Deflating wheel rolling without slipping

Question

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.

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.

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?

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

Answer 1

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.