Tham khảo tài liệu 'principles of engineering mechanics (2nd edition) episode 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 16 Kinematics of a particle in plane motion s m area m2 s 2 z m s 1 2 area D J2 0-5 0-10 0-15 0-20 0-25 0-30 0-35 0-40 Since t2 tỵ l p ds the area under the graph of l z versus 5 will give the required time. Corresponding values are given below and are plotted in Fig. . o Area 5-6s Figure 3D 30 40 5 m l z sm-1 s m l z s m-1 s m 0 25 5 30 10 35 15 20 40 The time taken is found to be approximately s. Example At a particular instant a point on a mechanism has a speed of m s and a tangential acceleration of magnitude m s2. If the magnitude of the total acceleration is m s2 what is the radius of curvature of the path being traced out by the point at this instant Solution Choice of co-ordinates is not difficult for this problem since radius of curvature is featured only in path co-ordinates. In these co-ordinates the total acceleration a see Fig. is given by a atet anen set v2 p en see equations Figure The magnitude of a is V .v2 y2 p 2 and substitution of the numerical values gives V 22 52 p 2 and p m Example See Fig. . The centre c of the wheel of radius m has a constant velocity of m s to the right. The angular velocity of the wheel is constant and equal to 6 rad s clockwise. Point p is at the bottom of the wheel and is in contact with a horizontal surface. Points Q and R are as shown in the figure. Determine a whether or not the wheel is slipping on the surface b the velocities and accelerations of the points p and Q and c the velocity and acceleration of the point R. Solution Usually the simplest way of dealing with the motion of a point on a wheel which is rotating and translating is to determine the motion of the wheel centre and add on the motion of the point relative to the centre. So for an arbitrary point A and centre c we can make use of c VAJC see equation .
