Question

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axis that lies in the plane of the plate, passes through the center of the plate, and is parallel to the side with length b.

Answer 1

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is

`dI =r^2*dm`
where
`dm =M(b*dr)/(ab)`
Now we find the moment of inertia by integrating from
`-a/2` to
`a/2`
The moment of inertia is

`I= \int\limits^(-a/2)_(a/2) {r^2*dm} = M \int\limits^(-a/2)_(a/2) r^2(b*dr)/(ab)=(M/a)(r^3/3)` (from (-a/2) to
`I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12` (a/2))