Mass fixed to a wheel - mechanics problem I am struggling with a problem from classical mechanics. Imagine a massless wheel (to make it simpler) with a mass $m$ fixed to it rolling without slipping on a horizontal ground. If we now try to find the equations of motion of the wheel (for instance the angle $\alpha$ it turns) we will find that all the forces are independent of velocity, so $\alpha ''=f(\alpha)$.
After doing that I decided to solve this problem using Euler-Langrange equations (since friction does no work). I came up with $L=\frac{1}{2} m R^{2} (\frac{d\alpha}{dt})^{2} (1+\cos \alpha)-mgR(1+\cos \alpha)$ which, upon solving, gives $\alpha ''$ as a function of both $\alpha$ and $\alpha '$. What is my problem?
 A: (Updated to reflect my better understanding of the problem, based on our discussion in the comments.)
In my comments above, I didn't understand the question. Now that I do, it seems to me that your Lagrangian is correct, except that the factor $1\over 2$ in the kinetic energy term doesn't belong. The equation of motion does involve $\dot\alpha$. I don't think I understand why that's a problem. I can't think of a general argument that proves that there shouldn't be any first derivatives in the Euler-Lagrange equations in a system like this.
For instance, consider a force-free particle moving in two dimensions. If we express things in polar coordinates, the Lagrangian is
$$
L={1\over 2}m(\dot r^2+r^2\dot\phi^2).
$$
The Euler-Lagrange equation for $\phi$ reduces to 
$$
2\dot r\dot\phi+r\ddot\phi=0.
$$
This involves first derivatives, even though there's no force at all (hence a fortiori no velocity-dependent force).
In subsequent comments (below) you explained your reason for not expecting there to be $\dot\alpha$ terms: calculating the torque about the instantaneous point of contact and setting that equal to $I\ddot\alpha$ yields an equation with no $\dot\alpha$'s. The reason that doesn't work is that the torque is equal to $d{\bf l}/dt=d(I\dot\alpha)/dt$. Since $I$ is a function of time, this is not equal to $I\ddot\alpha$. There's a bit more detail in the comments below. Anyway, the final equation of motion, derived either from the Euler-Lagrange equation or from torque considerations, comes out to
$$
2(1+\cos\alpha)\ddot\alpha=(\dot\theta^2+g/R)\sin\alpha.
$$
