A particle of mass m slides on the surface of a smooth bowl with equation p2 = az in cylindrical polar coordinates p, 0 and z, with the z-axis pointed

vertically upwards, where a is a positive constant. (i) Briefly explain why \frac{1}{2} m\left(\dot{\rho}^{2}+\rho^{2} \dot{\theta}^{2}+\dot{z}^{2}\right)+m g z=\text { constant } \rho^{2} \dot{\theta}=\text { constant. } ) The particle is initially at height a above the lowest point of the bowl, moving horizontally along the bowl with speed pé = v. Show that \dot{\rho}^{2}+\rho^{2} \dot{\theta}^{2}+\dot{z}^{2}+2 g z=v^{2}+2 g a \rho^{2} \dot{\theta}=a v Hence deduce that \left(1+\frac{a}{4 z}\right) \ddot{z}-\frac{a \dot{z}^{2}}{8 z^{2}}-\frac{a v^{2}}{2 z^{2}}+g=0 ) If v² > 2ag will the particle rise or fall initially? What is the particle's maximum height in the bowl?