[Math] Circle revolutions rolling around another circle

circlesgeometryinfinity

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.

Best Answer

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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$

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