222 UNIVERSITY OF COLORADO STUDIES 



3. To find the center of mass of a segment of a circle. 



Let ADB (Fig. 2), be the segment, and let the construction be 

 as there made, except that G and G ' are now the Centers of Mass of 

 the segments ADB and A'DB'. The portion transferred is now the 

 infinitely small triangle AMA' to BMB'. Since the Center of Mass 

 of a triangle is § of the median from the vertex, the distance gg' 

 traveled by the Center of Mass of this triangle is, in the limit, f AB. 

 The area of this triangle is ^AM. A'M sin a, which in the limit 



becomes -J I —r- ] ■^ *"• 



GG' 



i(i!)'" 



|AB segment 

 OGXa AB'Xa 



0G= 



|AB 8 X segment 

 cube of chord 



12 times area of segment 



2r sin= 



or 0G= (5) 



3(^— sin e cos 6) ' 



26 being the angle of the segment, and a segment being easily cal- 

 culated as the difference between a sector and a triangle. 



For a semicircular lamina this becomes — as before. 



Stt 



4. The Center of Mass of a zone of a hollow sphere is found 

 in an elementary manner by the elegant method of comparison with 

 the right cylinder circumscribed to the sphere which seems due to 

 Collignon. (See Collignon Statiq%ie p. 299; Loney Statics and Dy- 

 namics, etc.) The Center of Mass is on the axis of the zone and 

 half way between the parallel planes. For a spherical cap, a partic- 

 ular case of zone, this result may easily be expressed in the form: — 

 Distance of Center of Mass from center of sphere 



2(1— cos ») ^ ' 



