Question

A particle of mass $2kg$ 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

Given
$m=2kg$
$\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{j})$
$\overrightarrow{v}=2(\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})=(2\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k})$
The angular momentum is given by
$\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
$\overrightarrow{L}=m\left[\right(\overrightarrow{i}+\overrightarrow{j})\times (2\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}\left)\right]$
$\overrightarrow{L}=2[2\hat{i}-2\hat{j}-4\hat{k}]=$ $4\hat{i}-4\hat{j}-8\hat{k}$

Thus angular momentum about z-axis $L_z=-8$ $kg.m^2.s^{-1}$