Question

A particle having mass $m=2\text{kg}$ is moving with velocity $(2\hat{i}+3\hat{j})\text{m/s}$. Find angular momentum of the particle about the origin when it is at $(1,1,0)$ [in ${\text{kg m}}^2\text{/s}$].

• A. $2\hat{k}$
• B. $4\hat{k}$
• C. $6\hat{k}$
• D. Zero

Answer 1

The correct option is A $2\hat{k}$

Given,
Mass of particle $m=2\text{kg}$
Velocity of particle $\overrightarrow{v}=(2\hat{i}+3\hat{j})\text{m/s}$
Position vector about origin $\overrightarrow{r}=(1-0)\hat{i}+(1-0)\hat{j}+(0-0)\hat{k}$
$\overrightarrow{r}=(\hat{i}+\hat{j})\text{m}$
As we know,
Angular momentum $\left(\overrightarrow{L}\right)=m(\overrightarrow{r}\times \overrightarrow{v})$
$\overrightarrow{L}=2\left[\right(\hat{i}+\hat{j})\times (2\hat{i}+3\hat{j}\left)\right]$
$\overrightarrow{L}=2\left[3\right(\hat{i}\times \hat{j}+2(\hat{j}\times \hat{i})\left)\right]\frac{{\text{kg m}}^2}{\text{s}}$
$[\because \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=0]$
$\overrightarrow{L}=2\times [3\hat{k}+2(-\hat{k}\left)\right]$
$\overrightarrow{L}=2\hat{k}$ ${\text{kg m}}^2\text{/s}$