Question

The time dependence of the position of a particle of mass $m=2\text{kg}$ is given by $r\left(t\right)=2t\hat{i}-3t^2\hat{j}\text{m}.$ Its angular momentum (in ${\text{kg m}}^2/\text{s}$), with respect to the origin, at time $t=2\text{s}$ is

• A. $36\hat{k}$
• B. $-34(\hat{k}-\hat{i})$
• C. $-48\hat{k}$
• D. $48(\hat{i}+\hat{j})$

Answer 1

The correct option is C $-48\hat{k}$

The position of particle is,

$\overrightarrow{r}=2t\hat{i}-3t^2\hat{j}$

where, $t$ is instantaneous time.

Velocity of particle is

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=2\hat{i}-6t\hat{j}$

Now, angular momentum of particle with respect to origin is given by

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

$=m\left\{\right(2t\hat{i}-3t^2\hat{j})\times (2\hat{i}-6t\hat{j}\left)\right\}$

$=m(-12t^2(\hat{i}\times \hat{j})-6t^2(\hat{j}\times \hat{i}\left)\right)$

As, $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=0$

$\Rightarrow \overrightarrow{L}=m(-12t^2\hat{k}+6t^2\hat{k})$

As, $\hat{i}\times \hat{j}=\hat{k}\text{and}\hat{j}\times \hat{i}=-\hat{k}$

$\overrightarrow{L}=-m\left(6t^2\right)\hat{k}$

So, angular momentum of particle of mass $2\text{kg}$ at time $t=2\text{s}$ is

$\overrightarrow{L}=-48\hat{k}{\text{kg-m}}^2/\text{s}$

Hence, option $\left(\text{C}\right)$ is correct.