Solving a Pendulum Problem Using the Lagrangian Approach

In summary, the homework statement states that a mass is attached to one end of a light rod of length l, which can swing in a plane. The other end of the rod is pivoted so that the rod can swing in a circle. The pivot rotates in the same plane at angular velocity \omega in a circle of radius R. The "pendulum" behaves like a simple pendulum in a gravitational field g=\omega ^2 R for all values of l and amplitudes of oscillations.
  • #1
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Homework Statement


A mass m is attached to one end of a light rod of length [itex]l[/itex]. The other end of the rod is pivoted so that the rod can swing in a plane. The pivot rotates in the same plane at angular velocity [itex]\omega[/itex] in a circle of radius R. Show that this "pendulum" behaves like a simple pendulum in a gravitational field [itex]g=\omega ^2 R[/itex] for all values of l and amplitudes of oscillations.

Homework Equations



[itex]L=T-V[/itex].
I used the trigonometric formula [itex]\sin \theta _1 \sin \theta _2 + \cos \theta _1 \cos \theta _2 =\cos (\theta _1 - \theta _2)[/itex] for a part.

The Attempt at a Solution


I place my coordinate system (Cartesian) in the center of the circle that the pivot describes. [itex]\vec r = x \hat i +y \hat j=(R \cos \theta _1 +l \cos \theta _2 ) \hat i +(R \sin \theta _1 + l \sin \theta _2 ) \hat j[/itex].
Since [itex]T=\frac{m}{2} ( \dot { \vec r} )^2[/itex], I calculated it to be worth [itex]\frac{m}{2}[\omega ^2 R^2+ \dot \theta _2 ^2 l^2+2Rl\omega \dot \theta _2 \cos (\theta _1 - \theta _2)]=L[/itex].
I've got the Euler-Lagrange equation to be (for [itex]\theta _2[/itex]) [itex]l^2 \ddot \theta _2 -Rl\omega \sin (\theta _1 - \theta _2 )(\omega - \dot \theta _2 )-Rl \omega \dot \theta _2 \sin (\theta _1 - \theta _2 )=0[/itex]. I don't think I can simplify this anymore.
However the answer should be [itex]ml^2 \ddot \theta =-mR\omega ^2 l \sin \theta[/itex] where their [itex]\theta[/itex] is worth my [itex]\theta _2[/itex].
I don't know what I did wrong.
The given answer starts with [itex]x=R\cos ( \omega t )+l \cos (\omega t + \theta )[/itex]. I don't really understand why the argument of the second cos term is this way though. Also I don't see why my "method" failed.
 
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  • #2
You've done everything correctly so far. You just need to go a little further. You can simplify your equation (hint - expand the brackets). And then you have the answer. You might not see instantly that you have the answer, but you really do. Their [itex]\theta[/itex] is not the same as your [itex]\theta_2[/itex], because their [itex]\theta[/itex] is the angle of the pendulum as seen from a coordinate system which is going round with the pivot. Also, they give the argument of the second cos term in that way they did, which is the explicit way to write it, but your way of writing it is exactly the same (since your [itex]\theta_1[/itex] is a function of time, the same as their explicit form).
 
  • #3
BruceW said:
You've done everything correctly so far. You just need to go a little further. You can simplify your equation (hint - expand the brackets). And then you have the answer. You might not see instantly that you have the answer, but you really do. Their [itex]\theta[/itex] is not the same as your [itex]\theta_2[/itex], because their [itex]\theta[/itex] is the angle of the pendulum as seen from a coordinate system which is going round with the pivot. Also, they give the argument of the second cos term in that way they did, which is the explicit way to write it, but your way of writing it is exactly the same (since your [itex]\theta_1[/itex] is a function of time, the same as their explicit form).

Thank you very much, your words are really encouraging.
I reach [itex]\ddot \theta _2 -\frac{g\sin (\theta _1 - \theta _2 )}{l}=0[/itex]. (I used the fact that [itex]g= \omega ^2 R[/itex].)
According to wikipedia the DE of a simple pendulum is [itex]\ddot \theta +\frac{g\sin (\theta )}{l}=0[/itex].
So to get this result, I'd have to call [itex]\phi =\theta _1 - \theta _2[/itex] and then show that [itex]\ddot \theta _2 = -\ddot \phi[/itex].
So let [itex]\phi =\theta _1 -\theta _2[/itex]. I have that [itex]\dot \phi = \omega - \dot \theta _2[/itex] and thus [itex]\ddot \phi = - \ddot \theta _2[/itex].
This implies that [itex]\ddot \phi + \frac{g \sin \phi }{l}=0[/itex] as required.
 
  • #4
I probably would have left it as [itex]\ddot \theta _2 -\frac{g\sin (\theta _1 - \theta _2 )}{l}=0[/itex]. But you did better to show that [itex]\ddot \theta _2 = -\ddot \phi[/itex] and get the equation into exactly the required form, nice!
 
  • #5


I would like to commend you on your attempt at solving this problem using the Lagrangian approach. It shows that you are thinking critically and trying to apply the concepts you have learned. However, I would like to offer a few suggestions and explanations to help you understand where you may have gone wrong.

Firstly, your use of the trigonometric formula is not necessary in this problem. It is true that the pendulum's motion can be described using trigonometric functions, but it is not needed to solve the problem using the Lagrangian approach. The Lagrangian approach is a more general and powerful method, and it is best to try to use it without relying on other formulas.

Secondly, your choice of coordinate system is not ideal. It is better to choose a system that is more closely related to the motion of the pendulum. In this case, a polar coordinate system centered at the pivot point would be more appropriate. This would make the calculations simpler and more intuitive.

Thirdly, your calculation of the kinetic energy is incorrect. The correct expression for the kinetic energy is T = (1/2) m (l^2 \dot{\theta}_2^2 + 2lR \dot{\theta}_2 \cos{\theta_2}), where \theta_2 is the angle that the rod makes with the vertical. This can be derived using the polar coordinate system mentioned earlier.

Lastly, the equation you have derived using the Euler-Lagrange equation is also incorrect. The correct equation is (l \ddot{\theta}_2 - R \omega^2 \sin{\theta_2}) = 0, where \theta_2 is again the angle that the rod makes with the vertical. This can be derived using the correct expression for the kinetic energy and the potential energy.

In conclusion, your approach was on the right track, but there were some errors in your calculations. I would suggest reviewing the Lagrangian approach and trying to solve the problem again using the suggestions I have mentioned. I hope this helps and good luck with your future scientific endeavors!
 

1. What is a Lagrangian?

A Lagrangian is a mathematical function used to describe the dynamics of a physical system. It is typically used in classical mechanics and is named after the French mathematician Joseph-Louis Lagrange.

2. How is a Lagrangian different from other mathematical functions?

A Lagrangian is unique in that it takes into account both the position and velocity of a system, rather than just one or the other. This allows for a more comprehensive understanding of the system's behavior.

3. What is a weird pendulum?

A weird pendulum, also known as a double pendulum, is a physical system consisting of two pendulums connected by a joint. It exhibits complex and chaotic behavior, making it a popular subject for studying nonlinear dynamics.

4. How does the Lagrangian describe the motion of a weird pendulum?

The Lagrangian of a weird pendulum takes into account the positions and velocities of both pendulums, as well as the forces acting on them. By solving the equations of motion derived from the Lagrangian, we can predict the motion of the double pendulum over time.

5. What are the practical applications of studying Lagrangian and weird pendulums?

The study of Lagrangian and weird pendulums has practical applications in fields such as robotics, aerospace engineering, and biomechanics. Understanding the dynamics of these systems can help improve the design and control of various mechanical systems.

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