Find the moment of inertia of a ring of mass m and radius r about an axis passing through the tangent to the circle of the ring.

Explanation

Let m be mass of a ring and r be its radius. Then, the moment of inertia of the ring about the axis passing through its centre and perpendicular to its plane ( z - axis in Fig ) is, If `I_(x)` and `I_(y)` be the moments of inertia of the ring respectively about two perpendicular diameters on the plane of the ring ( along x and y-axes ) then from the perpendicular -axes theorem,
`I_(z)= (I_(x)+I_(y))`
`= 2 I_(x)=mr^(2)" " [I_(x)=I_(y),`from symmetry ]
or, `I_(x) = (1)/(2) mr^(2)`
Now from the parallel -axes theorem the moment of inertia about the axis AB, which is parallel to the x-axis and tangential to the ring is
`I=I_(x)+mr^(2)=(1)/(2)mr^(2)+mr^(2)=(3)/(2)mr^(2)`