Question

A thin ring of mass 0.5 kg has a radius of 0.25 m. calculate the moment of inertia about (a) an axis passing through its center and perpendicular to its plane. (b) an axis passing through a point on its circumference and perpendicular to its plane. (c) diameter as the axis.

Answer 1

$\left(1\right)$MOI about an axis passing through its center and perpendicular to its plane$=M{R}^2=0.5\times 0.25\times 0.25=31.25\times {10}^{-3}kg/m^2$
$\left(2\right)$MOI about an an axis passing through a point on its circumference and perpendicular to its plane$=M{R}^2+m{d}^2$
$=2M{R}^{2}since,(d=R)$
$=0.0625kg/m^2$
$\left(3\right)$MOI about diameter as the axis$=\frac{M{R}^2}{2}=0.015625kg/m^2$