Perpendicular Axes Theorem
The MI of a plane lamina about an axis perpendicular to its plane is equal to the sum of MI’s of the lamina about two normal axes, in the plane of lamina and intersecting each other at the point where perpendicular axis passes through lamina.
That is, the theorem states that if Ix and Iy are moments of inertia of a plane lamina about two perpendicular axes OX and OY (in the plane of the lamina) intersecting at point O, then the MI of the lamina about an axis which passes through O and is perpendicular to the plane of the lamina is given by
I = Ix + Iy
Proof: Consider plane of the lamina as XY-plane, O as origin and axes OX and OY as X and Y-axes. Take a mass element dm at point P (x, y) at distance r from O. Then, the concerned moments of inertia of the element are given by
dI = r2 dm . dIx = x2 dm, dIy = y2 dm
Since, r2 = x2 + y2, we get
dI = dIx + dIy
Something (or integrating) over complete lamina, we get
I = Ix + Iy
Let us illustrate above theorem by considering following examples.
Thin circular disc: Let us determine MI of a uniform thin disc about one of its diameters. By symmetry, it must be same about any other diameter. Hence, we take any two perpendicular diameters on the plane of the disc (lamina). The MI of the disc about central axis (line passing through center, perpendicular to plane of the disc) is therefore related to moment of inertia Idia about the diameter as,
Icentral axis = Idia + Idia
Rectangular lamina: Let us take a uniform rectangular lamina of length l and breadth b. The MI of the lamina about an axis passing through its center perpendicular to the plane of the lamina, according to above theorem, is given by
I = Ix + Iy
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