158 



KINEMATICS. 



[280. 



HI, we obtain, in the place of ar', the components ap at right 

 angles to IP and ar ' perpendicular to r Q f . 



As of the four components o> 2 /, ap, o)V , ar Q ' the last two are, 

 by (6), equal and opposite, it follows that the acceleration of 

 P has only the two components, aPp along PI, and ap perpen- 

 dicular to IP. 



280. The total acceleration of any point P is therefore 

 proportional to the distance / of this point from the centre of 

 acceleration /, viz. 



j=p V<* 2 + &> 4 ; (7) 



and the angle -ty it makes with this distance IP, being given by 

 the relation 



tan^ = -^ 2 , (8) 



is the same for all points. By (5), this angle i|r is equal to the 

 angle CHI. 



All points oft^a circle described about / as centre have 

 accelerations of equal magnitude but of different directions. 

 All points on a straight line drawn through 7 have accelerations 

 that are parallel but differ in magnitude. 



281. Returning to the resolution of the acceleration into 



three components coV, ar, wu, 

 as given in Art. 276, let us take 

 the common tangent of the 

 centrodes as axis of x, their 

 normal as axis of y (Fig. 75), 

 and let x, y be the co-ordinates 

 of any point P whose distance 

 from C is CP r. 



As the direction cosines of 



Fig. 75. wV, ar, <*u are respectively 



-x/r, y/r\ y/r,x/r; o, I, 



we have for the components of the acceleration / parallel to 



the axes : 



