CIECULAR AEG 



125 



proportional to x dx, and as we have just found, this at once leads 



to the result _ 9 



x = I a, 



where a is the length of the median. This is exactly the result 

 previously obtained. 



97. Center of gravity of a circular arc. The same method can 

 be used to find the center of gravity of a wire bent into the form 

 of a circular arc PQ. Let be the center of the circle, and A the 

 middle point of the arc, and let the whole arc subtend an angle 

 2 a; at the center. Consider a small element cd of the half PA of 

 the wire. Let the angle dOA be 6, 

 and cOA be 6 + dO, so that the 

 element subtends the angle dO at 

 the center. If a is the radius of 

 the circle, the length of this ele- 

 ment is a d6, so that if w is the 

 mass of the wire per unit length, 

 the mass of the element will be 

 wa d6. This and the similar ele- 

 ment c'd' in the half AQ of the 

 wire form a pair of equal particles 

 equidistant from the central line 

 OA. They may be replaced by a 

 single particle of mass 2 wa dO at their center of gravity. This 

 center of gravity is in OA, at the point at which the line joining 

 the two elements meets OA, and hence at a distance a cos 0. from 0. 

 Denoting this by x, and the mass 2 wa d6 by ra, we have, for the 

 distance x of the center of gravity of the whole wire from 0, 



x = 



/< 



IwadB 



