Lagrangian Mechanics, bead on a hoop

In summary: 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...##.
  • #1
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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.jpgDoes 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
 
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  • #2
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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  • #3
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
voko said:
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?
 
  • #5
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
voko said:
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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  • #7
dumbperson said:
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 ##?
 
  • #8
voko said:
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
 
  • #9
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
voko said:
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?
 
  • #11
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
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
dumbperson said:
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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  • #14
voko said:
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

$$k^2=\frac{3g^2}{R^2\omega^2}-\omega^2$$
 
  • #15
dumbperson said:
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 ##.
 
  • #16
dumbperson said:
$$k^2=\frac{3g^2}{R^2\omega^2}-\omega^2$$

I do not think it is correct. I obtain a similar expression, yet a different one.
 
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  • #17
voko said:
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).
 
  • #18
voko said:
I do not think it is correct. I obtain a similar expression, yet a different one.

I got this one now:

$$k^2=\omega^2(1-(\frac{g}{R\omega})^2),$$
 
  • #19
dumbperson said:
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.
 
  • #20
By the way, how did you analyze the case ##\theta_e = 0## and ## R\omega^2 = g ##?
 
  • #21
voko said:
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)
 
  • #23
voko said:
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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  • #24
dumbperson said:
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.

dumbperson said:
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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  • #25
voko said:
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?
 
  • #26
dumbperson said:
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.
 
  • #27
voko said:
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
 

1. What is Lagrangian Mechanics?

Lagrangian Mechanics is a mathematical formalism used to describe the motion of a system of particles or objects. It is based on the principle of least action, which states that the true path of a system is the one that minimizes the total energy of the system.

2. How is Lagrangian Mechanics different from Newtonian Mechanics?

While Newtonian Mechanics describes the motion of a system using forces and acceleration, Lagrangian Mechanics uses a more general approach by considering the total energy of the system. This allows for a more elegant and concise description of the system's motion.

3. What is a bead on a hoop in Lagrangian Mechanics?

A bead on a hoop is a common example used to illustrate the concepts of Lagrangian Mechanics. It refers to a small object (the bead) that is free to move along a circular track (the hoop) without any friction. This simple system allows for a clear understanding of the principles of Lagrangian Mechanics.

4. How is the motion of a bead on a hoop described using Lagrangian Mechanics?

In Lagrangian Mechanics, the motion of a bead on a hoop is described by the Lagrangian function, which is the difference between the kinetic and potential energies of the system. This function is then used to derive the equations of motion for the bead and to determine its path along the hoop.

5. What are some real-world applications of Lagrangian Mechanics?

Lagrangian Mechanics has many practical applications, such as in physics, engineering, and astronomy. It is used to model the motion of objects in space, analyze the behavior of mechanical systems, and even in the development of new technologies like robotics and control systems.

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