# Lagrangian Mechanics, bead on a hoop g

1. Oct 5, 2013

### dumbperson

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

This is the problem:
http://i.imgur.com/OJyzfhz.png?1

2. Relevant equations

$$\frac{dL}{dq_i}-\frac{d}{dt}\frac{dL}{d\dot{q_i}}=0$$
$$L=\frac{1}{2}mv^2-U(potential energy)$$

3. The attempt at a solution

This is my attempt at question A):

http://i.imgur.com/IeJVGm3.jpg

Does this look right? If not, am I atleast in the right direction. If I am right, how do I tackle B? I have no clue how to even begin on B.

Thanks

2. Oct 6, 2013

### voko

Your solution of A is generally correct, but there is a sign error in it.

To find the equilibrium positions, use the main property of equilibrium: everything stays constant. That means that all the dynamical variables are constant, and their derivatives are zero.

3. Oct 6, 2013

### dumbperson

Alright, thanks. I see I need a minus sign before the first term of my final answer. I'll try B with your tip!

4. Oct 6, 2013

### dumbperson

at C, do you think they want me to do a taylor expansion around theta_equilibrium?

5. Oct 6, 2013

### voko

Yes, the idea is that you should end up with $\ddot \theta + \omega ^2 \theta = 0$, where $\omega$ is to be determined from the expansion.

6. Oct 6, 2013

### dumbperson

I actually have no idea where/how to do the expansion. Shouldn't I do it on x(theta), y(theta) and z(theta) ?

Alsol, this is how I did B):

Eq. of motion is

$$\frac{g}{R}\sin{\theta}+\ddot{\theta}-\omega^2\sin{\theta}\cos{\theta}=0$$

the 2nd derivative is zero at equilibrium so i get

$$\theta_{eq} = arccos(\frac{g}{R\omega^2})$$

Last edited: Oct 6, 2013
7. Oct 6, 2013

### voko

You do it in the equation of motion you obtained.

Does it exist for all values of $\omega$?

What about $\sin \theta = 0$?

8. Oct 6, 2013

### dumbperson

It exists for
$$R\omega^2\geq g$$

Expansion of sine around my theta_eq??

$$\sin(\theta)=\sqrt{1-(\frac{g}{R\omega^2})^2}+\frac{g}{R\omega^2}\cdot (\theta -\arccos(\frac{g}{R\omega^2})-0.5\sqrt{1-(\frac{g}{R\omega^2})^2}\cdot (\theta -\arccos(\frac{g}{R\omega^2})^2+..$$

I'm not really sure about what you meant with ''what about sin(theta)=0 ?'' . Sorry

9. Oct 6, 2013

### voko

My question was whether $\sin \theta = 0$ also gives equilibrium points.

Your expansion is correct, but you do not need terms that are above linear in $\theta$.

10. Oct 6, 2013

### dumbperson

Oh, ofcourse! I forgot sin(theta). theta = 0 is an equilibrium point too.(well and pi but I guess that's not very reasonable here)

So I just have to expand sin(theta)cos(theta) now and then write it out?

11. Oct 6, 2013

### dumbperson

Also, how do you type in tex inside your normal sentences? The way I do it ([ tex] [/ tex]) , my text comes on a new line.

12. Oct 6, 2013

### BruceW

you can use itex instead. (sorry for jumping in on this thread). Don't be too quick to dismiss theta=Pi. What do questions b) and c) actually ask you for?

13. Oct 6, 2013

### voko

Do not guess. Your task is to find out whether $\theta = \pi$ is a stable equilibrium.

Yes - about each equilibrium. Note that in all cases except $\theta_{e} = 0$, you will have to change your variable to $\theta^* = \theta - \theta_{e}$, so that the resultant equation have the canonical form $\ddot {\theta ^*} + k^2 \theta ^* = 0$.

14. Oct 6, 2013

### dumbperson

I think I have it now. I did the expansion around theta_eq=pi, and the differential equation was the one of an exponential function(so not stable).

Around theta_eq=arccos(g/(Rω^2)), I found

$$k^2=\frac{3g^2}{R^2\omega^2}-\omega^2$$

15. Oct 6, 2013

### voko

At $\theta = 0$, the linearized equation is $\ddot \theta + (\frac g R - \omega^2) \theta = 0$. Clearly, the nature of its solution depends critically on $(\frac g R - \omega^2)$, and three cases are possible. You should analyze all of them.

Same for $\theta = \pi$.

16. Oct 6, 2013

### voko

I do not think it is correct. I obtain a similar expression, yet a different one.

17. Oct 6, 2013

### dumbperson

Thanks again! forgot about that. Still get that with theta_eq=pi, it is unstable in every case. I also see I made a mistake with determining k in my previous post, I got a K now that makes it stable for the allowed \omega (minus the end point).

18. Oct 6, 2013

### dumbperson

I got this one now:

$$k^2=\omega^2(1-(\frac{g}{R\omega})^2),$$

19. Oct 6, 2013

### voko

This is not self-consistent dimensionally. $(\frac{g}{R\omega})^2$ has the dimension of frequency squared, while 1, from which it is subtracted, is dimensionless.

20. Oct 6, 2013

### voko

By the way, how did you analyze the case $\theta_e = 0$ and $R\omega^2 = g$?