Question

A particle of mass $20\text{gm}$) slides along the frictionless surface. When the particle reaches point $B$, its angular momentum (in ${\text{kg m}}^2/\text{s}$) about $O$ will be


[Take $g=10{\text{m/s}}^2$]

Answer 1

Let the speed of particle be $v$ when it reaches point $B$.


As there are no non-conservative forces acting on the particle, hence on applying law of conservation of mechanical energy,
$M.E_i=M.E_f$
$\Rightarrow K{E}_i+P{E}_i=K{E}_f+P{E}_f$
Taking $PE=0$ at the reference level,
$\Rightarrow (\frac{1}{2}\times 0.02\times 5^2)+(0.02\times 10\times 10)=\frac{1}{2}\times 0.02\times v^2+0$
$\Rightarrow v=15\text{m/s}....\left(1\right)$
Angular momentum about $O$
$L=mv{r}_{\perp }$
Since at $B$, $r_{\perp }=20\text{m}$ about $O$.
$\Rightarrow L=0.02\times 15\times 20$
$\Rightarrow L=6{\text{kg m}}^2/\text{s}$
The angular momentum of the particle about point $O$ when it reaches $B$ is $6{\text{kg m}}^2\text{/s}$.