A disc is free to rotate in a horizontal plane about a vertical axis through its centre. The moment of inertia of the disc about this axis is \(mk^{2}.\) Along one diameter is a narrow groove in which a particle of mass \(m\) slides freely. At time \(t=0,\) the disc is rotating with angular speed \(\Omega,\) and the particle is at a distance \(a\) from the axis and is moving towards the axis with speed \(V\), where \(k^{2}V^{2}=\Omega^{2}a^{2}(k^{2}+a^{2}).\) Show that, at a later time \(t,\) while the particle is still moving towards the axis, the angular speed \(\omega\) of the disc and the distance \(r\) of the particle from the axis are related by \[ \omega=\frac{\Omega(k^{2}+a^{2})}{k^{2}+r^{2}}\qquad\mbox{ and }\qquad\frac{\mathrm{d}r}{\mathrm{d}t}=-\frac{\Omega r(k^{2}+a^{2})}{k(k^{2}+r^{2})^{\frac{1}{2}}}. \] Deduce that \[ k\frac{\mathrm{d}r}{\mathrm{d}\theta}=-r(k^{2}+r^{2})^{\frac{1}{2}}, \] where \(\theta\) is the angle through which the disc has turned at time \(t\). By making the substitution \(u=1/r\), or otherwise, show that \(r\sinh(\theta+\alpha)=k,\) where \(\sinh\alpha=k/a.\) Hence, or otherwise, show that the particle never reaches the axis.