# Unrolling Toilet Paper

In his Dec. 17 Action-Reaction blog post titled “Falling Rolls,” one of my heroes of physics instruction, Frank Noschese, details an exercise from Robert Ehrlich’s book Why Toast Lands Jelly-Side Down.

The exercise, a rotational motion problem that challenges students to find the ratio of heights at which you can drop two identical toilet paper rolls, one dropped regularly, the other dropped by holding onto the end of the paper and letting it unroll, such that the two rolls hit the ground at the same time.  It’s a terrific, easy-to-replicate and demonstrate problem that pulls together a great number of rotational motion skills –> finding the moment of inertia, applying the parallel-axis theorem, identifying forces and torques from free body diagrams, and converting angular acceleration to linear acceleration. My students dove into the challenge with zest!

To begin the exercise, we set our variables (H=height for dropped roll, h=height for unrolled roll, r = inner diameter, R = outer diameter), then identified the time it takes for the dropped roll to hit the ground using standard kinematics:

${t_{drop}} = \sqrt {{{2H} \over g}}$

Next, we did the same thing for the unrolling toilet paper roll:

${t_{unroll}} = \sqrt {{{2h} \over a}}$

Of course, if we want them to hit at the same time, the times must be equal, therefore we can show:

${H \over h} = {g \over a}$

Obviously, what we really need to focus our efforts on is finding the linear acceleration of the unrolling roll. To save ourselves some time, we started by looking up the moment of inertia for a cylinder:

$I = {\textstyle{1 \over 2}}M({r^2} + {R^2})$

Using the parallel-axis theorem to account for the unrolled roll rotating about its outer radius we find:

$I = {\textstyle{1 \over 2}}M({r^2} + {R^2}) + M{R^2} = {\textstyle{1 \over 2}}M({r^2} + 3{R^2})$

Next, we can use a free body diagram to identify the net torque on the roll as MgR, and use Newton’s 2nd Law for Rotational Motion to find the angular acceleration:

${{\tau }_{net}}=I\alpha \Rightarrow \alpha =\frac{{{\tau }_{net}}}{I}=\frac{MgR}{0.5*M({{r}^{2}}+3{{R}^{2}})}=\frac{2gR}{{{r}^{2}}+3{{R}^{2}}}$

Since linear acceleration can be found from angular acceleration multiplied by the radius of rotation (R):

$a = \alpha R = {{2g{R^2}} \over {{r^2} + 3{R^2}}}$

Finally, since we’re looking for the ratio of the dropped height to the unrolled height:

${H \over h} = {g \over a} = {g \over {{{2g{R^2}} \over {{r^2} + 3{R^2}}}}} = {{{r^2} + 3{R^2}} \over {2{R^2}}} = {3 \over 2} + {{{r^2}} \over {2{R^2}}}$

This conflicts with the results from Noschese’s class, where they derived $\frac{H}{h}=2+\frac{{{r}^{2}}}{{{R}^{2}}}$, however, their demonstration based on their results is very convincing.  Let’s take a look at the difference in ratios using the two derivations:

For a toilet paper roll of inner diameter .0095m and outer diameter R=.035m (our school rolls from the janitor supply closet):

$\frac{H}{h}=2+\frac{{{r}^{2}}}{{{R}^{2}}}=2+\frac{.0095{{m}^{2}}}{.035{{m}^{2}}}=2.074$ $\frac{H}{h}=\frac{3}{2}+\frac{{{r}^{2}}}{2{{R}^{2}}}=1.5+\frac{.0095{{m}^{2}}}{2*.035{{m}^{2}}}=1.54$

It appears that our discrepancies aren’t just differing mathematical representations of the same formula, but that we have a significant difference in our derivations.

In looking over our assumptions, we assumed no air resistance, and also that the unrolling toilet paper roll rotates about its outer radius (is this really true)? I wonder what assumptions were made in Noschese’s class that may account for these differences. It will be interesting to get his class’s perspective on the problem, and provides a great practical study for our students of different approaches to a problem, and the importance of understanding the ramifications of assumptions made in beginning a problem solving exercise!

Update: it appears our calculations are correct.  Check out our high-speed video confirmation!

Slow Motion Toilet Paper Falling

## 26 thoughts on “Unrolling Toilet Paper”

1. Looks like we made an algebra error. I ran through our derivation again (which uses energy instead of torque) and now I get the same solution as you, which is also the same solution in Erlich’s book. Thanks for trying it out with your students and catching our mistake! (In my video, it looks like our students do not release the rolls at exactly the same time, which may have compensated for our incorrect solution.)

• I never even thought about doing it through a conservation of energy approach… now we have a follow-up to start on when the kids return on Monday — thanks for the great idea!

• How exactly could we derive a formula using the conservation of energy? Help!

2. Awesome video! That is a perfect release and landing! Well done!

3. Can you explain why you decided that the roll was rotating around its outer radius. It seems that this is not what is demonstrated by various other resources I have come across. Take a look at https://www.youtube.com/watch?v=k4h-lkcnRWI from Professor Lewin of MIT. I would really appreciate if you could explain. Thanks!

• Hi! If you watch it drop in the video, it appears to rotate about the outer radius. However, if you did analyze from the center, I imagine you would work yourself to the same answer! You’ll notice in the post I note that we assumed it rotated about the outer radius — which is, of course, just an assumption.

• Thanks so much for the reply! I’m going to work out the problem and see what I get. Cheers.

• You were right. I got the same answer! Cool.

• Steven,

Any chance you can tell us how you solved the problem using kinematics/dynamics without modifying the inertia first? I cannot for the love of me figure out how the “analysis from the center” will end up with the “same answer”, and I’ve tried many routes. My ideas are revolving around the fact that a point mass will have MR^2 rotational inertia from a center of rotation, which is exactly what is added to the overall inertia, but I am clueless about how this comes about. Most likely I am incorrectly using alpha/a conversions, trying to combine a torque with a force when they can’t be, or my free body diagram is incorrect from the get-go:(.

It is lovely however, that using kinetic energy to solve the problem does come out to exactly the same answer, though! And for this, you don’t have to use the modified inertia!

• Nevermind, I also figured it out (while working on an unrelated review problem with a student of mine, as it just so happens).

One reason why you add an additional MR^2 to the object’s inertia is because gravity and its torque on the TP is acting effectively on a point mass with this force the radius away from where the toilet paper is held. For a point mass particle, its inertia can be calculated as MR^2. However, the TP already has a rotational inertia of 1/2 m(r^2 + R^2) due to the torque acting on it from the held paper, so both have to be added together, giving 1/2 m(r^2 + 3R^2).

• Your parallel axis theorem use isn’t justified here. Remember the toilet paper roll is rotating about its axis not by its our radius. The tension applied to the unrolling toilet paper is applied from the outer radius but it is obviously not rotating about that .

• It is justified when you consider that when you first drop the toilet paper roll while holding on to the paper, it initially rotates about an axis at the edge of the roll.

• I’m still struggling with the justification of the rotation about the outer radius. Doesn’t the 1/2m(r^2 + R^2) account for the rotation of the outer radius?

Thanks for the help!

4. Correct me if I’m wrong, but it seems there might be two other assumptions, or rather oversimplifications:

The outer diameter is constant.
The mass of the rolling paper is constant.

If you tried this over greater heights these might become an issue and the solution would then involve some complex calculus dealing with the changing mass and radius.

Great idea though, I’m probably going to do this with my AP Physics C kids tomorrow. We’ve moved on to SHM but they struggled with rotation and I want to spiral back to it.

5. Can you explain how you got the net torque as equal to MgR?

I get a force of tension that creates a CCW torque and the force of gravity creating a CW torque. So the net torque expression I get is (Ft * R) – (mg * R) = I * alpha.

Any help on where I’m going wrong would be much appreciated. Thanks!

• Hey Chris, thanks for the reply!
The definition of torque is force times the radius, so therefore, MgR makes perfect sense.
Secondly, quick inquiry… what does CCW and CW mean????????!

Thanks!