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. . . . .

A body of mass $5 \mathrm{~kg}$ moving with a uniform speed $3 \sqrt{2} \mathrm{~ms}^{-1}$ in $\mathrm{X}-\mathrm{Y}$ plane along the line $\mathrm{y}=\mathrm{x}+4$.The angular momentum of the particle about the origin will be______________ $\mathrm{kg}\  \mathrm{m} \mathrm{s}^{-1}$.

$A$ block of mass $m$ moves on a horizontal rough surface with initial velocity $v$. The height of the centre of mass of the block is $h$ from the surface. Consider a point $A$ on the surface.

The direction of the angular velocity vector along

$A$ paritcle falls freely near the surface of the earth. Consider $a$ fixed point $O$ (not vertically below the particle) on the ground.

A disc of mass $M$ and radius $R$ moves in the $x-y$ plane as shown in the figure. The angular momentum of the disc at the instant shown is