Obtain $\tau = I\alpha$ from the angular momentum of a rigid body.

(N/A) The total angular momentum of a rigid body rotating about a fixed axis is given by $\vec{L} = \vec{L}_{z} + \vec{L}_{\perp}$.
Differentiating with respect to time $t$:
$\frac{d\vec{L}}{dt} = \frac{d\vec{L}_{z}}{dt} + \frac{d\vec{L}_{\perp}}{dt}$.
Since the rotation is about a fixed axis,the component of angular momentum perpendicular to the axis $\vec{L}_{\perp}$ remains constant,so $\frac{d\vec{L}_{\perp}}{dt} = 0$.
The rate of change of angular momentum along the axis is equal to the external torque $\vec{\tau}$,so $\frac{d\vec{L}_{z}}{dt} = \vec{\tau}$.
Thus,$\frac{d\vec{L}}{dt} = \vec{\tau}$.
Since $\vec{L}_{z} = I\omega\hat{k}$,where $I$ is the moment of inertia and $\omega$ is the angular velocity:
$\vec{\tau} = \frac{d}{dt}(I\omega\hat{k}) = I\frac{d\omega}{dt}\hat{k}$.
Since $\alpha = \frac{d\omega}{dt}$,we get $\vec{\tau} = I\vec{\alpha}$.
In scalar form,this is $\tau = I\alpha$,which is the rotational analogue of Newton's second law $F = ma$.