Problem 1-3 Coin Rotation

Problem Posts

Problem 1-3 (Geometry) [Difficulty 1] [PDF]

QUESTION
(Original) Three identical cylindrical barrels with radius $\sqrt{3}$ are placed tangent to each other (represented by circles $c_{1}$ , $c_{2}$ , $c_{3}$ .) A metal sheet $AB$ is placed just touching $c_{1}$ and $c_{3}$. A coin $c$ with radius $2-\sqrt{3}$  is placed on the floor tangent to $c_{2}$ and $c_{3}$  (see the diagram), and rolled without slipping about the barrels (so that the coin is rotating clockwise), going through plank $AB$, until it returns to its starting point. How many radians has the coin rotated? (For example, half a turn is $\pi$ .)

Overview
Separate the coin's rotation into two 'components'.



Discussion
This piece will demonstrate a particularly rare method applicable to a limited class of problems. We begin with

Subproblem 1-3.1. (Coin Rotation Paradox)

The problem begins with two identical coins, one is rotated around the other without slipping; so that it ends up on the opposite side of the other coin from where it began. It has made a single rotation yet has rolled a distance equal to half of its circumference. (Wikipedia)

What's happening here? It so turns out that you have to consider two "components" of rotation:

• The first is the one caused by rotation due to path (the circumference of the stationary coin), which is $\pi$.
• The other one is rotation about itself. Imagine if the coin were totally slippery. This contributes another $\pi$ radians of rotation and resolves the paradox.

With these two components in mind, the original problem is straightforward.

Solution [show]

We first take the sum of the total path length that the coin comes into contact with: $2\sqrt{3}\pi+2\sqrt{3}+\frac{4}{3}\sqrt{3}\pi$ , and divide that with the circumference: $\frac{\frac{10\sqrt{3}}{3}\pi+2\sqrt{3}}{2-\sqrt{3}}$  rotations, which is $2\pi\cdot\frac{\frac{10\sqrt{3}}{3}\pi+2\sqrt{3}}{2-\sqrt{3}}$ . This is the rotation due to the path. Next, we add in the rotation “due to position”. This is$\frac{4}{3}\pi+2\cdot\pi=\frac{10}{3}\pi$ . All in all, we now have $2\pi\left(\frac{5}{3}+\frac{\frac{10\sqrt{3}}{3}\pi+2\sqrt{3}}{2-\sqrt{3}}\right)$.