o (1.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).65) aP = aT + aR + aC 1. that the system of coordinates is motionless yields expression for the relative acceleration. the absolute position vector rP in this case is rP = ro + rP.3 Motion in terms of translating and rotating system of coordinates. (1.64) Now.o . 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.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. 22 Introducing Eq. relative acceleration and Coriolis acceleration.o rP ro y o O ω Y x X Figure 22 . the term ε × rP is called tangential acceleration of transportation and the term ω × (ω × rP ) is called normal acceleration of transportation. 1. Hence..66) z P Z rP. aC = 2 ω × r0 (1. 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.63) The last term in equation 1. Assumption. aR = r00 (1.22). we can state that the absolute acceleration is composed of acceleration of transportation.43 and developing the last term we have aP = ¨ rP = r00P + ε × rP + ω × (ω × rP ) + 2 ω × r0P (1. o + 2 ω × r0P .oy + krP.o + ω × rP.o where r˙ o + ω × rP.o + ω × (r0P .o = irP.68) .67) According to the previously developed rules.oz (1.o + r00P. rP.o Here: ¨ ro + ε × rP. the absolute velocity of the point P is r˙ P = r˙ o + r0P.o = aC – Coriolis acceleration. 23 Let us assume that the relative position vector rP.o + ω × rP.o + ω × ω × rP.o + ε × rP.MOTION IN TERMS OF THE NON-INERTIAL FRAMES (RELATIVE MOTION).o = aR – relative acceleration 2 ω × r0P . (1.o = aT – acceleration of transportation r00P.o – velocity of transportation r0P.o + ω × ω × rP.o + ω × r0P .o ) ¨ rP = ¨ = ¨ ro + ε × rP.ox + jrP.o – relative velocity The second derivative yields the absolute acceleration ro + r00P.o is determined by its components along the translating and rotating system of coordinates xyz. System of coordinates x1 y1 z1 is rigidly attached to the frame 1 and system x2 y2 z2 is body 2 system of coordinates.PROBLEMS. 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 . 1.69) Its first vector derivative with respect to time yields absolute angular acceleration. Calculate components of the absolute angular velocity ω2 of the disc 2 and its absolute acceleration ε2 . 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. Solution. ¯ ¯ ¯ i1 j1 k1 ¯ ¯ ¯ (1.3 24 PROBLEMS. Absolute angular velocity of the body 2 is ω2 = k1 ω1 + j1 ω21 (1.70) ε2 = ω˙ 2 = ω02 + ω 1 × ω2 = k1 0 + j1 0 + ¯¯ 0 0 ω1 ¯¯ = −i1 ω 1 ω21 ¯ 0 ω21 ω1 ¯ . the body 2 system of coordinates x2 y2 z2 the magnitude of the angular velocity and acceleration of the antenna. along the inertial XY Z system of coordinates 2.. the magnitude of velocity and acceleration of the probe P . 25 Problem 2 O o1 . . 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. the components of velocity and acceleration of the probe P along the body 2 system of coordinates.PROBLEMS. 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 . 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. ⎤ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤⎡ ⎤⎡ 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. Fig. the rotating system of coordinates x1 y1 z1 and the rotating system of coordinates x2 y2 z2 . 24 permits to produce matrices of direction cosines between the inertial system of coordinates XY Z. The angular velocity of the body 2 is determined by the following vector equation. 1.73) B. A.72.71) 0 0 1 Z z1 Z ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎤ ⎡ x1 1 0 0 x1 x2 ⎣ y2 ⎦ = ⎣ 0 cos α sin α ⎦ ⎣ y1 ⎦ = [C1→2 ] ⎣ y1 ⎦ (1.76) ω2x2 = aA cos At ω2y2 = ω o sin(a sin At) ω 2z2 = ω o cos(a sin At) (1. Matrices of direction cosines.PROBLEMS.75) o ωo cos α Since α = asinAt (1. ⎤⎡ ⎤ ⎡ ⎡ ⎤ 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. Angular velocities. 26 Solution.74) Its components along the system of coordinates x2 y2 z2 may be calculated with help of Eq.77) the above components are . ω 2 = ω o k1 + αi ˙ 1 (1. 80) . the space cone and body cone. 1. They were computed for the following data a = 1[m]. 1.76 into Eq.79) Both. are presented in Fig. These components may be obtained with help of the equation 1.78) ωo Introduction of Eq. ω2X = aA cos At cos ω o t ω 2Y = aA cos At sin ω o t ω 2Z = ω o (1.PROBLEMS.78 yields the parametric equations of the space cone. (1. ⎤⎡ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 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. 25. Angular acceleration. 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.71. 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. ω o = 1[1/s]. ¯ ¯ ¯ i2 ¯ j2 k2 ¯ ¯ 0 (1. (1.87) (1. Velocity of the point P .82) where α˙ = aA cos At α ¨ = −aA2 sin At (1.75.PROBLEMS. The absolute acceleration of point P may be obtained by differentiation of the vector of its absolute velocity. 1. Position of the point P is determined by the position vector l.86) where α and α˙ are determined by formulae 1. its magnitude is aP = q a2Px2 + a2Py2 + a2Pz2 ¯ ¯ ¯ ¯ ¯ ¯ (1. Acceleration of the point P . 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. magnitude of velocity of the point P is p ˙ 2 vP = (ω o l sin α)2 + (lα) (1. E.83) D. 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.83.88) .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. ¨ + j2 ω o α˙ cos α − k2 ω o α˙ sin α ε2 = ω˙ 2 = ω02 + ω 2 × ω 2 = i2 α (1.76 and 1.84) l = k2 l Its first derivative with respect to time yields the absolute velocity of the point P .81) Hence. 28 The absolute angular acceleration of the antenna can be obtained by differentiation of the vector ω2 Eq. . Calculate magnitude of velocity and acceleration of the particle 4 travelling without slipping for the position defined by the distance L = 5m. 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. 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.PROBLEMS. 92) The first derivative of the absolute velocity yields the absolute acceleration of the particle.PROBLEMS. 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. ¯ ¯ ¯ ¯ i j k ¯ ¯ 0 ˙ = R + ω 2 × R = jL˙ + ¯ 0 ω sin β ω cos β ¯ = −iLω cos β + jv (1.89) Position vector which determines a position of the particle 4 is R = jL (1. 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.91) R ¯ ¯ ¯ ¯ 0 L 0 Hence.76 m/s |R| (1. ¯ ¯ ¯ ¯ 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. 30 Solution.93) .90) Its first derivative with respect to time produces the absolute velocity of the particle 4. Axes xyz are fixed to the rotating table 2 and have its origin at O. magnitude of the velocity is p p ˙ = (Lω cos β)2 + v2 = (5 · 1 · cos30o )2 + 22 = 4. 94) . 31 Its magnitude may be calculated according to the following formula.PROBLEMS. 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. PROBLEMS. The wheel rolls without slipping on the horizontal plane. as a function of its angular position α. the magnitudes of velocity and acceleration of the shown in the Fig. 28 point A . 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. . ω. l. Given are: ρ. Determine. PROBLEMS. Since the cone CEF may by considered as the body 2 cone and the cone CEG may be considered as the space cone. according to the imposed constraints. Axis x1 y1 z1 are rigidly attached to the axle 1 and form the body 1 system of coordinates.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 . form the inertial system of coordinates. in Fig.96) ω 2 = ω1 + ω 21 Direction of the relative angular velocity ω21 . 33 Solution. the absolute angular velocity of the body 2 mast have direction of the common generating line EC. Its absolute angular velocity ω 2 is assembled of the absolute angular velocity ω 1 and the relative velocity ω21. vector of the absolute angular velocity ω 2 is L ω 2 = i1 ω + k1 (−ω ) ρ (1. Its axis x1 coincides axis X of the inertial system of coordinates. coincide axis z2 .98) . Therefore ω21 = ω 1 cot β = ω1 L ρ (1. 29.97) Since the vector ω 21 has opposite direction to the positive direction of axis z1 . Therefore the angular speed of this system of coordinates is (1. (1. 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. the magnitude of absolute acceleration is q aA = (v˙ Ax1 )2 + (v˙ Ay1 − ωvAz1 )2 + (v˙ Az1 + ωvAy1 )2 (1. the first derivative of the absolute velocity yields the absolute acceleration of the point A.102) Similarly.PROBLEMS.104) (1. 34 Position vector of the point A is r = L + ρ = k1 L + i2 ρ (1.105) .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 (1.103) where. ¯ ¯ ¯ 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 ) (1.100) 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. according to (1.101) vAx1 = −ρα˙ sin α vAy1 = ρα˙ cos α − ωL vAz1 = ωρ sin α Hence. Answer: v = 7. 30 is revolving about vertical axis with the constant angular speed ω = 1rad/s in the direction shown.07m/s . Simultaneously the boom is being lowered at the constant angular speed Ω = 0.PROBLEMS.5rad/s. 35 Problem 5 P β ω Ω l Figure 30 A crane shown in Fig. 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. 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. When the barrel is in position defined by angles αt and αb a shell leaves the barrel with muzzle velocity vs and acceleration as . 31) rotates about the vertical axis at angular speed ω t and the barrel is being raised at a constant angular speed ωb . 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 . The tank has constant forward speed v. 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.PROBLEMS. the components of the absolute acceleration of the point P along the body system of coordinates x1 y1 z1 . 32 performs the rotational motion about the absolute axis Z. R. 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. Point P belongs to the body 2. γ. α(t) Produce the expression for: 1. Given are: l. the components of the linear absolute velocity of the point P along the body 1 system of coordinates x1 y1 z1 . Its instantaneous position is determined by the angle α.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. ω 21 . The link 2 rotates with respect to the link 1 with the constant relative angular velocity ω21 . 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 . Produce 1. 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. rotates about the vertical axis Z and its instantaneous angular position is determined by the angle α. 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) . . 33. shown in Fig. 38 Problem 8 Z.z 1 1 R 2 β y1 Figure 33 The link 1.PROBLEMS. 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.z 1 α y1 X x1 Y Z. 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 o ) = (2.P + ω × (ω × rQ. If position of the centre of mass is defined by a vector rG.o .o + ω × rP.o along the system of coordinates xyz are constant.1) be defined by a position vector rP.2) Hence.P ) The first term atQP = ω˙ × rQ.o + ω × (ω × rP.o = r˙ o + r0P .o ) = (2.o ) vGz = k · (˙ro + ω × rG.o (2. 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.is always 0.7) is called the tangential relative acceleration and the second one is called the normal relative acceleration.o (2.o ) vGy = j · (˙ro + ω × rG. the relative velocity of P with respect to the body system of coordinates.9) Absolute angular acceleration ε may be obtained as follow ε = ω˙ = ω0 + ω × ω = ω0 (2.o = r˙ o + ω × rP.o .P The absolute acceleration of P is aP = ¨ rP = ¨ ro + ω˙ × rP.P ) (2.o − ω × (ω × rP. r0P .o ) − ¨ ro − ω˙ × rP.o ) = ω × (rQ. Since components of the vector rP. the relative velocity of the point Q with respect to the point P is vQP = vQ − vP = r˙ o + ω × rQ.1) Similarly.o ) (2.P (2.o .GENERAL MOTION 42 Fig. the components of its absolute velocity along the body axes are vGx = i · (˙ro + ω × rG. Therefore its absolute velocity is vP = r˙ P = r˙ o + r˙ P.8) The components of angular velocity along the body coordinates are ωx = i · ω ωy = j · ω ωz = k · ω (2.o − rP.10) .o (2.6) anQP = ω × (ω × rQ. The absolute velocity of the point Q is vQ = r˙ o + ω × rQ.3) ω × rQ.o − (˙ro + ω × rP.o + ω × (ω × rQ.5) = +ω˙ × rQ.4) The relative acceleration of the point Q with respect to the point P is aQP = aQ − aP = ¨ ro + ω˙ × rQ. 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 (2.12) . 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.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. 2). it is said that the body performs rotational motion.11) rP = ω˙ × rP + ω × (ω × rP ) aP = ¨ (2. This motion is determined by the following equation α = αo sinωt. ω. Given are: αo .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. 2. Determine: 1. The centre of gravity G of the rotor is displaced from its axis of rotation by distance µ. 3. components of the absolute velocity of the centre of gravity of the rotor along the same system of coordinates. The axis of relative rotation of the rotor 2 is fixed at the constant angle β with respect to the horizontal plane. performs an oscillatory motion about the vertical axis Z of the inertial system of coordinates XY Z. β. 2. L. . Ω. components of the absolute angular velocity of the rotor along a system of coordinates fixed to the rotor. The rotor 2 rotate with a constant angular velocity Ω in the direction shown. µ.PROBLEMS. z12 is rigidly attached to the base 1 and it is turned by angle β about axis x11 . z11 is rigidly attached to the base 1 and rotates about the vertical axis Z of the inertial system of coordinates XY Z. Its axis y2 coincides axis y12 and its instantaneous position is determined by the angular displacement Ωt. 4.PROBLEMS. is RG = L + µ = j21 L + k2 µ = j2 L + k2 µ (2. System of coordinates x2 . 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. 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. y2 . z2 is fixed to the rotor 2. 45 Solution. y11 . according to Fig.15) . 4.13) Position vector of the centre of gravity G. System of coordinates x21 .14) Hence. The angular displacement α determines uniquely its instantaneous position. 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. system of coordinates x11 . y12 . Its link 1 can move along axis X and is free to rotate about that axis. 4. The link 3 can move in plane Y Z along axis which is parallel to Y . 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. Distance c between the point B and axis Y . 3. Given are: 1. Angular velocity of the link 2. 3. Length of the link 2 lAB .PROBLEMS. 2. 2. Determine: 1. Motion of the point A (its position vector rA ). . Linear velocity of the link 3. Angular velocity of the link 1. 5 shows the kinematic diagram of a spatial mechanism. 46 Problem 11 Z 3 B 1 lAB O c Y 2 A X Figure 5 Fig. Positions of individual links. 16 yields matrix of direction cosines between the body 2 system of coordinates and the inertial one.PROBLEMS. 47 Solution. 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. 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. Since the link 2 can rotate with respect to the link 1 about axis z1 only.18) = z2 sin α sin β sin α cos β cos α . 2. 2. Matrices of direction cosines.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 . may be determined by angle α. the relative position of the body 2 system of coordinates x2 y2 z2 is uniquely determined by angle β. ⎤ ⎡ ⎤ ⎡ ⎤⎡ x1 cos β − sin β 0 x2 ⎣ y1 ⎦ = ⎣ sin β cos β 0 ⎦ ⎣ y2 ⎦ (2.17) z1 z2 0 0 1 Introduction of Eq. A. Z 3 z1 . ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤⎡ 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 (2.17 into Eq. From Fig. Its angular position with respect to the inertial system of coordinates XY Z. J.27) Hence. Angular velocities of individual links. C.28) ω 2 = ω1 + ω 21 (2.19) Kc + JrBC = IrA + j2 lAB (2.26) (2.21) The dot products. To find positions of individual links let us consider the following vector equation rC + rBC = rA + rBA (2. can be taken directly from matrix 2. Positions of individual links.18. The three above equations determine position of all links for any instant of time.26 with respect to time. ω 1 = Iα˙ (2. K.PROBLEMS. Since the link 1 performs rotational motion about axis X. 48 B.25) (2.22) (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) . 0 = rA + I · j2 lAB rBC = J · j2 lAB c = K · j2 lAB (2. its absolute angular velocity ω 1 may be obtained by differentiation of 2.23) (2. 0 = rA − lAB sin β rBC = lAB cos α cos β c = lAB sin α cos β (2.24) β = arcsin(rA /lAB ) α = arcsin(c/lAB cos β) rBC = lAB cos α cos β (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.29) ω 21 = k2 β˙ (2. appearing in the above equations.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. 27 with respect to time.33) . 49 where α˙ and β˙ are derivatives of expressions 2.26 respectively. Velocity of the point B.PROBLEMS. D.25 and 2. r˙ B = vB = Jr˙BC = JlAB (−α˙ sin α cos β − β˙ cos α sin β) (2. Velocity of the point B can be obtained by direct differentiation of expression 2. 2.PROBLEMS.motion of the body 1 is uniquely determined by the function ZA . 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 . According to the described constraints. Produce: 1. This slide is by a apart from axis Y . Components of the linear velocity of the point B and the point C. The point C is in a constant contact with the plane XY . 50 Problem 12 The point A of the body 1 shown in Fig. 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. 7 can move along the vertical slide 2 which is located along the axis Z of the inertial system of coordinates XY Z. 34) JrBY + Ka = KZA (t) + ka (2.PROBLEMS.37) (2. 8).38) (2.39) .36) a a Multiplication of the above vector equation by the units vectors associated with the inertial system of coordinates I. A.35) or Hence ZA rBY +K(1 − ) (2. rB = rA + rBA (2. 51 Solution. J and K yields the direction cosines between the axis z and axes of the inertial system of coordinates XY Z. 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. 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. 48) It follows that Multiplication of the above equation by unit vectors I.49) √ 2a.47) rCB = rB − rC (2. 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. Hence (2.42) or Since according to the above equation j=I rCX rCY ZA +J −K a a a (2.45) 2 2 + rCY + ZA2 = a2 rCX (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. 52 The last relationship permits the unknown component rBY to be determined as the explicit function of time q rBY = 2aZA − ZA2 ) (2.PROBLEMS. 7 the length of the vector rCB is equal to 2 2 2 + rCBY + rCBZ = 2a2 rCBX (2.50) .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 (2.44) The following property of the direction cosines cos2 ∠jI + cos2 ∠jJ + cos2 ∠jK = 1 (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. J and K respectively yields components of the vector rCB along the initial system of coordinates. 53 Therefore the second equation for determination of the components rCX and rCY takes form ¶2 µq 2 2 2aZA − ZA −rCY + a2 = 2a2 (2.35 0.3 0.53) 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) (2.52) These equations allow the trajectory of the point C to be computed. This trajectory is shown in Fig. They are s 2(a − ZA ) rCX = a 2a − ZA ZA (a − ZA ) rCY = p ZA (2a − ZA ) rCZ = 0 (2.15 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.05 0 0 0.54) (2.PROBLEMS.25 r CY 0. i=j×k (2. 9 for a = 1 0.51) 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.1 0.2 0.55) . 54 Hence. differentiation of the vector rC yields velocity of the point C.56) B.40 q (2.59) 2a − ZA ZA (2a − ZA ) Hence (2. The angular velocity of the link 1 may be obtained from the following vector relationship vB = vA + ω × rBA (2. 2. Linear velocities The linear velocity of the point B can be obtained by differentiation of the position vector rB which according to Eq.PROBLEMS.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. The position vector rC according to Eq.62) or ω × rBA = vB − vA (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 ) (2.64) . 2.57) rB = J 2aZA − ZA2 + Ka Hence a − ZA vB = J p Z˙ 2 A 2aZA − ZA (2.58) Similarly.52 s 2(a − ZA ) ZA (a − ZA ) rC = Ia + Jp (2. 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 (2.61) Angular velocities. 2. ⎡ ⎤ I J K ⎣ ωx ωy ωz ⎦ = I(vCX − 0) + J(vCY − 0) + K(0 − Z˙ A ) rCX rCY −ZA (2. 2.66) −Z˙ A ωx = p (2.Introducing them into Eq.70) where rCX and rCY are given by equation 2.68) (2.44) permits the components of the relative position vector to be obtain as an explicit function of time.52. the equation 2.69) ω × rCA = vC − vA The developed earlier direction cosines (Eq. 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.PROBLEMS.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 1.63 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 ) (2.72 −ω y ZA − ωz rCY = vCX ω z rCX + ω x ZA = vCY ωx rCY − ω y rCX = −Z˙ A Hence. or vC = vA + ω × rCA (2. the wanted components of the angular velocity are vCY − ωx ZA ωz = rCX Z˙ A + ωx rCY ωy = rCX (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. 55 Hence.71) = I(vCX ) + J(vCY ) + K( − Z˙ A ) that is equivalent to three scalar equation of form 2.72) .69 one can get the following vector equation. s21 . 2. The link 2 can move along the vertical slide of the base and its relative position is determined by s21 . absolute angular velocity and acceleration of the link 3 along a body 3 system of coordinates. linear velocity of the point P along the same system of coordinates. The link 3 is hinged to the link 2 and the angle α32 determines its relative position.PROBLEMS. α32 are given functions of time and a. derive expressions for components of: 1. Upon assuming that α1 . l. 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. are given parameters. 10 rotates about the vertical Z and its angular position is determined by the angle α1 . . 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. 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. 57 Solution. 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 3.75) . 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. according to Fig.PROBLEMS.74) ) The position vector of the point P . 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. 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. 11. 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.PROBLEMS.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. r˙ = r0 + ω3 × r (2. 58 Its first derivative with respect to time represents the wanted velocity of the point P .78) +k3 [(s21 sin α32 + a cos α32 + l)α˙ 32 ] Upon adding the two above expression together.79) .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. The axis Z3 is fixed in the plane XZ and its position is determined by angle β. Given are: ω− angular velocity of the link 1.PROBLEMS. 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 shows the kinematic scheme of a mechanism. β− angle between axis Z and Z3 . Derive the analytical expression for the linear velocity of the link 3. l− length of the link 2. 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. The link 3 is free to slide along and to rotate about axis Z3 . r− distance between points O and P . 12. Its link 1 rotates with the constant angular velocity ω about the horizontal axis Z of the motionless system of coordinates XY Z. 60 Solution.Introduction of the above expressions into Eq.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. j1 and k1 respectively offers three scalar equations.PROBLEMS.83) where γ x .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 (2.90) Hence . 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.84 by the unit vectors i1 .86) (2. γ y and γ z are angles between the axis z2 and axes of the system of coordinates x1 y1 z1 .89) (2.85) (2.80) The above vectors can be expressed as follows. 2. 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. 13 one can see that rQ = rP + rQP (2.81) (2.82) (2.84) Multiplication of the equation Eq.88) (2. 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. 2. 93) Hence.15 takes form 2. 2 + (−2r cos α sin β)rQ + (r2 − l2 ) = 0 rQ (2. 2. One corresponds to sign ’+’ and the other corresponds to sing ’-’.91) yields (rQ cos α sin β − r)2 + (−rQ sin α sin β)2 + (rQ cos β)2 = l2 (2. 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.88..89 and 2.PROBLEMS. (2. the instantaneous position of the point Q is determined by the following expression.92) After simple manipulation. 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. 61 The direction cosines have to fulfil the following relationship cos2 γ x + cos2 γ y + cos2 γ z = 1 (2. . 2. the fist derivative with respect to time of the above expression yields absolute velocity of the point Q.90 into Eq. (2. the equation 2.93. 14.91) Introduction of Eq’s.95) There are two possible solution.94) rQ = r cos α sin β ± (r2 cos2 α sin2 β − r2 + l2 )1/2 Since axis Z3 is motionless. The relative angular velocity of the wheel 2 with respect to the base 1 is constant and is equal to ω 21 . l. Produce expression for components of: 1. 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. The system of coordinates x2 y2 z2 is rigidly attached to the wheel 2. The instantaneous position of the seat 3 with respect to the wheel 2 is determined by the angular displacement β. The system of coordinates x1 y1 z1 is rigidly attached to the base 1. α(t). 15. The instantaneous position of this base is determined by the angular position α. Its base 1 oscillates about the horizontal axis X of the XY Z inertial system of coordinates. ω 21 . . 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. s. The system of coordinates x3 y3 z3 is rigidly attached to the seat 3.PROBLEMS. the absolute angular velocity of the seat 3 along the system of coordinates x3 y3 z3 2. β(t). The seat 3 is hinged to the wheel at the point A. (2.100) .96) ω1 = i1 α˙ The absolute angular velocity of the system of coordinates x2 y2 z2 . 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 .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 = (2.97) (2.PROBLEMS.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. 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. PROBLEMS. 64 The absolute angular velocity of the seat 3 is equal to ω 3 . ˙ 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. 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) (2.103) . 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.102) The first derivative of this vector yields the absolute liner velocity of the point P. ω 32 (t). Answer: vP x2 = 0 vP y2 = −bβ˙ vP z2 = −aα˙ − bα˙ cos β 3. Produce 1. System of coordinates x2 y2 z2 is rigidly attached to the arm 2. The arm 2 rotates with respect to the base 1 about axis that is parallel to z1 . 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 . b. 17 shows a diagram of a wheel excavator. 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. the expressions for the components of the absolute linear velocity of the point P along system of coordinates x2 y2 z2 . Its base 1 rotates about the vertical axis Y of the inertial system of coordinates XY Z. Answer: ³ ´ ˙ ω3x2 = α˙ sin β ω 3y2 = α˙ cos β ω3z2 = β + ω 32 2. The wheel 3 rotates with respect to the arm 2 about axis that is parallel to z1 with the angular velocity ω32 . the expressions for components of the absolute acceleration of the point P along system of coordinates x2 y2 z2 . Its relative angular position is determined by an angle β. Its instantaneous position is determined by the angle α. Given are: α(t). the expressions for the components of the absolute angular velocity of the wheel 3 along system of coordinates x2 y2 z2 . β(t). System of coordinates x1 y1 z1 is rigidly attached to the base 1.PROBLEMS. a. the components of the absolute velocity of the point G along the body 3 system of coordinates x3 y3 z3 . 18 shows the kinematic scheme of a mechanism. Its link 1 rotates with a constant angular velocity ω 1 about the vertical axis Z. 66 Problem 17 Z z1 y3 ω1 4 z3 B 1 G β y1 3 rA 2 l/2 A l Figure 18 Fig. 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 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. Answer: vGx1 = − 2l ω 1 cos β vGy1 = −r˙A sin β vGz1 = −r˙A cos β + 2l β˙ . rA = −k1 rA Given are: ω1 − angular velocity of the link 1 rA (t)− position of the point A as a function of time. The link 2 can translate with respect to the link 1 and its motion is determined by position vector of the point A.PROBLEMS. l− length of the link 3 Produce expressions for 1. γ(t). The wheel 2 of radius R performs rotational motion about axis x1 with respect to the base 1. Derive expressions for the components of the absolute linear velocity of the point P along system of coordinates x2 y2 z2 . 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.PROBLEMS. 19 rotates about the vertical axis Z of the inertial system of coordinates XY Z. 1. Derive expressions for the components of the absolute angular acceleration of the seat 3 along the system of coordinates x2 y2 z2 . The system of coordinates x1 y1 z1 is rigidly attached to this base. 2. Given are: α(t). R. This relative motion is determined by function β(t). Its angular position is determined by the angular displacement α(t) which is a given function of time. The relative angular position of the seat 3 with respect to the wheel 2 is determined by angle γ(t). The seat 3 is hinged to the wheel 2. The system of coordinates x2 y2 z2 is rigidly attached to the wheel 2. L . The system of coordinates x3 y3 z3 is rigidly attached to the seat 3. β(t). Its instantaneous position is determined by the angle α1 . the absolute angular acceleration of the link 2 along the body 2 system of coordinates 2.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. The link 2 can rotates with respect to the link 1 and its relative angular position is determined by the angle α21 . Given are: α1 . The system of coordinates is rigidly attached to the link 1.L. Derive expression for the components of 1. R . 20 is free to rotate about axis X of the inertial system of coordinates XY Z. the absolute linear velocity of the point P along the body 2 system of coordinates . α21 . Produce: 1. The distance between the hinge and the rotor axis is c.PROBLEMS. 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. 21. 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. 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 . 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 components of the absolute angular acceleration of the blade 2 along the system of coordinates x2 y2 z2 . The relative position of the blade 2 with respect to the rotor 1 is determined by the angular displacement α. The system of coordinates x1 y1 z1 is rigidly attached to the rotor 1. the components of the absolute velocity and the absolute acceleration of the point P along the system of coordinates x2 y2 z2 2. z1 as a function of the angular displacements α and β. Answer: εx1 = −αω ˙ 2. Answer: vP x1 = −rβ˙ sin β − bα˙ vP y1 = α(R ˙ + a + r cos β) vP z1 = −rβ˙ cos β 2. y1 .PROBLEMS. Answer: aP x1 = v˙ P x1 − αv ˙ P y1 aP y1 = v˙ P y1 − αv ˙ P x1 aP z1 = v˙ P z1 3. y1 . Components of the absolute acceleration of the tip P along the system of coordinates x1 . Components of the angular acceleration of the propeller 2 along the system of coordinates x1 . Components of the absolute velocity of the tip P of the propeller 2 along the system of coordinates x1 . Its propeller 2 rotates with the constant angular velocity ω2. Its longitudinal axis E − F is always tangential to the path of the point A. follows the circular path of radius R with the constant linear velocity vA . Produce: 1.1 with respect to the plane 1. y1 . 70 Problem 21 0 α R E F vA A a Y ω 2. This path belongs to the horizontal plane of the inertial space XY Z. z1 as a function of the angular displacements α and β. z1 is rigidly attached to the plane 1. shown in Fig. System of coordinates x1 . z1 as a function of the angular displacements α and β. 22.1 y1 z1 β b X x1 1 r o1 o1 P 2 x1 Figure 22 The point A of the plane 1. y1 .1 εy1 = 0 εz1 = 0 . α3 (t). 23a) shows the self-steering mechanism of the yacht 1. y3 . y3 . y3 . 23b)). y1 . the components of the absolute velocity of the point A shown in Fig. 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. y3 . b. z2 is attached to this shaft. The link 3 is hinged to the link 2 at the point o. Produce the expression for 1. the components of the absolute angular velocity of the link 3 along the system of coordinates x3 . z1 is attached to the yacht. z3 Answer: ω3x3 = α˙ 2 cos α3 ω 3y3 = −α˙ 2 sin α3 ω3z3 = α˙ 3 2. v(t). z3 . 23a) along the system of coordinates x3 . y2 .PROBLEMS. The system of coordinates x1 . z2 is determined by the angle α3 . Answer: vAx3 = v cos α3 −bα˙ 2 sin α3 vAy3 = −v sin α3 −aα˙ 3 −bα˙ 2 cos α3 vAz3 = −aα˙ 2 sin α3 . The system of coordinates x2 . y2 . The instantaneous position of the body 3 system of coordinates x3 . z3 with respect to the system of coordinates x2 . 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. Given are: a. The yacht is travelling along a straight line with velocity v. the components of the absolute angular acceleration of the link 3 along the system of coordinates x3 . α2 (t). The shaft 2 of this mechanism is free to rotate about the axis x1 . Its instantaneous position is determined by the angle α2 (see Fig. PROBLEMS. y2 . The relative position of the housing with respect to the ship is given by the angular displacement β. y1 . z2 3. z1 is rigidly attached to the ship. y1 . 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. y2 . The gyroscope 3 rotates about axis with respect to the housing. 24) rotates about the axis Y of the absolute system of coordinates XY Z. Its relative angular velocity is Ω. y2 . components of the absolute linear velocity of the point P along the system of coordinates x2 . z2 2. components of the absolute angular velocity of the gyroscope along the system of coordinates x2 . z2 is fixed to the housing. z1 . System of coordinates x1 . The housing 2 performs the rotational motion about axis x1 . components of the absolute angular acceleration of the gyroscope along system of coordinates x2 . Produce: 1. y2 . components of the absolute linear velocity of the point P along the system of coordinates x1 . System of coordinates x2 . z2 4. Its instantaneous position is determined by the angle α. Its instantaneous position is determined by the angular displacement α. The system of coordinates x2 y2 z2 is rigidly attached to the wheel 2. 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 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) . 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.PROBLEMS. 25 shows the sketch of a Ferris Wheel. 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. 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 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. Its base 1 oscillates about the horizontal axis X of the inertial system of coordinates XY Z. Produce 1.PROBLEMS. The link 2 is free to rotate about the axis y1 and its relative angular position is determined by the angle β. the components of the absolute angular velocity of the link 2 along the system of coordinates x2 y2 z2 . The system of coordinates x1 y1 z is attached to the link 1. Its instantaneous position is determined by the angular displacement α. The dimensions f and L locate position of the point P with respect to this system of coordinates. . 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. Answer: ˙ + k2 (α˙ cos β) ω2 = i2 (−α˙ sin β) + j2 (β) 2. 26 rotates about the horizontal axis Z of the inertial system of coordinates XY Z. The system of coordinates x2 y2 z2 is attached to the link 2. the components of the absolute angular acceleration of the link 2 along the system of coordinates x2 y2 z2 . the components of the absolute linear velocity of the point P 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. Answer: ˙ − f α˙ cos β) + j2 (Lα˙ sin β) + k2 (−f α˙ sin β) vP = i2 (βL . Its relative angular position is determined by the angle β(t). The ladle 2 rotates about the axis y2 of the body 2 system of coordinates x2 y2 z2 . 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). Answer: ω2 = i2 α˙ cos β + j2 β˙ + k2 α˙ sin β 2. Produce: 1.PROBLEMS. the components of the absolute velocity of the point P along the system of coordinates x2 y2 z2 . The point o1 of the carriage 1 moves along the horizontal axis Y of the inertial system of coordinates XY Z. . the expression for the components of the absolute angular acceleration of the ladle 2 along the system of coordinates x2 y2 z2 . the expression for the components of the absolute angular velocity of the ladle 2 along the system of coordinates x2 y2 z2 . 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. Answer: ε2 = i2 (¨ α cos β − α˙ β˙ sin β) + j2 β¨ + k2 (¨ α sin β + α˙ β˙ cos β) 3. 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 ¯ ¯ ¯ ¯ ¯ ¯ . The link 2 is hinged to the link 1 at the point A. The system of coordinates x 1 y 1 z 1 is rigidly attached to the link 1. Its instantaneous position is given by the absolute angular displacement α. Produce: 1. 28 shows the physical model of a mechanical system. 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 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 . The other end of this link P always stays in contact with the cylindrical surface 3 of radius R (b>R). 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. 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 .PROBLEMS. I y2 and I z2 . The system of coordinates x 2 y 2 z 2 is attached to the body 2 and coincides with its principal axes. 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. Its centre of gravity G is located by the distance c. The link 1 of this system rotates about the vertical axes Z of the inertial frame XYZ. 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) 1 82 MOTION OF CENTRE OF MASS .10) The last formula permits to formulate the following statement.9 can be rewritten in form ˙ = P PN i=1 Fi = Fex (3.MOTION OF CENTRE OF MASS . 3.9) i=1 j=1 Taking into account that according to Eq. 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 (3.7) mi r˙ i is called inertial linear momentum of the system of particles. . Z mi G ri rG O Y X Figure 3 DEFINITION: The following vector P= PN i=1 (3.LINEAR MOMENTUM. 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.6 Fij = −Fji and Fii = Fjj = 0 the formula 3. The first derivative of the linear momentum is ˙ = P N X (3. 3. The motion of an individual particle mi is defined by the position vector ri (Fig.LINEAR MOMENTUM. Let us consider system of N particles in an inertial space. 3.MOMENT OF MOMENTUM.is the total mass of the system of particles.2 MOMENT OF MOMENTUM. 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. Relative angular momentum about a moving point in the inertial space. 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. 3. 1.10 is The last result permits the following statement to be formulated. 3. 83 Let vector rG be the position vector of the centre of mass of a system of particles (Fig. Angular momentum about a moving point in the inertial space.12) r˙ i mi = P i=1 Second differentiation yields ˙ ¨ rG m = P (3. always three kinds of moment of momentum (angular momentum) are introduced. 3 ).1 Moment of momentum about a fixed point in the inertial space. DEFINITION: The following vector HO = PN i=1 ri × (mi r˙ i ) (3. according to Eq. 4). Considering system of particles.15) is called angular momentum about the fixed point O. 2. Angular momentum about a fixed point in the inertial space.16) . 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. 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 .14) Hence. In this paragraph the three above types of angular momentum are defined and discussed. Differentiating the equation 3.11 with respect to time one can obtain r˙ G m = N X (3.13) ¨ rG m = Fex (3.2. 84 Z mi ri O Y X Figure 4 Applying the second Newton law to the equation 3.MOMENT OF MOMENTUM.20) . 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 (3.16 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 (3.19) Taking into account the above equation. one may say that N N X X i=1 j=1 ri × Fij = 0 (3.17) To show that the last term is equal to zero.18) Since vectors rij and Fij are parallel then ri × Fij + rj × Fji = 0 (3. let us consider two particles mi and mj (Fig. MOMENT OF MOMENTUM. the relative motion of the particle mi with respect to the point C is determined by the formula (3. Its first derivative is ˙C= H N X i=1 r˙ i. ri. Z mi r i.2 Moment of momentum about a moving point in an inertial space.C × (mi r˙ i ) (3. 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.21 represents resultant moment MO of all external forces acting on the system of particles with respect to the fixed point O. 85 and the equation 3.2.25) . 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.C = ri − rC (3.C × (mi¨ ri ) (3. we may conclude this paragraph with the following statement.C ri C rC O Y X Figure 6 If the motion of the particle mi is determined by vector ri .23).21) The left hand side of equation 3.17 takes form ˙O= H N X i=1 ri × Fi (3. Hence. ˙ O = MO H (3.22) 3.C × (mi r˙ i ) + N X i=1 ri.23) DEFINITION: The following vector HC = PN i=1 ri.24) is called angular momentum about the moving point C. 3. P = r˙ G m. According to consideration curried out in chapter 2 section 1.MOMENT OF MOMENTUM. the above formula yields ˙ G = MG H (3.C × Fji = (ri.j r i.28) If the arbitrarily chosen point C coincides the gravity centre G (˙rC = r˙ G ).C × Fi = −˙rC × P + MC (3.C rj.is resultant moment of all external forces about the moving point C.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. 3.26) j=1 But PN r˙ i × (mi r˙ i ) = 0 since r˙ i is parallel to mi r˙ i.is the linear momentum of the system considered P= N i=1 PN i i MC = i=1 ri.25 and taking advantage from the second Newton law. the first derivative of the angular momentum can be written as follow.C × Fi .C − rj.23 into Eq. . PN Pi=1 N i=1 ri.27) P (m r˙ ) .C × (Fi + N X Fij ) (3. 86 Introducing Eq.C ) × Fij = rij × Fij = 0 (see Fig. Hence ˙ C = −˙rC × r˙ G m + MC H (3.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 × Fij + rj. ˙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. 7) mj ri.j Z mi rj Fi.C × j=1 Fij = 0 because ri. 2.31 is equal to 0. according to Fig 8.C × (¨ ri. 8 and the arbitrarily chosen point C.33 into it yields h˙ C = N X i=1 = N X i=1 = N X i=1 ri. introduction of Eq.C mi ) i=1 (ri.C = ri −rC (3.C r G.C mi ) (3.C ri.C C rC O Y X Figure 8 DEFINITION: The following vector hC = PN × (˙ri.C = r˙ i − r˙ C ¨ ri.C × (Fi + N X i=1 N X j=1 Fij ) + ¨ rC × ri.33) The first term in the right hand side of equation 3. 3. Z G mi ri r i.30) is called moment of relative momentum.MOMENT OF MOMENTUM.34) .C mi ) i=1 (3.C × (¨ rC mi ) N X N X (ri. ri. 87 3.C = ¨ ri − ¨ rC and (3.C × (˙ri.C × Fi + ¨ rC × = MC + ¨ rC × mrG.C mi ) + N X i=1 ri.31) But. Hence.C (3.32) Hence r˙ i. Let us once more consider a system of particles shown in Fig. Its first derivative is h˙ C = N X i=1 r˙ i.C mi ) i=1 ri.C × (¨ ri mi ) − ri.3 Moment of relative momentum. Taking into account that the right hand sides of equations 3. moment of the relative angular momentum about the system’s centre of gravity is h˙ G = MG (3.37) ˙ G = MG ˙ O = MO H H h˙ G = MG Depending on the dynamic problem to be solved.29 are equal.C = 0). 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.EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS. 3.35) The above considerations allows to formulate the following statement.3. 3. that these problems can only be solved.36) The last expression is obvious because for motionless system of particles both moments of momentum are equal to 0. we can choose one of these three pairs of equations. In the previous sections of this chapter four vectorial equations has been derived. ½ ½ ½ ˙ =F ˙ =F ˙ =F P P P (3. It means.1 Conservation of momentum principle. Not all of them are independent. 88 If C always coincides the centre of gravity G (rG. Z F λ O Y X Figure 9 .35 and 3. Each of them is equivalent to six scalar equations. 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. The independent pairs of equations are. Very small range of dynamic problems fulfil these requirements. which have six scalar unknown only. without additional equations expressing interaction between individual particles. we arrive to conclusion that ˙ G = h˙ G H and HG = hG (3.3 EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS. results in the The scalar multiplication of equation H following scalar relationship ˙ O = λ · MO = 0 λ·H (3.EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS.2 Conservation of angular momentum principle. STATEMENT: If component of a resultant moment MO of all external forces along a fixed in an inertial space axis is equal to zero.38) Hence. its angular momentum is conserved along that axis. λ · P = const (3.40) λ · HO = const (3. 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.41) Hence. 89 ˙ = F by the unit vector λ yields the following scalar The scalar multiplication of P equation ˙ =λ·F=0 λ·P (3. its momentum is conserved along that axis (λ · P = const). Z MO λ O Y X Figure 10 ˙ O = MO by the unit vector λ.39) STATEMENT: If component of a resultant of all external forces along a fixed in an inertial space axis is equal to zero. 3.3. . 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.45) to .42) ˙ O = MO ˙ G = MG H H h˙ G = MG can be integrated with respect to time.46) to .43 yields ∆P = UF . ∆HO = UMO . to to to to Z t Z t Z t Z t ˙ ˙ HG dt = MG dt. MO dt.47) to . MG dt hG dt = to to to (3. If the resultant force and moment of all external forces are explicit functions of time. 90 3.increment of the linear momentum Z t ˙ O dt = HO (t) − HO (to ) = ∆HO H (3. .49) MG dt = UMG (3. the derived equations of motion ½ ½ ½ ˙ =F ˙ =F ˙ =F P P P (3.43) to The following expression are called respectively Z t ˙ = P(t) − P(to ) = ∆P Pdt (3.EQUATIONS OF MOTION AND THEIR FIRST INTEGRALS.3.44) to . Z t Z t Z t Z t ˙ ˙ O dt = Pdt = H Fdt.50) to -the angular impulse about point O Z t to -the angular impulse about point G Introduction of the above notations into 3.48) to .the linear impulse Z t MO dt = UMO (3. ∆HG = ∆hG = UMG (3.increment of the angular momentum about G Z t h˙ G dt = hG (t) − hG (to ) = ∆hG (3.3 Impulse – momentum principle.increment of the relative angular momentum about G Z t Fdt = UF (3.increment of the angular momentum about point O Z t ˙ G dt = HG (t) − HG (to ) = ∆HG H (3. when a food was placed outside the table at F .At the instant t = 0.4 91 PROBLEMS. 3. Problem 28 F B S B S Figure 11 On a massless and free to rotate about the vertical axis Z table. . 11. The dog B has a mass mB greater then that of the dog S (mB > mS ). two dogs take position at B and S as shown in Fig. Show.PROBLEMS. that regardless of relative velocities developed by the dogs with respect to the table. the dogs as well as the table were motionless. the small dog S will reach the food first. 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. RB B F . . the angular momentum of the system considered about fixed in the inertial space point O (see Fig. RS RB RS S O Y X Figure 12 According to the introduced definition (see Eq.15). the absolute velocity of the small dog R˙ S is always greater then the mS absolute velocity of the big dog R˙ B .PROBLEMS. Taking into account that RS = RB = R.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. 3. 92 Solution . we have mB ˙ R˙ S = RB mS (3.53) B Since m > 1. Each of the two nozzles has the exit area equal to A. The diameter d of the nozzle as well as D are small as compare with the distance R shown in the drawing 13. 13 distributes water of density at the volume rate Q.PROBLEMS. 93 Problem 29 1 2 D A ω R d α Figure 13 The sprinkler shown in Fig. . 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. All the above specified forces should be classified as external with respect to the system considered. 14b) presents position of these particles after a small increment in time ∆t. Let us produce expression for the linear momentum of particles in the volume B.PROBLEMS. one can assume B that all particles that belong to the volume B have the same absolute velocity vA . Axes XY Z forms the inertial system of coordinates and axes xyz are attached to the part 1. 14a) corresponds to an arbitrarily chosen instant of time t. 14a). 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. Position of the particles shown in the Fig.z Z.54) . The forces F represents external forces acting on the system of particles due to its interaction with the stationary part 2.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. it is not shown in this diagram. This system is assembled of the rotating part 1 and water that is shadowed in Fig. 94 Solution F ω F O O x x G p E Z. Therefore the increment in the linear momentum is B ∆PB = mB vA (3. If the diameter of the nozzle is small with respect to the distance R. The atmospheric pressure is evenly distributed over the entire outer surface of the system of particles. 14a) by the dash-dot line. 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. Fig. The resultant force due to gravitation is denoted by G. Since its resultant is always equal to zero. 62 and 3.55) 2 B is equal to sum of the velocity of transportation vTB and the The absolute velocity vA B relative velocity vR .66) .61) ∆HO =∆HO + ∆HO + ∆HO = k m(R ω − R cos α) A where m is m = mB + mC = Q ∆t (3.64) Introduction of 3. 95 where mB stands for total mass of particles in the volume B and is equal to 1 mB = Q∆t (3.62) To solve the problem. ∆HO = UMO (3.65) A Hence the wanted angular speed of the sprinkler is ω= Q2 R cos α A Q −M R2 (3.60) O = 0 Hence the total increment in the angular momentum is ¶ µ Q B C E 2 (3.PROBLEMS. ¯ ¯ ¯ ¯ 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 (3.58) A In the same manner one may produce expressions for increment in the angular momentum of particles associated with the volume C and E.61.57) 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) (3.63 yields Q Q ∆t(R2 ω − R cos α)∆t = −M∆t (3.angular impulse principle. one may take advantage of the angular momentum .63) where UMO = −kM∆t (3. 3.59) ∆HO = k m (R ω − R cos α) A ∆HE (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 − ) (3.64 into 3. ¶ µ Q C C 2 (3. Hence.o . 1).o = ri. the total mass of the rigid body can be expressed as follow Z ZZZ ρ(xyz)dxdydz = dm (4.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.1 LINEAR AND ANGULAR MOMENTUM. The mass of an individual element dm is (4.3) m m .Chapter 4 KINETICS OF RIGID BODY.1) dm = ρ(xyz)dxdydz where ρ is density. 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. i z Z ω ri.o dm (4.2) m= V m The centre of gravity of the rigid body. (Fig. According to consideration presented in chapter 2. with respect to the body system of coordinates is Z 1 rG. according to the above notations. 4. o dm = ri.o + ω × ri. ri.o = ix + jy + kz (4.o × (r0i . namely ω = iω x + jω y + kωz and ri.o + ω × rG.o · ri.o × (ω × ri.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.o (4.7) (4. the total angular relative momentum has form Z ho = ri.LINEAR AND ANGULAR MOMENTUM.o × (+ω × ri.5) where Introducing Eq.o × (ω × ri.o ) − ri.4) r˙ G = r˙ o + r˙ G.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.o = r˙ o + r0G. 4.6) Angular relative momentum of the element dm with respect to the origin o is dho = ri.o = r˙ o + ω × rG.o )dm Hence.o )m (4.10) the triple product is as follows.o )dm = ri.o )dm (4.9) If both vectors involved are given by their components along the body system of coordinates. 97 According to consideration in the previous chapter linear momentum P is P = r˙ G m (4. ri.o · (ri.8) m Let us calculate the triple cross product.4 one can obtain P = (˙ro + ω × rG.5 into Eq.o × (+ω × ri.o × r˙ i.o ) = ω · (ri. 4.12) .o · ω) (4. according to 4. Iy . Iyz .16 produces components of vector ho along body system of coordinates.17) HO = dHO = ri × r˙ i dm = ri ×(ri ×ω)dm m m m . The above relations can be written in the following matrix form. R Iz = (x2 + y 2 )dm m Ixz = m R xz dm (4. Iyz = m R yz dm. ⎡ ⎤ ⎡ ⎤⎡ ⎤ hox ωx Ix −Ixy −Ixz ⎣ hoy ⎦ = ⎣ −Iyx Iy −Iyz ⎦ ⎣ ω y ⎦ (4. 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.LINEAR AND ANGULAR MOMENTUM. Matrix {ω} is assembled of components of angular velocity along body system of coordinates.14) where: Ix . 2) the angular momentum is Z Z Z (4. R Iy = (x2 + z 2 )dm. m Ixy = R 98 m xy dm.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. Introducing notations R Ix = (y 2 + z 2 )dm.15) hoz −Izx −Izy Iz ωz or shorter {ho } = [I]{ω} (4.16) Matrix [I] is called inertia matrix.13) m the components of vector ho . Ixz −are called products of inertia. Iz −are called moments of inertia and Ixy . The formula 4. 2 (4. 99 or in the matrix form {HO } = [I]{ω} 4.18) PROPERTIES OF MATRIX OF INERTIA. The introduced definitions of moment of inertia e.PROPERTIES OF MATRIX OF INERTIA. 3.20) yz dm m permit elements of the inertia matrix to be calculated for bodies of a simple geometrical shape like a cylinder. 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 (4.19) m and product of inertia e.g.g. Iyz = Z (4. Z Ix = (y 2 + z 2 )dm (4. sphere.21) . rectangular block etc. As an example let as calculate the moments and products of inertia for the rectangular block shown in Fig. that the inertia matrix of a body is known about the body system of coordinates xyz (Fig.1 Parallel axis theorem. . Let xG. For bodies having more complicated shape (see Fig.PROPERTIES OF MATRIX OF INERTIA. 4. The result of such calculations is collected in appendix B. ⎡ ⎤ Ix −Ixy −Ixz [I] = ⎣ −Iyx Iy −Iyz ⎦ (4. c be coordinates of the centre of gravity G of the body considered. 5).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. zG be the body system of coordinates parallel to xyz having its origin at G. z ∆ mi zi yi y xi x Figure 4 Then. b.24) −Izx −Izy Iz Let a. the inertia matrix with respect to the arbitrarily chosen system of coordinates may be analytically established. 100 In a similar manner the remaining elements may be calculated. yG. 4) a division into small elements have to be carried out.2. ⎡ ⎤ Ix −Ixy −Ixz [I] = ⎣ −Iyx Iy −Iyz ⎦ (4. Let us assume. 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. let us consider a product of inertia.PROPERTIES OF MATRIX OF INERTIA.28) . m Z Z 2byG dm + m Z 2czG dm m (4. 1 m m Z yG dm (4.25) zG dm m zG dm. which is actually 0. according to the previously introduced definition. 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. Ix = IxG + m(b2 + c2 ) Now. 1 m Z xG dm. 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.27) Z m bcdm = IyzG + bcm (4.26) m represents components of the distance between origin G and the centre of gravity. Hence. 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. 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. z2 which has the same origin O. Introduce new body system of coordinates x2 . one may derive expressions for the remained products and moments of inertia. z1 . y2 . The relative position of the system x2 y2 z2 can be uniquely determined by the matrix of direction cosines [C1→2 ]. z1 (see Fig.32) hoz1 ωz1 .31) where e. zG or vise versa. but it is turned with respect to the first one. All these equations can be written in the following matrix form. 102 Similarly. 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. y1. yG.g. Let us assumed that the matrix of inertia of a body is known about the system of coordinates x1. according to Eq.2 Principal axes.16. Inertia matrices about axes having the same origin. ⎤ ⎡ 2 −ac b + c2 −ab −bc ⎦ [IG ]=[I]-m⎣ −ba a2 + c2 (4. components of its angular momentum along axes x1 . 4. y1 . 6). ⎤ ⎡ 2 −ac b + c2 −ab −bc ⎦ [I] = [IG ] + m ⎣ −ba a2 + c2 (4.2. are ⎡ ⎤ ⎡ ⎤ hox1 ωx1 ⎣ hoy1 ⎦ = [I]1 ⎣ ωy1 ⎦ (4.PROPERTIES OF MATRIX OF INERTIA.t23 = cos(∠y2 z1 ) If the body rotates with an angular velocity ω.30) 2 −ca −cb a + b2 4. 35 into 4. z1 is known.33 into the above formula we have [C1→2 ][I]1 {ω}1 = [I]2 [C1→2 ]{ω}1 (4. y1 . y2 . ⎤ ⎡ 0 Ixp 0 [I]p = ⎣ 0 Iyp 0 ⎦ (4.34) Similarly But {ho }2 = [C1→2 ]{h}1 and {ω}2 = [C1→2 ]{ω}1 (4. 4.35) Hence.33) {ho }2 = [I]2 {ω}2 (4. introducing Eq. that all products of inertia about these axes are equal to 0.38) The above formula permits to calculate the matrix of inertia along the system of coordinates x2 . 103 Or in shorter form. 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. {ho }1 = [I]1 {ω}1 (4. Principal axes.34 yields [C1→2 ]{h}1 = [I]2 [C1→2 ]{ω}1 (4.39) 0 0 Izp .36) Upon introducing 4. turned with respect to the arbitrarily chosen system x1 y1 z1 .37) and after a little manipulation one may obtain [I]2 = [C1→2 ][I]1 [C1→2 ]T (4. 7). z2 when the matrix of inertia about the system of coordinates x1 .PROPERTIES OF MATRIX OF INERTIA. ⎫ ⎧ ⎫ ⎧ ⎨ 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 (4.45) . Hence. [I]1 t22 .41) or or ⎫ ⎧ ⎫ ⎧ ⎫⎤ ⎡ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎤ ⎡⎧ ⎨ t11 Ixp ⎬ ⎨ t21 Iyp ⎬ ⎨ t31 Izp ⎬ ⎨ t11 ⎬ ⎨ t21 ⎬ ⎨ t31 ⎬ ⎣ t12 Ixp t12 . t32 Izp ⎦ = ⎣[I]1 ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ t13 Ixp t23 Iyp t33 Izp t13 t23 t33 (4. 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 (4.42) The above relationship is fulfilled if and only if the corresponding columns are identical.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. [I]1 t32 ⎦ .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 (4.40) [C1→p ]T [I]p = [I]1 [C1→p ]T (4. t22 Iyp .PROPERTIES OF MATRIX OF INERTIA.43) ⎩ ⎭ ⎩ ⎭ t33 t33 where ⎡ ⎤ 1 0 0 [1] = ⎣ 0 1 0 ⎦ 0 0 1 The relationships 4. 46) The last equation.46 Ixp . if developed. (Ix1 − Iyp )t21 + (−Ix1 y1 )t22 + (−Ix1 z1 )t23 = 0 (−Iy1 x1 )t21 + (Iy1 − Iyp )t22 + (−Iy1 z1 )t23 = 0 t221 + t222 + t223 = 1 (4.50) Solutions of the above equations determine uniquely a position of the principal axes. Izp are always real and positive.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. In similar manner one may obtain equations which allows the other direction cosines to be determined. or ¯ ¯ Ix1 − Ip −Ix1 y1 −Ix1 z1 ¯ ¯ −Iy1 x1 Iy1 − Ip −Iy1 z1 ¯ ¯ −Iz1 x1 −Iz1 y1 Iz1 − Ip 105 ¯ ¯ ¯ ¯=0 ¯ ¯ (4. It may be proved that for any matrix of inertia [I]1 . Hence.44 becomes linearly dependent.47) Since the direction cosines have to fulfill the following relationship t211 + t212 + t213 = 1 (4.PROPERTIES OF MATRIX OF INERTIA. For each of these roots the sets of equations 4. t12 . t13 . Iyp . .47 and 4. t13 .48 form set of equation which determines the direction cosines t11 .48) Equations 4. Only two of them within each set are independent and may be used to identify the unknown direction cosines t11 . the roots of equation 4. it has three roots. t12 . (Ix1 − Ixp )t11 + (−Ix1 y1 )t12 + (−Ix1 z1 )t13 = 0 (−Iy1 x1 )t11 + (Iy1 − Ixp )t12 + (−Iy1 z1 )t13 = 0 (4. forms an algebraic equation of third order. m = 12kg. Each of the bars have a length l and mass m. 106 4. Use the following numerical values. l = 1m. Calculate the principal moments of inertia of the assembly along axes through its centre of gravity. The bars were welded together at right angles at A.2.PROPERTIES OF MATRIX OF INERTIA.3 Problems Problem 30 l A l Figure 8 The figure above shows an element assembled of two identical thin and uniform bars. . ⎤ ⎡ 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.52) Inertia matrix about axes x1 y1 z1 for the element 2.PROPERTIES OF MATRIX OF INERTIA.1 = ⎣ 0 0 0 ⎦ 2 0 0 ml3 (4. 9) l 4 Inertia matrix about axes x1 y1 z1 for the element 1.53) Inertia matrix about axes x1 y1 z1 for the whole assembly.55) . ⎡ ml2 ⎤ 0 0 3 I1. ⎡ 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. 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. 107 Solution. ⎤ ⎡ 0 0 0 2 0 ⎦ I1.2 = ⎣ 0 ml3 ml2 0 0 3 (4.51) xG = yG = (4. 56) = −2 0 2 z2 0 0 1 Hence. ⎡ ⎤ ⎤⎡ ⎤ ⎡ 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. Transfer matrix C2→3 between axes x2 y2 z2 and x3 y3 z3 is.57) . 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. x3 y3 z3 are principal axes. 108 Since axis x3 is axis of symmetry of the body considered.PROPERTIES OF MATRIX 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.PROPERTIES OF MATRIX OF INERTIA. 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.radius of the ball 1 r2 = 5m.mass of the ball 2 r1 = √5/2 m .radius of the ball 2 R1 = 1m R2 = 2m l = 1m .mass of the ball 1 m2 =√2kg . 109 Problem 31 r1 R1 z O1 O y l l r2 R2 O2 Figure 10 The rigid body. Assume the following numerical data: m1 = 1kg . 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. shown in the figure 10. 60) Hence.58) Moment of inertia of the ball 2 about axes x2 y2 z2 2 Ix2 = Iy2 = Iz2 = m2 r22 = 4 kgm2 5 (4.5 kgm2 5 (4. l. 110 Solution.5 0 0 = ⎣ 0 1. 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 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. −l. −R2 ) (4.PROPERTIES OF MATRIX OF INERTIA. 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.59) Coordinates of the points O1 and O2 with respect to system of coordinates xyz are as follow O1 (0. 11) 2 Ix1 = Iy1 = Iz1 = m1 r12 = 0.61) 0 1 1. R1 ) O2 (0.5 1 ⎦ kgm2 (4.62) = ⎣ 0 12 4 ⎦ kgm2 0 4 6 . matrix of inertia of the whole assembly about system of coordinates xyz is ⎡ ⎤ 16.5 − Ip ) − 25] = 0 (4.4926 t13 = 0.70) They are fulfilled for t11 = 1 t12 = 0.5 − Ipx )t11 + (0)t12 + (0)t13 = 0 (0)t11 + (13.86700 t31 = 0 t32 = −0.5 − Ipx )t12 + (5)t13 = 0 t211 + t212 + t213 = 1 (4.71) .63) 0 5 7.5 − Ipy )t21 + (0)t22 + (0)t23 = 0 (0)t21 + (13.5 − Ipz )t31 + (0)t32 + (0)t33 = 0 (0)t31 + (13.5 − Ipy )t22 + (5)t23 = 0 t221 + t222 + t223 = 1 (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.PROPERTIES OF MATRIX OF INERTIA.67) zp t31 t32 t33 z The corresponding equations for its elements take form (16.0000 t23 = 0.69) (16.67kgm2 (4.68) (16.8700 (4.5 0 0 [I] = [I1 ] + [I2 ] = ⎣ 0 13.5 5 ⎦ kgm2 (4.65) Root of the above equation are: Ipx = 16.5kgm2 .5 − Ip 0 0 ¯ ¯ ¯=0 ¯ 0 13.5 − I 5 p ¯ ¯ ¯ 0 5 7.33kgm2 .00000 t21 = 0 t22 = 0. 111 Hence.5 Principal moments of inertia are roots of the following equation ¯ ¯ ¯ ¯ 16.4926 t33 = 0.5 − Ip ¯ (4. Ipy = 16.5 − Ip )[(13.5 − Ipz )t32 + (5)t33 = 0 t231 + t232 + t233 = 1 (4. Ipz = 4.5 − Ip )(7.64) or in a more developed form (16. 88o 0 −0. 112 Hence.72) Position of the principal axes with respect to the system of coordinates xyz is shown in Fig 12.75o cos 29.87 (4.PROPERTIES OF MATRIX OF INERTIA.4926 0.48 o yp 119. matrix of directional cosines Cxyz→xp yp zp is ⎤ ⎡ ⎤ ⎡ cos 90o cos 90o 1 0 0 cos 0o Cxyz→xp yp zp = ⎣ 0 0.75 o O 29.88 o y x xp Figure 12 .88o cos 60. zp z 60.48o ⎦ cos 90o cos 119.867 0.4926 ⎦ = ⎣ cos 90o cos 29. . 13 about axes x.PROPERTIES OF MATRIX OF INERTIA. 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. 113 Problem 32 y O x z Figure 13 Matrix of inertia of a flat object shown in Fig. 114 Solution.79) (4.6 = √2 13 − 169 − 20 = 0.885 (4. x)0 = arccos(t011 ) = 152.78) Introduction of Eq.76 one can get t11 = −5 t12 = −1. 4. the principal axis zp must coincide the axis z.7 r 1 t12 ” = − = −0.73) ¯ ¯ ¯ ¯ 0 0 13 − Ip which can be developed as follow (Ip2 − 3Ip − 10Ip + 30 − 25)(13 − Ip ) = 0 (4.7 t012 Equation 4. The principal moments of inertia can be obtained by solving the following equation.76) (4. 4.78 yields (4.4 = 2 = 13kgm2 13 + (4.5o ∠(xp .7t212 + t212 = 1 Hence r 1 = 0.5o (4.83) ∠(xp . ¯ ¯ ¯ 10 − Ip −5 ¯ 0 ¯ ¯ ¯ −5 ¯=0 0 3 − Ip (4.81) t011 = −0.6)t11 + (−5)t12 = 0 t211 + t212 = 0 (4.46 =+ 4.80) (4. 4.77) From Eq. The directional cosines between the principal axis xp and axes x and y may be obtained from the following set of equations.74) Hence Ipx Ipy Ipz √ 169 − 20 = 12.82) t11 ” = 0.46 4.92t12 2. (10 − 12.885 (4.77 yields 3.78 into Eq.84) Corresponding angles are .PROPERTIES OF MATRIX OF INERTIA.6 (4.75) Since the object is flat. y)0 = arccos(t012 ) = 62. y)” = arccos(t12 ”) = 117. x)” = arccos(t11 ”) = 27.5o ∠(xp . 14 .PROPERTIES OF MATRIX OF INERTIA.119 have to be neglected since those angles determine left handed system of coordinates and this is in disagreement with the adopted assumptions.85) yp 117.5 o x O z 27. 115 ∠(xp .5o y (4.5o xp Figure 14 The solution 4. The only possible position of axes xp and yp with respect to system of coordinates xy is shown in Fig. b .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 . Determine moment of inertia of the plate about this axis of rotation.PROPERTIES OF MATRIX OF INERTIA.length of sides of the plate. α . 15. rotates about axis x − x.angle between side a and the axis of rotation x − x. 116 Problem 33 y b C y a x α x Figure 15 The rectangular plate. m . shown in Fig. Given are: a. Compute the principal moments of inertia of the cam about axes through the point O. uniform and rigid body of mass M. Given are: r = 1[m]. flat.625kgm2 Ipz = 38kgm2 . 117 Problem 34 2r r O Figure 16 The cam shown in Fig.375kgm2 Ipy = 21.PROPERTIES OF MATRIX OF INERTIA. 16 can be considered as a thin. M = 20[kg] Answer: Ipx = 16. IyG = IzG = 1 2 ml 12 . Their masses are m1 and m2 respectively.PROPERTIES OF MATRIX OF INERTIA. Moments of inertia of the uniform rod of length l and mass m. 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. shown in Fig. 17a). IxG = 0. 17b. Determine matrix of inertia of this body about axis xyz. about axis through its centre of gravity G are as follows. 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. Mass of the rod and the particle is m1 and m2 respectively. the golf club shown in Fig. 119 Problem 36 yG l1 G xG 1 2 l2 Figure 18 To find its matrix of inertia. 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 ¯ ¯ . 18 was approximated by the slender rod 1 and the particle 2. Produce 1. expression for the matrix of inertia of the golf club about axes xG yG zG through its centre of gravity G.PROPERTIES OF MATRIX OF INERTIA. 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 . 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 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. Produce: 1. yG = 2(mm11+m 2) 2) 2. Answer: l2 l1 xG = 2(mm12+m . B2. 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. The expression for the coordinates of the centre of gravity G of the rigid body.PROPERTIES OF MATRIX OF INERTIA. 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 . The matrix of directional cosines that locate the principal axes through the point O in relation to the system of coordinates XY Z. Produce 1.PROPERTIES OF MATRIX OF INERTIA. The principal moments of inertia about axes through the point O 3. . 20. 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. The matrix of inertia of the body about axes XY Z 2. ⎤ ⎡ 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.PROPERTIES OF MATRIX OF INERTIA.87) . ⎤ ⎡ 0 0 0 I1O = ⎣ 0 13 0 ⎦ ml2 0 0 13 The matrix of inertia of the rod 2 about axes XY Z.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.b. 123 Solution. Z 3 G3(a.86) (4. ⎤ ⎡ 1 0 0 3 I2O = ⎣ 0 0 0 ⎦ ml2 0 0 13 The matrix of inertia of the rod 3 about axes XY Z. 459ml2 3 Taking into consideration the matrix of inertia IO . IXp = 1. 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. The two last equations if and only if t12 = t13 = 0.666)t13 = 0 3 (4.89) p 3 2 ¯ ¯ 5 1 2 2 ¯ ¯ 0 − 2 ml ml − I p 3 They are 5 2 ml = 1. 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 (4.5) t13 = 0 3 5 (0) t11 − (0.91) m Therefore the axis X coincides with the principal axis Xp and the root associated with the first raw of the determinant 4.92) Indeed.5) t12 + ( − 1.666ml2 (4. one can see that Z Z XY dm = 0 and XZdm = 0 m (4.93) the equations for determination of its elements associated with the above root are 5 ( − 1.666ml2 : 1.666)t11 + (0) t12 + (0) t13 = 0 3 2 (0) t11 + ( − 1. 124 The matrix of inertia of the body about axes XY Z.873ml2 : 0.PROPERTIES OF MATRIX OF INERTIA. Because t211 + t212 + t213 = 1 (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 (4.666)t12 − (0.91 ( 53 ml2 = 1. t12 and t13 .95) .666ml2 ) stands for the moment of inertia about the principal axis Xp .90) (4.94) The first equation is fulfilled for any magnitudes of t11 . 102) These equations are fulfilled for t31 = 0.101) The axis Z is located by the directional cosines linked with third raw of 4.99) They are fulfilled for Hence the angles between axis Yp and XY Z are h(Yp X) = 90o . They are 5 ( − 0.5) t33 = 0 3 t231 + t232 + t233 = 1 (4.5) t23 = 0 3 t221 + t222 + t223 = 1 (4.3823.100) h(Yp X) = 90o .93.103) Hence the angles between axis Zp and XY Z are h(Zp X) = 90o .459)t31 + (0)t32 + (0) t33 = 0 3 2 (0) t31 + ( − 0.105) The analysis offers two acceptable solutions shown in Fig. h(Xp Y ) = 90o h(Xp Z) = 90o (4.97) The axis Y is located by the directional cosines linked with second raw of 4.873ml2 and IZp = 0.9236 (4.52o (4. Hence h(Xp X) = 0.93.92360.98) t21 = 0. h(Yp Y ) = 67. t33 = ±0.873)t22 − (0.45 h(Zp Z) = 112.47o h(Yp Z) = 22.PROPERTIES OF MATRIX OF INERTIA.52o h(Yp Z) = 157. They are 5 ( − 1.54o h(Zp Z) = 67. t32 = ±0. t22 = ±0. 22 .45 (4.3823 (4.104) h(Zp X) = 90o .96) Let us assign the remaining roots as follows IY p = 1. h(Yp Y ) = 112.459ml2 (4. t23 = ∓0.459)t32 − (0. h(Zp Y ) = 22.873)t21 + (0)t22 + (0) t23 = 0 3 2 (0) t21 + ( − 1.54 (4. 125 the magnitude of t11 must be 1. h(Zp Y ) = 157.47o (4. 0 0.6667 0.3823 ⎡ ⎤ 1.9236 ⎦ ⎣ 0 23 − 12 ⎦ 0 0.52o Zp(1) X Y 22.459ml2 (4.0 0.0 0.923 6 0.0 1.8723 0.9236 ⎦ (4.PROPERTIES OF MATRIX OF INERTIA.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 0.108) .873ml2 and IZp = 0. Z into Xp .3823 −0.92360 0.923 6 0 − 12 35 0 −0.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 −0.382 3 0.45919 As one can see this transformation results in the diagonal matrix having elements equal to the principal moments of inertia IXp = 1.382 3 0 0.0 ⎦ = (4. Zp .666ml2 .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.0 ⎣ 0. Y. IY p = 1.107) 0. Yp . 126 Z Yp(2) 67.106) t31 t32 t33 0 0. DEFINITION: It is said that a rigid body performs the rotational motion if one point of the body considered. 4.110) (ri × r˙ i )dm = dHO Hence 1 dT = ω·dHO 2 Upon integrating of Eq. ωy .3.111) (4. ω z ][I] ⎣ ω y ⎦ ωz (4. according to consideration in the previous chapter (4.109) 2 2 where ri is the position vector of the particle. 23) is 1 1 dT = r˙ 2i dm = r˙ i · r˙ i dm (4.112) where ωx .111 over the entire body we are finally getting ⎤ ⎡ ωx T = 12 ω · HO = 12 [ωx . due to its constraints. 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.KINETIC ENERGY.3 127 KINETIC ENERGY. 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. 4. The kinetic energy of a particle of mass dm (see Fig. . 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.1 Rotational motion. 4. is motionless with respect to the inertial space. ω y . G dm) (4.G dm + 2˙rG · r˙ i.G )2 dm = (˙r2G dm + r˙ 2i.G Introducing Eq. Kinetic energy of a particle dm of a rigid body (see Fig. 4.KINETIC ENERGY.G G rG y O x Y X Figure 24 (4.G dm) 2 2 (4. 24) is 1 dT = r˙ 2i dm 2 (4.116) 2 m m m .113) In the general motion its position vector ri may be defined as follows z Z ω dm i ri ri.113 one can obtain that 1 1 dT = (˙rG + r˙ i.G dm + 2˙rG · r˙ i.3.2 General motion. 128 4. 4. DEFINITION: It is said that a rigid body performs the general motion if its motion can not be classified as the rotational one.114) ri = rG +ri.114 into Eq.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 · (ω × ri.G dm m = 2˙rG · (ω× Z ri. ωx .G )dm m ri. The last formula permits to formulate the following statement. 4. 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).G × r˙ i.117 into Eq.G dm = ω · hG Z ω × ri.118) where: [I] .G )dm = Z m r˙ i. . ωz ][I] ⎣ ωy ⎦ ωz ⎡ (4.is inertia matrix about system of coordinates through centre of gravity of the body.G dm = Z m = ω· 2˙rG · Z m r˙ i.G m) = 0 (4.G = 0.116 gives the following formula for kinetic energy.117) m The last relation becomes obvious if we notice that for the chosen system of coordinates rG. 129 But Z m Z dm = m m r˙ 2i.G dm) = 2˙rG · (ω × rG. Implementation of Eq.G × r˙ i.G dm = 2˙rG · Z ω · (ri. 4. ω y . ω z -are components of the angular velocity of the body along that system of coordinates. ⎤ ωx T = 12 r˙ 2G m + 12 ω · hG = 12 r˙ 2G m + 12 [ω x .KINETIC ENERGY. ωy . Mass of the link 2 is equal to m. 25. Moment of inertia of the link 1 about axis Z is I1Z .3 Problems Problem 39 y1 Z.KINETIC ENERGY. . z1 A y1 x1 z1 A l a β 1 2 G x2 Figure 25 Fig. Distance between the centre of gravity G and the point A is equal to l. I2y2 . Its link 1 is free to rotate about the vertical axis Z of the inertial system of coordinates XY Z. y2 . z1 X α Y x1 z2 Z. Derive expression for the kinetic energy of the system as a function of angles α and β. I2z2 respectively. Distance between the point A and axis of rotation Z is a. The link 2 of the system is hinged to the link 1 at the point A as it is shown in Fig. z2 are principal axis of inertia of the link 2 and the principal moments of inertia about these axes are I2x2 . 130 4. 25 shows a mechanical system.3. Axes x2 . Therefore its kinetic energy is 1 T1 = I1Z ω 21 2 (4. T = T1 + T2 (4. ω 2z2 are components of its absolute angular velocity.KINETIC ENERGY. ⎦ (4. ω2z2 ] ⎣ 0 I2y2 0 ⎦ ⎣ ω 2y2.123) . Therefore its kinetic energy is determined by the following equation. z1 X α Y x1 z2 Z. ⎤⎡ ⎤ ⎡ 0 ω 2x2 I2x2 0 1 1 2 T2 = m2 vG + [ω 2x2.119) The link 1 performs the rotational motion about axis Z which is fixed in the inertial space. ω 2y2. 131 Solution y1 Z.122) The absolute angular velocity of the link 2 is ω2 = ω1 + ω 21 = Kα˙ + j1 β˙ (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 ω2x2.120) The link 2 performs the general motion in the inertial space. 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 . ω2y2. Angular velocity of the link 1 is ω 1 = Kα˙ (4. 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) (4.122.124 and 4. 4.125) Hence. 132 Its components along the body system of coordinates x2 .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 £ ¤ ⎡ . rG = rA + rGA = j2 a − k2 l (4. y2 . 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.KINETIC ENERGY.124) The absolute velocity of the centre of gravity of the link 2 may be obtained by integration of its position vector.126) Taking into consideration Eq’s. 4.. Produce expressions for kinetic energy function of the mixing tank 1. b. Its motion is determined by the angular displacement α. combined. Iy1 . Iz1 and its mass is m. 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 . α.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. are considered as rigid body (see Fig. Given are: a. At the same time the housing 3 rotates about the axis X.KINETIC ENERGY. c. 27). m. Iy1 . Ω. 133 Problem 40 x1 Ωt y3 x3 O. Iz1 . Its principal moments of inertia about axes through its centre of gravity are Ix1 . The tank rotates with the constant angular velocity Ω about axis z3 relatively to the housing 3. Ix1 . KINETIC ENERGY. 28b) through centre of gravity G are IxG = T = mR2 . 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). derive expression for the kinetic energy of the cylinder 4. 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. a.3 and 4 forms one rigid body. b. Given are: H(t). the arm of the robot shown in Fig. α(t). m The principal moments of inertia of a cylinder (see Fig. 2 IyG = IzG = Answer: 1 + b2 α˙ 2 ) + 12 α˙ 2 12 m(3R2 + L2 ) 1 m(H˙ 2 1 m(3R2 + L2 ) 12 . Upon assuming that the elements 2. R. L. Produce kinetic energy function for the link 2 as a function of α(t) and β(t). 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. rotates about the vertical axis Z of the inertial system of coordinates XY Z. 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 . The link 2 can be considered as a rigid cylinder of length l. The instantaneous position of this base is determined by the angular displacement α. radius r and mass m attached rigidly to the massless element AB.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. 29a). The system of coordinates x1 y1 z1 and x2 y2 z2 are rigidly attached to the links 1 and 2 respectively.KINETIC ENERGY. The principal moments of inertia of the cylinder shown in Fig. 30b) are 1 1 IxG = IyG = mr2 + ml2 4 12 1 IzG = mr2 2 .KINETIC ENERGY. The principal moments of inertia for the sphere shown in Fig. These elements are join together by means of massless elements as it is shown in Fig. Produce expression for the kinetic energy of the rotor of the exciter. 30a). 30b) are 2 IxG = IyG = IzG = mr2 5 The principal moments of inertia for the cylinder shown in Fig. The oscillatory motion of the housing 1 is determined by the angular displacement α. 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. This rotor rotates with respect to its housing 1 with the constant angular velocity ω 2. 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 .1 r r Z Figure 30 The exciter shown in Fig.1 . z3 is rigidly attached to the barrel 3 and coincides with its principal axes. 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 . Produce expression for the kinetic energy of the barrel when the barrel passes the position defined by angles αt and αb . Mass of the barrel is m and its centre of gravity is by l apart from o1 . Iy3 and Iz3 respectively. The system of coordinates x3 . 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. The tank has the constant forward linear velocity v. 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 ) . Moments of inertia about these axes are Ix3 .KINETIC ENERGY. y3 . The axes of the system of coordinates x2 y2 z2 coincide with principal axes of the boom. The distance l determines the position of the centre of gravity. At the same time its boom 2 of mass m is being lowered. Produce the expression for the kinetic energy of the boom. 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. Its origin coincides with the boom’s centre of gravity G.KINETIC ENERGY. 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 . The system of coordinates x1 y1 z1 is attached to the base. Iy . Iz respectively. 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. 32 rotates with the constant angular velocity ω about the vertical axis Z of the inertial system of coordinates XY Z. The principal moments of inertia about these axes are Ix . 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. 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 . 33) is free to rotate about the horizontal axis Y of the inertial system of coordinates XY Z. Its angular position is determined by the angle α.KINETIC ENERGY. ω21 ] ⎣ 0 − 12 mal 13 ml2 ω 21 . The system of coordinates x1 y1 z1 is rigidly attached to the link 1. 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. The link 2 is hinged to the link 1 at the point O and it is supported by the slide 0. It is parallel to the axis Y and its position is determined by the dimensions a and b. 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. The system of coordinates x 1 y 1 z 1 is rigidly attached to the link 1. 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. Produce: 1. I y2 . The link 1 rotates about the vertical axis Z and its instantaneous position is given by the angular displacement α. Answer: ω2 = i2 (−α˙ sin β) + j2 β˙ + k2 (α˙ cos β) 2. The distance l identifies the position of the center of gravity G of the link 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. I z2 and its mass is m. The slide 0 is motionless with respect to the inertial system of coordinates XYZ . . The angular displacement β defines the relative position of the link 2 with respect to the link 1.KINETIC ENERGY. The principal moments of inertia of the link 2 are I x2 . The system of coordinates x 2 y 2 z 2 is rigidly attached to the link 2 and coincides with its principal axis. The expression for the angular displacement β as a function of α. 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: ¡ ¢ β = arctan ab cos α . 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. Answer: ˙ v2 = i2 (0) + j2 lα˙ cos β + k2 (−lβ) 4. 141 Answer: ε2 = i2 (−¨ α sin β − α˙ β˙ cos β) + j2 β¨ + k2 (¨ α cos β − α˙ β˙ sin β) 3.KINETIC ENERGY. Its centre of gravity G is located by the distance c. . The system of coordinates x 1 y 1 z 1 is rigidly attached to the link 1. The expression for the angular displacement β as a function of α. Produce: 1. Its instantaneous position is given by the absolute angular displacement α. 5. 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 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 . The link 2 is hinged to the link 1 at the point A. The system of coordinates x 2 y 2 z 2 is attached to the body 2 and coincides with its principal axes. 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. 35 shows the physical model of a mechanical system. The kinetic energy of the link 2 as a function of α and β. The other end of this link P always stays in contact with the cylindrical surface 3 of radius R (b>R). 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.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. The link 1 of this system rotates about the vertical axes Z of the inertial frame XYZ. 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 . I y2 and I z2 . 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. 143 Solution 1.KINETIC ENERGY. 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 2.127) . 36) is rP = j1 a + j2 b (4. 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. 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.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 (4.129) 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 . 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. −→ −−→ −−→ CO + OD = CD (4.134) or (a + b cos β)2 + 2a cos α (a + b cos β) − (R2 − a2 ) = 0 (4.130) or Ja+j1 (a + b cos β) = I (−R sin γ) + JR cos γ (4. 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.KINETIC ENERGY.135) Introducing the following substitution z = (a + b cos β) (4.132) or − sin α (a + b cos β) = −R sin γ a + cos α (a + b cos β) = R cos γ (4.136) z 2 + (2a cos α) z − (R2 − a2 ) = 0 (4.138) the above equation is Hence Taking advantage of the substitution 4. 37.139) b a .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 (4.137) µ ¶ q 1 2 2 2 (−2a cos α) ± (2a cos α) + 4(R − a ) z= 2 (4. To develop this relationship let us consider the vector equation shown in Fig.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. 144) .140) X ri.142) If components of velocity vG along the body system of coordinates are known vG = ivGx + jvGy + kvGz (4. its motion is governed by the same equations which govern any system of particles.EQUATIONS OF MOTION 4. The body system of coordinates has its origin at the body centre of gravity G and its axes coincides principal axes of inertia. To derive the formula for the linear momentum P let us introduce a body system of coordinates xyz.141) h˙ G = MG = Under influence of a set of external forces Fi (see Fig.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. 38) the considered body moves Fi Z ω z dm i ri ri. according to considerations in the previous section. is always possible. X ˙ = F= P Fi (4.143) the first derivative of the linear momentum is 0 ˙ = m(vG P + ω × vG ) (4.G × Fi (4. The linear momentum of the body considered is P = mvG (4.4 146 EQUATIONS OF MOTION 4.1 Euler’s equations of motion General case Each rigid body can be considered as a rigidly tied together system of particles.4. Hence. Such selection of the body system of coordinates. 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. 4. 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. we obtained six independent equations which allow to obtain either kinematic parameters (ω and vG ) in case all the external forces are known.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 ¯⎞ ¯ ¯ ¯⎠ ¯ ¯ (4.145 into Eq. 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 (4.145) ω = iωx + jωy + kω z Introduction of Eq. in the case considered may be adopted in the following form. 4. 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 147 where 0 vG = iv˙ Gx + jv˙ Gy + kv˙ Gz and (4.141 may be transformed to form 4.147) Components of moment of the relative momentum. according to Eq.149) Now. one can formulate three additional scalar equations. the equation 4.150) Eventually. .149.15.148) 0 0 IGz hGz ωz IGz ω z Taking into account the above relationship. ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ 0 ωx hGx IGx 0 IGx ω x ⎣ hGy ⎦ = ⎣ 0 IGy 0 ⎦ ⎣ ωy ⎦ = ⎣ IGy ω y ⎦ (4. X ˙ O = MO = H ri ×Fi (4.EQUATIONS OF MOTION 148 Rotational motion In a case. The wanted equations may be obtained from equation 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.151 may be rewritten as follows ˙ O = H0O + ω × HO = MO H (4. 39) it can be considered as system with three degree of freedom. 4. when the rigid body performs the rotational motion about the point O which is fixed in the inertial space (see Fig.154) The three unknown components of the reaction R in the pivot O are determined by Eq.155) .153) In a similar way to that shown in the previous paragraph. Hence only three equations are necessary to describe its motion. 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 (4.18.147 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. the equation 4. 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. one may transform the above equation into three equivalent equations known as Euler’s equations. 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.156) The external force F comprises both. As an example of possible application let us consider rotor of a turbo-compressor of a jet shown in Fig. 41. the known driving forces Fd and the unknown reaction R.157) 4.4. 2. F = R + Fd (4. 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. This system of coordinates possesses its own angular . The body has an axis of symmetry (see Fig.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.EQUATIONS OF MOTION 149 where vG = ω × rG (4. 40). 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. G ×Fi h˙ G = MG = (4.161) hGzb hGz .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. are ⎡ ⎤ ⎡ ⎤ hGxb ω Ax b ¤ £ ⎣ hGyb ⎦ = IG ⎣ ωAyb ⎦ (4. ωAxb . 42b). Hence. all the above requirements are fulfilled. ω y Fi i dm yb β y ri.159) Components of the angular momentum along the body system of coordinates xb . Taking into consideration that ⎡ ⎤ ⎤ ⎡ hGxb hGx ⎣ hGyb ⎦ = [Cxyz→xb yb zb ] ⎣ hGy ⎦ (4.zb is matrix of inertia of the body about the body system of coordinates.zb hGzb ωAzb where [IG ]xb. ω Ayb . 42) is govern by equations X ˙ = F= P Fi X ri. equations of motion of a rigid body may be formulated in terms of kinematic parameters of the system xyz (vG . 4. ωAzb are components of the absolute angular velocity of the body along the body system of coordinates. ω) and the relative angular velocity of the body Ω. zb (see Fig.EQUATIONS OF MOTION 150 speed ω. yb .160) xb.158) (4. yb . yb .15(see page 98). The rotor of the turbo-compressor itself rotates with angular velocity Ω about axis of symmetry z. according to Eq. In such cases. 163 to be simplified.167) ω A = ω + Ω = iω x + jω y + k(ω z + Ω) (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.zb hGz ω Az £ ¤ Generally.169) . 42 a that the absolute angular velocity of the body ω A is Introduction of Eq. yb . 4.z [Cxyz→xb yb zb ] b.168) It is easy to see from Fig.168 ⎤ ⎡ ⎡ hGx IG ⎣ hGy ⎦ = ⎣ 0 0 hGz into Eq. ⎤ ⎡ ⎤⎡ ⎤ ⎡ 0 ω Ax hGx IG 0 ⎣ hGy ⎦ = ⎣ 0 IG 0 ⎦ ⎣ ωAy ⎦ 0 0 IGz hGz ω Az (4.160 can be rewritten as follows ⎤ ⎤ ⎡ ⎡ ω Ax hGx £ ¤ ⎣ hGy ⎦ = ([Cxyz→xb yb zb ]T IG [Cxyz→xb yb zb ]) ⎣ ωAy ⎦ (4.166) 0 0 IGz The Eq.166 allows the relationship 4. the product [Cxyz→xb yb zb ]T IG x y .164) 0 0 1 and because of symmetry of the body £ Hence IG ¤ xb. b. in the case considered ⎡ ⎤ cos β(t) sin β(t) 0 [Cxyz→xb yb zb ] = ⎣ − sin β(t) cos β(t) 0 ⎦ (4. 4.163) xb.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. b b But.zb ⎤ 0 IG 0 = ⎣ 0 IG 0 ⎦ 0 0 IGz ⎡ (4.165) £ ¤ [Cxyz→xb yb zb ]T IG x y . 4.z [Cxyz→xb yb zb ] is a function of time. 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. yb . 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. treated in the same way as in the previous section. z is always chosen at a point of rotation O.171) The equation 4.158. z Ω Z R G rG ω Fi y ri X O x Figure 43 Y .172) Rotational motion. y. In case of rotational motion (see Fig. 43) the origin of the rotating system of coordinates x.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. 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. 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) 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 (4.184 and 4.57 and 4. In the case considered it is equal to zero. 4.183 vGx3 = 0 ω x3 = −ω vGy3 = −aω ω y3 = 0 vGz3 = 0 ω z3 = 0 (4.180 and 4. 4.187 into Eq’s.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.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. .186) (4. 4.188) The first five equations allow for determination of the unknown reaction at A and B.189) The last equation offers the driving moment of the tank. According to Eq’s.187) Introducing Eq’s. 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. The uniform rod AB of mass m and length l is hinged at A to the link CA. The link CA rotates with a constant angular speed ω about its vertical axis. 48. . 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. 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.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 (4.193) 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.EQUATIONS OF MOTION 161 Solution.196) . 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 (4.195) Since the link 2 performs rotational motion about the point A. 1y2 =0 R 2.1y2 1 M2.198) (4.1x2 1 My2 = mgl cos α 2 Mz2 = M21z2 (4.200 allows the interaction moment M21z2 to be obtained.199 presents the wanted equation of motion.1z2 A R 2.1x2 l/2 α 2 A R 2. . 1 Iα ¨ + Iω2 sin α cos α − mgl cos α = 0 2 (4.1x2 1 Iα ¨ + Iω 2 sin α cos α = mgl cos α 2 −Iω α˙ sin α − I αω ˙ sin α = M21z2 (4. 50.199) (4.1y2 z2 R 2.1z2 x1 M2.1x2 M2. 4. Ix = 0 Iy = Iz = I = ml2 3 Mx2 = M2. 4.1x2 M2.197 into Eq. the equation 4.197) Upon introducing Eq’s.1x2 G x2 mg x2 Figure 50 were according to Fig.200) Equation 4.EQUATIONS OF MOTION 162 z1 Z ω y2 R 2.201) After solving it with respect to the unknown function α.195 and 4.196 one may obtain 0 = M2. Upon assuming that the platform can rotate about a vertical axis only. The gyroscope 3 is symmetrical with respect to its axis of relative rotation z2 and its moments of inertia are Ix2 = Iy2 = I. The gyroscope rotates. The damper c produces a moment which can be approximated by a linear function Mc = −cϑ˙ (4.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. of a stiffness k. and Iz2 .202) The ring 2 may be considered as massless. The shown spring. keeps the ring 2 in the horizontal position (ϑ = 0) if the platform does not rotate (α˙ = 0). 51. . relative to the ring 2. with the constant angular velocity Ω. derive the relationship between its constant angular velocity α˙ = p and angular position of the gyroscope ϑ. 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) (4.208) Because for a constant angular velocity α˙ = α˙ o the angle ϑo is supposed to be constant. ω1 = j1 α˙ = j2 α˙ cos ϑ − k2 α˙ sin ϑ (4.207) into the first equation of the set 4. The system of coordinates x1 y1 z1 .205) and (4.which is attached to the platform. 4. the modified Euler’s equations may be used to derive its equations of motion.210) . the above equation yields (Iz2 − I)(−α˙ 2o sin ϑo cos ϑo ) + Iz2 Ωα˙ o cos ϑo = −kϑo (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. rotates with respect to the inertial frame XY Z with angular velocity ω 1 .EQUATIONS OF MOTION 164 Solution.206 produces equation of motion of the gyroscope.206) Since the ring 2 is massless its equilibrium conditions yields Mx2 = −kϑ − cϑ˙ (4. 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) (4.209) If angular velocity ϑo << Ω. I ϑ˙ + (Iz2 − I)(−α˙ 2 sin ϑ cos ϑ) + Iz2 Ωα˙ cos ϑ = −kϑ − cϑ˙ (4.205) Since the gyroscope 3 rotates about its axis of symmetry z2 with relative velocity Ω. the first term of Eq.207) Introduction of Eq’s. 4.209 can be omitted and the required relationship has the following explicit form α˙ o = − kϑo Iz2 Ω cos ϑo (4.204) Therefore. EQUATIONS OF MOTION 165 Problem 53 A 1 2 β Z l Figure 52 Link 1 of the mechanical system shown in Fig. Link 2. is hinged to the link 1 at point A. Its motion is determined by angle α which is a known function of time. Given are: α(t)− equation of motion of the link 1 m− mass of the link 2 l− length of the link 2 . 52 performs rotational motion about the horizontal and motionless axis Z. which can be considered as a uniform and rigid rod of mass m and length l. Derive the equation of motion of the link 2. Since the link 2 performs rotational motion about the point A. 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 .214) . Its instantaneous position is determined by the angle α. ω2x2 = i2 · ω2 = β˙ ω 2y2 = j2 · ω 2 = α˙ sin β ω2z2 = k2 · ω2 = α˙ cos β (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 . the following Euler equations of motion can be utilized. 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. 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.212) Hence. 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.EQUATIONS OF MOTION 166 Solution.213) Moment M2 is due to the interaction force R21 and interaction moment M21 (see Fig. 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. M2 = M21 +M2G (4. Angle β determines its instantaneous position with respect to the system of coordinates x1 y1 z1 . 54) in the constrain A and the gravity force G. 218) . 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 (4.1y2 M 2. 4.214 into first equation of the set 4.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 = = = = (4.1y2 R2.211 one can obtain equation of motion in the following form (4.EQUATIONS OF MOTION 167 M 2.1x2 =0 z2 A M 2.213 and 4. component of the interaction moment along axis x is M21x2 = 0 (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.1x2 x2 M 2.1y2 R2.1z2 R2.216) Upon introducing Eq’s.215) Moment produced by the gravity force G is a vector product of position vector rG and the gravity force G.1y2 y2 y2 y1 Figure 54 Since the link 1 is free to rotate about axis x2 .1z2 ω1 Z z1 A x1 x2 rG l 1 β G 2 R2. 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) . The wheel rolls without slipping on the horizontal plane.5kgm2 . Iy1 = Ix1 =10kgm2 – principal moments of inertia of the wheel . Determine the reaction between the wheel and the horizontal plane.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.5m R =2m m =100kg ω =1rad/s Iz1 =12. Given are: r =0. 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 .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 .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. 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 . Rotor of the turbine has mass m and its centre of gravity G is at a distance a from the axis X. 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 .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. m. b. Ω. . 59) a turbine is installed. a. I2z1 . ω. Axis z1 is axis of symmetry of the rotor and its principal moments of inertia I2x1 = I2y1 = I2 and I2z1 are known. 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. produce the expression for the dynamic reactions in these constraints as a function of the bar angular position β. 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 . Answer: Solution of the folowing set of equations yields the wanted reactions. Upon assuming that there is no friction in the constraints A and B and all the motion is in the vertical plane Y Z. Its end A is moving with constant velocity v along the horizontal axis Y of the inertial system of coordinates XY Z. 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. .EQUATIONS OF MOTION 175 Problem 57 B X. Iz .angle between axis z2 and the horizontal plane Hint: The body 3 performs rotational motion about the point O.mass of the body 3 I.x1 O Y T y1 y2 X.angular velocity of the link 1 α . 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.distance between the point of rotation O and centre of gravity G of the body 3 r . Given are: m .radius of the ring 4 ω .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.moments of inertia of the body 3 l . 61) is driven by means of the massless links 1 and 2 that are hinged to each other at the point O. Derive the expression for the components of the reaction N and T between the body 3 and the ring 4. 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. Upon assuming that the mass of the housing is negligible. . 62 shows the physical model of a shaker frequently used to excite the rolling motion of a ship. derive an expression for the components of the moment transmitted to the board 1 of the ship.EQUATIONS OF MOTION 176 Problem 58 z2 Z Z α Ω 3 Y G3 y2 2 1 X x2 Figure 62 Fig. The housing 2 of the gyroscope 3 performs oscillatory motion about axis Y of the inertial system of coordinates XY Z. rotates with the constant angular velocity Ω. The gyroscope 3 of mass m and moments of inertia I3x2 = I3y2 = I3 and I3z about axes x2 y2 z2 respectively. 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. The relative angular velocity of the gyroscope 3 with respect to . the car 1 was forced to move along the coordinate α (see Fig. To test this gyrostabilizer. 63a) shows a gyrostabilizer for stabilization of the monorail car 1. 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. Given are: A.angular velocity of the gyroscope 3 with respect to the housing 2 a.principal moments of inertia of the gyroscope 3 about axes through its centre of gravity. b. dimensions shown in Fig. c.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. XY Z is the inertial system of coordinates. Produce expressions for interaction forces between the gyroscope 3 and its housing 2. x2 y2 z2 is the body 2 system of coordinates. Iz . x1 y1 z1 is the body 1 system of coordinates.63a) m.EQUATIONS OF MOTION 178 the housing 2 was Ω. f . 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 .mass of the gyroscope 3 Ix = Iy = I. The instantaneous position of the arm 1 is determined by the angular displacement α.dimensions shown m . l.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.moments of inertia of the wheel 2 about axes perpendicular to the axis of relative rotation α(t) . Given are: a.the angular displacement of the arm 1 with respect to the plane.mass of the wheel 2 Iz . 64 is being retracted while the aircraft 0 moves with constant velocity along a horizontal straight line. Ω . b.moment of inertia of the wheel 2 about its axis of relative rotation I = Ix = Iy . The wheel 2 rotates with the constant relative angular velocity Ω with respect to the arm 1.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 . . . The relative angular velocity of the rotor 2 with respect to the base 1 is constant and its magnitude is ω 21 . 65 presents the physical model of a ventilator. z2 Z X 2 2 Y y1 y2 x2 1 ω 21 t O o1 x1 Figure 65 Fig. The system of coordinates x1 y1 z1 is attached to this base. Its base 1 rotates with the constant angular velocity ω about the vertical axis Y of the inertial system of coordinates XY Z. Answer: Rx1 = −maω 2 Ry1 = mg Rz1 = −bmω 2 Mx1 = Iz2 ωω21 My1 = 0 Mz1 = Md = 0 . 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 .EQUATIONS OF MOTION 180 Problem 61 1 a O ωt o1 b x1 z1 . The system of coordinates x2 y2 z2 is attached to the rotor 2. Iz2 respectively. The rotor 2 is free to rotate about the axis z1 . 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. were joined together to form the link 2 of this system. Two uniform bars. each of length a and mass m. The differential equation of motion of the link 2 in terms of the variable α. 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. 2.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. The free body diagram for the link 2. The system of coordinates x 1 y 1 z 1 is rigidly attached to this link. 3. The equations for the determination of the interaction forces between the link 2 and 1 in the kinematic constraints A and B.66 rotates about the vertical axis Z of the inertial system of coordinates XYZ with a constant angular velocity ω. The moment of inertia of a uniform bar of mass m and length l about the axis through its centre of gravity is . The system of coordinates x 2 y 2 z 2 is rigidly attached to the link 2. Physical model. Its centre of gravity is located on axis z at the point G. Let Ω be the angular velocity of the body with respect to xyz.5. 4. Let us consider motion of a rigid body about a fixed in an inertial space point O (Fig. Φ Θ G Φ y1 Φ O X y .5 182 MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. we shell assume that axis z is axis of symmetry of the body. 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.220) .MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. The angle Φ. For further consideration. 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. Since the body is free to rotate about axis z it has three degree of freedom.1 Modelling. The body has mass m. 67). 1 ml2 12 IG = 4. Its position is given by a distance r. IOx = IOy = IO and IOz represent principal moment of inertia of the body about axis xyz. z Z z1 Θ Ω r G . Kinematic analysis Angular velocity of the system of coordinates xyz is ˙ ˙ +Θ ˙ = KΦ˙ + iΘ ω=Φ (4. 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. General solutions of the mathematical model 4. 4.2 Analysis.222) Components of moment MO = kr × K(−mg) (4.224 and 4.222 one can obtain equations of motion in the following form.MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. . very often.225 can not be obtain by means of any analytical methods.225) 4. But.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. The introduced assumptions allow to take advantage from the modified Euler’s equations. ¨ + (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. ˙ 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. 4.221) Equations of motion. we can procure a number of particular solutions by guessing their form.221 into Eq. 183 Scalar multiplication of the above equation by unit vectors i. and k respectively yields its components along the rotating system of coordinates xyz.224) = −(k · (k × K))rmg = −rmg ¯¯ 0 ¯ 0 sin Θ cos Θ ¯ Upon introducing Eq. j.5. Let us adopt. Ωo are constant values.230) p1. if the equation 4. for further analysis.226) where p. 4.015m r { Figure 68 .015m m=0.226 may be considered as an solution. The relationship 4. 68 R b G R=0. Let us predict the particular solution of the set of equations 4. Ωo or for any set of parameters p.228 is fulfilled. Hence Φ˙ = p ¨ =0 Φ ˙ =0 Θ ¨ =0 Θ Ω˙ = 0 (4. Θo .226 into Eq.225 in the following form Φ = pt Θ = Θo Ω = Ωo (4.04m O b=0. which corresponds to sign + is called slow precession to distinguish it from fast precession p2 corresponding to sign −. θo .2 = 2(IOz − IO ) cos Θo The precession p1 .6kg 0. Ωo satisfying the following relationship p −IOz Ωo ± (IOz Ωo )2 + 4(IOz − IO )rmg cos Θo (4.227) Upon introducing Eq.225 one can see that second and third equation is fulfilled for any instant of time.025m r= 0.228) The predicted solution 4. It is fulfilled for Θo = 0 Θo = π (4.230 permits to compute so called static characteristic of a gyroscope.MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. The first equation yields (IOz − IO )p2 sin Θo cos Θo + IOz Ωo p sin Θo − rmg sin Θo = 0 (4. 184 Particular solutions.229) regardless of magnitude p. dimensions of the gyroscope shown in Fig. 4. IO = 0. Those solution were obtained for initial ˙ i = 8s−1 .170s−1 . Θ −1 Φi = 0.51rad. 10. 73B corresponds to situation when the gyroscope. For all cases the initial speed of the gyroscope was assumed to be Ωi = 15s−1 . Due to the following initial out of its motion at t = 0 with initial velocity Θ ˙ i = 8s−1 . Ωi = 15s−1 respectively. position (Θo = π) caused by the following initial conditions Θi = 2.51. Θ Fig.226. 185 For r = 0.For high angular velocity Ωo = 25s−1 (see Fig. Fig. Θ −1 ˙ i = 8s .000388kgm2 < IOz = 0. Θ performs small oscillation around regular precession. Θi = 2. equilibrium position Θo = 0(Θi = 0. Fig 73A presents solution for initial conditions chosen in vicinity of upper ˙ i = 0. performing regular ˙ o = 0. Ωi = 15s . Fig. Φi = 0. Φi = 0. Φ˙ o = p1 = 25s−1 . Figures 73D and 73E show solution of disturbed motion of the gyroscope about ˙ slow (Φo = 16.51rad (see Fig. Φi = 0. Ωi = 15s−1 the gyroscope conditions Θi = 1. If angular velocity Ωo is slower e. 73F presents motion of the gyroscope in vicinity of the lower equilibrium ˙ i = 20s−1 .025m. Φ˙ i = −77.17s−1 ) and fast (Φ˙ o = −77.01s−1 . 70) the gyroscope can perform the fast and slow regular precession p for any mutation angle Θo . Φi = 0.26rad.1s−1 ) precession which correspond to angle of mutation Θo = 2. Φ˙ o = p2 = 136. 20 and 25s−1 ) is shown in Fig.000482kgm2 .000482kgm2 the static characteristic of the gyroscope for a few values of angular velocity (Ωo = 0. 70).26rad. In this case IO = 0. 70) was pushed precession (Θo = 1. 71 present static characteristics for r = 0. 73C. Ωi = 15s−1 and conditions Θi = 2. Φ˙ i = 0. 70).51rad.MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM. Φ˙ i = 0. More particular solution can be obtained by numerical integration of the mathematical model 4.26. 15. This situation is (Θo = 1.∂ Numerical solution of equations of motion. the gyroscope can perform regular precession only for certain range of angle of mutation Θo . Φ˙ i = 0. 69.225. Θ presented in Fig.015m.025m. 5. 73.000629kgm2 > IOz = 0. Ωo = 15s−1 . Ωo = 15 see Fig. it execute large oscillations. Fig. The diagrams in this figure show instantaneous positions yG of the point G of the gyroscope versus its angle of precession Φ. Ωo = 15s−1 see Fig. 70).1s−1 . Θ ˙ i = 8s−1 . Ωi = 15s−1 ). presents a few numerical solutions carried out for r = . Φ˙ i = 16. Θ Since the gyroscope can not perform regular precession (see Fig. . 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.g. Similar behavior can be observed if it is pushed out of its regular fast precession ˙ o = 0. 186 Slow precession p1 versus angle of nutation Θ ο r=0.14 Ωο = 25 [rad] Fast precession p versus angle of nutation Θ ο 2 r=0.14 Θ o [rad] Ωο = 0 Ωο = 5 Ωο = 10 Ωο = 15 Figure 69 Ωο = 20 Ωο = 25 [rad] .88 Ωο = 15 2.51 3.025 [m] 60 50 40 30 p 1 [rad/s] 20 10 0 0 0.26 Ωο = 10 Θ o [rad] 1.88 2.63 1.MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.51 Ωο = 20 3.63 Ωο = 0 Ωο = 5 1.26 1.025 [m] 400 300 200 p 100 2 [rad/s] 0 -100 -200 -300 -400 0 0. 26 1.14 [rad] Θο r=0.88 Θ slow precession o 2.MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.14 [rad] fast precession p 2 .63 1.26 slow precession Θ 1.025 [m] 200 150 100 50 p [rad/s] 0 -50 -100 -150 -200 0 0. Ω ο = 25 [rad/s] Slow & fast precession for 187 Θο versus angle of nutation r=0.63 1.025 [m] 200 150 100 50 p [rad/s] 0 -50 -100 -150 -200 0 0.51 3.88 o p 1 Figure 70 2.51 p fast precession 1 p 2 Ω = 15 [rad/s] versus angle of nutation ο Slow & fast precession for 3. MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.51 Ωο = 20 3.14 Ωο = 25 [rad] Fast precession p versus angle of nutation Θ ο 2 r = 0.015 [m] 60 50 40 p 30 1 [rad/s] 20 10 0 0 0.51 3.26 Ωο = 10 Θ o [rad] 1.26 1.63 1.14 Θ o [rad] Ωο = 0 Ωο = 5 Ωο = 10 Ωο = 15 Figure 71 Ωο = 20 Ωο = 25 [rad] .63 Ωο = 0 Ωο = 5 1.015 [m] 400 300 200 100 p 2 [rad/s] 0 -100 -200 -300 -400 0 0. Slow precession p1 versus angle of nutation 188 Θο r = 0.88 Ωο = 15 2.88 2. 26 1.14 [rad] fast precession p 2 .51 o p fast precession 1 p 2 Ω = 10 [rad/s] versus angle of nutation ο Slow & fast precession for 3.88 Θ slow precession 2. Ω ο = 20 [rad/s] Slow & fast precession for versus angle of nutation 189 Θο r = 0.63 1.63 1.51 3.26 Θ slow precession 1.88 o p 1 Figure 72 2.015 [m] 200 150 100 50 p [rad/s] 0 -50 -100 -150 -200 0 0.14 [rad] Θο r = 0.MOTION OF THE GYROSCOPE WITH THREE DEGREE OF FREEDOM.015 [m] 200 150 100 50 p [rad/s] 0 -50 -100 -150 -200 0 0. 04 190 .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. (5. Vz .4) DOT OR SCALAR PRODUCT of two vectors P and Q is defined as scalar . z V k i j o Vz y Vx x Vy The vector quantities are printed in boldface type V or.2) DIRECTION COSINES l. Thus l = cos ]Vi = Vx V n = cos ]Vj = Vy V m = cos ]Vk = Vz V (5. n are the cosines of angles between a vector V and axes xyz.3) Useful relations l2 + n2 + m2 = 1 (5.Chapter 5 APPENDIXES 5. m. should always be indicated by symbol V to distinguish them from the scalar quantities V. SCALAR MAGNITUDE of a vector V is V = q Vx2 + Vy2 + Vz2 (5.. in handwriting.1 APPENDIX 1. REVISION OF THE VECTOR CALCULUS NOTATION.1) V = iVx + jVy + kVz were i. Vector quantities are usually defined in the right-handed system of coordinates by its scalar components Vx . k are unit vectors of the system of coordinates xyz. j. Vy . APPENDIX 1.14) (P × Q) · R (5.8) (5.6) i·i=j·j=k·k=1 (5.11) Q × P = −P × Q (5. REVISION OF THE VECTOR CALCULUS 192 magnitude P · Q =P Q cos α (5.16) (5.5) where α stands for the angle between the two vectors.15) TRIPLE CROSS-SCALAR PRODUCT is defined as . Px Q Q α P Useful relations P× (Q + R) = P × Q + P × R (5.9) (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 (5.13) i×i=j×j=k×k=0 ¯ ¯ ¯ i j k ¯¯ ¯ P × Q = ¯¯ Px Py Pz ¯¯ ¯ Qx Qy Qz ¯ (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. P α Q Useful relations P·Q= £ Px P·Q=Q·P (5.12) i×j=k j×k=i k×i=j (5. 19) P× (Q × R) = Q· (R · P) − R· (P · Q) (5.20) µ ¶ dP ∆P ˙ = lim =P ∆t→0 dt ∆t (5.22) (5.24) .APPENDIX 1.17) P× (Q × R) (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 (5.23) (5. 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 ¯ (5.21) (5. APPENDIX 2. CENTRE OF GRAVITY. 194 5. VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES.2 APPENDIX 2. 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. CENTRE OF GRAVITY. VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES.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 . 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 . VOLUME AND MOMENTS OF INERTIA OF RIGID BODIES.APPENDIX 2. CENTRE OF GRAVITY.
Report "Mechanics of Rigid Body by Janusz Krodkiewski Melbourne Text Book"