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AUBURN MECH 6710 - Velocity and Acceleration Analysis

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I.4 Velocity and Acceleration Analysis 0Contents4 Velocity and Acceleration Analysis 14.1 Kinematics of the Rigid Boby . . . . . . . . . . . . . . . . . . 14.2 Driver Link . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.3 RRR Dyad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.4 RRT Dyad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.5 RTR Dyad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.6 TRT Dyad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34I.4 Velocity and Acceleration Analysis 14 Velocity and Acceleration Analysis4.1 Kinematics of the Rigid BobyThe 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 O0x0y0z0(primary reference frame). The unitvectors ı0, 0, and k0of the primary reference frame are constant with respectto time. Let ı,  and k be the unit vectors of a mobile orthogonal Cartesianreference frame Oxyz (Fig. 4.1). A reference frame that moves with the rigidbody is a body fixed (or mobile) reference frame. The unit vectors ı, , and kof the body fixed reference frame are not constant, because they rotate withthe body fixed reference frame. The location of the point O is arbitrary.The position vector of a point M (M ∈(RB)), with respect to the fixedreference frame O0x0y0z0is 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 b etween thevectors r1, r and r0is given byr1= rO+ r = rO+ xı + y + zk, (4.1)where x, y and z represent the projections of the vector r on the mobilereference frame. The magnitude of the vector r = rOMis a constant as thedistance between the points O and M is constant (O ∈(RB) and M ∈(RB)).Thus, the x , y and z components of the vector r with respect to the mobilereference frame are constant. The unit vectors ı,  and k are time-dependentvector functions. The vectors ı,  and k are the unit vector of an orthogonalCartesian reference frame, the following relations can be writtenı · ı = 1,  ·  = 1, k · k = 1, (4.2)ı ·  = 0,  · k = 0, k · ı = 0. (4.3)I.4 Velocity and Acceleration Analysis 2Velocity of a point on the rigid bodyThe velocity of an arbitrary point M of the rigid body with respect tothe fixed reference frame Ox0y0z0, is the derivative with respect to time ofthe position vector r1v =dr1dt=˙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 p osition, their velocityrelative to the mobile reference frame Oxyz is zero: ˙x = ˙y = ˙z = 0.The velocity of the 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) the following convention is introduced˙ı ·  = −˙ · ı = ωz,˙ · k = −˙k ·  = ωx,˙k · ı = −˙ı · k = ωy, (4.7)where ωx, ωyand ωzcan be considered as the projections of a vector ωω = ωxı + ωy + ωzk.To calculate˙ı,˙,˙k the following formula is introduced for an arbitrary vector,d,d = dxı + dy + dzk = (d · ı) ı + (d · )  + (d · k) k. (4.8)Using Eq. (4.8) and the results from Eqs .(4.5) and (4.6) it results˙ı = (˙ı · ı) ı + (˙ı · )  + (˙ı · k) k= (0) ı + (ωz)  − (ωy) k=ı  kωxωyωz1 0 0= ω × ı,I.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) the velocity of M isv = 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) the components of the velocity arevx= vOx+ zωy− yωz,vy= vOy+ xωz− zωx,vz= vOz+ yωx− xωy.I.4 Velocity and Acceleration Analysis 4Acceleration of a point on the 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.(4.13)The acceleration of the point O with respect to the fixed reference frameO0x0y0z0isaO=˙vO=¨rO. (4.14)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 (4.15)= 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 expression the Poissonformulas have been used.Using Eqs. (4.13), (4.14) and (4.15) the acceleration of the point M isa = aO+ α × r + ω × (ω × r) . (4.16)The components of the acceleration areax= aOx+ (zαy− yαz) + ωy(yωx− xωy) …


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