Unrolling Toilet Paper

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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

Physics in a Winter Wonderland

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Over the river and through the words, to grandmother’s house we go…
the horse knows the way to carry the sleigh through the white and drifting snow – oh!

As part of our family’s holiday season festivities, we went on a horse-drawn sleigh ride through the woods in northwest Pennsylvania. It was a terrific time, with low winds, just a very light dusting of now coming down, and 28 degree temperatures. As  Miss Micro-APlusPhysics (aged 16 months) drove the sleigh, I couldn’t help but think what a terrific multi-faceted physics problem our trip would make… finding the force of friction the horses had to overcome to keep us moving at a constant velocity through the woods, the power supplied, and the energy consumed.

Of course, being a physics teacher, I couldn’t just leave it there:

With nine people on the sleigh, all bundled up, I think we can estimate an average mass of about 70 kg per person (we had a couple lightweights, including the baby.) So, the mass on the sleigh was probably on the order of 650kg. The sleigh itself was made out of fairly solid boards with steel runners, and a quick attempt at lifting up a corner provided a feel for its weight — let’s estimate the sleigh at 550kg, giving us a total load of 1200kg. The weight of the load, then, settles in a 12,000N.

The horses pulled the sleigh from a horizontal tether, so that given the equilibrium condition of the sleigh, we know the normal force had to offset the weight, so the normal force of the snow on the sleigh is 12,000N. Now, to estimate the coefficient of friction.  From the NY Physics Regents Reference Table, we find the coefficient of kinetic friction for a waxed ski on snow as 0.05. This seems like a reasonable esimate for the frozen runner on the snow. Using {F_f} = \mu N we find the force of friction as 600N.

For most of the 20-minute (1200s) journey the horses pulled us at a leisurely constant speed of approximately 1.5 m/s. Therefore, we can assume the applied force of the two LARGE Belgian horses as 600N. The power supplied can be calculated from P=Fv, or (600N)*(1.5 m/s) = 900W. And since they applied that power for roughly 1200s, the work done by the horses can be found from W=P*t=(900W)(1200s)=1,080,000 Joules, or the equivalent of 258 food calories (roughly the nutritional equivalent of one slice of pizza)!

A fun holiday activity providing another opportunity to highlight physics in the world around us.

50 Learning Goals For Physics Students

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What are the “big items” I want my students to take away from my class from each year?  It’s a big question… of course I want them to do a great job on their exams and understand our course content, but I realize that a vast majority of them will forget a majority of physics concepts shortly after leaving the classroom.  What are the enduring understandings and learnings that really matter? Here’s a list of my top 50. What key learnings are missing or overvalued?

  1. Learn to teach yourself.
  2. Think critically.
  3. Appreciate the beauty and patterns in the world.
  4. Be confident in your ability to attack an unfamiliar problem.
  5. Utilize the scientific method.
  6. Learn how to use a spreadsheet.
  7. Act like a professional
  8. Work productively in diverse groups.
  9. The universe is big.
  10. We aren’t.
  11. Trigonometry is useful.
  12. Calculus is just slopes and areas.
  13. Forces come in pairs.
  14. Doing work transfers energy.
  15. Ohms Law V=IR.
  16. Examine skeptically.
  17. Use a word processor.
  18. Learn to recognize what you don’t know (metacognition).
  19. Learn how to teach.
  20. Use and understand the metric system.
  21. Love learning.
  22. Be passionate about something.
  23. Estimate using orders of magnitude.
  24. Work productively, even when your team includes idiots.
  25. Forces cause accelerations.
  26. Mass/energy is always conserved.
  27. Waves transfer energy.
  28. Learn to create and analyze graphs.
  29. Use the Internet as a learning resource.
  30. Write coherently.
  31. Learn to study productively and efficiently.
  32. Velocity and acceleration are not the same thing.
  33. Learn from your mistakes.
  34. Draw and use free body diagrams.
  35. Gravity is an attractional force between masses.
  36. Momentum is conserved in any closed system.
  37. Understand the difference between electrical current and electrical potential.
  38. Transfer theoretical concepts to practical applications.
  39. Read and understand a technical text.
  40. Power is the rate at which you do work.
  41. Charge cannot be created or destroyed.
  42. Isaac Newton revolutionized our understanding of the world.
  43. Objects changing direction are accelerating.
  44. Reflect on your performance, and adjust your future habits accordingly.
  45. Horizontal and vertical motion are independent.
  46. Apply problem-solving methodologies in unfamiliar contexts.
  47. Learn to create and present effectively using Powerpoint.
  48. Take responsibility for your own learning.
  49. In the absence of air resistance, all objects fall at the same rate.
  50. There is nothing you cannot accomplish if you set your heart and mind to it.