### Constant Angular Acceleration

When we looked at constant linear acceleration we were able to derive a set of equations to relate the positions, velocities and accelerations:

- v = v
_{0}+ a*t - p = p
_{0}+ v_{0}t + ½ a t^{2} - v
^{2}= v_{0}^{2}+ 2 * a * p

It would be very tempting to just replace the linear values in these equations with their angular counterparts and use those for the equations of constant angular acceleration, unfortunately this would be wrong.

Unlike the linear case, angular orientation behaves differently from the angular velocity and the angular acceleration. Angular accelerations and velocities are combined by vector addition but angular orientations are combined by quaternion or matrix multiplication. In the quaternion case the two different types are related as follows:

d q(t) /dt = ½ * W(t) q(t)

Where:

- q(t) = normalised quaternion representing orientation as a function of time
- W(t) = angular velocity vector represented as quaternion with zero scalar part, i.e
- W (t ) = (0, W x (t ), W y (t ), W z (t ))

- t = time

this is explained on this page.

The above case replaces

dx/dt = v

for the linear case, so we can see that the rotational equations are going to be a more complicated.

At least angular acceleration and angular velocity are both vectors (bivectors) so we can start off, in the same way that we did for the linear case, by integrating the constant acceleration:

α = dω/dt = constant

integrating both sides:

∫α dt = ω_{f} - ω_{i} = α * t

where: | |||

symbol |
description |
type |
units |

α | instantaneous angular acceleration about centre. | bivector | s^{-2} |

ω_{i} |
initial instantaneous angular velocity about centre. | bivector | s^{-1} |

ω_{f} |
final instantaneous angular velocity about centre. | bivector | s^{-1} |

t | time | scalar | s |

But now we want to relate this to the orientation which we do by integrating again, but this time we have to be careful, using the earlier equation:

d q(t) /dt = ½ * W(t) q(t)

substituting W(t) = α * t gives:

d q(t) /dt = ½ * α * t * q(t)

I am not sure how to continue this to get the angular equivalent of p = p_{0} + v_{0} t + ½ a t^{2} because in the above case q(t) appears twice.

### Dynamics Issues

Although this page is not really about the dynamics it is worth thinking about when this constant angular acceleration will occur.

In the linear case, constant acceleration happens in a constant gravity field where there is no friction, due to F=m*a.

So when would constant angular acceleration be likely to happen? We can imagine cases where the torque may be made constant, for example:

- An aeroplane with control surfaces.
- A spacecraft with maneuvering jets.

However unlike the linear case, the relationship between torque and angular acceleration is given by:

T = [I]α

where [I] is the inertia tensor (as explained on this page), this inertia tensor may be a function of orientation and therefore of time which would also make α a function of time.