Show that kinetic energy of rotating body is `1/2 I sigma^2`

Give the expression for the kinetic energy of a rotating body.

A body having a moment of inertia about its axis of rotation equal to 3 kg m^(-2) is rotating with angular velocity of 3 rad s^(-1) . Kinetic energy of this rotating body is same as that of a body of mass 27 kg moving with velocity v. The value of v is

If the linear momentum of a body is increased by 50% then the kinetic energy of that body increases by

Two discs of moments of inertia I_(1) and I_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω_(1) and ω_(2) are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take omega_(1)neomega_(2)

Obtain an expression for common velocity and loss in the kinetic energy for a moving body m_(1) colliding against another at rest.