Unwinding: Physics of a spool of string

It’s been a long day. Let’s unwind with a physics problem.

This problem was on the pre-entrance exam I took before arriving at Caltech for my freshman year. I’ve seen it from time to time since, and here I hope to find an intuitive solution.

You have a spool of thread, already partially unwound. You pull on the thread. What happens?

Here it is in side view. The dashed circle is the inside of the spool and the green line is the thread. Take a minute to see if you can tell how it works. Does the spool go right or left?


The usual method is to work it out with torques. The forces you must account for are the force of tension from the string and the force of friction from the table.

Torques are actually a pretty easy way to solve this problem, especially if you calculate the torque around the point of contact between the spool and the table (since in that case friction has no moment arm and exerts no torque).

This method is direct, but it’s useful to find another viewpoint if you can.

Let’s first examine a different case where the string is pulled up rather than sideways.


In this case, even if the first situation was unclear, you probably know that the spool will roll off to the left. To see why, let’s imagine that the thread isn’t being pulled by your hand, but by a weight connected to a pulley.


I put a red dot on the string to help visualize its motion.

The physics idea is simply that the weight must fall, so the red dot must come closer to the pulley. Which way can the spool roll so the red dot moves upward?

When the spool rolls (we assume without slipping), the point at the very bottom, where it touches the table, is stationary. The spool’s motion can be described, at least instantaneously, as rotation around that contact point.

Googling, I found a nice description of this by Sunil Kumar Singh at Connexions. This image summarizes the point:

 Rolling motion  (rm1a.gif)

If the spool rolls to the right, as above, the point where the string leaves the spool (near point B), will have a somewhat downward motion. This will pull the red dot down and raise the weight. That’s the opposite of what we want, so what really happens is that the spool rolls to the left, the string rises, and the weight falls.

With this scenario wrapped-up (or unwrapped, I suppose), let’s return to the horizontal string segment.


Again, the weight must fall and so the red dot must go towards the pulley.

If we check out Mr. Singh’s graphic, we’re now concerned with the motion of a point somewhere near the bottom-middle, between points A and C. As the spool rolls to the right, this point also moves to the right. This is indeed what happens as the weight falls.

Notice that the red point actually moves more slowly than the spool as a whole. This means the spool catches up to the string as we move along – the spool winds itself up. If the inside of the spool is 3/4 as large as the outside (like it is in my picture), the spool rolls 4 times as fast the string moves, and so for every centimeter the weight falls, the spool rolls four centimeters.

Here’s a short video demonstration:

Tags: , ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: