Question

Moment of inertia of a uniform circular disc about a diameter is $l$. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be:

Answer 1

We know, moment of inertia of a uniform circular disc of mass M

and radius R about an axis passing through centre is given by,

$I_{center}=\frac{1}{2}M{R}^2$

from perpendicular axis theorem,

moment of inertia of disc about a diameter ,

$I_{diamenter}=\frac{Icenter}{2}$

a/c to question, it is given that, moment of inertia of

disc about a diameter is I

then,

$I=\frac{1}{2}{I}_{center}=\frac{1}{4}M{R}^2$

so, $M{R}^2=4I$ ........(1)

now, we have to find moment of inertia about an axis perpendicular

to its plane and passing through a point on its rim.

use parallel axis theorem,

$I_{rim}=I_{center}+M{R}^2$

$I_{rim}=\frac{1}{2}M{R}^2+M{R}^2$

= $\frac{3}{2}M{R}^2$

so, $I_{rim}=\frac{3}{2}4I=6I$