Derive the expresssion for moment of inertia of a rod about its centre and perpendicular to the rod.

Explanation

Consider a uniform rod of mass (M) and length (l) as shown in the figure. To find an expression for moment of inertia of this rod about an axis that passes through the centre of mass and perpendicular to the rod. First we have to find an origin fixed for the coordinate system. So that it coincides with the centre of mass, that is also geometric centre of the rod. THe rod is now along the x axis. Consider an infinitesimally small mass (dm) at a distance (x) from the origin.

`dI=I(dm)x^(2)`
As the mass is uniformly distributed the mass per unit length `(lamda)` of the rod is `lamda=M/l`.
The (dm) mass of the infinitesimally small length is given by `dm=lamdadx=M/l dx`.
The moment of inertia (I) of the entire rod is dI,
`I=intdI=int(dm)x^(2)=(M/l dx)x^(2)`
`I=M/l intx^(2) dx`
As the mass is distributed on either side of the origin, the limits for integration are taken from `-l//2` to `l//2`
`I=M/lint_((-1)/2)^(1/2)x^(2)dx=M/l[(x^(3))/3]_((-1)/2)^(1/2)`
`I=M/l[(l^(2))/24-(-(l^(2))/24)]=M/l[(l^(3))/24+(l^(3))/24]`
`I=M/l[2((l^(3))/24)]`
`I=1/12ml^(2)`