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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.

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  • $\begingroup$ For $\alpha$ radians revolution the distance the center takes from it's zero position is $x= R\alpha$.x is the projection of $arc=R\alpha$ on x axis. For one revolution $\alpha=2\pi$ and $x=2\pi R$. $\endgroup$
    – sirous
    Commented Sep 9, 2021 at 11:24
  • $\begingroup$ An approach using physics: Let the wheel of radius $R$ be in pure rolling motion with translational velocity $v$. Angle covered by any point P on the wheel in 1 time period $T=2\pi \Rightarrow$ Angluar velocity of wheel $=\omega=\frac{2\pi}T=\frac vR \Rightarrow$ Distance covered by centre of wheel in time $T=vT=2\pi R$ $\endgroup$
    – RiverX15
    Commented Sep 9, 2021 at 15:31
  • $\begingroup$ @RiverX15 Is it trivial that the center has the same velocity as the point P? $\endgroup$
    – mathslover
    Commented Sep 9, 2021 at 16:08
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    $\begingroup$ @mathslover No it's not trivial. In pure rolling motion, the speed of every point on the wheel remains same and is equal to $\omega R$. I think this question would be handled better on Physics Stack Exchange. $\endgroup$
    – RiverX15
    Commented Sep 9, 2021 at 16:25

4 Answers 4

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What you described is not a rule, it's only the case when you have rolling without slipping.

These two lengths you've described are usually independent on each other.

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Special case

When a Circle radius R ( circumference = $ 2 \pi R $) rolls on a straight line without slipping, a point on its rim traces out a cycloid when the length traversed by the Circle center is also the same$=2 \pi R$.

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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 it's as if 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.

enter image description here

In the right hand diagram the wheel is shown from the point of view of an observer travelling with the wheel. 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 $$vt = \frac{2 \pi rv}{v} = 2 \pi r$$

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  • $\begingroup$ John Hunter, we are saying that time is constant across different frame of references, but according to Einstein time ticks differently in different frame of references, will the logic still work in Einstein’s world of physics. I am not much experienced in physics, question might not make sense. $\endgroup$ Commented Dec 1, 2023 at 10:42
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As @hellofriends pointed out, rotation and translation are generally independent things.

In the special case when a circle in the plane is rolling without slipping on a curve on the plane (like a coin is rolling around another coin on the table), the speed of the point of the circle in which the circle touches the curve is zero, hence the speed of the center of the circle is $\omega r$ where $\omega$ is the instantaneous angular velocity of the circle.


Proof. The motion of the circle is the action of a one-parameter subgroup of the Euclidean group of the plane. The velocity field of its motion is $p\mapsto A(p-o)+v$ where $o$ is the position vector of the center of the circle and $A$ is an antisymmetric linear transformation. If we regard the plane as a subspace of $\mathbb R^3$ then we can define the angular velocity vector $\omega$ as the Hodge dual of $A$ and this means that $\omega$ is perpendicular to the plane, and the speed of a point $p$ on the plane is $\omega\times(p-o)+v$. Suppose that a point $p$ on the circle has velocity $0$, that is, $\omega\times (p-o)+v=0$. Then $v=-\omega\times(p-o)$, so the velocity of $o$ is $v_o=\omega\times (o-o)+v=-\omega\times(p-o)$. Since $\omega$ is perpendicular to $p-o$ and $\|p-o\|=r$ (the radius of the circle), $\|v_o\|=\omega r$.


Without loss of generality, we can always roll the circle with a constant angular speed $\omega$. Since the center of the circle moves in every moment with speed $\omega r$, the path (arc length) it traveled in time $t$ is $\omega r t$. During time $T=\frac{2\pi}{\omega}$ the circle makes one turn, and meanwhile, its center travels $\omega r T=\omega r \frac{2\pi}{\omega} = 2\pi r$ and this is just the circumference of the circle.

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