Average torque on a projectile of mass $m$, initial speed $u$ and angles of projection $ heta$, between initial and final position $P$ and $Q$ as shown in figure about the point of projection is [image:4_image29.emf]

Question

Average torque on a projectile of mass $m$, initial speed $u$ and angles of projection $	heta$, between initial and final position $P$ and $Q$ as shown in figure about the point of projection is [image:4_image29.emf]

Dr. Abubacker · Accepted Answer

(c) Time of flight. $T = \frac{2u\sin	heta}{g}$
Horizontal range, $R = \frac{u^{2}\sin{2	heta}}{g}$
Change in angular momentum,
$\left| d\overrightarrow{\mathbf{L}} ight| = \left| {\overrightarrow{\mathbf{L}}}_{f} - {\overrightarrow{\mathbf{L}}}_{i} ight|$ about point of projection
$= (mu\sin	heta) 	imes \frac{u^{2}\sin{2	heta}}{g}$
$= \frac{mu^{3}\sin{	heta\sin{2	heta}}}{g}$
$Torque\ \left| \overrightarrow{	au} ight| = \frac{change\ in\ angular\ momentum}{time\ of\ flight}$
$= \left| \frac{d\overrightarrow{\mathbf{L}}}{T} ight|$

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

Explanation

Reviewed by Dr. Abubacker

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