3 A uniform rod AB of mass \(m\) and length \(4 a\) is hinged at a fixed point C , where \(\mathrm { AC } = a\), and can rotate freely in a vertical plane. A light elastic string of natural length \(2 a\) and modulus \(\lambda\) is attached at one end to B and at the other end to a small light ring which slides on a fixed smooth horizontal rail which is in the same vertical plane as the rod. The rail is a vertical distance \(2 a\) above C . The string is always vertical. This system is shown in Fig. 3 with the rod inclined at \(\theta\) to the horizontal. \begin{figure}[h] \end{figure}

1. Find an expression for \(V\), the potential energy of the system relative to C , and show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = a \cos \theta \left( \frac { 9 } { 2 } \lambda \sin \theta - m g \right)\).
2. Determine the positions of equilibrium and the nature of their stability in the cases
   (A) \(\lambda > \frac { 2 } { 9 } m g\),
   (B) \(\lambda < \frac { 2 } { 9 } m g\),
   (C) \(\lambda = \frac { 2 } { 9 } m g\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{cb86219c-e0b1-4f75-b8b2-50b5a233aa54-3_522_755_342_696} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{figure}
3. Show, by integration, that the moment of inertia of the cone about its axis of symmetry is \(\frac { 3 } { 10 } M a ^ { 2 }\). [You may assume the standard formula for the moment of inertia of a uniform circular disc about its axis of symmetry and the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.] A frustum is made by taking a uniform cone of base radius 0.1 m and height 0.2 m and removing a cone of height 0.1 m and base radius 0.05 m as shown in Fig. 4.2. The mass of the frustum is 2.8 kg . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{cb86219c-e0b1-4f75-b8b2-50b5a233aa54-3_391_517_1352_813} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure} The frustum can rotate freely about its axis of symmetry which is fixed and vertical.
4. Show that the moment of inertia of the frustum about its axis of symmetry is \(0.0093 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). The frustum is accelerated from rest for \(t\) seconds by a couple of magnitude 0.05 N m about its axis of symmetry, until it is rotating at \(10 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
5. Calculate \(t\).
6. Find the position of G , the centre of mass of the frustum. The frustum (rotating at \(10 \mathrm { rad } \mathrm { s } ^ { - 1 }\) ) then receives an impulse tangential to the circumference of the larger circular face. This reduces its angular speed to \(5 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
7. To reduce its angular speed further, a parallel impulse of the same magnitude is now applied tangentially in the horizontal plane through G at the curved surface of the frustum. Calculate the resulting angular speed.