Question

Three partricles $A,B$ and $C$ of different masses $1\text{kg},2\text{kg}$ and $3\text{kg}$ are moving with different velocities $v_A=\hat{i}+2\hat{j},v_B=2\hat{i}+4\hat{j}$ and $v_C=3\hat{j}$. At time $t$, particles $A,B$ & $C$ are at points $(2,3),(3,2)$ & $(1,4)$ respectively. The total angular momentum of the system of $3$ particles about the origin [in ${\text{kg m}}^2\text{/s}$] is

• A. $26\hat{k}$
• B. $9\hat{k}$
• C. $10\hat{k}$
• D. $5\hat{k}$

Answer 1

The correct option is A $26\hat{k}$

As we know,
Angular momentum of a system of particles
${\overrightarrow{L}}_{system}={\sum }_{i=1}^n{m}_i(\overrightarrow{r_i}\times \overrightarrow{v_i})$
Hence position vectors for $A,B\&C$ particle about origin at time $t$
$\overrightarrow{r_A}=(2-0)\hat{i}+(3-0)\hat{j}=2\hat{j}+3\hat{j}$
$\overrightarrow{r_B}=(3-0)\hat{i}+(2-0)\hat{j}=3\hat{i}+2\hat{j}$
$\overrightarrow{r_C}=(1-0)\hat{i}+(4-0)\hat{j}=\hat{i}+4\hat{j}$

${\overrightarrow{L}}_{\text{system}}=m_A(\overrightarrow{r_A}\times \overrightarrow{v_A})+m_B(\overrightarrow{r_B}\times \overrightarrow{V_B})+m_C(\overrightarrow{r_C}\times \overrightarrow{V_C})$
Given, $m_A=1\text{kg},m_B=2\text{kg},m_C=3\text{kg}$
and $\overrightarrow{V_A}=\hat{i}+2\hat{j}$, $\overrightarrow{V_B}=2\hat{i}+4\hat{j}$, $\overrightarrow{V_C}=3\hat{j}$

Hence, ${\overrightarrow{L}}_{\text{system}}=1\left[\right(2\hat{i}+3\hat{j})\times (\hat{i}+2\hat{j}\left)\right]+2\left[\right(3\hat{i}+2\hat{j})\times (2\hat{i}+4\hat{j}\left)\right]+3\left[\right(\hat{i}+4\hat{j})\times (3\hat{j}\left)\right]$
$\Rightarrow {\overrightarrow{L}}_{system}=[4\hat{k}+3(-\hat{k}\left)\right]+2[12\hat{k}+4(-\hat{k}\left)\right]+[3\times 3\hat{k}]$
[$\because \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=0$
$\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{i}=-\hat{k}$]
${\overrightarrow{L}}_{\text{system}}=\hat{k}+16\hat{k}+9\hat{k}$
${\overrightarrow{L}}_{\text{system}}=26\hat{k}$