Question 7.3: Reconsidering Example 1.17, obtain the Hamiltonian and study......
Reconsidering Example 1.17, obtain the Hamiltonian and study its conservation, as well as that of the energy.
As seen in Example 1.17, the Lagrangian for the system is
L = T = \frac{m}{2} \left(\dot{r}^{2}+\omega ^{2}{r}^{2} \right), (7.26)
which equals the bead’s kinetic energy. In the present case, p_{r} = m\dot{r} and
H =\frac{p^{2}_{r}}{2m}-\frac{m\omega ^{2}}{2}{r}^{2}, (7.27)
which is not the total (purely kinetic) energy of the bead. However, H is a constant of the motion because ∂H/∂t = 0. On the other hand, the total energy
E = T =\frac{p^{2}_{r}}{2m}+\frac{m\omega ^{2}}{2}{r}^{2} (7.28)
is not conserved because the constraint force does work during a real displacement of the particle (verify this qualitatively). It is left for the reader to show that the Hamiltonian (7.27) is the “total energy” in the rotating reference frame.