Derive the expression for moment of inerita of a uniform disc about an axis passing through the centre and perpendicular to the plane.

Explanation

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 ther axis. We take an infinitesimally small mass (dm) at a distance (x) fromn the origin. The moment of inertia (dl) of this mass (dm) about the axis is,
`dI= (dm)x^(2)`
As the mass is uniformly distributed, the mass per unit length (`lambda`) of the rod is, `lambda= (M)/(l)`
The (dm) mass of the infinitesimally small length as `d(m)= lambdadx= (M)/(l)dx`
The moment of inertia (i) of the entire rod can be found by integrating dl,
`I= int (dI)= int (dm)x^(2)= int((M)/(l)dx^(2))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 to `l//2` to `l//2`.
`int_(-l//2)^(l//2)x^(2)dx= (M)/(l)[(x^(3))/(3)]_(-l//2)^(l//2)`
`I= (M)/(l)[(l^(3)/(24)-(-(l^(3)/(24))]= (M)/(l)[(l^(3)/(24)+(l^(3)/(24))]
`I= (M)/(l) 2[(l^(3)/(24)] , I= (1)/(12)Ml^(2)`