# Homework Help: Oscillation of disturbed constrained circular orbit

1. Jan 11, 2009

### Math Jeans

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

Consider a particle of mass m constrained to move on the surface of a paraboloid whose equation (in cylindrical coordinates) is $$r^2=4az$$. If the particle is subject to a gravitational force, show that the frequency of small oscillations about a circular orbit with radius $$\rho=\sqrt{4az_0}$$ is

$$\omega=\sqrt{\frac{2g}{a+z_0}}$$

2. Relevant equations

The frequency for disturbed circular orbits is given by the equation:

$$\omega^2=\frac{3g(\rho)}{\rho}+g'(\rho)$$

or

$$\omega^2=\frac{3l^2}{\mu^2r_0^2}-\frac{1}{\mu}[\frac{dF}{dr}]_r_0$$

3. The attempt at a solution

I'm having problems doing a couple of things on this problem:
I can't seem to visualize in my mind what this problem looks like, so I can't get an image that I can quantify.

My main issue is that I can't connect the problem's parameters with the forulas given above.

Its probably alot easier than I'm making it, but I've been wrestling with this problem for a couple months, and giving up isn't an option in my position.

2. Jan 12, 2009

### chrisk

For small circular oscillations, find an expression for the potential of the system then expand it using a Taylor expansion about the equilibrium point. The second term of the expansion is equivalent to k, and for circular motion the angular frequency of small radial oscillations is w = sqrt(k/m) or w2 = k/m. Since the only force acting on the mass is gravity, use the gravitational potential and expand this to find k. Then insert the given radius into k and solve for w.

3. Jan 12, 2009

### Math Jeans

That would work if there wasn't a paraboloid constraining the mass.

The paraboloid is exerting a force as well.

My most recent attempt at the problem involved me finding a differential equation for the motion of the mass using the Hamilton-Lagrange formula.

I keep getting really really close using this method, but no matter how I format it, I can't get the correct answer to come out of it.

I don't have enough time right now, but a little later I will add one of the attempts to this thread.

4. Jan 13, 2009

### turin

There is another contribution to the potential, the inextricable effective potential that comes from angular momentum. You must combine this with the gravitational potential to obtain the total potential.

Instead of going all the way to the EOM, just stick with the Hamiltonian, but remember to use the total potential instead of just the gravitational potential. From this total potential, you can get an effective K. And, from the remaining kinetic terms, you can get an effective M.

Last edited: Jan 13, 2009
5. Jan 13, 2009

### Math Jeans

I never went all of the way to the EOM, I just used the Hamiltonian to find a differential equation the equates to oscillations.

I tried your suggestion of including the rest of the potential into my Lagrangian. Unfortunately, while this method canceled out the pesky angular momentum term in my kinetic energy, it didn't change my answer.

I'm currently trying the Taylor expantion method (I don't know if these should be mixed or not), and I'll get back to you soon.

6. Jan 14, 2009

### Math Jeans

Nope. No dice.

This is going to drive me crazy.

7. Jan 14, 2009

### gabbagabbahey

Hmmm... why don't you post some of your attempt so we can get an idea of where you may be going wrong

8. Jan 14, 2009

### Math Jeans

I don't have a whole lot of time to do this, so I'll go into as much detail as I can.

I start with the definition of the total squared velocity for systems in cylindrical coordinates:

$$v^2=\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2$$

Substituting for Z, and inserting into the formula for kinetic energy, I get:

$$T=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{mr^2\dot{r}^2}{8a^2}$$

My potential energy is the following:

$$V=U+\frac{l^2}{2r^2\mu}=mgr+\frac{1}{2}mr^2\dot{\theta}^2$$

My Laplacian is then equal to:

$$L=\frac{1}{2}m\dot{r}^2+\frac{mr^2\dot{r}^2}{8a^2}-mgr$$

Evaluating $$\frac{\partial L}{\partial r}$$ and $$\frac{d}{dt}\frac{\partial L}{\partial\dot{r}}$$ gives me the following ODE:

$$\ddot{r}+\frac{\dot{r}^2r}{4a^2+r^2}+\frac{4a^2g}{4a^2+r^2}=0$$

I'm just about out of time now, but the the rest of it involves me substituting r for $$r=\rho+x$$ where x is the infinitesimal difference in r that causes the oscillation.

I mess around with that for a while, but no matter what I do, its the differential equation above that is flawed, and not my work after that.

I appreciate all of the responses that I have gotten so far. I am almost positive that my problem lies in my potential energy, and I'm also fairly sure that one of your responses has solved that problem in some way. However, I can't seem to be able to get the point that you are trying to convey into my head, and maybe this will allow for more computationally direct answers.

I appreciate everything you have done so far to help.

Thanks,
Jeans

9. Jan 14, 2009

### gabbagabbahey

There is no need to include the angular term in your potential. The constraint equation you have for z automatically takes care of any pseudo forces and effective potentials that arise.

Also, it isn't clear to me from your problem statement what direction gravity is acting in. My guess would have been that it acts in the negative z-direction, so that

$$V=mgz=\frac{mgr^2}{4a}$$

Errrm you mean Lagrangian not Laplacian right?

Sounds like a good plan of attack to me (as long as you treat x as a function of time; which I assume you do). I think if you fix your Lagrangian you should be able to arive at the correct expression.

Last edited: Jan 14, 2009
10. Jan 14, 2009

### Math Jeans

Actually, including the constraints into the potential was something that I just recently started doing.

$$U=mgz$$ was the first thing that I tried, however, my answer always turns up as:

$$\omega=\sqrt{\frac{g}{2(a+z_0)}}$$.

My process for getting that answer is the same as what I posted above, but I can assure you that it doesn't work.

11. Jan 14, 2009

### gabbagabbahey

Well it should; if you post your work for that method I should be able to point out where you are going wrong.

12. Jan 14, 2009

### Math Jeans

I don't have time to post it at the moment, but I don't think it will work mainly because I tried that particular method countless times over...and then re-did it a few more times with a physics buddy, all getting the same answer.

I'll see what I can do.

13. Jan 14, 2009

### turin

Sorry, I didn't mean to confuse you. Don't add the effective potential by hand or else you will double count it in error as you did. What I meant was that, once you get the effective potential from the orbital kinetic term, you must include this as part of the potential that you expand in Taylor series. So, you should have two kinetic energy terms and two potential energy terms.

$$H=T_{\rho}+T_{z}+T_{\phi}+V_{g} \rightarrow H=T_{\rho}+T_{z}+V_{eff}+V_{g}$$

Then, you reduce to one dimension. I used the axial distance. The effective M will depend on the actual mass and a factor that depends on the slope. The effective K will depend on a number of things, including the angular momentum. The angular momentum depends on all of the following independent parameters: m, g, z_0, a.

I was actually struggling with this one for over a day using the Lagrangian method, but I couldn't get it to work. I got the same results as y'all using that method. So, I used the Hamiltonian instead of the Lagrangian. The principle I used was simply conservation of energy rather than least action.

Last edited: Jan 14, 2009
14. Jan 14, 2009

### gabbagabbahey

I worked through it using the Lagrangian method and got the desired result. I think the source of your error is probably how you are handling the constant terms in your Taylor expansion for $$\ddot{x}(t)$$ as well what you are doing with your angular momentum term $$l\equiv mr^2\dot{\theta}=\text{some constant}$$.

Try first showing that $l$ is a constant using the EOM for theta. Then substitute your $r(t)=\rho+x(t)$ expression into your EOM for $r(t)$ and solve for $$\ddot{x}(t)$$. Then take the taylor series. Neglect all terms of order $x^2$ or higher as well as all terms of order $$\dot{x}^2$$ or higher (You can do this because you want $$\ddot{x}(t)\approx-\omega^2 x(t)$$ and therefor you know $$\dot{x}(t)\sim x(t)$$).

Then use the fact that you require the equation to look like $$\ddot{x}(t)\approx-\omega^2 x(t)$$ to solve for $l$ (since the constant term must be zero) and then finally solve for $\omega$)

15. Jan 14, 2009

### Math Jeans

I appologize, but I am utterly and totally confused.

The main thing that I have had a hard time wrapping my head around is how you factor the Taylor series into this.

I'm just having trouble reproducing what you are describing as I can't seem to visualize what you are describing.

Last edited: Jan 14, 2009
16. Jan 14, 2009

### gabbagabbahey

Start by deriving the equations of motion from your Lagrangian:

$$L=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta }^2+\frac{mr^2\dot{r}^2}{8a^2}-\frac{mgr^2}{4a}$$

$$\frac{d}{dt}(mr^2\dot{\theta})=0$$

Which tells you that $$mr^2\dot{\theta}$$ is a constant. Call that constant (whatever it is) $l$ and use that definition to replace $$\dot{\theta}$$ in your Lagrangian:

$$L=\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}+\frac{mr^2\dot{r}^2}{8a^2}-\frac{mgr^2}{4a}$$

Then calculate the EOM for $r$ and solve for $$\ddot{r}$$...post your result so that I can check for accuracy, and I'll help you proceed from there.

Last edited: Jan 15, 2009
17. Jan 15, 2009

### Math Jeans

Ok, I took the Lagrangian that you gave me and evaluated the EOM, and recieved the following:

$$\ddot{r}=-\frac{\dot{r}^2r}{4a^2+r^2}-\frac{4a^2l^2}{4a^2m^2r^3+m^2r^5}-\frac{2agr}{4a^2+r^2}$$

18. Jan 15, 2009

### gabbagabbahey

Good

Now make the substitutions : $$r(t)=\rho+x(t)$$, $$\dot{r}=\dot{x}$$ and $$\ddot{r}=\ddot{x}$$ to get an expression for $$\ddot{x}$$.

Then take the Taylor series of that expression and throw away all terms of order $x^2$ and higher.

19. Jan 15, 2009

### Math Jeans

Now this is where I get a little stuck as I haven't had to really do anything with Taylor series since the beginning of my calculus instruction, so my knowledge of them is very limited .

20. Jan 15, 2009

### gabbagabbahey

Well to start with, I'm referring to the taylor series about the point x=0. To first order the equation for that is:

$$f(x)\approx f(0)+f'(0)x$$

So as long as you know how to take derivatives you should be fine

21. Jan 15, 2009

### Math Jeans

And I appologize but just one more clarification:

What exactly do you mean by throwing away the terms?

22. Jan 15, 2009

### gabbagabbahey

No need to apologize; it's always a good idea to ask for clarification on topics you don't fully understand

The full taylor series of a function is:

$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\ldots=\sum_{n=0}^\infty \frac{f^n(0)}{n!}x^n$$

So when I say "throw away all terms of order x^2 or higher", I mean that since the oscillations are 'small', you expect x to be small and so x^2,x^3,.... are really really small, so you can neglect all the terms with powers of x higher than one and therefor you use:

$$f(x)\approx f(0)+f'(0)x$$

23. Jan 15, 2009

### Math Jeans

Ok, I took the Taylor expansion and got this:

$$\ddot{x}\approx\frac{4a^2l^2(12a^2m^2\rho^2+5m^2\rho^4)}{(4a^2m^2\rho^3+m^2\rho^5)^2}x-\frac{4a^2l^2}{4a^2m^2\rho^3+m^2\rho^5}-\frac{2ag\rho}{4a^2+\rho^2}-\frac{2ag(4a^2-\rho^2)}{(4a^2+\rho^2)^2}x$$

24. Jan 15, 2009

### gabbagabbahey

I think you have a couple of small errors; I get

$$\ddot{x}\approx\frac{4a^2l^2(12a^2m^2\rho^2+m^2\rho^4)}{(4a^2m^2\rho^3+m^2\rho^5)^2}x-\frac{4a^2l^2}{4a^2m^2\rho^3+m^2\rho^5}-\frac{2ag\rho}{4a^2+\rho^2}-\frac{2ag(4a^2-3\rho^2)}{(4a^2+\rho^2)^2}x$$

Now, the fact that you want stable oscillations of frequency $\omega$ should tell you that you require $$\ddot{x}=-\omega^2 x$$. The only way the above result can exhibit that behavior is if

$$-\frac{4a^2l^2}{4a^2m^2\rho^3+m^2\rho^5}-\frac{2ag\rho}{4a^2+\rho^2}=0$$

AND

$$\frac{4a^2l^2(12a^2m^2\rho^2+5m^2\rho^4)}{(4a^2m^2\rho^3+m^2\rho^5)^2}-\frac{2ag(4a^2-3\rho^2)}{(4a^2+\rho^2)^2}=-\omega^2$$

If you solve that system of equations, you should end up with your desired result for omega (solve the first one for $$l^2$$ and then substitute it into the second equation and solve for omega).

Last edited: Jan 15, 2009
25. Jan 15, 2009

### Math Jeans

Hmm, well I caught the sign problem on my last term, however, we must have started with different expressions as my calculator keeps telling me that my answer is as I posted it (without the sign change).