Constant angular acceleration An object with known initial orientation and angular velocity $\omega_0$ is subjected to constant angular acceleration $\alpha$. Its final orientation at time $t$ is some rotation of its initial orientation. How can I find this rotation?
Angular velocity and acceleration are 3D vectors. Note, $\alpha$ is not necessarily parallel to $\omega_0$, so you can't just use the rotation $\omega_0 t+\alpha t^2$ about a common axis.
EDIT: That should be $\omega_0 t+(\alpha /2) t^2$
Thanks.

I assume you mean $\vec{\omega}_0$ and $\vec{\alpha}$ as vector quantities. Unlike rotations, angular speeds and accelerations are vectors that can be added like
$$ \mathrm{d}\vec{\omega} = \vec{\alpha}\,\mathrm{d}t $$
I suppose the end result in speed is
$$ \vec{\omega} = \vec{\omega}_0 + \vec{\alpha}\,t $$
Now for the orientations things are more complicated. They do not add up as vectors. What you have to do is decompose the orientation to three euler angles $\varphi$, $\psi$ and $\theta$ and integrate their derivatives to find the final orientation. If the 3x3 rotation matrix is $R=R_y(\varphi)\,R_x(\psi)\,R_z(\theta)$ then
$$ \vec{\omega} = \hat{j}\dot{\varphi}+R_y(\varphi)\left(\hat{i}\dot{\psi}+R_x(\psi)\,\hat{k}\dot{\theta}\right)$$
which is a result of the sequencial rotations about $y$, $x$ and $z$.
The above can be expressed as $\vec{\omega}=\mathrm{J}\,\pmatrix{\varphi,\psi,\theta}\cdot\begin{bmatrix}\dot{\varphi} \\ \dot{\psi} \\ \dot{\theta}\end{bmatrix}$ with $\mathrm{J}$ the jacobian matrix.
Next the acceleration needs to be decomposed by
$$\vec{\alpha}=\mathrm{J}\,\pmatrix{\varphi,\psi,\theta}\cdot\begin{bmatrix}\ddot{\varphi} \\ \ddot{\psi} \\ \ddot{\theta}\end{bmatrix}+\mathrm{\dot{J}}\,\pmatrix{\varphi,\psi,\theta}\cdot\begin{bmatrix}\dot{\varphi} \\ \dot{\psi} \\ \dot{\theta}\end{bmatrix}$$
and solved as
$$ \begin{bmatrix}\ddot{\varphi} \\ \ddot{\psi} \\ \ddot{\theta}\end{bmatrix} = \vec{F}\pmatrix{\varphi,\psi,\theta,\dot{\varphi},\dot{\psi},\dot{\theta}} $$
and integrated (analytically not likely, but numerically).
Good luck!
