What is the Hamiltonian for a bead on a rotating rod with fixed z and R?

[gluon1988]

Homework Statement

A bead of mass, $$m$$ is threaded on a frictionless, straight rod, which lies in the horizontal plane and is forced to spin with constant angular velocity, $$\omega$$, about a fixed vertical axis through the midpoint of the rod. Find the Hamiltonian for the bead and show that it does not equal $$T+U$$

Homework Equations

$$\mathcal{L}=T-U$$

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{q}}=\frac{\partial\mathcal{L}}{\partial{q}}$$

$$p=\frac{\partial\mathcal{L}}{\partial\dot{q}}$$

$$\mathcal{H}=\sum p_{i}\dot{q}_{i}-\mathcal{L}$$

The Attempt at a Solution

The bead is threaded onto the road, so the radius is fixed. So is the z-axis. The only degree of freedom for the bead is $$\phi$$, the angle it is located at in reference to a starting point.

$$\mathcal{L}=\frac{1}{2}m(\dot{\phi} R)^{2}-U(\phi)$$

$$p=\frac{\partial\mathcal{L}}{\partial\dot{\phi}}=m{R^2}{\dot{\phi}}$$

$$\mathcal{H}=m{R^2}{{\dot{\phi}}^2}-\frac{1}{2}m(\dot{\phi} R)^{2}-U(\phi)=\frac{1}{2}m(\dot{\phi} R)^{2}+U(\phi)\rightarrow\mathcal{H}=T+U$$??

I'm assuming I am getting the kinetic energy wrong but I really don't see what it could be with z and R fixed. Please help

 

[Mindscrape]

Huh, I don't see any physical reason why H≠T+U. The two conditions seem to be met: the system is conservative (potential energy is velocity independent), and the equations of transformation connecting the rectangular and generalized coordinates are independent of time.

While it doesn't change the two points I listed above, why are you saying that z does not change? Seems to me like z is able to change.

 

[gluon1988]

Huh, I don't see any physical reason why H≠T+U. The two conditions seem to be met: the system is conservative (potential energy is velocity independent), and the equations of transformation connecting the rectangular and generalized coordinates are independent of time.

While it doesn't change the two points I listed above, why are you saying that z does not change? Seems to me like z is able to change.

The problem just stated that H≠T+U. I think they were trying to show that not every form of general coordinates will work for non-inertial frames. Regardless, I was missing a term of $${{\omega}^2}{sin}^{2}(\phi){\phi}^2$$ in the kinetic energy term, which does in fact lead to H≠T+U.

The problem was listed on the easier section of the text, so I assumed they weren't going to vary z, which they didn't thankfully.

Thanks for the reply

 

[Mindscrape]

Oh, I see now what the problem was describing. It's written in the usual cryptic, non-explicit description that classical mechanics problems tend to have.

The equations of transformation connecting the rectangular and generalized coordinates are not independent of time after all.

 

[paranormal]

!



Your attempt at the solution is correct. The Hamiltonian for this system is indeed equal to the sum of the kinetic and potential energies, i.e. \mathcal{H}=T+U. The mistake you made is in your calculation of the kinetic energy. The correct expression for the kinetic energy of a bead on a rotating rod is given by T=\frac{1}{2}m(\dot{\phi} R)^{2}, as you have correctly written in your Lagrangian. This can be derived by considering the rotational motion of the bead and using the parallel axis theorem to account for the fixed radius of the rod. Therefore, your final expression for the Hamiltonian is correct and does not need to be modified.