# Lagrangian Mechanics, bead on a hoop

## Homework Statement

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

## Homework Equations

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

## 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

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.

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

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.

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

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.

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.

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

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

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

You do it in the equation of motion you obtained.

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

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

What about ## \sin \theta = 0 ##?

You do it in the equation of motion you obtained.

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

What about ## \sin \theta = 0 ##?

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

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##.

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##.

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?

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.

BruceW
Homework Helper
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?

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)

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

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

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 ##.

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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 ##.

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

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

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).

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 ##.

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

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

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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 ##.

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).

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

I got this one now:

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

I got this one now:

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

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.

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

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.

The omega within the square should also be squared, that's my answer(just forgot to write the square here)

Woops

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

The motion is linear(theta=b*t+c) , so not stable. Is this correct?

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The omega within the square should also be squared, that's my answer(just forgot to write the square here)

Good. That is the solution I get as well.

I had that the motion will then be quadratic in time(0.5at²), not stable. Is this correct?

You obtained this equation from the power series expansion. That the first member of the expansion turns out to be zero does not really mean that a small oscillation regime is impossible. You should just continue the expansion and get the first member that isn't zero.

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Good. That is the solution I get as well.

You obtained this equation from the power series expansion. That the first member of the expansion turns out to be zero does not really mean that a small oscillation regime is impossible. You should just continue the expansion and get the first member that isn't zero.

If you mean what I think you mean, I expanded it further and the first nonzero term is a cubic one?

$$0.5\theta^3\frac{g}{R}+\ddot{\theta}=0$$

but isn't this term super small that we ignore it?

If you mean what I think you mean, I expanded it further and the first nonzero term is a cubic one?

$$0.5\theta^3\frac{g}{R}+\ddot{\theta}=0$$

but isn't this term super small that we ignore it?

We ignore it when there are bigger terms. But in this case there are not, and if we ignore this, we reach the wrong conclusion that this equilibrium point is not stable. While it is now seen (I hope) that it is stable.

We ignore it when there are bigger terms. But in this case there are not, and if we ignore this, we reach the wrong conclusion that this equilibrium point is not stable. While it is now seen (I hope) that it is stable.

Thank you very very much for all the help