C. Barns — Repulsion of Two Metallic Disks. 97 



'for the llow through a ring whose section is ydr, if rj is the 

 viscosity of the gas and V the volume of tiuid crossing per 

 second. If the flow is steady so that dp/dt — 0, for all dis- 

 tances from the center and if the liquid is virtually incom- 

 pressible, i. e., V independent of r, the problem may be solved 

 without difficulty. Neither of these conditions is quite true. 

 The second, however, inasmuch as the average pressure incre- 

 ment is exceedingly small relative to atmospheric pressure, may 



be admitted. Suppose, therefore, that at any time, y and V 

 are constant relatively to r, and integrate the equation. Hence 

 the pressure excess at r is 



If R is the radius of the disks, p = at r = R, and the equa- 

 tion may be considered to hold short of /' = 0. Thus the 



thrust / 2 7T r dr. p becomes 

 o 



P = ZrjRV/y* 



But V— — 7rR 2 dy/dt, whence on integrating Pdt 



Pt= 2 7 r V E 3 ( 1 --~) 



\y yj 



But for the horizontal pendulum the force P is proportional 

 to y — y', which may be written 



P^PAv-y') 



so that —Pdt becomes 



-P dt = 2 7 rr } R>dy/y'(y-y') 

 This, on Anal integration, reduces to 



P.rT l\y. y/ + og i-y'/y I 



natural logarithms being in question. Since y — y' = AiV/2, 

 equation for P, corresponds in case of the large disks to P = 

 F'k = 2 X 65-2AiT/2. Hence, P = 2 X 65-2. Furthermore, 

 in case of this apparatus in the present experiments, the fol- 

 lowing data may be entered : 



y' = 1/15 cm.; y = -10 cm.; R = 10'2 cm.; rj= 190 X 10" 6 



The following summary contains some corresponding values 

 of y = AiV/2 and t computed in this way. 



Am. Jour. Sci. — Fourth Series, Vol. XXXIX, No. 229.— January, 1915. 



7 



