Question

A particle having mass $m$ is projected with a speed $v$ at an angle $\alpha$ with the horizontal ground. Find the torque of the weight of the particle about the point of projection when the particle reaches the ground.

• A. $2m{v}^2\sin 2\alpha$
• B. $\frac{m{v}^2\sin 2\alpha }{2}$
• C. $m{v}^2\sin 2\alpha$
• D. $0$

Answer 1

The correct option is C $m{v}^2\sin 2\alpha$


As we know,
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
$\overrightarrow{\tau }=r_{\perp }F$
$r_{\perp }=\text{Range}=\frac{v^2\sin 2\alpha }{g}$
Hence, $\overrightarrow{\tau }=\frac{v^2\sin 2\alpha }{g}\times mg$
$\overrightarrow{\tau }=m{v}^2\sin 2\alpha$