SECTOR OF A CIRCLE 



129 



We find, on putting a = ~ that the center of gravity of a semi- 



4 

 circle is at a distance - a from the center. 



O 7T 



100. Center of gravity of a sector of a circle. The center of 

 gravity of a sector of a circle can be found by regarding the sector 

 as made up of a triangle and a segment. The center of gravity of 

 the triangle and of the seg- 

 ment both being known, it 

 is easy to find the center of 

 gravity of the whole figure. 

 A simpler way is the 

 following: We can divide 

 the sector by a series of 

 radii into a great number o< 

 of very narrow triangles. 

 The weight of each triangle 

 may be replaced by the 

 weight of a particle placed 

 at its center of gravity. 

 Now, in the limit, when the 

 triangles become of infini- 

 tesimal width, the center of 

 gravity of each is on its median at a distance from the center of 

 the circle equal to f a, where a is the radius of the circle. Thus 

 all the particles lie on a circle of radius f a. 



The weight of any particle must be equal to the weight of the 

 triangle OPQ which it replaces. It must, therefore, be propor- 

 tional to the base PQ of the triangle, and this again is proportional 



P to pq, the piece of the circle of 

 radius J a which is inclosed by 

 the triangle. Thus the weight of 

 the particle which has to be 

 placed in the small element pq of this circle is proportional to the 

 length pq. On passing to the limit, and making the number of 

 triangles infinite, we find that the string of particles may be 



FIG. 72 







FIG. 73 



