Laura Mann has a background in the humanities. She's currently in a blended MS/BS computer science program where she's learned HTML, XHTML, mySQL, asp.net, OOP, MVC, Ruby, ROR, Python, UNIX, Java and other mind blowingly abstract things. She thoroughly enjoys infrastructural projects and books: you can see some her suggestions:

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\chapter{Lab 9}
\Large
Experiment 9:\\
Conservation of Angular Momentum\\
Date: 3/06/2013\\
Partners: Gerardo Nunez, Juan Herandez\\
\section{Objectives}
\paragraph{objective} \hspace{0 pt}
\normalsize
The purpose of this experiment to study and test
the Law of Conservation of Angular Momentum for a
system in rotational motion. Part A of this experiment
was skipped because we lacked the equipment. Part B
is a quantitative section.\\

\section{Apparatus}
\paragraph{Items} \hspace{0 pt}
Apparatus similar to one used in uniform circular
motion experiment. It includes a horizontal bar
centred on the vertical rotating shaft. It has
end clamps, springs, sliding masses (needs to be
greased), string anchor, shaft, pasco photogate,
pulley, bearing and a weight that is
used with smart pulley software. Veriner caliper\\

\section{Data}
\large
\paragraph{Initial Measurements} \hspace{0 pt}
\\
Before the experiment we recorded the weight of
the hanging mass and the radius of the shaft
(with a vernier caliper).\\
\\
The hanging mass is \textbf{m} = .1 kg $\pm .0001 kg$\\
The Diameter is = .032 m $\pm .001 m$,\\
So, are radius \textbf{R} = .016 m $\pm .001 m$\\
\\
\paragraph{Extended configuration of the sliding masses} \hspace{0 pt}
\\

After we ran the experiment the line that the computer processed
was fairly linear and there was not a notable decrease in the slope of the line so we used the data from the computer for $\omega$.
\\
Here, $\omega$ = .104 $\frac{rad}{s}$

\paragraph{sliding masses tied together and then released} \hspace{0 pt}
\\
For this part of the experiment we found the values for
the sliding masses compressed together and then after
they are released by burning the string in motion.\\
These measurements are first taken from the computer
graph and then matched to the values from the computer's
chart.\\
\begin{table}[H]
\begin{tabular}{| c | c | c | c | c | c | c | }
\hline
& $t_{i} (s)$ & $t_{f} (s)$ & $\omega_{i} (rad/s)$ & $\omega_{f} (rad/s)$ \\

\hline
graph & 4.2 - .2 = 4 & 7-4.8 = 2.2 & 3.6-1.3 = 2.3
& 3.1-1.9 = 1.2 \\

\hline
matched to data & 4.27-.24 = 4.03 & 7.08-4.77 = 2.31 &
3.5-1.31 = 2.19 & 2.94-1.91 = 1.03 \\
\hline

\end{tabular}
\caption{Measurements of before and after string is burned.
The numbers from the graph data are $\pm .2$ }
\label{tab:template}
\end{table}

\paragraph{Values of $\alpha$ and the moment of inertia} \hspace{0 pt}
\\
Here, $\alpha$ is the slope of the line generated by
the two slopes taken from the computer data graph compared
to the data matching it in the chart from the computer.
Here we calculate the moment of initial and final inertia
and compare the results using ,
$I= \frac{(mg-mR\alpha)R}{\alpha}$
\begin{table}[H]
\begin{tabular}{| c | c | c | c | c | c |}
\hline
& $\alpha_i (rad/s^2)$
& $\alpha_{f} (rad/s^2)$ & $I_{i} (kg*m^2)$ & $I_{f} (kg*m^2)$ & $I_{i}\omega_{i} = I_{i}\omega_{i} (kg*m^2s^-1)$ \\
\hline
graph & .575 & .545 & .027 & .029 & .0155 = .0158\\
\hline
matched to data & .543 & .446 & .029 & .035 & .0157 = .0156 \\
\hline
\end{tabular}
\caption{Measurements of graph are $\pm .2$. The moment
of inertia is $\pm .0022$ and the inertia is
$\pm .0022$ for the matched data results.}
\label{tab:template}
\end{table}

\section{Analysis}
\paragraph{Question one} \hspace{0 pt}
Using the more accurate measurements from the table of values
rather then the eyeballed graph data, the angular momentum
just before the string is burned is :\\
$L_{total, I}= I_{I}W_{I} = .0157 (kg*m^2s^-1)\pm .0022 (kg*m^2s^-1)$
\paragraph{Question two} \hspace{0 pt}
Using the more accurate measurements from the table of values
rather then the eyeballed graph data, the angular momentum
just after the string is burned is :\\
$L_{total, F}= I_{F}W_{F} = .0156 (kg*m^2s^-1)\pm .0022 (kg*m^2s^-1)$
\paragraph{Question three} \hspace{0 pt}
\\
Comparing the initial and final values of angular momentum
by their percent difference is equal to:
$\frac{|L_{total, F} - L_{total, I} | }
{ \frac{1}{2}(L_{total,I} + L_{total,F}) } X 100
= \frac{|.0156 - .0157 |}{\frac{1}{2} (.0156+.0157) } x100 =
.64 \%
$

\paragraph{Question four}
Calculate the average rotational deceleration of the "T":
The average rotational deceleration of the "T"=
$\overline{\alpha} = \frac{\omega_{F} - \omega_{I}}{\Delta t}
= \frac{1.03-2.19} {2.31-4.03} = .67 (rad/m^2)$

The average is very close to our percent difference
between the initial and final angular accelerations
which makes sense because we think about angular
momentum like a area under a curve then in order
to be conserved the sum of the area of the
moment before and after should. Since that wasn't
quite true in our experiment there is a residual.
cancel one another out.
\paragraph{General Question}
Suppose the pulley has a friction equivalent to 2 gmwt decrease in
the tension of the string as it passes over it. Would this
increase, decrease, or not affect the calculation of
the total moments of inertia in the two configurations of
the sliding masses?
\\
This is a tricky question, but I think this has to do
with the conversation of angular momentum in that
if there are no external forces acting on the system
the total angular momentum is conserved. So, the
string that wraps around our axis causing it to
rotate counter clockwise (? I think)
pulling to the right without friction creates a
torque about the axis. Since there's a mass on
the pulley friction (static I think) is
a external force creating torque. So, the
string now has two different tensions. Therefore,
I think the force exerted on the axis by the
weighted pulley is less due to friction and
thusly, the moments of inertia are decreased.

\section{Results}
\paragraph{general conclusions} \hspace{0 pt}
I think that some of our measurements were off because
the "T" needed to be greased and I think we would have seen
a more dramatic decrease in angular acceleration if it