Moment of Inertia. 233 



would describe the arc GG' perpendicular to FG ; through the 

 point G' let the line G'K be drawn parallel and equal to GF. 

 Instead of supposing the body to turn about F, we may imagine 

 it carried parallel to itself with a velocity equal to GG 7 , and that 

 at the same time its several parts turn about a moveable point G 

 with such a velocity that by taking G'K = GF, the point K 

 would describe the arc KF, equal to G'G ; for, on this supposition, 

 the/ point F of the body L would still remain stationary. Now 

 the body in this case being free, the resultant of all the motions of 

 rotation about the moveable point G is zero. Consequently the 79. 

 resultant of all the motions with which the body is actually urg- 

 ed is no other than that which the body L would have, impress- 

 ed with the velocity GG / ; that is, this force must be perpendio 

 ular to -FG, and equal to 



L X GG', 



the mass of the body being denoted by L f Now since the parts 

 of the body describe similar arcs, we have 



*$& FM : FG :: v : GG' = 



therefore, the resulting force of all the motions of rotation about 

 the point F, is 



L x FG x v 



FM~ 



But although this resultant is the same as if, the body being 

 free, the centre of gravity had received the velocity GG', still it 

 will be seen that this resultant does not pass through G, but 

 through some point O of FG produced ; since, the more remote 

 parts having the greater force, the resultant, while it falls on the 

 same side of F with the centre of gravity, must pass at a greater 

 distance from F than this centre. Designating this distance FO 

 at which the resultant passes by D 7 , we shall have for the mo- 



ment of the resultant L *j?f X * X >'. 

 r M 



If now, at the instant when the forces M'KTC, &c., above consider- 

 ed, begin to act upon the parts of the body, there be opposed to 

 them, at the distance Z)', a force equal to that just determined; that is, 



Mech. 30 



