Transforming coordinates for a vertical hoop

[Loxias]

1. The problem statement, all variables and given/known data
A bead of mass m slips without friction an a vertically oriented hoop of radius R.
The hoop rotates around the z axis at a constant rate w.

(question 4 if anyone wants to see the figure)
http://phstudy.technion.ac.il/~wn114101/hw/wn2010_hw05.pdf (http://phstudy.technion.ac.il/~wn114101/hw/wn2010_hw05.pdf)


2. Relevant equations

Write the Lagrangian for the system

3. The attempt at a solution

I need to transform coordinates.
I thought about transforming to spherical ones.

constraints : r = R, \phi = wt

x = R sin(\theta) cos(wt), y = R sin(\theta) sin(wt) , z = R cos(\theta)

does this seem right? because when I try to write the velocity I get a very ugly equation

\dot{x}^2 + \dot{y}^2 + \dot{z}^2 = \dot{\theta}^2R^2 + w^2R^2sin^2(\theta) - 2\dot{\theta}wR^2 sin(\theta)sin(wt)cos(wt)cos(\theta) +2\dot{\theta}wR^2cos(\theta)sin(wt)sin(\theta)cos (wt) + \dot{\theta}R^2sin^2(\theta)

which simply doesnt look right..

[Loxias]

never mind.. i just noticed it..

[jdwood983]

3. The attempt at a solution

I need to transform coordinates.
I thought about transforming to spherical ones.

constraints : r = R, \phi = wt

x = R sin(\theta) cos(wt), y = R sin(\theta) sin(wt) , z = R cos(\theta)

does this seem right? because when I try to write the velocity I get a very ugly equation

\dot{x}^2 + \dot{y}^2 + \dot{z}^2 = \dot{\theta}^2R^2 + w^2R^2sin^2(\theta) - 2\dot{\theta}wR^2 sin(\theta)sin(wt)cos(wt)cos(\theta) +2\dot{\theta}wR^2cos(\theta)sin(wt)sin(\theta)cos (wt) + \dot{\theta}R^2sin^2(\theta)

which simply doesnt look right..

That doesn't look ugly! There's two terms that cancel and reduces the problem.

[Loxias]

yea.. thanks.. i noticed it..

a question though

is
\frac{m}{2}\left(\dot{\theta}^2R^2 + w^2R^2sin^2(\theta)\right) + mgRcos(\theta)

the energy in the system?

[jdwood983]

yea.. thanks.. i noticed it..

a question though

is
\frac{m}{2}\left(\dot{\theta}^2R^2 + w^2R^2sin^2(\theta)\right) + mgRcos(\theta)

the energy in the system?

This would be the Lagrangian, but not the energy. Your potential should be negative due to the use of \cos\theta.

[Loxias]

the Lagrangian is

L = \frac{m}{2} \left(\dot{\theta}^2R^2 + w^2R^2sin^2(\theta) + \dot{\theta}R^2sin^2(\theta)\right) - mgRcos(\theta)

from which i've derived the hamiltonian according to

H = \frac{\partial L}{\partial \dot{\theta}}\dot{\theta} - L

which is what i wrote below. what i'm wondering is, is this the energy?

[jdwood983]

the Lagrangian is

L = \frac{m}{2} \left(\dot{\theta}^2R^2 + w^2R^2sin^2(\theta) + \dot{\theta}R^2sin^2(\theta)\right) - mgRcos(\theta)

from which i've derived the hamiltonian according to

H = \frac{\partial L}{\partial \dot{\theta}}\dot{\theta} - L

which is what i wrote below. what i'm wondering is, is this the energy?


You seem to be adding something here. Where did \dot{\theta}R^2sin^2(\theta) come from? Your kinetic energy should be:


T=\frac{1}{2}m\left(\dot{R}^2+R^2\dot{\theta}^2+R^ 2\sin^2\theta\dot{\phi}^2\right)=\frac{1}{2}m\left (R^2\dot{\theta}^2+R^2\sin^2\theta\dot{\phi}^2\rig ht)


Also, your kinetic energy should be negative, so in the Lagrangian it should be positive: L=T-V=T+mgR\cos\theta.

[Loxias]

T = \frac{m}{2} (\dot{x}^2 + \dot{y}^2 + \dot{z}^2)
and according to my first post, you can see where all the terms came from.
also, \dot{R} = 0
a small mistake, the last term should be \dot{\theta}^2R^2sin^2(\theta)

[jdwood983]

T = \frac{m}{2} (\dot{x}^2 + \dot{y}^2 + \dot{z}^2)
and according to my first post, you can see where all the terms came from.
also, \dot{R} = 0

I know that R-dot is zero, I did cancel it in my equation. But I think you are confusing \dot{\mathbf{r}}\cdot\dot{\mathbf{r}} with [tex]\dot{r}^2[/itex]. Remember that both your position and velocities are vectors, so when you see the square term, you must think to take the dot product. Your kinetic energy should be as I wrote it above.

I probably should have caught it in the first post, but I did just wake up about an hour and a half ago...

[Loxias]

I'm trying to see what i did wrong and i can't find it. Can you see it?

[jdwood983]

I get:


\dot{x}^2=\dot{\theta}^2R^2\cos^2\theta\cos^2\phi+ \dot{\phi}^2R^2\sin^2\theta\sin^2\phi-2\dot{\phi}\dot{\theta}R^2\sin\theta\sin\phi\cos\t heta\cos\phi



\dot{y}^2=\dot{\theta}^2R^2\cos^2\theta\sin^2\phi+ \dot{\phi}^2R^2\sin^2\theta\cos^2\phi+2\dot{\theta }\dot{\phi}R^2\sin\theta\sin\phi\cos\theta\cos\phi



\dot{z}^2=\dot{\theta}^2R^2\sin^2\theta


So adding them together


\dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\co s^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\si n^2\phi-2\dot{\phi}\dot{\theta}R^2\sin\theta\sin\phi\cos\t heta\cos\phi+\dot{\theta}^2R^2\cos^2\theta\sin^2\p hi



+\dot{\phi}^2R^2\sin^2\theta\cos^2\phi+2\dot{\thet a}\dot{\phi}R^2\sin\theta\sin\phi\cos\theta\cos\ph i+\dot{\theta}^2R^2\sin^2\theta


Clearly those two cross terms cancel,



\dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\co s^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\si n^2\phi+\dot{\theta}^2R^2\cos^2\theta\sin^2\phi+\d ot{\phi}^2R^2\sin^2\theta\cos^2\phi+\dot{\theta}^2 R^2\sin^2\theta


Some trig identities ensue:



\dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\co s^2\theta+\dot{\phi}^2R^2\sin^2\theta+\dot{\theta} ^2R^2\sin^2\theta


A few more trig identities:


\dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2+\d ot{\phi}^2R^2\sin^2\theta

[Loxias]

indeed.. i saw where i went wrong. thanks alot!!

so the hamiltonian here is the energy of the system?

[jdwood983]

indeed.. i saw where i went wrong. thanks alot!!

so the hamiltonian here is the energy of the system?

If there is no explicit time dependence in the Lagrangian (as is the case here), the Hamiltonian is also the total energy of the system.

[Loxias]

thanks :smile:

[jdwood983]

thanks :smile:

Not a problem, glad I could help