Generalized Coordinates: Double Pendulumby Pythagorean Tags: coordinates, double, generalized, pendulum 

#1
Mar1807, 06:25 PM

PF Gold
P: 4,180

1. The problem statement, all variables and given/known data
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) 2. Relevant equations The Lagrangian d/dt[dL/(dq/dt)]  dL/dq = 0 [tex]\frac {d}{dt} \frac {\partial L}{\partial d \theta_k[/tex] 3. 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 timedependence) 



#2
Mar1807, 07:03 PM

P: 689

what is your Lagrangian for the problem?
if you are sure that the Lagrangian is correct, then to find derivative of sin(xy), you can either use product rule+chain rule, or expand sin(xy)=sinx*cosycosx*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
Mar1807, 07:13 PM

PF Gold
P: 4,180

Are you implying that I should take the small angle approximation before I take the lagrangian derivatives? 



#4
Mar1807, 07:28 PM

PF Gold
P: 4,180

Generalized Coordinates: Double Pendulum
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
Mar1807, 10:36 PM

P: 689

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. 



#6
Mar1807, 11:11 PM

PF Gold
P: 4,180

[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 derivativefirst. 



#7
Mar1807, 11:36 PM

P: 689

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? 



#8
Mar1907, 02:56 AM

PF Gold
P: 4,180

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. 



#9
Mar1907, 03:28 AM

PF Gold
P: 4,180

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] 



#10
Mar1907, 09:06 AM

P: 689

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. 


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