Choice of Generalized Coordinates

Question

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:

Answer 1

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 $C_3$.

Answer 2

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

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

Answer 3

$ \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

Answer 4

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*}