Question
State and prove: Law of conservation of angular momentum.

Answer

Statement:
The angular momentum of a body remains constant if the resultant external torque acting on the body is zero.
$I_1 \omega_1=I_2 \omega_2($ when $\tau=0)$
Here I is the moment of inertia and ω is angular. velocity.
Proof:
Consider a particle of mass $m$, rotating about an axis with torque ' $\tau$ '.
Let $\vec{p}$ be the linear momentum of the particle and $\vec{r}$ be its position vector.
$\therefore$ Angular momentum, $\vec{L}=\vec{r} \times \vec{p}$
Differentiating equation (1) with respect to time t, we get,
$\frac{d \vec{L}}{d t}=\frac{d}{d t}(\vec{r} \times \vec{p})=\vec{r} \frac{d \vec{p}}{d t}+\vec{p} \frac{d \vec{r}}{d t}$
We know that, $\frac{d \vec{p}}{d t}=\vec{F}, \frac{d \vec{r}}{d t}=\overrightarrow{ v }, \vec{p}=m \overrightarrow{ v }$
$\therefore \frac{d \vec{L}}{d t}=\vec{r} \times \vec{F}+m(\overrightarrow{ v } \times \overrightarrow{ v })$
$\therefore \frac{d \vec{L}}{d t}=\vec{r} \times \vec{F} \ldots \ldots \ldots(\because \overrightarrow{ v } \times \overrightarrow{ v }=0)$
$\therefore \frac{d \vec{L}}{d t}=\vec{\tau} \ldots \ldots \ldots \ldots(\because \vec{r} \times \vec{F}=\vec{\tau})$
Now, If the $\vec{\tau}=0$, then
$
\frac{d \vec{L}}{d t}=0
$
$\therefore \vec{L}$ is constant. Hence angular momentum remains conserved.
Example:
An athlete diving off a high springboard can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.

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