Question

A system of mass- pulley is shown in figure. Consider the pulley as a solid disc of radius $R$ and mass $m$. If the velocity of the masses is $V$, then the angular momentum of the system about point $\text{O}$ is $\frac{nmVR}{2}$, where $n$ is

Answer 1

As we know, angular momentum $L=mVR$
Then the angular momentum of mass (1) with respect to point $\text{O}$ is
$L_1=mVR...\left(i\right)$
The direction of $L_1$ is into the plane (according to right hand thumb rule)
The angular momentum of mass (2) with respect to $\text{O}$ is
$L_2=mVR...\left(ii\right)$
The direction of $L_2$ is into the plane of paper (same as $L_1$)
The angular momentum of pulley about point $\text{O}$
$L_3=I\omega$
where $I=MOI$ for pulley (solid dic) $=\frac{m{R}^2}{2}$
and $\omega =\frac{V}{R}$
So, $L_3=\frac{m{R}^2}{2}\times \left(\frac{V}{R}\right)$$=\frac{mVR}{2}...\left(iii\right)$
(Pulley is rotating in clockwise direction. Hence the direction of $L_3$ will also be into the plane of paper).

Thus, the total angular momentum of the system at this instant is given as
$L=L_1+L_2+L_3$
$L=mvR+mvR+\frac{mvR}{2}=\frac{5mvR}{2}$
(direction into the plane)