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Obtain $\tau = I\alpha $ from angular momentum of rigid body.
Solution
The total angular momentum of a motion of rigid body about a fixed axis $\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{L}_{z}}+\overrightarrow{\mathrm{L}_{\perp}}$
Differentiating w.r.t. time
$\frac{d \overrightarrow{\mathrm{L}}}{d t}=\frac{d \overrightarrow{\mathrm{L}}_{z}}{d t}+\frac{d \overrightarrow{\mathrm{L}_{\perp}}}{d t}$
but $\frac{d \overrightarrow{\mathrm{L}_{z}}}{d t}=\tau \hat{k}$ and $\frac{d \overrightarrow{\mathrm{L}_{\perp}}}{d t}=0$
$\therefore \frac{d \overrightarrow{\mathrm{L}}}{d t}=\tau \hat{k}$
but $\overrightarrow{\mathrm{L}_{\mathrm{Z}}}=\mathrm{I} \omega \hat{k}$
$\therefore \frac{d(\mathrm{I} \omega)}{d t}=\tau \hat{k}$
$\mathrm{I} \frac{d \omega}{d t}=\tau$
$\therefore \quad \overrightarrow{\mathrm{I} \alpha}=\vec{\tau} \Rightarrow \vec{\tau}=\mathrm{I} \vec{\alpha}$
Scalar form $\tau=\mathrm{I} \alpha$ is similar to $\mathrm{F}=m a$ in linear motion.
