Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ $(I_B > I_A)$ have equal kinetic energy of rotation. If $L_A$ and $L_B$ are their angular momenta respectively,then:

$A$ uniform thin rod of length $l$ and mass $m$ is suspended from one of its ends and is rotated at $f$ rotations per second. The rotational kinetic energy of the rod will be

$A$ kathak dancer is standing on a horizontal surface with folded hands. Initially, the dancer is rotating about his central axis, and his kinetic energy is $K$. The dancer then stretches his arms such that the moment of inertia becomes three times its initial value and the angular velocity becomes one-third of its initial value. The kinetic energy of the dancer now is:

Write the formula for rotational kinetic energy.

Rotational kinetic energy of a given body about an axis is proportional to

The moments of inertia of two bodies about their axes of rotation are $I$ and $2I$ respectively. If their rotational kinetic energies are equal,what is the ratio of their angular velocities?