Figure 111 — Diagram for 

 calculation of the lift. 



(J and far removed from the cylinder; see Figure 111. In time 8t the boundaries of this fluid 

 are displaced through a distance (J 8t into new positions QQ ^, Q,'Q,(. At the end of the dis- 

 placement, the part of the fluid that lies between PP^, and Q'lQ^'has the same momentum as 

 had the part of the fluid that occupied this position originally, since the motion is steady. 

 Hence this region may be disregarded; and the net change in momentum of the mass of fluid 

 under discussion is equal to the momentum present in the newly occupied layer of space 

 QQ.PP. minus the momentum that has disappeared, from the layer Q'Q( P'P^, which has 

 been vacated. 



In calculating the momentum in these two layers, use may be made of the approximate 

 description of the motion given in the last section. Here F = 0, since there is no outflow 

 from the cylinder. The uniform motion at velocity -V contributes nothing to the difference 

 in momentum between the two layers. The velocity gp gives rise, on the whole, to no momen- 

 tum having the direction of f , because of symmetry, but it does give rise to transverse 

 momentum. 



Take the axis of polar coordinates in the direction of V. Then an element of cross- 

 sectional area of the layer QP Q P '\s & parallelogram of height U St on a base of 

 rd 9/ {-cos d), as in Figure 111, and, using [72b] for g'p, the transverse momentum in this 

 layer is 





Vdt 



rd 9 



cos 9 



pTVSt. 



Equal but oppositely directed transverse momentum was initially present in the layer 

 Q'Q'P'P' The net change in transverse momentum of the original mass of fluid is thus 



163 



