Tham khảo tài liệu 'principles of engineering mechanics (2nd edition) episode 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 56 Kinematics of a rigid body in plane motion Figure Returning to Fig. and assuming that ÚAB is anticlockwise and of given magnitude we can place the points a b and d on the diagram Fig. . Note that a and d are the same point as there is no relative velocity between A and D. To construct the point c we must view the motion of c from two vantage points namely D and B. Since DC is of fixed length the only motion of c relative to D is perpendicular to DC hence we draw de perpendicular to DC. Similarly the velocity of c relative to B is perpendicular to CB hence we draw be perpendicular to BC. The intersection of these two lines locates c. The angular velocity of CB is obtained from z c b CB clockwise . The direction of rotation is determined by observing the sense of the velocity of c relative to B and remembering that the relative velocity is due only to the rotation of CB. Note again that angular velocity is measured with respect to a plane and not to any particular point on the plane. Instantaneous centre of rotation Another graphical technique is the use of instantaneous centres of rotation. The axes of rotation of DC and AB are easily seen but BC is in general plane motion and has no fixed centre of rotation. However at any instant a point of zero velocity may be found by noting that the line joining the centre to a given point is perpendicular to the velocity of that point. Figure In Fig. the instantaneous centre for member BC is found to be the intersection of AB and DC since the velocity of B is perpendicular to AB and the velocity of c is perpendicular to CD. If the velocity of B is known then Ve i2e VB vc Ú BC I2B - I2C Each point on link CB is instantaneously rotating about I2. Velocity image If the velocity diagram has been constructed for two points on a rigid body in plane motion then the point on the velocity diagram for a third point on the link is found by constructing a triangle on the vector diagram similar to that
