Question

State an expression for moment inertia of a thin uniform disc about an axis passing through its centre and perpendicular to its plane.

Answer 1

moment inertia of a thin uniform disc about an axis passing through its centre and perpendicular to its plane.

Let a disc of mass ‘ m ’ and radius R is given , cut an element of thickness dx , at x distance from the centre of the disc as shown in figure .
so, mass of element , $dm=\frac{m}{πR^2}2\pi x.dx$
or, $dm=\frac{2m}{R^2}x.dx$

actually, element seems as ring,
so, moment of inertia of ring of radius x about its axis , $I=(dm)x^2$

$I=\left(\frac{2m}{R^2}x.dx\right).x^2$

$I=\frac{2m}{R^2}\int\limits^R_0{x^3}\,dx$

$I=\frac{2m}{R^2}\left[\frac{R^4}{4}\right]^R_0$

$I=\frac{2m}{R^2}\frac{R^4}{4}$

$I=\frac{mR^2}{2}$