Moment of Inertia (Square Laminas)

Question

If I have a uniform square lamina of side length 2a and intend to find its Moment Of Inertia about a perpendicular axis to its plane, is there a general formula for this? If there isn't, I have tried splitting the shape into two rectangles of sides 2a and a; finding the MOI for each rectangle is easy, but adding these MOIs up doesn't seem right.

please note, the perpendicular axis is through the centre of the shape

Please, any suggestions (along the line of topic) will help; I just need another way of thinking about this. Thanks

Answer 1

If you are using standard results, and not direct integration, to obtain the MI about the $z$ axis perpendicular to the plane through the centre, then firstly, the MI of the square lamina of mass $m$ and sides $2a$ about a diameter of the square (the $x$ and $y$ axes) is $$\frac 13ma^2$$ Then apply the Perpendicular Axes Theorem so that $$I_z=I_x+I_y$$ so the answer is $$\frac 23ma^2$$

Answer 2

If the axis passes through the center of the lamina, then you just need to evaluate $$ \int_{-a}^a\int_{-a}^a(x^2+y^2)\rho\ dxdy, $$ which is quite easy (density $\rho$ is constant).

Answer 3

Hint we can have another axis in the same plane . so let the given axis in plane be $I_x$ and we draw a new axis which is in the same plane of the lamina lets say $I_y$ thus the axis perpendicular to the lamina be $I_z$ thus by perpendicular axis theorem we have $2I_z=I_x+I_y$ now we know moment of inertia about $X-axis,Y-axis $ which is $\frac{1}{3}ML^2$ now can you get $I_z$