Centre of Gravity in particular Bodies. 61 



With respect to GG", since y = \/aa: x 2 , we shall have, 



or f (ax dx x 2 dx 

 ~~ or * 33 



but 



we have, therefore, 



GG" = A 



If the question related to the entire segment, since it is evident 

 that the centre of gravity , must be in the radius C^, which bi-^'g* 

 sects the arc, and that it must be at the same distance from JVW 

 as the centre of gravity of each of the two semi-segments APM, 

 APM, we shall have 



CE = ^ 3 - A X 8 X PM 3 A MM' 3 

 ~ APM APM APM 



that is, the distance of the centre of a circle from the centre ofgravi* 

 ty f an y one f it s segments, is equal to the twelfth part of the cube 

 of the chord, divided by the surface of this segment. 



107. The centre of gravity of a sector CMAM' may be easily Fig ' 45< 

 found, by observing that E, the centre of gravity of the segment 

 MdM', G that of the sector, and F that of the triangle, are all 

 in the radius G/2; and that, according to the principle of mo- 

 ments, the moment of the sector must be equal to the moment of 

 the segment plus that of the triangle. We have then 



CMAM x CG = MAM x CE -f CM, M X CF. 





Now we have just found CE = %Tnj, which may be chang- 



We know, moreover, that CMM = PM X CP, and that 



CF = | CP, 



so that G/Jf-M' X CFis reduced to | PM X CP 2 . Substituting, 

 therefore, these values, we have 



CMAM x CG = | plr 3 4- | PJIf x CP 2 

 = | PM(PM 2 4- GP 2 ) 



