3
\includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-2_439_444_1318_822}
A uniform rod \(A B\) has mass \(m\) and length \(4 a\). The rod can rotate in a vertical plane about a smooth fixed horizontal axis passing through \(A\). One end of a light elastic string of natural length \(a\) and modulus of elasticity \(\lambda m g\) is attached to \(B\). The other end of the string is attached 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 \(3 a\) above \(A\). The string is always vertical and the rod makes an angle \(\theta\) radians with the horizontal, where \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\) (see diagram).
- Taking \(A\) as the reference level for gravitational potential energy, find an expression for the total potential energy \(V\) of the system, and show that
$$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 m g a \cos \theta ( 4 \lambda ( 1 + 2 \sin \theta ) - 1 ) .$$
Determine the positions of equilibrium and the nature of their stability in the cases
- \(\lambda > \frac { 1 } { 12 }\),
- \(\lambda < \frac { 1 } { 12 }\).
\includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-3_392_689_269_671}
The diagram shows the curve with equation \(y = \frac { 1 } { 2 } \ln x\). The region \(R\), shaded in the diagram, is bounded by the curve, the \(x\)-axis and the line \(x = 4\). A uniform solid of revolution is formed by rotating \(R\) completely about the \(y\)-axis to form a solid of volume \(V\). - Show that \(V = \frac { 1 } { 4 } \pi ( 64 \ln 2 - 15 )\).
- Find the exact \(y\)-coordinate of the centre of mass of the solid.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{57323af2-8cf3-4721-b2c8-a968264be343-4_385_741_269_646}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{figure}
Fig. 1 shows part of the line \(y = \frac { a } { h } x\), where \(a\) and \(h\) are constants. The shaded region bounded by the line, the \(x\)-axis and the line \(x = h\) is rotated about the \(x\)-axis to form a uniform solid cone of base radius \(a\), height \(h\) and volume \(\frac { 1 } { 3 } \pi a ^ { 2 } h\). The mass of the cone is \(M\). - Show by integration that the moment of inertia of the cone about the \(y\)-axis is \(\frac { 3 } { 20 } M \left( a ^ { 2 } + 4 h ^ { 2 } \right)\). (You may assume the standard formula \(\frac { 1 } { 4 } m r ^ { 2 }\) for the moment of inertia of a uniform disc about a diameter.)
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{57323af2-8cf3-4721-b2c8-a968264be343-4_501_556_1238_726}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{figure}
A uniform solid cone has mass 3 kg , base radius 0.4 m and height 1.2 m . The cone can rotate about a fixed vertical axis passing through its centre of mass with the axis of the cone moving in a horizontal plane. The cone is rotating about this vertical axis at an angular speed of \(9.6 \mathrm { rad } \mathrm { s } ^ { - 1 }\). A stationary particle of mass \(m \mathrm {~kg}\) becomes attached to the vertex of the cone (see Fig. 2). The particle being attached to the cone causes the angular speed to change instantaneously from \(9.6 \mathrm { rad } \mathrm { s } ^ { - 1 }\) to \(7.8 \mathrm { rad } \mathrm { s } ^ { - 1 }\). - Find the value of \(m\).
\includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-5_534_501_255_767}
A triangular frame \(A B C\) consists of three uniform rods \(A B , B C\) and \(C A\), rigidly joined at \(A , B\) and \(C\). Each rod has mass \(m\) and length \(2 a\). The frame is free to rotate in a vertical plane about a fixed horizontal axis passing through \(A\). The frame is initially held such that the axis of symmetry through \(A\) is vertical and \(B C\) is below the level of \(A\). The frame starts to rotate with an initial angular speed of \(\omega\) and at time \(t\) the angle between the axis of symmetry through \(A\) and the vertical is \(\theta\) (see diagram). - Show that the moment of inertia of the frame about the axis through \(A\) is \(6 m a ^ { 2 }\).
- Show that the angular speed \(\dot { \theta }\) of the frame when it has turned through an angle \(\theta\) satisfies
$$a \dot { \theta } ^ { 2 } = a \omega ^ { 2 } - k g \sqrt { 3 } ( 1 - \cos \theta ) ,$$
stating the exact value of the constant \(k\).
Hence find, in terms of \(a\) and \(g\), the set of values of \(\omega ^ { 2 }\) for which the frame makes complete revolutions.
At an instant when \(\theta = \frac { 1 } { 6 } \pi\), the force acting on the frame at \(A\) has magnitude \(F\). - Given that \(\omega ^ { 2 } = \frac { 2 g } { a \sqrt { 3 } }\), find \(F\) in terms of \(m\) and \(g\).
\section*{END OF QUESTION PAPER}