Question

Find out the moment of inertia of a ring having uniform mass distribution of mass $M$ and radius $R$ about an axis which is tangent to the ring and (a) in the plane of the ring, (b) perpendicular to the plane of the ring.

Answer 1

The moment of inertia of a ring about an axis passing through centre an perpendicular to plain of ring is $I_z=M{R}^2$ because of symmetry we can say $I_x=I_y$ and using perpendicular axis theorem
$I_z=I_x+I_y\Rightarrow M{R}^2=2I_x$
$\Rightarrow I_x=\frac{M{R}^2}{2}=I_0$ (in figure)
For case (a): Using parallel axis theorem
$I_1=I_0+M{R}^2=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$
For case (b): Using parallel axis theorem
$I_z=I_x+M{R}^2=M{R}^2+M{R}^2=2M{R}^2$