Question

A particle of mass m is projected with velocity v making an angle of 45º with the horizontal. The magnitude of the angular momentum of projectile about the axis of projection when the particle is at maximum height is

& \text{A}\text{. Zero} \\\ & \text{B}\text{. }\dfrac{m{{v}^{3}}}{4\sqrt{2}g} \\\ & \text{C}\text{. }\dfrac{m{{v}^{3}}}{\sqrt{2}g} \\\ & \text{D}\text{.}\dfrac{m{{v}^{3}}}{4g} \\\ \end{aligned}$$

Answer 1

Using angular momentum formula we can solve this problem.

Formula used: Angular momentum $L=\dfrac{mv}{\sqrt{2}}\dfrac{{{v}^{2}}}{4g}=\dfrac{m{{v}^{3}}}{4\sqrt{2}g}$where, m = mass, v = velocity and h = height.
And, maximum height $h=\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}$ where,= the angle with the horizontal.

Complete step by step solution:
Let us consider v as the velocity of the projection and angle of projection given is 45º.
$L=mvr\sin \theta$
At maximum point, velocity $v=v\cos \theta =\dfrac{1}{\sqrt{2}}v$ direction towards horizontal and no vertical velocity is present).
The maximum height reached will be $h=\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}=\dfrac{{{v}^{2}}{{\sin }^{2}}\left( {{45}^{\circ }} \right)}{2g}=\dfrac{{{v}^{2}}}{4g}............\left( ii \right)$
Now, let us substitute equation (ii) in (i) we get,
$L=\dfrac{mv}{\sqrt{2}}\dfrac{{{v}^{2}}}{4g}=\dfrac{m{{v}^{3}}}{4\sqrt{2}g}$

Therefore, the required answer is $\dfrac{m{{v}^{3}}}{4\sqrt{2}g}$i.e., option B.

Note: Angular momentum is an important quantity in physics as it is a conserved quantity. The angular momentum is the rotational equivalent of linear momentum.