\section{M-12~~Torsion Pendulum and Shear Modulus}


\noindent {OBJECTIVES}:  To study a torsion pendulum and to measure a shear 
modulus. \\

\noindent {APPARATUS}: 
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}} 
Wall mounted torsion pendulum 
(large disk plus suspension assembly); rods 
of different lengths, diameters and material; 
a ring and two cylindrical masses; timer, tally counter.
\end{enumerate}

\noindent {INTRODUCTION}:  
\begin{enumerate}
\item[] \latex{\vspace*{-.05in}}If a torque $\tau$ twists a rod, then the 
angle $\theta$ through which 
the rod twists is proportional to the torque if the elastic limit is not 
exceeded:  Thus $\tau_{\mbox{\tiny applied}} = \kappa \theta$ where $\kappa$ 
is the torsion constant.

\begin{figure}[!h] \centering
\includegraphics[width=4.8in]{figs/m12-1.eps}
\caption{(a) A thin-walled hollow cylinder. (b) The cylinder is twisted by
applying a torque. (c) The angle of shear $\phi$ and the
angle of twist $\theta$.}
\end{figure}

The equal and opposite reaction torque is 
$$\tau_{\mbox{\tiny{el}}} \; = \; - \kappa \theta \; = \; I \alpha \; = \; 
I \frac{d^2\theta}{dt^2} \;.$$
Since this is the same mathematical form as 
$$F_{\mbox{\tiny el}} \;= \; -kx \;= \; ma \; = \; m\frac{d^2x}{dt^2}, $$ 
it has a similar solution: $\theta = \theta_0 \sin \omega t$ where 
$\omega = \sqrt{\kappa/I} = 2 \pi f = 2 \pi/T$. Thus the system 
undergoes angular simple harmonic motion, SHM, with 
$T = 2 \pi \sqrt{I/ \kappa}$
and $T$ is independent of amplitude, (in contrast to a simple pendulum where
$T$ is only approximately independent of amplitude.)  For simple 
geometries, one readily calculates the rotational inertia $I$.  Hence
measuring $T$ can determine precisely the torsion constant, $\kappa$.
\noindent To relate the torsion constant $\kappa$ to the shear modulus $M_s$, 
consider first 
a tangential force $dF_t$ applied to the end of a thin ($dr$ thick) hollow 
tube. This tangential force/area is the shear 
stress and equals $dF_t/(2\pi r~dr$). The resultant shear strain 
is the angle $\phi$ (see Fig.~1(c)):



$$\phi =  \frac{a}{L} = \frac{r\theta}{L}~~.$$ 

Multiplying the stress by $r/r$ gives 
$$\frac{rdF_t}{2\pi r^2dr} = \frac{d\tau}{2\pi r^2dr} = \mbox{shear stress.}$$ 
Hence the shear modulus $M_s$ becomes:
$$M_s = \frac{\mbox{shear stress}}{\mbox{shear 
strain}} = \frac{d\tau / 2\pi r^2dr}{r \theta/L} = \frac{Ld \tau}{
\theta (2\pi r^3)dr} \hspace{.5in} \mbox{[tube]}$$
and $Ld\tau = M_s \theta(2\pi r^3 dr)$. If one integrates from 0 to $R$, 
the tube then becomes a solid rod:
$$L \tau = M_s \theta \; 2\pi \left( \frac{R^4}{4}\right)
\hspace{.5in} \mbox{[solid rod]}.$$
Thus
$$M_s =\frac{2L}{\pi R^4} \ \frac{\tau}{\theta} = 
\left(\frac{2L}{\pi R^4}\right) \kappa~~.$$
\end{enumerate}

\noindent{SUGGESTED EXPERIMENTS:} (only one of which need be performed)
\begin{enumerate}
\item \latex{\vspace*{-.05in}} 
Determine the shear modulus of one or more materials by measuring
the period of a torsion pendulum with the materials as the suspension.
\item Use the result of Exp.~1 (or from a  table of shear moduli) 
to calculate the 
period of a torsion pendulum with the ring (or the two cylindrical 
masses) placed on the disc.  Compare with the measured value of the period.
\item Check the expression for shear modulus,
$$M_s = \frac{2L\tau}{\pi R^4 \theta} =  \frac{2L}{\pi R^4} 
\kappa  .$$
where $\kappa = \tau/\theta$ is the torsion constant, R and L are the 
radius and length 
of the rod, $\tau$ is the torque and $\theta$ is the angular displacement.  
Use rods of the same material but different lengths and diameters.
\item Determine how the period of the torsion pendulum depends on the length 
and diameter of the rod; also on the rotational inertia of the suspended 
mass.  Check that the period is independent of angular amplitude (as long as 
one is below the elastic limit).  Compare with theory.
\end{enumerate}
\noindent {QUESTION:}  \begin{enumerate}
\item[] \latex{\vspace*{-.05in}}Is it reasonable to ignore the rotational 
inertia of the 
suspending rod?  Check your answer by estimating the ratio of the rod's 
rotational inertia to that of the large disc.
\end{enumerate}

