How to determine rectangular and polar radii of gyration?

1. Apr 25, 2015

sagarbhathwar

1. The problem statement, all variables and given/known data

2. Relevant equations
dIx = y2dA
Ix = ∫y2dA

dA = ydx
A = ∫(dA

3. 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.

2. Apr 25, 2015

BvU

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: Apr 25, 2015
3. Apr 26, 2015

sagarbhathwar

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?

4. Apr 26, 2015

BvU

$dI_x = y^2 dA$ 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.