Hello i was wondering if anyone could explain how i should treat the anguluar velocity in the following problem, and what form the kinetic energy should take, i assume that the potential energy is just $\frac{1}{2}kx^2$ for this case? and the lagrangian is then calculated by $L=T-U$ where $T $ is the kinetic energy and then $U$ is the potential energy.
i am also unsure on how to calculate the hamiltonian so if someone could show me how to do that it would be great, thanks
see the attatched photo for the problem.
$\mathbf v=\omega r\hat \theta+\dot r\hat r=\omega(l+x)\hat \theta+\dot x\hat r$
$T=(1/2)m\omega^2(l+x)^2+(1/2)m\dot x^2$
$L=T-V=(1/2)m\omega^2(l+x)^2+(1/2)m\dot x^2-(1/2)kx^2$
Note that even though the system is open (the bead is forced to move at some angular velocity and this entails that some mechanism is giving or removing energy), the process is equivalent to the acting of some conservative force: the centrifugal one $F_c=m\omega^2r\implies V'_c=-(1/2)m\omega^2r^2$. With this idea, the hamiltonian is the energy
$p_x=\dfrac{\partial L}{\partial\dot x}=m\dot x\implies \dot x=p_x/m$
$T'=p_x^2/2m$ and $V'=V'_c+V'_e=-(1/2)m\omega^2(l+x)^2+(1/2)kx^2$
$H=T'+V'=(1/2m)p_x^2-(1/2)m\omega^2(l+x)^2+(1/2)kx^2$
With the general definition for the hamiltonian:
$H=\dot xp_x-L=(1/2m)p_x^2-(1/2)m\omega^2(l+x)^2+(1/2)kx^2$
For the Hamilton's equations,
$\dot x=\dfrac{\partial H}{\partial p_x}=p_x/m$
$\dot p_x=-\dfrac{\partial H}{\partial x}=m\omega^2(l+x)-kx$
As expected.