288 



MOTION OF EIGID BODIES 



angles to the plane but in the direction opposite to that of the 

 former motion. Thus the total displacement of the particle is 



LNcodt-LN'oo'dt. (116) 



Since L is on the diagonal of the parallelogram, we see that the 

 area of the triangle PLQ is equal to that of the triangle PLQ f , so 



thafc 



Again, since PQ, PQ' are in the ratio of GO : &/, this equation may 

 be written in the form 



and on comparing with expression (116) we see that the displace- 



ment of the particle L vanishes. 



Thus the resultant of the two angular velocities is a motion such 



that the points P and L both remain at rest. It is therefore an 



angular velocity having PR, the diagonal of the parallelogram, as 



axis of rotation. 



We must next find the magnitude of this angular velocity. 



Let us denote it by fl. From Q draw perpendiculars QX, Q Y to 

 Q p PQ'au&PR. The displace- 



ment of the particle Q in 

 time dt will be QY-ldt 

 at right angles to the plane. 

 This displacement, however, 

 can also be obtained by com- 

 pounding the displacements 

 produced by the two angular 



velocities a>, ft/. That produced by the former is nil, since Q is on 



the axis of rotation ; that produced by the latter is QXco' dt. Thus 



QY-Cldt = QX- ft/ dt. (117) 



We have QY- Pit = QX- PQ', 



each being equal to the area of the parallelogram, and on combin- 

 ing this relation with (117), we find 



FIG. 139 



