Moment of Inertia of a Rectangle

1. Nov 11, 2013

physicskid123

I have been trying to do the moment of inertia of a rectangle and I have it figured out when we have the center of the rectangle as the center of the rotation.
The equation is ∫∫ρ(x^2+y^2)dy dx where the first integral is from -b/2 to b/2 (if b is the height) and the second integral is -a/2 to a/2 (if a is the width).
I can't seem to figure out how to change the parameters of the integrals for if the rotation of the rectangle is at any other point, say a corner. Or what if it was on the bottom but still in the center of width, but the bottom of height. How do I adjust the integrals for these parameters?

2. Nov 11, 2013

tiny-tim

hi physicskid123! welcome to pf!
i] they're limits not parameters

ii] your limits are the same, it's your integrand that needs changing !

3. Nov 11, 2013

physicskid123

Thanks, I'm new to the calculus lingo. So, for example, if the center of rotation is the bottom of the rectangle (its thin btw) so the center point would be (a/2,0) if the bottom left of the rectangle were the origin, how would I do the moment of inertia for this?

I'm still solving for every dm in area dA aka dy*dx but I don't know how to make the integral work.

Also I know that my original integral works for when the center of rotation is the center of the rectangle because it gives the formula 1/12*M(a^2+b^2)=I which is the correct moment of inertia.

4. Nov 12, 2013

tiny-tim

hi physicskid123!
your r has to be the distance from (a/2,0)

(or, generally, from (xo,yo))

you can choose your origin to be at the centre, a corner, or (a,b) itself, whichever you think is most convenient

5. Nov 12, 2013

voko

It would also be useful to specify not the point, but the axis of rotation, to avoid any possible confusion.

6. Nov 12, 2013

dauto

I think it's simpler to keep the integrand and change the limits. Both methods are possible. For instance, the the axis passes by the corner, you can simply change the limits to (0, a) and (0, b).