Question

A particle of mass m is projected with a velocity v making an angle of 30° 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

Answer 1

Answer: The magnitude of angular momentum is $L = \dfrac{mv^{3}\sqrt{3}}{16g}$

Explanation:

Given that,

Angle $\theta= 30^{\circ}$

Maximum height = h

We know that,

The angular momentum is defined as:

$L = mvr$

Where, r =h = distance of momentum from the fixed point

$L = mv\cos\theata\times h$

Here, it is clear that the vertical component becomes zero.

So, The angular momentum will be

$L = mv\cos30^{\circ}\times\dfrac{v^2sin^2\theta}{2g}$

$L = mv\cos30^{\circ}\times\dfrac{v^2sin^{2}30^{\circ}}{2g}$

$L = mv\times\dfrac{\sqrt{3}}{2}\times\dfrac{v^{2}}{8g}$

$L = \dfrac{mv^{3}\sqrt{3}}{16g}$

Hence, The magnitude of angular momentum is $L = \dfrac{mv^{3}\sqrt{3}}{16g}$