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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.

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Thanks. You're right. But only if the axes are parallel. –  SteveB Jan 16 '11 at 2:29
    
$\phi = \phi_0+ \omega_{0}t + \frac{1}{2} \alpha t^2$. –  PEV Jan 16 '11 at 2:29
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Thanks Trevor, but about what axis? –  SteveB Jan 16 '11 at 2:33
    
Is the acceleration axis fixed to the body or fixed to the global frame? That makes a difference. –  Tpofofn Jan 16 '11 at 2:58
    
The acceleration vector is fixed in the global frame. –  SteveB Jan 16 '11 at 3:11
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2 Answers

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!

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I think I understand, but I'm not sure about the last step. The Euler angles and their rates in F are unknown functions of t, so how do I go about integrating them? Do you mean to consider it as a system of differential equations to be solved? –  SteveB Jan 16 '11 at 3:36
    
@SteveB - yes the angles and rates are initial conditions for each step. –  ja72 Jan 16 '11 at 16:28
    
Caution when $\psi=\pi/2$ because $\vec{F}$ becomes singular. –  ja72 Jan 16 '11 at 16:42
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If being considered numerically, it's going to be easier I think to just find angular velocity vectors $\omega_t=(\omega_0+\alpha t_i)$ for each timestep, construct a matrix to rotate about each vector by the appropriate amount, and multiply them together. But I was hoping for something a bit faster and more elegant. Is an algebraic answer impossible, or just difficult?

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