Hello dear we have already learnt the parallel axis theorem and its application. According to parallel axis theorem,
The M.I. of a body about any axis is equal to the sum of its M.I. about a parallel axis passing through its canter of mass & the product of the mass of the body & square of distance between the two axes.
i.e. Io =Ic + Mh2
In this article we are going to learn about another important theorem for study of moment of inertia of different objects about different axes of rotation.
Let’s discuss the concept of perpendicular axis theorem in details……………!
Statement:
The M.I. of a plane lamina about an axis perpendicular to its plane us equal to the sum of its M.I. about two mutually perpendicular axes in the plane of the lamina & intersecting at the point where perpendicular axes cuts the lamina.
i.e. Iz = Ix + Iy
Proof: Consider a plane lamina in the horizontal XY plane. OX & OY are two mutually perpendicular axes in the plane of lamina. Axis OZ is perpendicular to the plane of the lamina. Let Ix, Iy& Iz be the M.I of the lamina about x, y & Z axes respectively.
Consider a small element of mass dm is situated at point ‘P’. Join OP & Perpendicular from P to X & Y aces.
The element ‘dm’ is at distance y from X axis.
Then by the definition of moment of moment of inertia we can write,
∴ M.I. of element about X – axis = dm.y2
∴ M.I. of lamina about X – axis= Ix = ∫ dm.y2 ………………(1)
Similarly, moment of inertia of the same particle about Y and Z axes can be obtained as,
The element ‘dm’ is at distance x from Y – axis
∴ M.I. of element about Y – axis = dm.x2
∴ M.I. of lamina about Y- axis = Iy = ∫ dm.x2 ………………(2)
The element ‘dm’ is at distance OP from Z – axis
∴ M.I. of element about Z- axis = dm.OP2
∴ M.I. of lamina about Z-axis = Iz = ∫ dm.r2 ………………(3)
From fig. Using Pythagoras theorem,
r2 = x2 + y2
Multiplying the above equation by ‘dm’ on both sides and then integrating we get
∫ dm.r2 = ∫ dm.x2 + ∫ dm.y2
∴ from equation (1), (2) and (3) we can say that,
∴ Iz = Iy + Ix
Some applications of perpendicular axes theorem…..!
M.I. of disc about any diameter:
Any axis passing through centre & in the plane of the disc is coinciding with diameter.
Consider two such axes X & Y in the plane of the disc.
Let Io be the M.I. of disc about tangent Let Ix, Iy & Iz be the M.I. of disc about X, Y & Z axes respectively.
Hence according to principle of perpendicular axes,
∴ Iz= Ix+ Iy
Mass of disc is distributed equally on both sides of X & Y axes.