Lagrangian equations of particle in rotational paraboloid

[Oomph!]

Hello. I solve this problem:

1. Homework Statement
The particles of mass m moves without friction on the inner wall of the axially symmetric vessel with the equation of the rotational paraboloid:

where b>0.

a) The particle moves along the circular trajectory at a height of z = z(0).

express:
- Lagrangian
- the equation of motion for the polar coordinate r
- energy of the particle (with m, z(0), b and g only)
- angular momentum of the particle (with m, z(0), b and g only)

b) We slightly deflect the particles downwards. Find the frequency of small oscillations around the original intact trajectory.

2. Relevant equation
Lagrangian equations, equations for energy in conservative field and angular momentum.

The Attempt at a Solution

a) I didn't have any problem with Lagrangian and equation of motion for the polar coordinate r. Here is the result, I know how to do it:
- Lagrangian:

(1)
- equation of motion for the polar coordinate r:

(2)

I have problem to express the energy and angular momentrum od particle. I show you my attempt:

So, the problem is that I don't know, how to express the time derivation of Θ.
Could I just say, that the time derivation of Θ is the (gb)^(1/2) because the equation (2) is in standart form where ω^2=gb?

b) Well, I don't have any idea. I just have the result:

And it doesn't make sense if I told that ω^2=gb.

So, please, could you tell me what is wrong and what to do?

Thank you.

 

[MathematicalPhysicist]

You should have two equations of motion, one for the $\phi$ coordinate and another for $r$.

You should also have $\ddot{\phi} = 0$ in you EOM.

Edit: your equation (2) seems wrong to me you should be getting:

$$0 = (1+b^2r^2)\ddot{r} + 2b^2 r\dot{r}^2+gbr - r\dot{\phi}^2=0$$

But I don't see how you can find $\dot{\phi}$.

 

[TSny]

Check the sign of the second term.

Apply the $r$ equation of motion to the special case of circular motion. (What are $\dot r$ and $\ddot r$ for this case?)

Note: Be sure not to confuse the angular velocity of the circular motion of part (a) with the angular frequency of small oscillations in part (b).

 

[mhelb]

So first you can find a solution for your velocity of a circular orbit. Therefore you know $\dot{r} = \ddot{r} = 0$. From your corrected equation 2 (what TSny wrote) you get:

$$r\dot\varphi = gbr \Rigtharrow v_0=\sqrt{gb}r_0$$

From the other equation you would get, if you've derived the Lagrangian for $\varphi$ you get, that you angular momentum $L$ is constant (central, conservative force).

$$L = m\dot\varphi r^2 = mv_0r_0 \Rightarrow \dot \varphi = \frac{\sqrt{gbr_0^4}}{r^2}$$

This is your expression of $\dot\varphi$ written in terms of the current radius $r(t)$ and the radius $r_0$ of the stable orbit. Putting this into eqaution 2 we get:

$$\ddot r(1+b^2r^2) - b^2r\dot r^2 - \frac{gbr_0^4}{r^3} + gbr = 0$$

Furthermore, for small oscillations we can say: $r(t) = r_0 + \epsilon(t)$ with $\dot r = \dot\epsilon$ and $\ddot r = \ddot \epsilon$ and $\epsilon << r_0$. Furthermore every term with an $\epsilon$ of higher order than 1 is neglegable. This means $\epsilon^2 = 0$, $\dot\epsilon^2 = 0$,$\dot\epsilon\epsilon = 0$ ... Doing this we get:

$$\ddot \epsilon (1+b^2r_0^2) - \frac{gbr_0^4}{(r_0+\epsilon)^3} + gbr_0 + gb\epsilon$$

With $(r_0 + \epsilon)^3 = r_0^3+3r_0^3\epsilon + 0 +0 \approx r_0^3$ because $r_0^3 >> 3r_0^2\epsilon$ which leads to the differential equation:

$$\ddot\epsilon -\frac{gb}{1+b^2r_0^2}\cdot r = 0$

So we get

$$\omega = \sqrt{\frac{gb}{1+b^2r_0^2}}$$ = \sqrt{\frac{gb}{1+2bz_0}}

I know I'm missing a 2 in front of the squareroot. I don't know where I lost it

 

[kuruman]

Please note that this thread is more than five years old and that the OP has not been seen for four years. It is unlikely that your response will be of help at this point but thank you for submitting it.