[Math] Lagrangian and hamiltonian problem involving mass constrained to frictonless parabolic surface

Question

Hello i was wondering if anyone could help me answer the following question specifically calculating the position of the mass m on the particle, i then assume i use the cylindrical polar coordinate formulation for the kinetic energy, is the potential energy given by $U=mgh$ where $h$ is the distance from the origin to the position of the mass? i'm not too sure what to do for parts b,c and d see the attached image for the problem

Answer 1

We have $z = kr^2$. Thus, $U = mgkr^2$ ($h$ should be height, not distance from the origin), and $$v = (\dot{r}\cos(\phi)-\dot{\phi}r\sin(\phi), \dot{r}\sin(\phi)+\dot{\phi}r\cos(\phi), 2k\dot{r}r)$$ giving us $$v^2 = (1+4k^2r^2)\dot{r}^2+r^2\dot{\phi}^2$$ Therefore, $$\mathcal{L} = T-U = \frac{1}{2}m\left((1+4k^2r^2)\dot{r}^2+r^2\dot{\phi}^2\right) - mgkr^2$$ The next parts essentially just involve using the right formulas. The generalized momenta are defined by $$p_r = \frac{\partial \mathcal{L}}{\partial \dot{r}} = m(1+4k^2r^2)\dot{r}$$ $$p_\phi = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = mr^2\dot{\phi}$$ and the Hamiltonian is given by $$\mathcal{H} = \dot{r}p_r+\dot{\phi}p_\phi-\mathcal{L}$$ (you'll have to algebraically rewrite $\mathcal{H}$ to get rid of the dotted terms)

For (c), Hamilton's equations of motion are given by $$\dot{p}_i = -\frac{\partial \mathcal{H}}{\partial i}$$ $$\dot{i}= \frac{\partial \mathcal{H}}{\partial p_i}$$ where $i=r,\phi$. Finally, (d) is essentially saying that $\dot{z} = 0$; you need to find the resulting relation between $r$ and $\dot{\phi}$ (Hamilton's equations of motion will be useful here).