Q.
A small particle of mass m is projected at an angle $\theta$ with the x-axis with an initial velocity $v_{0}$ in the x-y plane as shown in the figure. At a time $t < \frac{v_{0}\sin\theta}{g}$, the angular momentum of the particle is
Where $\widehat{i},\ \widehat{j}\ $and$\ \widehat{k}$ are unit vectors along x, y and z-axis respectively.
A) $\frac{1}{2}mgv_{0}t^{2}\cos\theta\widehat{i}$
B) $- mgv_{0}t^{2}\cos\theta\widehat{j}$
C) $mgv_{0}t\cos{}\widehat{k}$
D) $- \frac{1}{2}mgv_{0}t^{2}\cos\theta\widehat{k}$
Explanation
(d) $\because$ Angular momentum $\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p}$
Where
$\overrightarrow{r} = v_{0}\cos\theta t\widehat{i} + \left( v_{0}\sin{\theta t} - \frac{1}{2}gt^{2} \right)\widehat{j}$
$\overrightarrow{p} = m\lbrack v_{0}\cos\theta\widehat{i} + (v_{0}\sin\theta - gt)\widehat{j}\rbrack$
$\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} = - \frac{1}{2}mgv_{0}t^{2}\cos\theta\widehat{k}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.