Motion of Particle in Polar Coordinates

Theorem
Consider a particle $p$ of mass $m$ moving in the plane under the influence of a force $\mathbf F$.

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

Let $\mathbf F$ be expressed as:
 * $\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$

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 $p$
 * $F_r$ and $F_\theta$ are the magnitudes of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively.

Then the second order ordinary differential equations governing the motion of $m$ under the force $\mathbf F$ are:

Proof
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as:
 * $(1): \quad \mathbf r = r \mathbf u_r$

From Velocity Vector in Polar Coordinates, the velocity $\mathbf v$ of $p$ can be expressed in vector form as:


 * $\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$

From Acceleration Vector in Polar Coordinates, the the acceleration $\mathbf a$ of $p$ can be expressed as:


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

We have:
 * $\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$

and from Newton's Second Law:
 * $\mathbf F = m \mathbf a$

Hence:
 * $\mathbf F = m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$

Equating components:

Hence the result: