The magnitude of torque on a particle of mass $1 \mathrm{~kg}$ is 2.5 Nm about the origin. If the force…
The magnitude of torque on a particle of mass $1 \mathrm{~kg}$ is 2.5 Nm about the origin. If the force acting on it is $1 \mathrm{~N},$ and the distance of the particle from the origin is $5 \mathrm{~m},$ the angle between the force and the position vector is (in radians):
$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{8}$
$\frac{\pi}{4}$
Solution
Torque about the origin $=\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$
$=r F \sin \theta \Rightarrow 2.5=1 \times 5 \sin \theta$
$\sin \theta=0.5=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{6}$