Statement:
The angular momentum of a body remains constant, if resultant external torque acting on the body is zero.
Proof:
a. Consider a particle of mass \(m\), rotating about an axis with torque ' \(T\) '.
Let \(\vec{p}\) be the linear momentum of the particle and \(\vec{r}\) be its position vector.
b. By definition, angular momentum is given by, \(\vec{L}=\vec{r} \times \vec{p} \ldots(i)\)
c. Differentiating equation (i) with respect to time \(t\),
$$
\begin{aligned}
& \frac{\vec{d} \mathrm{~L}}{\mathrm{dt}}=\frac{d}{d t}(\vec{r} \times \overrightarrow{\mathrm{p}}) \\
\therefore \quad & \frac{\overrightarrow{d \mathrm{~L}}}{\mathrm{dt}}=\overrightarrow{\mathrm{r}} \times \frac{\overrightarrow{\mathrm{dp}}}{\mathrm{dt}}+\overrightarrow{\mathrm{p}} \times \frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}
\end{aligned}
$$
d. But, \(\frac{\overrightarrow{d r}}{d t}=\vec{v}, \frac{\overrightarrow{d p}}{d t}=\vec{F}\) and \(\vec{p}=m \vec{v}\)
\(\therefore \quad\) Equation (ii) becomes,
$$
\frac{\overrightarrow{d I}}{d t}=\vec{r} \times \vec{F}+0\left[\because \vec{v} \times \vec{v}=v^{2} \sin 0^{\circ}=0\right]
$$
e. Also, \(\vec{t}=\vec{r} \times \vec{F}\)
$$
\therefore \quad \frac{\overrightarrow{\mathrm{dL}}}{\mathrm{dt}}=\vec{\tau}
$$
F. If resultant external torque (T) acting on the particle is zero, then \(\frac{d L}{d t}=0\).
\(\therefore \vec{L}=\) constant
\(\therefore\) lw \(=\) constant
Hence, angular momentum remains conserved.
Examples:
1. The angular velocity of revolution of a planet around the sun in an elliptical orbit increases, when the planet comes closer to the sun and vice-versa.
2. A person carrying heavy weights in his hands and standing on a rotating platform can change the speed of the platform.
3. A diver performs somersaults by jumping from a high diving board keeping his legs and arms out stretched first, and then curling his body.
