Question

A particle having mass $2\text{kg}$ is moving along a straight line $3x+4y=5$ with speed $8\text{m/s}$. Find angular momentum of the particle about the origin.

• A. $20\text{kg}\frac{{\text{m}}^2}{\text{s}}$
• B. Zero
• C. $16\text{kg}\frac{{\text{m}}^2}{\text{s}}$
• D. $\frac{80}{3}\text{kg}\frac{{\text{m}}^2}{\text{s}}$

Answer 1

The correct option is C $16\text{kg}\frac{{\text{m}}^2}{\text{s}}$


As we know,
$L=mv{r}_{\perp }$
where $r_{\perp }$ is the perpendicular distance of the line of velocity from the origin.
As we know, perpendicular distance of a line from a point is given as
$r_{\perp }=\frac{|A{X}_1+B{Y}_1+C|}{\sqrt{A^2+B^2}}$
Substituting $(X,Y)=(0,0)$
$r_{\perp }=\left|\frac{3\left(0\right)+4\left(0\right)-5}{\sqrt{3^2+4^2}}\right|=1$
Hence,
$L=mv{r}_{\perp }$, where $m=2\text{kg},v=8\text{m/s},r_{\perp }=1\text{m}$
$L=2\times 8\times 1\text{kg}\frac{{\text{m}}^2}{\text{s}}$
$L=16\text{kg}\frac{{\text{m}}^2}{\text{s}}$