Derive the equation of angular momentum in the case of rotational motion about a fixed axis.

(N/A) For a rigid body rotating about a fixed $Z$-axis,the angular momentum of an individual particle $i$ is given by $\vec{l}_i = \vec{r}_i \times \vec{p}_i$.
The total angular momentum $\vec{L}$ of the rigid body is the vector sum of the angular momenta of all its particles:
$\vec{L} = \sum_{i=1}^{n} \vec{l}_i = \sum_{i=1}^{n} (\vec{r}_i \times \vec{p}_i)$.
Since $\vec{p}_i = m_i \vec{v}_i$ and for rotational motion $\vec{v}_i = \vec{\omega} \times \vec{r}_i$,we have:
$\vec{L} = \sum_{i=1}^{n} [\vec{r}_i \times (m_i (\vec{\omega} \times \vec{r}_i))]$.
For rotation about a fixed axis,the angular velocity $\vec{\omega}$ is constant for all particles and directed along the axis of rotation (let it be the $Z$-axis,$\vec{\omega} = \omega \hat{k}$).
The component of angular momentum along the $Z$-axis is:
$L_Z = \sum_{i=1}^{n} m_i r_i^2 \omega = (\sum_{i=1}^{n} m_i r_i^2) \omega$.
Since the moment of inertia $I = \sum_{i=1}^{n} m_i r_i^2$,we get:
$L_Z = I \omega$.
In vector form,$\vec{L} = I \vec{\omega}$,which is analogous to $\vec{p} = m \vec{v}$ in linear motion.