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Obtain the relation between angular momentum of a particle and torque acting on it.
Solution
By definition of angular momentum of a particle, $\vec{l}=\vec{r} \times \vec{p}$
Differentiating this equation w.r.t. time $\mathrm{t}, \frac{d \vec{l}}{d t}=\vec{r} \times \frac{d \vec{p}}{d t}+\frac{d \vec{r}}{d t} \times \vec{p}$
But, $\frac{d \vec{p}}{d t}=$ rate of change of linear momentum $=$ force $\overrightarrow{\mathrm{F}}$ and $\frac{d \vec{r}}{d t}=\vec{v}$
$\therefore \quad \frac{d \vec{l}}{d t}=\vec{r} \times \overrightarrow{\mathrm{F}}+\vec{v} \times \vec{p}$
But as $\vec{v}$ and $\vec{p}$ are in the same direction,
$\vec{v} \times \vec{p}=0$
$\therefore \quad \frac{d \vec{l}}{d t}=\vec{r} \times \overrightarrow{\mathrm{F}}=\vec{\tau}$
Thus, "the time rate of change of angular momentum is equal to the torque".
