The Elementary Principles of Mechanics volume 2
The Elementary Principles of Mechanics volume 2
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In our eReader you can find the full English version of the book. Read The Elementary Principles of Mechanics volume 2 Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book The Elementary Principles of Mechanics volume 2
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To quickly assess the difficulty of the text, read a short excerpt:
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'.
What to read after The Elementary Principles of Mechanics volume 2?
You can find similar books in the "Read Also" column, or choose other free books by A Jay Augustus Jay Du Bois to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
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'.
What to read after The Elementary Principles of Mechanics volume 2?
You can find similar books in the "Read Also" column, or choose other free books by A Jay Augustus Jay Du Bois to read onlineMoreLess
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