Question

A particle of mass $2\text{kg}$ is moving with a uniform speed of $10\text{m/s}$ in $x-y$ plane along the line $x=6\text{m}$. Find the magnitude of the angular momentum of the particle about the origin.

• A. $0{\text{kg-m}}^2/\text{sec}$
• B. $60{\text{kg-m}}^2/\text{sec}$
• C. $120{\text{kg-m}}^2/\text{sec}$
• D. $20{\text{kg-m}}^2/\text{sec}$

Answer 1

The correct option is C $120{\text{kg-m}}^2/\text{sec}$


Angular momentum of particle having mass $\left(m\right)$ is given by $\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
or $\overrightarrow{L}=mv{r}_1$
where, $r_1$ is the perpendicular distance from origin to path.
Here, $r_1$ will remain same, so angular momentum of the particle will be constant.
We have, $r_1=6\text{m},m=2\text{kg},v=10\text{m/s}$
So, $\overrightarrow{L}=mv{r}_1=2\times 10\times 6=120{\text{kg-m}}^2/\text{sec}$
$\therefore$ Magnitude of the angular momentum is $120{\text{kg-m}}^2/\text{sec}$