﻿250 



Prof. J. G. Gray 



on 



tin 



about ee'. If a is the angle bOa, and I the length of the 

 solid, we have for the sectional area 



\t 2 {2it — ct^ + r 2 sin -^a cos \a. 



And since A = 2lr sin ^a, and K 2 = %r 2 sin 2 -Ja, we have 



OG 



r sin 3 -ia 



IT- 



^a + sin ^acos \a 



Similarly, if G' is the centroid of the upper solid, we have 



ia — Sin fa COS fa 



The positions of the centroids of the solids obtained by 

 dividing a right circular cylinder into two portions by means 

 of planes Oa, Ob (fig. 6) are easily determined. If G is the 

 centroid of the larger solid, we obtain at once, by supposing 



Fig. 6. 



the solid turned through a small angle 6 about the axis of 

 figure of the complete cylinder, so that a arrives at a', and 



b at &', . 



VxOG = 2AK 2 sini<*, 



where A is the area represented by Oa, and K is its radius 

 of gyration about the axis of figure. We have, if I is the 

 length of the cylinder. 



OG 



2ZrJr 2 sin£a 

 P(2fl—a)7 



4 r sin ia 



3 2ir-cc ' 

 If the cylinder is divided into two equal parts, we obtain 



