How to Determine the Period of Oscillations of a Particle Inside a Smooth Cone?

[joberr]

Homework Statement

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.

Homework Equations

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

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.

 

[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.

 

[joberr]

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

 

[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.

 

[joberr]

http://img197.imageshack.us/img197/3584/probn.th.jpg
What constraint am I missing I can't get full answer I have an unknown constant B

 

[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.

 

[joberr]

http://img41.imageshack.us/img41/2464/prob2u.th.jpg
I get this set of coupled differential equations but when I try to solve for them I get
$$\omega * r^2 = 1$$ and r is a quadratic with respect to time which doesn't make sense

 

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

 

[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 don't 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

 

[gabbagabbahey]

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

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

Yikes!

How did you end up with that monstrosity?

Start with the euler-lagrange equations...what do you get for those?

 

[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.

 

[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...

 

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

 

[gabbagabbahey]

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

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

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

 

[joberr]

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

 

[gabbagabbahey]

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

 

[joberr]

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

 

[joberr]

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

 

[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 $$r^2\dot{\theta}$$?

 

[joberr]

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

 

[gabbagabbahey]

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

Hmmm, don't you mean $$r^2 \dot{\theta}$$ is conserved - constant? (you forgot the 'dot')

If so, then you are right. So, why not call that constant say $l_0$ and use that to replace $$\dot{\theta}^2$$ in your other Euler-Lagrange equation so that you get an equation involving only $r$ (and its derivatives) and the constant $l_o$...what do you get if you do that?

 

[joberr]

$$m \cdot csc^2 \alpha \cdot \ddot r = \frac{m l_0^2 }{r^3} - \frac{mg}{tan \alpha}$$

if I make $$R= R+ \epsilon$$
then
approximate $$\frac{1}{(R + \epsilon)^3}$$
as $$\frac{1}{R^3} + 3 \frac{\epsilon}{R^4}$$
I get treating R as a constant
$$m \cdot csc^2 \alpha \cdot \ddot \spsilon= m l_0^2 (\frac{1}{R^3} + 3 \frac{\epsilon}{R^4)- \frac{mg}{tan \alpha}$$

 

[gabbagabbahey]

Right, so

$$\ddot{r}=\frac{l_0^2\cos^2\alpha}{r^3}-\frac{\cos^2\alpha g}{\tan\alpha}$$

Now, since you expect $r=R+\epsilon(t)$ where $|\epsilon(t)|\ll R$, why not substitute this into the above equation and Taylor expand the RHS for small epsilon.

 

[joberr]

That gives me
$$\omega_{dist}^2 = \frac{3 l_0^2 }{R^4 \csc^2 \alpha}$$

where $$l_0 = R \omega_0$$

and
$$\omega_{dist}^2 = \frac{3 g \cos^2 \alpha }{h}$$

 

[joberr]

I checked in book and its right thanks so much gabbagabba I've gotten rusty with problem solving

 

[gabbagabbahey]

You're welcome