1891.] Mr GJiree, On thin rotating isotropic disks. 



205 



The constants in (11), (13) and (15) are not all arbitrary, being 

 connected through the identity (1). From it we find 



a x = A- 2a, 



1 m + n n 4 



6 = 0. 



Thus the solution we have arrived at is 



8 = A + G (z 2 - Y) - !«> 7(m + n), \ 



w = (A-2*)z- g Cz> + I (^ G - 8?) (2* 3 - Szr 2 ), 



u = — + ar + £(4*V - r s ) + ~ f — C- 



o) p 



m + n 



.(16), 



- + ar+^(4>z 2 r-r 3 ) + - - 



where all the constants are independent. 



To determine the constants we have the surface conditions 

 (4)-(7). 



From (4) and the expression for zz in terms of the strains we 

 easily find 



(m + n)A- 4wa + I 2 {(m + n)C- 16»f } 



+ ^| 8 «C-l(3m + n) 67-1^^ = 0.. .(17). 



Since this holds for all values of r between a' and a, the constant 

 part and the coefficient of r 2 must separately vanish. Thus we 

 get 



(m + >i) 4 - 4na + Z 2 {(m + n) (7 - 16<} = (18), 



8nK-USm + n)G-~ co 2 o = (19). 



2 2 in + n r v ' 



From (5) and the expression for rz in terms of the strains we 

 at once obtain 



16n£-(m + n)C=0 (20). 



The equations (18), (19) and (20) are satisfied by 



m + n . 



a = — A A, 



4m 



n \ in — n ,, 



O = — t ; CO'O, y 



2 m (m + n) 

 y 1 m — n 2 



.(21). 



