83 Proceedings of the Royal Irish Academy. 



We thus have shown that wheu S = 0, according as /x is greater or less 

 than fii, the representative point moves along the line of no sliding, or 

 along a line inclined at tan"' ^ to the axis of R, whose projection on i2 = 

 makes with the axis of P the angle associated with the negative value 

 of ^. 



The value of ju, can also be obtained by noting that as the circle touches 



the hyperbola for that value of fx, we have F(6) = and — -r^-^ = 0, 



do 



so that 

 a - b 



^ /u sin 2^ + ^ sin - /cos 9=0, (a - h) jj, cos 29 + r/ cos + / sin 6 = 0. 



Solving for / and g 



f = {a-h)n sin' 0, g = (b - a) fi cos' 6, 

 .: {a-lfy} = {f^f)\ 



10. Assuming now that S„ is not zero, and that ^o is not a root of 



F{6) = 0, sliding takes place initially, and we have 



dS 



-r=, = - Z7, cos ^ - V-, sin = - <t,{6 . 



dB ^ 



i' T7i = f^. sin e - U, cos e = F(d), 

 dM 



1 dS MB) F'(e) + p(0) ^ ,„ . ,^ , 



■"■ STe^ ~ Fm " Fid) ' ^ -^^^ ^ a/ism^e + h^icos'e. 



Integrating from 0„ to 6, 



dE S S,F{e,) . ( ce p{e) 



■ de F(6) \F(0)]^ ^^^- I ]e,'. 



de\ 



.F(9) 



As dli is always positive, it follows that 6 increases continually if F{6n) 

 is positive, and decreases continually if F(9o) is negative, until it becomes 

 equal to a root a of F(d) = 0. 



By the above equations, S and R both remain finite until nearly 

 equals a, and it is most important to find if they remain finite when = a. 



Putting 6 = a - i, 



so that integrating from a small value ■»; of t to j, and then making i in- 

 definitely small, the important part of 



-J 



p{e)d{9) 



FiO) 



IS 



f 2}(a)dt ^ _ 2'(a) ■ J_ 

 r, tF\a) F\a) ^ V 



