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

Explanation

Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (d) of this small ring is
` dl= (dm)r^(2)`
As the mass is uniformly distributed, the mass per unit area (a) is `sigma=("mass")/("area")=(M)/(piR^(2))`
The mass of the infinitesimally small ring is,
`dm- sigma 2pi rdr=(M)/(pi R^(2)) 2pi rdr`
where, the term `(2pi rdr)` is the area of this elemental ring `(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 () of the entire dise is
`I= int dI`
`I= overset(R) underset(0) int (2M)/(R^(2))r^(3) dr=(2M)/(R^(2)) overset(R) underset(0) int r^(3) dr`
`I=(2M)/(R^(2))[(r^(4))/(4)]_(0)^(R)=(2M)/(R^(2))[(R^(4))/(4)-0]`
`I=(1)/(2)MR^(2)`