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Is the expression $$\frac{v^2}{r}$$ for centripetal acceleration,only valid for uniform circular motion, or is it also valid even when particle is moving in a non-uniform circular motion?

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    $\begingroup$ It applies to any smooth path. And the specific case of non-uniform circular motion is explicitly discussed here. $\endgroup$ – lemon Jan 12 '18 at 10:34
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With little differential geometry, velocity and acceleration can be written as:

\begin{align*} \mathbf{v} &= \dot{s} \, \mathbf{T} \\ &= v \, \mathbf{T} \\ \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\ &= \frac{d^2s}{dt^2} \mathbf{T} + \frac{v^2}{\rho} \mathbf{N} \\ &= v\frac{dv}{ds} \mathbf{T} + \frac{v^2}{\rho} \mathbf{N} \\ \end{align*}

where $s$ is the arclength, $\kappa$ is the curvature and $\rho$ is the radius of curvature.

Acceleration will include the tangential component (along $\mathbf{T}$) and normal component (along $\mathbf{N}$) which is the centripetal acceleration but is given by radius of curvature instead.

See also answer 1 and answer 2 of mine.

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