DYNAMICS OF MACHINES

Transcription

1 DYNAMICS OF MACHINES Modelling Simulation Visualization Verification Ilmar Ferreira Santos, Professor, Dr.-Ing., Dr.Techn., Livre-Docente Department of Mechanical Engineering Technical University of Denmark Building 44, Room 4 28 Lyngby Denmark Phone: ifs@mek.dtu.dk February 215

2 Contents 1 Introduction to Multibody Dynamics Introduction From a Physical System and to a Mechanical Model Mass and Inertia Element Interaction Forces represented by Stiffness Element Interaction Forces represented by Damping Element Steps of the Mathematical Modelling Kinematics of Particles (Lumped Masses) Inertial Reference System Moving Reference System Illustrating Some Fundamentals Through Examples Differentiating the coordinate transformation matrix Differentiating vectors in terms of direction Differentiating vectors in an inertial reference frame and in a moving reference frame Two ways for calculating the absolute acceleration Dynamics of One Single Particle in 3D Examples One Equation of Motion and Two Dynamic Reaction Forces Three Dynamic Reaction Forces and No Equation of Motion Two Equations of Motion and One Dynamic Reaction Force Dynamics of a System of Particles in 2D Example Solution Using One Single Moving Reference Frame Solution Using Two Moving Reference Frames Computational Implementation and Numerical Simulation Dynamics of One Single Rigid Body in 3D An Example Solution Steps Equations of Motion and Dynamic Reactions Computational Implementation and Numerical Simulation Getting equations of motions using the software MATHEMATICA

3 Chapter 1 Introduction to Multibody Dynamics 1.1 Introduction In many practical situations, for example when designing machines elements which focus on reliability, information about forces and moments are of crucial importance. Such an information allows engineers to calculate strain and stresses, select material, optimize dimension and shape of machine components. Machine elements can be considered and treated as a system of particles, of rigid bodies or of flexible bodies, depending on the magnitude and frequency of the applied loads. These elements or bodies are normally attached to each other, and the interaction (forces and moments) among these bodies can be represented by spring and damping elements. In static cases, reaction forces and moments can be directly determined using information related to external loading and geometric properties of the bodies. Such cases are extensively treated in many STATICS books. In dynamic cases however, reaction forces and moments can not be directly determined using only loading and geometric properties of the bodies. In the dynamic cases, the reaction forces and moments are also depending on the movements described by the machine elements. Thus, in dynamic cases, reaction forces and moments have to be determined using information related to the external loading, the geometric properties of the bodies and the associated motions described by the bodies. It means that before the student starts with the DY- NAMICS itself, part of the Mechanics responsible for describing the causes of the movements, it is necessary to describe the geometry of the movements, and count with the KINEMATICS, part of the Mechanics responsible for describing the movement of the bodies from the viewpoint of the geometry. For achieving reaction forces and moments simultaneously with equation of motions one will use the axioms and principles postulated by NEWTON and EULER. In many other practical situations, for example stability analysis of satellites, comfort analysis in vehicles, vibration analysis in machines and structures, design of control system for mechatronic systems, only equations of motion are required to conduct such studies. In these dynamic cases, it is not necessary to calculate the reaction forces and moments among the bodies, and the equation of motion can be directly achieved by using Jourdain s principle of virtual power. 3

4 1.2 From a Physical System and to a Mechanical Model 4 Another possibility is also to use the energy method and Lagrange s equation. Both methods will be introduced in this section. 1.2 From a Physical System and to a Mechanical Model Mass and Inertia Element particle rigid body flexible body Interaction Forces represented by Stiffness Element Springs and Flexible Couplings Elasticity Theory Journal Bearings Fluid Dynamics Air Bearings Fluid Dynamics Seals Fluid Dynamics Magnetic Bearings Electromagnetism EXPERIMENTAL linear and nonlinear coefficients Interaction Forces represented by Damping Element Journal Bearings Fluid Dynamics Air Bearings Fluid Dynamics Seals Fluid Dynamics Magnetic Bearings Electromagnetism EXPERIMENTAL linear and nonlinear coefficients 1.3 Steps of the Mathematical Modelling A description of the steps for achieving the reaction and equation of motions are presented. With this one intents to offer a helpful guidance to students on the way to correctly use the Newton-Euler, Newton-Euler-Jourdain and Lagrange methods. The steps required to obtain equations of motion and reaction forces and moments of a mechanical system composed of N rigid bodies are summarized below: KINEMATICS

5 1.3 Steps of the Mathematical Modelling 5 1. DefininginertialreferencesystemsI andmovingreferencesystemsb 1,B 2,..., B N. Moving systems arealways attachedtobodies1, 2... N (Shabana1989). Ifthemechanical system is composed of N bodies, N moving reference systems will be used, that is, XYZ, X 1 Y 1 Z 1, X 2 Y 2 Z 2,..., X N Y N Z N, with their corresponding unit vectors (i,j,k), (i 1, j 1, k 1 ),... (i N,j N,k N ). 2. Defining coordinate transformation matrices among moving reference frames and inertial T i and, conversely, T T i, with (i 1,2,...N). 3. Defining position vectors and constraint equations. 4. Calculating physical parameters, such as vectors of absolute angular velocity of moving referencesystems Bi Ω i andvectorsofabsolutelinearacceleration Bi Ω i,basedongeometric relations of body motions, according to their physical restrictions (constraint equations). It should be stressed that these vectors are conveniently represented in their own moving reference system B i, attached to the ith body. 5. Calculating physical parameters, such as vectors of absolute angular velocity of bodies B i ω and vectors of absolute linear acceleration Bi ω, based on geometric relations of body motions, according to their physical restrictions (constraint equations). It should be mentioned that these vectors are conveniently represented in their own moving reference system B i. 6. Calculating physical parameters, such as vectors of absolute linear velocity of the center of mass of bodies v i and vectors of absolute linear acceleration of the center of mass of bodies a i, based on geometric relations of body motions, according to their physical restrictions (constraint equations). It should be stressed that these vectors are usually represented in the inertial reference system I, I v i, I a i, or in the moving reference system B i, Bi v i, Bi a i. Bi v i Bi v O + Bi ω i Bi r i + B2 v rel Bi a i Bi a O + Bi ω i ( Bi ω i Bi r i )+ Bi ω i Bi r i +2 Bi ω i B2 v rel + B2 a rel GEOMETRIC PROPERTIES OF THE BODIES 7. Defining geometrical properties of the various bodies that compose the mechanical system, such as total mass of each body m i and the inertia tensor I Oi with respect to point O composed of moments and products of inertia. DYNAMICS 8. Calculating vectors of linear momentum I J i of the ith body as a function of mass and velocity of the center of mass of the body, m i Iv i.

6 1.3 Steps of the Mathematical Modelling 6 9. Calculating the vector of angular momentum Bi H Oi of the ith body with respect to point O based on the body angular velocity Bi ω i and the body inertia tensor I Oi with respect to point O. 1. Representing the p 1 active forces and p 2 active moments of excitation acting on each of the rigid bodies in the vector form, p 1 j1 F I E j and p 2 j1 M I E j. (Free-Body Diagram) 11. Representing the p 3 reaction forces and p 4 reaction moments of excitation acting on each of the rigid bodies in the vector form, p 3 j1 F I R j and p 4 j1 M I R j. (Free-Body Diagram) 12. Calculating differential equations of motion and dynamic reactions by Newton s method (particles) or by Newton-Euler s method (rigid bodies). NEWTON p 1 F I E j + j1 p 3 (i 1,...,N) F I R j +m i Ia i (1.1) j1 EULER (i 1,...,N) p 2 j1 p 4 B i M EOi + j1 B i M ROi I d ( ) Oi B dt i ω i +Bi Ω i ( ) I Oi Bi ω i +mi Bi r Oi Bi a Oi (1.2) where N is the total number of bodies that compose the mechanical system. ) T and rotation ( Bi ω i q ) T are defined as a function of q, the 13. If only equations of motion are desired, reaction forces and moments are eliminated and only differential equations of motion are directly calculated, based on the Principle of Virtual Power postulated by Jourdain (Bremer 1988, Schiehlen 1986, Pfeiffer 1989). Jacobians of translation ( I v i q vector which contains the minimal coordinates of velocity, and eq.(1.3) is used as follows NEWTON-EULER-JOURDAIN ) N T p ( I v 1 ( i I q F Bi E j +m i Ia i + ω ) T i p 2 q i1 j1 j1 B i M EOi +I Oi d ( ) B dt i ω i + + Bi Ω i ( I Oi Bi ω i ) +mi Bi r Oi Bi a Oi )] (1.3) LAGRANGE where t ( ) Ekin q i ( ) Ekin q i + E kin N i1 [ 1 2I vt i m i Iv i E pot is the potential energy, F i p 1 j1 ( ) Epot ( q i [ ( I ) v j T ( F ) ] q I E j + p 2 j1 N min is the minimal coordinates. F i (i 1,...,N min ) (1.4) B i ω T i I CMi Bi ω i )] is the kinetic energy, [ ( I ) ω i T ( M ) ] q I E j is the generalized forces,

7 1.4 Kinematics of Particles (Lumped Masses) 7 NUMERICAL METHODS 14. Solving differential equations of motion numerically using Runge-Kutta, Newton-Raphson, etc. DYNAMICAL ANALYSIS 15. Analyzing the reaction forces and moments and the motions described by the multibody system. VIBRATION ANALYSIS 16. Linearizing differential equations of motion for the multibody system about an equilibrium position, and constructing mass, stiffness, damping and gyroscopic matrices. Analyzing vibration amplitudes, natural frequencies, vibration modes, stability, etc. 1.4 Kinematics of Particles (Lumped Masses) This chapter reviews the concepts related to inertial I and moving B 1 reference systems, transformation matrices from moving coordinates to inertial coordinates T T or from inertial to moving T. The definition of physical vectorial magnitudes as the position vector r, the velocity vector v and the acceleration a can be described both in the inertial reference system as in the moving reference system. The calculation of these vectorial magnitudes as a function of the system geometry (KINEMATICS), taking into account the restriction conditions, is performed so that the causes for the movement (forces) can be expressed in the form of equations using Newton law. Hence, both the KINEMATICS and the DYNAMICS of the mechanical system can be fully described. KINEMATICS corresponds to the study of the movement of the bodies without trying to explain its causes. The analysis is performed from the geometrical point of view. In other words, the positions, velocities and accelerations, lineal and angular of a body, are described as a function of the geometry of the mechanical system. 1.5 Inertial Reference System To describe the movement of a particle it is firstly necessary to define a reference system where the vectors such as position, velocty and acceleration, as well as forces and moments can be represented. The description of the movement is then performed on the chosen reference system. A reference system es defined as a vectorial base, represented by unit vectors or cursors. This reference system can be inertial or moving.

8 1.5 Inertial Reference System 8 Position vector The figure on the side shows the representation of a position vector (for the particle A) related to inertial system I (X Y Z). In vectorial notation it is written as: r I OA x o y o z o (1.5) or I r OA x o i + y o j + z o k The base used for representing the position vector is defined by the cursors or unit vectors i, j, k. The scalar magnitudes x o, y o, z o indicate the amplitude of this vector on the different directions Position vector for the particle A represented on the inertial reference system I (X Y Z). Velocity vector The absolute velocity vector is defined as the derivative in time of the position vector. It is very important to stress that the derivative of the position vector must be calculated when the position vector is represented in the inertial reference frame in order to obtain the absolute velocity vector. If one derives the position vector when it is represented in a moving reference frame, the obtained result does not include all information. This fact will be illustrated in an example later. Hence, the absolute velocity vector is defined as: I v A d dt ( I r OA ) d ( x dt o ) d ( y dt o ) d ( z dt o ) ẋ o ẏ o ż o (1.6) or I v A ẋ o i + ẏ o j + ż o k Acceleration vector The absolute acceleration vector is defined as the second derivative on time of the position vector. Once again, the second derivative must be calculated with the position vector represented on the inertial reference system, when absolute results are desired. or I a A d2 dt 2 ( I r OA ) d ( dt I v A ) d 2 dt 2 ( x o ) d 2 dt 2 ( y o ) d 2 dt 2 ( z o ) ẍ o ÿ o z o (1.7) I a A ẍ o i + ÿ o j + z o k

9 1.6 Moving Reference System Moving Reference System In many practical applications, the description of the movement of a particle can be performed in an easier way using a moving reference system. The objective of using moving reference systems in Kinematics is to enable an easier representation of complex movements, by dividing them in several simpler movements, which can be added in order to obtain the absolute movement. Here, it is possible to distinguish two types of moving reference systems: (a) pure translation, (b) pure rotation. All other kind of movements can be always described by decompossing them into a set of these two kind of movements: rotation and translation. Therefore, it is fundamental to establish a relationship between several reference systems (inertial and moving), to enable the transformation of a vector from a moving to an inertial system and viceversa. This can be done by introducing the concept of the coordinate transformation matrix, used to transform a vector from one base to another (for instance, from moving to inertial). (a) Translational moving reference system Relationship between unit vectors i, j, k from the inertial system and i 1, j 1, k 1 from the moving system. An inertial base I (X Y Z) represented by the cursors i, j, k. A moving system B1 (X 1 Y 1 Z 1 ) represented by the cursors i 1, j 1, k 1. To start with, it is necessary to establish a relationship between the cursors of the inertial system and the moving system. This relationship will be given by the coordinate transformation matrix. Since in this case the moving reference system is only translating, the cursors from the inertial and the moving system will be always parallel to each other, hence:

10 1.6 Moving Reference System 1 i, j, k i 1, j 1, k 1 therefore, i 1 j 1 k i j k B1s I Is Is I 1 B1s When the moving reference system is only translating, the coordinate transformation matrix from one base to the other (for instance, from I to B 1 ) is the identity matrix, with the characteristics of being constant e time invariant. Givenavector I r OB whichdescribes thepositionofparticleb withrespect toaninertial system, it is possible to rewrite it as the sum of two vectors, one of them represented in the inertial system and the other in the moving system: I r OB I r OA + I B1 r AB where I r OA isthevector thatdescribesthepositionofpointa, theoriginofthemovingreference frame, with respect to the point O. This vector is described directly in the inertial system. The vector B1 r AB describes the position of point B with respect to point A, and its representation is usually performed using the moving system B1. When adding these two vectors, it is mandatory that they are represented in the same base or referencial system. The vector B1 r AB can be transformed to the inertial system using the transformation matrix, hence: I r AB I. B1 r AB By using the definition of absolute velocity, which is the derivative in time of the position vector, when this vector is defined in the inertial system, one obtains: I v B d dt ( I r OB ) d dt ( I r OA + I B1 r AB ) v d I B dt ( r )+ d I OA dt ( I ) Ir AB +I d }{{} dt ( r ) B1 AB I v B I v A + I. B1 v rel I v OA + I v rel

11 1.6 Moving Reference System 11 Analogously, by using the definition of absolute acceleration, which is the second derivative in time of the position vector, when this vector is described in the inertial system, one obtains: I a B d2 dt 2 ( I r OB ) d2 dt 2 ( I r OA + I B1 r AB ) I a B I a A + I. B1 a rel I a A + I a rel (b) Rotating (moving) reference system AninertialbaseI (X Y Z)representedbythecursorsi,j, k. AmovingbaseB1(X 1 Y 1 Z 1 ) represented by the cursors i 1, j 1, k 1. First of all, it is necessary to obtain a relationship between the cursors for the inertial system and the moving reference system, given by a coordinate transformation matrix. Since the moving system is rotating, the cursors fron the inertial and the moving system are no longer parallel, hence the relationship between the cursors is given as a function of the angle θ(t). The first step in this analysis must be to define the relationship between the cursors of the inertial and the moving system. Inertial system I and moving reference system B1, represented by the cursors i, j, k and i 1, j 1, k 1 : moving system rotating with an angular velocity θ(t) around the axis Z of the inertial system. Vector r OA represented in the inertial system I and vector r AB represented in the moving system B1. Assuming that moving system rotates around the axis Z in positive direction according to the right hand rule. Hen, it is possible to write the angular velocity and acceleration vectors as:

12 1.6 Moving Reference System 12 ω I θ(t) ω I θ(t) By projecting the cursors from the moving base into the inertial base, the followwing relationship is obtained: i 1 cosθ i + sinθ j + k j 1 sinθ i + cosθ j + k k 1 i + j + 1 k Projection of the moving sytem unit vectors into the inertial system. By rewriting these 3 equations in matrix form, one obtains: i 1 cosθ sinθ i s T j 1 B1 θ. I s sinθ cosθ j k 1 1 k s I T 1 θ. B1 s Hence, any vector described in the base I or B1 can be expressed in the base B1 or I, simply by multiplying it with the transformation matrixt θ, or T 1 θ. It is important to point out that the transformation matrix has some relevant properties, for example its determinant is always equal to one and its inverse is equal to its tranposed: T 1 θ T T θ I s T T θ. B1 s For positive rotations around the axis Y e X the transformation matrices T θ are defined as: Positive rotation around axis Y: cosθ sinθ ω I θ T θ 1 sinθ cosθ B1 s T θ. I s I s T 1 θ. B1 s T T θ. B1 s Positive rotation around axis Y: I ω { θ } T.

13 1.6 Moving Reference System 13 Positive rotation around axis X: θ ω I T θ 1 cosθ sinθ sinθ cosθ B1 s T θ. I s I s T 1 θ. B1 s T T θ. B1 s Positive rotation around axis X I ω { θ } T. The coordinate transformation matrix T θ is time-depending and can be used to transform the representation of a vector described in the inertial system into the moving reference frame B1. On the other hand, its transpose transforms the representation of such a vector described in the moving base into the inertial base. Imagine that an inertial reference system I fixed in the center of the Sun, point O. Simultaneously, a moving reference system B1 is fixed in the center of the Earth, point A. The point A (origin of the moving system attached to Earth) has a translational movement with respect to the Sun. Also, the moving reference system has an absolute angular velocity I ω composed by several consecutive rotations 1, which will be explained in detail further in this chapter. If one wants to describe the movement of a car, particle B with respect to the sun, absolute position vector I r OB, when it moves in the surface of the Earth. The movement of the car can be decompossed in several simpler movements, as the movement of the Earth center around the Sum, and the movement of the car with respect to the center of the Earth. The distance between the center of the Sun and the Earth is known, I r OA.It is also known the variation of this distance according to time. The representation of this vector can be obtained in the inertial reference system I. The position vector B1 r AB describes the movement of the car with respect to the center of the Earth, and it is expressed according to an observer located in the center of the Earth. Therefore, the rotating baseé a B1 is chosen to describe this vector. In order to work simultaneously with the bases I e B1 it is necessary to define the respective coordinate transformation matrixes. By doing this, it is possible to obtain the position, velocity and acceleration vector in the desired base (of the reference system). Position vector: The position vector is given by: r r + I OB I OA TT θ. r B1 AB }{{} I r AB I r OA + I r AB 1 The rotation of the Earth is described by precession, nutation and spin. The spin can be verified by the existence of the day and the night. The precession si related to the seasons: spring, winter, autumn and winter. The nutation is related to inclination of the rotation axis of the Earth and it can be related to the Glacial ages. However, this movement is really slow, having a period of thousand of years, so it is difficult for a human being to feel its effects.

14 1.6 Moving Reference System 14 Velocity vector The absolute velocity vector is obtained by deriving the position vector in time, when such vector is represented in the inertial system : I v B d dt I v B d dt ( ( I r OB I r OA ) d ( dt I r OA +TT θ B1 r AB ) ) d ( ) + T T dt θ. r + B1 AB TT θ. d dt v v + ω ( T T I B I A I θ. r B1 AB }{{} I r AB ) + T T θ. B1 v Rel }{{} I v Rel ( B1 r AB d The verification that ( dt TT θ ). r ω ( T T B1 AB I θ. r ) will be donde in example 1 B1 AB of this chapter. Hence, the expression for the absolute velocity of particle B, I v B, described in the inertial system, is given by: )

15 1.6 Moving Reference System 15 I v B I v A + I ω I r AB + I v Rel (1.8) or v v + ω ( T T I B I A I θ. r ) B1 AB }{{} I r AB + T T θ. d dt ( B1 r AB ) }{{} B1 v Rel }{{} I v Rel I v A is the absolute linear velocity of point A (point where the moving reference system is located) represented in the inertial reference system I. I ω I r AB is the cross product between the angular velocity vector of the moving reference system and the position vector I r AB, both of them described in the inertial reference frame. The vector I r AB has its origin in point A and points toward the studied point B. It must be stressed that B1 r AB is usually described in the moving reference system B1, in order to simplify the description of the movement. When this vector is multiplied by the coordinate transformation matrix T T θ, one obtains its representation in the inertial system. I v Rel is the relative velocity of point B with respect to point A. This velocity is obtained by deriving the position vector B1 r AB with respect to time, when it is represented in the moving reference system, which absolute angular velocity is I ω. Deriving this vector in the moving system, entails obtaining its representation in the moving system B1. Moreover, deriving this vector in the moving system, entails that you obtain a relative quantity, i.e. B1 v Rel. Then, it is necessary to multiply this vector by the tranformation matrix T T θ to obtain its representation in the inertial reference frame. Afterwards, it is possible to obtain the sum of all the terms in the eq.(1.8) expressed in the same reference frame. The reader must focus on a important fact: many students are confused between the concept of representing the absolute velocity in the moving system and the concept of relative velocity vector. It must be stressed that the absolute velocity vector will always be an absolute velocity vector. Any vector can be expressed both in an inertial system as in a moving reference system, hence the absolute velocity vector represented in the moving reference frame will still be the absolute velocity vector. A relative velocity vector can also be represented in a moving or in an inertial reference frame, but expressing it into the inertial system does not imply that it is transformed into an absolute velocity vector. Frequently, it is more convenient to represent the velocities in the moving reference frame B1 and equation (1.8) can be rewritten as: B1 v B B1 v A }{{} T θ. I v A + B1 ω B1 r AB + d dt ( B1 r AB ) }{{} B1 v Rel (1.9)

16 1.6 Moving Reference System 16 Acceleration vector The acceleration vector is obtained by deriving two times the position vector with respect to time: I a B d2 dt 2 ( I r OB ) d 2 dt 2 ( I r OA + TT θ. B1 r AB ) I a B d dt [ d ( dt I r OA ) d ( ) + T T dt θ. r + B1 AB TT θ. d ( dt B1 r AB ) ] I a B d dt [ I v A + I ω ( T T θ B1 r AB) + T T θ. d ( dt B1 r AB ) ] I a B d dt ( I v A ) + d dt [ I ω ( T T θ B1 r AB [ )] d + T T θ dt. d ( dt B1 r AB ) ] I a B I a A + d dt ( I ω) [( T T θ B1 r AB )] + I ω [ ] d ( T T dt θ ) r B1 AB + + I ω [ T T θ d ] dt ( r ) B1 AB + d dt ( T T d ( θ ) dt B1 r AB ) + T T θ d2 dt 2 ( B1 r AB ) I a B I a A + I ω ( T T θ B1 r AB) + I ω [ I ω ( T T θ B1 r AB)] + I ω ( T T θ B1 v rel) + + I ω ( T T θ B1 v rel ) + T T θ B1 a rel

17 1.6 Moving Reference System 17 I a B I a A + I ω I r AB + I ω ( I ω I r AB ) + 2. I ω I v Rel + I a Rel (1.1) or I a B I a A + I ω ( T T θ B1 r AB) + I ω [ I ω ( T T θ B1 r AB)] I ω [ T T θ ] d dt ( r ) B1 AB + T T θ d2 dt 2 ( B1 r AB ) I a A is the absolute linear acceleration vector of the point A, where the origin of the moving reference system is located, represented in the inertial reference system I. I ω I r AB the cross product between the absolute anuglar acceleration of the moving reference frame and the position vector I r AB, both of them described in the inertial system. The vector I r AB has its origin in the point A and it is pointing toward the studied point B. This term is directly related to the tangential acceleration, due to the absolute variation on time of the vector I ω. I ω ( I ω I r AB ) is the cross product between the absolute angular velocity vector of the moving reference system and the vector resulting from the operation I ω I r AB. This vector is related to the change in direction of the vector I ω I r AB. The vector I ω I r AB rotates with the angular speed I ω. 2 I ω I v Rel is the cross product of the absolute angular velocity vector of the moving reference system and the relative velocity of point B with respect to point A, both of them expressed in the inertial system I. This term is known as the Coriolis acceleration, expressing the change in direction of the relative velocity vector I v Rel. This vector rotates in space with an angular velocity I ω. Example 5 of this chapter deals with the mathematical and physical meaning of this term. I a Rel is the relative acceleration of point B with respect to point A (origin of the moving reference system). This acceleration is obtained when deriving twice the position vector r with respect to time, represented in the moving reference system, whose angular B1 AB velocity is given by é I ω. By deriving this vector in the moving system, one obtains its representation in the moving system B1.Then, it is necessary to multiply it by the transformation matrix T T θ to obtain the representation of this vector in the inertial system, in order to obtain the sum of all the terms of eq.(1.1) expressed in the same base. Frequently, it is more convenient to represent the accelerations in the moving reference frame B1 and equation (1.1) can be rewritten as: B1a B B1 a A + B1 ω B1 r AB + B1 ω ( B1 ω B1 r AB )+2 B1 ω d dt ( B1 r AB )+ d2 dt 2 ( B1 r AB ) Some examples are now presented and solved, with the objective of giving the reader further insight into the already presented concepts: moving reference system; representation of position vectors in different reference systems; calculation of velocity and acceleration vectors, relative

18 1.7 Illustrating Some Fundamentals Through Examples 18 or absolute, depending on the reference system where the derivation operation is performed; physical meaning of these vectors and their derivatives. 1.7 Illustrating Some Fundamentals Through Examples Differentiating the coordinate transformation matrix EXAMPLE 1: Verify that d ( dt TT θ ). B1 r AB I ω ( ) T T θ. B1 r AB, meaning that, the scalar product between the time derivative of the coordinate transformation matrix, and the vector r, described according to the reference system B1 rotating with angular speed ω, is equal B1 AB I to the cross product between the angular speed vector and the projection of the vector B1 r AB in the inertial reference system, I r AB. SOLUTION: Given a generic vector B1 r AB { r X r Y r Z } T e and an angular speed vector I ω { θ } T, one can write the first term of the equality: d dt ( TT θ ). B1 r AB d dt θsinθ θcosθ θcosθ θsinθ cosθ sinθ sinθ cosθ 1 r X ry r Z. B1 r AB θ( r X sinθ +r Y cosθ ) θ( r X cosθ r Y sinθ ) (1.11) For the second term of the equality one can obtain: I ω ( TT θ. B1 r AB ) I ω I ω r X cosθ r Y sinθ r X sinθ+r Y cosθ r Z θ( r X sinθ +r Y cosθ ) θ( r X cosθ r Y sinθ ) cosθ sinθ sinθ cosθ 1 r X ry r Z i j k θ ( r X cosθ r Y sinθ ) ( r X sinθ +r Y cosθ ) r Z Comparing the equations (1.11) e (1.12), the initial statement can be verified. (1.12)

19 1.7 Illustrating Some Fundamentals Through Examples Differentiating vectors in terms of direction EXAMPLE 2: The objective of this example is to demonstrate that a position vector with constant modulus I r, rotating with constant angular velocity θ exhibits both velocity v, as well as acceleration a. SOLUTION: A vector is always represented by a modulus and a direction. When the vector is differentiated in the inertial system, its derivative is composed by the change in the magnitude or modulus and the change in direction. If a position vector has a constant modulus but it is rotating with constant angular speed I ω, its derivative in terms of magnitude is null, but its derivative regarding the change in direction is not. Such derivative can be easily calculated by the cross product between the angular speed I ω, and the position vector, as follows: Position vector in four different times I r cte; d dt ( I r) I ω I r I v The cross product between the position vector I r and the angular speed vector I ω, delivers the velocity vector v, which presents a constant magnitude. But this velocity vector is also changing, in terms of its direction. Hence, in order to determine the derivative of the velocity vector in terms of the change in direction, one just applies the cross product between the angular speed vector I ω and the velocity vector. This operation gives an acceleration vector I a, which exhibits a constant modulus: Velocity vector in four different times I v cte; d dt ( I v) I ω I v I ω I ω I r I a The velocity vector represents the tangential velocity of the particle, and the acceleration vector represents the centrifuge acceleration. It can be clearly seen that, despite the fact that the position vector does not change its modulus, it exhibits variations in its direction, generating velocity and acceleration vectors. To differentiate once the position vector according to its position, one calculates its cross product with the angular speed vector I ω. To differentiate it twice, one applies the cross product twice. Acceleration vector in four different times

20 1.7 Illustrating Some Fundamentals Through Examples Differentiating vectors in an inertial reference frame and in a moving reference frame EXAMPLE 3: The objective of this example is to clarify the relationship between the vectorial calculus and its physical interpretation regarding kinematics. For example, the derivative of a vector with respect to time must be always performed in the inertial reference system, so that the corresponding velocities and accelerations are absolute. Such derivative must contain both the variation of the vector in time in terms of modulus, as well as direction. If a vector is differentiated relative to a moving reference frame, then the information regarding the change in direction is lost, entailing that only the change in modulus is taken into account. Then, it is important to stress that for any given vector s, represented in an inertial reference frame I and a moving reference frame B1: d dt ( s ) d I }{{} dt ( s ) B1 }{{} v Abs v Rel and d 2 dt 2( s ) d 2 I }{{} dt 2( s ) B1 }{{} a Abs a Rel Let s think about an hydraulic piston with a mass m located at its end, rotating with angular speed θ relative to the axis Z (inertial frame). A moving reference frame X 1 Y 1 Z 1, fixed to the piston, also rotates with angular speed θ. An observer is sitting at the origin of the moving reference frame (point O A). For any given time t, the origin point coincides with the origin of the inertial frame, since the moving frame does not exhibit translation. Then, the observer sitting on top of the moving frame perceives that the mass m exhibit motion in the positive direction of axis X 1. An observer sitting on the moving reference frame measuring the movement of the mass m (particle B).

21 1.7 Illustrating Some Fundamentals Through Examples 21 SOLUTION: (a) Definition of the reference frames: Inertial system I (X Y Z), represented by the unity vectors i j k Moving system B (X 1 Y 1 Z 1 ), represented by the unity vectors i 1 j 1 k 1 (b) Coordinate transformation matrix: i 1 cosθ i+sinθ j+ k j 1 sinθ i+cosθ j+ k Positive rotation of the system B1 around the axis Z (inertial) k 1 i+ j+1 k i 1 j 1 k 1 B1 s T I s I s TT B1 s cosθ sinθ sinθ cosθ 1 i j k (c) Angular velocity and acceleration of the moving reference frame, described in the inertial system I: ω I θ ω I θ (d) Position vector L described in the moving frame B1 and the inertial frame I: L B1 L The vector B1 L can be represented in the inertial reference frame I by using the coordinate transformation matrix T T, as follows: cosθ sinθ L Lcosθ L I TT L B1 sinθ cosθ Lsinθ 1 (e) Linear velocity vector it is calculated when the position is differentiated. Firstly, one differentiates the position vector in the inertial frame:

22 1.7 Illustrating Some Fundamentals Through Examples 22 d ( dt I L ) Lcosθ L θsinθ Lsinθ +L θcosθ (1.13) Then, one differentiate the same position vector in the moving frame: d ( dt B 1 L ) L (1.14) Then, one can clearly observe in the equations (1.13) and (1.14) that, by differentiating the vector L in the inertial frame I and in the moving frameb1, different results are obtained. It becomes necessary to give a physical interpretation to these results. When the position vector is differentiated in the inertial reference frame I, one obtains the absolute velocity of the particle, represented in the inertial reference frame: d ( dt I L ) Lcosθ L θsinθ Lsinθ +L θcosθ I v B (1.15) When the position vector is differentiated in the moving reference frame B1, one obtains the relative velocity of the particle represented in the moving reference frame B1: d ( dt B 1 L ) L B1 v Rel (1.16) From equation(1.16), it can be inferred that if the observer is sitting on top of the moving reference frame, the only velocity that is detected is B1 v Rel, in other words, one can only detect whether the position vector increases or decreases its magnitude L(t). By multiplying the vector B1 v Rel with coordinate transformation matrix T T, one obtains the representation of the relative velocity in the inertial frame: I v Rel T T B1 v Rel I v Rel cosθ sinθ sinθ cosθ 1 L Lcosθ Lsinθ (1.17) By comparing the two vectors given in equation (1.17) and equation (1.13), after they have been expresded in the same reference frame, it can be verified that the vector I v Rel is a component of the vector I v B. This part is directly related to the variation of the vector in terms of magnitude, L. Hence, by differentiating a vector in a moving reference frame one can only obtain the information related to the modification of the magnitude of such vector. The variations related to the change in direction of the vector, associated to the angular speed θ, can not be detected when the differentiation is performed in the moving reference frame. The previous example illustrates the idea that the derivative of any given vector r, in terms of direction, is given by the cross product with the angular speed vector: I ω I r. Then, by adding

23 1.7 Illustrating Some Fundamentals Through Examples 23 up equation (1.17) with the derivative of vector L in terms of its direction, I ω I L, one obtains the absolute velocity vector, composed by the derivatives in terms of magnitude and direction: v + I Rel I ω I L Lcosθ Lsinθ + i j k θ Lcosθ Lsinθ Lcosθ L θsinθ Lsinθ+L θcosθ (1.18) By comparing equations (1.18), (1.17) and (1.13) it can be verified that by differentiating any vector in a moving reference frame, the information relative to its change in direction is lost, ω r. In order to compensate for this when working with a moving reference frame B1, I I the term I ω I L must be added up, as it provides the variation of the vector L in terms of its direction: d ( I dt L ) T T d ( B1 }{{} dt L ) + ω L (1.19) I I }{{}}{{} variation in direction I v A I v Rel In a more general case, where the point A is also translating relative to the point O, with a linear velocity I v A, one obtains: I v B I v A + I ω I L + I v Rel (f) Linear acceleration vector It is calculated by differentiating twice the position vector according to time. Firstly, the position vector is differentiated twice in the inertial reference frame, in other words, the vector is differentiated according to time, when it is represented in the inertial reference frame: d 2 dt 2( L ) d2 I dt 2 d 2 dt 2( I L ) d 2 dt 2( I L ) Lcosθ Lsinθ d dt Lcosθ L θsinθ Lsinθ+L θcosθ Lcosθ L θsinθ L θsinθ L θsinθ L θ 2 cosθ Lsinθ + L θcosθ+ L θcosθ+l θcosθ L θ 2 sinθ Lcosθ Lsinθ + 2 L θsinθ L θsinθ L θ 2 cosθ 2 L θcosθ +L θcosθ L θ 2 sinθ I a B (1.2) Next, the position vector is differentiated twice in the moving reference frame, in other words, the vector is differentiated when it is represented in the moving frame: d 2 dt 2( L ) d2 B1 dt 2 L L B1 a Rel (1.21)

24 1.7 Illustrating Some Fundamentals Through Examples 24 By comparing the equations (1.2) and (1.21), it can be seen that the second derivative of the position vector in the inertial and the moving reference frame are completely different. The acceleration vector given in equation (1.21) is given in the moving reference frame B1 can only be compared with the one given in equation (1.2) if it is transformed into the inertial frame. Then, one obtains: T T d2 dt 2( B1 L ) cosθ sinθ sinθ cosθ 1 L Lcosθ Lsinθ I a Rel (1.22) By comparing equations (1.2) and (1.21), one can notice the differences between these two acceleration vectors. It can be seen that by differentiating the position vector twice in the moving frame some information is lost, regarding the angular speed θ and acceleration θ, in other words, the variations of the vector in terms of direction are not taken into account. Hence, when the position vector is differentiated twice in the moving reference frame one obtains the relative acceleration of the particle B1 a Rel, represented in the moving reference frame. By multiplying this vector with the coordinate transformation matrix, one obtains the relative acceleration vector in the inertial reference frame I a Rel Two ways for calculating the absolute acceleration EXAMPLE 4: The objective of this example is to state clearly that there are two ways of calculating the acceleration of a particle: by differentiating twice the position vector according to time, when it is represented in the inertial reference frame (see equation (1.2)), or by using equation (1.1). If one refers to example 3, one can demonstrate the two ways of calculation as follows: SOLUTION: 1 st way: I a B d2 dt 2( I L ) d2 dt 2 Lcosθ Lsinθ Lcosθ 2 L θsinθ L θsinθ L θ 2 cosθ Lsinθ +2 L θcosθ+l θcosθ L θ 2 sinθ (1.23) 2 nd way: a a + I B I A }{{} I ω I ω I L + I ω I L +2. I ω I v Rel + I a Rel With the terms given as: ω I θ, ω I θ, L I Lcosθ Lsinθ e v I Rel Lcosθ Lsinθ one can calculate the four terms of the acceleration equation.

25 1.8 Dynamics of One Single Particle in 3D Examples 25 Term related to normal acceleration: i j k ω ω L θ I I I L θsinθ L θcosθ L θ 2 cosθ L θ 2 sinθ Term related to tangential acceleration: ω L i j k I I θ Lcosθ Lsinθ L θsinθ L θcosθ Term related to Coriolis acceleration: i j k 2. I ω I v Rel 2. θ Lcosθ Lsinθ 2 θ Lsinθ 2 θ Lcosθ Term related to relative acceleration: I a Rel T T θ. d 2 dt 2( L ) B1 Lcosθ Lsinθ By adding up the four terms, one obtains the absolute acceleration: I a B Lcosθ 2 L θsinθ L θsinθ L θ 2 cosθ Lsinθ+2 L θsinθ+l θcosθ L θ 2 sinθ (1.24) Equations (1.23) and (1.24) state the two methods for calculating the absolute acceleration. The second method is the most commonly used, when one desires to express the absolute acceleration of a particle in a moving reference frame. 1.8 Dynamics of One Single Particle in 3D Examples The procedure of achieving equations of motion for one single particle is illustrated. It is important to highlight that when working with one single particle in 3D you will always deal with 3 equations. Depending on the motion restrictions applied to the particle, one can achieve: a) 3 dynamic reaction forces and no equation of motion; b) 2 dynamic reaction forces and 1 equation of motion; c) 1 dynamic reaction force and 2 equations of motion and finally d) three equations of motions and no reaction force, if the particle is completely free in the space One Equation of Motion and Two Dynamic Reaction Forces Figure illustrates a simple pendulum of length l [m] and tip mass m [Kg], mounted on the extremity of a disk of radius r [m]. The mass of the arm is negligible when compared to the tip mass. The disk is mounted on the extremity of an arm of length b [m]. Distance between the center of disk to the arm extremity is c [m]. The disk as well the arm rotate at constant angular

26 1.8 Dynamics of One Single Particle in 3D Examples 26 Figure 1.1: Simple pendulum, mounted on the extremity of a rotating system Spatial movements of a particle performing three consecutive rotations. velocity of α [rad/s] e β [rad/s], respectively. It is important to mention that all bodies (arms and disk) are rigid. (a) Calculate the coordinate transformation matrices, which facilitate to transform the representation of the different vectors (displacement, velocity, acceleration, force, moment etc.) from the inertial reference frame to the moving reference frames or vice-versa. The inertial frame I is attached to the base, point O. The moving reference frame B1 is attached to the extremity of the arm, point B. The moving reference frame B2 is attached to the disk center, point C. The moving reference frame B3 is attached to the extremity of the second arm, point D. (b) Calculate the position vectors between points OA, AB, BC, CD and DE, representing them with help of the most convenient reference. Indicate how to write the position vector between points OE with help of the inertial reference frame I. Such a vector will be useful for describing the trajectory followed by the pendulum mass E. (c) Determine the absolute angular velocity of the reference frames B1, B2 and B3, representing such vectors with help of the respective reference frames, i.e. B1 ω 1, B2 ω 2 and B3 ω 3. (d) Determine the absolute angular acceleration of the reference frames B1, B2 and B3, representing such vectors with help of the respective reference frames, i.e. B1 ω 1, B2 ω 2 and B3 ω 3. (e) Determine the absolute linear acceleration of the point B (arm extremity), representing such a vector with help of the reference frame B1, i.e. B1 a B.

27 1.8 Dynamics of One Single Particle in 3D Examples 27 (f) Determine the absolute linear acceleration of the point C (disk center), representing such a vector with help of the reference frame B2, i.e. B2 a C. (g) Determine the absolute linear acceleration of the particle E, representing such a vector with help of the reference frame B3, i.e B3 a E. (h) Draw the forces which are acting on the particule E, i.e. elaborate a free-body diagram. Afterwards, write the force vectors with help of the most convenient reference frame. (i) Write the equations responsible for describing the dynamic equilibrium of the particle E. What is the result of the set of equations?

28 1.8 Dynamics of One Single Particle in 3D Examples 28 SOLUTION: (a) Coordinate transformation matrices. Coordinate transformation matrices between I and B1: cosα sinα T α sinα cosα 1 First rotation around the Z-axis of inertial reference frame. B1 s T α Is I s TT α B1 s Coordinate transformation matrices between B1 and B2: cosβ sinβ T β sinβ cosβ 1 Second rotation around the Z 1 -axis of the moving reference frame B1. B2 s T β B1 s B1 s TT β B2 s Coordinate transformation matrices between B2 and B3: T ψ 1 cosψ sinψ sinψ cosψ Third rotation around the X 2 -axis of the moving reference frame B2. B3 s T ψ B2 s B2 s TT ψ B3 s

29 1.8 Dynamics of One Single Particle in 3D Examples 29 (b) Position vectors. position vector I r OA : r { a } T I OA position vector B1 r AB : a [m] B1 r AB { b } T [m] position vector B2 r BC : B2 r BC { c } T [m] position vector B2 r CD : B2 r CD { r h } T [m] position vector B3 r DE : B3 r DE { l } T [m] The position vectors were written with help of the most convenient reference frames, it means using the reference frame in which their representation are the most compact one. For describing the particle trajectory is necessary to write the vector I r OE, i.e. or I r OE I r OA + I r AB + I r BC + I r CD + I r DE I r OE I r OA +TT α B1 r AB +T T α T T β ( B2 r BC + B2 r CD )+T T α T T β T T ψ B3 r DE (c) Absolute angular velocity of the moving reference frames. Absolute angular velocity of the moving reference frame B1: cosα sinα ω T B1 1 α I α sinα cosα ω B1 1 1 α Absolute angular velocity of the moving reference frame B2: ω B2 2 T β T α I α+t β B1 β α [rad/s] cos(α+β) sin(α+β) sin(α+β) cos(α+β) 1 α + cosβ sinβ sinβ cosβ 1 β

30 1.8 Dynamics of One Single Particle in 3D Examples 3 ω B2 2 α+ β [rad/s] Absolute angular velocity of the moving reference frame B3: ω B3 3 T ψ T β T α I α+t ψ T β B1 β +Tψ B2 ψ T ψ ( B2 ω 2 + B2 ψ) 1 cosψ sinψ sinψ cosψ ψ α+ β ψ ( α+ β)sinψ ( α+ β)cosψ [rad/s] (d) Absolute angular acceleration of the moving reference frames. Absolute angular acceleration of the moving reference frame B1: ω d B1 1 dt ( ω )+ ω ω B1 1 B1 1 B1 1 }{{} ω B2 2 d dt ( ω B2 2)+ B2 ω 2 B2 ω }{{} 2 α ω B1 1 Absolute angular acceleration of the moving reference frame B2: α+ β ω B2 2 Absolute angular acceleration of the moving reference frame B3: ω B3 3 d dt ( ω B3 3)+ B3 ω 3 B3 ω }{{} 3 [rad/s2 ] [rad/s2 ] ψ ω B3 3 ( α+ β)sinψ +( α+ ( α+ β)cosψ ( α+ β) ψcosψ ω B3 3 β) ψsinψ ψ ( α+ β) ψcosψ [rad/s2 ] ( α+ β) ψsinψ It is important to point out that α β! It is given in the beginning of the example, when the problem was formulated. (e) Absolute linear acceleration of the point B. a a + B1 B B1 O }{{} B1 ω 1 B1 ω 1 B1 r OB + B1 ω 1 }{{} B1 r OB +2 B1 ω 1 B1 v rel + }{{} B1 a }{{ rel }

31 1.8 Dynamics of One Single Particle in 3D Examples 31 ω B1 1 i 1 j 1 k 1 α b a i 1 j 1 k 1 α b α b α 2 a B1 B b α 2 [m/s2 ] (f) Absolute linear acceleration of the point C. B1 a C B1 a B B2 a C T β B1 a B cosβ sinβ sinβ cosβ 1 b α 2 a B2 C b α 2 sinβ b α 2 cosβ [m/s2 ] (g) Absolute linear acceleration of the particle E. a a + B3 E B3 D }{{} B3 ω 3 B3 ω 3 B3 r DE + }{{} B3 ω 3 B3 r DE + 2 }{{} B3 ω 3 B3 v rel + }{{} B3 a }{{ rel } (I) (II) (III) (I) B2 a D a B2 C }{{} see item (f) + B2 ω 2 B2 ω 2 B2 r }{{ CD + } B2 ω 2 B2 r CD +2 }{{} B2 ω 2 B2 v }{{ rel } (IV) + B2 a rel }{{} (IV) ω B2 2 i 2 j 2 k 2 ( α+ β) r h i 2 j 2 k 2 ( α+ β) r( α+ β) r( α+ β) 2 B2 a D b α 2 sinβ b α 2 cosβ r( α+ β) 2 [rad/s2 ]

32 1.8 Dynamics of One Single Particle in 3D Examples 32 B3 a D T ψ B2 a D 1 cosψ sinψ sinψ cosψ b α 2 sinβ b α 2 cosβ r( α+ β) 2 a B3 D b α 2 sinβ (b α 2 cosβ +r( α+ β) 2 )cosψ (b α 2 cosβ +r( α+ β) 2 )sinψ [rad/s2 ] (II) ω B3 3 i 3 j 3 k 3 ψ ( α+ β)sinψ ( α+ β)cosψ l i 3 j 3 k 3 ψ ( α+ β)sinψ ( α+ β)cosψ l( α+ β)sinψ l ψ l ψ( α+ β)cosψ l( α+ β) 2 sinψcosψ l( ψ 2 +( α+ β) 2 sin 2 ψ) (III) ω B3 3 B3 r DE i 3 j 3 k 3 ψ ( α+ β) ψcosψ ( α+ β) ψsinψ l l( α+ β) ψcosψ l ψ Adding the terms recently calculated, i.e. (I), (II) and (III), one achieves the absolute linear acceleration of the particle E, representing such a vector with help of the reference frame B3: a B3 E b α 2 sinβ 2l ψ( α+ β)cosψ (b α 2 cosβ +r( α+ β) 2 )cosψ l( α+ β) 2 sinψcosψ +l ψ (b α 2 cosβ +r( α+ β) 2 )sinψ +l( ψ 2 +( α+ β) 2 sin 2 ψ) [rad/s2 ] (h) Free-body diagram of the particle E and vector representation of the forces acting on the particle.

33 1.8 Dynamics of One Single Particle in 3D Examples 33 Gravity force acting on the particle E: I P { mg } T Reaction force between particle and arm(normal force acting on the arm longitudinal direction): B3 T { T } T Reaction force between particle and arm (shear force acting perpendicularly to the arm): B3 R { R } T (i) Dynamic equilibrium of the particle E according to the second Newton s law: B3 F T ψt β T α d dt (m I v E ) T ψt β T α ( ṁ }{{} B3 F B3 R + B3 T + T ψt β T α I P }{{} B3 P I v E +m I a E ) T ψt β T α I a E m B3 a E B3 R + B3 T+ B3 P m B3 a E where B3 P 1 cosψ sinψ sinψ cosψ cosβ sinβ sinβ cosβ 1 cosα sinα sinα cosα 1 mg mgsinψ mgcosψ The dynamic equilibrium of the particle E can be achieved by the equation: R b α 2 sinβ 2l ψ( α+ β)cosψ mgsinψ m (b α 2 cosβ +r( α+ β) 2 )cosψ l( α+ β) 2 sinψcosψ +l ψ mgcosψ +T (b α 2 cosβ +r( α+ β) 2 )sinψ +l( ψ 2 +( α+ β) 2 sin 2 ψ) The final result of the application of the second Newton s law is the achievement of three equations, i.e. two for calculating the dynamic reaction forces R and T, and one additional equation responsible for describing the movement of the particle E as a function of the time, ψ(t). Dynamic reaction forces: direction X 3 direction Z 3 : R m(b α 2 sinβ +2l ψ( α+ β)cosψ) : T m{gcosψ +[b α 2 cosβ +r( α+ β) 2 ]sinψ +l[ ψ 2 +( α+ β) 2 sin 2 ψ]}

34 1.8 Dynamics of One Single Particle in 3D Examples 34 Equation of motion for the particle E (direction Y 3 ) g direction Y 3 : ψ + l sinψ (b α2 cosβ +r( α+ β) 2 ) cosψ ( α+ l β) 2 sinψcosψ It is important to emphasize that the angle ψ(t) is defined in the moving reference frame B3. The equation of motion is a non-linear second order differential equation, which has to be numerically solved. After solving the equations by means of a Runge-Kutta integrator, computer animations are created, aiming at facilitating the visualization of the movement for different initial conditions of displacement ψ() and velocity ψ(). Figure illustrates one of frozen picture of the trajectory described by the particle E, when the arm and the disk rotate withaconstant angularvelocity of π [rad/s], starting fromtheinitial positionsα β [rad]. 4 The initial condition for the particle E are: ψ π [rad] e ψ [rad/s]. 4 Figure 1.2: Trajectory described by the particle E when the arm and the disk rotate with a constant angular velocity of π [rad/s], starting from the initial positions α β [rad], 4 ψ π [rad] e ψ [rad/s] Three Dynamic Reaction Forces and No Equation of Motion Figure illustrates a particle of tip mass m [Kg] and connected to an arm of length l [m], which is mounted on the extremity of a disk of radius r [m]. The mass of the arm is negligible when compared to the tip mass. The disk is mounted on the extremity of an arm of length b [m]. Distance between the center of disk to the arm extremity is c [m]. The disk as well the

35 1.8 Dynamics of One Single Particle in 3D Examples 35 arm rotate at constant angular velocity of α [rad/s] and β [rad/s], respectively. It is important to mention that all bodies (arms and disk) are rigid. Figure 1.3: A particle, mounted on the extremity of a rotating system Spatial movements of a particle performing two consecutive rotations. (a) Calculate the coordinate transformation matrices, which facilitate to transform the representation of the different vectors (displacement, velocity, acceleration, force, moment etc.) from the inertial reference frame to the moving reference frames or vice-versa. The inertial frame I is attached to the base, point O. The moving reference frame B1 is attached to the extremity of the arm, point B. The moving reference frame B2 is attached to the disk center, point C. The moving reference frame B3 is attached to the extremity of the second arm, point D, and it is fixed without performing any time-dependent rotation. (b) Calculate the position vectors between points OA, AB, BC, CD and DE, representing them with help of the most convenient reference. Indicate how to write the position vector between points OE with help of the inertial reference frame I. Such a vector will be useful for describing the trajectory followed by the particle of mass E. (c) Determine the absolute angular velocity of the reference frames B1, B2 and B3, representing such vectors with help of the respective reference frames, i.e. B1 ω 1, B2 ω 2 and B3 ω 3. (d) Determine the absolute angular acceleration of the reference frames B1, B2 and B3, representing such vectors with help of the respective reference frames, i.e. B1 ω 1, B2 ω 2 and B3 ω 3. (e) Determine the absolute linear acceleration of the point B (arm extremity), representing such a vector with help of the reference frame B1, i.e. B1 a B.

36 1.8 Dynamics of One Single Particle in 3D Examples 36 (f) Determine the absolute linear acceleration of the point C (disk center), representing such a vector with help of the reference frame B2, i.e. B2 a C. (g) Determine the absolute linear acceleration of the particle E, representing such a vector with help of the reference frame B3, i.e B3 a E. (h) Draw the forces which are acting on the particle E, i.e. elaborate a free-body diagram. Afterwards, write the force vectors with help of the most convenient reference frame. (i) Write the equations responsible for describing the dynamic equilibrium of the particle E. What is the result of the set of equations?

37 1.8 Dynamics of One Single Particle in 3D Examples 37 SOLUTION: (a) Coordinate transformation matrices. Coordinate transformation matrices between I and B1: cosα(t) sinα(t) T α sinα(t) cosα(t) 1 First rotation around the Z-axis of inertial reference frame. B1 s T α Is I s TT α B1 s Coordinate transformation matrices between B1 and B2: cosβ(t) sinβ(t) T β sinβ(t) cosβ(t) 1 Second rotation around the Z 1 -axis of the moving reference frame B1. B2 s T β B1 s B1 s TT β B2 s Coordinate transformation matrices between B2 and B3: T ψ 1 cosψ sinψ sinψ cosψ Third rotation around the X 2 -axis of the moving reference frame B2. B3 s T ψ B2 s B2 s TT ψ B3 s

38 1.8 Dynamics of One Single Particle in 3D Examples 38 (b) Position vectors. position vector I r OA : r { a } T I OA position vector B1 r AB : a [m] B1 r AB { b } T [m] position vector B2 r BC : B2 r BC { c } T [m] position vector B2 r CD : B2 r CD { r h } T [m] position vector B3 r DE : B3 r DE { l } T [m] The position vectors were written with help of the most convenient reference frames, it means using the reference frame in which their representation are the most compact one. For describing the particle trajectory is necessary to write the vector I r OE, i.e. or I r OE I r OA + I r AB + I r BC + I r CD + I r DE I r OE I r OA +TT α B1 r AB +T T α T T β ( B2 r BC + B2 r CD )+T T α T T β T T ψ B3 r DE (c) Absolute angular velocity of the moving reference frames. Absolute angular velocity of the moving reference frame B1: cosα sinα ω T B1 1 α I α sinα cosα ω B1 1 1 α Absolute angular velocity of the moving reference frame B2: ω B2 2 T β T α I α+t β B1 β α [rad/s] cos(α+β) sin(α+β) sin(α+β) cos(α+β) 1 α + cosβ sinβ sinβ cosβ 1 β

39 1.8 Dynamics of One Single Particle in 3D Examples 39 ω B2 2 α+ β [rad/s] Absolute angular velocity of the moving reference frame B3: B3 ω 3 T ψ T β T α I α+t ψ T β B1 β +Tψ B2 ψ T ψ ( B2 ω 2 + B2 ψ) 1 cosψ sinψ sinψ cosψ ψ α+ β ( α+ β)sinψ ( α+ β)cosψ [rad/s] It is important to point out that ψ! The arm connected to the particle is rigidly attached to point D, i.e. it is not executing a pendular movement! (d) Absolute angular acceleration of the moving reference frames. Absolute angular acceleration of the moving reference frame B1: ω B1 1 d dt ( ω B1 1)+ B1 ω 1 B1 ω }{{} 1 ω B2 2 d dt ( ω B2 2)+ B2 ω 2 B2 ω }{{} 2 α ω B1 1 Absolute angular acceleration of the moving reference frame B2: α+ β ω B2 2 Absolute angular acceleration of the moving reference frame B3: ω d B3 3 dt ( ω )+ ω ω B3 3 B3 3 B3 3 }{{} [rad/s2 ] [rad/s2 ] ψ ω B3 3 ( α+ β)sinψ +( α+ ( α+ β)cosψ ( α+ β) ψcosψ ω B3 3 β) ψsinψ [rad/s2 ] It is important to point out that α β ψ! It is given in the beginning of the example, when the problem was formulated. (e) Absolute linear acceleration of the point B.

40 1.8 Dynamics of One Single Particle in 3D Examples 4 a a + B1 B B1 O }{{} B1 ω 1 B1 ω 1 B1 r OB + B1 ω 1 }{{} B1 r OB +2 B1 ω 1 B1 v rel + }{{} B1 a }{{ rel } ω B1 1 i 1 j 1 k 1 α b a i 1 j 1 k 1 α b α b α 2 a B1 B b α 2 [m/s2 ] (f) Absolute linear acceleration of the point C. B1 a C B1 a B B2 a C T β B1 a B cosβ sinβ sinβ cosβ 1 b α 2 a B2 C b α 2 sinβ b α 2 cosβ [m/s2 ] (g) Absolute linear acceleration of the particle E. a a + B3 E B3 D }{{} B3 ω 3 B3 ω 3 B3 r DE + }{{} B3 ω 3 B3 r DE + 2 }{{} B3 ω 3 B3 v rel + }{{} B3 a }{{ rel } (I) (II) (I) B2 a D a B2 C }{{} see item (f) + B2 ω 2 B2 ω 2 B2 r }{{ CD + } B2 ω 2 B2 r CD +2 }{{} B2 ω 2 B2 v }{{ rel } (III) + B2 a rel }{{} (III) ω B2 2 i 2 j 2 k 2 ( α+ β) r h i 2 j 2 k 2 ( α+ β) r( α+ β) r( α+ β) 2 B2 a D b α 2 sinβ b α 2 cosβ r( α+ β) 2 [rad/s2 ]

41 1.8 Dynamics of One Single Particle in 3D Examples 41 B3 a D T ψ B2 a D 1 cosψ sinψ sinψ cosψ b α 2 sinβ b α 2 cosβ r( α+ β) 2 a B3 D b α 2 sinβ (b α 2 cosβ +r( α+ β) 2 )cosψ (b α 2 cosβ +r( α+ β) 2 )sinψ [rad/s2 ] (II) ω B3 3 i 3 j 3 k 3 ( α+ β)sinψ ( α+ β)cosψ l i 3 j 3 k 3 ( α+ β)sinψ ( α+ β)cosψ l( α+ β)sinψ l( α+ β) 2 sinψ cosψ l(( α+ β) 2 sin 2 ψ ) Adding the terms recently calculated, i.e. (I) and (II), one achieves the absolute linear acceleration of the particle E, representing such a vector with help of the reference frame B3: a B3 E b α 2 sinβ (b α 2 cosβ +r( α+ β) 2 )cosψ l( α+ β) 2 sinψ cosψ (b α 2 cosβ +r( α+ β) 2 )sinψ +l(( α+ β) 2 sin 2 ψ ) [rad/s2 ] (h) Free-body diagram of the particle E. Gravity force acting on the particle E: I P { mg } T Reaction force between particle and arm (normal force acting on the arm longitudinal direction): B3 T { T } T Reaction force between particle and arm (shear force acting perpendicularly to the arm): B3 R { R 1 R 2 } T

42 1.8 Dynamics of One Single Particle in 3D Examples 42 (i) Dynamic equilibrium of the particle E according to the second Newton s law: B3 F B3 R + B3 T + T ψt β T α I P }{{} B3 P B3 R + B3 T+ B3 P m B3 a E where B3 P 1 cosψ sinψ sinψ cosψ cosβ sinβ sinβ cosβ 1 cosα sinα sinα cosα 1 mg mgsinψ mgcosψ R 1 mgsinψ +R 2 mgcosψ +T m b α 2 sinβ (b α 2 cosβ +r( α+ β) 2 )cosψ l( α+ β) 2 sinψ cosψ (b α 2 cosβ +r( α+ β) 2 )sinψ +l(( α+ β) 2 sin 2 ψ ) The final result of the application of the second Newton s law is the achievement of three equations, i.e. three for calculating the dynamic reaction forces R 1, R 2, and T. Dynamic reaction forces: direction X 3 : R 1 mb α 2 sinβ direction Y 3 : R 2 m(gsinψ (b α 2 cosβ +r( α+ β) 2 )cosψ l( α+ β) 2 sinψ cosψ ) direction Z 3 : T m{gcosψ +[b α 2 cosβ +r( α+ β) 2 ]sinψ +l[( α+ β) 2 sin 2 ψ ]} Two Equations of Motion and One Dynamic Reaction Force Figure illustrates a tip mass m [Kg] located at point E. The tip mass is attached to point D by a linear spring of stiffness constant k [N/m] and initial length (non-deformed) l o [m]. The tip mass can simultaneously execute pendular and linear movements. The pendular motion executed by the tip mass follows the rotation of the arm around the point D and the linear motion occurs when the tip mass slides along the arm with negligible friction between mass and arm. The masses of the rotative arm and spring can be regarded negligible when compared to the tip mass. The pendular motion is described by the angular coordinate ψ(t) and the linear motion by the coordinate l(t) [m], referred from the spring length l o along the axis Z 3. The spring-mass system (pendulum) is mounted on the extremity of a disk-arm system of radius r [m] and hight h. The disk-arm system is mounted on the extremity of another rotating arm of length b [m]. The distance between the center of disk to the arm extremity is c [m]. The disk as well the arm rotate at constant angular velocity of α [rad/s] and β [rad/s], respectively. It is important to mention that all bodies (arms and disk) are rigid. Only the linear spring is considered flexible. The inertial reference frame I is attached to the base (point O). The moving reference frame B1 is attached to the extremity of the arm (point B). The moving

43 1.8 Dynamics of One Single Particle in 3D Examples 43 reference frame B2 is attached to the disk center (point C). The moving reference frame B3 is attached to the extremity of the second arm (point D). (a) Calculate the coordinate transformation matrices, which facilitate to transform the representation of the different vectors (displacement, velocity, acceleration, force, moment etc.) from the inertial reference frame to the moving reference frames or vice-versa. (b) Calculate the position vectors between points OA, AB, BC, CD and DE, representing them with help of the most convenient reference frames. Indicate how to write the position vector between points OE with help of the inertial reference frame I. Such a vector will be useful for describing the trajectory followed by the mass E. (c) Determine the absolute angular velocity of the reference frames B1, B2 and B3, representing such vectors with help of the respective reference frames, i.e. B1 ω 1, B2 ω 2 and B3 ω 3. (d) Determine the absolute linear velocity of the particle E, representing such a vector with help of the moving reference frame B3, i.e. B3 v E. (e) Determine the absolute angular acceleration of the reference frames B1, B2 and B3, representing such vectors with help of the respective reference frames, i.e. B1 ω 1, B2 ω 2 and B3 ω 3. (f) Determine the absolute linear acceleration of the particle E, representing such a vector with help of the reference frame B3, i.e B3 a E. (g) Draw the forces which are acting on the particle E, i.e. elaborate a free-body diagram. Afterwards, write the force vectors with help of the most convenient reference frame. (h) Write the equations responsible for describing the dynamic equilibrium of the particle E. What is the result of the set of equations? How many equations of motion and how many dynamic reaction forces? (i) Write a computational program in Matlab with the aim of solving the set of non-linear differential equations of motion obtained in (h). Choose 3 different initial conditions of motion and plot the variables l(t) and ψ(t) as a function of time. Plot the trajectory of the particle for the different initial conditions. Plot the dynamic reaction force as a function of time for the 3 different initial conditions.

44 1.8 Dynamics of One Single Particle in 3D Examples 44 Figure 1.4: Spring-mass system (pendulum) attached to the rotating reference frame B3 performing three consecutive rotations.

45 1.8 Dynamics of One Single Particle in 3D Examples 45 SOLUTION: (a) Coordinate transformation matrices. Coordinate transformation matrices between I and B1: cosα sinα T α sinα cosα 1 First rotation around the Z-axis of inertial reference frame. B1 s T α Is I s TT α B1 s Coordinate transformation matrices between B1 and B2: cosβ sinβ T β sinβ cosβ 1 Second rotation around the Z 1 -axis of the moving reference frame B1. B2 s T β B1 s B1 s TT β B2 s Coordinate transformation matrices between B2 and B3: T ψ 1 cosψ sinψ sinψ cosψ Third rotation around the X 2 -axis of the moving reference frame B2. B3 s T ψ B2 s B2 s TT ψ B3 s

46 1.8 Dynamics of One Single Particle in 3D Examples 46 (b) Position vectors. position vector I r OA : I r OA { a } T [m] position vector B1 r AB : B1 r AB { b } T [m] position vector B2 r BC : B2 r BC { c } T [m] position vector B2 r CD : B2 r CD { r h } T [m] position vector B3 r DE : B3 r DE { (l +l) } T [m] The position vectors were written with help of the most convenient reference frames, it means using the reference frame in which their representation are the most compact one. For describing the particle trajectory is necessary to write the vector I r OE, i.e. or I r OE I r OA + I r AB + I r BC + I r CD + I r DE I r OE I r OA +TT α B1 r AB +TT α TT β ( B2 r BC + B2 r CD )+TT α TT β TT ψ B3 r DE (c) Absolute angular velocity of the moving reference frames. Absolute angular velocity of the moving reference frame B1: cosα sinα ω B1 1 T α I α sinα cosα ω B1 1 1 α α [rad/s] Absolute angular velocity of the moving reference frame B2: B2 ω 2 T β T α I α+t β B1 β cos(α+β) sin(α+β) sin(α+β) cos(α+β) 1 α + cosβ sinβ sinβ cosβ 1 β

47 1.8 Dynamics of One Single Particle in 3D Examples 47 ω B2 2 α+ β [rad/s] Absolute angular velocity of the moving reference frame B3: B3 ω 3 T ψ T β T α I α+t ψ T β B1 β +Tψ B2 ψ T ψ ( B2 ω 2 + B2 ψ) 1 cosψ sinψ sinψ cosψ ψ α+ β ψ ( α+ β)sinψ ( α+ β)cosψ [rad/s] (d) Absolute linear velocity of the particle E. Absolute linear velocity of point B B1 v B B1 ω 1 B1 r AB i 1 j 1 k 1 α b b α [m/s] Absolute linear velocity of point C v v B1 C B1 B cosβ sinβ sinβ cosβ 1 B2 v C T β B1 v B b α v B2 C b αcosβ b αsinβ [m/s] Absolute linear velocity of point D v v + ω B2 D B2 C B2 2 B2 r CD + B2 v }{{ rel } b αcosβ b αsinβ + i 2 j 2 k 2 α+ β r h b αcosβ r( α+ β) b αsinβ [m/s]

48 1.8 Dynamics of One Single Particle in 3D Examples 48 Absolute linear velocity of the particle E v v + B3 E B3 D }{{} B3 ω 3 B3 r }{{ DE } (I) (II) (I) + B3 v rel }{{} (III) v T B3 D ψ B2 v D B3 v D 1 cosψ sinψ sinψ cosψ b αcosβ r( α+ β) b αsinβ b αcosβ r( α+ β) b αsinβcosψ b αsinβsinψ [m/s] (II) B3 ω 3 B3 r DE i 3 j 3 k 3 ψ ( α+ β)sinψ ( α+ β)cosψ (l +l) (l +l)( α+ β)sinψ (l +l) ψ (III) B3 v rel d dt ( B3 r DE ) l (I)+(II)+(III) v B3 E b αcosβ r( α+ β) (l +l)( α+ β)sinψ b αsinβcosψ +(l +l) ψ b αsinβsinψ l [m/s] (e) Absolute angular acceleration of the moving reference frames. Absolute angular acceleration of the moving reference frame B1: ω B1 1 d dt ( ω B1 1)+ B1 ω 1 B1 ω }{{} 1 ω B2 2 d dt ( ω B2 2)+ B2 ω 2 B2 ω }{{} 2 α ω B1 1 Absolute angular acceleration of the moving reference frame B2: α+ β ω B2 2 [rad/s2 ] [rad/s2 ]

49 1.8 Dynamics of One Single Particle in 3D Examples 49 Absolute angular acceleration of the moving reference frame B3: ω B3 3 d dt ( ω B3 3)+ B3 ω 3 B3 ω }{{} 3 ψ ω B3 3 ( α+ β)sinψ +( α+ ( α+ β)cosψ ( α+ β) ψcosψ ω B3 3 β) ψsinψ ψ ( α+ β) ψcosψ [rad/s2 ] ( α+ β) ψsinψ It is important to point out that α β. It is given in the beginning of the example, when the problem was formulated. (f) Absolute linear acceleration of the point E. Absolute linear acceleration of the point B. a a + B1 B B1 O }{{} B1 ω 1 ( B1 ω 1 B1 r OB )+ B1 ω 1 }{{} B1 r OB +2 B1 ω 1 B1 v rel + }{{} B1 a }{{ rel } ω B1 1 i 1 j 1 k 1 α b a i 1 j 1 k 1 α b α b α 2 a B1 B b α 2 [m/s2 ] Absolute linear acceleration of the point C. B1 a C B1 a B B2 a C T β B1 a B cosβ sinβ sinβ cosβ 1 b α 2 a B2 C b α 2 sinβ b α 2 cosβ [m/s2 ] Absolute linear acceleration of the particle D.

50 1.8 Dynamics of One Single Particle in 3D Examples 5 a a + ω B2 D B2 C B2 2 ( B2 ω 2 B2 r CD ) + }{{} B2 ω 2 B2 r CD +2 }{{} B2 ω 2 B2 v rel + }{{} B2 a }{{ rel } (I) (I) ω B2 2 i 2 j 2 k 2 ( α+ β) r h i 2 j 2 k 2 ( α+ β) r( α+ β) r( α+ β) 2 B2 a D b α 2 sinβ b α 2 cosβ r( α+ β) 2 [rad/s2 ] B3 a D T ψ B2 a D 1 cosψ sinψ sinψ cosψ b α 2 sinβ b α 2 cosβ r( α+ β) 2 a B3 D b α 2 sinβ (b α 2 cosβ +r( α+ β) 2 )cosψ (b α 2 cosβ +r( α+ β) 2 )sinψ [rad/s2 ] Absolute linear acceleration of the particle E. a a + ω B3 E B3 D B3 3 ( B3 ω 3 B3 r DE ) + }{{} B3 ω 3 B3 r DE + 2 }{{} B3 ω 3 B3 v rel + }{{} B3 a }{{ rel } (II) (III) (IV) (V) (II) ω B3 3 i 3 j 3 k 3 ψ ( α+ β)sinψ ( α+ β)cosψ (l +l) i 3 j 3 k 3 ψ ( α+ β)sinψ ( α+ β)cosψ (l +l)( α+ β)sinψ (l +l) ψ (l +l) ψ( α+ β)cosψ (l +l)( α+ β) 2 sinψcosψ (l +l)( ψ 2 +( α+ β) 2 sin 2 ψ)

51 1.8 Dynamics of One Single Particle in 3D Examples 51 (III) ω r B3 3 B3 DE i 3 j 3 k 3 ψ ( α+ β) ψcosψ ( α+ β) ψsinψ (l +l) (l +l)( α+ β) ψcosψ (l +l) ψ (IV) 2 i 3 j 3 k 3 ψ ( α+ β)sinψ ( α+ β)cosψ l 2 l( α+ β) ψsinψ 2 l ψ (V) B3 a rel d2 dt 2 ( B3 r DE ) l Adding the terms determined in (I), (II), (III), (IV) and (V), one achieves the absolute linear acceleration of the particle E represented in the reference frame B3: a B3 E b α 2 sinβ (l +l) ψ( α+ β)cosψ 2 l( α+ β)sinψ (b α 2 cosβ +r( α+ β) 2 )cosψ (l +l)( α+ β) 2 sinψcosψ +(l +l) ψ +2 l ψ (b α 2 cosβ +r( α+ β) 2 )sinψ +(l +l)( ψ 2 +( α+ β) 2 sin 2 ψ) l [m/s2 ] (g) Free-body diagram of the particle E. Gravity force acting on the particle E: I P { mg } T Force between the particle and the linear spring acting on the arm longitudinal direction (considering the linear stiffness coefficient of the spring k ): B3 T { kl } T Reaction force between particle and arm (shear force acting perpendicularly to the arm): B3 R { R } T

52 1.8 Dynamics of One Single Particle in 3D Examples 52 (h) Dynamic equilibrium of the particle E according to the second Newton s law: B3 F B3 R + B3 T + T ψt β T α I P }{{} B3 P B3 R + B3 T+ B3 P m B3 a E B3 P 1 cosψ sinψ sinψ cosψ cosβ sinβ sinβ cosβ 1 cosα sinα sinα cosα 1 mg mgsinψ mgcosψ m R mgsinψ mgcosψ +kl b α 2 sinβ (l +l) ψ( α+ β)cosψ 2 l( α+ β)sinψ (b α 2 cosβ +r( α+ β) 2 )cosψ (l +l)( α+ β) 2 sinψcosψ +(l +l) ψ +2 l ψ (b α 2 cosβ +r( α+ β) 2 )sinψ +(l +l)( ψ 2 +( α+ β) 2 sin 2 ψ) l The final result of the application of the second Newton s law is the achievement of three equations, i.e. one for calculating the dynamic reaction forces R and two additional equations responsible for describing the movement of the particle E as a function of the time, ψ(t) and l(t). Dynamic reaction force (direction X 3 ) : R m(b α 2 sinβ +(l +l) ψ( α+ β)cosψ +2 l( α+ β)sinψ) Equations of motion for the particle E (directions Y 3 and Z 3 ) ψ g (l +l) sinψ + (b α2 cosβ +r( α+ β) 2 ) cosψ +( α+ (l +l) β) 2 2 l ψ sinψcosψ (l +l) l k m l+gcosψ +[b α2 cosβ +r( α+ β) 2 ]sinψ +(l +l)[ ψ 2 +( α+ β) 2 sin 2 ψ] (i) Computational routine implemented using Matlab and numerical results. clear all close all % parameters of the mechanical system a.5; % [m] b.5; % [m] c.5; % [m] r.1; % [m] h.2; % [m]

53 1.8 Dynamics of One Single Particle in 3D Examples 53 lo.1; % [m] m.2; % [kg] k 2; % [N/m] g 9.82; % [m/s^2] % initial conditions alfa(1) % [rad] beta(1) % [rad] alfad 2*pi/2 % [rad/s] betad 2*pi*4 % [rad/s] l(1).; % [m] ld(1) ; % [m/s] psi(1) pi/6; % [rad] psid(1) ; % [rad/s] % defining the position vectors with respect to desired reference frames Iroa [; ; a]; B1rab [; b; ]; B2rbc [; ; c]; B2rcd [; r; h]; B3rde [; ; -(lo+l(1))]; % defining the transformation matrices Talfa [cos(alfa(1)) sin(alfa(1)) -sin(alfa(1)) cos(alfa(1)) 1]; Tbeta [cos(beta(1)) sin(beta(1)) -sin(beta(1)) cos(beta(1)) 1]; Tpsi [1 cos(psi(1)) sin(psi(1)); -sin(psi(1)) cos(psi(1))]; % creating the position vector which describes mass m s trajectory in the inertial frame roe Iroa + Talfa *B1rab + Talfa *Tbeta *(B2rbc+B2rcd) + Talfa *Tbeta *Tpsi *B3rde; rx(1) roe(1); ry(1) roe(2); rz(1) roe(3); % calculating the reaction force R represented in the X3 direction Reaction(1) m*( b*alfad^2*sin(beta(1))... + (lo+l(1))*psid(1)*(alfad+betad)*cos(psi(1)) *ld(1)*(alfad+betad)*sin(psi(1))); % numerical solution % time step deltat.5;

54 1.8 Dynamics of One Single Particle in 3D Examples 54 % number of integration points n_int 6; for i2:n_int, %tempo(i-2)*deltat t_int(i-1) (i-2)*deltat; % accelerations psidd(i-1) -g/(lo+l(i-1))*sin(psi(i-1)) +... (b*alfad^2*cos(beta(i-1))+r*(alfad+betad)^2)/(lo+l(i-1))*cos(psi(i-1)) +.. (alfad+betad)^2*sin(psi(i-1))*cos(psi(i-1)) *ld(i-1)*psid(i-1)/(lo+l(i-1)); ldd(i-1) -k/m*l(i-1)... + (g*cos(psi(i-1))+(b*alfad^2*cos(beta(i-1)))... + r*(alfad+betad)^2)*sin(psi(i-1))... + (lo+l(i-1))*(psid(i-1)^2+(alfad+betad)^2*sin(psi(i-1))^2); % velocities alfa(i) alfa(i-1) + alfad*deltat; beta(i) beta(i-1) + betad*deltat; psid(i) psid(i-1) + psidd(i-1)*deltat; ld(i) ld(i-1) + ldd(i-1)*deltat; % displacements psi(i) psi(i-1) + psid(i-1)*deltat; l(i) l(i-1) + ld(i-1)*deltat; % reaction force Reaction(i) m*( b*alfad^2*sin(beta(i))... + (lo+l(i))*psid(i)*(alfad+betad)*cos(psi(i)) *ld(i)*(alfad+betad)*sin(psi(i))); % transformation matrices Talfa [ cos(alfa(i)) sin(alfa(i)) -sin(alfa(i)) cos(alfa(i)) 1]; Tbeta [ cos(beta(i)) sin(beta(i)) -sin(beta(i)) cos(beta(i)) 1]; Tpsi [1 cos(psi(i)) sin(psi(i)) -sin(psi(i)) cos(psi(i))]; % defining the position vectors with respect to desired reference frames Iroa [; ; a]; B1rab [; b; ];

55 1.8 Dynamics of One Single Particle in 3D Examples 55 end B2rbc [; ; c]; B2rcd [; r; h]; B3rde [; ; -(lo+l(i))]; % observe that only B3rde is time dependent! % creating the position vector which describes mass m s trajectory in the inertial frame roe Iroa + Talfa *B1rab + Talfa *Tbeta *(B2rbc+B2rcd) + Talfa *Tbeta *Tpsi *B3rde; rx(i) roe(1); ry(i) roe(2); rz(i) roe(3); nplotn_int-1; figure(1) plot(t_int(1:nplot),l(1:nplot)) grid xlabel( time [s] ) ylabel( l(t) [s] ) figure(2) plot(t_int(1:nplot),psi(1:nplot)) grid xlabel( time [s] ) ylabel( \psi(t) [s] ) figure(3) plot3(rx,ry,rz) axis([ ]) xlabel( x [m] ) ylabel( y [m] ) zlabel( z [m] ) grid figure(4) plot(t_int(1:nplot),reaction(1:nplot)) grid xlabel( time [s] ) ylabel( R(t) [N] ) Figure illustrates the results obtained aided by the computational routine implemented using Matlab.

56 1.8 Dynamics of One Single Particle in 3D Examples l(t) [s].15 ψ(t) [s] (a) time [s] (b) time [s] (c) (d) Figure 1.5: (a) Angle ψ(t) as function of time; (b) displacement l(t) as function of time; (c) trajectory described by the particle E when the arm and the disk rotate with a constant angular velocity given in the Matlab routine; (d) reaction force R(t) as function of time.

57 1.9 Dynamics of a System of Particles in 2D Example Dynamics of a System of Particles in 2D Example The aim of this example is to illustrate the procedure to achieving the equation of motion and the dynamic reaction forces in a system of particles. In this case, the system is composed of 2 particles of masses m 1 and m 2. The dynamic reaction forces can only be predicted when the equations of motion are solved. Such equations are non-linear second-order differential equations and are responsible for describing the dynamic behavior of the particle in the time domain. Such equations are numerically solved for some conditions and the results are presented in graphics and computer animations. Afterwards, some of the movements simulated and animated can be qualitatively compared to those described by a prototype of double pendulum. Two ways of achieving the equations of motion will be illustrated: firstly using only one moving reference frame and afterwards using two moving reference frames. Such an example will clearly show the compactness and small size of the mathematical expressions when using several moving reference frames and representing acceleration and force vectors with help of the appropriate frames. The results can easily be applied to optimize the utilization of softwares, like MAPLE or MATHEMATICA, for achieving equations of motion. (a) (b) Figure 1.6: Double pendulum (a) physical system with a moving reference frame attached to mass m 1 ; (b) mechanical model composed of two particles of masses m 1 and m Solution Using One Single Moving Reference Frame The double pendulum shown in Figure 1.9 is built by two masses m 1 [Kg] and m 2 [Kg]. Observing the absolute movement of particle 2, one gets an impression that such movement is extremely complicated. Nevertheless, using the moving reference frame attached to mass 1, one can see that the movement of mass 2 is a composition of two planar circular movements, i.e. mass 1 rotates around point O (origin of the inertial frame) and mass 2 rotates around mass 1. Such an example clarifies, why moving reference frames are so useful: they facilitate the representation of complicated movements accomplished by particles and bodies.

58 1.9 Dynamics of a System of Particles in 2D Example 58 (1) Aiming at describing the movements of the particles, two coordinate reference frames are used, as it can be seen in Figure??: Inertial reference frame I, represented by the unit vectors i, j, k; Moving reference frame B1, represented by the unit vectors i 1, j 1, k 1. The moving reference frame B1 is attached to the mass m 1, rotating at the angular velocity of θ 1 [ras/s] around the Z-axis. The position of the mass m 1 can be defined as function of the the angle θ 1 (t) and the length l 1. It is taken into consideration that the arm l 1 is rigid and massless. (2) Once one works with two reference frames, it is useful to calculate the coordinate transformation matrix, which makes possible to transform the representation of a vector from the inertial reference frame to the moving one: i 1 cosθ 1 i+sinθ 1 j+ k j 1 sinθ 1 i+cosθ 1 j+ k Positive rotation of the moving reference frame B1 around the Z-axis. k 1 i+ j+1 k i 1 j 1 k 1 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 B1 s T θ 1. I s I s T T θ 1 B1 s i j k (3) After defining the reference frames and transformation matrix, the vectors can be easily written. Firstly, the position vectors l 1 and l 2 are written. The vector B1 l 1 can be easily written with help of the moving reference frame B1: l B1 1 l 1 Transforming its representation into the inertial frame by using the transformation matrix, one gets: I l 1 T T θ 1. B1 l 1 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 l 1 l 1 sinθ 1 l 1 cosθ 1 The vector B1 l 2 can be easily written using the moving reference frame B1: l 2 sinθ 2 l B1 2 l 2 cosθ 2

59 1.9 Dynamics of a System of Particles in 2D Example 59 Transforming its representation into the inertial frame by using the transformation matrix, one gets: l 2 (sinθ 2 cosθ 1 +cosθ 2 sinθ 1 ) l 2 sin(θ 1 +θ 2 ) l I 2 T T. θ B1 l 2 l 2 ( sinθ 2 sinθ 1 +cosθ 2 cosθ 1 ) 1 l 2 cos(θ 1 +θ 2 ) Following the absolute angular velocity and acceleration vectors of the moving reference frame B1 is defined. The absolute angular velocity of B1 can be easily written, knowing that it rotates around the Z-axis of the inertial frame: I θ 1 θ 1 The absolute angular acceleration of B1 can be written as: I θ 1 θ 1 The relative angular velocity of the mass m 2 can be easily written with help of the reference frame B1, knowing that it rotates around the Z1-axis of the moving reference frame. Afterwards its representation can be transformed into the inertial frame: B1 θ 2 θ 2 I θ 2 T T θ 1. B1 θ2 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 θ 2 Similarly, the absolute angular acceleration vector of the particle m 2 can be described with help of the reference frame B1 and afterwards its representation can be transformed into the inertial frame: B1 θ 2 θ 2 θ I 2 T T. θ B1 θ2 1 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 θ 2 θ 2 θ 2 (4) After defining the position vectors and angular velocity of the the reference frame, the absolute linear velocity vectors of the particles m 1 and m 2 can be calculated. The positions of the masses m 1 and m 2 are coincident with the position of the points B and A, respectively. v v + θ I A I B I 1 I l 2 + I v RelAB v θ I B I 1 I l 1 i j k θ1 l 1 sinθ 1 l 1 cosθ 1 θ 1 l 1 cosθ 1 θ 1 l 1 sinθ 1

60 1.9 Dynamics of a System of Particles in 2D Example 6 I θ 1 I l 2 I v RelAB TT θ 1 d dt ( B1l 2 ) I v A i j k θ1 l 2 sin(θ 1 +θ 2 ) l 2 cos(θ 1 +θ 2 ) cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 θ 2 l 2 cosθ 2 θ 2 l 2 sinθ 2 d dt θ 1 l 2 cos(θ 1 +θ 2 ) θ 1 l 2 sin(θ 1 +θ 2 ) l 2 sinθ 2 l 2 cosθ 2 θ 1 l 1 cosθ 1 θ 1 l 2 cos(θ 1 +θ 2 ) θ 2 l 2 cos(θ 1 +θ 2 ) θ 1 l 1 sinθ 1 θ 1 l 2 sin(θ 1 +θ 2 ) θ 2 l 2 sin(θ 1 +θ 2 ) θ 2 l 2 cos(θ 1 +θ 2 ) θ 2 l 2 sin(θ 1 +θ 2 ) (5) After defining the absolute linear velocity vectors, v and v, the absolute linear I A I B acceleration vectors of the masses m 1 and m 2 can be calculated, coinciding with the points B and A, respectively. a a + θ I A I B I 1 θ1 I I l 2 + I θ1 I l θ1 I I v RelAB + I a RelAB Calculating each of the terms of the acceleration equation, one achieves: a θ I B I 1 θ1 I I l 1 + I θ1 I l 1 θ 1 2l 1sinθ 1 θ 1 l 1 cosθ 1 θ 1 2l 1cosθ 1 θ 1 l 1 sinθ 1 a Bx a By I θ 1 I θ1 I l 2 θ 2 1 l 2sin(θ 1 +θ 2 ) θ 2 1l 2 cos(θ 1 +θ 2 ) I θ 1 I l 2 2. θ1 I I v RelAB I a RelAB TT θ 1 d 2 θ 1 l 2 cos(θ 1 +θ 2 ) θ 1 l 2 sin(θ 1 +θ 2 ) dt 2 ( B1l 2 ) 2 θ 1 θ2 l 2 sin(θ 1 +θ 2 ) 2 θ 1 θ2 l 2 cos(θ 1 +θ 2 ) cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 d2 dt 2 l 2 sinθ 2 l 2 cosθ 2

61 1.9 Dynamics of a System of Particles in 2D Example 61 θ 2 2 l 2sin(θ 1 +θ 2 ) θ 2 l 2 cos(θ 1 +θ 2 ) θ 2 2 l 2cos(θ 1 +θ 2 ) θ 2 l 2 sin(θ 1 +θ 2 ) Adding all the terms, a very long mathematical expression is achieved for representing the absolute linear acceleration of the particle A. Such expression is written with help of the inertial reference frame: a I A θ 2 1 l 1sinθ 1 θ 1 l 1 cosθ 1 + θ 2 1 l 2sin(θ 1 +θ 2 ) θ 1 l 2 cos(θ 1 +θ 2 ) θ 2 1l 1 cosθ 1 θ 1 l 1 sinθ 1 θ 2 1l 2 cos(θ 1 +θ 2 ) θ 1 l 2 sin(θ 1 +θ 2 ) 2 θ 1 θ2 l 2 sin(θ 1 +θ 2 )+ θ 2 2 l 2sin(θ 1 +θ 2 ) θ 2 l 2 cos(θ 1 +θ 2 ) 2 θ 1 θ2 l 2 cos(θ 1 +θ 2 ) θ 2 2l 2 cos(θ 1 +θ 2 ) θ 2 l 2 sin(θ 1 +θ 2 ) a Ax a Ay (6) Calculation of the reaction forces and equations of motion based on the second and third Newton s laws. Drawing the free-body diagram of the two particles, the forces can be represented in the vector form: weight force vector represented with help of the inertial reference frame I: P I B m 1 g P I A m 2 g Free-body diagram of the two particles. Reaction force vectors acting on the pendulum arms, represented with help of the moving reference frame B1: For mass m 1 : T B1 1 T 1 T I 1 T T. θ B1 T 1 1 T 1 sinθ 1 T 1 cosθ 1 For mass m 2 : T B1 2 T 2 sinθ 2 T 2 cosθ 2 T I 2 T θ1. B1 T 2 T 2 sin(θ 1 +θ 2 ) T 2 cos(θ 1 +θ 2 ) Using the second Newton s law for each one of the particles, one achieves: F m a I B 1 I B I P B + I T 1 I T 2 m 1 I a B

62 1.9 Dynamics of a System of Particles in 2D Example 62 T 1 sinθ 1 T 2 sin(θ 1 +θ 2 ) m 1 g T 1 cosθ 1 +T 2 cos(θ 1 +θ 2 ) m 1 a Bx a By (1.25) F m a I A 2 I A m 2 g + I P A + I T 2 m 2 I a A T 2 sin(θ 1 +θ 2 ) T 2 cos(θ 1 +θ 2 ) m 2 a Ax m 2 a Ay (1.26) Solving the four equations obtained from eq.(1.25) and (1.26), one gets the analytical expressionsfortheforcesactingonthearmsofthependulumt 1 andt 2,andadditionallytwoequations of motion, θ 1 and θ 2, which will be depending on the time and on the initial conditions of the movements of the masses m 1 and m 2. It is important to emphasize that the solution of the problems using the inertial reference frame avoids mistakes while differentiating the position vectors for obtaining the absolute acceleration vectors. Nevertheless, such a solution results in mathematical expressions for the absolute acceleration vectors extremely long, difficult to be handled later on. In most practical situations, solving the problems with help of the moving reference frames, attached to each one of the particles, allows the achievement of shorter and more compact mathematical expressions for the acceleration vectors. Such shorter expressions are much easier to be handled, interpreted and understood. Moreover, when working with software of symbolic manipulation, like MAPLE and MATHEMATICA, it is very advantageous to describe such vectors with help of the moving reference frames. The shorter mathematical expressions for the acceleration vectors, and also the shorter mathematical expressions for the reaction force vectors contribute to increase the rapidity of the computational calculations of equation of motion.

63 1.9 Dynamics of a System of Particles in 2D Example Solution Using Two Moving Reference Frames Solving the problem with help of two reference frames, one can illustrate the compactness of the mathematical expressions. Let us define two moving reference frames: B1 attached to the mass 1 (point B) and B2 attached to mass 2 (point A), as it can be seen in figure on your right hand side. Reference frames: I Inertial reference frame defined by the unit vectors i j k; B1 Moving reference frame attached to the mass m 1, defined by the unit vector i 1 j 1 k 1 with origin in the point B; B2 Moving reference frame attached to the mass m 2, defined by the unit vector i 2 j 2 k 2 with origin in the point B; Coordinate transformation matrices: From the inertial reference frame I into B1 and vice-versa: T θ1 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 B1 s T θ 1. I s I s TT θ 1 B1 s From the moving reference frame B1 into B2 and vice-versa: T θ2 cosθ 2 sinθ 2 sinθ 2 cosθ 2 1 B2 s T θ 2. B1 s B1 s TT θ 2 B2 s Position vectors: l B1 1 l 1 l B2 2 l 2 Absolute angular velocity vector of the moving reference frames B1 and B2: angular velocity vector of B1 cosθ 1 sinθ 1 ω B1 1 T θ1 I θ1 sinθ 1 cosθ 1 1 θ 1 θ 1

64 1.9 Dynamics of a System of Particles in 2D Example 64 angular velocity vector of B2 B2 ω 2 T θ2 T θ1 I θ1 +T θ2 B1 θ2 cosθ 2 sinθ 2 sinθ 2 cosθ 2 1 cosθ 1 sinθ 1 sinθ 1 cosθ 1 1 θ cosθ 2 sinθ 2 sinθ 2 cosθ 2 1 θ 2 θ 1 + θ 2 Absolute angular acceleration vectors of the moving reference frames B1 e B2: absolute angular acceleration of B1: ω B1 1 θ 1 absolute angular acceleration of B2: ω B2 2 θ 1 + θ 2 Absolute linear acceleration vectors of the masses m 1 and m 2 (points B and A), represented with help of the reference frames B1 and B2, respectively: a a B1 B B1 }{{} + B1 ω 1 B1 ω 1 B1 l 1 + B1 ω 1 B1 l B1 ω 1 B1 v Rel + B1 a Rel }{{} a ω ω l B1 B B1 1 B1 1 B1 1 + B1 ω 1 B1 l 1 l 1 θ1 l 1 θ2 1 a a + ω B2 A B2 B B2 2 B2 ω 2 B2 l 2 + B2 ω 2 B2 l B2 ω 2 B2 v Rel + B2 a }{{ Rel } B2 a A T θ 2 B1 a B + B2 ω 2 B2 ω 2 B2 l 2 + B2 ω 2 B2 l 2 B2 a A l 1 ( θ 1 cosθ 2 + θ 2 1 sinθ2 ) l 2 ( θ 1 + θ 2 ) l 1 ( θ 1 sinθ 2 θ 2 1 cosθ2 ) l 2 ( θ 1 + θ 2 ) 2

65 1.9 Dynamics of a System of Particles in 2D Example 65 Reaction forces acting on the masses m 1 and m 2 (points B and A), represented with help of the reference frames B1 and B2, respectively: weight forces P T B1 B θ 1 m 1 g P T B2 A θ 2 T θ1 m 2 g m 1 g sinθ 1 m 1 g cosθ 1 m 2 g sin(θ 1 +θ 2 ) m 2 g cos(θ 1 +θ 2 ) reaction forces acting on the arm of length l 1, represented with help of the moving reference frame B1: T B1 1 T 1 reaction forces acting on the arm of length l 2, represented with help of the moving reference frame B2: T B2 2 T 2 Dynamic equilibrium between the two particles: For mass m 1 (particle B) B1 F B m 1 B1 a B m 1 g sinθ 1 T 2 sinθ 2 m 1 g cosθ 1 T 1 +T 2 cosθ 2 B1 P B + B1 T 1 B1 T 2 m 1 B1 a B m 1 l 1 θ1 l 1 θ2 1 (1.27) For mass m 2 (particle A) F m a B2 A 2 B2 A m 2 g sin(θ 1 +θ 2 ) m 2 g cos(θ 1 +θ 2 ) T 2 B2 P A + B2 T 2 m 2 B2 a A m 2 l 1 ( θ 1 cosθ 2 + θ 2 1 sinθ2 ) l 2 ( θ 1 + θ 2 ) l 1 ( θ 1 sinθ 2 θ 2 1 cosθ2 ) l 2 ( θ 1 + θ 2 ) 2 (1.28)

66 1.9 Dynamics of a System of Particles in 2D Example 66 Rewriting eq.(1.27) and (1.28) in a matrix form, one achieves: sinθ 2 m 1 l 1 1 cosθ 2 m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 1 m 2 l 1 sinθ 2 T 1 T 2 θ 1 θ 2 m 1 gsinθ 1 m 1 l 1 θ1 2 m 1gcosθ 1 m 2 [gsin(θ 1 +θ 2 )+l 1 θ1 2sinθ 2] m 2 [gcos(θ 1 +θ 2 )+l 1 θ1 2cosθ 2 +l 2 ( θ 1 + θ 2 ) 2 ] Solving the system by Cramer, one obtains the dynamic reaction forces acting on the arms of the pendulum, T 1 and T 2, and additionally the non-linear second order differential equations of motion, θ 1 and θ 2, which describe the movements of the particles in the time domain: REACTION FORCE T 1 T 1 m 1 gsinθ 1 sinθ 2 m 1 l 1 m 1 l 1 θ1 2 m 1 gcosθ 1 cosθ 2 m 2 [gsin(θ 1 +θ 2 )+l 1 θ 1sinθ 2 2 ] m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 m 2 [gcos(θ 1 +θ 2 )+l 1 θ1cosθ 2 2 +l 2 ( θ 1 + θ 2 ) 2 ] 1 m 2 l 1 sinθ 2 sinθ 2 m 1 l 1 1 cosθ 2 m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 1 m 2 l 1 sinθ 2 or, T 1 ( θ 2 l2l 1 1 2m 2 m 1 2 gl 1 l 2 m 2 m 1 2cosθ 1 θ 2 l 1 1l 2 m 2 1m 2 cosθ θ 1 θ2 l 1 l 2 m 2 1m 2 cosθ 2 2 θ 2 l 2 1l 2 m 2 1m 2 cosθ 2 2 θ 2 1 l2l 1 2m 1 m 2 2 cos2 θ 2 gl 1 l 2 m 1 m 2 cosθ 2 2cos(θ 1 +θ 2 ) gl 1 l 2 m 1 m 2 cosθ 2 2sinθ 1 sinθ 2 θ 2 l2l 1 1 2m 1 m 2 sin2 θ 2 2 gl 1 l 2 m 1 m 2 cosθ 2 1sin 2 θ 2 )/ ( l 1 l 2 m 1 m 2 l 1 l 2 m 2 sin2 θ 2 2 ) (1.29) REACTION FORCE T 2 T 2 m 1 gsinθ 1 m 1 l 1 1 m 1 l 1 θ1 2 m 1 gcosθ 1 m 2 [gsin(θ 1 +θ 2 )+l 1 θ 1sinθ 2 2 ] m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 m 2 [gcos(θ 1 +θ 2 )+l 1 θ1cosθ 2 2 +l 2 ( θ 1 + θ 2 ) 2 ] m 2 l 1 sinθ 2 sinθ 2 m 1 l 1 1 cosθ 2 m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 1 m 2 l 1 sinθ 2

67 1.9 Dynamics of a System of Particles in 2D Example 67 or, T 2 ( θ 2 1 l 1l 2 2 m 1m θ 1 θ2 l 1 l 2 2 m 1m 2 2 θ 2 2 l 1l 2 2 m 1m 2 2 θ 2 1 l2 1 l 2m 1 m 2 2 cosθ 2 gl 1 l 2 m 1 m 2 2 cos(θ 1 +θ 2 ) gl 1 l 2 m 1 m 2 2 sinθ 1sinθ 2 )/( l 1 l 2 m 1 m 2 l 1 l 2 m 2 2 sin2 θ 2 ) (1.3) FIRST EQUATION OF MOTION θ 1 θ 1 sinθ 2 m 1 gsinθ 1 1 cosθ 2 m 1 l 1 θ 1 2 m 1 gcosθ 1 m 2 [gsin(θ 1 +θ 2 )+l 1 θ 1sinθ 2 2 ] m 2 l 2 1 m 2 [gcos(θ 1 +θ 2 )+l 1 θ 1cosθ 2 2 +l 2 ( θ 1 + θ 2 ) 2 ] sinθ 2 m 1 l 1 1 cosθ 2 m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 1 m 2 l 1 sinθ 2 or, θ 1 (gl 2 m 1 m 2 sinθ 1 θ 2 1 l2 2 m2 2 sinθ 2 2 θ 1 θ2 l 2 2 m2 2 sinθ 2 θ 2 2 l2 2 m2 2 sinθ 2 θ 2 1 l 1l 2 m 2 2 cosθ 2sinθ 2 gl 2 m 2 2 cos(θ 1 +θ 2 )sinθ 2 )/( l 1 l 2 m 1 m 2 l 1 l 2 m 2 2 sin2 θ 2 ) (1.31) SECOND EQUATION OF MOTION θ 2 θ 2 sinθ 2 m 1 l 1 m 1 gsinθ 1 1 cosθ 2 m 1 l 1 θ 1 2 m 1 gcosθ 1 m 2 (l 1 cosθ 2 +l 2 ) m 2 [gsin(θ 1 +θ 2 )+l 1 θ 1sinθ 2 2 ] 1 m 2 l 1 sinθ 2 m 2 [gcos(θ 1 +θ 2 )+l 1 θ 1cosθ 2 2 +l 2 ( θ 1 + θ 2 ) 2 ] sinθ 2 m 1 l 1 1 cosθ 2 m 2 (l 1 cosθ 2 +l 2 ) m 2 l 2 1 m 2 l 1 sinθ 2 or,

68 1.9 Dynamics of a System of Particles in 2D Example 68 θ 2 ( gl 2 m 1 m 2 sinθ 1 gl 1 m 1 m 2 cosθ 2 sinθ 1 + θ 2 l2m m sinθ + θ l2m2sinθ θ 1 θ2 l 2 m2sinθ θ 2 l2m2sinθ θ 2 l 1 1l 2 m 2 cosθ 2 2sinθ 2 +2 θ 1 θ2 l 1 l 2 m 2 cosθ 2 2sinθ 2 + θ 2 l l m2cosθ sinθ + θ l2m2cos2 θ sinθ 2 +gl 2 m 2 cos(θ +θ )sinθ gl 1 m 2 cosθ 2 2cos(θ 1 +θ 2 )sinθ 2 + θ 2 l2m2sin3 θ gl 1 m 1 m 2 sin(θ 1 +θ 2 )+ gl 1 m 2 sin2 θ 2 2 sin(θ 1 +θ 2 ))/( l 1 l 2 m 1 m 2 l 1 l 2 m 2 sin2 θ 2 2 ) (1.32) Computational Implementation and Numerical Simulation program pend2; uses crt; {DEFINITION OF THE VARIABLES} var m1, m2 : real; l1, l2 : real; t1, t2 : real; dt1, dt2 : real; ddt1, ddt2 : real; Tr1, Tr2 : real; t, tempo, deltat, freq : real; h1x, h1y, h2x, h2y : real; arq1, arq2, arq3, arq4 : text; ni,n,fim,nj,na : integer; {DEFINITION OF THE CONSTANTS} const g 9.81; {gravity acceleration [m/s 2 ]} {DEFINITION OF THE FUNCTIONS} function power(base:real;exp:integer):real; var iii:integer; pow:real; begin pow:1; for iii:1 to exp do begin pow:pow*base; end; power:pow; end; {BEGIN OF PROGRAM} begin

69 1.9 Dynamics of a System of Particles in 2D Example 69 clrscr; {PARAMETERS RELATED TO THE NUMERICAL INTEGRATION} t :.; {initial time [s]} tempo : 4.; {total time [s]} n : 25; {number of integration points} na : 1; {number of integration repetition} deltat : (tempo/na)/n; {integration step [s]} fim : 1; {criterium for closing the files} {NAME OF THE FILES WHERE THE RESULTS WILL BE SAVED} assign(arq1, arqt1.res ); assign(arq2, arqt2.res ); assign(arq3, arqh1.res ); assign(arq4, arqh2.res ); {PARAMETERS OF THE DOUBLE PENDULUM} m1 :.46; {mass 1 [Kg]} m2 :.46; {mass 2 [Kg]} l1 :.7; {length of the arm 1 [m]} l2 :.7; {length of the arm 2 [m]} {MOVEMENT INITIAL CONDITIONS} t1 : /4.; {initial value of theta 1 [rad]} t2 : /4.; {initial value of theta 2 [rad]} dt1 :.; {initial angular velocity of mass 1 [rad/s]} dt2 :.; {initial angular velocity of mass 2 [rad/s]} h1x : (-l1 * sin (t1)); {initial position of mass 1 in X-direction} h1y : -(l1 * cos (t1)); {initial position of mass 1 in Y-direction} h2x : (-l1 * sin (t1) - l2 * sin (t1 + t2)); {initial position of mass 2 in X-direction} h2y : -(l1 * cos (t1) + l2 * cos (t1 + t2)); {initial postion of mass 2 in Y-direction} {OPENING FILES} rewrite(arq1); rewrite(arq2); rewrite(arq3); rewrite(arq4); {SAVING INITIAL CONDITIONS IN THE FILE} writeln(arq1,t,,t1); writeln(arq2,t,,t2); writeln(arq3,h1x,,h1y); writeln(arq4,h2x,,h2y); {BEGINNING OF THE INTEGRATION IN TIME DOMAIN} for nj:1 to na do begin for ni:2 to n do begin t : t + deltat; {time change} {ACCELERATION OF MASS 1} ddt1:(g*l2*m1*m2*sin(t1) - Power(dt1,2)*Power(l2,2)*Power(m2,2)* Sin(t2) - 2*dt1*dt2*Power(l2,2)*Power(m2,2)*Sin(t2) - Power(dt2,2)*Power(l2,2)*Power(m2,2)*Sin(t2) - Power(dt1,2)*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) -

70 1.9 Dynamics of a System of Particles in 2D Example 7 g*l2*power(m2,2)*cos(t1 + t2)*sin(t2))/ (-(l1*l2*m1*m2) - l1*l2*power(m2,2)*power(sin(t2),2)); {ACCELERATION OF MASS 2} ddt2:(-(g*l2*m1*m2*sin(t1)) - g*l1*m1*m2*cos(t2)*sin(t1) + Power(dt1,2)*Power(l1,2)*m1*m2*Sin(t2) + Power(dt1,2)*Power(l2,2)*Power(m2,2)*Sin(t2) + 2*dt1*dt2*Power(l2,2)*Power(m2,2)*Sin(t2) + Power(dt2,2)*Power(l2,2)*Power(m2,2)*Sin(t2) + 2*Power(dt1,2)*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) + 2*dt1*dt2*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) + Power(dt2,2)*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) + Power(dt1,2)*Power(l1,2)*Power(m2,2)*Power(Cos(t2),2)*Sin(t2) + g*l2*power(m2,2)*cos(t1 + t2)*sin(t2) + g*l1*power(m2,2)*cos(t2)*cos(t1 + t2)*sin(t2) + Power(dt1,2)*Power(l1,2)*Power(m2,2)*Power(Sin(t2),3) + g*l1*m1*m2*sin(t1 + t2) + g*l1*power(m2,2)*power(sin(t2),2)*sin(t1 + t2))/ (-(l1*l2*m1*m2) - l1*l2*power(m2,2)*power(sin(t2),2)); {DYNAMIC REACTION FORCE ACTING ON ARM 1} Tr1:(-(Power(dt1,2)*Power(l1,2)*l2*Power(m1,2)*m2) - g*l1*l2*power(m1,2)*m2*cos(t1) - Power(dt1,2)*l1*Power(l2,2)*m1*Power(m2,2)*Cos(t2) - 2*dt1*dt2*l1*Power(l2,2)*m1*Power(m2,2)*Cos(t2) - Power(dt2,2)*l1*Power(l2,2)*m1*Power(m2,2)*Cos(t2) - Power(dt1,2)*Power(l1,2)*l2*m1*Power(m2,2)*Power(Cos(t2),2) - g*l1*l2*m1*power(m2,2)*cos(t2)*cos(t1 + t2) - g*l1*l2*m1*power(m2,2)*cos(t2)*sin(t1)*sin(t2) - Power(dt1,2)*Power(l1,2)*l2*m1*Power(m2,2)*Power(Sin(t2),2) - g*l1*l2*m1*power(m2,2)*cos(t1)*power(sin(t2),2))/ (-(l1*l2*m1*m2) - l1*l2*power(m2,2)*power(sin(t2),2)); {DYNAMIC REACTION FORCE ACTING ON ARM 2} Tr2:(-(Power(dt1,2)*l1*Power(l2,2)*m1*Power(m2,2)) - 2*dt1*dt2*l1*Power(l2,2)*m1*Power(m2,2) - Power(dt2,2)*l1*Power(l2,2)*m1*Power(m2,2) - Power(dt1,2)*Power(l1,2)*l2*m1*Power(m2,2)*Cos(t2) - g*l1*l2*m1*power(m2,2)*cos(t1 + t2) - g*l1*l2*m1*power(m2,2)*sin(t1)*sin(t2))/ (-(l1*l2*m1*m2) - l1*l2*power(m2,2)*power(sin(t2),2)); {ANGULAR VELOCITIES OF THE MASSES 1 AND 2} dt1 : dt1 + deltat * ddt1; dt2 : dt2 + deltat * ddt2; {ANGULAR POSITION OF THE MASSES 1 AND 2} t1 : t1 + deltat * dt1; t2 : t2 + deltat * dt2; {TRAJECTORY ACCOMPLISHED BY THE MASSES 1 AND 2} h1x : (-l1 * sin (t1)); h1y : - (l1 * cos (t1)); h2x : (-l1 * sin (t1) - l2 * sin (t1 + t2)); h2y : - (l1 * cos (t1) + l2 * cos (t1 + t2)); {PRINTING RESULTS ON THE SCREEN}

71 1.9 Dynamics of a System of Particles in 2D Example 71 gotoxy(1,1); writeln( **********, ni*nj); writeln( t:,t); writeln; writeln(t1:5:5,,t2:5:5); writeln(dt1:5:5,,dt2:5:5); writeln(ddt1:5:5,,ddt2:5:5); {SAVING RESULTS IN THE FILES} writeln(arq1,t,,t1); writeln(arq2,t,,t2); writeln(arq3,h1x,,h1y); writeln(arq4,h2x,,h2y); end; end; close(arq1);close(arq2);close(arq3);close(arq4); {PROGRAM END} end.

72 1.9 Dynamics of a System of Particles in 2D Example 72 Theoretical and Experimental Results Qualitative Comparison Initial conditions of movement. Following some simulation results are presented. Such results can be qualitatively compared to the prototype of a double pendulum designed and built. The figure on your left hand side illustrates the initial conditions of the movement for the masses A and B: positions θ 1 45 o.785 rad and θ 2 45 o.785 rad; velocities θ 1 rad/s and θ 2 rad/s. The above figure illustrates simultaneously the behavior of the angles θ 1 and θ 2 as a function of the time and the trajectories accomplished by the masses A and B, due to the initial conditions. The results were obtained by solving the equations of motion with help of the computational program previously presented arqt1.res arqh1.res Behavior of the angle θ 1 as a function of the time and trajectory accomplished by the mass B arqt2.res arqh2.res Behavior of angle θ 2 as a function of the time and trajectory accomplished by the mass A.

73 1.9 Dynamics of a System of Particles in 2D Example 73 Trajectory accomplished by the mass attached to the extremity of the arm, considering the following initial conditions: θ 1 45 o.785 rad, θ 2 45 o.785 rad, θ 1 rad/s e θ 2 rad/s. Trajectory accomplished by the mass attached to the extremity of the arm, considering the following initial conditions: θ 1 9 o π/2 rad, θ 2 9 o π/2 rad, θ 1 rad/s e θ 2 rad/s.

74 1.9 Dynamics of a System of Particles in 2D Example 74 Trajectory accomplished by the mass attached to the extremity of the arm, considering the following initial conditions: θ 1 18 o π rad, θ 2 45 o.98 π rad, θ 1 rad/s e θ 2 rad/s.

75 1.1 Dynamics of One Single Rigid Body in 3D An Example Dynamics of One Single Rigid Body in 3D An Example The aim of this example is to illustrate the application of the Euler s angle for describing the movements of a top, a rigid body, shown in Figure 1.7. In the example precession, nutation and spin angles are defined with help of different moving reference frames. Afterwards it is possible to calculate the absolute angular velocity vectors of the reference frame attached to the top and of the top itself. (a) (b) Figure 1.7: Top as an example of rigid body (a) physical system; (b) mechanical model illustrating also the reference frames. The momentum vector is obtained, and its representation is done with help of the moving reference frame attached to the top. Such a vector is derived referred to time and equals to the sum of external moments acting on the top, i.e. Euler s equation. The sum of the forces acting on the top is equaled to the inertia forces. In that way, the non-linear differential equations of second order, responsible for describing precession, nutation and spin of the top, are achieved. Additionally, the dynamic reaction forces (contact forces between top and floor) are obtained. Finally, the equations are numerically solved, taking into consideration that the top does not slip on the floor. Some theoretical results are illustrated graphically as well as by means of computer animations. Qualitative experimental results are also presented in the form of a video Solution Steps Equations of Motion and Dynamic Reactions Steps towards obtaining the equations of motion and dynamic reaction forces: KINEMATICS (1) Definition of the reference frames (inertial and moving). (2) Calculation of the coordinate transformation matrices.

76 1.1 Dynamics of One Single Rigid Body in 3D An Example 76 (3) Calculation the absolute angular velocity vector of the moving reference frame attached to the top, Ω. (4) Calculation of the absolute angular velocity vector of the top, ω. (5) Calculation of the absolute angular acceleration vector of the top, ω. (6) Calculation of the absolute linear velocity vector of the top center of mass, v. (7) Calculation of the absolute linear acceleration vector of the top center of mass, a. DYNAMICS (8) Free-body diagram. (9) Moments and products of inertia, inertia tensor I. (1) Dynamic equilibrium [NEWTON-EULER]. NUMERICAL METHODS AND ANALYSIS (11) Computational program for solving the equations of motion. (12) Theoretical and qualitative experimental results. (13) Using the program MATHEMATICA for achieving the equations of motion. SOLUTION: (1) Reference frames: Inertial reference frame I X, Y, Z and unity vectors i, j, k. Moving reference frame B1 X 1, Y 1, Z 1 and unity vectors i 1, j 1, k 1. Moving reference frame B2 X 2, Y 2, Z 2 and unity vectors i 2, j 2, k 2. (2) Coordinate transformation matrices: ϕ - rotation of the reference frame B1 around Z-axis of the inertial frame. θ - rotation of the reference frame B2 around X 1 -axis of the moving reference frame B1. φ - rotation of the top around Z 2 -axis of the moving reference frame B2. T ϕ T θ cosϕ sinϕ sinϕ cosϕ 1 1 cosθ sinθ sinθ cosθ B1 S T ϕ I S B2 S T θ B1 S cosϕ sinϕ T θ.t ϕ cosθsinϕ cosθcosϕ sinθ B2 S T θ.t ϕ I S sinθsinϕ sinθcosϕ cosθ (3) Absolute angular velocity of the moving reference B2, Ω, in which the momentum and moments vectors will be represented later, and in which the inertia tensor remains constant:

77 1.1 Dynamics of One Single Rigid Body in 3D An Example 77 Ω ϕ + θ B2 B2 B2 ϕ I ϕ T.T ϕ B2 θ ϕ I ϕ B1 B2 θ φ θ φ θ T B2 θ B1 θ Ω ϕ + θ T B2 B2 B2 θ.t ϕ I ϕ + T θ B1 θ θ ϕsinθ ϕcosθ (4) Absolute angular velocity of the top, ω, represented with help of the moving reference frame B 2, in which the momentum and moments vectors will be represented later, and in which the inertia tensor remains constant: ω ϕ + θ + B2 B2 B2 B2 φ ω ϕ + θ + B2 B2 B2 B2 φ Tθ.T ϕ I ϕ + T θ B1 θ + B2 φ θ ϕsinθ ϕcosθ+ φ (5) Absolute angular acceleration of the top: B2 ω d dt ( B2 ω) + B2 Ω B2 ω d dt ( B2 ω) θ ϕsinθ + ϕ θcosθ ϕcosθ ϕ θsinθ + φ B2 Ω B2 ω i 2 j 2 k 2 θ ϕsinθ ϕcosθ θ ϕsinθ ϕcosθ + φ ϕ φsinθ θ φ B2 ω θ + ϕ φsinθ ϕsinθ + ϕ θcosθ θ φ ϕcosθ ϕ θsinθ + φ

78 1.1 Dynamics of One Single Rigid Body in 3D An Example 78 (6) Absolute linear velocity vector of the top center of mass: B2 v B2 v }{{} o Ω B2 B2 r + B2 Ω B2 r + B2 v rel }{{} i 2 j 2 k 2 θ ϕsinθ ϕcosθ h h ϕsinθ h θ B2 v h ϕsinθ θ (7) Absolute linear acceleration vector of the top center of mass: B2 a B2 a }{{} o + B2 Ω B2 Ω B2 r + B2 Ω B2 r + 2Ω v rel }{{} + a rel }{{} B2 Ω B2 Ω B2 r h θ ϕcosθ h ϕ 2 cosθsinθ h ϕ 2 sin 2 θ h θ 2 B2 Ω B2 r h ϕsinθ + h ϕ θcosθ h θ B2 a h 2 θ ϕcosθ + ϕsinθ ϕ 2 cosθsinθ θ ϕ 2 sin 2 θ θ 2

79 1.1 Dynamics of One Single Rigid Body in 3D An Example 79 (8) Free-body diagram and forces acting on the top: weight force I P, represented with help of the inertial reference frame I P { mg}t Reaction force I R, which is represented with help of the inertial reference frame I R { R X R Y R Z } T In case of necessity, such force vectors can be represented using other reference frames. Free-body diagram weight and dynamic reaction forces acting in the contact point between top and floor. (9) Tensor of inertia: Mass moments of inertia referred to the top center of mass: I CM CM I xx CM I yy CM I zz, where CM I xx 1 4 mr ml2 CM I yy 1 4 mr ml2 CM I zz 1 2 mr2 Mass moments of inertia referred to the contact point between top and floor: I O I xx I yy I zz, where I xx CM I xx + mh 2 I yy CM I yy + mh 2 I zz CM I zz (1) Dynamic equilibrium [NEWTON-EULER]: NEWTON: F ṁ B2 }{{} EULER: B2 M O B2 I O B2 v + m B2 a d dt ( B2 ω) + B2 Ω ( B2 I O. B2 ω ) + B2 r O CM m B2 a O }{{}

80 1.1 Dynamics of One Single Rigid Body in 3D An Example 8 Forces action on the top: R e P In this case, the most appropriate reference frame for representing the reaction forces between top and floor is the inertial one. The force components in X and Y directions of R, shall not overcome the static friction forces, otherwise the top will slip on the floor. R R X R I Y R Z P I mg R T T R B2 θ ϕ I P T T P B2 θ ϕ I R X cosϕ+r Y sinϕ R X cosθsinϕ+r Y cosθcosϕ+r Z sinθ R X sinθsinϕ R Y sinθcosϕ+r Z cosθ mgsinθ mgcosθ M h P + B2 O B2 B2 M i 2 j 2 k 2 h mgsinθ mgcosθ + M mghsinθ M I O. d dt ( ω) B2 I xx I yy I zz θ ϕsinθ+ ϕ θcosθ ϕcosθ ϕ θsinθ+ φ I xx θ I yy ( ϕsinθ+ ϕ θcosθ) I zz ( ϕcosθ ϕ θsinθ+ φ) B2 Ω ( I O. B2 ω ) B2 Ω I xx I yy I zz θ ϕsinθ ϕcosθ+ φ B2 Ω I xx θ I yy ϕsinθ I zz ( ϕcosθ+ φ) i 2 j 2 k 2 θ ϕsinθ ϕcosθ I xx θ Iyy ϕsinθ I zz ( ϕcosθ+ φ) B2 Ω ( I O. B2 ω ) ϕ 2 (I zz I yy )sinθcosθ+i zz φ ϕsinθ ϕ θ(i xx I zz )cosθ I zz φ θ ϕ θ(i yy I xx )sinθ B2 M O I O. d dt ( B2 ω) + B2 Ω ( I O B2 ω ) + B2 r O CM m B2 a O }{{} Substituting the mass moments of inertia, item (9), one achieves:

81 1.1 Dynamics of One Single Rigid Body in 3D An Example 81 mghsinθ M ϕ [ φ CM I zz sinθ + ϕ 2 sin2θ ( CM I zz CM I yy mh 2 )] + θ ( CM I xx + mh 2 ) θ ϕ CM I zz + θ ϕ cosθ [ CM I xx CM I zz + CM I yy + 2mh 2 ] + ϕ sinθ ( CM I yy + mh 2 ) ϕ θ sinθ [ CM I yy CM I xx CM I zz ] + ( ϕ cosθ + φ) CM I zz Based on such a system of non-linear differential equations of second order it is possible to describe de movements of the top: Precession ϕ θ sinθ ( CM I xx + mh 2 ) [ φ CM I zz + ϕcosθ ( CM I zz CM I xx CM I yy 2mh 2 )] Nutation θ sinθ I CM xx +mh {mgh + ϕ [ I 2 CM zz φ + ϕcosθ ( CM I yy CM I zz + mh 2 )]} Spin φ θ tanθ ( CM I yy + mh 2 ) [ I CM zz φ + ϕcosθ ( CM I xx + CM I yy CM I zz + 2mh 2 )] + 1 CM I zz [M + ϕ θsinθ ( CM I xx CM I yy + CM I zz )] In order to obtain the dynamic reaction forces between the top and floor, the equation has to be solved. B2 F m B2 a B2 R + B2 P T θ T ϕ I R + T θ T ϕ I P Substituting all the above calculated terms into Newton s equation, one gets: T θ T ϕ R X R Y R Z mgsinθ mgcosθ +m h 2 θ ϕcosθ + ϕsinθ ϕ 2 cosθsinθ θ ϕ 2 sin 2 θ θ 2 R X R Y R Z TT ϕ TT θ mgsinθ mgcosθ +m h 2 θ ϕcosθ + ϕsinθ ϕ 2 cosθsinθ θ ϕ 2 sin 2 θ θ 2

82 1.1 Dynamics of One Single Rigid Body in 3D An Example 82 After handling the equations, one achieves the three components of the reaction force: Component R x of the dynamic reaction force: R x cosϕ [m θh( ϕcosθ+ φ)+mh( ϕsinθ + ϕ θcosθ)] + + sinϕ [ mgsinθ m ϕhsinθ( ϕcosθ+ φ)+mh θ] Component R y of the dynamic reaction force: R y cosθ sinϕ [m θh( ϕcosθ+ φ)+mh( ϕsinθ+ ϕ θcosθ)] + + cosθ cosϕ [mgsinθ +m ϕhsinθ( ϕcosθ+ φ) mh θ] + + sinθ [ mgcosθ+mh( θ 2 + ϕ 2 sin 2 θ)] Component R z of the dynamic reaction force: R z sinθ sinϕ [m θh( ϕcosθ+ φ)+mh( ϕsinθ+ ϕ θcosθ)] + + sinθ cosϕ [mgsinθ+m ϕhsinθ( ϕcosθ+ φ) mh θ] cosθ [ mgcosθ +mh( θ 2 + ϕ 2 sin 2 θ)] The components R x and R y, which can be calculated as a function of the top movements (precession, nutation and spin) are time-depending, i.e. R x (t) e R y (t). In order to avoid slipping, the friction force has to be bigger than the resultant force R 2 x +R2 y. Mathematically, R 2 x +R 2 y µ R z.

83 1.1 Dynamics of One Single Rigid Body in 3D An Example Computational Implementation and Numerical Simulation (11) Computational Program in Pascal Language: program top; uses crt; {DEFINITION OF THE PROGRAM VARIABLES} var psi,phi,teta : real; Dpsi,Dphi,Dteta : real; DDpsi,DDphi,DDteta : real; t,tempo,deltat,freq,r,mp,h,l,cgixx,cgiyy: real; Ixx,Iyy,Izz,cgIzz,Mz,mom,hx,hy,hz : real; arq1,arq2,arq3,arq4,arq5,arq6,arq7,arq8 : text; i,n,fim,j,a : integer; {DEFINITION OF THE PROGRAM CONSTANTS} const g 9.81; {aceleracao da gravidade [m/s 2 ]} {BEGINNING OF THE PROGRAM} begin clrscr; {PARAMETERS REFERRED TO THE TIME INTEGRATION} t :.; {initial time [s]} tempo : 1; {total time [s]} n : 3; {number of points} a : 1; {repetition of the integration process} deltat: (tempo/a)/n; {time step [s]} fim : 1; {criterium to close the files} {NAME OF THE FILES WHERE THE RESULTS WILL BE SAVED} assign(arq1, arqpsi.res ); assign(arq2, arqphi.res ); assign(arq3, arqteta.res ); assign(arq4, arqh.res ); assign(arq5, arqdpsi.res ); assign(arq6, arqdphi.res ); assign(arq7, arqdteta.res ); assign(arq8, arqhz.res ); {GEOMETRIC AND MASS PARAMETERS OF THE TOP} mp :.582; {total mass [kg]} l :.215; {thickness of the disk [m]} r :.5; {disk radius [m]} h :.558; {length of the arm [m]} Ixx :.29393; {moment of inertia Ixx around O} Iyy :.29393; {moment of inertia Iyy around O} Izz :.2258; {moment of inertia Izz around O} {INITIAL CONDITIONS} psi :.; {initial precession angle} teta: (5/18)*2* ; {initial nutation angle} phi :.; {initial spin angle}

84 1.1 Dynamics of One Single Rigid Body in 3D An Example 84 Dpsi: (/18)*2* ; {initial velocity of precession} Dteta:.; {initial velocity of nutation} Dphi: 4*2* ; {initial spin} Mz : -.1; {moment applied to the top in Z2-direction} hx : h*sin(teta)*sin(psi); {initial position of the top center of mass direction X} hy : -h*sin(teta)*cos(psi); {initial position of the top center of mass direction Y} hz : h*cos(teta); {initial position of the top center of mass direction Z} {OPENING THE FILES} rewrite(arq1); rewrite(arq2); rewrite(arq3); rewrite(arq4); rewrite(arq5); rewrite(arq6); rewrite(arq7); rewrite(arq8); {SAVING THE INITIAL CONDITIONS IN THE FILES} writeln(arq1,t,,psi); writeln(arq2,t,,phi); writeln(arq3,t,,teta); writeln(arq4,hx,,hy); writeln(arq5,t,,dpsi); writeln(arq6,t,,dphi); writeln(arq7,t,,dteta); writeln(arq8,t,,hz); {BEGINNING OF THE TIME INTEGRATION} for j:1 to a do begin for i:2 to n do begin t : t + deltat; {variacao do tempo} {CALCULATION OF THE ANGULAR ACCELERATIONS} DDteta: (1/Ixx)*(mp*g*h*sin(teta)+Sqr(Dpsi)*(-Izz+Iyy)*sin(teta)* cos(teta)-izz*dphi*dpsi*sin(teta)); DDpsi: (1/(Iyy*sin(teta)))*(Izz*Dpsi*Dteta*cos(teta)+ Dpsi*Dteta*cos(teta)*(-Ixx-Iyy)+Izz*Dphi*Dteta); DDphi: (1/Izz)*(-Izz*DDpsi*cos(teta)+Izz*Dpsi*Dteta*sin(teta)+ Dpsi*Dteta*(-Iyy+Ixx)*sin(teta)+Mz); {CALCULATION OF THE ANGULAR VELOCITIES} Dpsi : Dpsi + deltat * DDpsi; Dphi : Dphi + deltat * DDphi; Dteta: Dteta + deltat * DDteta; {CALCULATION OF THE ANGLES} psi : psi + deltat * Dpsi; phi : phi + deltat * Dphi; teta : teta + deltat * Dteta; {CALCULATION OF THE TOP CENTER OF MASS} hx : h*sin(teta)*sin(psi); hy :-h*sin(teta)*cos(psi); hz : h*cos(teta); {PRINTING THE RESULTS ON THE SCREEN} writeln( **********, i);

85 1.1 Dynamics of One Single Rigid Body in 3D An Example 85 writeln( t:,t, psi:,psi); writeln( t:,t, phi:,phi); writeln( t:,t, teta:,teta); writeln( ); writeln(ddpsi:5:5,,ddteta:5:5,,ddphi:5:5); {SAVING THE RESULTS IN THE FILES} writeln(arq1,t,,psi); writeln(arq2,t,,phi); writeln(arq3,t,,teta); writeln(arq5,t,,dpsi); writeln(arq6,t,,dphi); writeln(arq7,t,,dteta); writeln(arq4,hx,,hy); writeln(arq8,t,,hz); {CRITERIUM TO STOP THE INTEGRATION} if teta > (PI/2) - arctan(r/.442) then begin writeln( THE DISK TOUCHED THE FLOOR ); close(arq1);close(arq2);close(arq3);close(arq4); close(arq5);close(arq6);close(arq7);close(arq8); j:a; i:n; fim:; readln; end; end; end; if fim < > then begin close(arq1);close(arq2);close(arq3);close(arq4); close(arq5);close(arq6);close(arq7);close(arq8); end; end. (12) Theoretical results for different initial conditions:

86 1.1 Dynamics of One Single Rigid Body in 3D An Example arqh.res.2.1 y [m] x [m] Trajectory of the top center of mass projected into the XY-plane as a function of the time, until the disk touches the floor Initial conditions given in the case (a) arqhz.res CASE (a) INITIAL CONDITIONS: psi :.; teta: (15/18)*2* ; phi :.; Dpsi: (2/18)*2* ; Dteta:.; Dphi: 7*2* ; Mz : -.25; h [m] t [s] Behavior of the height of the top center of mass as a function of the time, until the disk touches the floor Initial conditions given in the case (a). The above illustrations show the XY-trajectory and the height of the top center of mass as a function of the time, until the disk touches the floor. The initial conditions are given in the case (a). The resistance moment Mz is relative big, which forces the top spin to reduce rapidly. As consequence, the disk touches the floor after 2,5 [s]. The behavior of the precession and nutation angles, simultaneously with the top spin in time domain are visualized in the next figures, for the same initial conditions, case (a).