Motion on a Paraboloid: Reduce to Quadratures

[neworder1]

Homework Statement

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.

Homework Equations

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...

 

[siddharth]

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...

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?

 

[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:

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

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

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

Using polar transformation leads to:

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

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

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

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).

 

[siddharth]

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).

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.