180 T. H. Havelock 
3. We now calculate the forces on the cylinder from the expression 
2 
RAAT & si f(r) dz. (7) 
We write (6) in the form 
dw imtinlA 
as, =c+y a + 2B 2", (8) 
(oe) ( = Aye P*(k) e-2ka — i" F*( = k) e—2kb Ke 
j [ed nl 
where [oy = | 
of(=1)e (6) Hit (Ket Ke 
=| g=z ni ek (9) 
From (7) and (8), we obtain 
X—iY = —27p(icAy+ Li7+1n! A, B,). (10) 
If I’ is the circulation round the cylinder, we have J’ = 277A); further, 
using (9), we easily obtain X = 0, and 
(ee) * —2ka _ i738 *( —2kb 
¥ = pel’+2np| NEN i sone (ae dk. (11) 
; => 
For the moment about the origin we have 
dw\? 
eo TO ake 
M ph b2( 72) dz (12) 
= 2np Ri{cA,+21"(n+1)!4,4,B,}, (13) 
where & denotes the real part. 
Using (9), this may be expressed in the form 
co Fl * —2ka __ F'( _ Eo fae —2kb 
M= 2npRil cA, + | Hats aisle ; ae eee ae 
3 = 
c {B"(k) F(=k) = F'(54) F(x) ar | . (14) 
0 1 — ed 
To complete the solution of the problem in any given case we have to 
determine the function F(«) so that the boundary condition of zero normal 
velocity is satisfied over the contour C. 
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