Substituting into equations 6, 9 and 10, we obtain T ror+Tou UP=vo+u×ro〃 From equations 6, 9 and 10, replacing P with O", we have that To=ro+rom, vo=von+w xro ao+×r”+山×(u×ro). Therefore, we can write, ao〃+山xrP+×( These equations show that if the velocity and acceleration of point P are referred to point o"rather than point O, then r'ptrp, vof uow, and aof aom, although the angular velocity and acceleration vectors w and a, remain unchanged Instantaneous Center of rotation We have established that the motion of a solid body can be described by giving the position, velocity and acceleration of any point in the body, plus the angular velocity and acceleration of the body. It is clear that if we could find a point, C, in the body for which the instantaneous velocity is zero, then the velocity of the body at that particular instant would consist only of a rotation of the body about that point (no translation). If we know the angular velocity of the body, w, and the velocity of, say, point O, then we could determine the location of a point, C, where the velocity is zero. From equation 9, we have 0=υo+ Point C is called the instantaneous center of rotation. Multiplying through by w, we have -wX vo (w x rc), and, re-arranging terms which shows that rc and vo are perpendicular, as we would expect if there is only rotation about C. can be determined geometrically as the intersection of the lines which go through points P and P and are perpendicular to up and vpr. From the above expression, we see that when the angular velocity, w, is very mall, the center of rotation is very far away, and, in particular, when it is zero(i. e. a pure translation), the enter of rotation is at infiniSubstituting into equations 6, 9 and 10, we obtain, rP = rO′ + r ′ O′′ + r ′′ P (11) vP = vO′ + ω × r ′ O′′ + ω × r ′′ P (12) aP = aO′ + ω˙ × r ′ O′′ + ω˙ × r ′′ P + ω × (ω × r ′ O′′ ) + ω × (ω × r ′′ P ) . (13) From equations 6, 9 and 10, replacing P with O′′, we have that rO′′ = rO′ + r ′ O′′ , vO′′ = vO′ + ω × r ′ O′′ , and aO′′ = aO′ + ω˙ × r ′ O′′ + ω × (ω × r ′ O′′ ). Therefore, we can write, rP = rO′′ + r ′′ P (14) vP = vO′′ + ω × r ′′ P (15) aP = aO′′ + ω˙ × r ′′ P + ω × (ω × r ′′ P ) . (16) These equations show that if the velocity and acceleration of point P are referred to point O′′ rather than point O′ , then r ′ P 6= r ′′ P , vO′ 6= vO′′ , and aO′ 6= aO′′ , although the angular velocity and acceleration vectors, ω and α, remain unchanged. Instantaneous Center of Rotation We have established that the motion of a solid body can be described by giving the position, velocity and acceleration of any point in the body, plus the angular velocity and acceleration of the body. It is clear that if we could find a point, C, in the body for which the instantaneous velocity is zero, then the velocity of the body at that particular instant would consist only of a rotation of the body about that point (no translation). If we know the angular velocity of the body, ω, and the velocity of, say, point O′ , then we could determine the location of a point, C, where the velocity is zero. From equation 9, we have, 0 = vO′ + ω × r ′ C . Point C is called the instantaneous center of rotation. Multiplying through by ω, we have −ω × vO′ = ω × (ω × r ′ C ), and, re-arranging terms, we obtain, r ′ C = 1 ω2 (ω × vO′ ) , which shows that r ′ C and vO′ are perpendicular, as we would expect if there is only rotation about C. Alternatively, if we know the velocity at two points of the body, P and P ′ , then the location of point C can be determined geometrically as the intersection of the lines which go through points P and P ′ and are perpendicular to vP and vP ′ . From the above expression, we see that when the angular velocity, ω, is very small, the center of rotation is very far away, and, in particular, when it is zero (i.e. a pure translation), the center of rotation is at infinity. 4