#set document(
    title: "Challenge Problem: Toilet Paper Roll",
    author: "<NAME>",
    date: datetime.today(),
)
#set page(paper: "us-letter")
#set text(font: "New Computer Modern")
#set quote(block: true)
#show quote: set pad(x: 2em)

#align(center)[
    #text(size: 24pt)[Challenge Problem: Toilet Paper Roll]
]

= Purpose
The purpose of this experiment is to predict the falling rate of an unrolling
toilet paper roll. The success of the calculations will be determined by a
video recording of an unrolling roll and a falling roll dropped from different
heights.

= Materials
In solving this problem, the following materials were used:
- Two toilet paper rolls
- Meter stick
- Scale
- 30 FPS Camera

= Procedure
+ Measure the outer radius of the toilet paper roll. Record the value in
    @measurements.
+ Measure the inner (hollow) radius of the toilet paper roll. Record the value
    in @measurements.
+ Measure the mass of the toilet paper roll using the scale. Record the value in
    @measurements.
+ Use the measured values to calculate the height to drop the held roll from to
    match the fall time of the free-falling roll.
+ Drop the two rolls from their respective heights. Record the fall using the
    camera.
+ Determine the maximum possible $Delta t$ between the two rolls impacting the
    ground using the difference in frames from the video.

= Calculations
== Part 1
#grid(
    columns: (auto, auto),
    rows: 30%,
    quote(
        attribution: [from the problem handout],
    )[
        A toilet paper roll, dropped while holding the paper will #emph()[not] fall
        with an acceleration of $g=9.8"m/s/s"$. Using your new knowledge of torque and
        using the extended free body diagram for the roll, determine the actual linear
        acceleration that the roll would experience. Write your result in terms of $g$
        and numerical constants such as measured dimensions of the toilet paper roll.
    ],
    figure(caption: [Diagram from Problem Handout])[
        #image("assets/toilet-paper-challenge-problem/diagram.png")
    ],
)

=== Rotational Inertia
First, the rotational inertia of the toilet paper roll must be calculated for
the next part.
$ I = integral x^2 dif m $

For a partially-hollow cylinder, rotational inertia can be calculated using
volumetric density:
$ phi = m/V $
$ I = integral x^2 phi dif V $

Substituting an expression for change in volumetric density is as follows:
$ I = integral x^2 phi dot 2 pi x h dif x $

Since the mass of a partially-hollow cylinder is contained within two rings,
the bounds of integration are from the inner ring to the outer ring.
$ I &= integral_r^R x^2 phi dot 2 pi x L dif x \
  &= 2 phi pi L integral_r^R x^3 dif x $

Integrating yields the following:
$ I &= 2 phi pi L dot (1/4 R^4 - 1/4 r^4) \
  &= 1/2 dot m/V dot pi dot L dot (R^4 - r^4) \
  &= 1/2 dot m/(pi L (R^2 - r^2)) dot pi L dot (R^4 - r^4) \
  &= 1/2 m dot (R^4 - r^4)/(R^2 - r^2) \
  &= 1/2 m dot ((R^2 + r^2)(R^2 - r^2))/(R^2 - r^2) \
  &= 1/2 m (R^2 + r^2) $

=== Acceleration Function <accel-function>
Applying Newton's 2nd Law for both the translational and rotational motion
yields the following two equations:
$ a = F_"net"/m "and" alpha = tau_"net"/I $

Linear acceleration can be solved from rotational acceleration:
$ alpha = a/R $
$ a = alpha R = (tau_"net" R)/I $

Since the tension force from the held toilet paper is the only torque exerted
on the rotating system, $tau_"net" = limits(T)^arrow R$. // TODO there has to be a better way to do vector notation?
$ a = (limits(T)^arrow R^2)/I $

The translational acceleration equation can then be solved for $limits(T)^arrow$.
$ a = F_"net"/m = (m g - limits(T)^arrow)/m $
$ limits(T)^arrow = m(g-a) $

$limits(T)^arrow$ can then be substituted into the earlier equation.
$ a = (m(g-a) R^2)/I $

Solving for $a$ yields the following:
$ a = (m g R^2)/(I + m R^2) $

Finally, substituting in rotational inertia results in the following:
$ a &= (m g R^2)/(1/2 m (r^2 + R^2) + m R^2) \
  &= (2 g R^2)/(r^2 + R^2 + 2R^2) \
  &= (2 g R^2)/(r^2 + 3R^2) $

== Part 2 <drop-height-calculations>
#quote(
    attribution: [from the problem handout],
)[
    If another roll is dropped without being held (accelerating under free fall)
    from $1"m "$, from what height should the held roll be dropped so that, if
    released at the same time, they hit the ground simultaneously?
]

Since the falling toilet paper roll experiences uniform acceleration and
begins falling from rest, the time which it takes to reach the ground can be
calculated using the following basic kinematic equation.
$ Delta y_"free" = 1/2 g t^2 $

The equation can then be solved for $t$.
$ t = sqrt((2 Delta y_"free")/g) $

Since the drop height of the held roll also experiences near-uniform
acceleration, the drop height can be calculated using the same equation as
above, substituting in the time from the above drop.
$ Delta y_"held" &= 1/2 a t^2 \
               &= 1/2 a (sqrt((2 Delta y_"free")/g))^2 \
               &= a/2 dot (2 Delta y_"free")/g \
               &= (a Delta y_"free")/g $

= Results
== Measurements
#figure(caption: [Toilet Paper Roll Measurements])[
    #table(
        columns: (auto, auto),
        align: (right, left),
        [Outer Radius ($R$)],
        [0.054m],
        [Inner Radius ($r$)],
        [0.020m],
        [Mass ($m$)],
        [0.175kg],
    )
] <measurements>

== Calculations
=== Part 1
Substituting measured values into the #link(label("accel-function"))[previously calculated acceleration function] results
in the following:
$ a &= (2 dot 9.8"m/s"^2 dot 0.054^2"m "^2)/(0.020^2"m "^2 + 3 dot 0.054^2"m "^2) \
  &= 6.249 "m/s"^2 $

=== Part 2
Substituting measured values as well as the acceleration calculated above into
the #link(label("drop-height-calculations"))[drop height equation] yields the
following:
$ Delta y_"held" &= (6.249 "m/s"^2 dot 1"m ")/(9.8"m/s"^2) \
               &= 0.638"m " $

== Drops
Both toilet paper rolls were dropped from their respective calculated heights
to validate the results obtained in this experiment. As predicted, both rolls
impacted the floor nearly simultaneously—impact occurred on the same frame of
the 30 FPS video captured, satisfying the goal of this experiment.

@impact-frame shows the frame of the video during which both rolls impacted
the floor. The full video is available online #link(
    "https://drive.google.com/file/d/1hEyK2Em6v7i22vvdASuWQ_-FS7laz5oN/view?usp=drive_link",
)[here].

Since both rolls impacted the floor within one frame of a 30 FPS video, the
margin of error between their contact, $Delta t$, is at most $33.3"ms"$.

#figure(
    caption: [Frame of Impact for Both Rolls],
)[
    #image("assets/toilet-paper-challenge-problem/impact.png", height: 50%)
] <impact-frame>
