Question

Average torque on a projectile of mass $m$, initial speed $u$ and angle of projection $\theta$ between initial and final positions P and Q as shown in the figure about the point of projection is:

• A. $\frac{m{u}^{2}sin2\theta }{2}$
• B. $m{u}^{2}cos\theta$
• C. ${\mu }^{2}sin\theta$
• D. $\frac{m{u}^{2}cos\theta }{2}$

Answer 1

The correct option is A $\frac{m{u}^{2}sin2\theta }{2}$

$\overrightarrow{{\tau }_{av}}\cdot \Delta t=\Delta \overrightarrow{L}$
$|\Delta \overrightarrow{L}|=|{\overrightarrow{L}}_f-{\overrightarrow{L}}_i|$ above point of projection.
$=\left(musin\theta \right)\left(Range\right)$
$=\frac{\left(musin\theta \right)\left(u^{2}sin2\theta \right)}{g}=\frac{m{u}^{3}sin\theta sin2\theta }{g}$
$|\overrightarrow{{\tau }_{av}}|=\frac{\Delta \overrightarrow{L}}{\Delta t}=\frac{m{u}^{3}sin\theta sin2\theta }{g}\frac{g}{2usin\theta }$
$=\frac{m{u}^{2}sin2\theta }{2}$