Motion of Particle in Polar Coordinates

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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 $\polar {r, \theta}$.

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:

\(\ds F_\theta\) \(=\) \(\ds m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} }\)
\(\ds F_r\) \(=\) \(\ds m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}\)


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:

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


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

$\mathbf a = \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} } \mathbf u_\theta + \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2} \mathbf u_r$


We have:

$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$

and from Newton's Second Law of Motion:

$\mathbf F = m \mathbf a$


Hence:

$\mathbf F = m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} } \mathbf u_\theta + m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2} \mathbf u_r$

Equating components:

\(\ds F_r \mathbf u_r\) \(=\) \(\ds m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2} \mathbf u_r\)
\(\ds F_\theta \mathbf u_\theta\) \(=\) \(\ds m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} } \mathbf u_\theta\)

Hence the result:

\(\ds F_\theta\) \(=\) \(\ds m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} }\)
\(\ds F_r\) \(=\) \(\ds m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}\)

$\blacksquare$


Sources

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