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 $kg \cdot m^2/s$)

$A$ sphere is rolling without slipping on a fixed horizontal plane surface. In the figure,$A$ is the point of contact,$B$ is the centre of the sphere,and $C$ is its topmost point. Then:
$(i) \vec{V}_C - \vec{V}_A = 2(\vec{V}_B - \vec{V}_C)$
$(ii) \vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$
$(iii) |\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|$
$(iv) |\vec{V}_C - \vec{V}_A| = 4|\vec{V}_B|$

$A$ disc is rolling without slipping on a surface. The radius of the disc is $R$. At $t=0$,the top most point on the disc is $A$ as shown in the figure. When the disc completes half of its rotation,the displacement of point $A$ from its initial position is

$A$ body is rolling on a horizontal surface without slipping. Its rotational kinetic energy is equal to its translational kinetic energy. The body is a .......

In the figure,$E$ and $v_{cm}$ represent the total energy and speed of the centre of mass of an object of mass $1 \ kg$ in pure rolling. The object is

$A$ uniform disc of mass $0.5\,kg$ and radius $r$ is projected with velocity $18\,m/s$ at $t = 0\,s$ on a rough horizontal surface. It starts off with a purely sliding motion at $t = 0\,s$. After $2\,s$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2\,s$ will be $..............J$ (given,coefficient of friction is $0.3$ and $g = 10\,m/s^2$).