Pendulum problem using Lagrangian approach

In summary, the conversation is about a physics problem involving a mass attached to a pivot by a rod. The goal is to find the angle at which the force in the rod changes from compression to tension. The problem is to be solved using a Lagrangian approach. The conversation also includes a discussion about attaching a file and a mistake in the first reply. The final part of the conversation involves an explanation of how to solve the problem using integration.
  • #1
Luminous Blob
50
0
Hi,

I'm trying to do a problem that goes something like this:

There is a mass (m) attached to one end of a massless rod (length l). The other end of the rod is attached to a frictionless pivot. The rod is released from rest at an angle F0 < pi/2. At what angle F does the force in the rod change from compressive to tension?

Attached is a diagram of the situation, as well as what I have done so far. I apologize for not putting my working out in the body of my post, but I haven't used LaTex before, and I don't have the time at the moment to learn how to use it.

Another thing: The problem should be done using a Lagrangian approach (although it isn't entirely necessary).

Any help would be highly appreciated!
 
Physics news on Phys.org
  • #2
Oops, forgot to attach the file. And now that I've tried, I realize that I don't know how. The "Manage Attachments" button under "Additional Options" doesn't seem to do anything, and I can't see any other button/link for attachments. The FAQ says:

"To attach a file to a new post, simply click the [Browse] button at the bottom of the post composition page, and locate the file that you want to attach from your local hard drive."

Only problem is, I can't find any "Browse" button.

I know it's probably right in front of me, but I can't see it anywhere. Can anyone help me out here? My "Posting Rules" thing does say that I may post attachments.

Edit: Here is the attachment. I just realized I have my browser configured to block popups. I know, I am an idiot :)
 

Attachments

  • question.doc
    31.5 KB · Views: 327
Last edited:
  • #3
I can't find the attachment, but I would assume that the angle is measured in a right-handed system with [tex]\frac{\pi}{2}[/tex] the direction of the upwards vertical (antiparallell to the direction of the force of gravity)

Since the trajectory is circular, we may write:
[tex]\vec{r}(t)=l\vec{i}_{r},\vec{i}_{r}(t)=\cos\theta(t)\vec{i}+\sin\theta(t)\vec{j}[/tex]
The acceleration may then be written as:
[tex]\vec{a}=-l{\omega}^{2}\vec{i}_{r}+l\dot{\omega}\vec{i}_{\theta},\omega=\dot{\theta}[/tex]

Newton's laws of motion may then be written as:
[tex]T-mg\sin\theta=-ml\omega^{2}[/tex] (radial component)
[tex]-mg\cos\theta=ml\dot{\omega}[/tex] (transverse component)

Multiplying the last equation by omega, and integrating, yields:
[tex]-g(\sin\theta(t)-\sin\theta_{0})=\frac{l\omega(t)^{2}}{2}\rightarrow\omega(t)^{2}=\frac{2g}{l}(\sin\theta_{0}-\sin\theta(t))[/tex]

Clearly, T changes from compression to tension (at [tex]\theta=\hat{\theta}[/tex])when it is zero:
[tex]2\sin\theta_{0}=3\sin\hat{\theta}\rightarrow[/tex]
[tex]\hat{\theta}=\sin^{-1}\frac{2}{3}\sin\theta_{0}[/tex]
 
Last edited:
  • #4
Major error in first reply edited.
 
  • #5
arildno said:
I can't find the attachment, but I would assume that the angle is measured in a right-handed system with [tex]\frac{\pi}{2}[/tex] the direction of the upwards vertical (antiparallell to the direction of the force of gravity)

Since the trajectory is circular, we may write:
[tex]\vec{r}(t)=l\vec{i}_{r},\vec{i}_{r}(t)=\cos\theta(t)\vec{i}+\sin\theta(t)\vec{j}[/tex]
The acceleration may then be written as:
[tex]\vec{a}=-l{\omega}^{2}\vec{i}_{r}+l\dot{\omega}\vec{i}_{\theta},\omega=\dot{\theta}[/tex]

Newton's laws of motion may then be written as:
[tex]T-mg\sin\theta=-ml\omega^{2}[/tex] (radial component)
[tex]-mg\cos\theta=ml\dot{\omega}[/tex] (transverse component)

Multiplying the last equation by omega, and integrating, yields:
[tex]-g(\sin\theta(t)-\sin\theta_{0})=\frac{l\omega(t)^{2}}{2}\rightarrow\omega(t)^{2}=\frac{2g}{l}(\sin\theta_{0}-\sin\theta(t))[/tex]

Clearly, T changes from compression to tension (at [tex]\theta=\hat{\theta}[/tex])when it is zero:
[tex]2\sin\theta_{0}=3\sin\hat{\theta}\rightarrow[/tex]
[tex]\hat{\theta}=\sin^{-1}\frac{2}{3}\sin\theta_{0}[/tex]

Thanks for that, and sorry I took so long to get back to this. I get everything you did except for the integration bit. Could you explain in more detail how that was done?

Thanks a lot.
 
  • #6
All right:
1. Initial conditions are:
[tex]\theta(0)=\theta_{0},\dot{\theta}(0)=\omega(0)=0[/tex]

That is, we start from rest.

2. Using transverse component:
Remember that [tex]\omega=\dot{\theta}[/tex]
i.e omega is the temporal derivative of the angle.

By multiplying the transverse component with omega, we have:
[tex]-mg\dot{\theta}\cos\theta=ml\omega\dot{\omega}[/tex]

Hence, we see that both sides are exactly a temporal derivative (by using the chain rule of differentiation); hence, we may integrate both sides from t=0, to an arbitrary time value.
Using the initial conditions yields the answer.
 
  • #7
Aaahh, now I get it.

You've been a great help, thank you very much.
 

1. What is the Lagrangian approach in solving pendulum problems?

The Lagrangian approach is a mathematical method used in classical mechanics to analyze the motion of a system. It is based on the principle of least action, which states that the path a system takes between two points in time is the one that minimizes the action integral. For pendulum problems, the Lagrangian approach is used to derive the equations of motion for the pendulum, taking into account any constraints or external forces.

2. How is the Lagrangian approach different from other methods of solving pendulum problems?

The Lagrangian approach is different from other methods, such as the Newtonian approach, because it considers the entire system as a whole and takes into account any constraints or external forces. It does not require the use of force equations or free body diagrams, making it a more elegant and efficient method for solving complex systems.

3. What are the advantages of using the Lagrangian approach in pendulum problems?

One of the main advantages of using the Lagrangian approach is that it simplifies the analysis of complex systems by reducing the number of equations needed to describe the motion. It also allows for the inclusion of constraints and external forces, which can be difficult to incorporate in other methods. Additionally, the Lagrangian approach is based on fundamental principles and is applicable to a wide range of problems in classical mechanics.

4. Are there any limitations to using the Lagrangian approach in pendulum problems?

One limitation of the Lagrangian approach is that it can be more difficult to apply in systems with non-conservative forces, such as friction. In these cases, additional considerations and modifications may be needed to accurately model the system. Additionally, the Lagrangian approach may not be as intuitive for beginners compared to other methods, such as the Newtonian approach.

5. How can the Lagrangian approach be applied to other problems in physics?

The Lagrangian approach is a powerful tool in classical mechanics, but it can also be applied to other areas of physics, such as quantum mechanics and field theory. It is a general mathematical framework that can be adapted to different systems and has been used to solve a wide range of problems in physics, including celestial mechanics, fluid dynamics, and electromagnetism.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
630
  • Introductory Physics Homework Help
Replies
7
Views
740
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Replies
1
Views
708
  • Mechanical Engineering
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
27
Views
644
Back
Top