Question

A particle of mass 2 kg is located at the position $(\hat{i}+\hat{j})$ m has a velocity $2(+\hat{i}-\hat{j}+\hat{k})$ m/s. Its angular momentum about z-axis in $kg-m^2/s$ is:

• A. zero
• B. $8$
• C. $12$
• D. $-8$

Answer 1

Answer: (D)

$-8$

$\begin{array}{cc}\text{Given -}& \ast \text{Mass of the particle}M=2kg\\ & \ast \text{Position vector}\overrightarrow{r}=\hat{i}+\hat{j}m\\ & \ast \text{velocity vector}\overline{v}=2(\hat{i}\hat{j}+\hat{k})m/8\end{array}$

$\begin{array}{c}\text{we know that}\\ \text{angular momentum is given by}\\ \begin{array}{cc}\overrightarrow{L}& =m\left(\overrightarrow{r}x\overrightarrow{v}\right)=2x\left[\right(\hat{i}+\hat{j})\times 2(\hat{i}\hat{j}+\hat{k}\left)\right]\\ & =4\left[\right(\hat{i}+\hat{j})\times (\hat{i}\hat{j}+\hat{k}\left)\right]\\ & =4[\hat{i}-\hat{j}-2\hat{k}]\\ & =4\hat{i}-4\hat{j}-8\hat{k}\\ \overrightarrow{L}=-8& \text{units about}z-axis\end{array}\end{array}$