246.] 



MOMENTS OF INERTIA. 



135 



(10) Solid sphere, for a diametral plane. 



(n) Solid ellipsoid, for the three principal planes. 



(12) Area of ring bounded by concentric circles of radii a ly a z , for a 

 diameter. 



(13) Area of the cross-section of a JL-iron : (a) for its line of 

 symmetry; (b) for its base. (Dimensions as in Fig. 8, Part II., p. 18.) 



(14) A rectangular door of width b and height h has a thickness 8 to 

 a distance a from the edges, while the rectangular panel (whose dimen- 

 sions are b 2 a, h 2 a) has half this thickness. Find the moment 

 of inertia for a line through the centroid parallel to the side b. 



245. The moment of inertia of any mass M for a point can 

 easily be found if the moments of inertia of the same mass 

 are known for any line passing through the 

 point, and for the plane through the point 

 perpendicular to the line. Let O (Fig. 29) 

 be the point, / tbe line, TT the plane ; r, q, p 

 the perpendicular distances of any particle of 

 mass m from O, /, TT, respectively. Then 

 we have, evidently, r 2 =g> 2 -\-fl 2 . Hence, mul- 

 tiplying by m, and summing over the whole 

 mass M, 



or, putting 



are the radii of inertia for O, /, TT, 



Fig. 29. 



, where r , 



(i) 



246. The moment of inertia of any mass M for a line is 

 equal to the sum of the moments of inertia of the same mass 

 for any two rectangular planes passing through the line. Thus, 

 in particular, the moment of inertia for the axis of x in a 

 rectangular system of co-ordinates is equal to the sum of the 

 moments of inertia for the sar-plane and ;rj/-plane. This fol- 

 lows at once by considering that the square of the distance 

 of any point from the line is equal to the sum of the squares 



