Given $V_{CM} = 2\; m/s$,$m = 2\; kg$,$R = 4\; m$. Find the angular momentum of the ring about the origin if it is in pure rolling. (in $kg \cdot m^2/s$)

$A$ disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is

$A$ wheel of radius $2 \ cm$ is at rest on a horizontal surface. $A$ point $P$ on the circumference of the wheel is in contact with the horizontal surface. When the wheel rolls without slipping on the surface,the displacement of point $P$ after half a rotation of the wheel is:

$A$ spherical shell of $1 \, kg$ mass and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane (as shown in the figure). The magnitude of the angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} \omega$. The value of $a$ will be ..............

$A$ solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is $:$

$A$ disc of radius $r$ is rotating about its centre with an angular speed $\omega_0$. It is gently placed on a rough horizontal surface. After what time will it be in pure rolling?