Question

A circular lamina of radius $a$ and centre O has a mass per unit area of $k{{x}^{2}}$, where $x$ is the distance from O and $k$ is a constant. If the mass of the lamina is $M$, find in terms of $M$ and $a$, the moment of inertia of the lamina about an axis through O and perpendicular to the lamina.

Answer 1

The moment of inertia of an elementary mass $dM$ having mass per unit area $\sigma$ is ${{x}^{2}}\sigma dA$. The moment of inertia of the circular lamina is the integral of ${{x}^{2}}\sigma dA$ over the area of the entire lamina. The total mass of the circular lamina is the integral of the elementary mass over the area of the entire lamina.

Formula used: $I=\int{{{x}^{2}}dM=\int{{{x}^{2}}\sigma dA}}$
where,
I is the moment of inertia
$\sigma$ is the mass per unit area
dA is the elementary area
dM is the elementary mass

Complete step by step solution:
For the circular lamina,
$\begin{aligned} & \sigma =k{{x}^{2}} \\\ & dA=2\pi xdx \\\ \end{aligned}$
The moment of inertia is,
$I=\int\limits_{0}^{a}{{{x}^{2}}(k{{x}^{2}})2\pi xdx=2\pi k\int\limits_{0}^{a}{{{x}^{5}}dx}}=\pi k\dfrac{{{a}^{6}}}{3}$
Now the mass is,
$M=\int{dM=\int\limits_{0}^{a}{\sigma dA}}=\int\limits_{0}^{a}{k{{x}^{2}}(2\pi x)dx=\pi k\dfrac{{{a}^{4}}}{2}}$
Hence,
$I=\dfrac{2}{3}M{{a}^{2}}$

Additional information: Moment of inertia of a rigid body determines the torque needed for angular acceleration about an axis analogous is to how mass determines the force needed for linear acceleration. It is an extensive property of the body.

Note: The moment of inertia of a body about an axis is the product of its mass and square of the distance from the axis. A continuous mass distribution can be thought of as a collection of infinitesimal mass particles. The total moment of inertia will be the sum of the moment of inertia of all these mass particles and will be given by integration over the area of the entire lamina. $\sigma$ is not constant and it should be taken into consideration while integrating.
The moment of inertia of a planar lamina about an axis perpendicular to the plane of lamina passing through its centre is equal to the sum of moment of inertia of the lamina about two mutually perpendicular axis in the plane of the lamina, intersecting at the point where the perpendicular axis passes through. This is known as the perpendicular axis theorem.
Hence, the moment of inertia of this circular lamina about its diameter is
${{I}_{|}}_{|}=\dfrac{1}{3}M{{a}^{2}}$