Question

A particle of mass $m$ is projected with a velocity $V$ making an angle of ${30}^{\circ }$ with the horizontal. The magnitude of the angular momentum of the projectile motion about the point of projection when the particle is at maximum height $H$ is given by which of the following expression(s)?

• A. $\frac{\sqrt{3}}{2}mVH$
• B. $\frac{\sqrt{3}}{16}\frac{m{V}^3}{g}$
• C. $m\sqrt{6g{H}^3}$
• D. $\frac{16}{\sqrt{3}}m{V}^{3}g$

Answer 1

Answer: (A), (B), (C)

$m\sqrt{6g{H}^3}$


We know that,
Angular momentum = Linear momentum $\times$ perpendicular distance between line of action and axis of rotation.
Given that
mass of particle $=m$
velocity of particle $=V$
So, Angular momentum about $\text{O}$ at highest point is
$L=M{V}_{x}H=mV\cos {30}^{\circ }\times H$
$\Rightarrow L=\frac{\sqrt{3}}{2}mVH...\left(1\right)$
For highest point,
$H=\frac{V^2(\sin {30}^{\circ }{)}^2}{2g}=\frac{V^2}{8g}$
$[\because \sin {30}^{\circ }=\frac{1}{2}]$
So eq (1) becomes
$L=\frac{\sqrt{3}}{2}mV.\frac{V^2}{8g}=\frac{\sqrt{3}}{16}\frac{m{V}^3}{g}$

Also, we know from conservation of mechanical energy,
$\frac{1}{2}m{V}^2+0=\frac{1}{2}m(V\cos {30}^{\circ }{)}^2+mgH$
$\Rightarrow \frac{1}{8}m{V}^2=mgH\Rightarrow V^2=8gH$
Substituting in (1),
$L=\frac{\sqrt{3}}{2}mH.\sqrt{8gH}=m\sqrt{8gH\times H^2\times \frac{3}{4}}=m\sqrt{6g{H}^3}$
Thus, options (a), (b) and (c) are correct.