$A$ flywheel is making $\frac{3000}{\pi}$ revolutions per minute about its axis. If the moment of inertia of the flywheel about that axis is $400 \ kg \ m^2$,its rotational kinetic energy is

$A$ 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

$A$ flywheel is in the form of a solid circular disc with a mass of $72 \ kg$ and a radius of $0.5 \ m$. If it rotates at $70 \ rpm$,its rotational kinetic energy is ........ $J$.

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:

The moment of inertia of a body about a given axis,rotating with an angular velocity of $1 \ rad/s$,is numerically equal to '$P$' times its rotational kinetic energy. The value of '$P$' is:

$A$ ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is