506 G. Barns — Application of the Displacement 



normal distance of the center of gravity from the axis. The 

 rise of the latter above its lowest position is 



y = h (1 — cos 6) sin </> = 2h sin <£ sin 2 — (1 ) 



and the energy potentialized, if the total mass is JH, will be 



W = 2Mgh sin </> sin 2 - (2) 



which for small displacements corresponds to the torque Fh. 

 at the angle 6. This torque is 



9 W 



-—- = Mgh sin <f> sin 6 = 3Igh $6, (3) 



do 



nearly ; whence, the total force F acting at the center of 

 gravity, or F/3f])er gram of mass If, is 



JJ=^ e (4) 



In the above apparatus 3£ = 1215 grams, h = 80 centi- 

 meters. Hence per vanishing interference ring, since the 

 gilding moves, if AN is the displacement of the micrometer to 

 bring back the center of ellipses to the' fiducial sodium line 



AN/2_AN 

 6 --R--2R < o) 



where i? is the distance of the point of the grating at the line 

 of light corresponding to the slit, from the axis of rotation. 

 In the apparatus H = 111 centimeters. Hence the angle 

 corresponding to a vanishing ring is, since AN = 30 X 10 -6 , 



30 X 10" 6 _s 



6 — - = 13 X 10 radians = -028 ' 



2 X HI 



Furthermore if cf> = 1° = *0175 radians 



F/M — 981 X -OlVo X 13 X 10" 8 = 2-3 X 10~ 6 dynes 



per vanishing ring, per gram mass at the center of gravity of 

 the pendulum. 



The total pull of the center of gravity of the above pendu- 

 lum is thus 



F= 2-3 X 10~ 6 X 1245 = 2-9 X 10~ 3 dynes 



per vanishing ring, on one side. By lengthening the radius 

 from htoB this may be decreased to about 2 X 10 -3 or less. 

 Hence in case of gravitational attraction at one centimeter of 



