# Motion of a bead on a rolling wheel (with gravity) [closed]

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?

• It looks to me your Lagrangian is only for the bead. What about the wheel? – user154997 Jun 19 '17 at 22:50
• @LucJ.Bourhis You mean we should have a 0.5Iw^2 in the Lagrangian? I believe wheel's angular velocity is included in the velocity of the bead isn't it? Correct me if I'm wrong. – Alireza Jun 20 '17 at 6:48
• You have a constraint, that the $\theta$ for the bead is the same as that of the wheel, indeed. But that does not mean you can just forget about the dynamics of the wheel. Another way to look at it: the bead pushes on the wire and therefore applies a torque to the wheel. If you were to model it all with Newton laws, you would have an equation for the wheel involving its moment of inertia and that torque. – user154997 Jun 20 '17 at 7:29
• There is also one point in need of clarification. You say your wheel is rolling. Not rolling on the ground, isn't it? Otherwise, the centre of the wheel would move, and your kinematic is then wrong. But if you did indeed not mean that, then you should make it clear it can only turn about its centre. – user154997 Jun 20 '17 at 7:33
• Rolling means the point of contact P of the wheel on the ground has a null velocity wrt to the ground. Then the velocity of the centre is related by $\vec{v}_O=\vec{v}_P+\vec{\omega}\times\overrightarrow{PO}$ where $\vec{\omega}$ is the angular speed of the wheel (i.e. $\dot{\theta}\vec{k}$ where $\vec{k}$ is perpendicular to the drawing). Then $\vec{v}_\text{bead}=\vec{v}_O + \dot{r}\vec{u}_r + r\dot{\theta}\vec{u}_\theta$ where $\vec{u}_r$ is along the wire and $\vec{u}_\theta$ is perpendicular to it (cylindrical coordinates). – user154997 Jun 20 '17 at 21:49

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.