Question

A particle of mass $13\text{kg}$ moves with $5\text{m/s}$ in $y-z$ plane along the path $5y=12z+60$. Here $y$ and $z$ are in metre. Magnitude of angular momentum of the particle about origin (in ${\text{kg m}}^2{\text{s}}^{-1}$), when the particle is at $z=10\text{m}$, is

[Your answer must be an integer]

Answer 1

Given, $5y=12z+60$ ... (1)
or $5\frac{dy}{dt}=12\frac{dz}{dt}$ ... (2)
As $v=5\text{m/s}$,
$\therefore v_y^2+v_z^2=25$
$\Rightarrow 25={\left(\frac{dy}{dt}\right)}^2+{\left(\frac{dz}{dt}\right)}^2$
$\Rightarrow 25=\frac{144}{25}{\left(\frac{dz}{dt}\right)}^2+{\left(\frac{dz}{dt}\right)}^2$, from (2)
$\therefore \left(\frac{dz}{dt}\right)=\frac{25}{13}$
and so, $\frac{dy}{dt}=\frac{25}{13}\times \frac{12}{5}=\frac{60}{13}$, from (2)
$\therefore v=\frac{60}{13}\hat{j}+\frac{25}{13}\hat{k}$
As $z=10,$
$\therefore y=\frac{180}{5}=36$
i.e., $P\equiv (0,36,10)$
$\therefore \overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$
$=(\frac{60}{13}\hat{j}+\frac{25}{13}\hat{k})\times 13(36\hat{j}+10\hat{k})$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 60& 25\\ 0& 36& 10\end{array}\right|=-\hat{i}(900-600)=-300\hat{i}$
$\therefore \left|\overrightarrow{L}\right|=300$