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I am struggling to understand how to determine whether the constraints are nonholonomic.

For an example:

"Consider a disk which rolls without slipping across the horizontal plane, what is the best generalized coordinates that may be used, note that the plane of the disk remains vertical, and free to rotate about the vertical axis. What is the differential equation that describe the rolling constraints, then check if this equation can be integrated or not. Finally check that the constraint holonomic or not."

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We find the constraints to be: \begin{equation}dx\cos(\theta) + dy\sin(\phi) = Rd\theta \end{equation} and \begin{equation} \tan(\phi) =\frac{dy}{dx}\end{equation} then they say "its obvious that these equations are not integrable".

Question 1: why is it obvious that these equations are not integrable? is it determined from just looking at these equations?

One person said that because we can reach the same point with different values of $\theta$ by taking different paths means that finding the equation of constraint $f(x,y, \theta, \phi) = 0$ is not possible

Question 2: how does $f(x,y, \theta, \phi) \neq 0$ relate to the fact that the equations are not integrable(or taking different paths and end up having different values of $\phi$)

enter image description here

Question 3: I made my own system up: "suppose the small disk is constrained to roll on the larger disk without slipping and held onto the outer rim by a link of length a. So as were rotate the link around, the smaller disk rolls along the larger one"

By using the idea of $\phi$ would be different if we were to take different paths to the same point p, the constraint is \begin{equation}hd\phi=ad\theta \end{equation} where the constraint is non holonomic as we can get to point p by revolving the small disk backwards(eg: path1) or going the long way round(eg: path2)?

enter image description here

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  • $\begingroup$ Shouldn't your constriant in the third question be $hd\phi=(a-h)d\theta?$ $\endgroup$
    – Kksen
    Aug 29, 2021 at 7:25
  • $\begingroup$ @Kksen yes you are right, I made a mistake $\endgroup$
    – Reuben
    Aug 29, 2021 at 7:28
  • $\begingroup$ I think that the constraint equation is \begin{equation}hd\phi=(a+h)d\theta \end{equation} this is holonomic constraint $\endgroup$
    – Eli
    Aug 29, 2021 at 11:00

2 Answers 2

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$\mathbf{Q1}$
There are 2 kinematic constraints for this system, using your notation, $$dx=R\sin\phi\, d\theta,\tag{1}\label{1}$$ $$dy=-R\cos\phi\,d\theta,\tag{2}\label{2}$$ which become, after divided by $dt$ on both sides, $$\dot x=R\sin\phi\;\dot\theta,$$ $$\dot y=R\cos\phi\;\dot\theta,$$ and you can use $v=R\,\dot\theta$ to write the constraint equations more compactly. The reason Eq.\eqref{1},\eqref{2} are called nonholonomic constraints is that both of them cannot be reduced to a definite functional form $$F(x,y,\phi,\theta)=0\tag{3}\label{3}$$ to which we refer as geometric constraint. The distinguishing property of Eq.\eqref{3} is, if it exists, that it imposes restriction on the possible values of the generalized coordinates $x,y,\phi$ and $\theta$. Mathematically speaking, Eq.\eqref{3}, if it exists, represents a 3-dimensional surface in the 4-dimensional configuration space of the rolling disk just like $x^2+y^2+z^2=1$ represents a 2-sphere in the 3-dimensional Euclidean space, which we know, prevent arbitrary values of $x,y$ and $z$ to be taken. This means although kinematic constraints \eqref{1} and \eqref{2} exist, they impose no restriction on the values of the 4 generalized coordinates and we can make the disk to move from a given starting position to any other position, i.e. you can even bring the disk from $(x_0,y_0,\phi_0,\theta_0)$ to $(x_0,y_0,\phi_0,\theta_1)$ where $\theta_0\neq\theta_1$ and this clearly violates Eq.\eqref{3} since it assigns a (unique) $\theta$ to a certain set $\{x,y,\phi\}$.

$\mathbf{Q2}$
It is not that $F(x.y,\phi,\theta)\neq0$ is related to the nonholonomicity of a system, it is the non-existence of Eq.\eqref{3} from the kinematic constraints prescribed by the nature of the system.

$\mathbf{Q3}$
@Eli has shown how this is a holonomic constraint mathematically. However, there might be concerns about the periodicity of the generalized coordinates and the configuration. In this question, we are only looking at the constraints without considering the dynamics, so gravity is ignored as usual. Let call the smaller disk of radius $r$ as $A$ and the bigger disk of radius $R$ as $B$. So you might wonder: Is the configuration of the system is the same after $A$ has rolled around $B$ a full revolution ($\theta$ from $0$ to $2\pi$)?

The answer is no. The reason lies in the fact that $\phi$ is not an integral multiple of $2\pi$. That means picking a representative point in $A$ and it has different positions before and after the revolution, therefore the configuration of the whole system is different. But there are 2 cases of interest: whether the ratio $r/R$ is a rational or an irrational number. Suppose $r/R=h$, then after one revolution around $B$, $\phi$ becomes $\frac{2\pi}{h}$. In order to have the same configuration as the original one at the very beginning, we require that when $\theta$ becomes $a(2\pi)$, where $a$ is an integer, $\phi$ must be $b(2\pi)$, where $b$ is an integer as well so that we can make the identification for $\theta=0$ to $\theta=a(2\pi)$ as well as $\phi=0$ to $\phi=b(2\pi)$. If $h$ is a rational number, namely $\frac{c}{d}$ where $c\, \&\, d$ are integers, $\phi$ becomes $\frac{2\pi d}{c}$ after one revolution around $B$ and that means $A$ just needs to rotate around $B$ for $c$ revolutions to have $\phi$ as an integral multiple of $2\pi$. If $h$ is an irrational number, however, the configuration of the whole system will never be the same, no matter how many revolutions $A$ rotates around $B$. There is no periodicity for the $2$ generalized coordinates in this case.

$\mathbf{Update}$
I have recently read a book that gives a rigorous (in the sense of a physicist) mathematical proof that the constraints \eqref{1} and \eqref{2} are not integrable, that is Analytical Mechanics by Nivaldo A.Lemos. The proof utilizes the Frobenius Integrability Theorem and requires the reader to be familiar with differential forms as prerequisite. However, most of the essential materials are included in Appendix B of his book. If you are mathematically oriented, it might be something you are after.

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  • $\begingroup$ thank you for you reply, for Q3, I wanted to know if the constraint $hdϕ=(a-h)dθ$ on the system was nonholonomic $\endgroup$
    – Reuben
    Aug 29, 2021 at 8:48
  • $\begingroup$ Interesting question. I will update the answer when I am free. The short answer is it is a holonomic constraint. $\endgroup$
    – Kksen
    Aug 29, 2021 at 14:38
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your constraint equations should be:

$$\frac{dy}{dx}=-\cot(\phi)\\ dx\,\sin(\phi)-dy\,\cos(\phi)=R\,d\theta$$

solving those equations for $dx~$ and $~dy$

(assuming that $~\phi~$ and $~\theta$ are the generalized coordinates)

$$dx=R\,\sin(\phi)\,d\theta \\ dy=-R\,\cos(\phi)\,R\,d\theta $$

those are the right equations that @KKsen wrote
if you integrate the constraint equations you obtain

$$x=R\,\int \sin(\phi)\,d\theta\\ y=-R\,\int \cos(\phi)\,d\theta$$

for holonomic constraint equations the results should be $x=x(\phi,\theta)~$ and $y=y(\phi,\theta)$ but this not the case. those non holonomic constraint equations are not integrable .

assume that $~\phi=\text{constant}=\phi_0~$, the integration of the constraint equations give you

$$x=R\, \sin(\phi_0)\,\theta\\ y=-R\, \cos(\phi_0)\,\theta$$

those are holonomic constraint equations $~x=x(\theta)~,y=y(\theta)$

take this constraint equation

$$h\,d\phi=(h+a)\,d\theta\\ h\,\int d\phi=(h+a)\,\int d\theta\\ \phi=\frac{h+a}{h}\theta$$

this is holonomic constraint equation $~\phi=\phi(\theta)$

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  • $\begingroup$ Shouldn't the constraint equation be $h\,d\phi=(h-a)\,d\theta$, instead of $(h+a)$? and why you tag me, instead of the OP?I am confused. $\endgroup$
    – Kksen
    Aug 30, 2021 at 4:12
  • $\begingroup$ I didn’t tag you I wrote that your non holonomic constraint equations are the right one. $\endgroup$
    – Eli
    Aug 30, 2021 at 6:15
  • $\begingroup$ Ohh, okay, thanks for doing that. Sorry, I didn't understand that. $\endgroup$
    – Kksen
    Aug 30, 2021 at 6:19
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    $\begingroup$ I made it clearly, thank you $\endgroup$
    – Eli
    Aug 30, 2021 at 6:24

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