Find the expression for moment of inertia of a thin uniform rod
about an axis passing through its centre and perpendicular to its length

Find the expression for moment of inertia of a thin uniform rod about an axis passing through its one end and perpendicular to its length.

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

Define the theorem of parallel axes and apply it to find the moment of inertia of a uniform rod about an axis passing through one of its ends and perpendicular to its length.

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 a solid cylinder about its axis of symmetry is equal to moment of inertia about an axis passing through its centre and perpendicular to its length.Show that L//r = sqrt3 .

A thin circular ring od fiameter 15 cm has a mass of 100g.Find its moment of inertia about an axis passing through its centre and perpendicular to its plane.

The moment of inertia of a rod about an axis through its centre and perpendicular to its length is ML^2//12 (where M is the mass and L is the lenth of the rod).The rod is bent in the middle so that the two halves make an angle 60^@ .The moment of inertia of the bent rod about the same axis would be :

Calculate the radius of gyration of a rod of mass 100 g and length 100 m about an axis passing through is centre of gravity and perpendicular to its length.