55, 56] 



ROTATIONS ABOUT PARALLEL AXES. 



201 



From (Fig. 41) lay off the translation vector 00' of length, t and 

 find a point C on the perpendicular bisecting 00' which makes the 

 angle OCO' equal to ca, the angle of rotation, and in the right sense. 

 Then if OC be rotated about through the angle to to C f and 

 then C" be moved by the translation it will 

 return to C. Therefore the point C remains 

 fixed, and is the center of rotation, and 

 thus the rotation co about C is equivalent 

 to the equal rotation about together with 

 the translation, 



1) r = 200 sin ~, 



and if p is the perpendicular from C to 00', 



a 2 Fig. 41. 



56. Rotations about two Parallel Axes. As before the 

 motion is uniplanar and is specified by two points. Let A and B (Fig. 42) 

 be the intersections of the axes with the plane of the paper perpen- 

 dicular to them. Turn about 

 A through the angle co 1 , 

 bringing B to B'. Then turn 

 about B' through the angle a> 2 , 

 bringing A to A'. Bisect o 1 

 by AC. B could be brought 

 to B' by rotation about any 

 point of AC, since all such 

 points are equidistant from 

 BB ! . Bisect w 2 by B'D. 

 A could be brought to A' 

 by rotation about any point 

 in B'D. Therefore the motion 

 of A and B could be produced 



by a rotation about 0, the intersection of AC and B'D. Triangle 

 AOA' is isosceles. 



Angle AOD = angle OAB' + angle AB'O = ^ + ^, 

 Angle AOA' =-- 2 - angle AOD = o^ + o> 2 , 



that is, two rotations about parallel axes compound into a rotation 

 equal to their algebraic sum about a parallel axis. To find the 

 position of this axis we have 



OS' OA AB 



sm '- 



*- sin 



2 ~ 2 2 



If the order of rotation is changed we obtain a different result. 



