the Problem of Columbus. 471 



And on the right-hand side they are contained in 



Picking the terms in -3 out of this we have 



AC ■„,. 





Therefore the term in — _ vanishes. 

 a 



Thus the term in — A in the absolute term disappears as the 



1 a 



•one in —. did. 

 a 



Accordingly if we multiply by a 2 we shall get rid of all 

 terms with a in the denominator. 



Now, in expanding the right-hand side, we find that //. is 

 never multipled by a power of a less than its own by more 

 than two, for if we multiply by a 2 we get 



2\?i{h+ari^^ + tf + 2 u U lfJ L+d 2 {{C-A)^ + A} 



and X and n only contain a once each in their denominators. 



Therefore if we expand as far as yu, 4 we keep all terms con- 

 taining a 2 ; and there is no expansion in terms of /x on the 

 left-hand side, and no power of fi higher than the third. 



It follows from this that if we neglect a 3 the equation can 

 be solved in elliptic functions. 



This gives a closer approximation to the case of the ordinary 

 top than the usual one where the lower end of the top is 

 taken as a point, i. e. a is neglected altogether. 



As such a top is usually spun, g) 3 is large and m small, and 

 Q is fairly small. In such a case the condition of art. 8 (7) is 

 nearly satisfied, so that such a top will stand nearly upright 

 even without slipping. 



1 2. The expressions for the reactions can be obtained from 

 the equations of motion. 



From (4) 



^ n . n[Aco*d-c( h +cos6>Yjtf 

 2 _ Co) 3 _ L \(( JJ 



M asintf , AC 4 . „ „/h \- \ 



.^+Adtf« + o(J+oo.«)} 



Clearly this can never become infinite. 



2 12 



