EOD OF VARYING DENSITY 123 



5. A right-angled isosceles triangle is cut out from a lamina in the shape of 

 an equilateral triangle so as to have the same base as the original triangle. Find 

 the center of gravity of the V-shaped piece left over. 



6. The center of gravity of a quadrilateral lies on one of its diagonals. Show 

 that this diagonal bisects the other diagonal. 



CENTERS OF GRAVITY OBTAINED BY INTEGRATION 



94. Center of gravity of a rod of varying density. Let AB be 



a rod of which the weight per unit length varies from point to 

 point, and let p denote the weight per unit length at any point. 



Let P, Q be two adjacent points, the distances of P, Q from the 

 point A being x and x + dx respectively. Then the length PQ is 

 dx, and its mass is pdx, where p is the P Q 



mass per unit length at this point. ' ' 



When dx is made vanishingly small, FlG - 68 



the distance of the center of gravity of PQ from A may be taken 

 to be x. Hence if x denotes the distance of the center of gravity 

 of the whole rod from A, 



X = 



m 



where m is the mass of any element such as PQ, and the summa- 

 tion is taken over all the particles of which the rod is formed. 

 Putting m = pdx, this becomes 



x 



or, in the notation of the integral calculus, 



I pxdx 



x = 



fpdx 



(28) 



where the integration extends in each case over the whole rod. 

 The variable p will be a function of x, and the integrations cannot 

 be performed until the exact form of this function is known. 



