10 PHILOSOPHICAL TRANSACTIONS. [aNNO 1713. 



by the equation c = cg'^ X (a + d) in prop. 3. Hence also will be given the 



parallelogram cg x go = - (corol. prop. 3) = cg^. By help of which, 



from the centre of gravity and the point of suspension being given, the centre 

 of oscillation is given by division only. Therefore, in any example, it will 

 always be most convenient to find this parallelogram first, by computing either 

 the quantity d, or c, by a proper assumption of the centre of suspension. 



It remains that we now illustrate this by some examples. 



Example 1. Let the figure proposed be the pyramid adc, fig. 8, whose 

 base is the parallelogram ad ; and let the motion of its centre of gravity be in 

 the plane passing through the vertex c, and the diameter ef of the base parallel 

 to the side ab. 



To perform the calculation most conveniently, let the vertex itself c be the 

 point of suspension. Then, after the manner of prob. 1, let the figure be 

 reduced to the physical plane of the isosceles triangle cef, fig. Q, in which 

 efy parallel to ef, represents a physical line composed of particles p. Put 

 CH = a, HF = b, and ch = x. Then, from the nature of the figure it will 

 be eh =. ' , and the particle j&, situated at the point z, will be as x. Or 



rather, making hz = t;, then vx will be the base of the elementary prism, and 

 p will be as vxx. Hence it will be c = cz^ X vxx = vxx^ + v^vv^^- There- 

 fore the sum of all the cz^ X /) in the line Az, will be vxx^ -\- ^vxv^\ and in 

 the line ef (putting — for v) that sum will be ——3 — X xx*. Hence again 



taking the fluent, and writing a for ar, it will be c = {Qba^ -f- 2^^) X rV"^- 

 But the pyramid itself is a = ^hd^, and the distance of the centre of gravity 

 G from the vertex c, is cg = ^. Hence 



CG'' = - = CG X GO = — . 



A A 80 



Exam. 1. Let the figure proposed be a right cone, described by the rota- 

 tion of the isosceles triangle ecf about the perpendicular ch. 



Here again taking the vertex c for the centre of suspension, and making 

 CH = o, HE = ^, ch = X, hz =- V, as above, it will be 



p = 2fiV (— ^' - t;"); hence c = 2i;iV (— ^' - v^) X (x'' + v""). Let 



•* ^ aa ^ aa / \ t / 



B be the segment of the circle described on the diameter ef which is adjacent 

 to the absciss hz = v, and ordinate V ( — x"^ — v^) : then the sum of all the 



^ aa ^ 



cz^ X pf in the right line hz, will be "^ x^ xb — -^vx (— x'^ — v^)^. And 



when V = eh. this sum will be -^-—r~ oc^ '^b ; whose double — r — x^xb is a 

 part of c in the right line ef. But the area b is as x^-y put therefore b = car'; and 



