How to determine rectangular and polar radii of gyration?

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The discussion focuses on determining the rectangular and polar radii of gyration for a given area. The user successfully calculated the radius of gyration about the y-axis but encountered an error with the x-axis calculation, initially obtaining an incorrect value of 1.705 instead of the correct answer of 0.75. The mistake was identified as a missing factor in the integration process for Ix, which should involve a double integral rather than a single one. Clarifications were requested regarding the specific errors in the calculations, emphasizing the need to correctly apply the equations for dIx and dA. The conversation highlights the importance of careful integration and understanding of the geometry involved in the problem.
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Homework Statement


[/B]
Determine the rectangular and polar radii of gyration of the shaded area about the axes shown.

Snapshot_zpsemlzcj6q.jpg


Homework Equations


dIx = y2dA
Ix = ∫y2dA

dA = ydx
A = ∫(dA

The Attempt at a Solution


dIx = y2dA
=(x6/16) *(x3/4) dx
=(x9/64)dx
Ix =(integrating from 1 to 2) ∫dIx = 1023/640

dA = ydx
A = (integrating from1 to 2)∫(x3/4)dx = 15/16

kx = √(Ix/A) = 1.705

I am getting the radius of gyration about y-axis correctly but that baout x-axis is wrong.
The actualy answer given is 0.75. I am unable to figure out where I am going wrong.
 
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No, my mistake, let me re-think, or rather: think about this one some more.
One at the time: start with Ix:
Yes: work it out dA bit more and you'll see you miss a factor 1/3 in Ix.

By the way, you confuse me with ##k_x =\sqrt{I_x/A} = 1.705##; I get ##1.306\;##; a typo ?​
 
Last edited:
Yes. Sorry I forgot to take the square root so yea a typo.
But to be honest, I don't get what you are saying. Could you be more specific in where I am going wrong?
 
##dI_x = y^2 dA## needs to be worked out to ##dI_x = y^2 dy dx##: it's a double integral. You seem to treat it as a single one.
 

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