A small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $\mathrm{v}_{0}$ in the $\mathrm{x}-\mathrm{y}$ plane as shown in the figure. At a time $\mathrm{t}<\frac{\mathrm{v}_{0} \sin \theta}{\mathrm{~g}}$, the angular momentum of the particle is

(C) $-\frac{1}{2} m g v_{0} t^{2} \cos \theta \hat{k}$

(A) $-\mathrm{mgv}_{0} \mathrm{t}^{2} \cos \theta \hat{\mathrm{j}}$

(B) $\mathrm{mgv}_{0} \mathrm{t} \cos \theta \hat{\mathrm{k}}$

(D) $\frac{1}{2} m g v_{0} t^{2} \cos \theta \hat{i}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Rotational Mechanics

A small particle of mass

m

is projected at an angle

θ

with the

x

-axis with an initial velocity

v_{0}

in the

x - y

plane as shown in the figure. At a time

t < \frac{v _{0} s i n θ}{g}

, the angular momentum of the particle is

Answer: (C)

Take the launch point as origin. At time

t

the particle is at

r = (v_{0} cos θ t) \hat{i} + (v_{0} sin θ t - \frac{1}{2} g t^{2}) \hat{j}

with velocity

v = v_{0} cos θ \hat{i} + (v_{0} sin θ - g t) \hat{j}

. The angular momentum about the origin is

L = m (r \times v)

, which lies along

\hat{k}

. Working out the cross product, the

v_{0}^{2} sin θ cos θ t

terms cancel and we are left with

L = - \frac{1}{2} m g v_{0} t^{2} cos θ \hat{k}

. The same follows quickly from torque: gravity alone acts about the launch point,

τ = - m g v_{0} cos θ t \hat{k}

, and integrating from zero reproduces this result.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option