\section{MC-8~~The Physical Pendulum}

\noindent Do either PART A or PART B but not both! 

\begin{center} {\bf PART A} \end{center}

\noindent OBJECTIVE: 
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}To measure the rotational inertia of a 
ring by swinging it as a pendulum from a point of the rim and to compare 
the value to that computed.
\end{enumerate}\latex{\vspace*{-.05in}}

\noindent APPARATUS:
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}{\it Basic equipment:} 
Metal rings, (full, half, and quarter rings); knife 
edge and 2-point supports.
\item[]  \latex{\vspace*{-.05in}}{\it Computer equipment:}   Personal computer set
to the MC-8 lab manual web-page; PASCO interface module;  photogate sensor and
extension jack(plugged into DIGITAL input \#1).
\end{enumerate}\latex{\vspace*{-.05in}}

\noindent INTRODUCTION: 
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}The period of a rigid body swinging on 
an axis as a pendulum is (for small amplitudes $\simeq 5 ^\circ$):
$$T=2\pi\sqrt{\frac{I_{0}}{Mgh}} $$
where
\begin{eqnarray*}
 T   & = & \mbox{period} \\
 I_0 & = & \mbox{rotational inertia relative to axis about which ring swings} \\
 M   & = & \mbox{mass of the body}  \\
 g   & = & \mbox{acceleration of gravity} \\
 h   & = & \mbox{distance from axis to center of mass.} 
 \end{eqnarray*}
\end{enumerate}

\noindent {REQUIRED INVESTIGATIONS:}
\begin{enumerate}
\item  \latex{\vspace*{-.05in}}Initiate the PASCO interface software by clicking on the telescope icon below (web version). There will be just a single table for recording the measured period.

\input{toolbar/m07_tb1}

\item Measure the period of the ring supported on the knife edge and 
swinging in its own plane.  From the period calculate $I_0$.  Use 
the parallel axis  theorem to also calculate $I_c$,  the rotational inertia about an axis through  the c.~of m.~and perpendicular to plane of the ring.  Compare this $I_c$ with the computed value of $I_c = M (r_1^2 + r_2^2)/2$ 
where the $r$'s are the inner and outer radii of the ring.  Do the two 
$I_c$'s agree within reasonable experimental error?  Explain.

\item Measure the period of the ring when supported on the two sharp points but
swinging perpendicular to its plane.  From this period calculate the rotational
inertia about an axis through those two points.  Then use the parallel axis
theorem to find the rotational inertia about a diameter of the ring.  How does
this rotational inertia compare with that about an axis perpendicular to the
plane of the ring and through the center of mass?

\begin{enumerate}
\item[] OPTIONAL PROBLEM:  
Prove that this relationship should exist by 
calculating the rotational inertias about the two orthogonal axes. 
\end{enumerate}

\item One circular hoop has been cut in sections.  Measure the period of 
a half-hoop when set at its midpoint on a knife edge; also the 
period of a quarter-hoop. (Be sure hoop sections have the same radius of 
curvature $r$).   
\end{enumerate}

\noindent{\em Proof that any section of a thin hoop has the same 
period if oscillating in 
the plane of the hoop:}

\latex{\vspace*{-.10in}}
\html{\begin{tabular}{p{4.8in}c}}
\latex{\parbox{0.60\linewidth}}
{
\html{  % Since equation arrays break in latex2html need two formats
\begin{enumerate}
\item[] Let
\begin{enumerate}
\item[$r~=$]  radius of the thin hoop            
\item[$O~=$]  axis about which it swings (knife edge)          	            
\item[$C~=$]  center of mass of the partial hoop             	    
\item[$m~=$]  mass of the partial hoop			    
\item[$h~=$ ] $OC =$ distance from c.~of~m.~to axis
\end{enumerate}
\item[] Note that $ I_A=  mr^2$ for any partial hoop.
\item[] The parallel axis theorem then gives:
$$ I_A = I_C  + m(r - h)^2 
\mbox{~~~or~~~} I_C = I_A - m(r - h)^2$$
\end{enumerate}
}
\begin{latexonly} % Since equation arrays breaks in latex2html need two formats
\noindent 
\begin{eqnarray*}
   \mbox{Let}~~~r & = & \mbox{radius of the thin hoop} \\                 
              O & = & \mbox{axis about which it swings (knife edge)} \\          	            
              C & = & \mbox{center of mass of the partial hoop}  \\                 	    
              m & = & \mbox{mass of the partial hoop} \\  			    
              h & = & OC = \mbox{distance from c.~of~m.~to axis}
\end{eqnarray*}
Note that $ I_A=  mr^2$ for any partial hoop.  \\ 
The parallel axis theorem then gives:
$$ I_A = I_C  + m(r - h)^2 
\mbox{~~~~~~or~~~~~~} I_C = I_A - m(r - h)^2$$
\end{latexonly}
}
\html{ & }
\latex{\parbox{0.40\linewidth}}
{\latex{\includegraphics[height=1.8in]{figs/m8-01.eps}}
\html{\includegraphics[height=2.0in]{figs/m8-01.eps}}}
\html{ \\ \end{tabular}}


\begin{enumerate}
\item[] \latex{\vspace*{-.15in}}Hence  \\
$$ I_C~=~ mr^2 - mr^2 + 2mrh - mh^2 ~=~ 2mrh - mh^2 ~. $$
When one substitutes this $I_C$ into
$$ T=2\pi \sqrt{I_0 /(mgh)} ~=~ 2\pi \sqrt{(I_C+mh^2)/(mgh)} $$
one finds
$$T = 2\pi \sqrt{(2mrh-mh^2+mh^2)/mgh} ~=~2\pi \sqrt{2r/g} $$
or a period independent of what fraction of a hoop is used!
\end{enumerate}\latex{\vspace*{-.05in}}


\noindent OPTIONAL PROBLEM:
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}For a thick hoop, the above relationship
does not hold exactly.  Show that the thickness of the laboratory hoops accounts for the small difference between $T$  for a whole hoop and $T$  for a half hoop. (This is a rather difficult problem.  For the older style laboratory hoops the finite thickness increases the period by $\simeq$ 1.8\%; the period of the half-hoop will be $\simeq$5.2\% larger, and the quarter-hoop over 22\% longer!)
\end{enumerate}\latex{\vspace*{-.05in}}

\begin{center}  {\bf PART B of M-8} \end{center}
\begin{center}  KATER'S REVERSIBLE PENDULUM  \end{center}
\noindent OBJECTIVE:  
To study conjugate centers of oscillation and to measure {\em g} accurately.\\

\noindent APPARATUS: 
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}Long rod with movable weights and knife edges; bearing 
surface (Kater's pendulum mount); infrared photogate \& support stand;
computer equipment as in PART A.
\end{enumerate}\latex{\vspace*{-.05in}}

\noindent INTRODUCTION: \smallskip 


\html{\begin{tabular}{p{0.6in}p{7.5in}}} %p{4.5in}
\latex{\parbox{0.20\linewidth}}
{
\html{\includegraphics[width=1.2in]{figs/m8-02.eps}}
\latex{\includegraphics[width=1.0in]{figs/m8-02.eps}}
}
\html{ & }
\latex{\parbox{0.75\linewidth}}
{
\begin{latexonly}
\noindent
When swung from O the period is:
\begin{eqnarray*} T_0 & = & 2\pi\sqrt{\frac{I_0}{mgh_0}}\; = \; 
2\pi \sqrt{\frac{I_C + mh_0^2}{mgh_0}}  \\
\mbox{When swung from P:}~~
 T_P &  = & 2\pi \sqrt{\frac{I_P}{mgh_P}}\; = 2\pi 
\sqrt{\frac{I_C +  mh_P^2}{mgh_P}}\ . 
\end{eqnarray*}
\end{latexonly}

\html{
\begin{enumerate}
\item[] When swung from O the period is:
\item[] $$ T_0~= 2\pi\sqrt{\frac{I_0}{mgh_0}}\; = \; 
2\pi \sqrt{\frac{I_C + mh_0^2}{mgh_0}}$$
\item[] When swung from P:
\item[] $$ T_P~=2\pi \sqrt{\frac{I_P}{mgh_P}}\; = 2\pi 
\sqrt{\frac{I_C +  mh_P^2}{mgh_P}}\ .$$ 
\end{enumerate}}

By substitution one easily verifies that $T_0 = T_P$ for two arrangements:
\medskip 

\latex{\hspace*{.2in}} 1) the trivial solution $h_0 = h_P$, and \medskip 

\latex{\hspace*{.2in}} 2) when $I_C = mh_0h_P$.  \medskip 

\noindent For the second case: \latex{\\}
\begin{equation} T_0 = T_P \; = \; 2\pi \sqrt{(h_0 + h_P)/g} \; = 
\; 2\pi \sqrt{L/g} 
\end{equation}
\noindent where $O$ and $P$ are called conjugate centers of oscillation.  
(See note at end of experiment.)  A 
measurement of the period ($T$) and distance 
($L = h_0 + h_P$) between knife edges then gives an 
accurate value of $g$. \smallskip

\noindent Empirically finding an $L$ to give $T_0 = T_P$ is tedious.  Instead, find an $L$ for which $T_0 \sim T_P$ and then eliminate $I_C$ from the first two equations above to give:}
\html{ \\ \end{tabular}}



\begin{enumerate}
\item[] \begin{equation}
\frac{4\pi^2}{g} \; = \; \frac{h_0 T_0^2 - h_P T_P^2}{h_0^2 - h_P^2}\; = 
\; \frac{T_0^2 + T_P^2}{2(h_0 + h_P)} + \frac{T_0^2 - T_P^2}
{2(h_0 - h_P)} ~~. \end{equation}

\item[] If one chooses an asymmetric geometry for the location of the weights, one can avoid $h_O \; \sim \; h_P$. Note then that for $T_0 \; \sim \;T_P$ the first term dominates, and an accurate value of $g$
results if one knows {\em accurately} $L = h_0 + h_P$, 
(the distance between the knife edges), and to much less accuracy the difference $ h_0 - h_P$.
\end{enumerate}

\noindent SUGGESTED PROCEDURE: 
\begin{enumerate}
\item \latex{\vspace*{-.08in}}Find the approximate c.~of~m.\ of the pendulum plus asymmetrically located weights.  Then set one knife edge as far from the c.~of~m.\ as feasible.  This avoids $h_0 \sim h_P$ which would make the last term of Eqn.~2 large.  \\{\it NOTE:  Keep this knife edge position fixed throughout the experiment}.

\item Determine the period of the pendulum, $T_0$, when swinging from this 
knife edge.  Keep amplitude small ($< 5^{\circ}$)  and use a photogate timer. 
\item Calculate the length, $L$, of a simple pendulum  to give the same period.
\item Set the second knife edge at this distance $L$ from the first knife  edge. 
\item Determine the new period of the pendulum, $T_P$, for swinging from the
second knife edge.  This period $T_P$ will {\em not quite equal} $T_0$ since
moving the second knife edge has changed slightly the c.~of m.~and hence $I_C,
\; h_0, \; h_P$, and also both $T_0$ and $T_P$.
\item Recalculate the length $L$ of the equivalent simple pendulum to give 
this new period $T_P$. Then reset the knife edge accordingly.
\item Redetermine the period about the first knife edge.
\item One can continue this iterative process (6 thru 7) until the two periods are arbitrarily close to each other, and hence $g$ is given by 
Eqn.~1.  However this is not necessary if one accurately finds the new c.~of~m.\ (e.g.\ by balancing the  pendulum on the knife edge for the hoops of M-8A) and then uses Eqn.~2.
\item Estimate your uncertainty in g and compare with accepted value (see M-4).

\end{enumerate}
\begin{center}
\hspace*{1in} \hrulefill \hspace*{1in}  \\ 
\end{center}

\begin{enumerate}
\item[]  NOTE:  These two conjugate centers of oscillation ($O$ and $P$) 
exist of course for any rigid physical pendulum, e.g.~{\em a baseball bat}: \\

Let one center of oscillation be where the batter grasps the bat.  The conjugate center of oscillation is then called the {\em center of percussion} because if the ball hits the bat at this point, the blow rotates the bat about the other center of oscillation, (i.e.~the batter's hands) and so the bat transmits no ``sting" to the hands.  However, if the ball hits very far from the {\em center of percussion}, the hands receive much of the blow and an unpleasant ``sting'' can result.
\end{enumerate}
