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 of mass $M$ and radius $R$ is rolling on a horizontal surface with an angular speed $\omega$. The angular momentum of the disc about the origin $O$ is:

The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$,then the ratio of their masses will be

$A$ solid sphere and a solid cylinder, each of mass $M$ and radius $R$, are rolling with a linear speed $v$ on a flat surface without slipping. Let $L_1$ be the magnitude of the angular momentum of the sphere with respect to a fixed point $O$ on the surface along the path of the sphere. Likewise, let $L_2$ be the magnitude of the angular momentum of the cylinder with respect to the same fixed point $O$ along its path. The ratio $\frac{L_1}{L_2}$ is

In a bicycle,the radius of the rear wheel is twice the radius of the front wheel. If $r_F$ and $r_r$ are the radii,and $v_F$ and $v_r$ are the speeds of the topmost points of the wheels respectively,then:

$A$ sphere cannot roll on a ..........