Rotational motion

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A particle of mass m is projected with a velocity v making an angle of 45⁰ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is $\frac{m v^{3}}{(\sqrt{x} g)}$ . Find the value of x.

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Solution

$\begin{array}{l} At the highest point, velocity = v \cos 45^{0} = \frac{v}{\sqrt{2}} in horizontal direction \\ ∴ Momentum = \frac{m v}{\sqrt{2}} L = Angular momentum = Momentum \times Perpendicular distance \\ \Rightarrow L = \frac{mv}{\sqrt{2}} \times h \\ Here h = \frac{v^{2} \sin^{2} 45^{0}}{2 g} = \frac{v^{2}}{4 g} \\ ∴ L = \frac{m v}{\sqrt{2}} \frac{v^{2}}{4 g} = \frac{m v^{3}}{4 \sqrt{2} g} = \frac{m v^{3}}{\sqrt{32} g} \end{array}$

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Rotational motion

Remember concepts with our Masterclasses.

A particle of mass m is projected with a velocity v making an angle of 450 with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is mv3(xg). Find the value of x.

At the highest point, velocity =vcos450=v2 in horizontal direction ​∴ Momentum =mv2​L= Angular momentum = Momentum × Perpendicular distance ​⇒L=mv2×h​ Here h=v2sin24502g=v24g​∴L=mv2v24 g=mv342 g=mv332 g

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A particle of mass m is projected with a velocity v making an angle of 45⁰ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is $\frac{m v^{3}}{(\sqrt{x} g)}$ . Find the value of x.