Ans.

The angular momentum of a particle is given by,

\( \vec{L}=\vec{r}\times\vec{p}\tag{1} \)

where \( \vec{p} \) is the linear momentum of a particle of mass m whose position vector with respect to the origin of a inertial reference frame is \( \vec{r} \). If \( \vec{v} \) be the velocity of the particle then, \( \vec{p}=m\vec{v} \) and \( \vec{v}=\frac{d\vec{r}}{dt} \)

Differentiating both side pf equation (1) with respect to time,

\( \frac{d\vec{L}}{dt}=\frac{d}{dt}\left(\vec{r}\times\vec{p}\right)\\=\frac{d\vec{r}}{dt}\times\vec{p}+\vec{r}\times\frac{d\vec{p}}{dt}\\=\left(\vec{v}\times{m}\vec{v}\right)+\vec{r}\times\frac{d\vec{p}}{dt}\\=\vec{r}\times\frac{d\vec{p}}{dt}\\=\vec{r}\times\vec{F}\)

where, \( \vec{v}\times\vec{v}=0 \) and \( \vec{F}=\frac{d\vec{p}}{dt} \) is the rate of change of linear momentum of the particle.

Again we know that the torque acting on the particle is \( \vec{\tau}=\vec{r}\times\vec{F} \).

Therefore, \( \displaystyle{\vec{\tau}=\frac{d\vec{L}}{dt}} \)