Last modified on 26 April 2007, at 07:23

# Coordinate systemsEdit

There are several ways of assigning coordinates to a system. Which one you choose depends on what is happening within the system.

## Fixed Rectangular coordinatesEdit

In this coordinate system, vectors are expressed as an addition of vectors in the x, and y, direction from a non-rotating origin. Usually $\vec i \, \!$ is a unit vector in the x direction, and $\vec j \, \!$ is a unit vector in the y direction.

The position vector, $\vec s \, \!$ (or $\vec r \, \!$), the velocity vector, $\vec v \, \!$, and the acceleration vector, $\vec a \, \!$ are expressed using rectangular coordinates in the following way:

$\vec s = x \vec i + y \vec j$

$\vec v = \dot {s} = \dot {x} \vec {i} + \dot {y} \vec {j}$

$\vec a = \ddot {s} = \ddot {x} \vec {i} + \ddot {y} \vec {j}$

Note: $\dot {x} = \frac{dx}{dt}$ , $\ddot {x} = \frac{d^2x}{dt^2}$

## Rotating coordinatesEdit

Unlike rectangular coordinates which are measured relative to an origin that is fixed and non rotating, the origin of these coordinates can rotate and translate - often following a particle on a body that is being studied.

This system of coordinates is based on three orthogonal unit vectors: the vector $\vec i$, and the vector $\vec j$ which form a basis for the plane in which the objects we are considering reside, and $\vec k$ about which rotation occurs.

### Derivatives of Unit VectorsEdit

The position, velocity, and acceleration vectors of a given point can be expressed using these coordinate systems, but we have to be a bit more careful than we do with fixed frames of reference. Since the frame of reference is rotating, we must take the derivatives of the unit vectors into account when taking the derivative of any of these vectors. If the coordinate frame is rotating at a rate of $\vec \omega \, \!$ in the counterclockwise direction (that's $\omega \vec k$ using the right hand rule) then the derivatives of the unit vectors are as follows:

$\dot \vec i = \omega \vec k \times \vec i = \omega \vec j$

$\dot \vec j = \omega \vec k \times \vec j = - \omega \vec i$

### Position, Velocity, and AccelerationEdit

Given these identities, we can now figure out how to represent the position, velocity, and acceleration vectors of a particle using this coordinate system.

#### PositionEdit

Position is straightforward:

$\vec s = x \vec i + y \vec j$

It's just the distance from the origin in the direction of each of the unit vectors.

#### VelocityEdit

Velocity is the time derivative of position:

$\vec v = \frac{d\vec s}{dt} = \frac{d (x \vec i)}{dt} + \frac{d (y \vec j)}{dt}$

By the chain rule, this is:

$\vec v = \dot x \vec i + x \dot \vec i + \dot y \vec j + y \dot \vec j$

Which from the identities above we know to be:

$\vec v = \dot x \vec i + x \omega \vec j + \dot y \vec j - y \omega \vec i = (\dot x - y \omega) \vec i + (\dot y + x \omega) \vec j$

or equivalently

$\vec v = (\dot x \vec i + \dot y \vec j) + (y \dot \vec j + x \dot \vec i) = \vec v_{rel} + \vec \omega \times \vec r$

where $\vec v_{rel}$ is the velocity of the particle relative to the coordinate system.

#### AccelerationEdit

Acceleration is the time derivative of velocity.

We know that:

$\vec a = \frac{d \vec v}{dt} = \frac{d \vec v_{rel}}{dt} + \frac{d (\vec \omega \times \vec r)}{dt}$

Consider the $\frac{d \vec v_{rel}}{dt}$ part. $\vec v_{rel}$ has two parts we want to find the derivative of: the relative change in velocity ($\vec a_{rel}$), and the change in the coordinate frame ($\omega \times \vec v_{rel}$).

$\frac{d \vec v_{rel}}{dt} = \vec a_{rel} + \omega \times \vec v_{rel}$

Next, consider $\frac{d (\vec \omega \times \vec r)}{dt}$. Using the chain rule:

$\frac{d (\vec \omega \times \vec r)}{dt} = \dot \vec \omega \times \vec r + \vec \omega \times \dot \vec r$

$\dot \vec r$ we know from above:

$\frac{d (\vec \omega \times \vec r)}{dt} = \dot \vec \omega \times \vec r + \vec \omega \times (\vec \omega \times \vec r) + \vec \omega \times \vec v_{rel}$

So all together:

$\vec a = \vec a_{rel} + \omega \times \vec v_{rel} + \dot \vec \omega \times \vec r + \vec \omega \times (\vec \omega \times \vec r) + \vec \omega \times \vec v_{rel}$

And collecting terms:

$\vec a = \vec a_{rel} + 2(\omega \times \vec v_{rel}) + \dot \vec \omega \times \vec r + \vec \omega \times (\vec \omega \times \vec r)$