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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 ®:

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

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 derivation is correct, per our visual confirmation with a high speed video camera:

You can follow the original blog response at Physics In Flux.

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