5 
From this we infer the velocity nL of the end of On equals K, i.e., 
nL = K ; 
and since the principal moment of the air resistance is perpendicular to 
xOx v then nL is perpendicular to same plane. 
Hence the angular velocity of xOxi round Oxi is fixed by the equation 
dv _ nL _ K 
dt ~ nT\ ~ nT x 
From the triangle nOTi, 
nTi—G- sin (8—y) 
dv _ K 
dt Gr sin (8—y) 
... (5) 
and Gr sin (8—y) — = K . ... (6) 
From (3) and (4), 
Gr sin y = B sin 8 ..(6i) 
From (3), (6), and (6 X ), 
Ap sin 8 ^ — B sin 8 cos 8 = K ... ... (7) 
Hence 
(±\ 2 = A P . . / ^ V ' K r8 x 
\ dt / 2 B cos 8 — v 4 B 2 cos 2 8 B sin 8 cos 8 ' ” ^ 
Now in a vacuum jfir=0, and therefore-^ =0, which shows the 
negative sign before the root need only be retained. 
In order that — should be real, 
dt 
A 0 0^1 PK 
A P = 4 SEs 
( 9 ) 
Therefore the minimum velocity is determined from the following 
equation 
A 2 p 2 = 4 
BK 
tan 8 
( 10 ) 
P 
2_ 
A 
v 
BK 
tan 8 
or 
(ii) 
