128 



CENTER OF GRAVITY 





99. Center of gravity of a segment of a circle. Suppose next 

 that we require to find the center of gravity of a segment PAQN 



of a circle, cut off by a chord PNQ, 



T5 



which subtends an angle 2 <z at 

 the center of the circle. Let us 

 divide the whole segment into thin 

 strips parallel to the chord, and 

 let cc'dd' in fig. 71 be a typical 

 strip bounded by chords cc 1 ', dd f . 

 Let the angle cOA be 0, and let 

 dOA be + dO. Then the width 

 of the strip is cd sin or a sin e0, 

 while its length is 2 c% or 2 a ski 0. 

 Thus the area is 2 a sin 2 e0. Its 

 mass may be supposed to be all 

 F ^ concentrated at n, of which the 



distance from is a cos 0. 



Thus if # is the distance from of the center of gravity of the 

 whole segment, we shall have 



h' 

 f d' 



f (a cos 0) (2 a sin 2 



f(2asi 



and the integration has to be taken from 6 = to = a. Simpli- 

 fying, we have 



f"sin 2 



J 



cos 0^0 



f sin 2 

 Jo 



l- (a sin a cos a) 



3 a sin a cos a 



