# Homework Help: Motion on a paraboloid

1. Dec 9, 2007

### neworder1

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

A body of mass M moves (in a gravitational field g) on the inner surface of given by equation:

$$z=\frac{1}{2a}(x^{2}+y^{2})$$

(a is positive)

Reduce the question of finding the motion to quadratures.

2. Relevant equations

3. The attempt at a solution

I used Lagrange equations (1st kind) to find relevant equations for x, y and z, and after separating variables, transformation to polar coordinates $$(r, \phi, z)$$ etc. I came up with the following equation (C is a constant dependent on initial conditions):

$$\ddot{r}-\frac{C}{r^{3}}=-\frac{1}{a^{2}}(r{\dot{r}}^{2}+r^{2}\ddot{r})-\frac{g}{a}r$$

I don't have any idea how to integrate this equation, but maybe I've done things in an unnecessarily complicated way...

2. Dec 9, 2007

### siddharth

I get a slightly different equation in r.

Since

$$L = \frac{m}{2}\left(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 \right) - mgz$$

and

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

$$\dot{z} = \frac{r \dot{r}}{a}$$

and using

$$\frac{\partial L}{\partial \theta} = 0 = \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}$$

$$\frac{\partial L}{\partial r} = \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}$$

I get

$$\frac{r \dot{r}}{a^2} -\frac{gr}{a} = \ddot{r} + \frac{2\dot{r}^2 + r^2 \ddot{r}}{a^2}$$

How does the constant depending on the initial condition (which should be the component of the angular momentum, I guess) come into the final equation for r?

3. Dec 9, 2007

### neworder1

The way I actually obtained that equation is the following: because the body moves on the surface given by:
$$f=z-\frac{1}{2a}(x^{2}+y^{2})=0$$

,the reaction force must be proportional to the constraint function's gradient, i.e.:

$$F_{r}=\lambda\nabla{f}$$ (Lagrange multipliers). So I get 3 equations:

$$m\ddot{x}=-\lambda\frac{x}{a}$$

$$m\ddot{y}=-\lambda\frac{y}{a}$$

$$m\ddot{z}=-mg+\lambda$$

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

$$\ddot{y}=\ddot{r}sin\phi+2\dot{r}\dot{\phi}cos\phi+r\ddot{\phi}cos\phi-r{\dot{\phi}}^{2}sin\phi$$
$$z=\frac{1}{2a}r^{2}$$

$$\ddot{z}=\frac{1}{a}({\dot{r}}^{2}+r\ddot{r})$$

After plugging these expressions into the original equations (plus eliminating $$\lambda$$ using $$z$$) I get two new equations (1) and (2) for $$r$$ and $$\phi$$. Our aim is to separate variables, so I do the following trick:

$$(1)cos\phi+(2)sin\phi$$
$$(1)sin\phi-(2)cos\phi$$

and I get another two equations:

$$(1') \ddot{r}-r{\dot{\phi}}^{2}=-\frac{1}{a^{2}}(r{\dot{r}}^{2}+r\ddot{r})-\frac{g}{a}r$$
$$(2') 2\dot{r}\dot{\phi}+r\ddot{\phi}$$

(2') is easily integrable and yields:
$$\dot{\phi}=\frac{C}{r^{2}}$$.

Plugging this into (1') results in the equation I've written in my first post.

Is there a mistake somewhere in these calculations? I know that this can be done quicker by writing apropriate lagrangians etc., but the problem's formulation was to do everything using this method (i.e. Lagrange multipliers).

Last edited: Dec 9, 2007
4. Dec 10, 2007

### siddharth

I've not checked the calculations yet, but I think the final equation which describes the motion of r should be independent of the method. Since using the appropriate Lagrangian is much less time consuming, check if you get the same equation. That way, you'll know if there's a calculation error.