A portion of a ring of radius R has been removed as shown in figure. Mass of the remainting portion is `m`. Centre of the ring is at origin O. Let `I_(A) and I_(O)` be the moment of inertia passing through points A and O are perpendicular to the plane of the ring. Then,

Find the moment of inertia of ring about an axis passing through the centre and perpendicular to its plane.

Calculate the moment of inertia of a ring about an axis passing through centre of the ring and perpendicular to the plane of the ring.

Calculate the moment of inertia of a ring about in axis passing through centre and perpendicular to its plane.

Find the moment of inertia of a thin circular ring about an axis passing through the centre and perpendicular to the plane of the ring

Moment of inertia of a uniform ring about its diameter is I. Then its moment of inertia about axis passing through its centre and perpendicular to its plane is

Derive an expression for moment of inertia of a thin circular ring about an axis passing through its centre and perpendicular to the plane of the ring.

The moment of inertia of a thin uniform ring of mass 1 kg and radius 20 cm rotating about the axis passing through the center and perpendicular to the plane of the ring is

Moment of inertia of a half ring of mass m and radius R about an axis passing through point A perpendicular to the plane of the paper is I_(A) . If I_(C ) is the moment of inertia of the ring about an axis perpendicular to the plane of paper and passing through point C , then

Moment of inertia of a half ring of mass m and radius R about an axis passing through point A perpendicular to the plane of the paper is I_(A) . If I_(C ) is the moment of inertia of the ring about an axis perpendicular to the plane of paper and passing through point C , then