126 CENTER OF GRAVITY 



where the integration is from 6 = to 6 a. Simplifying, we find 



r 



i/0 = 



f>6= a 



I 



Je = o 



a sn a 

 a 



(29) 



giving the position of the center of gravity. 



When a is very small, sin a and a become equal, so that for 

 very small values of a, formula (29) reduces to x = a, as it ought. 

 This simply expresses that as the curvature of the arc decreases, 

 the center of gravity approximates more and more closely to the 

 middle point of the arc. Finally, when a = 0, the arc becomes a 

 straight rod, and the center of gravity is, of course, found to be 

 exactly at the middle point. 



For an arc bent into a semicircle, we take a = > and obtain 



Lt 



x = 



= = .6366 a. 



7T 



98. The center of gravity of a cir- 

 cular arc PQ can be found in an in- 

 teresting manner, without the use of 

 the integral calculus. 



From symmetry it is clear that the 

 center of gravity of the arc AP must 

 lie in the radius which bisects the 

 angle A OP. Let p be this center of 

 gravity, and let q be the center of 

 gravity of the arc AQ. Then the 

 center of gravity or the whole arc PQ 

 must be N, the middle point of pq. 



Now since the angle pON = \ a, 

 we have 



ON = Op cos \ a . 



FIG. 70 



