Impulsive Motion of Electrified Systems. 139 



By the results of § 15 we obtain 



U=i M P+U =U ^=g )i =U (l+i»*+.. i ), 



T = W~iMP=U ^ ] ~, i =™ 2 U„(i+in 2 +...). 



It' we put ft = sin 7, we obtain 



U = J(sec 7 + eos 7) D , 

 m*=2T/w 2 =sec 7 . TJ /v 2 3 mi = 2d(£ju)/du=seG s y . U /v 2 , 



where m t is the fi transverse mass " and mj the " longitudinal 

 mass." 



§ 27. Circular disk moving in its own plane. — In this 

 instance the axis of symmetry o£ the system is not parallel t3 

 the direction of motion, as is the case in the other problems 

 Ave have considered. We therefore require two angular 

 coordinates, (p and yfr, where c/> is measured from the axis of 

 the disk and \jr is measured round the axis, starting from the 

 plane which contains the axis, and is parallel to the direction 

 of motion. Then, if the radius defined by <f> and yfr makes 

 the angle 6 with the direction of motion, we have 



cos # = sin <£cost/c. 

 Taking the disk, of radius b, as the limiting form of an ellip- 

 soid, we see that p, the thickness of the pulse, is equal to 

 '2b sin <£, and thus, by § 23, dQ/d: = Q 2b sin (/>. Now dco, 

 the element of solid angle, is equal* to sin cf> d<j) d\jr and thus. 

 by (3T), since dQ/dz is independent of z, 



w = A"^Q 2 CC sin 2 d< f> df 

 8w6 JJ (l-ncos0) 2 



. ptfQfCC (l-sin'»cos-*^) 



~~ *7t6 J J (1 - n sin </> cos f ) 2 ^ ^ ' 



where goes from to it and i/r from to 27r. 



Expanding the denominator and integrating term by term, 

 we find 



7^Q2 f 3/]V , 7/1.3y , 4 11/ 1. 3.5V , I 



"We can express this result in a finite form by means of the 

 first and second complete elliptic integrals F and E. If 

 iA" = 1 — n~ snr a, we have 



