52 APPLIED MECHANICS 
of surface; W=weight of body; then I,=I+ Ar? for the surface, and 
I, =1+ Wr’ for the body. 
Consider a small element P of the surface or body, and let m denote 
Fig. 51. Fig. 52. Fig. 53. 
its area or weight. Referring now to Figs. 51 and 52, let PM and PN 
be perpendiculars from P to the axes XX and X,X, respectively, and let 
PQ be the perpendicular to MN from P. Then, 
PN?=MN? + PM? — 2MN “MQ 
mPN?2 = SmMN?2 + =mPM2 - 2MN=mMQ, 
but 2m MQ =0, therefore I, =I+A7* for the surface, and I, =I+W7? 
for the body. 
The case which is of most importance, on account of its frequent 
occurrence in practice, is the simple one in which the surface EF is 
a plane figure (Fig. 53), and the parallel axes XX and X,X, are in the 
plane of the figure. In this case P and Q coincide. 
Corollary 1.—If & and k, are the radii of gyration about the axes XX 
and X,X, respectively, I= Ak? or Wh’, and I, =Ak; or Wk}. Hence 
ki =k? +72, 
Corollary 2.—The radius of gyration about a given axis passing 
through the centre of gravity is less than the radius of gyration about 
an axis parallel to the given axis, and the axis about which the radius of 
gyration is least must pass through the centre of gravity. 
69. Moment of Inertia—Fundamental Examples.—The graphical 
method of finding moments of inertia was explained in Article 67, p. 50, 
The analytical method will now be used, and in 
practice this is generally the most convenient. Tis Eee 
(1) Straight line, or straight and uniform —x- —+ 1 
slender rod (Fig. 54) about an axis XX, per-  y! EA 50 
pendicular to it, and passing through one end. » r 
Consider an element of length dx at a dis- Ae Bae 
tance w from the axis. Let w denote the weight of the rod per unit 
of length. The weight of this element is wdx, its moment of inertia 
is wa*dx, and the total moment of inertia 
1 Y 
I,= | warda = w| edz = 
0 0 
? 
wit _WE 
a De 
where W is the total weight of the rod. 
?2 
Radius of gyration squared =k; = 3° 
