Period of oscillations of a cone

1. Aug 21, 2009

joberr

1. The problem statement, all variables and given/known data
A particle of mass m moves on the inside surface of a smooth cone whose axis is vertical and whose half-angle is alpha . Find the period of small oscillations about a horizontal circular orbit a distance h above the vertex.

2. Relevant equations
Not sure. Lagrangian maybe F = ma
The answer is 2pi/tan(alpha) • sqrt(h/(3g))
not sure how to get it though.

3. The attempt at a solution
not sure what the question is . Initially i thought it was asking for the period of an oscillation in the vertical axis if disturbed by a small force while in a circular orbit than i thought it might mean what he period for the circular orbit is. I tried using a lagrangian after trying by Newtonian methods.

Last edited: Aug 21, 2009
2. Aug 21, 2009

kuruman

Yes, Lagrangian. Get the equations of motions in r and θ after using the constraint to eliminate z, then see what happens to the radial equation when the path is circular.

3. Aug 21, 2009

joberr

Is that for solving for the period of the circular orbit or a disturbance in while in a circular orbit.

4. Aug 22, 2009

kuruman

Both. Once you have the radial equation adapted to a circular orbit, you can get the period of the orbit from it. You also need this equation to find the period of small oscillations when the particle is slightly disturbed from the circular orbit it's in.

5. Aug 23, 2009

joberr

Last edited by a moderator: May 4, 2017
6. Aug 23, 2009

kuruman

Sorry, you need to start over again. You correctly say

r = z tanα which implies

$$z=\frac{r}{tan \alpha}$$

and

$$\dot{z}=\frac{\dot{r}}{tan \alpha}$$

Use these to eliminate z and z-dot in your Lagrangian.

7. Aug 23, 2009

joberr

Last edited by a moderator: May 4, 2017
8. Aug 23, 2009

gabbagabbahey

You missed a term when calculating $$\frac{\partial\mathcal{L}}{\partial r}$$ :

$$\frac{\partial}{\partial r} \left(\frac{1}{2}mr^2\dot{\theta}^2\right)\neq 0$$

Also, I find your handwritten notes difficult to read. In the future, please post your work using $\LaTeX$.

You can find an introduction to using $\LaTeX$ on these forums in this thread, and you can always click on any $\LaTeX$ image to see the code that generated it.

For example, your Lagrangian can be written clearly:

$$\mathcal{L}=\frac{1}{2}m\left(k\dot{r}^2+r^2\dot{\theta}}^2\right)-\frac{mgr}{\tan\alpha}$$

where $k\equiv1+\frac{1}{\tan^2\alpha}=\csc^2\alpha$

Last edited: Aug 23, 2009
9. Aug 23, 2009

joberr

$$\frac {\dot \theta^4}{\ddot \theta} = \frac{ \csc ^2 \alpha \ddot{r} + \frac{g}{tan \alpha}}{\dot r}$$

Known latex for a while just dont prefer it to using a pen and pencil

This is what I get as a differential equation after gabbagabbaheys correction

Dont know how to solve to get omega from it though

10. Aug 23, 2009

gabbagabbahey

Yikes!

How did you end up with that monstrosity?

11. Aug 23, 2009

joberr

I followed kuruman advice on the constraint to plug into the Lagrangian to eliminate the z variable then I used euler lagrange and then solve for r on both and set that equal.

12. Aug 24, 2009

gabbagabbahey

I'm sure that wasn't quite what kuruman had in mind....

Instead, start by posting the two DEs you get from the Euler-Lagrange equations...

13. Aug 24, 2009

joberr

That was my impression given he told me to eliminate z in the lagrangian (not euler-lagrange) using the equations he gave.

$$m \cdot csc^2 \alpha \cdot \ddot r = \frac{2 m r \dot \theta^2}{2} - \frac{mg}{tan \alpha}$$

and
$$0 = \frac{2 m r \dot \theta^2 \dot r}{2} + \frac{m r^2 2 \ddot \theta}{2}$$

14. Aug 24, 2009

gabbagabbahey

Okay, now what is the defining characteristic for a circular orbit....$r=$____?

15. Aug 24, 2009

joberr

you mean
$$-\frac{\ddot r}{\dot \theta^2} = r$$

16. Aug 24, 2009

gabbagabbahey

No, I mean...what makes a circle a circle?...A constant ____?

17. Aug 24, 2009

joberr

You mean
$$\dot r = 0 , r = R$$ that gives
$$\omega = \sqrt{\frac{g}{r \cdot \tan \alpha}}$$

18. Aug 24, 2009

joberr

How do I solve for the period of a small disturbance in this circular orbit.

19. Aug 24, 2009

gabbagabbahey

Well, for small disturbances of a circular orbit, instead of $r=R$ you expect $r=R+\epsilon(t)$ where $|\epsilon(t)|\ll R$

Start with your Euler-Lagrange equation for $\theta$,

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

What does that tell you about the quantity $$r^2\dot{\theta}$$?

20. Aug 24, 2009

joberr

$$r^2 \theta$$ is conserved - constant.