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Question
Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x2 + y2).
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Solution 1
The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m,in the x–y plane at (x, y) is shown in the following figure.
Moment of inertia about x-axis, Ix = mx2
Moment of inertia about y-axis, Iy = my2
Moment of inertia about z-axis, Iz = `m(sqrt(x^2 + y^2))^2`
Ix + Iy = mx2 + my2
= m(x2 + y2)
`= m(sqrt(x^2 + y^2))`
`I_x + I_y = I_z`
Hence the theorem is proved
Solution 2
The theorem of perpendicular axes: According to this theorem, the moment of inertia of a plane lamina (i.e., a two dimensional body of any shape/size) about any axis OZ perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually perpendicular axes OX and OY in the plane of lamina, meeting at a point where the given axis OZ passes through the lamina. Suppose at the point ‘R’ m{ particle is situated moment of inertia about Z axis of lamina
= moment of inertia of body about r-axis
= moment of inertia of the body about y-axis.
RELATED QUESTIONS
State an expression for the moment of intertia of a solid uniform disc, rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration:
i. about a tangent in the plane of the disc, and
ii. about a tangent perpendicular to the plane of the disc.
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(Hint: If the centre of mass is chosen to be the origin `summ_ir_i = 0`
Answer in brief:
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