I was asked a middle school physics homework problem by a "bear kid".

Jim was holding one end of a plank, and the other end of the plank was placed on a cylinder as shown in the figure (OK, let's not focus on why he had to do that). Jim pushed this plank forwards, and the cylinder began rolling and moving. The length of the plank was L, the radius of the cylinder is r, and in the whole process the cylinder was clinging to both the floor and the plank so that there's no slip at all.Question:how far did Jim move when theposition Preached the cylinder?

This problem suddenly reminds me another "rolling wheel" : two same round coins gear into each other, fix one, let the other roll around it for a complete revolution. Then the question is, for how many cycles has the rolling coin rotated?

Are you able to think out these two problems

I find it difficult. Many people find it difficult, too. If the answers easily come to your mind, perhaps you are very familiar to epitrochoids and hypotrochoids, or you have met these problems before, or, congratulations, you do have a powerful physics engine in your mind. Mm, this tweet won't discuss too much about math, but something that may be more interesting --

Before posting my guess, first putting the answers would be good.

The key point to the rolling wheel problems is "how much distance the wheel rolled, how far the center of the wheel has gone", i.e. when the wheel rotates for a $\theta$ angle (it can be larger than $2\pi$), the center of the wheel moves $r\theta$ far.

So,

Some of you may get it now ... but my imagination is poor, so I have to illustrate it for myself.

It's more clear now. And

We render this process, rolling a square, much more precisely and clearly in our mind.

"Yes, the square moves forward s distance, and the plank, 2s distance. No doubt." You may state it firmly, full of self-confidence, even without seeing the illustration above.

This will happen again when you imagine a square rolling around another square with the same size.

You can easily (at least more easily than thinking a rolling circle) get it: the rolling square (the blue one) rotates for two cycles when it comes back to the original place.

How amazingly the

However, the anchoring, is considered as a cognitive bias in many psychological researches. Maybe you may argue they are not the same concept, but at least they look alike, and very probably have the same cause. Or alternatively, my examples are the cause of the anchoring effect -- the "anchors" help humans understand how some nature works, so it becomes a part in our intuitive physics engines. Anyway, this conjecture needs to be examined by well-designed experiments.

Let $Q$ be the center of a regular $n$-sided polygon, $MO$ and $OP$ be two adjacent sides, so $QMOP$ is a part of the whole polygon. When the polygon rotates forward for

It's easy to prove that $\bigtriangleup QOQ'$, $\bigtriangleup Q'OP'$, $\bigtriangleup QMO$, and $\bigtriangleup QOP$ are congruent, and a useless but interesting fact, point $P$ just lies on $Q'P'$.

Thus, the trajectory and the displacement are respectively $$\begin{align}x_1 & = r \theta = \frac{2\pi}{n} r \\ x_2 & = s = 2r \sin \frac{\theta}{2} = 2r \sin \frac{\pi}{n} \text{.}\\ \end{align}$$

When the polygon rotates for one cycle, the total length of the trajectory of the center is $nx_1 = 2\pi r$ and the total displacement of the center is $n x_2 = 2rn \sin (\pi/n)$.

We can see that, for $n \rightarrow \infty$, i.e. the case of the polygon becoming a circle, the trajectory length keeps $2\pi r$, and the displacement is then $$2r \lim_{n \rightarrow \infty} n \sin \frac{\pi}{n} = 2 \pi r \text{.}$$ The limit can be calculated by Taylor series or by L'Hôpital's rule (after making $n \sin (\pi/n)$ to its continuous case $x \sin (\pi / x)$).

So, if we have to imagine a rolling wheel, we can first try to imagine a rolling square instead, as there are

Are you able to think out these two problems

**just by imagination**?I find it difficult. Many people find it difficult, too. If the answers easily come to your mind, perhaps you are very familiar to epitrochoids and hypotrochoids, or you have met these problems before, or, congratulations, you do have a powerful physics engine in your mind. Mm, this tweet won't discuss too much about math, but something that may be more interesting --

**why do many people feel difficult?**#### Answers to the two problems

Before posting my guess, first putting the answers would be good.

The key point to the rolling wheel problems is "how much distance the wheel rolled, how far the center of the wheel has gone", i.e. when the wheel rotates for a $\theta$ angle (it can be larger than $2\pi$), the center of the wheel moves $r\theta$ far.

So,

**for the first problem**, when the plank is $s$ distance ahead the cylinder, the cylinder must have rolled $s$ distance, and then at that moment the center will be $s$ distance away from the original point -- as a result, the plank will be moved $2s$ in total. Thus, the final answer should be $2L$.Some of you may get it now ... but my imagination is poor, so I have to illustrate it for myself.

It's more clear now. And

**for the second problem**, to finish a revolution, the center of the rolling coin has to move around a larger circle with a doubled radius, so it will rotate for $$\frac{2\pi(2r)}{2 \pi r} = 2$$ cycles. Again, an illustration.#### Interesting psychological effect

Something I find interesting is, if we don't roll a wheel in our mind, but a square, it will be much more easier. I guess the cause is**we can easily find an anchor**on a square, as a square is "angular".We render this process, rolling a square, much more precisely and clearly in our mind.

"Yes, the square moves forward s distance, and the plank, 2s distance. No doubt." You may state it firmly, full of self-confidence, even without seeing the illustration above.

This will happen again when you imagine a square rolling around another square with the same size.

You can easily (at least more easily than thinking a rolling circle) get it: the rolling square (the blue one) rotates for two cycles when it comes back to the original place.

How amazingly the

**anchor**helps our imagination!However, the anchoring, is considered as a cognitive bias in many psychological researches. Maybe you may argue they are not the same concept, but at least they look alike, and very probably have the same cause. Or alternatively, my examples are the cause of the anchoring effect -- the "anchors" help humans understand how some nature works, so it becomes a part in our intuitive physics engines. Anyway, this conjecture needs to be examined by well-designed experiments.

#### Rolling squares are the same as rolling wheels?

This section is for those who are interested in math.**To some extent**, yes, and in fact rolling all kinds of regular polygons are similar to rolling circles.Let $Q$ be the center of a regular $n$-sided polygon, $MO$ and $OP$ be two adjacent sides, so $QMOP$ is a part of the whole polygon. When the polygon rotates forward for

*one move*, $Q$ will reach $Q'$, and $P$ will reach $P'$. The length of one side of the polygon is $s$, and the circumradius is $r$. $\overline{QQ'}$ is**the displacement of the center in one move**, denoted by $x_2$; and $\overset{\frown}{QQ'}$ is**the trajectory of the center in one move**, denoted by $x_1$.It's easy to prove that $\bigtriangleup QOQ'$, $\bigtriangleup Q'OP'$, $\bigtriangleup QMO$, and $\bigtriangleup QOP$ are congruent, and a useless but interesting fact, point $P$ just lies on $Q'P'$.

Thus, the trajectory and the displacement are respectively $$\begin{align}x_1 & = r \theta = \frac{2\pi}{n} r \\ x_2 & = s = 2r \sin \frac{\theta}{2} = 2r \sin \frac{\pi}{n} \text{.}\\ \end{align}$$

When the polygon rotates for one cycle, the total length of the trajectory of the center is $nx_1 = 2\pi r$ and the total displacement of the center is $n x_2 = 2rn \sin (\pi/n)$.

We can see that, for $n \rightarrow \infty$, i.e. the case of the polygon becoming a circle, the trajectory length keeps $2\pi r$, and the displacement is then $$2r \lim_{n \rightarrow \infty} n \sin \frac{\pi}{n} = 2 \pi r \text{.}$$ The limit can be calculated by Taylor series or by L'Hôpital's rule (after making $n \sin (\pi/n)$ to its continuous case $x \sin (\pi / x)$).

So, if we have to imagine a rolling wheel, we can first try to imagine a rolling square instead, as there are

**on a square!***anchors*