Question

A particle of mass $3\text{kg}$ moves along a straight line $4y-3x=2$ ($x$ and $y$ are in $\text{m}$) with constant velocity $5\text{m/s}$. The magnitude of its angular momentum about the origin is

• A. $12{\text{kg m}}^2\text{/s}$
• B. $6{\text{kg m}}^2\text{/s}$
• C. $4.5{\text{kg m}}^2\text{/s}$
• D. $8{\text{kg m}}^2\text{/s}$

Answer 1

The correct option is B $6{\text{kg m}}^2\text{/s}$

Given, equation of the straight line, $4y-3x=2$

$\Rightarrow y=\frac{3}{4}x+\frac{1}{2}$

On plotting it on the graph we get,


Slope can be determined as,
$m=\tan \theta =\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}$

$\Rightarrow \theta ={37}^{\circ }$

From figure, $r_{\perp }=\frac{1}{2}\cos \theta =\frac{1}{2}\cos {37}^{\circ }=0.4\text{m}$

Now, angular momentum about origin, $L=mv{r}_{\perp }$

$\Rightarrow L=3\times 5\times 0.4$

$\Rightarrow L=6{\text{kg m}}^2\text{/s}$

Alternate:
Distance of line from the origin $r=\frac{|(-3\times 0)+(4\times 0)-2|}{\sqrt{4^2+(-3{)}^2}}=\frac{2}{5}\text{m}$
Now, angular momentum about origin, $L=mv{r}_{\perp }$

$\Rightarrow L=3\times 5\times 0.4$

$\Rightarrow L=6{\text{kg m}}^2\text{/s}$

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