Center of wheel travels the length of circumference in one revolution I asked the same question first on the mathematics forum here.

I was wondering if there is a more mathematical/rigorous way of seeing
that the wheel/circle/its center travels the length of wheel's
circumference in one revolution.
Intuitively, one could cover the wheel/circle with a string the length
of which is exactly equal to its circumference. Then in one revolution
the string would be spread so that we can see the center traveled the
length which is equal to the circle's circumference.

So if there's no slipping, the wheel's center travels the length of wheel's circumference in one round. To show that, I need the information that the center travels with the same speed as every point on the wheel revolves.
So is it indeed that this question is not purely mathematical so one needs the information from physics? And how could I justify this statement?
 A: In the left hand diagram the wheel is rolling along the ground, point $A$ is in contact with the ground and has zero speed.  For small changes in time the green line is being tilted, pivoted around point $A$,  so if point $O$ has a speed of $v$ m/s, then point $B$ has a speed of $2v$ m/s.

In the right hand diagram the wheel is shown from the point of view of an observer travelling with it.  Point $O$ is now stationary, point $A$ moves left at $v$ m/s and  point $B$ moves right at $v$ m/s.
If the observer watches a red spot of paint on the rim it takes time $$t=\frac{2 \pi r}{v}$$ to return to the bottom, the wheel has then made a complete turn.
In this time, as seen from the left hand diagram, the wheel would travel a distance $$vt = \frac{2 \pi rv}{v} = 2 \pi r$$
A: 

*

*Define the wheel moving forward with mode rolling without slipping:
A wheel is called rolling without slipping if the point (X in Figure) of the wheel which touched the ground has always a zero speed relative to the ground.


*The velocity of each point on the wheel $\vec v$  can be decomposed as sum of the velocity of the center $\vec V_c$ and the relative motion of this point to the center $\vec v'$.
$$ \vec v(\vec r) = \vec V_c + \vec v'(\vec r-\vec r_c)$$


*For a circular rotation motion with angular velocity $ \vec \omega(t)$, the instantaneous velocity of a point in the circumference $\vec R$ relative to the center is
$$\vec v'(\vec R-\vec r_c) = \vec \omega \times (\vec R-\vec r_c) $$


*The velocity of the point touched ground (along the x-direction) is always zero:
$$ v_x(t) =  V_c(t) - R \omega(t) =0.$$
where $\vec\omega$ in the $-\hat z$ direction, and $\vec \omega \times R (-\hat y) $ is in the $(-\hat x)$ direction, and assume $\vec V_c$ moves along the $(+\hat x)$ direction.


*Therefore, $ V_c(t) = R \omega(t)$.


*Finally, for one revolution of the circumference:
$$
\int_0^T \omega(t) dt = 2\pi.
$$
and the displacement of the center:
$$
s = \int_0^T V_c(t) dt = \int_0^T R\omega(t) dt = R \int_0^T \omega(t) dt = 2\pi R.
$$
A: Note that what you are trying to prove is only true for planar/linear movement. Therefore, I suppose that this is actually missing from your assumptions. If the wheel moves on a curved surface, the arc length that the center of the wheel moves is different from the arc length that the wheel contact point moves.
I provide a purely geometric proof, which might be preferable if you are mathematically inclined:


*

*The vector (red) connecting the wheel contact point with the center of the wheel is always perpendicular to the supporting plane, because at the contact point the wheel circumference and the plane touch each other (tangentially) and any radius vector is perpendicular to the circumference; since the plane has constant orientation, the said vector has a constant direction

*This vector has also always the same length, namely the radius of the wheel; so it is actually a constant vector in the narrower sense

*The contact point undoubtedly moves on a finite line (blue) by presumption

*Hence, the curve that the wheel center prescribes (green) can be obtained geometrically by translating every point of the line from (3) by the constant vector from (2); this effectively translates the whole line by said vector; but translating a line does not change its length

Consequently, the line that the center of the wheel prescribes has the same length as the line that the contact point prescribes. QED.
This proof does not depend on whether the wheel moves one reveolution or not. Note also, that the only reason why you would be interested in the no-slip condition is if you want to relate the points of the blue contact line 1:1 with the wheel circumference points/rotation angle (or physically, if you are interested in angular momentum). Otherwise the wheel could also be perfectly slippery (not rotate at all).
