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 as shown in the figure. At a time $t<\left(v_{0}sin\theta /g\right),$ the angular momentum of the particle is

• A. $-mg{v}_0{t}^{2}cos\theta \hat{j}$
• B. $mg{v}_{0}tcos\theta \hat{k}$
• C. $\frac{1}{2}mg{v}_0{t}^{2}cos\theta \hat{k}$
• D. $\frac{1}{2}mg{v}_0{t}^{2}cos\theta \hat{i}$

Answer 1

The correct option is C $\frac{1}{2}mg{v}_0{t}^{2}cos\theta \hat{k}$

The position vector of the particle from the origin at any time t can be calculated by using $\overrightarrow{r}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$
$\Rightarrow \overrightarrow{r}=v_{0}cos\theta t\overrightarrow{i}+(v_{0}sin\theta t-\frac{1}{2}g{t}^2)\hat{j}$
$\therefore$ Velocity vector at any time can be calculated by using $\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t,$
$\Rightarrow \overrightarrow{v}=v_{0}cos\theta \hat{i}+(v_{0}sin\theta -gt)\hat{j}$
The angular momentum of the particle about the origin is $\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$

$=m[v_{0}cos\theta t\hat{i}+(v_{0}sin\theta t-\frac{1}{2}g{t}^2\left)\hat{j}\right)\times (v_{0}cos\theta \hat{i}+(v_{0}sin\theta -gt\left)\hat{j}\right)]$
$=m[v_0^{2}cos\theta sin\theta t-v_{0}g{t}^{2}cos\theta )\hat{k}+(v_0^{2}sin\theta cos\theta t-\frac{1}{2}g{t}^2{v}_{0}cos\theta )(-\hat{k})]$
$=m[v_0^{2}sin\theta cos\theta t\hat{k}-v_{0}g{t}^{2}cos\theta \hat{k}-(v_0^{2}sin\theta cos\theta t\hat{k}+\frac{1}{2}{v}_{0}g{t}^{2}cos\theta \hat{k}]$
$=m[-\frac{1}{2}{v}_{0}g{t}^{2}cos\theta \hat{k}]=\frac{1}{2}mg{v}_0{t}^{2}cos\theta \hat{k}$