the Problem of Columbus. 465 



7. From (12) Ai/rsin 2 0=A,-Ca> 3 (~ + cos e J'> 



Cn(- + cos0 J 



When cos#=/z, is known in terms of the time, (16) will 

 give i/r and therefore yfr in terms of the time. Thus (16) 

 defines the precession. 



Now we see from (16) that if becomes zero we must 

 either have i/r sin infinite or 



*£ +1 ) 



Cn 



\a / 



= 0. 



/AC _/A a- 



But if ijrsinfl were infinite the kinetic energy would be 

 infinite, as we see from its expression in f 11) ; for the only 



negative term in that expression is — 2<*y^ sin 2 Ol ~ + cos # ) > 



which cannot become infinite, for i/rsin 2 # is not, neither 

 is « 3 . 



.'. the egg cannot stand up on end unless 



'AC" 



S + c (7+ 1 )'-^ +1 ) 



The right-hand side of (16) is an even function of 0, so 

 that if it be expanded in terms of we see that the terms 

 containing will disappear, and ^r sin will tend to the limit 



zero, and ^ to a finite limit if the above condition be 

 satisfied. 



8. The integral of the equation (15) cannot be expressed 

 in known functions, but the following considerations will 

 throw some light on the nature of the motion :— 



r ^^-^A» + W-An-+A-K)-a M {Jai + 0^ + ^ } ] * 



-An 2 + A 2 K-(X 2 + C,r) A ;-^ 

 v J a- a 2 M. 



V=± 



+ Mi+'V(&+A+O J; + 2C^ + (C-A)^) ^ 

 A | A ■ h* , 2h k * 



