A particle is moving in a circular path of radius $a,$ with a constant velocity $v$ as shown in the figure.The centre of circle is marked by $'C'$. The angular momentum from the origin $O$ can be written as

A thin rod of mass $M$ and length $a$ is free to rotate in horizontal plane about a fixed vertical axis passing through point $O$. A thin circular disc of mass $M$ and of radius $a / 4$ is pivoted on this rod with its center at a distance $a / 4$ from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4 \Omega$. The total angular momentum of the system about the point $O$ is $\left(\frac{ M a^2 \Omega}{48}\right) n$. The value of $n$ is. . . . .

Two rigid bodies $A$ and $B$ rotate with rotational kinetic energies $E_A$ and $E_B$ respectively.  The moments of inertia of $A$ and $B$ about the axis of rotation are $I_A$ and $I_B$ respectively. If $I_A = I_B/4 \,$and$ \, E_A = 100\ E_B$ the ratio of angular momentum $(L_A)$ of $A$ to the angular momentum $(L_B)$ of $B$ is

A disc of mass  $M$  and radius  $R$  is rolling with angular speed $\omega $ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

A particle of mass $20\,g$ is released with an initial velocity $5\,m/s$ along the curve from the point $A,$ as shown in the figure. The point $A$ is at height $h$ from point $B.$ The particle slides along the frictionless surface. When the particle reaches point $B,$ its angular momentum about $O$ will be ……… $kg – m^2/s$. [Take $g = 10\,m/s^2$ ]