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Torque and Angular Momentum

Finally we come to the formally trickiest transformation of the SECOND LAW, the one involving the vector product (or ``cross product'') of   $\mbox{\boldmath$\vec{F}$\unboldmath }$  with the distance   $\mbox{\boldmath$\vec{r}$\unboldmath }$  away from some origin11.15 ``O.'' Here goes:

\begin{displaymath}\mbox{\boldmath $\vec{r}$\unboldmath } \; \times \; \left[
 . . . 
 . . . $\unboldmath }
\times \mbox{\boldmath $\vec{F}$\unboldmath } \end{displaymath}

Now, the distributive law for derivatives applies to cross products, so

\begin{displaymath}{d \over dt} \, [ \mbox{\boldmath $\vec{r}$\unboldmath } \tim . . . 
 . . .  } \times
{d\mbox{\boldmath $\vec{p}$\unboldmath } \over dt} \end{displaymath}

but

\begin{displaymath}{d\mbox{\boldmath $\vec{r}$\unboldmath } \over dt} \; \equiv  . . . 
 . . . h } \; \equiv \; m \,
\mbox{\boldmath $\vec{v}$\unboldmath } \end{displaymath}


\begin{displaymath}\hbox{\rm so} \qquad
{d\mbox{\boldmath $\vec{r}$\unboldmath . . . 
 . . . h } \times
\mbox{\boldmath $\vec{v}$\unboldmath }) \; = \; 0 \end{displaymath}

because the cross product of any vector with itself is zero.11.16 Therefore

\begin{displaymath}{d \over dt} \, [ \mbox{\boldmath $\vec{r}$\unboldmath }
\t . . . 
 . . . unboldmath } \times
\mbox{\boldmath $\vec{F}$\unboldmath } . \end{displaymath}

If we define two new entities,

 \begin{displaymath}\mbox{\boldmath$\vec{r}$\unboldmath } \times
\mbox{\boldmat . . . 
 . . . math } \;
\equiv \; \mbox{\boldmath$\vec{L}$\unboldmath }_O,
\end{displaymath} (11.18)


\begin{displaymath}\hbox{\rm the {\sl Angular Momentum\/} about } O \end{displaymath}

and

 \begin{displaymath}\mbox{\boldmath$\vec{r}$\unboldmath } \times
\mbox{\boldmat . . . 
 . . . th } \;
\equiv \; \mbox{\boldmath$\vec{\tau}$\unboldmath }_O,
\end{displaymath} (11.19)


\begin{displaymath}\hbox{\rm the {\sl Torque\/} generated by }
\mbox{\boldmath $\vec{F}$\unboldmath }
\hbox{\rm ~about } O \, , \end{displaymath}

then we can write the above result in the form

 \begin{displaymath}{d \mbox{\boldmath$\vec{L}$\unboldmath }_O \over dt} =
\mbox{\boldmath$\vec{\tau}$\unboldmath }_O
\end{displaymath} (11.20)

This equation looks remarkably similar to the SECOND LAW. In fact, it is the rotational analogue of the SECOND LAW. It says that
``The rate of change of the angular momentum of a body about the origin  O  is equal to the torque generated by forces acting about  O.''

So what? Well, if we choose the origin cleverly this ``new'' Law gives us some very nice generalizations. Consider for instance an example which occurs very often in physics: the central force.



 
next up previous
Next: Central Forces Up: The Emergence of Mechanics Previous: Friction
Jess H. Brewer
1998-10-08