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Consider a particle of mass $m$ having linear momentum $\vec p$ at position $\vec r$ relative to the origin $O$ . Let $\vec L$ be the angular momentum of the particle with respect the origin. Which of the following equations correctly relate $(s)\, \vec r,\,\vec p$ and $\vec L$ ?
$\frac{{d\vec L}}{{dt}} + \vec r\, \times \frac{{\overrightarrow {dp} }}{{dt}} = 0$
$\frac{{d\vec L}}{{dt}}\, + \,\frac{{\overrightarrow {dr} }}{{dt}} \times \vec p = 0$
$\frac{{d\vec L}}{{dt}}\, - \,\frac{{\overrightarrow {dr} }}{{dt}} \times \vec p = 0$
$\frac{{d\vec L}}{{dt}}\, - \vec r \times \frac{{\overrightarrow {dp} }}{{dt}} = 0$
Solution
As $\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}$
Differentiate both sides with respect to time, we get
$\frac{\mathrm{d} \overrightarrow{\mathrm{L}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{\vec r} \times \overrightarrow{\mathrm{p}})$
$=\frac{d \vec r}{d t} \times \vec{p}+\vec{r} \frac{\overrightarrow{d p}}{d t}$
$=\overrightarrow{\mathrm{r}} \times \frac{\overrightarrow{\mathrm{dt}}}{\mathrm{dt}}\left(\because \frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}} \times \overrightarrow{\mathrm{p}}=0\right)$
$\frac{d \vec{L}}{d t}-\vec{r} \times \frac{\vec{d p}}{d t}=0$
