The rotational kinetic energy of an object is $E$ and its moment of inertia is $I$. The angular momentum of the object will be:

$A$ solid sphere of mass $M$ and radius $R$ is rotating about its diameter. $A$ solid cylinder of the same mass and radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($K_{\text{sphere}}$ to $K_{\text{cylinder}}$) will be:

$A$ body is rotating about its own axis. Its rotational kinetic energy is $x$ and its angular momentum is $y$. Hence,its moment of inertia about its own axis is:

Two objects have moments of inertia $I_1$ and $I_2$ $(I_1 > I_2)$ and their angular velocities are equal. If their rotational kinetic energies are $E_1$ and $E_2$ respectively,then:

$A$ thin uniform rod of mass $m$ and length $l$ is suspended from one end,which can oscillate in a vertical plane about the point of suspension. It is pulled to one side and then released. It passes through the equilibrium position with angular speed $\omega$. The kinetic energy while passing through the mean position is

If the rotational kinetic energy of a body is increased by $300\ \%$,then the percentage increase in its angular momentum will be .......... $\%$