# Centripetal acceleration for particle that moves in a circle at non-constant speed

How is $a_{R}=v_T^{2}/r$ derived for changing tangential speed? ($a_R$-centripetal component $r$-distance from axis $v_T$- tangential speed)

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Do you know how to derive it for constant tangential speed? Its the same thing, just applied for a movement of infinitesimal angle $d\theta$ – udiboy1209 Aug 21 '13 at 11:19
For constant tangential speed I can express the time interval as: $\Delta t=(2r\theta)/v$ thus, can be substituted into the centripetal acceleration: $(-vsin\theta-vsin\theta )/\Delta t)$ but in the case of changing speed I have different expressions- how would time interval be expressed? – user Aug 21 '13 at 11:44
@ udiboy Is changed speed suppose to be treated for infinitesimal angle as having the same speed? – user Aug 21 '13 at 14:07
Yes, you assume the speed doesn't change much for such a small angle, and then you use the constant speed formula. – udiboy1209 Aug 21 '13 at 18:55

We can express the velocity of the particle as:

$\vec{v}= \vec{\omega} \times \vec{r}$ where $\vec{\omega}$ is perpendicular to the plane in which $\vec{r}$ rotates and $\vec{\omega}$ can change only in magnitude i.e. the tangential speed can change.

We have, $\frac{d\vec{v}}{dt}=(\frac{d\vec{\omega}}{dt} \times \vec{r})+(\vec{\omega} \times \vec{v}).$

Now the first term of this equation points is the same/opposite direction as $\vec{v}$ because $\frac{d\vec{\omega}}{dt}$ as same direction as $\vec{\omega}$(We assumed $\vec{\omega}$ can change only in magnitude).So the contribution to the centripetal acceleration can come only from the second term which we can expand easily by the vector triple product identity.

$\vec{\omega} \times \vec{v}=\vec{\omega} \times (\vec{\omega} \times \vec{r})=\vec{\omega}(\vec{\omega} \cdot \vec{r})-\vec{r}(\vec{\omega} \cdot \vec{\omega}).$

Now first term is zero since $\vec{\omega}$ is perpendicular to plane of rotation i.e. $\vec{r}.$So finally we have the centripetal acceleration as $\vec{a_{R}}=-\vec{r}(\vec{\omega} \cdot \vec{\omega})=-\frac{v_{T}^2}{r}\hat{r}$ even in the case the tangential speed changes.

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