The moments of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. Given $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies,then:

$A$ body rotating with an angular velocity of $3 \ rad/s$ has a moment of inertia of $3 \ kg \cdot m^2$ about its axis of rotation. The rotational kinetic energy of this body is equal to the translational kinetic energy of another body of mass $27 \ kg$. The speed of the $27 \ kg$ mass body is .......... $m/s$.

$A$ 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 (in $\,kJ$)

$A$ rigid body rotates with an angular momentum $L$. If its rotational kinetic energy is made four times, its angular momentum will become

$A$ 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 the axis is

$A$ particle of mass '$m$' is performing uniform circular motion along a circular path of radius '$r$'. Its angular momentum about the axis passing through the centre and perpendicular to the plane is '$L$'. The kinetic energy of the particle is