The Velocities of the Contact Points of Two Rolling Curves are Equal at the Instant of Contact

Question

Consider two 2-Dimensional rigid bodies surrounded by two planar smooth curves. Suppose that the two bodies are in the same plane and in contact with each other such that they are rolling with respect to each other. To demonstrate the meaning of rolling, suppose that the points $C_1$ and $C_2$ in the figure below are the contact points at the present time and the points $B_1$ and $B_2$ are the points that will be in contact after some time. Then the rolling condition is defined as $s_1=s_2$. So our definition of rolling is

Definition. Two smooth curves are said to be rolling with respect to each other if the length of their contacted portions during a time interval is equal.

Now, the main question is to prove the following theorem

Theorem. Two smooth curves are rolling with respect to each other if and only if the velocity vectors of the contact points are equal to each other at the instant of contact.

So the theorem is expressing an equivalent condition for rolling. Simple versions of the theorem are rolling of a circle over a straight line, inclined line, another circle, ellipse or parabola (See the animation below). Taking a look at the links will help you to visualize better. Without loss of generality you can assume that curve $2$ is still and curve $1$ is rolling on it.

This is a well-known theorem that is taught to mechanical engineering students in a Machine Dynamics course without a proof! I was not able to find the proof anywhere in the engineering, physical or mathematical texts.

I would be happy to see a full detailed answer but I do not expect one. Also, I have not defined the problem rigorously so if you see flaws you can modify it as you wish. But I think you can imagine what I mean by the examples and links I provided. Any guidance, help or hint is welcome and appreciated.

This animation is made by J. M. and is included for better visualization.

Answer 1

Personally I think the theorem is intuitively obvious and requires no proof. But if you really want one. We prove the theorem in one direction and the other one is left as an exercise. So let us prove that the arc-length definition of rolling will result in the velocity definition.

Step 1: Change Coordinates

Since the statement is that the relative velocities are 0, what we can do is to change to a moving coordinate system so that body #2 remains stationary, with the point of contact (the one at which we will take the derivative) fixed at the origin, and that the tangent line to body #2 is horizontal at the origin (so it is the $x$ axis).

Step 2: $y$ or Normal Velocity (Optional)

It is clear that the $y$ velocity is zero, at least if we assume that the position of the "point" is required to be a $C^1$ function of time, since then the $y$ coordinate of the position of the point attains a local minimum there.

If you allow infinite rotation speeds (so that the position need not be a differentiable function of time), then your theorem doesn't even make sense. So the differentiability assumption is, I think, implicit in your theorem. However, if you don't like this reasoning the next step will take care of the whole velocity vector.

Step 3: $x$ or Tangential Velocity

We will approach this by assuming that the points of contact travels along body #2 with constant speed (this assumption is not essential since it can be gotten rid of by applying the chain rule), and tracking the motion of the point on body #1 which contacts body #2 at the origin. Also, this part proves that

First place the two curves so that they are in contact at the origin, with body #2 as in step 1. Let $s$ be the (signed) arclength parameter as measured from the origin, with $\gamma_2$ the curve for body #2 and $\gamma_1$ the curve for body #1. Let $\theta_i = \theta_i(s)$ be the functions defined by $$ \tan \theta_i(s) = \text{slope at } \gamma_i(s) $$ Note that by our normalization $\gamma_1(s) = \gamma_1(s) - \gamma_1(0)$ is the vector to move from $\gamma_1(0)$ (which is the origin) to $\gamma_1(s)$. (See the graph in the image above)

Then the position of the relevant point $C_1$ on body #1 as you roll can be parameterized by

$$ \gamma_2(s) - R(\theta_1(s) - \theta_2(s))[\gamma_1(s)] $$

where $R(\theta)$ denote a clockwise rotation by angle $\theta$. In fact, we do a translation by $\gamma_2(s)$ and then a clockwise rotation by $R(\theta_1(s) - \theta_2(s))[\gamma_1(s)]$ to obtain the current position of $C_2$. The amount the graph $\gamma_1$ has rotated is exactly $\theta_1 - \theta_2$ [notice that we take the signed angle so $\theta_2 < 0$ in the illustration].

Since $\theta_1(0) = \theta_2(0) = 0$ by assumption, and that $\gamma_1(0) = 0$, you have that the derivative of the above function, evaluated at $s = 0$, gives

$$ \dot{\gamma}_2(0) - \dot{\gamma}_1(0) = (1,0) - (1,0) = 0.$$

where $\dot{\gamma}$ is the derivative with respect to $s$.

Answer 2

Proof 1. By linearity of the derivative, one can assume that the point of contact is stationary (this involves viewing the tangent vector as a complex number). Let $C_1(0)=C_2(0)$ be the point of contact and $C_i(t)$ be the movement of the point $C_i$ as the bodies roll, where $t$ is the time parameter. The contact condition immediately implies that the vectors $C_i'(0)$ are proportional. Since by the contact condition the two rolled distances are equal, by the fundamental theorem of calculus the two speeds are equal, as well. Hence the velocities are equal.

Proof 2. Since the derivative is linear with respect to addition of vector functions, we can always subtract off the motion of the second body to make sure that the second body is fixed, and the first body is rolling along it. Therefore the question becomes to show that the velocity of the first body at the point of contact is always zero. By the hypothesis of rolling the velocity must be proportional to the normal vector of the curve. But a moment before contact it points in the opposite direction from the moment after contact. Hence at the time of contact itself it must be zero.

In more detail, the velocity will be calculated as $\frac{1}{ds}$ times the difference of the distinct endpoints of two infinitesimal arcs of equal length tangent at a common endpoint. This difference is necessarily perpendicular to the arcs (up to higher order terms), essentially because an infinitely thin isosceles triangle has right angles at the base up to higher order.

More formally, if two "touching" curves $\alpha(s)$ and $\beta(s)$ are parametrized by arclength, so that $\alpha(0)=\beta(0)$ and $\alpha'(0)=\beta'(0)$ then the difference $\alpha(s)-\beta(s)$ is negligible compared to $s$ because $\lim_{s\to0} \frac{\alpha(s)-\beta(s)}{s}=\lim \frac{\alpha(s)-\alpha(0)-(\beta(s)-\beta(0))}{s}=\lim \frac{\alpha(s)-\alpha(0)}{s}-\lim\frac{\beta(s)-\beta(0)}{s}=\alpha'(0)-\beta'(0)=0$ by hypothesis.

Answer 3

Engineering-wise (well actually physically) I would present the matter as follows.

1. the bodies are rigid, thus impenetrable, and thus at any fixed time, at the point of contact, the respective tangent lines shall be the same, and consequently also the normal line shall be the same;
2. by definition, during rolling the bodies remain in contact;
3. if the bodies keep in contact and are impenetrable then the normal relative speed at contact point shall be null, and being rigid, the speeds of the two whole bodies shall be the same in the normal direction;
4. define rolling as the relative motion of two bodies in contact, which at the contact point do not have slip, so also the tangential velocity at contact point shall be the same;
5. therefore, in a reference system placed at the various contact points, the two bodies can just have an instantaneous rotation around the origin;
6. then taking the reference system with origin in the contact point and the axes oriented in the tangent ($x$) and normal ($y$) directions, the curvature centers will be on the $y$ axis, and the angular speeds shall be such that, after $dt$, the traces of the "old" contact on the bodies have moved the same distance $dx$, meaning the same $ds$, thus angular speeds inversely proportional to the curvature radii.

So we have the two equivalent definitions for two rigid bodies rolling on each other, made with respect to a general reference system, and upon the hypothesis of having permanently at least a point of contact:

• the relative speeds of the two bodies at the point(s) of contact is null;
• the trace of the contact points along the contour of the two bodies has the same length vs. time.

Answer 4

The assumption here is that the common normal of contactin curves are always in a straight line. Else there would be a slipping component of of transmitted force and separation of gears and rolling with or without help of friction would not be the same.

You need to mention that the bodes have different fulcrums and velocity components along common normals should balance out. See for example Gear Vel_ vector

So the fundamental law of gearing requires above condition to be satisfied.

In addition if the force along common normal does not oscillate in direction, the involute which is tangent to base circle is the only solution.