A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner. (By double integration)

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Position the lamina in the x-y plane with one corner at the origin.

Consider a tiny rectangle with height dy and width dx. Its area is dydx and has mass ρdydx. This rectangle is at (x,y) and its centre of mass (CoM) is also at (x,y).

If h is the height of the lamina (y direction) and L its width (x direction) then the moment of inertia of the tiny rectangle is mass×(distance of CoM from corner)²=ρr²dydx where r²=x²+y².

The moment of inertia I of the lamina is ρ∫∫(x²+y²)dydx, that is, the sum of the moments of inertia of all the tiny rectangles constituting the lamina.

The limit of the inner integral for dy is [0,h] and the outer integral for dx is [0,L].

So ∫[0,h](x²+y²)dy=(x²y+y³/3)[0,h]=x²h+h³/3.

And I=ρ∫[0,L](x²h+h³/3)dx=ρ(x³h/3+h³x/3)[0,L]=

ρ(L³h/3+h³L/3)=⅓ρhL(L²+h²). ρhL is the mass of the lamina.

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