— -1888.] viscous rotating cylinder. 263 
The boundary conditions at the free surface r=a are to be 
found in just the same way as when there is no central nucleus, 
but the attraction of the nucleus must be taken into account. The 
values of V,, V, given by (14), (15) will be increased by 
Araldite 
— 21 (o—p)b log =. 
and instead of (29) we shall have to assume 
n—-1 8B /o : PB ere 
= ray | +5, iE " 1)} + (=) =~ o%...(68), 
We shall then obtain, resolving tangentially, 
2n (n —1) Aa” + 2n (n+ 1) A’a™” 
— B[{h’a? — 2n(n —1)} J, (ha) — 2haJ,, ,, (ha)] - 0 ...(69). 
— B' [{h’a* — 2n (n—1)} Y, (ha) — 2haY,,,,(ha)] 
Resolving normally 
Aa” {a (a+ 2a) + nk, —2n(n — 1) va/a?} 
+ A’a™ {a (— a+ 2v) + nk, + 2n (n +1) va/a’?} 
+ B (Qe + nk, + 2nva/a’) J, (ha) — 2nvala? . haJ,’ (ha)} 
+ BY {(2ea + nk, + 2nva/a*) VY, (ha) — 2nvala?. ha,’ (ha)} 
The elimination of the four constants A, A’, B, B’ from equa- 
tions (66), (67), (69), (70) leads to a somewhat complicated equation 
to determine the admissible values of a or h which after several 
reductions can be put into the form 
=0 (70). 
a(at2or)t+nk, —4n(n—1)va/a?, 2n(n—1) , b” ; 0 
a(—a+2e1)+nk, Ser O (eS 11) Oe , 2na"/b” 
2(n—l)a’J,,,(ha)/ha, 2had,,, (ha)—h'e'S, (ha), ab" J,(hb)—B"S, (ha), hbJ, ,,(hb) 
2(n—1)a’°Y,,,,(ha)/ha, 2haY,, (ha)—h?a°Y (ha), a"b” V, (hb)—b" Y,, (ha), hbY,, (hd) 
n+1 
By putting b=0 in this equation it reduces, as it should, to 
the form (33). 
Unfortunately, however, this equation does not admit of having 
its roots discussed with the same facility as does (35). It cannot 
mm general have real or conjugate complex roots unless w = 0, since 
by the results (ii), Gi) of § 10 this is impossible when b=0, 
moreover the general statements of § 11 shew that a purely 
imaginary root is impossible. 
