Circle revolutions rolling around another circle

Question

I just watched this video, and I'm a bit perplexed.

Problem:

The radius of Circle A is 1/3 the radius of Circle B.

Circle A rolls around Circle B one trip back to its starting point.

How many times will Circle A revolve in total?

The intuitive answer is 3, but the correct answer is 4. I understand the trick -- that the center of Circle A must travel a distance of $2\pi(r_B + r_A)$, not $2\pi r_B$ -- but I'm still confused on one item.

At the risk of sounding very un-mathematical, how do the (infinite set of) points on the circumference of each circle map to each other to accomplish this?

Consider Circle A rolling along a straight line the length of the circumference of Circle B. Then it will revolve 3 times. It's like the universe "knows" when to apply a different point mapping when you change the arrangement of matter.

Answer 1

Circle $A$, with radius $r$, gets back to its starting point when $A$'s centre completes one rotation (around the centre of circle $B$ with radius $R$). Clearly $A$'s centre traverses a circular path of radius = $R+r$.

Now, Physics to the rescue.

Let the speed of $A$'s centre = $v$

Let the angular speed of $A$'s rotation around its own centre = $\omega$

Since, $A$ rolls over $B$ $\implies$ $v = \omega r$ (Assuming $B$ is fixed, the condition of rolling is that the point of contact is at rest).

Let the time taken by $A$'s centre to complete one rotation be $t$.

Then, $2\pi (R+r) = vt$

$\implies t = \frac{2\pi (R+r)}{v}$

Total angular distance traversed by $A$ around its centre in the same duration: $\theta = \omega t$

Using the above results, we get $\theta = \frac{v}{r} \frac{2\pi (R+r)}{v} = \frac{2\pi (R+r)}{r}$

In this time $t$, $A$ completes, say, $N$ rotations around its centre.

$\implies N = \frac{\theta}{2\pi} = \frac{(R+r)}{r}$

Now, In your case $ r = \frac{R}{3}$

Using this information, $ N = 4$

Answer 2

Thank you everyone for your answers. They were all informative, but I wasn't able to intuitively understand them until I saw this visual:

https://www.geogebra.org/m/v3a437ux

The key for me was to see that there are two kinds of revolution happening:

1. Revolutions of Circle A with respect to Circle B, and
2. Revolutions of Circle A with respect to the (overhead) observer.

To the observer, the revolutions of the type (2) complete before revolutions of type (1). The revolutions of type (1) happen at the $r_B/r_A$ roots of unity, and the revolutions of type (2) happen at the $(r_B/r_A + 1)$ roots of unity. It was very helpful to see precisely where these revolutions occur, because it was mentally impossible to unify the two types of revolution without the visual. It's also useful to notice that at any given moment, with respect to the observer, the points of Circle A on the far side from Circle B are moving faster than the points of Circle A on the side that is touching Circle B.

David K's answer is great for understanding that the parametric mapping of time to point pairs is the same whether Circle A is rolling along a straight line or a circle, and that we are dealing with frames of reference. I simply didn't believe the mapping was the same until I saw the visual.

zoli and Hans Lundmark's answers are great for understanding that one extra revolution must occur at some point along the rolling path. The complete answer, of course, is that this last mysterious revolution does not happen all at once, but gradually along the entire roll.

The visual was discovered in the comments on this page, which is another discussion of the same problem:

https://plus.maths.org/content/circles-rolling-circles

Answer 3

As it is depicted on the figure below the smaller circle has to turn around $3$ times if it travels along a straight distance equaling the perimeter of the larger circle. In this case the center of the smaller circle takes the same distance.

Now imagine that the straight line of length $6\pi r_B$ takes a complete rotation around its center while the small circle rolls along. Also, suppose that the straight takes one full turn while the circle reaches the other end.

As a result the small circle takes a fourth revolution.

The number of rotations is not different if the path is circular. The circularity of the path was modeled in the case of the straight path by turning around the segment.

The video explains the same by claiming that the small circle has to travel a distance of $8\pi r_B$. This is the same as having to travel on the straight line of length $8\pi r_B$ which does not rotate about its center.

Answer 4

The mapping of points from one circle to the other is no different than when the small circle is rolling on a straight line; the extra revolution is purely due to bending.

It might be easier to see this in the case of gliding, when there is a single point $A$ on the small circle which is in contact with the line or the large circle throughout the process. When gliding on a line, the small circle doesn't revolve at all. But if it's supposed to glide along a large circle in such a way that the same point $A$ is in contact all the time, then it also has to revolve as at glides (and it will make one revolution as it glides one lap along the large circle).

Answer 5

I don't know what you mean by "the points map to each other". To "map" one set to another implies a function that maps each point in one set to a unique point in the other; to "map to each other" implies that the function is one-to-one. There is no one-to-one mapping between points: each point on the small circle meets three different points on the large circle. If you roll a circle of radius $2$ twice around the circle of radius $3$ then the correspondence between points that meet isn't even a function in one direction, let alone a one-to-one function. I think perhaps you mean there is a mapping from some parameter to the pairs of points along the two circles that are brought into contact; for example, if the parameter is time in seconds since the motion started, the mapping says which two points will be in contact at each time $t$.

I prefer a completely different approach than the one in the video. Seat yourself on a frame that is attached to the centers of the two circles. As the small circle rolls around the larger one, the motion of the center of the small circle rotates the frame (and you) around the center of the larger circle.

While sitting on the frame, what you see is the two centers of the circles remaining in the same place within your field of view, while the larger circle rotates around its center and the smaller circle also rotates without slipping against the larger circle. And of course you see the smaller circle rotate three times.

But someone who remained motionless relative to the larger circle, not sitting in your rotating frame, sees the smaller circle rotate four times: the three times that you observed, plus one rotation that you did not observe because you were doing a full rotation in the same direction yourself. If you step off the frame and "undo" the effect of its rotation upon you by turning yourself once in the opposite direction, you'll observe the fourth rotation of the small circle.

Answer 6

Nothing new, but a diffent way to look at the question:

First consider a cirlce with an inside hexagon (in general an n-figure). Then the circle will roll the distance of the circumference PULS an angle wihich is due to tilting at the edges. Thus, the additional angle (the $+1$ you look for) amounts in total to

$$6 \cdot \frac{2\pi}{6} = 2\pi, (n \cdot \frac{2\pi}{n} = 2\pi)$$

explaining the extra turn - no matter how many edges you take for your approximation.

Second consider an infinitisimal approach: For $x \in [0, R)$, $f(x) = \sqrt{R^2-x^2}$ is a circle with radius $1$. Later, we also need

$$f'(x) = \frac{-x}{\sqrt{R^2-x^2}}$$

A tangent vector $\vec{T}$ at $x$ is $$\vec{T}=\begin{pmatrix}1\\f'(x)\end{pmatrix}$$ and $\vec{T'}$ at $x+\Delta x$ is $$\vec{T}=\begin{pmatrix}1\\f'(x+\Delta x)\end{pmatrix} = \begin{pmatrix}1\\f'(x)+f''(x)\cdot \Delta x + \frac{f''(x)}{2}\Delta x^2\end{pmatrix}$$.

Now, the cosine of the angle $\Delta \alpha$ between the to vectors is

$$\cos(\Delta \alpha)=\frac{\vec{T}\cdot \vec{T'}}{\left|\vec{T}\right|\cdot \left|\vec{T'}\right|}$$

To first order this is

$$h(\Delta x) = \frac{1+\frac{f'f''\Delta x}{1+f'^2}+\frac{0.5f'f'''\Delta x^2}{1+f'^2}}{\left(1+\frac{f''^2\Delta x^2}{1+f'^2}+\frac{2f'f''\Delta x}{1+f'^2}+\frac{f'f'''\Delta x^2}{1+f'^2}\right)^{1/2}}$$

Use the following abbreviations for a Talor-series (which equals $\cos(x) = 1 - x^2/2$): $$N=\frac{1}{1+f'^2}, a = Nf'f'', b = 0.5Nf'f''', c = Nf''^2, d = c + 2b$$

Then

$$h(\Delta x) = \frac{1 + a\Delta x + b \Delta x^2}{\left(1 + 2a \Delta x + d \Delta x^2\right)^{1/2}}$$

$$h'(\Delta x) = \frac{(a^2-c)\Delta x + 3ab \Delta x^2 + bd \Delta x^3}{\left(1 + 2a \Delta x + d \Delta x^2\right)^{3/2}}$$ and

$$h''(\Delta x = 0) = \frac{(a^2-c)\Delta x + 3ab \Delta x^2 + bd \Delta x^3}{\left(1 + 2a \Delta x + d \Delta x^2\right)^3/2}$$

Finally we obtain

$$1 - 0.5\frac{f''^2}{\left(1+f'^2\right)^2} \Delta x^2 = 1 - \frac{\Delta \alpha^2}{2}$$

or

$$\int_{x_0}^{x_1} \frac{f''}{1+f'^2}dx = \arctan(f'(x_1)) - \arctan(f'(x_0))$$.

This again gives for a full circle $8\cdot\frac{pi}{4} = 2 \pi$. Furthmore we observe, that in the end, only the difference between the angles at $x_0$ and $x_1$ play a role.

Answer 7

I'm not sure if this is a novel argument, but the "flawed logic" used to suggest 3 is the same logic used to suggest 5, hence demonstrating the inconsistency:

Instead of viewing the small circle (radius r) moving around the outside of the large circle (radius 3r) we can equivalently view the small circle moving around the inside of the larger circle (radius 5r).

Thus the suggested logic is flawed because it would give the contradictory simultaneous answers 3 and 5, respectively.

(Best take their average, 4 haha.)

Answer 8

Given The radius of Circle A is 1/3 the radius of Circle B. Circle A rolls around Circle B one trip back to its starting point.

Question How many times will Circle A rotate about its center

*** Solution, Part 1 Begin by drawing (1) Circle B (2) Circle A where it initially contacts Circle B (3) Circle A where it contacts Circle B after rolling 1/3 the way around Circle B

Add to this drawing 3 lines Line 1 the line connecting the centers of Circles A and B when Circle A initially contacts Circle B

Line 2 the line connecting the centers of Circles A and B when Circle A has 1/3 the way around Circle B

Line 3 the line that is parallel to Line 1 and passes through the center of Circles A when Circle A has 1/3 the way around Circle B

*** Solution, Part 2 (1) The angle between Line 2 and Line 1 = 120 degrees, so (2) The angle between Line 2 and Line 3 = 120 degrees

So the total angle about Circle A's center through which the point of contact has rotated is 360 + 120 degrees

Answer 9

Really enjoyed this, thanks so much. I thought I would just contribute something of a more "mechanical" view of this problem. I've used clockfaces just to keep things familiar.

In the top image, image $A$, the two clockfaces are aligned next to each other, with the "$3$" on the left touching at the "$9$" on the right black clockface.

In the middle image, image $B$, I've started the rotation of the right black clockface around the left white clockface, lining up the $4$ on the left with the $8$ on the right.

In the bottom image, image $C$, I've done exactly the same transformation, only this time I've rotated both clockfaces. (If you tilt your head to the right, image $B$ will look like image C does with your head level.)

So what this shows, mechanically, is when we appear to rotate one circle around the other, this is an illusion. In fact we are always rotating both circles at the same time. Because of that, the total distance that needs to be travelled is the sum of the circumferences of both circles. In this case, as the circles are the same size, it requires $2$ apparent rotations of one of the circles to get around the other.

If you want to introduce a bit more maths into this, you can look at the tangents at $4$ and $8$ as they differ from the vertical perpendicular.

Answer 10

Let $\mathscr{F}$ be the overhead observer frame of reference, i.e. bird's-eye view. Let the circle's center be $O$ and its frame of reference be $\mathscr{G}$. Here, $\mathscr{G}$ doesn't rotate relative to $\mathscr{F}$, it only has translation. (So the circle is rotating relative to both $\mathscr{F}$ and $\mathscr{G}$). Let $P$ be a fixed point on the circle rolling along with it. (You can think the circle is made of stone and $P$ is a permanent engraving made on its edge). Freeze the motion at some instant in time, say $t=0$, precisely when the point of tangency is $P$. (Of course, the point of tangency is constantly changing, so $P$ is the point of tangency only when $t=0$, not immediately before nor after this instant).

Any solution to this problem must crucially use the fact that there's no slipping. One way to quantify this is: the instantaneous $\mathscr{F}$-velocity of $P$ is $0$ at time $t=0$. Any nonzero velocity would imply slipping along the curve we're rolling on. In fact, this is true for a circle rolling on any smooth curve, it doesn't have to be a bigger circle, and it can even roll at a nonconstant speed, as long as all motion is smooth.

So $P$ is $\mathscr{F}$-stationary at time $t=0$. This implies the $\mathscr{F}$-velocity of $O$ at $t=0$ equals its velocity relative to $P$ at $t=0$. The latter is just the additive inverse of the $\mathscr{G}$-velocity of $P$ at $t=0$. Let $v(t)$ denote the $\mathscr{F}$-velocity of $O$ at time $t$ and similarly let $w_Q(t)$ denote the $\mathscr{G}$-velocity of an arbitrary object $Q$ at time $t$. We have just proven that $v(0)=-w_P(0)$, so in particular $$\Vert v(t)\Vert=\Vert w_P(t)\Vert \hspace{1cm} \text{ at } t=0. \hspace{1cm} (\star)$$

Now, notice that for any two points on the circle $P'$ and $P''$, and any time $t$, we have $\Vert w_{P'}(t)\Vert=\Vert w_{P''}(t)\Vert$, by definition of rotation. So, as time changes and the point of tangency changes, the equation ($\star)$ still holds true for all $t$ even without changing $P$! (At any $t$, some point $P'$ will be the point of tangency, even if it's not $P$, so $\Vert v(t) \Vert = \Vert w_{P'}(t) \Vert = \Vert w_P(t) \Vert$).

Finally, we can integrate: let $T$ be a time interval, then $$\int_T \Vert v(t)\Vert dt = \int_T \Vert w_P(t)\Vert dt.$$ The former is just the distance $O$ traverses in $\mathscr{F}$ during $T$, and the latter is just $2\pi r$ times the number of revolution the circle does during $T$. We have proven:

Theorem: Let a circle of radius $r$ roll smoothly along another smooth curve for a time interval $T$. Then the distance traversed by the circle's center equals $2\pi r$ times the number of revolutions performed by the circle.

The original problem is a special case. If the bigger circle has radius $R$, then $O$ travels along a larger circle of radius $R+r$, so the total number of revolutions is $\frac{2\pi(R+r)}{2\pi r}=1+\frac{R}{r}$.

---

Note. I think the hardest part about this problem is we have a great intuitive understanding of no slipping, but we're not great at rigorously defining it. Even though the mathematical formulas for it might seem weird or out-of-the-blue, they are in fact definitions of what no slipping means, specially if you're a theoretical mathematician. Indeed, try defining "no slipping" in the pure math world (without using physics intuition or concepts). At least for me, it's exceedingly difficult, specially if you're in the most general setting of two arbitrary curves rolling along each other (see this).

In the case of a circle rolling on a curve, Physics has a more widely-known definition of slipping: $$ v =\omega R,$$ where $v$ is the object's speed, $\omega$ its angular velocity, and $R$ its radius. It's easy to prove this is equivalent to the other notion of "the point of tangency has zero instantaneous velocity". However, I personally like the latter definition more as it applies to the most general setting, not only circles.

Answer 11

The way you stated the question demonstrates the confusion. You are trying to figure out how to map the points on the circumference to each other. But that isn't the amount of turning. In terms of distance rolled along, the ratio is indeed 3 as you would expect. But that isn't the amount of turning, its just the gear ratio of the two circles the ratio of the circumferences. In a flat geometry the two measurements are the same.

Imagine now that you are sliding one circle around the other. How many times does it turn? In a flat geometry it turns 0 times. But in a circular geometry it turns exactly once, as evidenced by looking at the tangent or the normal vector and seeing how many times it turns.

This shows that the amount of turning is always 1 more than the gear ratio of the two circles. 1 more than the amount of turning in a flat geometry. The curvature of the second circle causes the small circle to turn faster.

But this is all relative perspective. If you are in the center of the large circle you will only notice 3 turns. The extra turn is gone because you are turning. The more complicated motion is analogous to the paths of the planets when earth is taken as the point of reference.

Answer 12

There are many good explanations above, but I think that I can display a view a little bit different about this classic question:

Look at the below figure:

When the blue circle rotates around the black circle the corresponding to the angle $\theta$, its center moves $\theta(R+r)$.

If one marks contact point P in the initial position, now it is in P’ and the new contact point is at Q. All translation movement in O is traduced in a rotation of P.

if O move to O’ in a infinitesimal segment part of a line forward, P in contact point moves backward to P' the same length and Q is the new contact point

So we can show that OO’ = P’Q.

To convince the reader that this happens, let's approximate the circle by a regular polygon with thousands of sides, which tends to a exact circle, when the number of sides tends to $\infty$.

To be easy to see, let's imagine, instead of a regular polygon with a multitude of sides, a regular hexagon.

Now drop B run over point P of the hexagon until B touched the floor in B'. Some visions of the hexagon's PB side trajectory are shown in the figure in dotted segments.

Note that the backward displacement of P to P' is the arc PP', than with a big number of sites can be approximated by the hexagon side PP'. It's the same forward displacement from B to B'. Similarly, the hexagon will have shifted horizontally exactly the length of the hexagon side CC'(and also the center of regular hexagon).

So the complete rotation of the hexagon to bring point P to the same position will correspond to an equal horizontal displacement of the perimeter of the hexagon. Such reasoning can be imagined for any regular polygon, however much the number of sides, which approaches the perfect circle as much as you want.

It doesn't matter if the surface on which the circle spins is a straight line , a circle or a irregular shape. In all cases, we divide it into equal segments the same size as the adopted surface of the regular hexagon.

Let's also suppose the 2 segments make an $\alpha$ angle. We can transfer the hexagon by the size of the side, as in the previous figure and then drop the end B' in angle $\alpha$.(*)

Note that the center C of the hexagon first assumes position B, as in the previous figure, then assumes the position C'. Note that the sector ABC' is equivalent to sector AB'B'' (2 equal sides and 1 angle), so the displacement from center C (CBC') is equal to travelled distance for contact point A, though B' and B'' (AB'B'').

If movement of which makes it coincide with P, the total extent of translation of the center of the smaller circle (not affected by the rotation of the smaller circle) moves $2\pi (R+r)$. As we consider $R = nr$, this corresponds to $2\pi(n+1)r$.

This is the same distance traveled by the point of contact of the smaller circle, as we saw above. Thus, as the perimeter of the smallest circle is $2\pi r$, one can calculate how many smaller circles have been traveled in the complete translation: $2\pi (n+1)r/2\pi r = n+1$

Note that the above solution applies to the circle spinning over any closed figure. The ratio between perimeters is what counts.

(*) There is a alternative way to move is hexagon, it's not showed here, for simplicity, but it's equivalent.

Answer 13

Since 'revolution' has a mathematical definition, i.e. $1 \ rev = 2 \pi r $ radians, where $r$ is the radius, this problem is easily explained. We're looking for how many revs of Circle A occur.

Circle A:

Note: $r$ = radius(Circle B)

$1 \ rev = \frac{2}{3}\pi r $ radians

$2 \ revs = \frac{4}{3}\pi r $ radians

$3 \ revs = 2\pi r $ radians = Circle B's circumference which Circle A traversed.

Answer 14

I don't know why people overthink these simple problems. The circumference of the larger circle is 3x the circumference of the smaller circle. Provided the small circle is not slipping, it will take 3 turns to cover the distance around the larger circle. Nothing magical here.

The reason you were perplexed by the video is because the uploader makes a mistake at 1:34 to 1:42. He counts this as a full revolution of the smaller circle, which it clearly is not. If he marked both circles with points on their circumferences, you'd see that it would take 3 revolutions. It's simply conservation of length. The rolling motion is completely irrelevant to the stated problem.

By the way, the applet you linked to was for a 4:1 scale factor, this is why in the applet it takes four revolutions.