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For a wheel of radius $R$ on an horizontal plane, the "rolling without slipping condition" is given by $$\Delta s =\Delta l$$ with $\Delta l$ being the distance travelled on the plane. $\Delta l$, because of the constance of $R$, equals the displacement $\Delta s$ of the center of mass. Normal

My question concerns a wheel deflating over time (with a fixed law $R(t)$). How to write down the rolling without slipping condition?

Edit

Citing JoDraX answer, I'm elaborating a little my question

The deflation of the wheel actually only affects the vertical component of the wheel's position, so $$dR^2+R^2d\theta^2=ds^2$$ but $dl=Rd\theta$ and $\dot{l}=R\dot{\theta}$ still, as it would with a wheel of fixed radius.

equivalent

with $l$ the distance travelled along the plane. Now, generalizing a little, does the wheel rolling on an incline of costant slope make any difference? I don't think so, because $dR^2$ , $dl^2$, $ds^2$ are constant with respect to global rotation.

Arbitrary ramp

Now, if the incline has a fixed shape (i.e. skating ramps) $\gamma(x)$, how to write the rolling without slipping condition?

Curved

So far, I have found what follows: $$dl^2=[1+(\frac{d\gamma}{dx})^2]dx^2$$ $$ds^2=dr^2+r^2d\theta^2$$

And $dl=rd\theta$ (hence $\dot{l}=r\dot{\theta}$) holds exactly as for the flat surface. I carried forward my question, because it is in fact a part of a bigger problem, presented here (I offered a bounty on it):

Solid of revolution rolling on an arbitrary incline

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  • $\begingroup$ I don't see why the question has been tagged as "homework-and-exercise". I haven't been able to find it on any book, any exercise list. Isn't that to be discussed as a question about classical mechanics, rather than a question on how to apply it? To me, it seems a question about the boundaries of standard concepts in classical mechanics (rolling and slipping). $\endgroup$ – Riccardo Buscicchio May 4 '17 at 12:24
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Since wheel remains circular as its radius changes, the point of contact is always directly beneath the center of the wheel. This means that the displacement of the center of the wheel in the $x$-direction is equal to $l$. We can write the position of the center of the wheel as $\vec{r} = \langle l, R(t) \rangle$, or $\textrm{d}\vec{r} = \langle \textrm{d}l, \textrm{d}R \rangle$. The deflation of the wheel actually only affects the vertical component of the wheel's position, so $\textrm{d}r^2 = \textrm{d}R^2 + R^2 \textrm{d}\theta$, but $\textrm{d}l = R \textrm{d}\theta$ and $\dot{l} = R \omega$ still, as it would with a wheel of fixed radius.

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  • $\begingroup$ I've edited my question with your suggestion, and I developed it further. Thanks! $\endgroup$ – Riccardo Buscicchio May 5 '17 at 10:06

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