33 2 Rotation of a Solid Body round a Fixed Point. 



gyration ; this is evident from the consideration, that these per- 

 pendiculars are reciprocal to the radii vectores of the ellipsoid 

 (3). In fact, the moment of inertia with respect to any axis 

 will be represented by the formula 



M = (a* cos 2 a + b 2 cos 2 /3 + c 2 cos 2 7) ju = juP 2 = -^ ; (5) 



J& 



(R, P] denoting the radius vector and perpendicular on tangent 

 plane of the ellipsoid (4) ; and (&',} the corresponding lines in 

 the ellipsoid (3). 



IV. To FIND THE MAGNITUDE, POSITION, AND DIRECTION OF 

 THE STATICAL COUPLE PRODUCED BY THE CENTRIFUGAL 

 FORCES. 



If from any point (x, ?/, z) of the body, a perpendicular be 

 let fall on the axis of rotation (a, /3, 7), the centrifugal force 

 will be represented by the product of the square of the angular 

 velocity and this perpendicular : the corresponding elementary 

 statical couple will be found by multiplying the centrifugal 

 force by the distance from the foot of the perpendicular to the 

 origin, which is represented by the quantity (x cos a + y cos /3 

 + z cos 7). The components of the elementary couple will be 

 proportional to the projections of the triangle formed by the 

 lines before mentioned. The components of the elementary 

 couples must be integrated for the entire extent of the body, 

 and the integrals thus found will be the components of the 

 couple produced by centrifugal force: the expressions aj-e as 



follows : 



o> 2 (x cos a + y cos j3 + z cos 7) (z cos j3 - y cos 7) dm ; 



to 2 (x cos a + y cos /3 + z cos 7) (x cos 7 - z cos a) dm ; 

 a> 2 (x cos a + y cos ft + z cos 7) (y cos a - x cos /3) dm. 



If the axes of co-ordinates be principal axes, these expressions, 

 when integrated, will become 



w- cos /3 cos y(B-C) = JT (B-C}\ 



tw 2 cos a cos 7 ( C- A)=pr (C - A] ; (6) 



w 2 cos a cos p(A-B) =pq (A-B): 



