Define Moment of Inertia. Derive an expression for the moment of inertia of thin uniform rod about an axis through it's center and perpendicular to its length.

 Answer: The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration, which is the sum of the product of the mass of every particle with its square of the distance from the axis of rotation.

Let us consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small Moment of inertia of uniform rod mass (dm) at a distance (x) from the origin. The moment of inertia (d) of this mass (du) about the axis is,
$dI=\left(dm\right)x^2$
As the mass is uniformly distributed, the mass per unit length $\left(\lambda \right)$ of the rod is $\lambda =\frac{M}{l}$
The (dm) mass of the infinitesimally small length as, $dm=\lambda dx=\frac{M}{l}dx$.
The moment of inertia (1) of the entire rod can be found by integrating dI.
$I=\int dI\int \left(dm\right)x^2=\int \left(\frac{M}{l}dx\right)x^2$
$I=\frac{M}{l}\int x^{2}dx$
As the mass is distributed on either side of the origin, the limits for integration are taken from to $-l/2$ to $l/2$.
$I=\frac{M}{l}\stackrel{l/2}{\underset{-1/2}{\int }}{x}^{2}dx=\frac{M}{l}{\left[\frac{x^3}{3}\right]}^{l/2}$
$I=\frac{M}{l}[\frac{l^3}{24}-(-\frac{l^3}{24})]=\frac{M}{l}[\frac{t^3}{24}+\frac{l^3}{24}]$
$I=\frac{M}{l}\left[2\left(\frac{l^3}{24}\right)\right]$
$I=\frac{1}{12}M{l}^2$