How to determine rectangular and polar radii of gyration?

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SUMMARY

The discussion focuses on calculating the rectangular and polar radii of gyration for a given shaded area using the equations for moment of inertia (Ix) and area (A). The user initially calculates Ix as 1023/640 and the area A as 15/16, leading to a radius of gyration (kx) of 1.705. However, they realize a mistake in their approach, specifically missing a factor of 1/3 in the calculation of Ix, which results in an incorrect value for kx. The correct radius of gyration about the x-axis is confirmed to be 0.75.

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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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