Lagrangian Problem - Degrees of Freedom & Solution

[maxtor101]

Homework Statement

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 [itex](r, \phi) [i/tex] <br /> <br /> the<br /> Lagrangian is <br /> <br /> $L = \frac{m}{2} \left( 2 \dot{r}^2 + r^2 \dot{ \phi}^2 - \frac{a^2}{a^2 + r^2} \dot{r}^2 -2g \sqrt{(a^2 + r^2)} \right )$<br /> <br /> <h2>Homework Equations</h2><br /> <br /> $$L = T - V$$<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Converting to cylindrical co-ordinates:<br /> <br /> $$x = r cos \phi$$<br /> $$y = r sin \phi$$<br /> <br /> $$\dot{x} = -r \dot{ \phi} sin \phi + \dot{r} cos \phi$$<br /> $$\dot{y} = r \dot{ \phi} cos \phi + \dot{r} sin \phi$$From the equation of the hyperboloid:<br /> <br /> $$z^2 = x^2 + y^2 + a^2$$<br /> $$z^2 = r^2 + a^2$$<br /> <br /> $$z = \sqrt{r^2 + a^2}$$<br /> <br /> $$\dot{z} = \frac{r}{r^2 + a^2} \dot{r}$$<br /> Now find Kinetic Energy:<br /> <br /> $$T = \frac{m}{2} \left ( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right )$$<br /> <br /> $$\dot{x}^2 = r^2 \dot{\phi}^2 \sin^2{\phi} - 2r\dot{r} \dot{\phi} \sin{\phi} \cos{\phi} + \dot{r}^2 \cos^2{\phi}$$<br /> <br /> $$\dot{y}^2 = r^2 \dot{\phi}^2 \cos^2{\phi} + 2r\dot{r} \dot{\phi} \sin{\phi} \cos{\phi} + \dot{r}^2 \sin^2{\phi}$$<br /> <br /> Therefore <br /> <br /> $$\dot{x}^2 + \dot{y}^2 = \dot{r}^2 + r^2 \dot{\phi}^2$$<br /> <br /> So<br /> $$T = \frac{m}{2} \left ( \dot{r}^2 + r^2 \dot{\phi}^2 + \frac{r^2}{r^2 + a^2} \dot{r}^2 \right )$$<br /> <br /> Now find Potential Energy:<br /> <br /> Not too sure about this but I'm guessing it is <br /> <br /> $$V = - mgz$$<br /> <br /> $$V = - mg \sqrt{r^2 + a^2}$$So the Lagrangian is:<br /> <br /> $$L = T - V$$<br /> <br /> $$L = \frac{m}{2} \left ( \dot{r}^2 + r^2 \dot{\phi}^2 + \frac{r^2}{r^2 + a^2} \dot{r}^2 \right ) + mg \sqrt{r^2 + a^2}$$<br /> <br /> This is not correct.<br /> Any ideas?Thanks in advance for your help![/itex]

 

[maxtor101]

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

 

[thebigstar25]

one mistake I noticed :

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

shouldnt it be :

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

 

[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