Rotation 2 Question 13
13. A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is
(a) zero
(b) $m v^{3} /(4 \sqrt{2} g)$
(c) $m v^{3} /(\sqrt{2} g)$
(d) $m \sqrt{2 g h^{3}}$
(1990, 2M)
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Answer:
Correct Answer: 13. (b)
Solution:
- $L=m \frac{v}{\sqrt{2}} r _{\perp}$
Here, $r _{\perp}=h=\frac{v^{2} \sin ^{2} 45^{\circ}}{2 g}=\frac{v^{2}}{4 g}$
$\therefore \quad L=m \frac{v}{\sqrt{2}} \quad \frac{v^{2}}{4 g}=\frac{m v^{3}}{4 \sqrt{2} g}$
