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
- \(\omega\) is its angular speed
- \(m\) is its mass
- \(h\) is the distance of its centre of mass from A
- \(\theta\) is the angle the rod makes with the downward vertical
- \(g\) is the acceleration due to gravity
- \(I\) is a quantity known as the moment of inertia of the pendulum.
\includegraphics{figure_6}
- Use the expression for \(E\) to deduce the dimensions of \(I\). [4]
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.
- Use dimensional analysis to find the values of \(\alpha\), \(\beta\) and \(\gamma\). [5]
A class experiment finds that, when all other quantities are fixed, \(T\) is proportional to \(\frac{1}{\sqrt{m}}\).
- Determine whether this result is consistent with your answer to part (ii). [1]