Euler's Angles for Rotating a Cube

[Cosmossos]

Homework Statement

What are the Euler's angles corresponding to the rotations of a cube in [tex]\theta[/tex] radians around each of its principal axes



around the x: [tex]\theta[/tex]=[tex]\theta[/tex]
[tex]\phi[/tex]=[tex]\psi[/tex]=0

around z:[tex]\psi[/tex]=[tex]\theta[/tex]=0
[tex]\phi[/tex]=[tex]\theta[/tex]

around y:[tex]\phi[/tex]=[tex]\theta[/tex]=[tex]\theta[/tex]
[tex]\psi[/tex]=-[tex]\theta[/tex]


is it correct? how can I make it more clear? It's very confusing...

 

[Cosmossos]

Help me please...

 

[Maxim Zh]

In order to formalize the solution you can use the rotation matrices: [tex] \hat{R}_x(\alpha) [/tex], [tex] \hat{R}_y(\alpha) [/tex], [tex] \hat{R}_z(\alpha) [/tex] and the matrix which describes the whole Euler's transform:

[tex]
\hat{R}(\theta, \phi, \psi) =
\hat{R}_z(\phi) \hat{R}_x(\theta) \hat{R}_z(\psi) \quad (1)
[/tex]

It's easy to get rotation around the x and z axes from (1) and your answers for these cases are right.

As for y-axis the condition

[tex]
\hat{R}(\theta, \phi, \psi) = \hat{R}_y(\alpha)
[/tex]

yields

[tex]
\theta = -\frac{\pi}{2} - \alpha;
[/tex]

[tex]
\phi = \psi = -\frac{\pi}{2}.
[/tex]