Motion of a bead on a rolling wheel (with gravity) I'm new to Lagrangian mechanics and I wanted to draw an x-y plot for motion of a small bead on a wire which is the radius of a wheel rolling on the ground.

I started by writing Lagrange equations in spherical coordinates
$$L=0.5mr'^2+0.5mr^2 \theta'^2-mgrsin(\theta)$$
Using the Lagrange method I got
$$r\theta''+2r'\theta'=-g*cos(\theta)$$
$$r''=r\theta'^2-g*sin(\theta)$$
Now I don't know if this can be solved or not. But let's simplify it and say the angular velocity of the wheel is constant so 
$$\theta''=0$$
$$2r'\theta'=-g*cos(\theta)$$
$$r''=r\theta'^2-g*sin(\theta)$$
How can I get the equations of motion from a point of view outside the wheel? (Like someone who is standing on the street where this wheel is rolling on)
Are my equations of motion right or I should include the kinetic energy of the wheel in the Lagrangian too?
 A: Your motion equations are perfectly right. However, it's very likely it has not analytic solution due to those $sin \theta$ and $cos \theta$ there. In pendulum motions, we usually consider small angle variations so that we can say $sin \theta \approx \theta$ and the equation becomes easily solvable.
As for how the radius constraint acts in the equations, it's simply a initial condition, necessary for finding a particular solution. Remember that for differential equations solutions usually one or more initial conditions are required.
Getting the motion for a wheel moving is quite complicated. The radius of the wheel would be different from the one of the polar coordinate system. That way, it's convenient to introduce a cycloid coordinate system:
$x = r(\theta - sin\theta) $ 
$y = r(1-cos\theta)$
$L = 0.5m[(r')^2(\theta^2 + 2) + (\theta')^2(2r^2)] - mgr(1-cos\theta)$
This would give us the following motion equations:
$2mr(\theta')^2 -mg(1-cos\theta) = mr''(\theta^2 +2) + 2mr'(\theta \theta')$
$m \theta(r')^2 -mgr(sin\theta)= 2mr^2 \theta'' + 4mrr'\theta'$
You see, that would give a very complex solution even harder than the first case. I don't think you can easily solve any of those numerically, they probably require way advanced methods.
