Question

A particle of mass $m=1\text{kg}$ is moving along the line $y=x+2$ with speed $2\text{m/s}$. The angular momentum of the particle about origin $\text{O}$ is

• A. $\overrightarrow{L}=0$
• B. $\overrightarrow{L}=2\sqrt{2}\left(\hat{k}\right)$
• C. $\overrightarrow{L}=2\sqrt{2}(-\hat{k})$
• D. $\overrightarrow{L}=\sqrt{2}(-\hat{k})$

Answer 1

The correct option is C $\overrightarrow{L}=2\sqrt{2}(-\hat{k})$

Given, $M=1\text{kg},v=2\text{m/s}$
Comparing $y=x+2$ with $y=mx+c$
{where $m=\tan \theta =$ slope, $c=$ intercept on $y$-axis}
$m=1,c=2$
$\therefore \tan \theta =1,\theta ={45}^{\circ }$


Hence, the velocity in vector form
$\overrightarrow{v}=2\cos \theta \hat{i}+2\sin \theta \hat{j}$
$=2\cos 45\hat{i}+2\cos 45\hat{j}$
$=\sqrt{2}\hat{i}+\sqrt{2}\hat{j}$
As we know, angular momentum $\overrightarrow{L}=M(\overrightarrow{r}\times \overrightarrow{v})$ remains constant on any point of the line, since the perpendicular distance of the line from the origin is constant.
For position vector $\left(\overrightarrow{r}\right)$, let us assume particle is at point (0,2)
Then, $\overrightarrow{r}=(0-0)\hat{i}+(2-0)\hat{j}=2\hat{j}$
Hence, $\overrightarrow{L}=M(\overrightarrow{r}\times \overrightarrow{v})$
$\overrightarrow{L}=1\times [2\hat{j}\times (\sqrt{2}\hat{i}+\sqrt{2}\hat{j})]$
$\overrightarrow{L}=2\sqrt{2}(-\hat{k})$
$[\because \hat{j}\times \hat{i}=-\hat{k},\hat{j}\times \hat{j}=0]$