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)

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