Home > Mathematics > Seeking the centre

## Seeking the centre

I have been thinking about motion in a circle and on curves. A lot can be gleaned from simple mathematical models of motion. In the following, I will assume we can define a reference frame and absolute time (i.e. ignore the complexities of rotating reference frame and relativity). I will use bold type face for vectors and dot notation for differentiation with respect to time. For the following considerations the origin of the reference frame will be the centre of the rotational motion (where the central force points). $\mathbf{r}$ will represent the position vector. We will consider the two dimensional case in polar coordinates. $\mathbf{\hat r}$ and $\mathbf{\hat\theta}$ will represent the unit radial and tangential vectors respectively. Progressively differentiating with respect to time:

$\begin{array}{ccc}\mathbf{ r} &= &r\mathbf{\hat r}\\\mathbf{\dot r}&=&\dot r \mathbf{\hat r} + r \dot\theta\mathbf{\hat\theta}\\\mathbf{\ddot r}&=&\ddot r \mathbf{\hat r}+ \dot r \dot\theta \mathbf{\hat\theta}+\dot r\dot\theta\mathbf{\hat\theta}+r\ddot\theta\mathbf{\hat\theta}-r{\dot\theta}^2\mathbf{\hat r}\end{array}$

(note: $\mathbf{\dot{\hat r}}=\dot\theta\mathbf{\hat\theta}$ and $\mathbf{\dot{\hat\theta}}=-\dot\theta\mathbf{\hat r}$This follows from noting: $\mathbf{\hat r} = <\cos\theta,\sin\theta>, \mathbf{\hat\theta}=<-\sin\theta,\cos\theta>$)

Gathering terms in the last equation of the array yields,

$\mathbf{\ddot r} =(\ddot r - r{\dot\theta}^2)\mathbf{\hat r} + (2 \dot r\dot\theta + r\ddot\theta)\mathbf{\hat\theta}$

This equation can be interpreted as decomposing acceleration into a radial and tangential component (first and second terms respectively).

A number of insights emerge from considering this equation.

•  circular motion:   In  this care $\dot r =0$, hence $\mathbf{r}\cdot\mathbf{\dot r}= r^2\dot\theta\mathbf{\hat r}\cdot\mathbf{\hat\theta} =0$. Hence, the position and velocity vector are orthogonal.  This is,perhaps,  more easily seen by just noting in circular motion that  $\mathbf{r} \cdot \mathbf{r} =\text{constant}$. Hence, $\frac{d(\mathbf{r}\cdot\mathbf{r})}{dt} =2 \mathbf{r}\mathbf{\cdot{\dot r}} =0$
• uniform  circular motion: when an object rotates around an axis at constant speed ($\dot\theta =\text{constant}, v=r\dot\theta=\text{constant}$): $\dot r =\ddot r = \ddot \theta =0$.   This yields the centripetal acceleration as:

$\mathbf{\ddot r}= -r{\dot\theta}^2\mathbf{\hat r}=-\frac{v^2}{r} \mathbf{\hat r}$

The negative sign indicating the acceleration and hence force is directed centrally to axis of rotation (hence the term centripetal) and the magnitude of centripetal force: $\frac{m v^2}{r}$

• Areal velocity under a centrally acting force: Keplers second law can be expressed as :A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.This can be seen to follow for any centrally acting force. In the case of central force the second term general equation above vanishes, i.e.$2 \dot r\dot\theta +r\ddot\theta =0$. Therefore, $2r\dot r\dot\theta +r^2\ddot\theta =0=\frac{d}{dt}r^2\dot\theta$. The right hand side implies $r^2\dot\theta/2=\text{constant}$, i.e. areal velocity is constant under a centrally acting force
• Central acting force leads to planar motion: Consider the plane defined by the position vector and the velocity vector. Motion will be planar if and only if the normal to this plane is constant or equivalently the normal is time-invariant. Let $\mathbf{n}$ represent this vector. Then

$\begin{array}{ccc}\mathbf{n}&=&\mathbf{r}\times\mathbf{\dot r}\\\mathbf{\dot n}&=& \mathbf{r}\times\mathbf{\ddot r}= r (\ddot r -r{\dot\theta}^2)\mathbf{\hat r}\times \mathbf{\hat r}=\mathbf{0}\end{array}$