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

Explanation

(a) (i) consider a unifrom disc of mass M and radius R as shown in figure.
(ii) The moment of inertia of the disc is found about the axis, which passes through its center and perpendicular to the plane. This disc is made up of many infinitesimally small rings
(ii) Now consider the moment of inertia of an ring, which has infinitesimal small mass 'dm' thickness dr and radius 'r' . which can be expressed as.
` dI = ( dm)r^(2)`
(iv) The moment of inertia (I) of the entire disc can be found be intergrating the equation (1) as.
` I = int dl = int_(0)^(2piR) ( dm) R^(2)`
Surface ( Mass) density `sigma = ( " Mass ")/(" area") , sigma = M/( pi R^(2))`
(v) If ` sigma ` is surface mass density ( i.e,` sigma = M/( 2 pi R)` the small mass dm can be written as
`dm= sigma 2 pi r dr = M/( pi R^(2)) 2 pi r dr = ( 2M)/ R^(2) d r `
(vi) Subsituting the 'dm' value in equation (2), wer get
` I = int_(0)^(R) ( (2M)/(R^(2)) rdr) r^(2) = ( 2M)/R^(2)int_(0)^(R) r^(3) dr `
` I = ( 2M)/R^(2) [ r^(4)/4]_(0)^(R) , I = M/(2pi) [ R^(4)/4 - 0] , I = 1/2 MR^(2)`