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]
\includegraphics[alt={},max width=\textwidth]{cb86219c-e0b1-4f75-b8b2-50b5a233aa54-2_387_613_1763_767}
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\caption{Fig. 3}
\end{figure}
- Find an expression for \(V\), the potential energy of the system relative to C , and show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = a \cos \theta \left( \frac { 9 } { 2 } \lambda \sin \theta - m g \right)\).
- 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} - 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. - 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 \(10 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
- Calculate \(t\).
- 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 \(5 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
- 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.