Two identical discs are positioned on a vertical axis as shown in the figure. The bottom disc is rotating at angular velocity `omega_(0)` and has rotational kinetic energy `K_(0)`. Change in the rotational kinetic energy of the system is

Two identical discs are positioned on a vertical axis as shown in the figure. The bottom disc is rotating at angular velocity omega_(0) and has rotational kinetic energy K_(0) .The top disc is initially at rest. It then falls and sticks to the bottom disc. The change in the rotational kinetic energy of the system is

The body is rotating with uniform angular velocity ( omega ) having rotational kinetic energy ( E) . Its angular momentum (L ) is :

A disc rotates with angular velocity omega and kinetic energy E . Then its angular momentum

What is the ratio of the rolling kinetic energy and rotational kinetic energy in the motion of a disc ?

Moment of inertia of a disc is 100 g. cm^(2) . If the disc rotates with an angular velocity of 2 rad .s^(-1) the rotational kinetic energy of the disc is

A thin circular disc of mass 12 kg and radius 0.5 m rotates with an angular velocity of 100 rad/s. The rotational kinetic energy of the disc is

A disc is rolling without slipping. The ratio of its rotational kinetic energy and translational kinetic energy would be -

A disc is rolling without slipping. The ratio of its rotational kinetic energy and translational kinetic energy would be -

A ballet dancer is rotating about his own vertical axis. Without external torque if his angular velocity is doubled then his rotational kinetic energy is