First slide

Angular momentum

Question

A particle of mass m is projected with a velocity u making an angle of 45^o 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

Easy

A

zero
B

${mu}^{3} / 4 \sqrt{2} g$
C

${mu}^{3} / \sqrt{2} g$
D

$m \sqrt{(2 {gh}^{3})}$

Solution

$u_{x} = ucos 45^{\circ} = u / \sqrt{2}$
Linear momentum = $mu / \sqrt{2}$
Maximum height = $u^{2} / 4 g$
Angular momentum of the particle
$= \frac{mu}{\sqrt{2}} \times \frac{u^{2}}{4 g} = \frac{{mu}^{3}}{4 \sqrt{2} g}$

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