Question

A 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 $t<\frac{v_0\sin \theta }{g}$, the angular momentum of the particle about origin is

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

Position vector of particle from origin at time $t$ is,
$\Rightarrow \overrightarrow{r}=x\hat{i}+y\hat{j}$
where
$x=v_0\cos \theta \times t$
$y=v_0\sin \theta \times t-\frac{1}{2}g{t}^2$
$\therefore \overrightarrow{r}=\left(v_0\cos \theta t\right)\hat{i}+(v_0\sin \theta t-\frac{1}{2}g{t}^2)\hat{j}$
We also know
$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$
Where $v_x$ and $v_y$ are the component of velocity of particle along $x$ and $y$ direction respectively.
$\Rightarrow v_x=v_0\cos \theta$
$v_y=v_0\sin \theta -gt$
$\Rightarrow \overrightarrow{v}=\left(v_0\cos \theta \right)\hat{i}+(v_0\sin \theta -gt)\hat{j}$
Hence angular momentum of the particle about origin is represented as:
$\therefore \overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{P}=m(\overrightarrow{r}\times \overrightarrow{v})\overrightarrow{L}=m[v_0\cos \theta t\hat{i}+(v_0\sin \theta t-\frac{1}{2}g{t}^2\left)\hat{j}\right]\times [v_0\cos \theta \hat{i}+(v_0\sin \theta -gt\left)\hat{j}\right]$
$\overrightarrow{L}=m{v}_0^2\cos \theta \sin \theta t\hat{k}-m{v}_0\cos \theta g{t}^2\hat{k}-m{v}_0^2\cos \theta \sin \theta t\hat{k}-(-\frac{1}{2}mg{v}_0\cos \theta t^2)$
$\therefore \overrightarrow{L}=-\frac{1}{2}mg{v}_0{t}^2\cos \theta \hat{k}$