Question

The angular momentum of a particle about the origin is varying as $L=4\sqrt{2}t+8$ (SI units) when it moves along a straight line$y=x-4$ ($x,y$ in meters). The magnitude of force (in $\text{N}$) acting on the particle would be

Answer 1

Angular momentum of particle about the origin is varying as $L=4\sqrt{2}t+8$ (SI units) when it moves along a straight line $y=x-4$ ($x,y$ in meters).
As we know,
$\tau =\frac{dL}{dt},L=(4\sqrt{2}t+8)$
$\tau =\frac{d}{dt}(4\sqrt{2}t+8)=4\sqrt{2}\text{Nm}-\left(1\right)$
and $\tau =|\overrightarrow{r}\times \overrightarrow{F}|=rF.\sin \theta$
$\Rightarrow \tau =F.\left(r\sin \theta \right)=F.r_{\perp }$
where $r_{\perp }$ is the perpendicular distance of force vector from origin.
Force is acting along a straight line. Hence, the perpendicular distance of the line from origin is given as
$r_{\perp }=\left|\frac{A{X}_1+B{Y}_1+C}{\sqrt{A^2+B^2}}\right|$
[putting $(X_1,Y_1)=(0,0$)]
$=\left|\frac{0-0-4}{\sqrt{1^2+1^2}}\right|=2\sqrt{2}$
Hence, $\tau =F.r_{\perp }$
$4\sqrt{2}=F.2\sqrt{2}$ (from (1))
$\Rightarrow F=2\text{N}$