# Proof 33: Angular momentum

If we define angular momentum as the cross-product $\vec L=\vec r \times \vec p$, its magnitude is $L=r_\perp p$ (see the Cross-product skill). For a spinning object/system, the particles it consists of rotate in a perfect circle about the axis-of-rotation, and so $r$ is always perpendicular to the momentum, $r=r_\perp$.

Each rotating particle of the object/system carries angular momentum:

$$L_\text{particle}=r p=r\underbrace{m v}_p=r m \underbrace{r\omega}_{v}=r^2m\omega$$

(The geometric bond $v=r\omega$ was used here.) All particles of the object/system rotate equally fast, if we **assume a rigid body**. Then $\omega$ is the same for them all. The total angular momentum $L$ for the object/system is the sum of all $L_\text{particle}$:

$$L=\sum L_\text{particle}=\sum \left(r^2m\right)\,\omega=\omega\underbrace{\sum r^2m}_I=I\omega$$

The rotational speed $\omega$ rotates about the same axis as the angular momentum $L$. Their vector versions therefore point along the same axis and we can write:

$$\vec L=I\vec \omega\qquad\blacksquare$$

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