Let us first calculate the moment of inertia of half square plate that is cut through the diagonal
and the axis of rotation is along the diagonal as shown in figure.1.
Let the axis of rotation is along Y-axis of cartesian coordinate system.
Let us consider a strip of mass dm whose length is 2y and its width is dx.
Moment of Inertia dI of this strip is given by, dI = dm × x2 ...............(1)
mass of strip dm = Area×density = 2y × dx × ρ = 2×[ ( a/√2 ) - x ] × dx × ρ .........................(2)
[ ΔAOB and ΔDEB are similar triangles. hence DE/EB = AO/OB =1, hence DE = EB = ( a/√2 ) - x ) ]
where ρ is density per unit surface area.
Hence eqn.(1) becomes, dI = 2×[ ( a/√2 ) - x ] × x2 × dx × ρ
Moment of Inertial of half square plate I =
Hence Moment of inertia of full square plate about the axis of rotation along its diagonal as shown in fig.(2) is given by,
Where M is mass of the given square plate.
If the axis of rotation is shifted to corner of the square as shown in fig.(3), then the required moment of inertia is given by,