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I Hamiltonian for mass on a smooth fixed hemisphere

  1. Nov 16, 2016 #1
    I am trying to figure out how to get the Hamiltonian for a mass on a fixed smooth hemisphere.

    Using Thorton from example 7.10 page 252

    My main question is about the Potential energy= mgrcosineθ is the generalized momenta Pdotθ supposed to be equal to zero because θ is cyclic? Or is Pdotθ= -∂H/∂θ= mgr sineθ

    Sorry for not being able to upload a pictuure or putting dots over P but I dont know how to do so
  2. jcsd
  3. Nov 18, 2016 #2


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    But in the link above you don't have a particle on a sphere but on a circle, and then of course ##\dot{r}=0##. So the correct Lagrangian is
    $$L=\frac{m}{2} r^2 \dot{\theta}^2+mgr \cos \theta.$$
    I also have more conveniently pointed the ##y## axis in direction of ##\vec{g}## such that the stable stationary state is ##\theta=0=\text{const}##. Then you have
    $$p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=mr^2 \dot{\theta} \; \Rightarrow \; \dot{p}_{\theta}=m r^2 \ddot{\theta}=\frac{\partial L}{\partial \theta}=-mgr \sin \theta,$$
    and you get the equation of motion for a mathematical pendulum, as you should:
    $$\ddot{\theta}=-\frac{g}{r} \sin \theta.$$

    For the more general case you should use spherical coordinates
    $$\vec{x}=\begin{pmatrix} r \cos \varphi \sin \vartheta \\ r \sin \varphi \sin \vartheta \\ r \cos \vartheta \end{pmatrix},$$
    again with ##r=\text{const}##. Now write down the Hamiltonian, and you'll get the equations of motion for a spherical pendulum, but this problem you should discuss in the homework forum since it's way better to get some guidance to solve the problem yourself than just a solution!
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