# Homework Help: Lagrangian problem

1. Feb 7, 2010

### maxtor101

1. The problem statement, all variables and given/known data

Consider a particle of mass m sliding under the influence of gravity, on the smooth inner surface of the hyperboloid of revolution of equation

$x^2 + y^2 = z^2 -a^2$

where (x,y,z) are the Cartesian co-ordinates of the particle (z > 0) and (a > 0) is a constant.
The particle is constrained to move on the surface defined above.

a) How many degrees of freedom does this system have?

b) Show that in terms of the independent cylindrical co-ordinates $(r, \phi) [i/tex] the Lagrangian is [itex] L = \frac{m}{2} \left( 2 \dot{r}^2 + r^2 \dot{ \phi}^2 - \frac{a^2}{a^2 + r^2} \dot{r}^2 -2g \sqrt{(a^2 + r^2)} \right )$

2. Relevant equations

$$L = T - V$$

3. The attempt at a solution

Converting to cylindrical co-ordinates:

$$x = r cos \phi$$
$$y = r sin \phi$$

$$\dot{x} = -r \dot{ \phi} sin \phi + \dot{r} cos \phi$$
$$\dot{y} = r \dot{ \phi} cos \phi + \dot{r} sin \phi$$

From the equation of the hyperboloid:

$$z^2 = x^2 + y^2 + a^2$$
$$z^2 = r^2 + a^2$$

$$z = \sqrt{r^2 + a^2}$$

$$\dot{z} = \frac{r}{r^2 + a^2} \dot{r}$$

Now find Kinetic Energy:

$$T = \frac{m}{2} \left ( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right )$$

$$\dot{x}^2 = r^2 \dot{\phi}^2 \sin^2{\phi} - 2r\dot{r} \dot{\phi} \sin{\phi} \cos{\phi} + \dot{r}^2 \cos^2{\phi}$$

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

Therefore

$$\dot{x}^2 + \dot{y}^2 = \dot{r}^2 + r^2 \dot{\phi}^2$$

So
$$T = \frac{m}{2} \left ( \dot{r}^2 + r^2 \dot{\phi}^2 + \frac{r^2}{r^2 + a^2} \dot{r}^2 \right )$$

Now find Potential Energy:

$$V = - mgz$$

$$V = - mg \sqrt{r^2 + a^2}$$

So the Lagrangian is:

$$L = T - V$$

$$L = \frac{m}{2} \left ( \dot{r}^2 + r^2 \dot{\phi}^2 + \frac{r^2}{r^2 + a^2} \dot{r}^2 \right ) + mg \sqrt{r^2 + a^2}$$

This is not correct.
Any ideas?

Last edited: Feb 7, 2010
2. Feb 7, 2010

### maxtor101

This question would be so much easier in spherical co-ordinates, but unfortunately that is irrelevant here :(

3. Feb 7, 2010

### thebigstar25

one mistake I noticed :

z = r r(dot) / (r^2 + a^2)

shouldnt it be :

r r(dot) / sqrt (r^2 +a^2) ?

4. Feb 7, 2010

### maxtor101

Ah it's cool I figured it out.

Rewrite:

$$\frac{r^2}{r^2 + a^2} \dot{r}^2$$

as

$$\left ( 1 - \frac{a^2}{a^2 + r^2} \right ) \dot{r}^2$$

and the potential should actually be

$$V = mg \sqrt{r^2 + a^2}$$

Thanks