183_notes:ap_derivation

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183_notes:ap_derivation [2014/11/20 17:54] caballero183_notes:ap_derivation [2014/11/20 18:06] caballero
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 ===== Derivation of the Angular Momentum Principle ==== ===== Derivation of the Angular Momentum Principle ====
 +
 +Consider a single particle (mass, $m$) that is moving with a momentum $p$. This particle experiences a net force $F_{net}$, which will change the particle's momentum based on the momentum principle,
  
 $$F_{net} = \dfrac{d\vec{p}}{dt}$$ $$F_{net} = \dfrac{d\vec{p}}{dt}$$
 +
 +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.
  
 $$\vec{r} \times \vec{F}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$ $$\vec{r} \times \vec{F}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$
 +
 +This cross product of the lever arm and the net force is the net torque about that chosen location,
  
 $$\vec{\tau}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$ $$\vec{\tau}_{net} = \vec{r} \times \dfrac{d\vec{p}}{dt}$$
 +
 +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.
  
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \dfrac{d\vec{r}}{dt} \times \vec{p}$$ $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \dfrac{d\vec{r}}{dt} \times \vec{p}$$
 +
 +The term on the far right is the cross product of the particle's velocity and momentum,
  
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \vec{v} \times \vec{p}$$ $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - \vec{v} \times \vec{p}$$
 +
 +which for an object that doesn't change identity is zero.
  
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - m \underbrace{\vec{v} \times \vec{v}}_{=0}$$ $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right) - m \underbrace{\vec{v} \times \vec{v}}_{=0}$$
 +
 +And thus, we have the angular momentum principle in its derivative form,
  
 $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right)$$ $$\vec{\tau}_{net} = \dfrac{d}{dt}\left(\vec{r} \times \vec{p}\right)$$
  
 $$\vec{\tau}_{net} = \dfrac{d\vec{L}}{dt}$$ $$\vec{\tau}_{net} = \dfrac{d\vec{L}}{dt}$$
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  • 183_notes/ap_derivation.txt
  • Last modified: 2014/11/20 18:06
  • by caballero