Question

A small particle of mass $m$ is projected at an angle $\theta$ with the $x-axis$ with an initial velocity $v_0$ in the $x-y$ plane shown in the figure . At a time $t<\frac{v_0\sin \theta }{g},$ the angular momentum of the particle is

Answer 1

at time $t=\frac{v_{0}sin\theta }{g}$

$v_y=v_{0}sin\theta -g\left(\frac{v_{0}sin\theta }{g}\right)$

$\Rightarrow v_y=0$

So,

at $t=\frac{v_{0}sin\theta }{g}$, the particle will be at maximum height

So,

for time $t<\frac{v_{0}sin\theta }{g}$

$v_y=v_{0}sin\theta -gt$

$v_x=v_{0}cos\theta$

$\Rightarrow \overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$\overrightarrow{v}=v_{o}cos\theta \hat{i}+(v_{0}sin\theta -gt)\hat{j}$

And,

$r_x=vcos\theta t$

and $r_y=vsin\theta -\frac{1}{2}g{t}^2$

$\Rightarrow \overrightarrow{r}=r_x\hat{i}+r_y\hat{j}$

$\overrightarrow{r}=vcost\hat{i}+(vsin\theta -\frac{1}{2}g{t}^2)\hat{k}$

Now, we know that angular momentm $\left(\overrightarrow{L}\right)$

$\overrightarrow{L}=\overrightarrow{P}\times \overrightarrow{r}$

$\overrightarrow{L}=m\overrightarrow{v}\times \overrightarrow{r}$

$=m(v_x\hat{i}+v_y\hat{j})\times (v_x\hat{i}\times r_y\hat{j})$

$=m[v_x{r}_y\hat{k}-v_y{r}_x\hat{k}]$

$\Rightarrow \overrightarrow{L}=m(v_x\times r_y-v_y{r}_x)\hat{k}$

$=m\left[v_{0}cos\theta \right(vsin\theta -\frac{1}{2}g{t}^2)-(v_{0}sin\theta -gt\left)v_{0}cos\theta t\right]\hat{k}$

$=m[v_0^{2}cos\theta -\frac{v_{0}cos\theta g{t}^2}{2}-v^{2}sin\theta cos\theta +v_{0}cos\theta g{t}^2]\hat{k}$

$\Rightarrow \overrightarrow{L}=m[v_{0}cos\theta g{t}^2-\frac{1}{2}{v}_{0}cos\theta g{t}^2]\hat{k}$

$\Rightarrow \overrightarrow{L}=\frac{1}{2}m{v}_{0}cos\theta g{t}^2\hat{k}$