# Hamiltonian of flyball governor

1. Dec 20, 2004

### shizzle

Hamilton equations of flyball governor

I'm trying to find
1. The Hamiltonian
2. The Hamilton equation of motion for the flyball governor shown in problem 2 here

http://www.srl.caltech.edu/phys106/1999/Homework3.pdf

This is what i have. Can someone tell me if i'm right?

$$\begin{gather*} L = ml^2\dot{\theta}^2+ ml^2\omega^2\sin^2\theta + 2Ml^2\dot{\theta}^2\sin^2\theta +2(m+M)gl\cos\theta\\ H = ml^2\dot{\theta}^2+ ml^2\omega^2\sin^2\theta + 2Ml^2\dot{\theta}^2\sin^2\theta -2(m+M)gl\cos\theta\\ \partial{L}/\partial\dot{\theta} = 2ml^2\dot{\theta} + 4Ml^2\dot{\theta}\sin^2\theta = P_\theta\\ \dot{\theta} =P_\theta/2ml^2\dot{\theta} + 4Ml^2\dot{\theta}\sin^2\theta\\ \partial{H}/\partial{P_\theta} = P_\theta/2ml^2\dot{\theta} + 4Ml^2\dot{\theta}\sin^2\theta = \dot{\theta}\\ \partial{H}/\partial\theta = 2ml^2\omega^2\sin\theta\cos\theta +4Ml^2\dot{\theta}^2\sin\theta\cos\theta + 2(m+M)gl\sin\theta = -\dot{P_\theta}\\ -\dot{P_\theta}=-(ml^2\ddot{\theta}+4Ml^2\ddot{\theta}\sin^2\theta \\ equating the two equations and solving for theta double dot, we get\\ \ddot{\theta} = \frac{ml\omega^2\sin\theta\cos\theta+2Ml\dot{\theta}^2\sin\theta\cos\theta-(m+M)g\sin\theta}{l(m+2M\sin^2\theta)}\\ \end{gather*}$$

Last edited: Dec 20, 2004
2. Dec 20, 2004

### da_willem

Loose the spaces in your [ tex ] and [ / tex ].

3. Dec 20, 2004

### dextercioby

Never mind the code.It's intelligible.At least,the first 2 lines...:tongue2: The Lagrange's function is correct.It's identically to the one given in Landau&Lifschitz,p.21 (ed.1966,Mir,fr.)(with different notations,of course).So that should be okay.
Define the canonical momenta and then write the Hamilton's function for the system.Noteamilton's function wil not depend on $\dot{\theta}$,but on $p_{\theta}$.I believe it won't be hard to write the kinetic part in the Hamiltonian.The Potential part is just the one in the Lagrangian but with reversed sign.If u've chosen "+"in the Lagrangian,then it will "-"in the Hamiltonian.

Daniel.

4. Dec 20, 2004

### dextercioby

Once u got the Hamiltonian right,then Hamiltonian equations of motion will be very easy to write.First of all write that Hamiltonian.U may want to define the Poisson bracket for this system and use the equations of motion in terms of fundamental Poisson brackets.Or,perhaps,that's too much and you'd stick to the usual form of the equations.
Feel free to post any problem u have with Analytical Mechanics as you wouldn't be the only one.I believe Cepheid is struggling with the same things...Though much simpler.

Daniel.

EDIT:Both for this and for the previous post.Apparently,among those signs,i could see that u defined the canonical momentum.But yet,the formula containg in the LFS the symbol "H" (for Hamiltonian i presume) and in the RHT generalized velocites is wrong.As i said before,i'll say it again.The Hamiltonian is the (invertible) Legendre transformation of the Lagrangian wrt to canonical momenta.It never depends on generalized velocities,for any case...

Last edited: Dec 20, 2004
5. Dec 20, 2004

### shizzle

Okay, Daniel, my post looks a bit more intelligible :yuck: now so please check it and give me some feedback aight:) I actually love doing physics this way. I mean physics is fun even if its hard but this is waaaay cool. I'm becoming a physics forum adict.

okay, back to the problem...BUT u must admit, my equations are almost elegant...i hope they're right :tongue2:

Last edited: Dec 20, 2004
6. Dec 20, 2004

### dextercioby

It's okay.I already told you that.

It's wrong.I already told u that it should be
$$H=H(\theta,p_{\theta})$$

This is alright.Now express $$\dot{\theta}=\dot{\theta}(\theta,p_{\theta})$$ and insert in the Hamiltonian's definition.

It's okay what you wrote,but it's not useful.You must isolate all expliticit dependence of $\dot{\theta}$.

I'm not sure of this,as long as u don't have (or at least u haven't written) the Hamiltonian.

Daniel.

Last edited: Dec 21, 2004
7. Dec 21, 2004

### pervect

Staff Emeritus
Since this is a homework problem, let's work this out in full, without taking any shortcuts, starting with the Lagrangian.

We have

L := m*l^2*tdot^2+ m*l^2*omega^2*sin(theta)^2 + 2*M*l^2*tdot^2*sin(theta)^2 +2*(m+M)*g*l*cos(theta);

where I've taken the liberty of changing the variables because I'm using a computer program (maple).

This is L(theta, tdot) where tdot = $$\frac{\partial theta}{\partial t}$$

Now we can write the energy function, h in the same variables. The energy function is the Hamiltonian's "little brother", it's numerically equal but defined in different variables.

$$h = tdot \frac{\partial L}{\partial tdot} - L$$

this is then

h(theta,tdot) = tdot*(2*m*l^2*tdot+4*M*l^2*tdot*sin(theta)^2)-m*l^2*tdot^2-m*l^2*omega^2*sin(theta)^2-2*M*l^2*tdot^2*sin(theta)^2-2*(m+M)*g*l*cos(theta)

The last step is to work out ptheta, and substitute it in the expression for h, a change of variable.

You've already observed $$ptheta = \frac{\partial L}{\partial tdot}$$

Unfortunately I'm being called away from the computer, but all you have to do at this point in substitute variables to change h(theta, tdot) into H(theta, ptheta).

8. Dec 21, 2004

### pervect

Staff Emeritus
OK, I'm back so

h := m*l^2*tdot^2+2*M*l^2*tdot^2-2*M*l^2*tdot^2*cos(theta)^2
-m*l^2*omega^2+m*l^2*omega^2*cos(theta)^2-2*g*l*cos(theta)*m
-2*g*l*cos(theta)*M

and
$${\it tdot}=1/2\,{\frac {{\it ptheta}}{ \left( m+2\,M \left( \sin \left( \theta \right) \right) ^{2} \right) {l}^{2}}}$$

so

H :=
$$1/4\,{\frac {m{{\it ptheta}}^{2}}{{l}^{2} \left( m+2\,M \left( \sin \left( \theta \right) \right) ^{2} \right) ^{2}}}+1/2\,{\frac {M{{ \it ptheta}}^{2}}{{l}^{2} \left( m+2\,M \left( \sin \left( \theta \right) \right) ^{2} \right) ^{2}}}-1/2\,{\frac {M{{\it ptheta}}^{2} \left( \cos \left( \theta \right) \right) ^{2}}{{l}^{2} \left( m+2\, M \left( \sin \left( \theta \right) \right) ^{2} \right) ^{2}}}-m{l}^ {2}{\omega}^{2}+m{l}^{2}{\omega}^{2} \left( \cos \left( \theta \right) \right) ^{2}-2\,gl\cos \left( \theta \right) m-2\,gl\cos \left( \theta \right) M$$

the last truncated part is

2gl cos(theta) M

It's a good thing we used the full, laborious approach to find H, because many of the usual shortcuts don't work on this problem.

Often, L = T-V, and H = T+V, but not in this case.