# Choice of Generalized Coordinates

I am working on solving problem 5.24 from the $$3^{\mathrm{rd}}$$ edition of Goldstein's Classical Mechanics:

A wheel rolls down a flat inclined surface that makes an angle $$\alpha$$ with the horizontal. The wheel is constrained so that its plane is always perpendicular to the inclined plane, but it may rotate about the axis normal to the surface. Obtain the solution for the two-dimensional motion of the wheel, using Lagrange’s equations and the method of undetermined multipliers.

I am having trouble choosing the correct generalized coordinates for this problem. My instinct is to choose $$\theta$$, the angle through which the wheel has rotated about its axis, and $$\phi$$, the angle about which the wheel has rotated normally to the surface. This way, the velocity of the center of mass of the wheel can be written down as something like

$$\begin{equation*} \mathbf{v} = a\dot{\theta}\cos{(\phi)}\hat{\boldsymbol{\imath}} + a\dot{\theta}\sin{(\phi)}\hat{\boldsymbol{\jmath}} \end{equation*}$$

where $$\hat{\boldsymbol{\imath}}$$ is parallel to the surface pointing downward and $$\hat{\boldsymbol{\jmath}}$$ is parallel to the surface and orthogonal to $$\hat{\boldsymbol{\imath}}$$.

The rotational kinetic energies due to $$\dot{\theta\,}$$ and $$\dot{\phi\,}$$ are similarly easy to write down.

My problem is in obtaining an expression for the potential energy. It depends on the distance the center of mass of the wheel has traveled down the ramp, but since the direction of motion of the wheel is (in general) always changing, it seems like it is not possible to obtain an expression for this distance without actually solving the problem. Are there better generalized coordinates to use here? Am I misunderstanding the problem? For example, if the wheel somehow never changed direction and rolled straight down the ramp and, independently, the wheel was able to also rotate normally to the surface, the problem would be easy, but that doesn't seem to me to be how the problem is worded.

Edit: Here is a drawing of the angles I mean:

• I am not sure, I understand the difference between $\theta$ and $\phi$. The wheel is rolling down an inclined plane with the constrained to not "tilt". I thought that is what is meant by "it may rotate about the axis normal to the surface". If I understood the problem correctly, I would just use one coordinate of the center of mass (the other is given by the indlined plane and the radius of the wheel) and one angle that describes the rotation of the wheel around its axis. What do I miss? Commented Jul 22 at 9:23
• @RefikMansuroglu From my understanding, the wheel can rotate about an axis parallel to the surface, i.e., the axis of the wheel, as well as about an axis normal to the surface. I have tried to illustrate that in my edit. Commented Jul 22 at 9:36
• I understand! This constraint seems very artificial, but ok :D It seems that you just need to find the vertical coordinate of your center of mass in terms of $\theta$ and $\phi$. You already did something similar with the velocity. I will give it a try and write down my thoughts in more detail. Commented Jul 22 at 10:04
• @RefikMansuroglu I was thinking the same thing, but I haven't found a way to write down the vertical component in terms of those angles. Of course, it is the time integral of the $\hat{\boldsymbol{\imath}}$ component of velocity, but that isn't much help. I hope you have some ideas! ;) Commented Jul 22 at 10:05
• What is the $a$ in the velocity expression? Commented Jul 22 at 12:38

If you want to use Lagrange multipliers as the exercise hints to, you can do the following: Start with an overdetermined parameterization of the system, say using $$\left( \theta, \dot \theta, \phi, \dot \phi, s_1, \dot s_1, s_2, \dot s_2 \right)$$, with the angles $$\theta, \phi$$ that you suggest and also a two-dimensional projection of the center of mass of the wheel onto the plane (or equivalently the touching point of the wheel and the plane) given by the two spatial coordinates $$s_1, s_2$$. Let $$s_2$$ parameterize the direction perpendicular to the inclination (this should be $$\bf j$$ direction in your notation). The overdetermination will be dealt with with Lagrange multipliers later.

Let us start by writing down the kinetic energy in these coordinates $$T = \frac{1}{2} I_{\theta} \dot\theta^2 + \frac{1}{2} I_{\phi} \dot\phi^2 + \frac{1}{2} (\dot s_1^2 + \dot s_2^2 ).$$

With some moments of inertia $$I_{\theta/\phi}$$, which can be easily looked up. I also set the mass of the wheel to one. Next, we write down the potential energy that will only depend on $$s_1$$ $$V = -g s_1 \sin(\alpha).$$ $$g$$ is the gravitatonal acceleration and $$\sin(\alpha)$$ singles out the vertical component. Finally, we write down the constraints that relate $$s_1, s_2$$ and the angles $$\theta, \phi$$. This, I can copy from your post as you correctly found the center of mass velocity dependent on $$\theta, \phi$$. In my notation, we have $$\dot s_1 = a \dot \theta \cos(\phi) \qquad \dot s_2 = a \dot \theta \sin(\phi).$$

Now, you can just use standard Lagrangian methods to solve the system. Let us define the Lagrange multipliers $$\lambda_1, \lambda_2$$ and write down the Lagrangian $$\mathcal{L} = T - V + \lambda_1 (\dot s_1 - a \dot \theta \cos(\phi)) + \lambda_2 (\dot s_2 - a \dot \theta \sin(\phi)).$$

Next, write down the Euler-Lagrange equations (please double check) \begin{align} (\theta):& \qquad I_\theta \ddot \theta - a \left( \dot \lambda_1 \cos(\phi) + \dot \lambda_2 \sin(\phi) - \lambda_1 \dot \phi \sin(\phi) + \lambda_2 \dot \phi \cos(\phi) \right) = 0 \\ (\phi):& \qquad I_\phi \ddot \phi + a \dot \theta \left( - \lambda_1 \sin(\phi) + \lambda_2 \cos(\phi) \right) = 0 \\ (s_1):& \qquad \ddot s_1 + \dot \lambda_1 - g \sin(\alpha) = 0 \\ (s_2):& \qquad \ddot s_2 + \dot \lambda_2 = 0 \\ (\lambda_1):& \qquad \dot s_1 = a \dot \theta \cos(\phi) \\ (\lambda_2):& \qquad \dot s_2 = a \dot \theta \sin(\phi). \end{align}

Maybe we can find a clever way to simplify these equations. First of all, we would like to get rid of the undetermined multipliers. From $$(s_1)$$ and $$(s_2)$$, we can write them as $$\lambda_1 = C_1 - \dot s_1 + g \sin(\alpha)t \\ \lambda_2 = C_2 - \dot s_2,$$ with constants $$C_1$$ and $$C_2$$ to be determined by initial conditions. This expression can be further simplified using $$(\lambda_1)$$ and $$(\lambda_2)$$ to $$\lambda_1 = C_1 - a \dot \theta \cos(\phi) + g \sin(\alpha)t \\ \lambda_2 = C_2 - a \dot \theta \sin(\phi).$$ Inserting into the remaining equations should give you two more equations of motion. If I did no mistake those should be $$(I_\theta + a^2) \ddot \theta \dot \theta - I_\phi \ddot \phi \dot \phi - ag \sin(\alpha) \dot \theta \cos(\phi) = 0 \\ (I_\theta - a^2) \dot \theta + a \cos(\phi) (C_1 + g \sin(\alpha)t) - a \sin(\phi) C_2 = C_3,$$ with another constant $$C_3$$.

• I don't think, it is wrong. Plugging in your angular velocity yields the same rotational energy as in my Lagrangian, if you use $I_x = I_y$ from the symmetry of the circle. However, I find your notation a little misleading, since it suggests that the rotation axis of the $\phi$ rotation changes over time, while it is constrained to be perpendicular to the plane by definition of the problem. I think, it is more instructive to work in a tilted coordinate system that captures the 2D-coordinates of the center of mass and the two angles as done in my answer. Commented Jul 23 at 19:27
• I can't thank you enough for your thorough and enlightening answer! I only have two questions--in your second equation, should $V = -gs_{1}\sin{(\alpha)}$ since the potential energy should decrease as the wheel rolls down the ramp? And in your third equation, should $\dot{s}_{1} = a\dot{\theta}\cos{(\phi)}$ (and so $\sin{(\phi)}$ for $\dot{s}_{2}$)? Commented Jul 24 at 7:55
• No problem! I think both your questions are conventional matters and can be changed by shifting coordinates accordingly. I wanted to follow your convention and I think you are right, I corrected my answer and hope that I did not mess up any signs along the way. Commented Jul 24 at 8:31
• Check out this wikipedia page. I think, this is a more general introduction to constrained mechanics. Commented Jul 24 at 8:59
• It's a good introduction, but it assumes that the constraints are of the form $f(\mathbf{r},t)$ rather than $f(\mathbf{r},\dot{\mathbf{r}},t)$. Commented Jul 24 at 9:03

You are correct that you are going to need some positional generalized coordinates also.

I would suggest the distance $$s$$ along the ramp from some horizontal reference plane, and the distance $$t$$ for the side-to-side displacement.

The height of the disk above the ground would then be $$y = s \sin \alpha$$

So the velocity of the center does not depend on it's spin, but on the rate of change of the positional coordinates.

## Position Vector

$${\bf r}(s,t)={\rm R}_{Z}(-\alpha)\begin{pmatrix}-s\\ a\\ t \end{pmatrix}=\begin{pmatrix}a\sin\alpha-s\cos\alpha\\ a\cos\alpha+s\sin\alpha\\ t \end{pmatrix}$$

where

$$\small {\rm R}_{Z}(-\alpha)=\begin{bmatrix}\cos\alpha & \sin\alpha\\ \text{-}\sin\alpha & \cos\alpha\\ & & 1 \end{bmatrix}$$ is the rotation vector to align coordinates to the ramp.

## Velocity Vector

The velocity vector is the direct time derivative of the position vector

$${\bf v}(s,t,\dot{s},\dot{t})=\begin{pmatrix}-\dot{s}\cos\alpha\\ \dot{s}\sin\alpha\\ \dot{t} \end{pmatrix}$$

## Orientation

The rotation matrix for the disk is a result of one fixed rotation (the ramp) and two varying rotations

$${\rm R}(\phi,\theta)={\rm R}_{Z}(-\alpha)\,{\rm R}_{Y}(\phi)\,{\rm R}_{Z}(\theta)$$

where

\small \begin{aligned}{\rm R}_{Y}(\phi) & =\begin{bmatrix}\cos\phi & & \sin\phi\\ & 1\\ \text{-}\sin\phi & & \cos\phi \end{bmatrix} & {\rm R}_{Z}(\theta) & =\begin{bmatrix}\cos\theta & \mbox{-}\sin\theta\\ \sin\theta & \cos\theta\\ & & 1 \end{bmatrix}\end{aligned}

## Rotational Velocity

Using the sequence of rotation one can derive the rotational velocity vector

$$\boldsymbol{\omega}(\phi,\theta,\dot{\phi},\dot{\theta})={\rm R}_{Z}(-\alpha)\left(\boldsymbol{\hat{j}}\dot{\phi}+{\rm R}_{Y}(\phi)\,\boldsymbol{\hat{k}}\dot{\theta}\right)$$

which yields

$$\boldsymbol{\omega}=\begin{pmatrix}\dot{\phi}\sin\alpha+\dot{\theta}\cos\alpha\sin\phi\\ \dot{\phi}\cos\alpha-\dot{\theta}\sin\alpha\sin\phi\\ \dot{\theta}\cos\phi \end{pmatrix}$$

# Appendix

If the orientation can be defined by a sequence of 1DOF rotations

$${\rm R}(q_{1},q_{2},q_{3},\ldots q_{n})={\rm rot}(\boldsymbol{\hat{u}}_{1},q_{1}){\rm rot}(\boldsymbol{\hat{u}}_{2},q_{2}){\rm rot}(\boldsymbol{\hat{u}}_{3},q_{3})\,\ldots$$

then angular velocity is found using the following recursive formula

$$\boldsymbol{\omega}(q_{1,}q_{2},\ldots\dot{q}_{1},\dot{q}_{2},\ldots)=\boldsymbol{\hat{u}}_{1}\dot{q}_{1}+{\rm rot}(\boldsymbol{\hat{u}}_{1},q_{1})\left(\boldsymbol{\hat{u}}_{2}\dot{q}_{2}+{\rm rot}(\boldsymbol{\hat{u}}_{2},q_{2})\left(\boldsymbol{\hat{u}}_{3}\dot{q}_{3}+\ldots\right)\right)$$

• That makes sense! But using $s$ and $t$, it seems like writing down the kinetic energy of the system (in particular the rotational kinetic energies of the wheel about the two axes the wheel can rotate) might be difficult. Commented Jul 22 at 12:47
• It is a feature of generalized coordinates to treat positional and velocity coordinates as independent. Commented Jul 22 at 13:12
• I agree, but I'm having trouble seeing how that clarifies things. Commented Jul 22 at 13:36
• See my update above. You had defined the velocity in terms of the angular speeds trying to bake-in the no-slip constraint. But if you consider the contact point position as part of the gen. coordinates then the velocity is only the time derivative of positions and the angles play no role here. A separate no-slip condition needs to be considered for the solution, and it will be implemented in the kinetic energy and not in the potential energy where you had the difficulty. Commented Jul 22 at 13:57
• I am not familiar in detail with this specific textbook. But it might be that $\phi$ is fixed, or that when it comes to velocities $\dot s$ and $\dot t$ are not independent as they are projections of the speed of the disk center which is constrained to be tangent the the path the contact point makes. These kind of problems and complex in general, and the author might expect some simplifications that are discussed in the book that I am not familiar with at this point. Commented Jul 22 at 14:13

$$\def \b {\mathbf}$$

The Rotation Matrix between the wheel coordinate system and inertial system is:

\begin{align*} &\b R=\b R_y(\phi)\,\b R_z(\theta) \end{align*} form here you obtain the angular velocity $$~\b\omega~$$ given in wheel system \begin{align*} &\b \omega= \left[ \begin {array}{c} \sin \left( \theta \right) \dot\phi \\\cos \left( \theta \right) \dot\phi \\\dot\theta \end {array} \right] \end{align*}

The rolling conditions

\begin{align*} &\b v_r=\begin{bmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \\ \end{bmatrix}+\b R\,\b\omega\times\left( \b R_y(\phi)\,\begin{bmatrix} 0 \\ -a \\ 0 \\ \end{bmatrix} \right)=\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\\ &\dot{x}+\cos(\phi)\,a\,\dot{\theta}=0\\ &\dot{y}=0\\ &\dot{z}-\sin(\phi)\,a\,\dot{\theta}=0 \end{align*}

where $$~\dot x~,\dot y~,\dot z~$$ are the velocities given in inertial system

the position vector of the wheel center given in inertial system is:

\begin{align*} &\b P=\begin{bmatrix} x \\ 0 \\ z \\ \end{bmatrix}+\b R_y(\phi)\,\b R_z(\alpha)\,\begin{bmatrix} 0 \\ -a \\ 0 \\ \end{bmatrix} =\begin{bmatrix} x \\ 0 \\ z \\ \end{bmatrix}+\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \alpha \right) a\\-\cos \left( \alpha \right) a \\-\sin \left( \phi \right) \sin \left( \alpha \right) a\end {array} \right] \end{align*} thus the potential energy $$~U=-m\,g\,\left(\b P\right)_y=-m\,g\,a\,\cos(\alpha)~$$

Notice

the contact point $$~p~$$ must "remain" on the plane ($$~\b R_z(\theta)\mapsto \b R_z(0)~$$)

equations of motion

with \begin{align*} &\b q=\begin{bmatrix} x \\ y \\ z \\ \phi \\ \theta \\ \end{bmatrix} \end{align*}

and the nonholonomic constraint equations $$~\b v_r=0~$$

you obtain the kinetic energy

$$T=\frac m2\,\b v\cdot\b v+\frac 12 \b\omega ^T\,\b I\,\b \omega+\b\lambda\,\cdot \b v_r$$

where $$~\b v=[~\dot x~,\dot y~,\dot z~]^T~$$ and $$~\b\lambda= [~\lambda_1~,\lambda_2~\lambda_3]^T~$$ Lagrange multipliers

I wanted to post my own answer, inspired by that of Refik Mansuroglu.

Let $$\theta$$ be the angle down the slope the wheel has rolled and let $$\phi$$ be the counterclockwise angle through which the wheel has rotated about the normal to the inclined surface. Further, let $$s_{1}$$ be the distance down the surface the wheel has rolled and $$s_{2}$$ be the distance on the surface the wheel has rolled normally to $$s_{1}$$. Then the kinetic energy of the system can be written

$$\begin{equation*} T = \frac{1}{4}ma^{2}\dot{\theta}^{2} + \frac{1}{8}ma^{2}\dot{\phi}^{2} + \frac{1}{2}m\dot{s}_{1}^{2} + \frac{1}{2}m\dot{s}_{2}^{2}. \end{equation*}$$

Note that in the preceding calculation we have approximated the wheel by a thin, uniform, solid disk of radius $$a$$. The potential energy can be written as

$$\begin{equation*} V = -mgs_{1}\sin{(\alpha)}. \end{equation*}$$

We immediately have two constraint equations that result from the assumption that the wheel rolls without slipping:

$$\begin{equation*} f_{1} = a\dot{\theta}\cos{(\phi)} - \dot{s}_{1} = 0 \end{equation*}$$

and

$$\begin{equation*} f_{2} = a\dot{\theta}\sin{(\phi)} - \dot{s}_{2} = 0. \end{equation*}$$

The Lagrangian for the system can be written

$$\begin{equation*} L = \frac{1}{4}ma^{2}\dot{\theta}^{2} + \frac{1}{8}ma^{2}\dot{\phi}^{2} + \frac{1}{2}m\dot{s}_{1}^{2} + \frac{1}{2}m\dot{s}_{2}^{2} + mgs_{1}\sin{(\alpha)} \end{equation*}$$

implying four Lagrange equations:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{\theta}}\right)-\frac{\partial L}{\partial \theta} + \mu_{1}\frac{\partial f_{1}}{\partial\dot{\theta}} + \mu_{2}\frac{\partial f_{2}}{\partial \dot{\theta}} &= \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{2}ma^{2}\dot{\theta}\right) + \mu_{1}a\cos{(\phi)} + \mu_{2}a\sin{(\phi)}\\ &= \frac{1}{2}ma^{2}\ddot{\theta} + \mu_{1}a\cos{(\phi)} + \mu_{2}a\sin{(\phi)}\\ &= 0 \end{align*}

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\phi}}\right)-\frac{\partial L}{\partial \phi} + \mu_{1}\frac{\partial f_{1}}{\partial \dot{\phi}} + \mu_{2}\frac{\partial f_{2}}{\partial \dot{\phi}} &= \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{4}ma^{2}\dot{\phi}\right)\\ &= \frac{1}{4}ma^{2}\ddot{\phi}\\ &= 0 \end{align*}

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{s}_{1}}\right)-\frac{\partial L}{\partial s_{1}} + \mu_{1}\frac{\partial f_{1}}{\partial \dot{s}_{1}} + \mu_{2}\frac{\partial f_{2}}{\partial \dot{s}_{2}} &= \frac{\mathrm{d}}{\mathrm{d}t}\left(m\dot{s}_{1}\right) - mg\sin{(\alpha)} - \mu_{1}\\ &= m\ddot{s}_{1} - mg\sin{(\alpha)} - \mu_{1}\\ &= 0 \end{align*}

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{s}_{2}}\right)-\frac{\partial L}{\partial s_{2}} + \mu_{1}\frac{\partial f_{1}}{\partial \dot{s}_{2}} + \mu_{2}\frac{\partial f_{2}}{\partial \dot{s}_{2}} &= \frac{\mathrm{d}}{\mathrm{d}t}\left(m\dot{s}_{2}\right) - \mu_{2}\\ &= m\ddot{s}_{2} - \mu_{2}\\ &= 0. \end{align*}

We immediately have that $$\ddot{\phi} = 0$$, so that $$\phi(t) = \phi_{0} + \dot{\phi}_{0}t$$. Also, we have that

$$\begin{equation*} \dot{s}_{1} = a\dot{\theta}\cos{(\phi)} = a\dot{\theta}\cos{\left(\phi_{0} + \dot{\phi}_{0}t\right)} \end{equation*}$$

so that

$$\begin{equation*} \ddot{s}_{1} = a\ddot{\theta}\cos{\left(\phi_{0} + \dot{\phi}_{0}t\right)} - a\dot{\theta}\dot{\phi}_{0}\sin{\left(\phi_{0} + \dot{\phi}_{0}t\right)}. \end{equation*}$$

Similarly,

$$\begin{equation*} \dot{s}_{2} = a\dot{\theta}\sin{\left(\phi_{0}+\dot{\phi}_{0}t\right)} \end{equation*}$$

so

$$\begin{equation*} \ddot{s}_{2} = a\ddot{\theta}\sin{\left(\phi_{0}+\dot{\phi}_{0}t\right)} + a\dot{\theta}\dot{\phi}_{0}\cos{\left(\phi_{0}+\dot{\phi}_{0}t\right)}. \end{equation*}$$

Using the equation for $$s_{2}$$, we obtain

$$\begin{equation*} \mu_{2} = m\ddot{s}_{2} = ma\ddot{\theta}\sin{\left(\phi_{0}+\dot{\phi}_{0}t\right)} + ma\dot{\theta}\dot{\phi}_{0}\cos{\left(\phi_{0}+\dot{\phi}_{0}t\right)} \end{equation*}$$

and using the equation for $$s_{1}$$ we obtain

$$\begin{equation*} \mu_{1} = m\ddot{s}_{1} - mg\sin{(\alpha)} = ma\ddot{\theta}\cos{\left(\phi_{0} + \dot{\phi}_{0}t\right)} - ma\dot{\theta}\dot{\phi}_{0}\sin{\left(\phi_{0} + \dot{\phi}_{0}t\right)} - mg\sin{(\alpha)}. \end{equation*}$$

Therefore, we have the equation for $$\theta$$:

\begin{align*} &\frac{1}{2}ma^{2}\ddot{\theta} + \left[ ma\ddot{\theta}\cos{\left(\phi_{0} + \dot{\phi}_{0}t\right)} - ma\dot{\theta}\dot{\phi}_{0}\sin{\left(\phi_{0} + \dot{\phi}_{0}t\right)} - mg\sin{(\alpha)}\right]a\cos{(\phi_{0}+\dot{\phi}_{0}t)}\cdots\\ &\hspace{1pc}\cdots+ \left[ma\ddot{\theta}\sin{\left(\phi_{0}+\dot{\phi}_{0}t\right)} + ma\dot{\theta}\dot{\phi}_{0}\cos{\left(\phi_{0}+\dot{\phi}_{0}t\right)}\right]a\sin{(\phi_{0}+\dot{\phi}_{0}t)}\\ &= 0 \end{align*}

or, simplifying,

$$\begin{equation*} \ddot{\theta} - \frac{2g}{3a}\sin{(\alpha)}\cos{(\phi_{0}+\dot{\phi}_{0}t)} = 0. \end{equation*}$$

This has solution

$$\begin{equation*} \theta(t) = c_{1} + c_{2}t - \frac{2g}{3a\dot{\phi}_{0}^{2}}\sin{(\alpha)}\cos{(\phi_{0}+\dot{\phi}_{0}t)}. \end{equation*}$$

Taking $$\dot{\theta}(0) = \dot{\theta}_{0}$$ and $$\theta(0) = \theta_{0}$$, we obtain

$$\begin{equation*} c_{1} = \theta_{0} + \frac{2g\sin{(\alpha)}\cos{(\phi_{0})}}{3a\dot{\phi}_{0}^{2}}\hspace{1pc}\mbox{ and }\hspace{1pc}c_{2} = \dot{\theta}_{0} - \frac{2g\sin{(\alpha)}\sin{(\phi_{0})}}{3a\dot{\phi}_{0}}. \end{equation*}$$

Therefore

$$\begin{equation*} \theta(t) = \theta_{0} + \frac{2g\sin{(\alpha)}\cos{(\phi_{0})}}{3a\dot{\phi}_{0}^{2}} + \left[\dot{\theta}_{0} - \frac{2g\sin{(\alpha)}\sin{(\phi_{0})}}{3a\dot{\phi}_{0}}\right]t - \frac{2g}{3a\dot{\phi}_{0}^{2}}\sin{(\alpha)}\cos{(\phi_{0}+\dot{\phi}_{0}t)}. \end{equation*}$$

Finally, we can write that

$$\begin{equation*} \dot{s}_{1} = a\dot{\theta}\cos{(\phi)} = a\left[\dot{\theta}_{0} - \frac{2g\sin{(\alpha)}\sin{(\phi_{0})}}{3a\dot{\phi}_{0}} + \frac{2g}{3a\dot{\phi}_{0}}\sin{(\alpha)}\sin{(\phi_{0}+\dot{\phi}_{0}t)}\right]\cos{\left(\phi_{0}+\dot{\phi}_{0}t\right)}. \end{equation*}$$

Taking $$s_{1}(0) = 0$$, we obtain

$$\begin{equation*} s_{1}(t) = \frac{2a\dot{\theta}_{0}}{\dot{\phi}_{0}}\cos{\left(\phi_{0} + \frac{1}{2}\dot{\phi}_{0}t\right)}\sin{\left(\frac{1}{2}\dot{\phi}_{0}t\right)} + \frac{4g\sin{(\alpha)}}{3\dot{\phi}_{0}^{2}}\cos^{2}{\left(\phi_{0} + \frac{1}{2}\dot{\phi}_{0}t\right)}\sin^{2}{\left(\frac{1}{2}\dot{\phi}_{0}t\right)}. \end{equation*}$$

The solution for $$s_{2}$$ is also readily obtained. It is worth noting that these hold only for $$\dot{\phi}_{0} \neq 0$$, but they are true in the limiting case as $$\dot{\phi}_{0}\to 0$$:

$$\begin{equation*} s_{1}(t) \to a\dot{\theta}_{0}\cos{(\phi_{0})}t + \frac{1}{3}g\sin{(\alpha)}\cos^{2}{(\phi_{0})}t^{2}. \end{equation*}$$

• I think your approach is reasonable. However, I do not agree with your Euler-Lagrange equations. Since the constraints depend on velocities, the minimal action principle will give you a term $-\frac{d}{dt} (\mu_i \frac{\partial}{\partial \dot \theta} f_i)$ and similar with $\dot \phi$. You should look up Variational calculus, which is the underlying framework for the derivation of the Euler-Lagrange equations. This term is important, as it makes the dynamics of the $\phi$ variable very non-trivial. Commented Jul 24 at 12:10
• @RefikMansuroglu Yes, I will review the calculus of variations! In the mean time, is there a physically intuitive reason why the motion in $\phi$ should be non-trivial? Commented Jul 24 at 14:30
• Well, I guess the unambiguous answer is because the Euler-Lagrange equations say so. But if you want an intuitive explanation you should think about the constraint force ensuring that the wheel stays perpendicular to the plane. The constraint force is, however, dependent on $\phi$ itself, so it will influence the acceleration of the wheel depending on the value of $\phi$. This rules out a constant zero acceleration $\ddot \phi = 0$ as you have it. Commented Jul 24 at 14:56