Question

The angular momentum of particle of mass $0.01Kg$ and position vector $\overrightarrow{r}=(10\hat{i}+6\hat{j})$ meter and moving with a velocity 5 $\hat{i}metre/sec$ about the origin will be :

• A. $-0.3\hat{k}Joulesec$
• B. $3\hat{k}joulesec$
• C. $1/3joulesec$
• D. $0.03\hat{k}Joulesec$

Answer 1

The correct option is A $-0.3\hat{k}Joulesec$

We know that,

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

$\overrightarrow{L}=m\left[\right(10\hat{i}+6\hat{j})\times (5\hat{i}\left)\right]$

$\overrightarrow{L}=m\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 10& 6& 0\\ 5& 0& 0\end{array}\right|$

$\overrightarrow{L}=m\left[\right[\left(10\right)\left(0\right)-\left(5\right)\left(6\right)]\hat{k}+[\left(6\right)\left(0\right)-\left(0\right)\left(0\right)]\hat{i}+[5\left(0\right)-10\left(0\right)\left]\hat{j}\right]$

$\overrightarrow{L}=\left(0.01\right)[-30\hat{k}+(0)\hat{i}+(0\left)\hat{j}\right]$

$\overrightarrow{L}=-0.3\hat{k}$ Joule sec

Option A is correct.