\input{labpreamble}
\begin{document}


\subsubsection*{PHSX 221L\hfill Accelerated Motion With Two Bodies\hfill Name:\underline{\hspace*{2in}}}

% \noindent
% \underline{Title:} Accelerated Motion: A Two Body Problem

\noindent
\underline{Object:} To measure the acceleration of an Atwood Machine
and that of a cart-pulley mass combination and compare the results
with theory.

\noindent
\underline{Theory:} 
\vspace{-.1in} 

\noindent
The acceleration of a system can be calculated from Newton's 2nd law,
$\boldsymbol{F}_{\rm net}=m\boldsymbol{a}$, if all the forces are known.  In this
lab free body diagrams on each block must be constructed and Newton's
2nd law applied to each.

\vspace{-.3in} 
\begin{center}
\begin{picture}(350,200)(-140,0)
\thicklines
\put(50,150){\line(1,0){200}}  % horizontal edge of table
\put(250,150){\line(0,-1){100}}  % vertical edge of table
\put(270,170){\circle{10}}  % pulley
\put(250,150){\line(1,1){20}}  % pulley support
\thinlines
\put(100,150){\framebox(30,40){$m_{1}$}}  % block on top
\put(260,75){\framebox(30,50){$m_{2}$}}  % hanging block
\put(130,175){\line(1,0){140}}  % horizontal string
\put(275,125){\line(0,1){45}}  % vertical string
\put(160,100){\makebox(0,0){Cart-Pulley Combination}} % label
\end{picture}
\end{center}


\vspace{-3.in} 
\begin{center}
\begin{picture}(350,200)(180,0) % Atwood Machine
\thicklines
\put(200,150){\circle{39}}  % pulley
\put(200,150){\line(0,1){35}}  % pulley support
\thinlines
\put(165,20){\framebox(30,40){$m_{1}$}}  % left hanging block
\put(205,60){\framebox(30,50){$m_{2}$}}  % right hanging block
\put(180,60){\line(0,1){90}}  % left vertical string
\put(220,110){\line(0,1){40}}  % right vertical string
\put(260,165){\makebox(0,0){Atwood}}
\put(260,150){\makebox(0,0){Machine}}
\put(250,40){\makebox(0,0){$m_{2}>m_{1}$}}
\end{picture}
\end{center}
\vspace{-.4in} 

\underline{Atwood Machine}
\vspace{-.15in} 

\noindent 
The Atwood Machine, named after British mathematician George Atwood, 
(see example 5.9 in Serway)
consists of two unequal masses (say $m_{2}>m_{1}$), one on each side
of a suspended pulley and connected by a massless string.  If we 
also assume the friction in the pulley and the mass of the pulley are 
negligible, then the tension is the same on both sides of the pulley, 
and the free body diagrams and equations are simple: 
\begin{equation}\label{one}
	T-m_{1}g = m_{1}a,
\end{equation}
\begin{equation}\label{two}
	m_{2}g-T = m_{2}a
\end{equation}
where the positive direction is chosen to be in the direction of the
acceleration for each mass, and since they are connected by the
string, the magnitudes of the accelerations are equal.  Eliminating
$T$ from these equations gives the simplified theoretical acceleration
of the system as
\begin{equation}\label{simplea}
	a={(m_{2} - m_{1})g\over m_{1} + m_{2}}.
\end{equation}

\noindent
If mass is transferred incrementally from $m_{1}$ to $m_{2}$ then the
numerator will change and the denominator won't.  The actual
acceleration can be measured (computed) in each case by timing the
fall of $m_{2}$ from rest a certain distance and using $y=\frac{1}{2}
at^{2}$.  Rearranging equation \ref{simplea} to look a little more
like Newton's Second Law as given above produces $(m_{2} - m_{1})g =
(m_{1} + m_{2}) a$ so plotting a graph of $(m_{2} - m_{1})g$ ({\it
i.e.} the net force) vs.~$a$ should give a straight line with a slope
of $m_{1} + m_{2}$ ({\it i.e.} the total mass).

\noindent
However, if the mass (inertia) of the pulley and friction are not
neglected, the tension is not the same on both sides of the pulley and
some other terms enter the equations.  Equation \ref{one} becomes
\begin{equation}\label{oneprime}
	T_{1}-m_{1}g = m_{1}a.
\end{equation}
and equation \ref{two} becomes
\begin{equation}\label{twoprime}
	m_{2}g-T_{2} = m_{2}a,
\end{equation}

\noindent
There is also now a third equation (good thing, too, since there are
more unknowns to solve for) for the pulley.  The forces on the pulley
will include $T_{1}$ and $T_{2}$ and $f$ (the force of friction of the
bearing on the pulley).  The right side of Newton's Second Law is
trickier, since the pulley isn't accelerating in a translational way,
but rather in a rotational way---this means the inertia of the pulley
is more complex than its simple mass.  The equivalent mass (inertia)
of the pulley, $m_{eq}$, is equal to 1/2 of the pulley's mass if the
pulley is a solid disk, {\it i.e.,} $m_{eq}=\frac{1}{2}m_{pulley}$. 
Newton's Second Law for the pulley is
\begin{equation}\label{pulley}
	T_{2} - T_{1} - f = m_{eq}a.
\end{equation}

\noindent
Eliminating $T_{1}$ and $T_{2}$ from equations \ref{oneprime}, 
\ref{twoprime}, and \ref{pulley} gives
\begin{equation}
	a = {(m_{2}-m_{1})g - f \over m_{1} + m_{2} + m_{eq}}.
\end{equation}

\noindent
Now a plot of $F_{\rm net}$ for the system (still $(m_{2}-m_{1})g$) vs. $a$ will 
produce a straight line with a slope of $ m_{2} + m_{1} + m_{eq}$ and 
an intercept of $f$.  See also example 10.13 in Serway.
	

\underline{Cart-Pulley Combination}
\vspace{-.15in} 

\noindent
This apparatus consists of a cart, which is free to roll or slide on 
a horizontal plane, connected by a massless string over a pulley to a hanging 
mass.  

\noindent
Before doing the experimental procedure, do mathematical derivations 
similar to the ones for the Atwood machine---once assuming a 
massless, frictionless pulley and a frictionless plane, and then again 
taking friction and the mass of the pulley into account.  (See example 
5.14 in Serway.)

\noindent
\underline{Procedure:}
\vspace{-.1in} 

\underline{Atwood Machine}
\vspace{-.25in} 
\begin{enumerate}

	\item Set up the Atwood apparatus with the figure as your guide. 
	Put the same amount of mass (about 150 g, including the hanger) on
	each side of the pulley, but use some small masses as part of
	$m_{1}$ so they can be transferred.  Use a pan balance to check that 
	$m_{1}$ and $m_{2}$ are equal to within a few hundredths of a gram.  
	If they aren't, adjust the lighter one with small pieces of paper or 
	tape until they are the same.
	
	\item Estimate the friction in the pulley by adding a small mass,
	$m_{f}$ (the exact amount to be determined by trial and error), to
	$m_{2}$ to unbalance the system and cause $m_{2}$ to descend at a
	constant rate.  Then $f=m_{f}g$.  Remove $m_{f}$ for the rest of
	the procedure.
	
	\item\label{transfer} Transfer one or two grams of mass from
	$m_{1}$ to $m_{2}$.  Then, holding $m_{1}$ on the cushion,
	simultaneously release $m_{1}$ and start the timer.  Stop the
	timer when $m_{2}$ hits the cushion; also immediately stop the
	pulley.  Repeat this measurement a few times.  Calculate $a$ from 
	$y=\frac{1}{2}at^{2}$.
	
	\item Repeat procedure \ref{transfer} for various larger mass 
	transfers.
	
\end{enumerate}


\underline{Cart-Pulley Combination}
\vspace{-.25in} 
\begin{enumerate}

\item Set up the horizontal track and adjust it to be level.  Pile some 
weights (masses) on the cart.

	\item Estimate the friction in the pulley by adding a small mass,
	$m_{f}$ (the exact amount to be determined by trial and error), to
	$m_{2}$ to cause $m_{2}$ to descend at a
	constant rate.  Then $f=m_{f}g$.  Remove $m_{f}$ for the rest of
	the procedure.

\item Shift mass from the rolling cart to the hanger in increments and 
measure the acceleration for each situation using $d = \frac{1}{2}a 
t^{2}$.  Keep the total mass (mass of cart plus mass on cart plus 
hanging mass) constant.

% \item Plot the weight of the hanging mass (minus the frictional force) 
% vs.~acceleration and compute the slope of the line (using a least 
% squares analysis).
% 
% \item For different values of the mass, measure the position of the 
% accelerating cart at different times and analyze this data (as in 
% experiment 2) to compute the acceleration.
% 
% \item Compare each result with that calculated from Newton's Second 
% Law.  A least squares fit and a graph would be helpful.

\end{enumerate}

\noindent
\underline{Analysis}
\vspace{-.1in} 

For each apparatus (both the \underline{Atwood Machine} and the
\underline{Cart-Pulley Combination}):

\vspace{-.25in} 
\begin{enumerate}
	\item Plot the net force vs.~acceleration.
	
	\item Use a least squares analysis to compute the slope and the 
	intercept of the line and interpret what these mean physically.
	
	\item Compare the least squares slope with the theoretical value.
\end{enumerate}


\noindent
\underline{Questions:} 
\vspace{-.15in}
\begin{enumerate}
	\item Instead of verifying Newton's Second Law, how could this 
	experiment be used to measure $g$ (as Atwood did)?  How accurate 
	would it be?
	\item Is there a way to make the Atwood Machine accelerate at a 
	higher rate than 9.8 m/s$^{2}$?
	\item The Atwood Machine experiment can be reconfigured to keep the force 
	($(m_{2}-m_{1})g$) constant while varying the accelerated mass ($m_{1} + 
	m_{2}$) by adding the same amount of mass to both sides.  What kind of 
	analysis would be done in this case?  What would be the abscissa 
	($x-$coordinate) and 
	ordinate ($y-$coordinate) and slope of the graph?
	\item Using  $y=\frac{1}{2}at^{2}$ assumes that $a$ is constant.  Is 
	it?  Is it a reasonable approximation?  How can you know?
	\item How important was it to account for friction and the mass of 
	the pulley?
\end{enumerate}


\noindent
\underline{Conclusions:} 
\vspace{-.15in}

Summarize and evaluate your experiment.  What did you learn?  Think 
hard.

% \newpage \ 
\end{document}
