Rotation matrix in yo-yo problem? I need to solve the yo-yo problem not in the normal sense. Instead, I need to include the position vector $r$ and rotation matrix $R$. Assume the yo-yo is rotating in the plane.
In the problem yo-yo is made of two identical cylinders of radius $R$, thickness $h$ and mass $M$, and the yo-yo is let go.
My attempt:
In the description of asteroid problem in space, where the  $r$ is in $R^3$ and transformation $\phi$ is  $\left( {\begin{array}{*{20}{c}}
0&{ - {x_3}}&{{x_2}}\\
{{x_3}}&0&{ - {x_1}}\\
{ - {x_2}}&{{x_1}}&0
\end{array}} \right)$.
The normal problem has 6 degrees of freedom. 
Here comes my problem, how to define transformation $\phi$ in yo-yo problem. 
Is the inertia tensor $J_{\Omega}$ is a $2\times 2$ or $3 \times 3$ matrix?
The problem has only two degrees of freedom, so I assign the origin of system in the center of yo-yo's first position, the position vector is in $R^1$. 
The rotation of the yo-yo is just rotation around the one-point. 
Can I just follow the method in the asteroid example in proving Euler's equation of solid body motion? does the lie algebra thing from which we derive the Euler's equation of solid body motion still works?  Is there anything I need to change for this example?
So can anyone help me with this? Thanks so much!
 A: Of course you can specialise the derivation of the three dimensional Euler equations for your special case. However, unless you are expressly asked to do this, I doubt that your teacher means for you to take this path because it is very much an overkill. The yoyo is spinning in a plane about its axis of symmetry. The inertia tensor is now a simple scalar $J$: the nett torque $\tau$ on the yoyo (assume anticlockwise about the +z axis is positive) is related to its angular speed $\omega$ (assume same sign convention) by $\tau = J\,\dot{\omega}$. The matrix you have written down is the time derivative of a rotation matrix (relative to the initial axes). Knock out one row and the corresponding column (because you're rotating only about one axis) and you're left with a matrix of the form $\omega\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$; it exponentiates to the the rotation matrix $\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)$ (this is what you need to use) where $\theta(t) = \int_0^t \omega(u)\,\mathrm{d}u$. Then the equations of motion $\tau = J\,\dot{\omega}$, $F_x=m\,\dot{x}$ and $F_y = m\,\dot{y}$, where $x,\,y$ define the position of the yoyo's centre of mass.

Questions from OP

Thanks for your answer! I don't understand why the inertial tensor J is a scalar? I thought it is a cylinder rotating about an axis and the form is scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html Could you please explain it for me? 

As you can see from your link, the inertia tensor is when one co-ordinate axis is along the cylinder's axis of symmetry is diagonal. Your problem constrains the cylinder to rotate about its axis of symmetry, so you can forget about the $xx$ and $yy$ components of the tensor and simply use $\tau = \frac{1}{2}\,M\,R^2\,\dot{\omega}$. Strictly speaking, though, it is, as you still a tensor, it's simply that the constrained motion lets you treat it as a scalar.
