Rotational inertia problem

1. Sep 15, 2006

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.

2. Sep 15, 2006

Galileo

a) and b) are correct, but your post is difficult to read. Check here:
$$I_z=\sigma\int \limits_{-a/2}^{a/2}\int \limits_{-b/2}^{b/2}(x^2+y^2)dxdy$$