Rotational inertia problem

[John O' Meara]

A thin rectangular sheet of steel is .3m by .4m and has mass 24kg. Find the moment of inertia about an axis (a) through the center, parallel to the long sides; (b) through the center parallel to the short sides. (c) through the center, perpendicular to the plane.
(a) We divide the sheet into N very narrow strips parallel to the axis and of width deltaXi and length L=.4m. The mass of each strip is deltaMi and which is a distance Xi from the axis. Now
deltaMi = k*L*deltaXi,
where k is the mass per unit area. Substituting this value into the expression for I (rotational inertia)
I= Sum k*L*Xi^2*deltaXi = k*L*Sum Xi^2*deltaXi
If we now pass to the limit deltaX -> 0 and N -> infinity. So the above sum can be replaced by an integral =>
I=2*k*L*[X^3/3]0,w/2 the limits of integration => I= 2*w^2, where w=.3m : I=.18 kg.m^2
(b) is similar to (a); I= .32 kg.m^2
(c) for this section I need a drawing please, on how to integrate for I. Do I divide the sheet up into N thin strips con-centric with the axis of rotation
like I would for a solid disk or cylinder? Many Thanks.

[Galileo]

a) and b) are correct, but your post is difficult to read. Check here:
http://www.physicsforums.com/showthread.php?t=8997

For c), the easiest way is to use the perpendicular axis theorem. Which makes it trivial. If you don't know that, a simple double integral will do. The distance of a mass element to the axis is x^2+y^2, so :

I_z=\sigma\int \limits_{-a/2}^{a/2}\int \limits_{-b/2}^{b/2}(x^2+y^2)dxdy

If you haven't had double integrals, you could also try adding the moments of inertia from each strip, like in parts a) and b), and use the parallel axis theorem to find these moments of inertia. If you don't know that theorem, brute force integration will give the same result.
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