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Let $L$ be the lamina of uniform density $\rho=1$ obtained by removing circle $A$ of radius $r$ from circle $B$ of radius $2 r$ (see figure).

Show that $M_x=0$ for $L$ .

Question

Let $L$ be the lamina of uniform density $\rho=1$ obtained by removing circle $A$ of radius $r$ from circle $B$ of radius $2 r$ (see figure).

Show that $M_y$ for $L$ is equal to $\left(M_y\right.$ for $\left.B\right)-\left(M_y\right.$ for $\left.A\right)$ .

Solution

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Moment of area is summation of area time's distance to an axis. It is a measure of the distribution of the area of a shape in relationship to an axis.

To summarize, for symmetric surfaces(along an axis) the moment of area along that axis( $M_{axis}$ ) will be zero because the area on one side of the axis will be exactly equal in magnitude and opposite in direction to the area on the other. For surfaces with higher area on one side of the axis(assuming equidistant from axis), the center will shift towards the side with higher area because that side will contribute more moment than the side with less area.