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 disc of mass M and radius R. It is found that this disc is made up of many infinitesimally small rings as shown in the figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inerita (dI) of this small ring is given by
`dI=(dm)r^(2)`
As the mass is uniformly distributed the mass per unit area `(sigma)` is `sigma=("mass")/("area")=M/(piR^(2))`

The mass of the infinitesimaly small ring is given by
`dm=sigma 2pirdr-M/(piR^(2)) 2pi r dr`
where tehh term `(2pir dr)` is the area of this elemental ring where `2pir` is the length and dr is the thickness.
`:.dm=(2M)/(R^(2))rdr`
`:.dI=(2M)/(R^(2))r^(3)dr`
The moment of inertia (I) of the entire disc is,
`I=intdI`
`I=int_(0)^(R)(2M)/(R^(2))r^(3)dr=(2M)/(R^(2))int_(0)^(R)r^(3)dr`
`K=(2M)/(R^(2))[(r^(4))/4]_(0)^(R)=(2M)/(R^(2))[(R^(4))/4-0]`
`I=1/2MR^(2)`