Question

Calculate moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation coinciding with its diameter and tangent perpendicular to its plane.

Answer 1

Moment of inertia of rind about an axis of rotation coinciding wit its diameter and tangent perpendicular to its plane

$I=\frac{3}{2}m{r}^2$

Given : $m=500gm=0.5kg$ $r=0.5m$

$⟹I=\frac{3\times 0.5\times {0.5}^2}{2}=0.1875kg{m}^2$