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--- 183_notes:ap_derivation [2014/11/20 17:54] – caballero
+++ 183_notes:ap_derivation [2014/11/20 18:06] (current) – caballero
@@ Line 1: / Line 1: @@
 ===== Derivation of the Angular Momentum Principle ====
-$$F_{net} = \dfrac{d\vec{p}}{dt}$$
+Consider a single particle (mass, $m$) that is moving with a momentum $\vec{p}$. This particle experiences a net force $\vec{F}_{net}$, which will change the particle's momentum based on the momentum principle,
+$$\vec{F}_{net} = \dfrac{d\vec{p}}{dt}$$
+Now, if we consider the cross product of the momentum principle with some defined lever arm (e.g., the origin of coordinates), $\vec{r}$, we can show this results in the angular momentum principle.
 $$\vec{r} \times \vec{F}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$
+This cross product of the lever arm and the net force is the net torque about that chosen location,
 $$\vec{\tau}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$
+The right hand-side of the equation can be re-written using the [[http://en.wikipedia.org/wiki/Chain_rule|chain rule]]. This gives the difference of two terms.
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \dfrac{d\vec{r}}{dt} \times \vec{p}$$
+The term on the far right is the cross product of the particle's velocity and momentum,
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \vec{v} \times \vec{p}$$
+which for an object that doesn't change identity is zero.
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - m \underbrace{\vec{v} \times \vec{v}}_{=0}$$
+And thus, we have the angular momentum principle in its derivative form,
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right)$$
 $$\vec{\tau}_{net} = \dfrac{d\vec{L}}{dt}$$