Question
A particle of mass $m$ is projected from the ground with an initial speed $u$ at an angle $\alpha$. Find the magnitude of its angular momentum at the highest point of its trajectory about the point of projection.
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We need to find the magnitude of its angular momentum at the highest point of its trajectory about the point of projection. Show more…
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A particle of mass $m$ is projected from origin $O$ with spced $u$ at an angle $\theta$ with positive $x$ -axis. Positive $y$ -axis is in vertically upward dircction. Find the angular momentum of particle at any time $t$ about $O$ before the particle strikes the ground again.
Rotational and Rolling Motion
Section A
A particle of mass $m$ is projected from the ground with an initial velocity $y$ at an angle $\theta$ with the horizontal. The magnitudc of angular momentum of the particle with respect to the point of projection, when it strikes the ground will be (a) $\frac{\left(2 m v^{3} \sin 2 \theta\right)}{g}$ (b) $\frac{\left(m y^{2} \sin 2 \theta\right)}{g}$ (c) $\frac{\left(m v^{3} \sin \theta \cos \theta\right)}{2 g}$ (d) $\frac{\left(2 m v^{3} \sin ^{2} \theta \cos \theta\right)}{g}$
Section B
A particle of mass $m$ is projected upwards from level ground with an initial velocity v, making an angle of $45^{\circ}$ with the horizontal. The angular momentum of the particle about the point of projection, when the particle is at its maximum height, is (a) $\frac{\mathrm{mv}^{3}}{\sqrt{2 \mathrm{~g}}}$ (b) $\frac{\mathrm{mv}^{3}}{\mathrm{~g} \sqrt{32}}$ (c) $\frac{\mathrm{m}^{2} \mathrm{v}}{\sqrt{2 \mathrm{~g}}}$ (d) $\frac{\mathrm{m}^{2} \mathrm{v}^{2}}{\sqrt{2} \mathrm{~g}}$
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