4 Velocity and Acceleration Analysis 0Contents4 Velocity and Acceleration Analysis 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Velocity Field for a Rigid Body . . . . . . . . . . . . . . . . . 24.3 Acceleration Field for a Rigid Body . . . . . . . . . . . . . . . 54.4 Motion of a Point that MovesRelative to a Rigid Body . . . . . . . . . . . . . . . . . . . . . 104.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 Velocity and Acceleration Analysis 14 Velocity and Acceleration Analysis4.1 IntroductionThe motion of a rigid body (RB) is defined when the position vector, velocityand acceleration of all points of the rigid body are defined as functions oftime with respect to a fixed reference frame with the origin at O0.Let ı0, 0and k0, be the constant unit vectors of a fixed orthogonalCartesian reference frame x0y0z0and ı,  and k be the unit vectors of a bodyfixed (mobile or rotating) orthogonal Cartesian reference frame xyz (Fig. 4.1).The unit vectors ı0, 0, and k0of the primary reference frame are constantwith respect to time.A reference frame that moves with the rigid body is a body fixed (or mo-bile) reference frame. The unit vectors ı, , and k of the body fixed referenceframe are not constant, because they rotate with the body fixed referenceframe. The location of the point O is arbitrary.The position vector of a point M (M ∈(RB)), with respect to the fixedreference frame x0y0z0is denoted by r1= rO0Mand with respect to themobile reference frame Oxyz is denoted by r = rOM. The location of theorigin O of the mobile reference frame with respect to the fixed point O0isdefined by the position vector rO= rO0O.Then the relation between the vectors r1, r and r0is given byr1= rO+ r = rO+ x ı + y  + z k, (4.1)where x, y and z represent the projections of the vector r = rOMon themobile reference frame: r = x ı + y  + z k.The magnitude of the vector r = rOMis a constant as the distance be-tween the points O and M is constant (O ∈(RB) and M ∈(RB)). Thus, thex, y and z components of the vector r with respect to the mobile referenceframe are constant. The unit vectors ı,  and k are time-dependent vectorfunctions.The vectors ı,  and k are the unit vector of an orthogonal Cartesian referenceframe, thus one can writeı · ı = 1,  ·  = 1, k · k = 1, (4.2)ı ·  = 0,  · k = 0, k · ı = 0. (4.3)Figure 4.14 Velocity and Acceleration Analysis 24.2 Velocity Field for a Rigid BodyThe velocity of an arbitrary point M of the rigid body with respect to thefixed reference frame x0y0z0, is the derivative with respect to time of theposition vector r1v =dr1dt=drO0Mdt=˙r1=˙rO+˙r = vO+ x˙ı + y˙ + z˙k + ˙xı + ˙y + ˙zk, (4.4)where vO=˙rOrepresent the velocity of the origin of the mobile referenceframe O1x1y1z1with respect to the fixed reference frame Oxyz. Because allthe points in the rigid body maintain their relative position, their velocityrelative to the mobile reference frame xyz is zero, i.e., ˙x = ˙y = ˙z = 0.The velocity of point M isv = vO+ x˙ı + y˙ + z˙k.The derivative of the Eqs.(4.2) and (4.3) with respect to time gives˙ı · ı = 0,˙ ·  = 0,˙k · k = 0, (4.5)and˙ı ·  +˙ · ı = 0,˙ · k +˙k ·  = 0,˙k · ı +˙ı · k = 0. (4.6)For Eq.(4.6) one can introduce the convention˙ı ·  = −˙ · ı = ωz,˙ · k = −˙k ·  = ωx,˙k · ı = −˙ı · k = ωy, (4.7)where ωx, ωyand ωzmay be considered as the projections of a vector ω,ω = ωxı + ωy + ωzk.To calculate˙ı,˙,˙k one can use the relation for an arbitrary vector vv = vxı + vy + vzk = (v · ı) ı + (v · )  + (v · k) k. (4.8)Using Eq.(4.8) and the results from Eqs.(4.5) and (4.6) one can write˙ı = (˙ı · ı) ı + (˙ı · )  + (˙ı · k) k= (0) ı + (ωz)  − (ωy) k=ı  kωxωyωz1 0 0= ω × ı,4 Velocity and Acceleration Analysis 3˙ = (˙ · ı) ı + (˙ · )  + (˙ · k) k= (−ωz) ı + (0)  + (ωx) k=ı  kωxωyωz0 1 0= ω × ,˙k =˙k · ıı +˙k ·  +˙k · kk (4.9)= (ωy) ı − (ωx)  + (0) k=ı  kωxωyωz0 0 1= ω × k.The relations˙ı = ω × ı,˙ = ω × ,˙k = ω × k. (4.10)are known as Poisson formulas.Using Eqs.(4.4) and (4.10) one can obtainv = vO+ xω × ı + yω ×  + zω × k = vO+ ω × (xı + y + zk) ,orv = vO+ ω × r. (4.11)Combining Eqs.(4.4) and (4.11) it results˙r = ω × r. (4.12)Using Eq.(4.11) one can write the components of the velocity asvx= vOx+ zωy− yωz,vy= vOy+ xωz− zωx,vz= vOz+ yωx− xωy.The relation between the velocities vMand vOof two points M and O onthe rigid body isvM= vO+ ω × rOM, (4.13)orvM= vO+ vrelMO, (4.14)4 Velocity and Acceleration Analysis 4where vrelMOis the relative velocity, for rotational motion, of M with respectto O and is given byvrelMO= vMO= ω × rOM. (4.15)The relative velocity vMOis perpendicular to the position vector rOM,vMO⊥ rOM, and has the direction given by the angular velocity vector ω.The magnitude of the relative velocity is |vMO| = vMO= ω rOM.4 Velocity and Acceleration Analysis 54.3 Acceleration Field for a Rigid BodyThe acceleration of an arbitrary point M ∈(RB) with respect to a fixedreference frame O0x0y0z0, represents the double derivative with respect totime of the position vector r1a =¨r1=˙v =dvdt=ddt(vO+ω×r) =ddtvO+ddtω×r+ω×ddtr =˙vO+˙ω×r+ω×˙r.(4.16)The acceleration of the point O with respect to the fixed reference frameO0x0y0z0isaO=˙vO=¨rO. (4.17)The derivative of the vector ω with respect to the time is the vector α givenbyα =˙ω = ˙ωxı + ˙ωy + ˙ωzk + ωx˙ı + ωy˙ + ωz˙k (4.18)= αxı + αy + αzk + ωxω × ı + ωyω ×  + ωzω × k= αxı + αy + αzk + ω × ω =αxı + αy + αzk.where αx= ˙ωx, αy= ˙ωy, and αz= ˙ωz.In the previous exprerssion the Poisson formulas˙ı = ω × ı,˙ = ω × ,˙k = ω × k,have been used.Using Eqs.(4.16), (4.17) and (4.18) one can write the acceleration of thepoint M asa = aO+ α × r + ω × (ω × r) . (4.19)Using Eq.(4.19) one can write the components of the acceleration asax= aOx+ (zαy− yαz) + ωy(yωx− xωy) + ωz(xωx− xωz) ,ay= aOy+ (xαz− zαx) + …