10, 11] 



CURVATURE. RADIAL AND TRANSVERSE ACCEL. 



17 



Differentiating again 



d*r 



dcp dr 

 lit lit 

 dtp dr 

 dt lit 



dt* ~ dt* 

 The direction cosines of the radius vector are: 



cos(rx) = cos (pj cos(ry) = sinqp, 

 so that we obtain by resolution, 



d*r 



34) 



a r = - cos (r x) 



being less than the scalar acceleration of the radial velocity by the 

 product of the radius vector and the square of the angular velocity. 



(If ,,- = 0, the motion is circular, and a r is the normal acceleration.) 



The direction cosines of a line perpendicular to r and in the 

 direction of increasing <p are, sin qp, cos qp, so that for the trans- 

 verse acceleration we obtain, 



dt* dt* dt* dt dt 



which may be written 



Of course we have 



11. Moment of Acceleration. 



The expression in the paren- 



dS 



thesis of 35) is by 23) equal to twice the sector velocity 



us call 



36) 



dt 



-T.2 



sector acceleration. 

 2 d*S 



ra m = 



r dt* 



^d*S 



dt* 



Suppose (Fig. 8) AS represents 

 the acceleration a, then AC per- 

 pendicular to r represents a y , 

 therefore ra v is twice the area 

 of the triangle OAR But that 

 is the moment of the acceleration 

 about the point 0. Accordingly 

 twice the sector acceleration is 

 equal to the moment of the accel- 

 eration about the origin, or 



WEBSTER, Dynamics. 



- . 



Thus in plane motion 



Fig. 8. 



d*S 

 -^=xa y -ya x 



d*y 



>L 



dt* 



Let 



