AN ELONGATED PROJECTILE. 
121 
Then, if OA be the equatoreal axis in the plane ZOC, 
c-yu = component momentum in the direction OA 
= Z cos ZOA — — Z sin a, 
c^w = component momentum in the direction OC 
= Z cos ZOO = Z cos A \ 
and therefore* if OT be the direction of motion of 0, the tangent to the 
helical path described by 0, 
tan COT = - - = tan a. 
w 
In consequence of the direction of motion OT not being in the 
direction of the axis OC, the body will experience a couple about the 
axis OB, perpendicular to the plane AOC, of magnitude. 
(<? 3 — c x ) uw — — (c x — c 3 ) w 2 tan a. * 
c i 
Since the rate of change of angular momentum is equal to the 
impressed couple* therefore 
G/jl sin (a — 0) = 3 (c x — c 3 ) w 2 tan a. 
( 1 ) 
But 0 cos 6 = component angular momentum about OC — c 6 r, 
— Cr sin 6 = „ „ „ u OA — c $; 
and p = — /x sin a, 
since the velocity of C, considered as due to the angular velocity about 
OA is jo, and due to the angular velocity about OZ is /* sin a, and these 
are in opposite directions. 
Therefore 
But* from (1)* 
G sin 0 = c 4 /x sin a, 
and tan 0 = sin a. 
c 6 r 
_ t sin (a — 6) 
Gfi sin (a — 0) — c 6 r[A 
cos 6 
— ctf'p, (sin a — cos a tan 6) 
= CqT/x sin a — <? 4 /x 2 sin a cos a 
= C? > (q — c 3 ) tan a ; 
* Thomson and Tait. “ Natural Philosophy,” sec. 324, new edition. 
