Question

A particle of mass $10\text{kg}$ is projected with a velocity $5\text{m/s}$ from a point on horizontal ground, making an angle ${37}^{\circ }$ with the horizontal. Find the total torque about the point of projection acting on the particle when it is at its maximum height.

• A. $120\text{Nm}$ along $\hat{k}$ direction
• B. $120\text{Nm}$ along $-\hat{k}$ direction
• C. $60\text{Nm}$ along $\hat{k}$ direction
• D. $60\text{Nm}$ along $-\hat{k}$ direction

Answer 1

The correct option is B $120\text{Nm}$ along $-\hat{k}$ direction

Given, mass of the projectile, $m=10\text{kg}$
Initial velocity of projectile, $u=5\text{m/s}$
Angle of projection, $\theta ={37}^{\circ }$

Torque about point of projection due to gravity at maximum height
$\overrightarrow{\tau }=\frac{R}{2}\hat{i}\times mg(-\hat{j})$
$\because$ Range $\left(R\right)=\frac{u^2\sin 2\theta }{g}$
$\Rightarrow \tau =\frac{mg\left(u^2\sin 2\theta \right)}{2g}$
$=\frac{m{u}^2\sin 2\theta }{2}=m{u}^2\sin \theta \cos \theta$
$=10\times (5{)}^2\times \sin {37}^{\circ }\times \cos {37}^{\circ }=120$
$\therefore \overrightarrow{\tau }=120\text{Nm}(-\hat{k})$