Fig. 6 shows a pendulum which consists of a rod AB freely hinged at the end A with a weight at the end B. The pendulum is oscillating in a vertical plane. The total energy, $E$, of the pendulum is given by

$$E = \frac{1}{2}I\omega^2 - mgh\cos\theta,$$

where
\begin{itemize}
\item $\omega$ is its angular speed
\item $m$ is its mass
\item $h$ is the distance of its centre of mass from A
\item $\theta$ is the angle the rod makes with the downward vertical
\item $g$ is the acceleration due to gravity
\item $I$ is a quantity known as the moment of inertia of the pendulum.
\end{itemize}

\includegraphics{figure_6}

\begin{enumerate}[label=(\roman*)]
\item Use the expression for $E$ to deduce the dimensions of $I$. [4]
\end{enumerate}

It is suggested that the period of oscillation, $T$, of the pendulum is given by $T = kI^\alpha(mg)^\beta h^\gamma$, where $k$ is a dimensionless constant.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Use dimensional analysis to find the values of $\alpha$, $\beta$ and $\gamma$. [5]
\end{enumerate}

A class experiment finds that, when all other quantities are fixed, $T$ is proportional to $\frac{1}{\sqrt{m}}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Determine whether this result is consistent with your answer to part (ii). [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major  Q6 [10]}}