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Centripetal Acceleration

  

Figure: Differences between vectors at slightly different times for a body in uniform circular motion.

From Fig. 10.3 we can see the relationship between the change in position $\Delta \vec{\mbox{\boldmath$\space r $\unboldmath }}$ and the change in velocity $\Delta \vec{\mbox{\boldmath$\space v $\unboldmath }}$ in a short time interval $\Delta t$. As all three get smaller and smaller, $\Delta \vec{\mbox{\boldmath$\space v $\unboldmath }}$ gets to be more and more exactly in the centripetal direction (along $-\hat{\mbox{\boldmath$\space r $\unboldmath }}$) and its scalar magnitude $\Delta v$ will always (from similar triangles) be given by

$${\vert\Delta \vec{\mbox{\boldmath $ v $\unboldmath }}\vert \o . . . . . . t\Delta \vec{\mbox{\boldmath $ r $\unboldmath }}\vert \over r}$$

where I have been careful to write $\vert\Delta \vec{\mbox{\boldmath$\space r $\unboldmath }}\vert$rather than $\Delta r$ since the magnitude of the radius vector, r, does not change! Now is a good time to note that, for a tiny sliver of a circle, there is a vanishingly small difference between $\vert\Delta \vec{\mbox{\boldmath$\space r $\unboldmath }}\vert$ and the actual distance $\Delta \ell$ travelled along the arc, which is given exactly by $\Delta \ell = r \Delta \theta$. Thus

$${\Delta \vec{\mbox{\boldmath $ v $\unboldmath }} \over v} \; . . . . . . - \hat{\mbox{\boldmath $ r $\unboldmath }} \; \Delta \theta .$$

If we divide both sides by $\Delta t$ and then take the limit as $\Delta t \to 0$, the approximation becomes arbitrarily good and we get

$${1 \over v} \left( d\vec{\mbox{\boldmath $ v $\unboldmath }} . . . . . . boldmath $ r $\unboldmath }} \left( d\theta \over dt \right) .$$

We can now combine this with the definitions of acceleration ( $\vec{\mbox{\boldmath$\space a $\unboldmath }} \equiv d\vec{\mbox{\boldmath$\space v $\unboldmath }}/dt$) and angular velocity ( $\omega \equiv d\theta/dt$) to give (after multiplying both sides by v) $\vec{\mbox{\boldmath$\space a $\unboldmath }} = - \hat{\mbox{\boldmath$\space r $\unboldmath }} \; \omega \; v$. We need only divide the equation $\Delta \ell = r \Delta \theta$ by $\Delta t$ and let $\Delta t \to 0$ to realize that $v = r \omega$. If we substitute this result into our equation for the acceleration, it becomes

$$\vec{\mbox{\boldmath$ a $\unboldmath }} \; = \; - \hat{\mbo . . . . . . \; = \; - \vec{\mbox{\boldmath$ r $\unboldmath }} \; \omega^2$$ (10.2)

which is our familiar result for the centripetal acceleration in explicitly vectorial form.