Question

If the moment of inertia of a disc about an axis tangential and parallel to its surface be l, then what will be the moment of inertia about the axis tangement but perpendicular to the surface?

Answer 1

Parallel axis theorem:

$I=\frac{M{R}^2}{4}-M{R}^2$

$=\frac{5}{4}M{R}^2$

$I=\frac{M{R}^2}{2}+M{R}^2$

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

$I^{\prime }=\frac{5}{4}\times \frac{4}{5}(\frac{3}{2})M{R}^2$

$=\frac{6}{5}I$