Lecture notes on Kinematics of Particles

kinematics of a particle force and acceleration, kinematics of a particle subjected to projectile motion, kinematics of a particle work and energy, kinematics of a moving particle pdf free download
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Published Date:21-07-2017
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Chapter 2 Kinematics 2.1 Basic Concepts Kinematics describes the motion of mechanical systems, without considering the forces that produce that motion. Kinematics deals with velocities and accelerations, which are defined for points of interest on the mechanical systems. The description of motion is relative in nature. Velocities and accelerations are therefore defined with respect to a reference frame. 2.2 Kinematics of a particle. Rectilinear and curvi- linear motion The particle is classically represented as a point placed somewhere in space. A rectilinear motion is a straight-line motion. A curvilinear motion is a motion along a curved path. 2.2.1 Position vector. Velocity vector. Acceleration vector The position vectorr(t) (see Fig. 2.1) of the particle P at a given instant of time t refers to its location relative to some reference point usually taken as the origin of a coordinate system. Note that every vector considered in section 2.2 may be projected onto the coordinate frame oxyz. As the particle moves along its straight- line path, its position changes with time. By definition the displacementr of the particle during a time intervalt is given by the change of its position during this time interval. r=r(t +t)r(t) (2.1) 16CHAPTER 2. KINEMATICS 17 z P r path of particle r( ) t r t+ t ( ) y o x Figure 2.1: Position vector 2.2.2 Average and instantaneous velocities The average velocity during the time intervalt is defined as r v= (2.2) av t The instantaneous velocity is given by rdr v=lim==r_ (2.3) t0 tdt 2.2.3 Average and instantaneous acceleration We need to learn how the velocity varies with time ,we define average acceleration by v a= (2.4) av t and the instantaneous acceleration is defined by letting the time intervalt ap- proach zero in the limit: vdv a=lim==v_ (2.5) t0 tdtCHAPTER 2. KINEMATICS 18 2.2.4 Absolute frame Let us express the position vectorr to point P on the path of the particle in terms P ofx,y,z components (see Fig. 2.2) z P r P k j y o i x Figure 2.2: Absolute frame r(t)=x(t)i+y(t)j+z(t)k (2.6) P dr P v==x_(t)i+y_(t)j+z_(t)k (2.7) P dt dv P a==x(t)i +y(t)j +z(t)k (2.8) P dt pp 222222 thus we have the magnitudesv=x_+y_+z_ anda=x +y +z .The PP velocity and acceleration vectors are obtained by successive time differentiation of position vector. Let us state now some notations which will be used.r andr PQ being the position vectors of two points P and Q, we have; r=rr (2.9) P=QPQ v=vv (2.10) P=QPQ 2.2.5 Tangential and normal coordinates In many plane problems dealing with the motion of a particle along a curve, it is convenient to express the acceleration in term of two components; one along theCHAPTER 2. KINEMATICS 19 tangent to the trajectory, and the second along the inward normal to the path. For this purpose we define two unit vectorsn andt respectively along the inward nor- mal and along the tangent to the path (see Fig. 2.3). Consider a particle moving along a curved path in a plane shown in Fig. 2.3 y P( ) t t s Δr t P( + ) Δt n s+ s Δ Path of r particle P j o x i k Figure 2.3: Plane motion of a particle As we have seen above; rdr v=lim= t0 tdt Thus we can write; rs v=lim t0st and sds lim= t0 tdt Ast goes to zero, the direction ofr approaches the tangent to the trajectory at positionr(t) and approachess in magnitude. Consequently, in the limit,r=s p becomes the unit vectort r lim=t t0s thus rsds v=lim=t t0 stdt Note thatds=dt is the magnitude of the velocity, ds v= dtCHAPTER 2. KINEMATICS 20 Let calculate now the two components of the acceleration dvdds a==t dtdtdt 22 dsdsdtdsdsdtds a=t+=t+ 22 dtdtdtdtdtdsdt Let evaluate the derivation oft with respect tos. Consider now the unit vectort at two positionss ands +s (see Fig. 2.4 (a)) y Path of particle P( ) t s t( ) s t P( + ) t Δ n s+ s Δ Δϕ t( ) s+ s Δ r P C Center of curvature j o x i (a) k Δϕ s t( ) s Δt( ) (b) t( ) s+ s Δ Figure 2.4: Plane motion of a particle dtt(s +s)t(s)t =lim=lim s0s0 dsss In the limit ass goes to zero, the vectort ends up in the plane normal to the path ats and directed toward the center of curvature, it is the direction of the unit vectorn (see Fig. 2.4 (b)).CHAPTER 2. KINEMATICS 21 Knowing the limiting direction oft, we next evaluate its limiting magnitude. Ac- cording to Fig. 2.4 (b) we can say that,whens0: s jtjjtj = R thus the magnitude and the direction are established in an approximate manner. s tn R and so dtt(s=R)nn =lim=lim= s0s0 dsssR y Path of particle P( ) t t a t v P n R a a n ( ) r t P C Center of curvature j o i k Figure 2.5: The acceleration components then the acceleration can be evaluated by: 2 2 ds(ds=dt) a=t+n (2.11) 2 dtR or a=a+a=at+an (2.12) tntn where 2 dsdv a== (2.13) t 2 dtdt 2 2 (ds=dt)v a== (2.14) n RRCHAPTER 2. KINEMATICS 22 Fora planecurvey=y(x), the radiusR of curvature is given by; 3 22 dy 1+ dx R= (2.15) 2 dy 2 dx 2.2.6 Rotation around a fixed point in a plane The center O of the fixed frame (see Fig. 2.6) is the center of rotation; the instanta- neous position and velocity of the point P are given by; r=OP=Rcosi+Rsinj P d v=Rsini+cosj (2.16) P dt 2323 0Rcos d 4545 =0Rsin (2.17) dt 10 ijk d =00 (2.18) dt RcosRsin0 d =kr (2.19) P dt =r (2.20) P y v P P r j P θ o x i k Figure 2.6: Rotation of a particle around a fixed pointCHAPTER 2. KINEMATICS 23 2.3 Kinematics of a rigid body The description of motion is relative. Any velocity or acceleration is expressed with regard to a specific reference frame. This fact induces specific notations that must be understood:v denotes the instantaneous velocity of the point P attached to P S=s the body S, relatively to the body s. A rigid body is considered to be composed of continuous of distribution of particles having fixed distances between each others. There are various types of rigid-body motion but the most important of them are translations and rotations. 2.3.1 Translation A2 A1 B2 B1 Figure 2.7: Rectilinear translation A2 A2 B2 A1 A1 B1 B2 B1 Figure 2.8: Curvilinear translation Amotionissaidtobea translation if any straight line defined inside the body keeps the same direction during the motion. In translation all particles move along parallel paths. We have rectilinear translation when the paths are straight lines as in Fig. 2.7 in other cases it is a curvilinear translation as in Fig. 2.8. Referring to Eq. 2.9, we have; r=r+r (2.21) BAB=A wherer=AB. B=A Let us differentiate the relation with respect tot. Since the vectorr =AB has a B=ACHAPTER 2. KINEMATICS 24 constant direction and a constant magnitude, its time derivative is zero: v=v BA a=a BA In a translation all particles of the rigid body have same velocity and same acceler- ation. 2.3.2 Rotation about a fixed axis If a part of a rigid body, or a hypothetical extension of the body, has zero velocity to some reference, the body is said to be in rotation.The axis of rotation is the line of stationary particles. Since the velocity of P is a vector perpendicular to the plane v P P B k r P j i S (s) Figure 2.9: Rotation about fixed axis (Fig. 2.9) containing the rotation axis andr . We can write referring, to Eq. 2.19: P dr P v= (2.22) P S=s dt =r (2.23) S=sP or in a condensed form dr v==r (2.24) dt z y o xCHAPTER 2. KINEMATICS 25 Note that the vector product can be computed as the determinant: vijk x v=v= (2.25) yxyz vxyz z And then yz vi=i=i(z y) xyz yz xz vj=j=j(zx) yxz xz xy vk=k=k(yx) zxy xy Since _ =k (2.26) _ =0, =0,= and the velocity is completely determined. We have xyz The accelerationa of P is now determined as dv P S=s a= (2.27) P S=s dt d =r (2.28) S=sP dt d dr S=sP =r+ (2.29) PS=s dtdt =kr+r (2.30) PS=sS=sP 2.3.3 Particular case: Motion in plane A Plane Motion is a motion in which all particles of the body move in parallel planes. Velocity in plane motion Given two particles A and B of a rigid body in plane motion the velocityv of B is B obtained from the velocity formula (referring to Eq. 2.10) v=v+v (2.31) BAB=A In relative motion about A , A is fixed (v=0). Thusv can be associated A=AB=A with the rotation of the body about A and is measured with respect to axes centered at A v=r (2.32) B=AB=ACHAPTER 2. KINEMATICS 26 v v A A A A A + = v B v v B BBB B/A Figure 2.10: Plane motion and v =AB (2.33) B=A =k is the angular velocity of the body, we note thatr=AB B=A v=v+AB (2.34) BA Acceleration in plane motion y x A (s) a v n B a S t B Figure 2.11: Plane motion dv B a= (2.35) B dt dr dv B=A A =+ (2.36) dtdt ddr B=A =a+r+ (2.37) AB=A dtdt _ =a+r+r (2.38) A B=AB=ACHAPTER 2. KINEMATICS 27 If A is the center of a fixed frame s (see Fig. 2.11) we havea=0 and then; A 2 a=_krr (2.39) B B=AB=A 2 =_kABAB (2.40) In the right hand of Eq. 2.40, the first term is perpendicular toAB and the second is parallel. Equiprojectivity For two points A and B of a given rigid body we can write v A A B v B Figure 2.12: Equiprojectivity v=v+r (2.41) BAB=A vr=vr+rr (2.42) BB=AAB=AB=AB=A vr=vr (2.43) BB=AAB=A vAB=vAB (2.44) BA Instantaneous center of rotation Considering a general plane motion of a body, at given instant, the velocities of various particles of the body could be expressed as the result of a rotation whose axis is perpendicular to the plane. This axis intersects the plane at a point called the instantaneous center of rotation. The position of this particular point can be defined in many ways. If the directions of the velocities of two particles A and B are known and if they are different, (Fig. 2.13, at left) the instantaneous center of rotation is obtained by drawing the perpendicular tov through A and the perpendicular tov ABCHAPTER 2. KINEMATICS 28 through B and finding the point in which these two lines intersect. If the velocitiesv andv are perpendicular to the line AB and if their magnitude AB are known, the instantaneous center of rotation can be found by intersecting AB with the line joining the extremities of the vector (Fig. 2.13, at right). C C A A B B Figure 2.13: Instantaneous center of rotation Kennedy’s theorem The Kennedy’s theorem states that the three instant centers shared by three rigid bodies in relative planar motion to another (whether or not connected) all lie on the same straight line. Application of Kennedy’s theorem S 2 S 1 S 3 S 0 Figure 2.14: Four-bar linkage The figure 2.14 shows four-bar linkage let us locate all instant centers. When the number of bodies is large, it is helpful to use some kind method to find the instant centers. Note thatS represents the stationary frame. 0 1. By inspection ,determine as many centers as possible, in the exemple the instant centersI ,I ,I ,I are easily placed. 01122303CHAPTER 2. KINEMATICS 29 2. Using Kennedy’s theorem with linksS ,S ,S the instant centerI must lie 01202 on the same straight line withI ,I but it must also lie on the line through 0112 I andI . The location is defined by the intersection of the two lines. 2303 3. The same reason can be used to locate the centerI . 13 I S 0 02 Revolute Revolute I 12 S 2 S S I 3 1 23 S 1 S 3 Revolute Revolute I S I I S 2 01 0 03 13 Figure 2.15: Locating instant center 2.3.4 General motion in space Z z P Y A S o y X (s) x Figure 2.16: General motion in space without rotating axis The most general motion of a rigid body in space is equivalent at any given instant to the combination of a translation and a rotation (as we have seen for planeCHAPTER 2. KINEMATICS 30 motion). Considering two particles A and P of the rigid body S, we have: v=v+v (2.45) PA P=A Wherev is the velocity of P relative to a frame attached to A, thusv= P=AP=A r orv=AP where is the angular velocity of the body S=sP=AP=AS=s S relative to the fixed frame s. The absolute velocity of a particle P belong to S is given by from above: v=v+AP (2.46) PAS=s S=sS=s The equation 2.46 allows the determination of the velocity of any point P of a body S with respect to another frame s,v , if the following variables are known: P S=s v : velocity of a point A of the body S with respect to s. A S=s : angular velocity of S with respect to s. S=s AP: position of the particle P with respect to A. The acceleration of P is obtained by differentiating the equation with respect to time. dv P a= (2.47) P S=s dt dv dAP A S=s P=s =+ (2.48) dtdt dAP _ =a+AP+ (2.49) AS=s S=s dt =a+_AP+AP (2.50) A=sS=sS=sS=s The equation 2.50 allows the determination of the acceleration of any point P of a body S with respect to another frame s,a , if the following variables are known: P S=s a : acceleration of the point A of the body S with respect to s. A S=s : angular velocity of S with respect to s. S=s _ : angular acceleration of S with respect to s. S=s AP: position of the particle P with respect to A. In some cases, (Fig. 2.17) it is needed to express either the velocity either the accel- eration in different frames. then the following equation can be used: For velocity: v=v+v (2.51) P=sP=SP S=sCHAPTER 2. KINEMATICS 31 z Z Y P A S X o y (s) x Figure 2.17: General motion of a rigid body in space with rotating axis assume to S and s are two frames, note that herev represents the velocity of the P S=s frame S with respect to the frame s. The acceleration is then given by: a=a+a+a (2.52) P=sP=SPcor S=s wherea is the Coriolis acceleration: cor a =2v (2.53) corS=sP=S The Coriolis acceleration has a zero value if: the point P has no relative velocity with respect to S (v=0); P=S the relative velocityv=0 is parallel to the angular velocity . P=SS=s (see 2 for demonstration) 2.3.5 Rolling without slipping The point of contact G between a cylinder and a the flat ground has instantaneously zero velocity (v=0) if the cylinder rolls without slipping (Fig. 2.18). G v=v+xGP PG S=sS=s v=0+xGP PS=s S=sCHAPTER 2. KINEMATICS 32 Z P z A Y y o X x G Figure 2.18: Rolling without slipping In particular for the center A of the cylinder we get from above v=vxGA AGS=s S=sS=s v=0xGA AS=s S=s thus xyz 00 v== Ry A 00R 2.4 Kinematics of systems of rigid bodies 2.4.1 Mechanism A mechanism is an collection of rigid bodies connected together by joints. Mecha- nisms transfer motion and mechanical work from one or more members to others. When several links are connected by joints,they form a kinematics chain with one link possibly fixed. The joints permit relative motion in some directions while con- straining motion in others. 2.4.2 Degrees of freedom The types of motion permitted are related to the degree of freedom (dof) also called mobility. This represents the number of input parameters which can be controlled independently in order to bring the device in a particular position. It is possible to determine the mobility of a mechanism by counting the number of links (including the fixed one) and the degrees of freedom constrained by each joint. For a plan motion, we have: X dof =3 (nn )+f (2.54) bjjCHAPTER 2. KINEMATICS 33 where n is the total number of rigid bodies including the fixed link; b n is the total number of joints possibly including the fixed link j f degree of freedom of relative motion between the bodies constrained by j the kinematical joint. For a three-dimensional motion X dof =6 (nn )+f (2.55) bjj 2.4.3 Lower pairs and higher pairs Relative Degree of Skecth Other Name view motion freedom (f ) symbol j Rigid 0 rotation 0 joint 0 translation Revolute 1 rotation 1 0 translation Prismatic 0 rotation 1 1 translation Helical 1 rotation (right) 1 1 translation (left) Cylindrical 1 rotation 2 1 translation Spherical 3 rotations 2 0 translation Planar 1 rotation 3 2 translations Figure 2.19: Lower pair joints We divide joints joints into two groups: A lower pair joint is one in which contact two rigid bodies occurs at every points of one or more surface segments (see Fig. 2.19).CHAPTER 2. KINEMATICS 34 Typical Description Degree of freedom form (f ) j Cylindrical surface on a plan 1 without slipping Cylindrical surface on a plan with 2 slipping Ball on a plan 3 without slipping Point on a plan with slipping 5 Figure 2.20: Higher pair joints Symbol Name s Fixed body A higher pair joint is one which contact occurs only at isolated points or along a line segments (see Fig. 2.20) represents that there is not slipping on the plane.CHAPTER 2. KINEMATICS 35 2.4.4 Kinematics exercises The MATLAB fileKexx.m can be executed by typingKexx in the interactive window of MATLAB . It provides an interface where the user may examine the numerical aspects of the exercises simply by pressing command buttons, corresponding to the various kinematics exercises. Each button calls the corresponding MATLAB file with an illustration of the exercise solution. It is also possible to see the solution of each exercise by calling the corresponding file, directly from the command line. Exercise 2.4.1 Piston connected to a nut (see Fig. 2.21) Kex1.m A single-threaded screw S , defining a pitchp (translation of the nut along the 0 screw axis for one turn), supports a nut S . A solid rod AB (lengthL) connects the 1 nut (point B) and to a piston S (point A). The piston S can only slide along the 33 Az axis in a straight slot (this axis is parallel to the screw axis Oz). The distance between the two axes isd. xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxd xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxx xxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx S3 xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxx xx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxx Axxxxxxxx xxxxxxxxxxxxxxxx S2 xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxx z L xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx O xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx x xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx y ω S1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxS0 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx S1 y A O B ω S2 x Figure 2.21: Piston connected to a nut.

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