An energy of $484\,J$ is spent in increasing the speed of a flywheel from $60\,rpm$ to $360\,rpm$. The moment of inertia of the flywheel is $.............\,kg\cdot m^2$.

Two bodies have their moments of inertia $I$ and $2I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal,their angular momentum will be in the ratio

$A$ solid sphere of mass $M$ and radius $R$ is rotating about its diameter. $A$ disc of the same mass and radius is also rotating about an axis passing through its centre and perpendicular to the plane,but its angular speed is twice that of the sphere. The ratio of the kinetic energy of the disc to that of the sphere is: (in $: 1$)

$A$ thin uniform rod of length $L = 2 \text{ m}$, cross-sectional area $A$, and density $d$ is rotated about an axis passing through its center and perpendicular to its length with angular velocity $\omega$. If the value of $\omega$ in terms of its rotational kinetic energy $E$ is $(\alpha E/Ad)^{1/2}$, then the value of $\alpha$ is:

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

$A$ 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