The Elementary Principles of Mechanics volume 2

Cover The Elementary Principles of Mechanics volume 2
The Elementary Principles of Mechanics volume 2
A Jay Augustus Jay Du Bois
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But since OX passes through the centre of mass of the area, 2my = (page 17), where m is the mass of an elementary area. But m - da, where a is the area and 5 the surface density. Hence 2day = S2ay = 0, or 2ay = 0. Therefore, since 2a = A = the entire area, we have I' = 2ay* + Ad* = 1+ Ad*.
272 APPLICATIONS OF STATICS. [CHAP. II.
That is, the moment of inertia of an area icith reference to an eccen- tric axis is equal to the moment of inertia with reference to a par- allel axis through the centr
...e of mass plus the area into the square of the distance between the two axes.
Radius of Gyration of an Area. The square root of the quotient obtained by dividing the moment of inertia of an area with refer- ence to any axis by the area is called the radius of gyration of the area with reference to that axis. We denote the radius of gyration by K. Then by definition K ' V ~A and K = V ~AI ^3. J\.
where K' and I' indicate an eccentric axis, and K and / an axis through the centre of mass. We have then AK" = I, or AK'* = I'.


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