Question

The position vector $\overrightarrow{r}$ of a particle, of mass $m$, is given by the following equation.
$\overrightarrow{r}=\alpha t^3\hat{i}+\beta t^2\hat{j}$
Where $\alpha =\frac{10}{3}{\text{m/s}}^3,\beta =5{\text{m/s}}^2$ and $m=0.1\text{Kg}$ at $t=1\text{s}$. The angular momentum $\overrightarrow{L}$ with respect to the origin is

• A. $\frac{5}{6}\hat{k}\text{N ms}$
• B. $\frac{-5}{3}\hat{k}\text{N ms}$
• C. $\frac{5}{3}\hat{k}\text{N ms}$
• D. $\frac{-5}{6}\hat{k}\text{N ms}$

Answer 1

The correct option is B $\frac{-5}{3}\hat{k}\text{N ms}$

Given, $m=0.1\text{kg}$

$\overrightarrow{r}=\alpha t^3\hat{i}+\beta t^2\hat{j}$

$\overrightarrow{V}=\frac{d\overrightarrow{r}}{dt}=3\alpha t^2\hat{i}+2\beta t\hat{j}$

${\overrightarrow{V}}_{t=1}=3\times \frac{10}{3}\times 1\hat{i}+2\times 5\times 1\hat{j}$

$=10\hat{i}+10\hat{j}$
Also,

${\overrightarrow{r}}_{t=1}=\frac{10}{3}\times 1^3.\hat{i}+5\times 1^2.\hat{j}=\frac{10}{3}\hat{i}+5\hat{j}$

Now, angular momentum is given by,

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$

$\overrightarrow{L}=0.1[\overrightarrow{r}\times \overrightarrow{v}]$

${\overrightarrow{L}}_{t=1}=0.1\left[\right(\frac{10}{3}\hat{i}+5\hat{j})\times (10\hat{i}+10\hat{j}\left)\right]$

${\overrightarrow{L}}_{t=1}=0.1\times [\frac{100}{3}\hat{k}+50(-\hat{k}\left)\right]=\frac{-5}{3}\hat{k}$

Hence, $\left(B\right)$ is the correct answer.