Relative velocity method_02
- 1. For the mechanism shown in the figure, determine the velocity of C, E and P. Crank OB rotates at 120 rpm. Solution: 1 =2 = 2 (30 ) 120 = 377 (a) Solve for the linear velocity of crank 2 60 (b) Draw the link using CAD: (c) The vector equation connecting known velocity vB to unknown velocity vC on the same link BC is: = + / (d) Sketch the velocity polygon, (scale 1:5) Select suitable position of pole OV, draw the first velocity vector VB, considering its magnitude and direction. Sketch velocity vC from OV, knowing only its direction, which is perpendicular to crank DC. = (114.0306)(5) = 570.153 Finally, sketch velocity VC/B from the tip of VB, the direction of VC/B is perpendicular to link BC = (77.2158)(5) = 386.079 The intersection of OvC and bc is the magnitude of each velocity: and / . (e) To obtain linear velocity of point P, we will use the principle of velocity image. From c, draw line ce, in the direction perpendicular to the line joining CE. Similarly, from Ov, draw a line Ove, in a direction perpendicular to the line joining ED.
- 2. The intersection of the two lines define velocity image of DCE. It can be verified that triangles DCE and Ov ce are similar. Line OVe on velocity polygon represents velocity ve. Line OVp on velocity polygon represents velocity vP. The sense of vP is decided by referring to the velocity equation: = + / The magnitude of velocity P is: = (137.1862)(5) = 685.931 The magnitude of velocity P is: = (141.3807)(5) = 706.9035 Summary of Answers: = . = . = . Reference: Mechanisms and Machine Theory by Ashok G. Ambekar © 2007 by Prentice Hall of India Private Limited, New Delhi