The combined motion of translation and rotation from AB to A’B’ may be assumed to be a motion of entirely rotation about a certain point 0 which is called the instantaneous centre of rotation.
The point 0 can be found out by following procedure.
Draw perpendicular bisectors of chords AA’ and BB’. This is PO and QO. This bisectors meet at 0 which is the instantaneous centre (of zero velocity).
Fig. 4.4 (a) shows different parts of a reciprocating engine. Point A executes linear motion along line AC and point B moves in a circle with ro as angular velocity.
Let w = angular velocity of link AB about 0 .
VA = Linear velocity of point A
= wr A = wAO
V s = Linear velocity of point B
= wr8 = w BO
VA / VB =AO / BO
VA is at right angles to AO at A.
VB is at right angles to BO at B.
Thus, if VA and VB are known, then the instantaneous centre of AB is obtained by drawing perpendiculars to the directions of the velocities at A and B. The point, where these two perpendiculars meet, is the instantaneous centre.