# Acceleration Vector in Polar Coordinates

## Theorem

Consider a particle $p$ moving in the plane.

Let the position of $p$ at time $t$ be given in polar coordinates as $\left\langle{r, \theta}\right\rangle$.

Then the acceleration $\mathbf a$ of $p$ can be expressed as:

$\mathbf a = \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$

where:

$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$
$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$

## Proof

Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as:

$(1): \quad \mathbf r = r \mathbf u_r$

 $\text {(2)}: \quad$ $\displaystyle \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}$ $=$ $\displaystyle \mathbf u_\theta$ $\text {(3)}: \quad$ $\displaystyle \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}$ $=$ $\displaystyle \mathbf u_r$
$\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$

where $\mathbf v$ is the velocity of $p$.

The acceleration of $p$ is by definition the rate of change in its velocity:

 $\displaystyle \mathbf a$ $=$ $\displaystyle \dfrac {\mathrm d \mathbf v} {\mathrm d t}$ $\displaystyle$ $=$ $\displaystyle r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \mathbf u_r} {\mathrm d t}$ Product Rule for Derivatives $\displaystyle$ $=$ $\displaystyle r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t}$ Chain Rule for Derivatives $\displaystyle$ $=$ $\displaystyle r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_r \dfrac {\mathrm d \theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_\theta \dfrac {\mathrm d \theta} {\mathrm d t}$ substituting from $(2)$ and $(3)$ $\displaystyle$ $=$ $\displaystyle \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$ gathering terms

$\blacksquare$