Derive an expression for the MI. of a disc,
(i) about an axis passing through its centre and perpendicular to its plane. (ii) about its diameter.

Derive an expression per the M.I. of a rectangular rod about an axis passing through its centre of mass and perpendicular to its length.

The moment of inertia of a copper disc, rotating about an axis passing through its centre and perpendicular to its plane

Radius of gyration of a uniform circular disc about an axis passing through its centre of gravity and perpendicular to its plane is

A circular disc of radius R and thickness (R)/(6) has moment of inertia about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is

A circular disc of mass 100 g and radius 10 cm' is making 2 rps about an axis passing through its centre and perpendicular to its plane. Calculate its kinetic energy.

The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is "4 kg m"^(2) . The diameter of the ring is

The moment of inertia of ring about an axis passing through its diameter is I . Then moment of inertia of that ring about an axis passing through its centre and perpendicular to its plane is

The moment of inertia of a uniform ring about an axis passing through its centre and perpendicular to its plane is 100kgm^(2) . What is the moment of inertia of the ring about its diameter ?

The moment of inertia of a uniform ring about an axis passing through its centre and perpendicular to its plane is 100kgm^(2) . What is the moment of inertia of the ring about its diameter ?

The surface mass density of a disc of radius a varies with radial distance as sigma = A + Br where A & B are positive constants then moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane