124] MOTION OF A SOLID OF REVOLUTION. 187 



in the usual notation of elliptic integrals. If we eliminate t 

 between (5) and (9), and then integrate with respect to 6, 

 we find 



the origin of x being taken to correspond to the position 6 = 0. 

 The path can then be traced, in any particular case, by means of 

 Legendre's Tables. See the curve marked I in the figure. 



If, on the other hand, the solid does not make a complete 

 revolution, but oscillates through an angle a on each side of the 

 position 6 0, the proper form of the first integral of (3) is 



, . ABQ ft> 2 , 10X 



where Bm i a= _J|._ (12). 



If we put sin 6 = sin a sin -\Jr, 



this gives & = . (1 sin 2 a sin 2 -vlr), 



sin 2 a v 



whence = F(sma, ty) (13). 



Transforming to i/r as independent variable, in (5), and integrating, 

 we find 



x = =- sin a . F(sm a, ty) -^-cosec a . E(siu a, -^r 

 Qco 



--- 



The path of the point is here a sinuous curve crossing the line 

 of the impulse at intervals of time equal to a half-period of the 

 angular motion. This is illustrated by the curves III and IV of the 

 figure. 



There remains a critical case between the two preceding, where 

 the solid just makes a half-revolution, 6 having as asymptotic 



