$A$ flywheel rotating about a fixed axis has a kinetic energy of $500 \text{ J}$. If its angular frequency is $5 \text{ Hz}$, calculate the moment of inertia of the wheel about the axis of rotation. (Take $\pi^2 = 10$) (in $\text{ kg m}^2$)

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:

Two bodies $A$ and $B$ have moments of inertia $I_A$ and $I_B$, and angular momenta $L_A$ and $L_B$ respectively. Both of them have the same kinetic energy of rotation. The ratio of $L_A$ to $L_B$ is:

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

In the given figure,two wheels $P$ and $Q$ are connected by a belt $B$. The radius of $P$ is three times as that of $Q$. In case of same rotational kinetic energy,the ratio of rotational inertias $\left(\frac{I_{P}}{I_{Q}}\right)$ will be $x: 1$. The value of $x$ will be $.....$ .

$A$ flywheel rotating about a fixed axis has a kinetic energy of $360 \ J$ when its angular speed is $30 \ rad/s$. The moment of inertia of the wheel about the axis of rotation is ...... $kg \ m^2$.