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One example of a nonholonomic constraint is a disk rolling around in the cartesian plane that is constrained to not be slipping.

These leads to the constraint $dx - a \sin\theta d\phi = 0$ and $dy - a\cos\theta d\phi = 0$ Where $\phi$ is the angle of how far the disk has rotated, $-\theta$ is angle that velocity makes with respect to $x$.

We know that these aren't exact differentials because to put it in form $Mdx + Ndy = 0$, we don't have $\partial M/\partial y = \partial N / \partial x$. But that doesn't mean we can't find some integrating factor to multiply it by and make it an exact differential? So I don't see how we know that is it nonintegrable?

If I take $f(x,\phi)[ dx - a \sin\theta d\phi] = 0$ and try to manipulate it to get something analogous to $\partial M/\partial y = \partial N / \partial x$, then I get a tricky PDE, which I don't know how to solve.

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You can explicitly build an example that shows that these constraints are not path-independent. –  yohBS Feb 18 '12 at 23:08

1 Answer 1

up vote 2 down vote accepted

1) We give a proof by contradiction.

2) Assume there is a time-independent holonomic constraint of the form

$$ f(x,y,\theta,\phi)~=~0.$$

3) Now, by performing a small loop of various sizes, we can (by all the time obeying the rolling condition) make the disc return to the same values of $x$, $y$ and $\theta$, but with an arbitrary value of $\phi$. We conclude that the $f$ constraint cannot depend on $\phi$, i.e.,

$$ f(x,y,\theta)~=~0.$$

4) By performing a small loop, we can return the disc to point in an arbitrary new direction $\theta$. We conclude that the $f$ constraint cannot depend on $\theta$ either, i.e.,

$$ f(x,y)~=~0.$$

5) This constrains where the disc can touch the table, which is absurd. Hence $f$ does not exist.

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