4 In this question you may assume without proof the standard results in Examination Formulae and Tables (MF2) for

\begin{itemize}
  \item the moment of inertia of a disc about an axis through its centre perpendicular to the disc,
  \item the position of the centre of mass of a solid uniform cone.
\end{itemize}

Fig. 4 shows a uniform cone of radius $a$ and height $2 a$, with its axis of symmetry on the $x$-axis and its vertex at the origin. A thin slice through the cone parallel to the base is at a distance $x$ from the vertex.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0166dd50-5069-47f4-a015-d01a9c54faf4-3_497_748_1283_699}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

The slice is taken to be a thin uniform disc of mass $m$.\\
(i) Write down the moment of inertia of the disc about the $x$-axis. Hence show that the moment of inertia of the disc about the $y$-axis is $\frac { 17 } { 16 } m x ^ { 2 }$.\\
(ii) Hence show by integration that the moment of inertia of the cone about the $y$-axis is $\frac { 51 } { 20 } M a ^ { 2 }$, where $M$ is the mass of the cone. [You may assume without proof the formula for the volume of a cone.]

The cone is now suspended so that it can rotate freely about a fixed, horizontal axis through its vertex. The axis of symmetry of the cone moves in a vertical plane perpendicular to the axis of rotation. The cone is released from rest when its axis of symmetry is at an acute angle $\alpha$ to the downward vertical. At time $t$, the angle the axis of symmetry makes with the downward vertical is $\theta$.\\
(iii) Use an energy method to show that $\dot { \theta } ^ { 2 } = \frac { 20 g } { 17 a } ( \cos \theta - \cos \alpha )$.\\
(iv) Hence, or otherwise, show that if $\alpha$ is small the cone performs approximate simple harmonic motion and find the period.

RECOGNISING ACHIEVEMENT

\hfill \mbox{\textit{OCR MEI M4 2011 Q4 [24]}}