$A$ 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

$A$ 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

$A$ molecule consists of two atoms each of mass $m$ and separated by a distance $d$. At room temperature,the average rotational kinetic energy is $E$. What is its angular frequency?

$A$ molecule consists of two atoms each of mass $m$ and separated by a distance $d$. At room temperature,if the average rotational kinetic energy is $E$,then the angular frequency is:

Three particles are situated on a light and rigid rod along the $Y$-axis as shown in the figure. If the system is rotating with an angular velocity of $2 \, rad/s$ about the $X$-axis,then the total kinetic energy of the system is ...... $J$.

Two identical particles each of mass $m$ are separated by a distance $d$. The axis of rotation passes through the midpoint of $d$ and is perpendicular to the length $d$. If $K$ is the rotational kinetic energy of the system,then the angular frequency $\omega$ is: