# Transforming coordinates for a vertical hoop

1. Nov 24, 2009

### 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" [Broken]

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..

Last edited by a moderator: May 4, 2017
2. Nov 24, 2009

### Loxias

never mind.. i just noticed it..

3. Nov 24, 2009

### jdwood983

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

4. Nov 24, 2009

### 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?

5. Nov 24, 2009

### jdwood983

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

6. Nov 24, 2009

### 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?

7. Nov 24, 2009

### jdwood983

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

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

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

8. Nov 24, 2009

### 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 $$\dot{\theta}^2R^2sin^2(\theta)$$

9. Nov 24, 2009

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 $$\dot{r}^2[/itex]. Remember that both your position and velocities are vectors, so when you see the square term, you must think to take the dot product. Your kinetic energy should be as I wrote it above. I probably should have caught it in the first post, but I did just wake up about an hour and a half ago... 10. Nov 24, 2009 ### Loxias I'm trying to see what i did wrong and i can't find it. Can you see it? 11. Nov 24, 2009 ### jdwood983 I get: [tex] \dot{x}^2=\dot{\theta}^2R^2\cos^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\sin^2\phi-2\dot{\phi}\dot{\theta}R^2\sin\theta\sin\phi\cos\theta\cos\phi$$

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

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

$$\dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\cos^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\sin^2\phi-2\dot{\phi}\dot{\theta}R^2\sin\theta\sin\phi\cos\theta\cos\phi+\dot{\theta}^2R^2\cos^2\theta\sin^2\phi$$

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

Clearly those two cross terms cancel,

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

Some trig identities ensue:

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

A few more trig identities:

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

12. Nov 24, 2009

### Loxias

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

so the hamiltonian here is the energy of the system?

13. Nov 24, 2009

### jdwood983

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.

14. Nov 24, 2009

### Loxias

thanks

15. Nov 24, 2009

### jdwood983

Not a problem, glad I could help