628 Prof. T. H. Havelock 
Comparing with (32) it follows that 
A,=2kva,?/a7{r~P+k(k+4)}. . . . (35) 
With this expression for the solving function, (28) gives 
= A 2kyN Me oie ea 
7 el ar, erate (ea - (86) 
By expanding r? by (33) and putting r=1, it can be 
shown that 
2(k+4) 
P+ k(k+A4) ” . e . ° (37) 
Carrying out the integration in (36) and using (37), we 
find 
> 
dQ, N 2kN GRE 
Eisai owen oame © CY) 
The angular acceleration has a finite limiting value in this 
case, the same as if the cylinder and enclosed liquid were 
rotating like a rigid body. We notice that in this case zero 
is excluded from the roots of the equation (29) for the 
exponents of the nucleus. 
Integrating (38) we obtain the angular velocity of the 
cylinder at any time; then, using the differential equation 
(26), we may complete the solution by writing down the 
angular velocity of the liquid. It is found to be given by 
‘au Nf nga wens 
on I+4pat By 12(k+4) 
_ ANS BPI Ar/aje let 
I > E+ ED}, (A) 
184 
