Proving Hamiltonian ≠ Energy for Rotating Ball

[M. next]

Consider a ball of mass m rotating around an axis Oz (vertical). This ball is on a circle whose center is the same O.
Given: Angular velocity of ring is d∅/dt = ω.
Mind explaining it so we can prove that Hamiltonian here is different from Energy?!

 

[vanhees71]

The question is, is your assertion true? Let's start from the Lagrangian, using \phi as the generalized coordinate. The motion on a circle is then described by

\vec{x}=\begin{pmatrix}<br /> r \cos \phi \\ r \sin \phi<br /> \end{pmatrix}<br />

with r=\text{const}. The Lagrangian is

L=T=\frac{m}{2} r^2 \dot{\phi}^2.

The Hamiltonian is then defined as

H(q,p)=\dot{q} p-L

with the canonical momentum

p=\frac{\partial L}{\partial \dot{\phi}}=m r^2 \dot{\phi}.

The Hamiltonian is thus

H(q,p)=\frac{p^2}{2 m r^2}.

Written in terms of \dot{q}=\partial_p H=p/(m r^2) one sees that H=T, and thus H is the energy of the system.

 

[M. next]

Thank you for your reply, very organized, and this is so true! But why did he ask us to prove that H different from E?

 

[M. next]

And why did he mention "Angular velocity of ring is d∅/dt = ω"?