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 figure, about the point of projection is:

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

Answer 1

Answer: (A)

${\bar{\tau }}_{av}\cdot \Delta t=\Delta \vec{L}\dots$ (i)
Here $\Delta t=$ time of flight $=\frac{2u\sin \theta }{g}$
Change in angular momentum about point of projection (initially it is zero)
$|\overrightarrow{\Delta L}|=\left|{\vec{L}}_f-{\vec{L}}_i\right|=(mu\sin \theta )$ Range
$=\frac{(mu\sin \theta )\left(u^2\sin 2\theta \right)}{g}=\frac{m{u}^3\sin \theta \sin 2\theta }{g}$
Now $\left|{\vec{\tau }}_{\alpha v}\right|=\left|\frac{\overrightarrow{\Delta L}}{\Delta t}\right|=\frac{m{u}^3\sin \theta \sin 2\theta }{g}\times \frac{g}{2u\sin \theta }$
$=\frac{m{u}^2\sin 2\theta }{2}$