Polar moment of inertia/polar radius of gyration via integration

  • Thread starter drunknfox
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Homework Statement



Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P.

http://imgur.com/8Kc1S

Homework Equations



Jp = Ix + Iy
Ix = &int y^2dA
Iy = &int X^2dA

The Attempt at a Solution



A = 2(a/2)(a) + (2)(1/2)(a/2)(a) = 3a^2/2

Jp = Ix + Iy

Ix = &int y^2dA = ??? Im having trouble with this next step. If someone could please help me with it and explain to me whats suppose to integrated I would be eternally grateful

I have 2 &int a-0
 

Answers and Replies

  • #2
jambaugh
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Firstly be warned are specific to planar objects of unit area or density 1/area.

Integrating y^2 dA involves expressing y and dA in terms of your independent variable (variable of integration) and its differential.

Looking at your region (a trapazoid?) it would appear that your best bet is to express the width at a point as a function of height i.e. express left and right line boundaries in terms of x as a function of y.

Left boundary x =f(y)= p y + q,
Right boundary x = g(y) = r y + s.

You are basically, in the Riemann sum, slicing the object up into horizontal strips with thickness dy and width x_right - x_left = g(y)-f(y), and so its area is:

dA = [g(y)-f(y)] dy

With this integrate y^2 dA between the appropriate limits.
 

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