Physics Help Forum moments of inertia

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 Dec 1st 2016, 09:38 PM #1 Junior Member   Join Date: Dec 2016 Posts: 2 moments of inertia There are two questions about the moments of inertia, but I am lack of the knowledge of calculus. Could anybody help me please? 1. Determine the moments of inertia of the shaded area shown with respect to the x and y axes. 2. Determine the polar moments of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.
 Dec 1st 2016, 09:41 PM #2 Junior Member   Join Date: Dec 2016 Posts: 2 I am sorry this picture should be no.2 and this one should be no.1
 Dec 2nd 2016, 06:17 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 I would assume that you have a list of formulas of moments of inertia for various geometric shapes, perhaps in your text or in class notes, correct? You don't need to know calculus if you have such a list - just apply the formulas adding and subtracting shapes as required. For example, for the second problem (the one that asks about moment of inertia about point P), you can consider the shaded area as a large rectangle of width 4a and height 2a, with a piece missing of width 2a and height a. The formula for moment of inertia about a point on the edge of a rectangle is m/12(4h^2 + w^2) -- more on this below; use this to calculate the moment of inertia of the large rectangle then subtract the moment of inertia of the missing piece. A handy thing to remember is that if you know the moment of inertia about the centroid of a shape, you can determine the moment of inertia about another point P' which is distance d away by using the formula $\displaystyle I_{P'} = I_P + md^2$ where 'm' is the mass of the shape in question. So, if your book does not have the formula for moment of inertia about a point on the edge of a rectangle, but does have it for the center of the rectangle, you can determine the proper equation as follows: $\displaystyle I_{edge} = I_{center} + md^2$ The formula for I_center is (m/12)(h^2+m^2), so for a point on the edge you have: $\displaystyle I_{edge} = \frac m {12} (h^2 + w^2) + m(\frac h 2)^2 = \frac m {12} (4h^2 + w^2)$ Hope this helps. topsquark likes this. Last edited by ChipB; Dec 2nd 2016 at 06:30 AM.

 Tags cmoments, inertia, moments