Motion In Two Dimensions Question 4
Question 4 - 2024 (30 Jan Shift 1)
A particle of mass $m$ projected with a velocity ’ $u$ ’ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is :
(1) $\frac{\sqrt{3}}{16} \frac{m u^{3}}{g}$
(2) $\frac{\sqrt{3}}{2} \frac{m u^{2}}{g}$
(3) $\frac{m u^{3}}{\sqrt{2} g}$
(4) zero
Show Answer
Answer: (1)
Solution:
$L=m u \cos \theta H$
$=m u \cos \theta \times \frac{u^{2} \sin ^{2} \theta}{2 g}$
$=\frac{m u^{3}}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^{2}=\frac{\sqrt{3} m u^{3}}{16 g}$
