## ROTATIONS IN THE PLANE

### INTRODUCTION

In Cartesian coordinates, the natural orthonormal basis is $\{\ii,\jj\}$, where $\ii\equiv\xhat$ and $\jj\equiv\yhat$ denote the unit vectors in the $x$ and $y$ directions, respectively. The position vector from the origin to the point ($x$,$y$) takes the form $$\rr = x \,\ii + y \,\jj$$ Note that $\ii$ and $\jj$ are constant.

A moving object has a position vector given by $$\rr(t) = x(t) \,\ii + y(t) \,\jj$$ Its velocity $\vv$ and acceleration $\aa$ are obtained by differentiation, resulting in \begin{eqnarray} \vv &=& \dot{\rr} = \dot{x} \,\ii + \dot{y} \,\jj \\ \aa &=& \ddot{\rr} = \ddot{x} \,\ii + \ddot{y} \,\jj \end{eqnarray} where dots denote differentiation with respect to $t$.

If we wish to describe things as seen from some point $(p,q)$ other than the origin, all we have to do is replace $\rr$ by $$\rr_{\subM} = \rr - \RR$$ where $$\RR = p \,\ii + q \,\jj$$ If the point is fixed ($\RR=\hbox{constant}$), then this has no effect on the velocity and acceleration: \begin{eqnarray} \vv_{\subM} &=& \dot{\rr}_{\subM} = \dot{\rr} - \dot{\RR} = \dot{\rr} = \vv \\ \aa_{\subM} &=& \ddot{\rr}_{\subM} = \ddot{\rr} - \ddot{\RR} = \ddot{\rr} = \aa \end{eqnarray} If, however, the reference point is moving ($\RR=\RR(t)$), then of course the *relative* velocity ($\vv_{\subM}$) will differ from the “true” velocity ($\vv$) previously computed by the velocity ($\dot{\RR}$) of the moving reference point, and similarly for the relative acceleration.

For instance, suppose an observer undergoes constant *linear
acceleration* in the $x$ direction, so that $$\RR = {1\over2} at^2 \,\ii$$ Such an observer would measure relative velocity and acceleration given by \begin{eqnarray} \vv_{\subM} &=& \dot{\rr} - \dot{\RR} = \vv -at \,\ii \\ \aa_{\subM} &=& \ddot{\rr} - \ddot{\RR} = \aa - a \,\ii \end{eqnarray}

### POLAR COORDINATES

In polar coordinates \begin{eqnarray} x &=& r \cos\theta \\ y &=& r \sin\theta \end{eqnarray} the natural orthonormal basis is $\{\rhat,\that\}$, where \begin{eqnarray} \rhat &=& \hphantom{-}\cos\theta \>\ii + \sin\theta \>\jj \\ \that &=& -\sin\theta \>\ii + \cos\theta \>\jj \end{eqnarray}

We wish to describe the *relative* position of an object with respect to an observer located at the (for now) *fixed* point ($R$,$\Theta$), *not* at the origin. How does this observer describe vectors? The natural basis is just $\{\rhat,\that\}$ *at the location of the
observer*, that is, with $\theta=\Theta$. Thus, the observer describes the object using the relative position vector $$\rr_{\subM}(t) = X(t) \,\rhat + Y(t) \,\that$$ for some functions $X$ and $Y$, which points from the observer to the object. 1) The observer's own position vector, described using the same basis, is of course 2) $$\RR = R \,\rhat$$

The “true” position vector (relative to the origin) is a combination of the observer's position and the object's position relative to the observer, that is $$\rr(t) = \RR + \rr_{\subM}(t)$$ However, just as before, so long as $\RR$ is constant, the relative velocity and acceleration \begin{eqnarray} \vv_{\subM} &=& \dot{X} \,\rhat + \dot{Y} \,\that \\ \aa_{\subM} &=& \ddot{X} \,\rhat + \ddot{Y} \,\that \end{eqnarray} will be the same as the “true” velocity and acceleration.

### ROTATING FRAME

Consider now a rotating observer whose position is given by \begin{eqnarray} r &=& R = \hbox{constant} \\ \theta &=& \Theta = \Omega t \end{eqnarray} Observers in this frame will naturally continue to use $\rr_{\subM}$, $\vv_{\subM}$, and $\aa_{\subM}$ to describe relative motion. And they will need to take into account the fact that $\RR$ is not constant in order to compare their description to the “true” values. But there is now an additional complication, since the basis vectors themselves change with time.

#### Velocity and Acceleration

The basis vectors now take the form \begin{eqnarray} \rhat &=& \hphantom{-}\cos(\Omega t) \,\ii + \sin(\Omega t) \,\jj \\ \that &=& -\sin(\Omega t) \,\ii + \cos(\Omega t) \,\jj \end{eqnarray} so that \begin{eqnarray} \rhatdot &=& -\Omega\sin(\Omega t) \,\ii + \Omega\cos(\Omega t) \,\jj \\ \thatdot &=& -\Omega\cos(\Omega t) \,\ii - \Omega\sin(\Omega t) \,\jj \end{eqnarray} Comparing these equations with the preceding ones, we see that 3) \begin{eqnarray} \rhatdot &=& \hphantom{-}\Omega \,\that \\ \thatdot &=& -\Omega \,\rhat \end{eqnarray}

We are finally ready to compare the relative and “true” velocities and accelerations. Differentiating $$\rr = \RR + \rr_{\subM} = (R+X) \,\rhat + Y \,\that$$ we obtain \begin{eqnarray} \vv = \dot{\rr} &=& \left( \dot{X} \,\rhat + \dot{Y} \,\that \right) + \left( (R+X)\Omega \,\that - Y\Omega \,\rhat \right) \end{eqnarray} Further differentiation yields \begin{eqnarray} \aa = \dot{\vv} &=& \left( \ddot{X} \,\rhat + \ddot{Y} \,\that \right) + 2 \left( \dot{X}\Omega \,\that - \dot{Y}\Omega \,\rhat \right) - \left( (R+X)\Omega^2 \,\rhat + Y\Omega^2 \,\that \right) \\ &=& \aa_{\subM} + 2\Omega \left( \dot{X} \,\that - \dot{Y} \,\rhat \right) -\Omega^2 \,\rr \end{eqnarray}

We therefore obtain the following equations for the relative velocity and acceleration: \begin{eqnarray} \vv_{\subM} &=& \vv - \left( (R+X)\Omega \,\that - Y\Omega \,\rhat \right) \\ \aa_{\subM} &=& \aa - 2\Omega \left( \dot{X} \,\that - \dot{Y} \,\rhat \right) + \Omega^2 \,\rr \end{eqnarray} The *effective* acceleration $\aa_{\subM}$ consists of 3 parts: the “true” acceleration $\aa$, the *centrifugal* acceleration $\Omega^2\,\rr$, and another term which we will discuss later. The centrifugal acceleration points in the direction of $\rr$ and is just (the opposite of) the acceleration of circular motion, as expected.

#### Cross Products

Introducing the angular velocity $$\om = \Omega \,\kk$$ it is easy to check directly that \begin{eqnarray} \rhatdot &=& \om \times \rhat \\ \thatdot &=& \om \times \that \end{eqnarray} But this means that for *any* relative vector of the form $$\FF(t) = X(t) \,\rhat(t) + Y(t) \,\that(t)$$ we have \begin{eqnarray} \dot{\FF} &=& \left( \dot{X} \,\rhat + \dot{Y} \,\that \right) + \left( X \,\rhatdot + Y \,\thatdot \right) \\ &=& \left( \dot{X} \,\rhat + \dot{Y} \,\that \right) + \left( X \,\om\times\rhat + Y \,\om\times\that \right) \\ &=& \left( \dot{X} \,\rhat + \dot{Y} \,\that \right) + \om \times \FF \end{eqnarray} Note that the first term is the “naive” derivative of $\FF$; this “naive” differentiation is precisely what was used to obtain $\vv_{\subM}$ and then $\aa_{\subM}$ starting from $\rr_{\subM}$.

We can use $\om$ to recompute “true” velocities and accelerations. Differentiating $$\rr = \RR + \rr_{\subM}$$ we obtain \begin{eqnarray} \vv = \dot{\rr} &=& \dot{\RR} + \dot{\rr}_{\subM} \\ &=& \om\times\RR + \left( \vv_{\subM} + \om\times\rr_{\subM} \right) \end{eqnarray} Further differentiation yields \begin{eqnarray} \aa = \dot{\vv} &=& \om\times\dot{\RR} + \left( \dot{\vv}_{\subM} + \om\times\dot{\rr}_{\subM} \right) \\ &=& \om\times(\om\times\RR) + \left( \left(\aa_{\subM} + \om\times\vv_{\subM} \right) + \om\times\left( \vv_{\subM} + \om\times\rr_{\subM} \right) \right) \\ &=& \om\times(\om\times\RR) + \aa_{\subM} + 2\,\om\times\vv_{\subM} + \om\times(\om\times\rr_{\subM}) \end{eqnarray}

Rewriting these expressions slightly, we obtain the following equations for the relative velocity and acceleration: \begin{eqnarray} \vv_{\subM} &=& \vv - \om\times\rr \\ \aa_{\subM} &=& \aa - 2\,\om\times\vv_{\subM} - \om\times(\om\times\rr) \end{eqnarray} As before, the *effective* acceleration $\aa_{\subM}$ therefore consists of 3 parts: the “true” acceleration $\aa$, the *centrifugal* acceleration $-\om\times(\om\times\rr)$ and the *Coriolis* acceleration $-2\,\om\times\vv_{\subM}$. The centrifugal acceleration points in the direction of $\rr$, since $$-\om\times(\om\times\rr) = \Omega^2 \rr$$ and is just (the opposite of) the acceleration of circular motion, as expected. Finally, for counterclockwise rotation ($\Omega>0$), the Coriolis acceleration always points *to the right* of the direction of motion $\vv_{\subM}$.

Note that the above expressions for $\vv_{\subM}$ and $\aa_{\subM}$ depend only on the angular velocity $\om$, not on the particular choice of basis $\{\rhat,\that\}$ nor the position of the rotating observer! Even though our derivation was basis-dependent, the result is therefore basis-independent, and the above expressions hold for *any* observer with angular velocity $\om$, *including one at the origin*.

*Cartesian*coordinates (with a particular orientation) centered at the observer's location; this is always true when working with orthonormal bases.