Generalized Coordinates: Double Pendulum

In summary: Standard double pendulum setup. A string with mass, connected to a string with a mass, mounted to the ceiling. Given is m1,m2,l1,l2a) choose a suitable set of coordinates and write a lagrangian function, assuming it swings in a single vertical plane (I did this, using L = T - U)b)write out lagrange's equations and show that they reduce to the equations for a pair of coupled harmonic oscillators. (here's where my problem arises)In summary, the Lagrangian for the problem is a function that involves three variables (dx1/dt, dx2/dt, and theta). The first term is
  • #1
Pythagorean
Gold Member
4,400
312

Homework Statement



Standard double pendulum setup. A string with mass, connected to a string with a mass, mounted to the ceiling. Given is m1,m2,l1,l2

a) choose a suitable set of coordinates and write a lagrangian function, assuming it swings in a single vertical plane (I did this, using L = T - U)

b)write out lagrange's equations and show that they reduce to the equations for a pair of coupled harmonic oscillators. (here's where my problem arises)

Homework Equations



The Lagrangian
d/dt[dL/(dq/dt)] - dL/dq = 0

[tex]\frac {d}{dt} \frac {\partial L}{\partial d \theta_k[/tex]

The Attempt at a Solution



My issue is really a implicit/explicit differentiation problem.

I come up with a term under the d/dt (first term) of the lagrangian that involves three variables (all degrees) in this form:

(dx1/dt)*sin(x1 - x2)

when I take the time derivative of this, how do I handle the x1 and x2 (which are actually angles theta in my written notation)

I realize the first term (by the product rule) would be:

(d^2x1/dt^2)*sin(x1 - x2)

but how do I handle the two angles under the sin term that have no explicit time dependence?

Thank you for your help.

LATEX VERSION BELOW (probably being updated, I'm slow at it)

[tex] \frac {d}{dt} \left \dot{\theta_1}sin(\theta_1 - \theta_2) \right

= \ddot{\theta_1}sin(\theta_1 - \theta_2) + \dot{\theta_1} (?) + \dot{\theta_1} (?)[/tex]

the above equation is what I have, where I don't know what to do for the (?) that involves taking the time derivative of theta (which has no explicit time-dependence)
 
Last edited:
Physics news on Phys.org
  • #2
what is your Lagrangian for the problem?

if you are sure that the Lagrangian is correct, then to find derivative of sin(x-y), you can either use product rule+chain rule, or expand
sin(x-y)=sinx*cosy-cosx*siny

usually, problems involving double pendulum implies the small angle approximation. doing so will leave you will all quadratic terms or products of the two angle terms (and their first order derivatives).
 
  • #3
tim_lou said:
what is your Lagrangian for the problem?

if you are sure that the Lagrangian is correct, then to find derivative of sin(x-y), you can either use product rule+chain rule, or expand
sin(x-y)=sinx*cosy-cosx*siny

usually, problems involving double pendulum implies the small angle approximation. doing so will leave you will all quadratic terms or products of the two angle terms (and their first order derivatives).

I'm fairly sure of my lagrangian, I've compared it to http://scienceworld.wolfram.com/physics/DoublePendulum.html" [Broken] solution.

Are you implying that I should take the small angle approximation before I take the lagrangian derivatives?
 
Last edited by a moderator:
  • #4
I asked the question better and more specifically in the math forum and I got the answer:

multivariable chain rule


So this way, I can take derivative first and then apply the approximation.

Thank you for you help.
 
  • #5
since in your posts, you said reduce the equation to "harmonic oscillators". It usually means the small angle approximation...

taking derivative first then approximate will probably work, but it is much messier, you get all kinds of product rule, chain rule and stuffs like that.

you can avoid all that and get linear differential equations very quickly by using

[tex]\cos\theta \approx 1-\frac{\theta^2}{2}[/tex]
and
[tex]\sin\theta \approx \theta[/tex]

and then dropping all terms with order higher than two. when differentiating, all the 2nd order terms become linear, and you got yourself a nice happy system of linear differential equations.

but anyway... I guess it would be nice if you do the problem using both ways, and see if they agree with each other. Hey, it doesn't hurt to practice more physics.:smile:
 
Last edited:
  • #6
tim_lou said:
since in your posts, you said reduce the equation to "harmonic oscillators". It usually means the small angle approximation...

taking derivative first then approximate will probably work, but it is much messier, you get all kinds of product rule, chain rule and stuffs like that.

you can avoid all that and get linear differential equations very quickly by using

[tex]\cos\theta \approx 1-\frac{\theta^2}{2}[/tex]
and
[tex]\sin\theta \approx \theta[/tex]

and then dropping all terms with order higher than two. when differentiating, all the 2nd order terms become linear, and you got yourself a nice happy system of linear differential equations.

but anyway... I guess it would be nice if you do the problem using both ways, and see if they agree with each other. Hey, it doesn't hurt to practice more physics.:smile:

hrm. I was fearful I'd lose information if I did that, but having worked out the solution, I got this for one of the accelerations:

[tex]\frac {g}{l_1} (\theta_1 + \theta_2)[/tex]

which is similar to the cho:

[tex]k_1 x_1 + k_2 (x_1 + x_2)[/tex]

which has one more term than mine...
I went back through it all a few times and couldn't find where I would have dropped such a term, so I think I'll take your advice and try it derivative-first.
 
  • #7
I don't know what you got for the Lagrangian, but I can tell you that the tricky part is the kinetic energy of the second pendulum (you should have a term calculated from dot product or law of cosine), and from that kinetic energy, you get coupling. Maybe you can tell me what you got for the kinetic energy of the second pendulum?

edit: I see you have a
[tex]\dot{\theta_1}\dot{\theta_2}\cos(\theta_1-\theta_2)[/tex]
in the Lagrangian, right?

using the small angle approximation, you can basically drop the cos, and make it 1, that would simplify things a lot. (the other terms have order higher than 2. the order of theta dot counts)

however, for gravitational potential energy, you should have a term involving
[tex]\cos\theta_1\approx 1-\frac{\theta_1^2}{2}[/tex]
that cosine cannot be ignored since you are using second order approximation. Maybe that's where the terms got dropped?
 
Last edited:
  • #8
tim_lou said:
I don't know what you got for the Lagrangian, but I can tell you that the tricky part is the kinetic energy of the second pendulum (you should have a term calculated from dot product or law of cosine), and from that kinetic energy, you get coupling. Maybe you can tell me what you got for the kinetic energy of the second pendulum?

edit: I see you have a
[tex]\dot{\theta_1}\dot{\theta_2}\cos(\theta_1-\theta_2)[/tex]
in the Lagrangian, right?

using the small angle approximation, you can basically drop the cos, and make it 1, that would simplify things a lot. (the other terms have order higher than 2. the order of theta dot counts)

however, for gravitational potential energy, you should have a term involving
[tex]\cos\theta_1\approx 1-\frac{\theta_1^2}{2}[/tex]
that cosine cannot be ignored since you are using second order approximation. Maybe that's where the terms got dropped?

ah, yes, that could very well be. I was only taking first-order approximations of the cos and sin terms.

and yes, I have:
[tex]\dot{\theta_1}\dot{\theta_2}\cos(\theta_1-\theta_2)[/tex]
In fact, I can imagine taking the gravitational to second approximation easily solving all my problems. I'll give it a shot. Though, the only trig terms that are with g are sin terms after I take the lagrangian derivatives, so it might be a lot simpler to approximate before taking the derivative.
 
Last edited:
  • #9
if I may ask, is there a particular reason gravitational potential goes to second order, while the kinetic terms do not?

This is what I have for one of the accelerations when I take the derivatives first then solve both lagrangians for theta_1, set them equal, and solve for theta_2: (the second mass hanging from the first).

This is also taking the approx (second order for grav pot) before taking the derivative:

[tex]\ddot{\theta_2} = \frac{g l_2}{l_2^2 - \gamma l_1^2} (\theta_2 - \theta_1)[/tex]Been working on this generalized coordinates homework set for days. So much for spring break. I'll see what the prof says tomorrows.

edit: oh yeah

[tex]\gamma = \frac{m_2}{m_1 - m_2}[/tex]

this is actually looking like it might work out. I should go to bed, but I can't help but pursue this further!ok... I can pull that out to:

[tex]\ddot{\theta_2} = \frac{g l_2 (m_1 \theta_2 + m_2 \theta_2 - m_1 \theta_1 - m_1 \theta_1)}{m_1 l_2^2 + m_2 l_2^2 - m_2 l_1^2}[/tex]
 
Last edited:
  • #10
because when you expand the kinetic term, you'll get things that look like
[tex]\dot{\theta}_1\dot{\theta}_2 \left [ 1-\frac{(\theta_1-\theta_2)^2}{2} \right ][/tex]

then, when you expand it, you'll get a 2nd order term, and 3 4th order term. the 4th order terms are neglected (mainly because it is asssumed that theta and theta dot are small... and the that when neglecting higher order terms, the partial derivatives become linear. So basically it is a 2nd order approximation in the Lagrangian, and a linear approximation in the derivatives.)

your answer looks good to me... though I cannot 100% guarantee that it is correct... I just know that the answers are not nice when the massese and the lengths aren't equal.
 

1. What are generalized coordinates in the context of a double pendulum?

Generalized coordinates are a set of variables used to describe the position and orientation of a system of particles or rigid bodies. In the case of a double pendulum, it refers to the angles and lengths that define the position of each pendulum rod.

2. How do generalized coordinates differ from regular coordinates?

Generalized coordinates are independent of the specific coordinate system used to describe the system, whereas regular coordinates are dependent on the choice of coordinate system. Generalized coordinates also take into account the constraints and degrees of freedom of the system, making them more efficient for solving complex systems.

3. Why are generalized coordinates useful in studying a double pendulum?

Generalized coordinates allow for a more comprehensive analysis of the double pendulum system, as they provide a complete description of the system's configuration and motion. They also simplify the equations of motion, making it easier to solve for the system's behavior.

4. How many generalized coordinates are needed to fully describe a double pendulum?

For a double pendulum, two generalized coordinates are needed to fully describe the system's configuration and motion. These are typically chosen to be the angles of the two pendulum rods with respect to a fixed reference point.

5. Can generalized coordinates be used for any type of system?

Yes, generalized coordinates can be used for any system, as long as the constraints and degrees of freedom are taken into account. They are particularly useful for systems with a large number of particles or degrees of freedom, as they simplify the equations of motion and make them easier to solve.

Similar threads

  • Advanced Physics Homework Help
Replies
12
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
764
  • Advanced Physics Homework Help
Replies
16
Views
867
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
3K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
1K
Back
Top